problems calculus.pdf

SCHAUM'S OUTLINE SERIES
Schaum's Outline of Theory and Problems of Beginning
Calculus
Second Edition
Elliott Mendelson, Ph.D.
Professor of Mathematics
Queens College
City University of New York

To the memory of my father, Joseph, and my mother, Helen
ELLIOTT MENDELSON is Professor of Mathematics at Queens College of the City University of
New York. He also has taught at the University of Chicago, Columbia University, and the University of
Pennsylvania, and was a member of the Society of Fellows of Harvard University. He is the author of
several books, including Schaum's Outline of Boolean Algebra and Switching Circuits. His principal
area of research is mathematical logic and set theory.
Schaum's Outline of Theory and Problems of
BEGINNING CALCULUS
Copyright © 1997,1985 by The McGraw-Hill Companies, Inc. All rights reserved. Printed in the United
States of America. Except as permitted under the Copyright Act of 1976, no part of this publication
may be reproduced or distributed in any form or by any means, or stored in a data base or retrieval
system, without the prior written permission of the publisher.
5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 PRS PRS 9 0 1 0 9
ISBN 0-07-041733-4
Sponsoring Editor: Arthur Biderman
Production Supervisor: Suzanne Rapcavage
Editing Supervisor: Maureen B. Walker
Library of Congress Cataloging-in-Publication Data
Mendelson, Elliott.
Schaum's outline of theory and problems of beginning calculus /
Elliott Mendelson, -- 2nd ed.
p. cm. -- (Schaum's outline series)
Includes index.
ISBN 0-07-041733-4 (pbk.)
1. Calculus. 2. Calculus--Problems, exercises, etc. I. Title.
QA303.M387 1997
515' .076--dc21 96-39852
CIP

Preface
This Outline is limited to the essentials of calculus. It carefully develops, giving all steps, the principles
of differentiation and integration on which the whole of calculus is built. The book is suitable for
reviewing the subject, or as a self-contained text for an elementary calculus course.
The author has found that many of the difficulties students encounter in calculus are due to weakness in
algebra and arithmetical computation, emphasis has been placed on reviewing algebraic and
arithmetical techniques whenever they are used. Every effort has been made—especially in regard to
the composition of the solved problems—to ease the beginner's entry into calculus. There are also some
1500 supplementary problems (with a complete set of answers at the end of the book).
High school courses in calculus can readily use this Outline. Many of the problems are adopted from
questions that have appeared in the Advanced Placement Examination in Calculus, so that students will
automatically receive preparation for that test.
The Second Edition has been improved by the following changes:
1. A large number of problems have been added to take advantage of the availability of graphing
calculators. Such problems are preceded by the notation . Solution of these problems is not
necessary for comprehension of the text, so that students not having a graphing calculator will not
suffer seriously from that lack (except insofar as the use of a graphing calculator enhances their
understanding of the subject).
2. Treatment of several topics have been expanded:
(a) Newton's Method is now the subject of a separate section. The availability of calculators makes
it much easier to work out concrete problems by this method.
(b) More attention and more problems are devoted to approximation techniques for integration, such
as the trapezoidal rule, Simpson's rule, and the midpoint rule.
(c) The chain rule now has a complete proof outlined in an exercise.
3. The exposition has been streamlined in many places and a substantial number of new problems have
been added.
The author wishes to thank again the editor of the First Edition, David Beckwith, as well as the editor of
the Second Edition, Arthur Biderman, and the editing supervisor, Maureen Walker.
ELLIOTT MENDELSON

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Contents
Chapter 1
Coordinate Systems on a Line
1
1.1 The Coordinates of a Point 1
1.2 Absolute Value 2
Chapter 2
Coordinate Systems in a Plane
8
2.1 The Coordinates of a Point 8
2.2 The Distance Formula 9
2.3 The Midpoint Formulas 10
Chapter 3
Graphs of Equations
14
Chapter 4
Straight Lines
24
4.1 Slope 24
4.2 Equations of a Line 27
4.3 Parallel Lines 28
4.4 Perpendicular Lines 29
Chapter 5
Intersections of Graphs
36
Chapter 6
Symmetry
41
6.1 Symmetry about a Line 41

Chapter 7
Functions and Their Graphs
46
7.1 The Notion of a Function 46
7.2 Intervals 48
7.3 Even and Odd Functions 50
7.4 Algebra Review: Zeros of Polynomials 51
Chapter 8
Limits
59
8.1 Introduction 59
8.2 Properties of Limits 59
8.3 Existence or Nonexistence of the Limit 61
Chapter 9
Special Limits
67
9.1 One-Sided Limits 67
9.2 Infinite Limits: Vertical Asymptotes 68
9.3 Limits at Infinity: Horizontal Asymptotes 70
Chapter 10
Continuity
78
10.1 Definition and Properties 78
10.2 One-Sided Continuity 79
10.3 Continuity over a Closed Interval 80
6.2 Symmetry about a Point 42

Chapter 11
The Slope of a Tangent Line
86
Chapter 12
The Derivative
92
Chapter 13
More on the Derivative
99
13.1 Differentiability and Continuity 99
13.2 Further Rules for Derivatives 100
Chapter 14
Maximum and Minimum Problems
104
14.1 Relative Extrema 104
14.2 Absolute Extrema 105
Chapter 15
The Chain Rule
116
15.1 Composite Functions 116
15.2 Differentiation of Composite Functions 117
Chapter 16
Implicit Differentiation
126
Chapter 17
The Mean-Value Theorem and the Sign of the Derivative
129
17.1 Rolle's Theorem and the Mean-Value Theorem 129
17.2 The Sign of the Derivative 130

Chapter 18
Rectilinear Motion and Instantaneous Velocity
136
Chapter 19
Instantaneous Rate of Change
143
Chapter 20
Related Rates
147
Chapter 21
Approximation by Differentials; Newton's Method
155
21.1 Estimating the Value of a Function 155
21.2 The Differential 155
21.3 Newton's Method 156
Chapter 22
Higher-Order Derivatives
161
Chapter 23
Applications of the Second Derivative and Graph Sketching
167
23.1 Concavity 167
23.2 Test for Relative Extrema 169
23.3 Graph Sketching 171
Chapter 24
More Maximum and Minimum Problems
179
Chapter 25
Angle Measure
185
25.1 Arc Length and Radian Measure 185
26.2 Directed Angles 186

Chapter 26
Sine and Cosine Functions
190
26.1 General Definition 190
26.2 Properties 192
Chapter 27
Graphs and Derivatives of Sine and Cosine Functions
202
27.1 Graphs 202
27.2 Derivatives 205
Chapter 28
The Tangent and Other Trigonometric Functions
214
Chapter 29
Antiderivatives
221
29.1 Definition and Notation 221
29.2 Rules for Antiderivatives 222
Chapter 30
The Definite Integral
229
30.1 Sigma Notation 229
30.2 Area under a Curve 229
30.3 Properties of the Definite Integral 232
31.1 Calculation of the Definite Integral 238
31.2 Average Value of a Function 239
31.3 Change of Variable in a Definite Integral 240
Chapter 31
The Fundamental Theorem of Calculus
238

Chapter 32
Applications of Integration I: Area and Arc Length
249
32.1 Area between a Curve and the y-axis 249
32.2 Area between Two Curves 250
32.3 Arc Length 251
Chapter 33
Applications of Integration II: Volume
257
33.1 Solids of Revolution 257
33.2 Volume Based on Cross Sections 259
Chapter 34
The Natural Logarithm
268
34.1 Definition 268
34.2 Properties 268
Chapter 35
Exponential Functions
275
35.1 Introduction 275
35.2 Properties of ax 275
35.3 The Function ex 275
36.1 L'Hôpital's Rule 284
36.2 Exponential Growth and Decay 285
Chapter 36
L'Hôpital's Rule; Exponential Growth and Decay
284

Chapter 37
Inverse Trigonometric Functions
292
37.1 One-One Functions 292
37.2 Inverses of Restricted Trigonometric Functions 293
Chapter 38
Integration by Parts
305
Chapter 39
Trigonometric Integrands and Trigonometric Substitutions
311
39.1 Integration of Trigonometric Functions 311
39.2 Trigonometric Substitutions 313
Chapter 40
Integration of Rational Functions; The Method of Partial Fractions
320
Appendix A
Trigonometric Formulas
329
Appendix B
Basic Integration Formulas
330
Appendix C
Geometric Formulas
331
Appendix D
Trigonometric Functions
332
Appendix E
Natural Logarithms
333
Appendix F
334
Answers to Supplementary Problem 335
Index 371

Chapter 1
Coordinate Systems on a Line
1.1 THE COORDINATES OF A POINT
Let 9 be a line. Choose a point 0 on the line and call this point the origin.
Now select a direction along 9;
say, the direction from left to right on the diagram.
For every point P to the right of the origin 0,let the coordinate of P be the distance between 0and P.
(Of course, to specify such a distance, it is first necessary to establish a unit distance by arbitrarily
picking two points and assigning the number 1to the distance between these two points.)
In the diagram
the distance is assumed to be 1,so that the coordinate of A is 1.The point B is two units away from
0;therefore, B has coordinate 2. Every positive real number r is the coordinate of a unique point on 9
to the right of the origin 0;namely, of that point to the right of 0 whose distance from 0 is r.
To every point Q on 9to the left of the origin 0,
-
we assign a negative real number as its coordinate; the number -Q0,the negative of the distance
between Q and 0.For example,in the diagram
the point U is assumed to be a distance of one unit from the origin 0;therefore, the coordinate of U is
-1. The point W has coordinate -4, which means that the distance is *. Clearly, every negative
real number is the coordinate of a unique point on 9 to the left of the origin.
The origin 0 is assigned the number 0 as its coordinate.
1

[CHAP. 1
2 COORDINATE SYSTEMS ON A LINE
This assignment of real numbers to the points on the line 9is called a coordinatesystem on 9.
Choosing a different origin, a different direction along the line, or a different unit distance would result
in a different coordinate system.
1.2 ABSOLUTE VALUE
For any real number b define the absolute value IbI to be the magnitude of b; that is,
b i f b 2 0
-6 i fb<O
l b l =
In other words, if b is a positive number or zero, its absolute value Ib I is 6 itself. But if b is negative, its
absolute value Ib I is the corresponding positive number -b.
EXAMPLES
Propertiesof the Absolute Value
Notice that any number r and its negative -r have the same absolute value,
Irl = 1-4 (1.1)
(14
lal = lbl implies a = +b (1-3)
la12 = a‘ (1.4)
An important special case of (1.1)results from choosing r = U - U and recalling that -(U - U ) = U - U,
lu - U1 = l u - U 1
If Ia I = Ib I, then either a and b are the same number or a and b are negatives of each other,
Moreover, since Ia I is either a or -a, and (-a)’ = a’,
Replacing a in (14)by ab yields
Iabl2 = (ab)2= a2b2= IU 1’ Ib 1’ = (IU IIb I)’
whence, the absolute value being nonnegative,
lab1 = lallbl
Absolute Value and Distance
Consider a coordinate system on a line 9and let A, and A2 be points on 9with coordinates a,
and a 2 .Then
Ia, - a2I = A,A2 = distance between A , and A2 (1.6)

CHAP. 13
EXAMPLES
COORDINATE SYSTEMS ON A LINE 3
0 1 2 3 4 5
I . . I .
. , , . . . ,Y
0 AI A
2
-3 -2 - 1 0 1 2 3 4
I 1 U U 1 1 I P
Y
A
2 0 AI
1 1 1 1 I 1 I I
- = 14 - (
4
1
= 14 4- 31 = 171 = 7 = A1A2
A special case of (1.6)is very important. If a is the coordinate of A, then
Ia I = distance between A and the origin
Notice that, for any positive number c,
lul Ic is equivalent to -c Iu Ic
1 1 1 *Y
1 1 1
-C 0 C
EXAMPLE lul S 3 ifand only if -3 I U I 3.
Similarly, lul < c is equivalent to -c < U < c (1.9)
EXAMPLE To find a simpler form for the condition Ix - 3 I < 5, substitute x - 3 for U in (1.9), obtaining
-5 <x - 3 < 5. Adding 3, we have -2 < x < 8. From a geometric standpoint,note that Ix - 3 I < 5 is equivalent
to saying that the distance between the point A having coordinatex and the point having coordinate 3 is less than 5.
I 1 I
I 1 1 Y
-2 3 8
It follows immediately from the definition of the absolute value that, for any two numbers a and b,
-1al I a Ilal and -161 I
6 I
lbl
(In fact, either a = Ia I or a = -1 a I.) Adding the inequalities,we obtain
(-14)+(-lbl) I
a +b 2
s I 4 + lbl
- ( l a l + 161)I
U +6 I l a l + lbl
and so, by(1.8),with u = a +6 and c = lal + Ibl,
l a +61 lal+ 161 (1.10)
The inequality (1.10)is known as the triangle inequality. In (1.20)the sign c applies if and only if a and
b are of oppositesigns.
EXAMPLE 13 +(-2)( = 11I = 1, but 131 + 1-21 = 3 +2 = 5.

4 COORDINATE SYSTEMS ON A LINE [CHAP. 1
SolvedProblems
1.1 Recalling that &always denotes the nonnegatiue square root of U, (a)evaluate f l ;
(b)evalu-
ate ,/m;
(c) show that ,
/
? = Ix I. (d)Why isn’t the formula p=
x always true?
(a)f l =
Jd = 3.
(d) By part (c),,
/
? = Ix I, but Ix I = x is false when x < 0. For example, ,
/
- = fi = 3 # -3.
(b)4- = fi = 3. (c) By (1.4),x2 = Ix 1’; hence, since Ix I2 0, = Ix I.
1.2 Solve Ix +3 I 5 5; that is, find all values of x for which the given relation holds.
By (1.8),Ix +31 5 5 if and only if -5 5 x +3 5 5.Subtracting 3, -8 5 x 5 2.
I 1 1
I 1 1
-8 0 2
1.3 Solve I3x +2 I < 1.
By (1.9),I3x +2 I < 1 is equivalent to -1 < 3x +2 < 1. Subtracting 2, we obtain the equivalent rela-
tion -3 < 3x < -1. This is equivalent, upon division by 3,to -1 < x < -3.
I 1 b
- 1 -113 0
1.4 Solve 15 - 3x I<2.
By (1.9), -2 < 5 - 3x < 2
.Subtracting 5, -7 < -3x < -3.Dividing by -3,; > x > 1.
~ ~ ~
ALGEBRA REVIEW Multiplying or dividing both sides of an inequality by a negative number reuerses the
inequality: if U bc.
To see this, notice that a 0. Hence, (b - U)C < 0, since the product of a positive
number and a negative number is negative. So bc - ac < 0, or bc < ac.
1.5 Solve
-
I 1 1 1
7
I I 1 . b
0 1 2 3
x + 4 < 2
x - 3
We cannot simply multiply both sides by x - 3,because we do not know whether x - 3 is positive or
negative.
Case 1: x - 3 > 0
.Multiplying (1) by this positive quantity preserves the inequality:
x +4 < 2~ - 6
4 < x - 6 [subtract x]
10 < x [add 61
Thus, if x > 3,(1)holds if and only if x > 10.
Case 2: x - 3 < 0.Multiplying (1) by this negative quantity reverses the inequality:
x +4 > 2~ - 6
4 > x - 6
10 > x [add 61
[subtract x]

CHAP. 1
3 COORDINATE SYSTEMS ON A LINE 5
Thus, if x < 3, (1)holds if and only if x < 10. But x < 3 implies that x < 10. Hence, when x < 3, (1)is true.
From cases 1and 2, (1)holds for x > 10and for x < 3.
0 3 10
1.6 Solve(x - 2Xx +3) > 0.
A product is positive if and only if both factors are of like sign.
Case 1: x - 2 > 0 and x +3 > 0. Then x > 2 and x > -3. But
x > 2 implies x > -3.
Case2: x - 2 < 0 and x+3<0. Then x < 2 and x<-3,
x < -3 implies x < 2.
Thus, (x - 2)(x +3) > 0 holds when either x > 2 or x < -3.
these are equivalent to
which are equivalent
x > 2 alone, since
to x < -3, since
- 3 - 2 - I 0 1 2
1.7 Solve I3x - 2 I 2 1.
Let us solve the negation of the given relation, I3x - 2 I < 1. By (2.9),
- 1 < 3 x - 2 < 1
1<3x<3 [add 21
+ < X < l [divide by 31
Therefore, the solution of I3x - 2 I 2 1is x 5 4 or x 2 1.
I
3
1
1.8 Solve(x - 3Kx - 1)(x +2) > 0
.
The crucial points are x = 3, x = 1
, and x = -2, where the product is zero. When x > 3, all three
factors are positive and the product is positive. As we pass from right to left through x = 3, the factor
(x - 3) changes from positive to negative, and so the product will be negative between 1and 3. As we pass
from right to left through x = 1, the factor (x - 1) changes from positive to negative, and so the product
changes back from negative to positive throughout the interval between x = -2 and x = 1. Finally, as we
pass from right to left through x = -2, the factor (x +2) changes from positive to negative, and so the
product becomes negative for all x < -2.
+ +
1 1 I
I I I Y
-2 I 3
Thus, the solution consists of all x such that x > 3 or -2 < x < 1.
Supplementary Problems
1.9 (a)For what kind of number u is I u I = -U? (b) For what values of x does I3 - x Iequal x - 3? (c) For
what values of x does 13 - x I equal 3 - x?

6
1.10
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.20
1.21
1.22
[CHAP. 1
COORDINATE SYSTEMS ON A LINE
(a) Solve I2x +3I = 4. (b) Solve I5x - 7 I = 1. (c) m Solve part (a) by graphing y, = (2x+ 3I and
y, = 4. Similarly for part (b).
Solve: (a) Ix - 11< 1 (b) 13x + 51 14 (c) I X +41 > 2
(4 1 2 ~ - 5 1 2 3 (e) I x 2 - 1 0 ( ~ 6 (f) 1?+31<1
X
(9)[H3 Check your answers to parts (a)-(f) by graphing.
(4 11+:1>2 (e) 1 < 3 - 2 x < 5 (f) 3 1 2 x + 1 < 4
(9)IE3 Check your answers to parts (a)-(f) by graphing.
Solve: (a) x(x +2) > 0 (b) (X - 1)(x +4) < 0 (c) X’ - 6~ + 5 > 0
(4 x 2 + 7 x - 8 < 0 (e) x2 < 3x + 4 (f)x(x - 1Nx + 1) > 0
(9) ( 2 ~
+ 1)(x - 3 ) ( ~
+7) < 0
(h) Check your answers to parts (a)-@) by graphing.
[Hints: In part (c),factor; in part (f),
use the method of Problem 1.8.1
Show that if b # 0, then - = -
1;l It;
Prove: (a) 1a21= lalZ (b)
Solve: (a) 12x - 31 = Ix +21
[Hint: Use ( I 4.1
a3I = Ia l3 (c) Generalize the results of parts (a)and (b).
(b) 17x-51=13x+41 (c) 2x- 1 = I x + 7 )
(d) Check your answers to parts (a)-@)by graphing.
Solve: (a) 12x - 31 < Ix + 21
(b) I3x - 2 I 5 Ix - 1I
[Hint: Consider the threecasesx 2 3, -2 I
x < 3,x < -2.1
(c) mCheck your solutions to parts (a)and (b) by graphing.
(a) Prove: [ a- bl 2 Ilal - Ibl I.
and Ibl I [ a- bl +[al.]
(b) Prove: la - 61 I lal + 161.
[Hint: Use the triangle inequality to prove that lal 5 l a - bl +lbl
Determine whether f l =
a
’ holds for all real numbers a.
Does f l < always imply that a < b?
Let 0,
I , A, B, C, D be points on a line, with respective coordinates0, 1,4, -1,3, and -4. Draw a diagram
showing these points and find: m,
AI,m,z
,
a+m,ID, +z
,
E.
Let A and B be points with coordinates a and b. Find b if: (a) a = 7, B is to the right of A, and
Ib - a1 = 3; (b) a = -1, Bis to the left of A, and Ib - a / = 4; (c)a = -2, b < 0, and Ib - a1 = 3.

CHAP. 11 COORDINATE SYSTEMS ON A LINE
1.23 Prove: (a) a < b is equivalent to a +c < b +c.
7
ALGEBRA a < b means that b - a is positive. The sum and the product of two positive numbers are posi-
tive, the product of two negative numbers is positive, and the product of a positive and a negative number
is negative.
a b
(b) If 0 < c, then a < b is equivalent to ac < bc and to - < -.
c c
1.24 Prove (1.6).[Hint: Consider three cases: (a)A, and A, on the positive x-axis or at the origin; (b)A, and A,
on the negative x-axis or at the origin; (c) A, and A, on opposite sides of the origin.]

Chapter 2
Coordinate Systems in a Plane
2.1 THE COORDINATES OF A POINT
We shall establish a correspondence between the points of a plane and pairs of real numbers.
Choose two perpendicular lines in the plane of Fig. 2-1. Let us assume for the sake of simplicity that
one of the lines is horizontal and the other vertical. The horizontal line will be called the x-axis and the
vertical line will be called the y-axis.
c
:L
‘I
1
- l t
Fig. 2-1
Next choose a coordinate system on the x-axis and one on the y-axis. The origin for both coordi-
nate systemsis taken to be the point 0,
where the axes intersect.The x-axis is directed from left to right,
the y-axis from bottom to top. The part of the x-axis with positive coordinates is called the positioe
x-axis, and the part of the y-axis with positive coordinates the positivey-axis.
Consider any point P in the plane. Take the vertical line through the point P, and let a be the
coordinate of the point where the line intersects the x-axis. This number a is called the x-coordinate of P
(or the a6scissa of P).Now take the horizontal line through P, and let 6 be the coordinate of the point
where the line intersects the y-axis. The number 6 is called the y-coordinate of P (or the ordinate of P).
Every point has a unique pair (a, b) of coordinates associated with it.
EXAMPLES In Fig. 2-2, the coordinates of several points have been indicated. We have limited ourselves to
integer coordinatesonly for simplicity.
Conversely,every pair (a, 6)of real numbers is associated with a unique point in the plane.
EXAMPLES In the coordinate system of Fig. 2-3, to find the point having coordinates (3, 2), start at the origin 0,
move three units to the right and then two units upward. To find the point with coordinates (-2, 4), start at the
origin 0,move two units to the left and then four units upward. To find the point with coordinates (- 1, -3), start
from the origin,move one unit to the left and then three units downward.
Given a coordinate system, the entire plane except for the points on the coordinate axes can be
divided into four equal parts, called quadrants. All points with both coordinates positive form the first
quadrant, quadrant I, in the upper right-hand corner (see Fig. 2-4). Quadrant I1 consists of all points
with negative x-coordinate and positive y-coordinate; quadrants I11 and IV are also shown in Fig. 2-4.
8

CHAP. 23
4 Y
(-2.4)
T
4
I
4
4 -
(-2.3) 3 - 0 (3.3)
2 - 0 (2.2) (5.2) +
I
(3.0) 1
- I + 1 - 1 -
I -
(-4,I)
1 1 1 1 1 I A I 1
-4 - 3 - 2 - 1 0 I 2 3 4 5 x %+*
‘ - 2 ,
4
0 (-3. -2) -2(C (0, -2)
-3 - 0 (4, -3) (-1,-3) 1 -3
COORDINATE SYSTEMS IN A PLANE
4 Y
4 -
3 -
2 - 9 (3*2)
1 -
3 4 5 x
-
9
11 3 -
(-1.2) 0 2
(-. +)
I 1 I
-3 -2 - 1 0
0 (-3,-1) -1
4 Y
I
(+*+)
-
I ’ 0 (2.1)
I 1 I
1 2 3
-
Fig. 2-4
The points having coordinates of the form (0, b) are precisely the points on the y-axis. The points
If a coordinate system is given, it is customary to refer to the point with coordinates (a, b) simply as
having coordinates (a,0)are the points on the x-axis.
“the point (a,b).”Thus, one might say: “The point (1,O) lies on the x-axis.”
2.2 THE DISTANCE FORMULA
Let P1and P, be points with coordinates (x,, y,) and (x,, y,) in a given coordinate system (Fig.
2-5). We wish to find a formula for the distancePIP2.
Let R be the point where the vertical line through P, intersects the horizontal line through P,.
Clearly, the x-coordinate of R is x2,the same as that of P,;and the y-coordinate of R is y,, the same as
that of P,.By the Pythagorean theorem,
-
= Fp2
+P,R2
Now if A, .and A, are the projections of P, and P2 on the x-axis, the segments P,R and A,&
are opposite sides of a rectangle. Hence, P,R = A,& But A,A2 = Ix, -x, I by (1.6). Thus,
- -
= Ix, - x2I. Similarly,P,R = Iy1 - y, I. Consequently,
PlP,* = Ix1 - x2 1, + IY l - Y, 1
, = (x, - x2), +011 - Y,),
PlP, = Jbl -x2I2 +b
1- Y,),
whence
[Equation (2.1) is called the distance formula.] The reader should check that this formula also holds
when P,and P, lie on the same horizontal line or on the same vertical line.
(2.1)

10 COORDINATE SYSTEMS IN A PLANE [CHAP. 2
1’
I I
I I
I I
X
Fig. 2-5
Y
I I I
I I I
I I I
X X
Fig. 2-6
EXAMPLES
(a) The distancebetween(3, 8) and (7, 11) is
J(3 - 7
)
’ +(8 - 11)’ = ,/(-4)’ +(-3)’ = J
W= JZ= 5
(b) The distancebetween(4, -3) and (2,7)is
,/(4 - 2)’ +(-3 - 7
)
’ = Jm
= ,
/
- = @
= JGZ = J2*f i = 2 f i
ALGEBRA For any positive numbers U and U, ,
/
& = fi &,since(Afi)2
= (&)’(&)’ = uu.
(c) The distancebetween any point (a,b)and the origin(0,O) is Jm.
2
.
3 THE MIDPOINT FORMULAS
Again consideringtwo arbitrary points Pl(xl,y,) and P2(x2,y2), we shall find the coordinates (x, y)
of the midpoint M of the segment P,P2 (Fig. 2-6). Let A, B, C be the perpendicular projections of P,,
M,P2 on the x-axis. The x-coordinates of A, B, C are x,, x, x2, respectively.
- -
Since the lines P,A, MB,
and P2C are parallel, the ratios P1M/MP2and B
/
B
Care equal. But P,M = MP,;hence AB = E.
SinceAB = x - x1 and
--
= x2 - x,
x - x1= x2 - x
2x = X I +x2
x1 + x 2
2
x=-
(The same result is obtained when P2 is to the left of P,,in which case AB = x1 - x and = x - x2.)
Similarly, y = (yl+y2)/2. Thus, the coordinates of the midpoint M are determined by the midpoint
formulas
Y l + Y 2
and y = -
x1+ x2
x=-
2 2
In words, the coordinates of the midpoint are the averages of the coordinates of the endpoints.

CHAP. 21 COORDINATE SYSTEMS IN A PLANE 11
EXAMPLES
(a) The midpoint of the segment connecting(1, 7)and (3, 5) is (F,
y)
= (2,6).
-
(b) The point halfway between (-2, 5) and (3, 3)is (*, =)
-
- (i,4).
2 2
Solved Problems
2.1 Determine whether the triangle with vertices A( -1,2),B
(
4
,7), C(-3,6) is isosceles.
-
AB = J(-1 - 4)2 +(2 - 7)2= J(-5)2 + ( - 5 ) 2 = ,/ET% = Jso
AC = J[-1 - (-3)12 +(2 - 6)2= Jw'
= JGZ = JZ
BC = J[4 - (-3)]' +(7 - 6)2= J7T+1z = JZGT = Jso
-
-
Since AB = E
,
the triangle is isosceles.
2.2 Determine whether the triangle with vertices A( -5, -3),B( -7, 3), C(2,6)is a right triangle.
Use (2.2)to find the squares of the sides,
-
AB2 = ( - 5 +7)2+(-3 - 3)2= 22 +(-6)2 = 4 + 36 = 40
BC2 = (-7 - 2)2+(3 - 6)2= 81 +9 = 90
AC2 = (-5 - 2)2+(-3 - 6)2= 49 + 81 = 130
-
-
SinceAB2 +m2= E2,
AABC is a right triangle with right angle at B.
GEOMETRY The converse of the Pythagorean theorem is also true: If x2
= AB2 +m2in AABC, then
<ABC is a right angle.
2
.
3 Prove by use of coordinates that the midpoint of the hypotenuse of a right triangle is equidistant
from the three vertices.
Let the origin of a coordinate system be located at the right angle C; let the positive x-axis contain leg
CA and the positive y-axis leg CB [see Fig. 2-7(a)].
Vertex A has coordinates (b, 0),where b = CA; and vertex B has coordinates (0, a), where a = E.
Let
M be the midpoint of the hypotenuse. By the midpoint formulas (2.2),the coordinates of M are (b/2, a/2).
-
(a)
Fig. 2-7

12 COORDINATE SYSTEMS IN A PLANE
Now by the Pythagorean theorem,
and by the distance formula (24,
[CHAP. 2
ALGEBRA For any positive numbers U, U,
- -
Hence, MA = MC. [For a simpler, geometrical proof, see Fig. 2-7(b);MD and BC are parallel.]
2.4 In Fig. 2-8, find the coordinates of points A, B,C, D, E, and F.
2.5 Draw a coordinate system and mark the points having the following coordinates: (1, -l), (4, 4), (-2, -2),
(3, -31, (0,2),(2,0),(-4, 1).
2.6 Find the distance between the points: (a) (2, 3) and (2, 8); (b) (3, 1) and (3, -4); (c) (4, 1) and (2, 1);
(4(-3,4) and (54).
2.7 Draw the triangle with vertices A(4, 7), B(4, -3), and C
(-1,7) and find its area.
2.8 If (-2, -2), (-2,4), and (3, -2) are three vertices of a rectangle, find the fourth vertex.
e F
E
Fig. 2-8

CHAP. 2) COORDINATE SYSTEMS IN A PLANE 13
2.9 If the points (3, 1) and (- 1,O) are opposite vertices of a rectangle whose sides are parallel to the coordinate
axes, find the other two vertices.
2.10 If (2, -l), (5, -l), and (3, 2) are three vertices of a parallelogram, what are the possible locations of the
fourth vertex?
2.11 Give the coordinates of a point on the line passing through the point (2,4)and parallel to the y-axis.
2.12 Find the distance between the points: (a)(2,6)and (7,3); (b)(3, -1)and (0,2);
(c)(43)and (- $, 3).
2.13 Determine whether the three given points are vertices of an isosceles triangle or of a right triangle (or of
both). Find the area of each right triangle.
(4 (-1, 21, (3, -21, (7, 6) (b) (4, I), (1, 2), (3, 8) (4 (4, 11, (1, -41, (-4, -1)
2.14 Find the value of k such that (3, k)is equidistant from (1,2)and (6,7).
2.15 (a) Are the three points A(l, 0),B(3,4), and C(7,8) collinear (that is, all on the same line)? [Hint: If A, B, C
form a triangle, the sum of two sides, AB +E,
must be greater than the third side, AC. If B lies
between A and C on a line,AB +
(b) Are the three points A(-5, -7), B(0, -l),and C(10, 11) collinear?
= x
.
1
2.16 Find the mid oints of the line segments with the following endpoints: (a)(1, -1) and (7, 5); (b)(3,4) and
(190);(4
($11 and (
5
9
3
)
.
2.17 Find the point (a,b) such that (3, 5) is the midpoint of the line segment connecting (a,b) and (1,2).
2.18 Prove by use of coordinates that the line segmentjoining the midpoints of two sides of a triangle is one-half
the length of the third side.

Chapter 3
3
1
- 1
- 2
- 3
- 4
Graphs of Equations
4 9
1512
2 6
912
0 3
312
0
-312
- 3
Consider the following equation involving the variables x and y:
2y - 3~ = 6 (9
Notice that the point (2, 6) satisfies the equation; that is, when the x-coordinate 2 is substituted for x
and the y-coordinate 6 is substituted for y, the left-hand side, 2y - 3x, assumes the value of the right-
hand side, 6. The graph of (i) consists of all points (a, b) that satisfy the equation when a is substituted
for x and b is substituted for y. We tabulate some points that satisfy (i) in Fig. 3-l(a), and indicate these
points in Fig. 3-l(b). It is apparent that these points all lie on a straight line. In fact, it will be shown
later that the graph of (i) actually is a straight line.
1 1 1 1
-4 -3 -2 - 1
-
- 1
I t 1 1 1 1 *
1 2 3 4 5 6 X
r
0 :i’
2
I
Fig. 3-1
In general, the graph of an equation involving x and y as its only variables consists of all points
(x, y) satisfyingthe equation.
EXAMPLES
(a) Some points on the graph of y = x2 are computed in Fig. 3-2(a)and shown in Fig. 3-2(b).These points suggest
that the graph looks like what would be obtained by filling in the dashed curve. This graph is of the type
known as a parabola.
(b) The graph of the equation xy = I is called a hyperbola. As shown in Fig. 3-3(b), the graph splits into two
separate pieces. The points on the hyperbola get closer and closer to the axes as they move farther and farther
from the origin.
(c) The graph of the equation
is a closed curve, called an ellipse (see Fig. 3-4).
14

CHAP. 31

t'
2
4
I I PW'l
GRAPHS OF EQUATIONS
I
I
6 - I
I
7 -
i
5 -
4 -
3 - I
I
I:; 1 1 -
I
f
9K
8 I
I

t
113
It4
-1/4
-113
-112
- 1
-2
-3
15
3
4
-4
- 3
-2
- 1
-112
-113
1

1
1 1
-2 -1
- 1

'
.
0 0
- 1
-2 4
T
-3 9

1 I
1
*
, i X
1
-
/'
0
-3 -2 - 1 "I 1 2 3 x
Fig. 3-2
I
- 4 - 3 -2 - 1
1 I 1 1
Fig. 3-3
- - 1
- -2
- -3
- - 4
/- I
Fig. 3-4

16 GRAPHS OF EQUATIONS [CHAP. 3
Circles
For a point P(x, y) to lie on the circle with center C(a,b) and radius r, the distance pc must be r
(Fig. 3-5). Now by (24,
-
PC = J(x -a)2+(y -
The standard equation,E2
= r2, of the circle with center (a,b) and radius r is then
(x - a)2+(y - b)2 = r2
x2 +y2 = r2
For a circle Entered at the origin,(3.1)becomes simply
t’
Fig. 3-5
EXAMPLES
(a) The circle with center (1,2)and radius 3 has the equation
(x - 1)2 +(y - 2)2 = 9
(b) The circle with center (- 1,4) and radius 6 has the equation
(X + 1)2 +(y - 4)2 = 36
(c) The graph of the equation (x - 3)2 +(y - 7)2= 16 is a circle with center (3, 7)and radius 4.
(d) The graph of the equation x2 +(y +2)2 = 1 is a circle with center (0, -2) and radius 1.
Sometimesthe equation of a circle will appear in a disguised form. For example,the equation
x2 +y2 - 6x +2y +6 = 0 (ii)
is equivalent to
(x - 3)2 +(y + 1)2= 4 (iii)
ALOBBRA Use the formulas (U+U)’ = u2 +2uo +o2 and (U - u ) ~
= u2- 2uo +o2 to expand the left-hand side
of (iii).
If an equation such as (ii) is given, there is a simple method for recovering the equivalent standard
equation of the form (iii) and thus finding the center and the radius of the circle. T
h
i
s method depends

CHAP. 3) GRAPHS OF EQUATIONS 17
0
0
0
0
0
0
0
0
on completing the squares; that is, replacing the quantities x2 +Ax and y2 +By by the equal quantities
A2 and ( y + : ) ’ - ~
B2
EXAMPLE Let us find the graph of the equation
x 2 + y 2 + 4 x - 2 y + 1 = o
( x + 2 ) 2 - 4 + ( y - 1 ) 2 - 1 + 1 = o
(x +2)2 +(y - 1)2= 4
Completing the squares, replace x2 +4x by (x +2)2- 4 and y2 - 2y by (y - 1)2- 1,
This is the equation of a circle with center (- 2, 1)and radius 2.
Solved Problems
3
.
1 Find the graph of: (a)the equation x = 2; (b) the equation y = -3.
(a) The points satisfying the equation x = 2 are of the form (2, y), where y can be any number. These
points form a vertical line [Fig. 3-6(a)].
(6) The points satisfying y = -3 are of the form (x, -3), where x is any number. These points form a
horizontal line [Fig. 3-6(6)].
I
Fig. 3-6
3
.
2 Find the graph of the equation x = y2.
2
:F
...... ..............
- 3 t
Plotting several points suggests the curve shown in Fig. 3-7. This curve is a parabola, which may be
obtained from the graph of y = x2(Fig. 3-2) by switching the x- and y-coordinates.

18
3 -
2 -
'
GRAPHS OF EQUATIONS
4
-
-
----
,
,
-
-/
[CHAP. 3
- 2
- 3 --
-44
--
-
-
3.3 Identify the graphs of:
(U) 3x2 + 3y2 - 6~ - y + 1 = 0 (6) x2 +y2 - 8~ + 16y +80 = 0
(c) x2 +y2 +20x - 4y + 120 = 0
(a) First, divide both sides by 3,
1 1
3 3
x2 +y2 - 2x - - y +- = 0
Complete the squares,
1
or
1 1 36 1 12 25
36 3 36 36 36 36
( x - 1 ) 2 + y - - = I + - - - = - + - - - = -
(
Hence, the graph is a circle with center (1,i) and radius 2.
(6) Complete the squares,
(x - 4)2+(y +8)2 +80 - 16 - 64 = 0 or (x - 4)2+(y +8)' = 0
Since (x - 4)22 0 and (y +8)2 2 0, we must have x - 4 = 0 and y +8 = 0. Hence, the graph consists
of the single point (4, -8).
(c) Complete the squares,
(x + 1
0
)
2+(y - 2)2 + 120 - 100 - 4 = 0 or (x + 10)2 +(y - 2)2 = -16
This equation has no solution, since the left-hand side is always nonnegative. Hence, the graph consists
of no points at all, or, as we shall say, the graph is the null set.
3.4 Find the standard equation of the circle centered at C(l, -2) and passing through the point
P(7, 4).
The radius of the circle is the distance
-
CP = J(7 - q2+[4 - (-2)12 = JZT-55 = J72
Thus, the standard equation is (x - 1)2 +(y + 2)2 = 72.

CHAP. 31 GRAPHS OF EQUATIONS 19
3.5 Find the graphs of: (a)y = x2 +2; (b)y = x2 - 2;(c) y = (x - 2)2;(6)Y = (x +2)2.

The graph of y = x2 +2 is obtained from the graph of y = x2 (Fig. 3-2)by raising each point two units
in the vertical direction [see Fig. 3-8(a)].
The graph of y = x2 - 2 is obtained from the graph of y = x2 by lowering each point two units [see
Fig. 3-8(b)].
The graph of y = (x - 2)2 is obtained from the graph of y = x2 by moving every point of the latter
graph two units to the right [see Fig. 3-8(c)]. To see this, assume (a, b) is on y = (x - 2)2. Then
b = (a - 2)2. Hence, the,point (a - 2, b) satisfies y = x2 and therefore is on the graph of y = x2. But
(a,b) is obtained by moving (a - 2, b) two units to the right.
The graph of y = (x +2)2is obtained from the graph of y = x2 by moving every point two units to the
left [see Fig. 3-8(d)]. The reasoning is as in part (c).
Parts (c) and (d) can be generalized as follows. If c is a positive number, the graph of the equation
/
;c I
F(x - c, y) = 0 is obtained from the graph of F(x, y) = 0 by moving each point of the latter graph c units to
the right. The graph of F(x +c, y) = 0 is obtained from the graph of F(x, y) = 0 by moving each point of

9

3

4
- I
1 1 I ' l l
b
- 3 - 2 0
the latter graph c units to the left.
F L
7,
6 - I
I
I
4 - I
I
5 -
2
: Ii
I - I
- b
I l l 1 1 1
I / 2
1 1 1 1 1
-2 - 1 0
;
I
I
I
I
I
I
I
9
I
/
/
/
#
1 -
1 1 1 1 1 ,
I 2 X
t
/ 2
I
?
I
;-3
-
1 -
1 1 I J l
-5 -4 -3 -2 - 1 0
I 1 1 1 1
I 2
x
-
i
I
I

t

t t
Fig. 3-8

20 GRAPHS OF EQUATIONS [CHAP. 3
3
.
6 Find the graphs of: (a)x = 0) - 2)2;(b)x = ( y +2)2.
(a) The graph of x = (y - 2)2 is obtained by raising the graph of x = y2 [Fig. 3-7(b)]by two units [see
Fig. 3-9(a)]. The argument is analogous to that for Problem 3 3 4 .
(b) The graph of x = (y +2)2 is obtained by loweringthe graph of x = y2 two units [see Fig. 3-9(b)].
These two results can be generalized as follows. If c is a positive number, the graph of the equation
F(x,y - c) = 0 is obtained from the graph of F(x, y) = 0 by moving each point of the latter graph c units
vertically upward. The graph of F(x, y +c) = 0 is obtained from the graph of F(x, y) = 0 by moving each
point of the latter graph c units vertically downward.
3
.
7
Fig. 3-9
I 2 3 4
I 1 1 1 +
X
Find the graphs of: (a)y = (x - 3)2 +2; (b)fix - 2) = 1.
(a) By Problems 3.5 and 3.6, the graph is obtained by moving the parabola y = x2 three units to the right
and two units upward [see Fig. 3-1qu)l.
(b) By Problem 3.5,the graph is obtained by moving the hyperbola xy = 1 (Fig. 3-3)two units to the right
[see Fig. 3-1qb)l.
Fig. 3-10

x2 y2 y2 x2
- 1; (b) - - - = 1.
9 4
3.8 Draw the graphs of: (a) -- --
9 4
X2 x2 y2 X2
y2 - 1, then - = -+ 1 2 1. So -2 1 and, therefore, Ix I 2 3. Hence, there are no points
If -- --
9 4 9 4 9
(a)
(x, y) on the graph within the infinite strip -3 < x < 3. See Fig. 3-ll(b) for a sketch of the graph,
which is a hyperbola.
(b) Switching x and y in part (a),we obtain the hyperbola in Fig. 3-12.
Fig. 3-11
Fig. 3-12
3.9 Draw the graphs of the followingequations:
(U) 3 y - x = 6 (b) 3 y + x = 6 (c) x = - 1
1
(4 y = 4 (e) y = x 2 - 1 (j)y = ; + ~
(9) Y = x (h) y = - x (i) y2 = x2
U) Check your answers on a graphing calculator.

22
3.10
3.11
3.12
3.13
GRAPHS OF EQUATIONS
On a single diagram, draw the graphs of:
[CHAP. 3
1 1
y = x 2 (b) y = 2x2 (c) y = 3x2 (d) y = j x 2 (e) y = j x 2
Check your answers on a graphing calculator.
Draw the graph of y = (x - 1)2.(Include all points with x = -2, -1, 0, 1, 2, 3, 4.) How is this graph
related to the graph of y = x2? Check on a graphing calculator.
. Check on a graphing calculator.
1
Draw the graph of y = -
x - 1
Draw the graph of y = (x + 1)2. How is this graph related to that of y = x2?[H3 Check on a graphing
calculator.
.
1
Draw the graph of y = -
x + l
Check on a graphing calculator.
Sketch the graphs of the followingequations. Check your answers on a graphing calculator.
(c) x2 - y2 = 1
x2 y2
(a) -+-= 1 (b) 4x2 +y2 = 4
4 9
(4 Y = x 3
[Hint:
Parts (c) and (f) are hyperbolas. Obtain part (e)from part (a).]
Find an equation whose graph consists of all points P(x, y) whose distance from the point F(0, p)
is equal to its distance PQ from the horizontal line y = - p (p is a fixed positive number). (See
Fig. 3-13.)
0
- p I
Fig. 3-13
3.14 Find the standard equations of the circles satisfying the given conditions: (a) center (4, 3), radius 1;
(b) center (- 1, 9, radius fi;(c) center (0, 2), radius 4; (d)center (3, 3), radius 3fi; (e) center (4, -1)
and passing through (2, 3); (f)
center (1,2) and passing through the origin.
3.15 Identify the graphs of the followingequations:
(U)
(c) x2 +y2 +3x - 2y +4 = 0
(e)
X’ +y2- 1 2 ~
+20y + 15 = 0 (b) x2 +y2 +30y +29 = 0
(d) 2x2 + 2y2 - x = 0
x2 +y2 +2x - 2y +2 = O (f)x2 +y2 +6x +4y = 36

3.16
3.17
3.18
3.19
3.20
(a) Problem 3.3 suggests that the graph of the equation x2 +y2 +Dx +Ey +F = 0 is either a circle, a
point, or the null set. Prove this.
(b) Find a condition on the numbers D, E, F which is equivalent to the graph’s being a circle. [Hint:
Completethe squares.]
Find the standard equation of a circle passin through the following points. (a) (3, 8), (9, 6),
and (13, -2); (b) (5, 5), (9, l), and (0, d.
[Hint: Write the equation in the nonstandard
form x2 +y2 +Dx +Ey +F = 0 and then substitute the values of x and y given by the three points. Solve
the three resultingequationsfor D, E, and F.]
For what value@)
of k does the circle (x - k)2 +0,-2k)2 = 10pass through the point (1, l)?
Find the standard equationsof the circles of radius 3 that are tangent to both linesx = 4 and y = 6.
What are the coordinates of the center(s)of the circle@)of radius 5 that pass through the points (- 1,7)and
(
-
2
.
9 6)?

Chapter 4
Straight Lines
4.1 SLOPE
If P,(x,, y,) and P2(x2,y2) are two points on a line 9,the number m defined by the equation
m=- Y2 -Y,
x2 - x1
is called the slope of 9.The slope measures the “steepness”of 9.It is the ratio of the changey, -y ,
in the y-coordinate to the change x2 -x, in the x-coordinate. T
h
i
s is equal to the ratio RPz/P,R in
Fig. 4l(a).
--
Fig. 4-1
Y Y
Notice that the value m of the slope does not depend on the pair of points P,, P, selected. If
another pair P3(x3,y3) and P4(x4,y4) is chosen, the same value of m is obtained. In fact, in Fig. 4-l(b),
AP,P4S is similarto M 1 P 2
R.
o m m y The angles at R and S are both right angles, and the angles at PI and P, are equal because they are
corresponding angles determined by the line 9 cutting the parallel lines P,R and P,S. Hence,AP, P, S is similar
to M,P,R because two angles of the first triangleareequal to two correspondingangles of the second triangle.
Consequently,sincethe corresponding sidesof similar triangles are proportional,
- -
Y2 - Y , - Y4-Y3
P,R P,S xz-x, x4-xg
RP, SP,
--- Or ---
---
that is, the slope determined from Pi and Pz is the same as the slope determined from P, and P I .
24

CHAP. 41 STRAIGHT LINES 25
EXAMPLE In Fig.4-2, the slope of the line connectingthe points (1,2) and (3, 5) is
5 - 2 3
3 - 1 2
1.5
-=-=
Notice that as a point on the line moves two units to the right, it moves three units upward. Observe also that the
order in which the points are taken has no effect on the slope:
2 - 5 -3
1 - 3 -2
-=-- - 1.5
In general,
Y2 -Y1 Yl - Y 2
X2 -x, XI -X2
-=-
Fig. 4 2
The slope of a line may be positive, zero, or negative. Let us see what the sign of the slope indicates.
Consider a line 9that extends upward as it extends to the right. From Fig. 4-3(a),we see that
y2 >y,; hence, y , - y , > 0.In addition, x2 > x, and, therefore, x2 - x, > 0. Thus,
m=- Y 2 - Y l ,o
x2 - x1
The slope of 9is positive.
Consider a line 9
’ that extends downward as it extends to the right. From Fig. 4-3(b), we see that
y , <y , ; therefore,y , - y , < 0.But x2 > x,, so x2 - x, > 0. Hence,
Y 2 - Y , < 0
m=-
*2 - x1
The slope of 9is negative.
Consider a horizontal line 9.
From Fig. 4-3(c),y , = y , and, therefore,y , - y , = 0. Since x2 > x,,
x2 - x, >0. Hence,
O O
Y2 - Y ,
x, -x, x, -x,
The slope of 9is zero.
Consider a vertical line 9.
From Fig. 4-3(4, y 2 > y,, so that y , - y , > 0. But x2 = x,, so that
x2 - x, = 0. Hence, the expression
m=-=-=
Y , - Y l
x2 - x1
is undefined. The concept of slope is not defined for 9.
(Sometimes we express this situation by
saying that the slope of 9is infinite.)

26 STRAIGHT LINES [CHAP. 4
f’
(4
Fig. 4-3
Now let us see how the slope varies with the “steepness” of the line. First let us consider lines with
positive slopes, passing through a fixed point P,(x,, yJ. One such line is shown in Fig. 4-4. Take
another point, P2(x2,y,), on A? such that x2 - x1 = 1. Then, by definition, the slope rn is equal to the
distance RP, .Now as the steepnessof the line increases, RP, increases without limit [see Fig. 4-5(a)J.
Thus, the slope of 9increases from 0 (when 9is horizontal) to +CO (when 9 is vertical). By a similar
construction we can show that as a negatively sloped line becomes steeper and steeper,the slope stead-
ily decreases from 0 (when the line is horizontal) to -CQ (when the line is vertical) [see Fig. 4-5(b)].
- -
t’ Y
Fig. 44

CHAP. 41 STRAIGHT LINES
4 Y
m = O m = O
27
(a1
Fig. 4-5
4.2 EQUATIONS OF A LINE
Consider a line 9
’ that passes through the point P,(x,, y,) and has slope rn [Fig. 4-6(a)]. For any
other point P(x,y) on the line, the slopern is, by definition,the ratio of y - y, to x - xl. Hence,
O
n the other hand, if P(x, y) is not on line 9 [Fig. 4-6(b)], then the slope (y -y,)/(x - x,) of the line
PP,is different from the slope rn of 9,so that (4.1)does not hold. Equation (4.1)can be rewritten as
Note that (4.2)is also satisfied by the point (x,, y,). So a point (x, y)is on line 14 if and only if it satisfies
(4.2);that is, 9 is the graph of (4.2).Equation (4.2)is called a point-slope equation of the line 9.
t Y
c
X
t Y
X
EXAMPLES
(U) A point-slope equation of the line going through the point (1,3) with slope 5 is
(b) Let 9 be the line through the points (1,4) and (- 1,2). The slope of 9 is
y -3 = 5(x - 1)
2 1
4 - 2
m=-=-=
1-(-1) 2

Therefore, two point-slopeequationsof 9 are
Y - 4 - x - 1 and y - 2 = ~ + 1
Equation (4.2) is equivalentto
y-y, =mx-mx, or y=mx+(y, -mx,)
Let b stand for the numbery, - mx,.Then the equation becomes
y = m x + b (4.3)
When x = 0, (4.3) yields the value y = b. Hence, the point (0,b) lies on 9.Thus, b is the y-coordinate of
the point where 9 intersects the y-axis (see Fig. 4-7). The number b is called the y-intercept of 9,and
(4.3)is called the slope-interceptequation of 9.
4 Y
Y
*
X
Fig. 4-7
EXAMPLE Let 14 be the line through points (1,3) and (2, 5). Its slopem is
5 - 3 2
-- - - = 2
2 - 1 1
Its slope-interceptequation must have the form y = 2x +b. Since the point (1, 3) is on line 14,(1,3) must satisfythe
equation
3 = 2(1) +b
So, b = 1, giving y = 2x + 1 as the slope-interceptequation.
An alternativemethod is to write down a point-slopeequation,
whence,
y -3 = 2(x - 1)
y - 3 = 2x -2
y = 2 x + 1
4 3 PARALLEL LINES
Assume that 9,and 9, are parallel, nonvertical lines, and let P, and P, be the points where 2Yl
and 9
, cut the y-axis [see Fig. 4-8(u)]. Let R, be one unit to the right of P,,and R, one unit to the
right of P,. Let Q, and Q2be the intersections of the vertical lines through RI and R2with 2Yl and
Y 2 .
Now AP,RIQ1is congruent to AP, R2Qz.
GEOMETRY U
s
e the ASA (angle-side-angle)congruencetheorem. S R I = SR, since both are right angles,
-
P1R-,=P2R2=1
<P1= %P,,since SP,and <P, are formed by pairs of parallel lines.

CHAP. 41
A Y
STRAIGHT LINES
t’
29
Fig. 4-8
- -
Hence, RIQl
= R2Q2,
and
- -
slope of Yl = -
RIQ1
-
--
R 2 Q 2-
- slope of 9 2
1 1
Thus, parallel lines have equal slopes.
Conversely,if different lines Yl and Y 2
are not parallel, then their slopes must be different. For if
Yl and 9, meet at the point P [see Fig. 4-8(b)] and if their slopes are the same, then Yl and Y 2
would have to be the same line. Thus, we have proved:
Theorem4.1: Two distinct lines are parallel if and only if their slopes are equal.
EXAMPLE Let us find an equation of the line 9through (3, 2) and parallel to the line A having the equation
3x - y = 2. The line A has slope-intercept equation y = 3x +2. Hence, the slope of A is 3, and the slope of the
parallel line 9also must be 3. The slope-intercept equation of 9 must then be of the form y = 3x +b. Since(3, 2)
lies on 9,2 = 3(3) +b, or b = -7. Thus, the slope-intercept equation of 9is y = 3x - 7. An equivalent equation
is 3x -y = 7.
4.4 PERPENDICULAR LINES
Theorem4.2: Two nonvertical lines are perpendicular if and only if the product of their slopes is -1.
Hence, if the slope of one of the lines’isrn, then the slope of the other line is the negative
reciprocal -l/m.
For a proof, see Problem 4.5.
4.1 Find the slope of the line having the equation 5x - 2y = 4. Draw the line and determine whether
the points (10,23)and (6, 12)are on the line.

Solve the equation for y,
5
y = - x - 2
2
Thus, we have the slope-intercept form; the slope is # and the y-intercept is -2. The line goes through the
point (0, -2). To draw the line, we need another point on the line. Substitute 2 for x in (I), obtaining y = 3.
Hence, (2, 3) is a point on the line (see Fig. 4-9). (We could have found other points on the line by substitut-
ing numbers other than 2 for x.) To test whether (10, 23) is on the line, substitute 10 for x and 23 for y and
see whether the equation 5x - 2y = 4 holds. The two sides turn out to be equal; so (10, 23) is on the line. A
similar check shows that (6, 12)is not on the line.
4.2 Find an equation of the line 9 which is the perpendicular bisector of the line segment connect-
ing the points A(-1, 1)land B(4, 3) (see Fig. 4-10).
9 must pass through the midpoint M of segment AB. By the midpoint formula, the coordinates of M
are (3,2).The slope of the line through A and B is
3 - 1 2
4-(-1) 5
-=-
Hence, by Theorem 4.2, the slope of 14 is
A point-slope equation of 9 is y - 2 = -3(x - 3).
- B
A

Fig. 4-9 Fig. 4-10
4.3 Determine whether the points A(-1, 6), B(5, 9), and C
(
7
,10) are collinear; that is, whether the
points all lie on the same line.
A, B, C will be collinear if and only if the line AB is the same as the line AC, which is equivalent to the
slope of AB being equal to the slope of AC.
The slopes of AB and AC are
10-6 4 1
and -=-=-
5-(-1) 6 2 7-(-1) 8 2
9 - 6 3 1
-=-=-
Hence, A, B, C are collinear.
4.4 Prove by use of coordinates that the diagonals of a rhombus (a parallelogram of which all sides
are equal) are perpendicular to each other.

Represent the rhombus as in Fig. 4-11. (How do we know that the x-coordinate of D is U +U?) Then
the slope of diagonal AD is
W
--
w - 0
m, =
u + u - 0 - u + u
and the slope of diagonal BC is
Hence,
w - 0 w
m2=-=-
U - U U - U
W2
m,m2= - - =-
(U: U
) u2 - u2
Since ABDC is a rhombus,AB = E.But AB = U and AC = ,
/
=
. So,
Jm-
= U or u2 +w2 = U ’ or w2 =U’ - u2
Consequently,
w2 u2 - u2
m , m , = m = p - J = - 1
and, by Theorem4.2,lines AD and BC are perpendicular.
t Y C(u.
w ) D(u+ U. w )
Fig. 4-11
4.5 Prove Theorem 4.2.
Assume that 9,
and 9,
are perpendicular nonvertical lines of respective slopes m, and m, .We shall
show that m1m2 = -1.
Let 1
4
: be the line through the origin 0 and parallel to 9,,
and let 1
4
: be the line through the origin
and parallel to 14, [see Fig. 4-12(a)]. Since 147 is parallel to 14,and 2’;
is parallel to 9,,
the slope of 147
is m, and the slope of 9
:
is m2 (by Theorem 4.1).Also 1
4
: is perpendicular to 9;
since9,
is perpendicu-
lar to 14,. Let R be the point on 9fwith x-coordinate 1, and let Q be the point on 9
3with x-coordinate
1 [see Fig. 4-12(b)]. The slope-interceptequation of 1
4
: is y = mlx, and so the y-coordinate of R ism, since
its x-coordinate is 1.Similarly,the y-coordinateof Q is m, .By the distance formula,
-
OQ = J(1 -0
)
’+(m,-0
)
’ = ,
/
-
OR = ,/(l - O), +(ml- 0
)
’= Jm
-
-
’ QR = &i - i)2 +(mz-ml)2 = ,
/
-
By the Pythagorean theoremfor the right triangle QOR,
-
QR~=OQZ+OR’
(m, - m,)’ = (1 +mf) +(1 +mf)
mf - 2m2m, +mf = 2 +mf+mi
-2m,m, = 2
mlm2 = -1
Conversely,assume that mlm2 = -1, where rn, and m, are the slopes of lines 9,
and 14,. Then 9,
is
not parallel to 9,.
(Otherwise,by Theorem 4.1,m: = -1, which contradicts the fact that a square is never

(b)
Fig. 4-12
negative.)Let Yl intersect Y 2at point P (see Fig. 4-13). Let LY3 be the line through P and perpendicular to
Y1.
If m3 is the slope of Y 3 ,
then, by what we havejust shown,
m1m3= -1 = m1m2 or m3= rn,
Since Lf2 and Y 3pass through P and have the same slope, they must coincide. Since LYl is perpendicular
to Y 3 ,
Yl is perpendicular to Y 2 .
4.6 Show that the line y = x makes an angle of 45" with the positive x-axis (that is, KPOQ in Fig.
4-14 contains 45").
Let P be the point on the line y = x with coordinates (1, 1).Drop a perpendicular PQ to the positive
x-axis. Then PQ = 1 and ?@= 1. Hence, XOPQ = XQOP,since they are base angles of isosceles triangle
QPO.Since 3c OQP contains90",
XOPQ + XQOP = 180" - XOQP 180" - 90"= 90"
Since SOPQ = <QOP,they each contain 45".
4.7 Sketch the graph of the equation Ix I +Iy I= 1.
Consider each quadrant separately. In the first quadrant, Ix I = x and Iyl = y. Hence, the equation
becomes x +y = 1; that is, y = -x + 1. This yields the line with slope -1 and y-intercept 1. This line
intersects the x-axis at (1,O). Hence, in the first quadrant our graph consists of the line segment connecting
(1, 0) and (0, 1) (see Fig. 4-15). In the second quadrant, where x is negative and y is positive, 1x1= -x
and Iy(= y, and our equation becomes -x +y = 1, or y = x + 1. This yields the line with slope 1 and
t' t'
Fig. 4-13 Fig. 4-14

y-intercept 1, which goes through the points (- 1,O) and (0, 1). Hen& in the second quadrant we have the
segment connecting those two points. Likewise, i
n the third quadrant we obtain the segment connecting
(- 1,O) and (0, -l),and in the fourth quadrant the segmentconnecting(0,-1) and (1,O).
t’
Fig. 4-15
4.8 Find a point-slope equation of the line through each of the followingpairs of points: (a) (85)and (- 1,4);
(b) (1,4)and the origin;(c)(7, -1) and (- 1,7).
49 Find the slope-interceptequation of the line: (a) through points (-2, 3) and (4, 8); (b) having slope 2 and
y-intercept -1; (c) through points (0, 2) and (3, 0);(a) through (1,4)and parallel to the x-axis; (e) through
(1,4)and rising fiveunits for each unit increase in x; (f) through (5, 1) and falling thrct units for each unit
increase in x; (g) through (- 1, 4) and parallel to the line with the equation 3x +4y = 2; (h) through the
origin and parallel to the line with the equation y = 1;(i) through (1,4)and perpendicular to the line with
the equation 2x - 6y = 5 ; 0
1 through the origin and perpendicular to the line with the equation
5x +2y = 1;(k)through (4,3) and perpendicular to the line with the equation x = 1; (0through the origin
and bisectingthe anglebetween the positive x-axis and the positive y-axis.
410 Find the slopesand y-interceptsof the lines given by the followingequations. Also find the coordinatesof a
point other than (0, b) on each line.
4.11 If the point (2, k)lies on the line with slopem = 3 passing through the point (1, a), find k.
4.12 D
o
e
sthe point (- 1, -2) lie on the line through the points (4,7)and (5,9)?
4.13 (a) Use slopes to determine whether the points 4 4 , l), B(7, 3), and C(3, 9) are the vertices of a right
triangle.
(b) U
s
e slopes to show that A(5,4),B(-4,2), C(-3, -3), and D(6, -1) are verticesof a parallelogram.
(c) Under what conditions are the points A(& U +w),B(o,U +w), and C(w,U +U) on the sameline?
(a) Determine k so that the points 4 7 , 5), B(-1, 2), and C(k, 0) are the vertices of a right triangle with
right angleat B.

4.14
4.15
4.16
4.17
4.18
4.19
4.20
4.21
4.22
4.23
4.24
4.25
Determinewhether the given lines are parallel,or perpendicular,or neither.
(a) y = 5x - 2 and y = 5x +3 (b) y = x + 3 a n d y = 2 x + 3
(c) 4x - 2y = 7 and lOx - 5y = 1
(e) 7x + 3y = 6 and 3x +7y = 14
(d) 4x - 2y = 7 and 2x +4y = 1
Temperature is usually measured either in Fahrenheit or in Celsius degrees.The relation between Fahren-
heit and Celsius temperaures is given by a linear equation. The freezingpoint of water is 0" Celsius or 32"
Fahrenheit, and the boiling point of water is 100"Celsius or 212" Fahrenheit. (a) Find an equation giving
Fahrenheit temperature y in terms of Celsius temperature x. (b) What temperature is the same in both
scales?
For what values of k will the line kx +5y = 2k: (a)have y-intercept 4; (b) have slope 3; (c) pass through
the point (6, 1); (d) be perpendicularto the line 2x - 3y = l?
A triangle has vertices A(1,2), B(8,0),C(5,3). Find equations of: (a) the median from A to the midpoint of
the opposite side;(b) the altitudefrom B to the oppositeside;(c) the perpendicular bisector of side AB.
Draw the line determinedby the equation 4x - 3y = 15. Find out whether the points (12,9)and (6,3)
lie on
this line.
(a) Prove that any linear equation ax +yb = c is the equation of a h e , it being assumed that a and b are
not both zero. [Hint: Consider separatelythe casesb # 0 and b = 0.)
(b) Prove that any line is the graph of a linear equation. [Hint: Consider separately the cases where the
line is vertical and where the line is not vertical.]
(c) Prove that two lines ulx +b,y = c, and a, x +b,y = c2 are parallel if and only ifa,b, = a, b,. (When
a, # 0 and b, # 0, the latter conditionis equivalent to a,/a, = bJb,.)
If the line 9 has the equation 3x +2y -4 = 0, prove that a point P(x, y) is above 9 if and only if
3x +2y - 4 > 0.
If the line .
M has the equation 3x - 2y - 4 = 0, prove that a point P(x, y) is below 4 if and only if
3x -2y - 4 > 0.
(a) Use two inequalities to define the set of all points above the line 4x +3y - 9 = 0 and to the right of
the line x = 1. Draw a diagram indicatingthe set.
(b) mUse a graphing calculatorto check your answerto part (a).
(a) The leading car rental company, Heart, charges $30 per day and 1% per mile for a car. The second-
ranking company,Bird, charges$32 per day and 126 per mile for the same kind of car. If you expect to
drivex miles per day, for what values of x would it cost less to rent the car from Heart?
(b) Solve part (a) using a graphing calculator.
Draw the graphs of the followingequations:(a) Ix I- Iy I= 1;(b)y = +(x +Ix I).
(c) mUse a graphing calculatorto solvepart (b).
Prove the followinggeometric theorems by using coordinates.
(a) The figure obtained by joining the midpoints of consecutive sides of any quadrilateral is a parallelo-
gram.
(b) The altitudesof any triangle meet at a common point.
(c) A parallelogramwith perpendiculardiagonalsis equilateral(a rhombus).
(d) If two medians of a triangle are equal, the triangle is isosceles.
(e) An angle inscribed in a semicircle is a right angle. [Hint: For parts (a), (b),and (c), choose coordinate
systemsas in Fig. 4-16.1

CHAP. 41 STRAIGHT LINES
D
35
D C
A Y
C
A B x
(61
Fig. 4-16
+ Y 4 Y
- - + + +
A B x A B X
4.26 Describegeometrically the families of the following lines, where m and b are any real numbers:
(a) y = m x + 2 (b) y = 3 x + b
4.27 The x-intercept of a line 9 is defined to be the x-coordinate of the unique point where 9 intersects the
x-axis. Thus, it is the number U for which (a,0)lies on 9.
(a) Which lines do not have x-intercepts?
(b) Find the x-intercepts of (i) 2x - 3y = 4, (ii)x + 7y = 14, (iii)5x - 13y = 1,(io) x = 3, (U)y = 1.
(c) If a and b are the x-intercept and y-intercept of a line, show that
X Y
- + - = I
a b
is an equation of the line.
(d) If (2) is an equation of a line, show that a is the x-intercept and b is the y-intercept.
4.28 In the triangle with vertices 4 3 , l), B(2, 7), and C(4, 10) find the slope-intercept equation of: (a) the
altitude from A to side BC;(b)the median from B to side AC; (c)the perpendicular bisector of side AB.

Chapter 5
Intersections of Graphs
The intersection of two graphs consists of the points that the graphs have in common. These points
can be found by solving simultaneouslythe equations that determine the graphs.
EXAMPLES
(a) To find the intersection of the lines 9 and A determinedby
9: 4 x - 3 ~ ~ 1 5 A: 3x+2y=7
multiply the first equation by 2 and the secondequation by 3,
8x - 6y = 30
9x +6y = 21
Now y has the coefficients -6 and 6 in the two equations,and we add the equations to eliminatey,
17x=51 or x = - =
51 3
17
From the eqilation for A,when x = 3,3(3) +2y = 7. Hence,
9 + 2 y = 7 or 2 y = - 2 or y = - 1
Thus,the only point of intersectionis (3, -1)(see Fig. 5-1).
t’
Fig. 5-1
(b) Let us find the intersection of the line 2’:y = 2x + 1 and the circle (x - 1)2+0,+ 1)2= 16. We must solve
the system
y = 2 x + 1 (1)
(2)
(X - 1)2 +(J + 1)2= 16
By (Z), substitute2x + 1for y in (2),
(X - 1)2+( 2 ~
+2)2 = 16
(x’ - 2~ + 1) +(4x2 +8x +4) = 16
5x2 +6x + 5 = 16
5x2 +6x- 11 = O
(5x + llXx - 1) = 0
Hence,either 5x + 11= 0 or x - 1 = 0; that is, either x = -11/5 = -2.2 or x = 1. By (Z), when x = 1, y = 3;
and when x = -2.2, y = -3.4. Thus, there are two intersection points, (1, 3) and (- 2.2, -3.4), as indicated in
Fig. 5-2.
36

CHAP. 51 INTERSECTIONS OF GRAPHS 37
(c) Tofind the intersectionof the line y = x +2 and the parabola y = x2, we solve the system
y = x + 2
y = x2
By (4),substitutex2for y in (3),
x 2 = x + 2
x2 - x - 2 = 0
(x -2xx + 1) =0
Hence, either x -2 = 0 or x + 1 = 0; that is, either x = 2 or x = -1. By (3) or (4,when x = 2
,y 5 4; and
when x = -1, y = 1. Thus,the intersectionpoints are (2,4) and (- 1, 1) (see Fig. 5-3).
t’
Fig. 5-2 Fig. 5-3
Solved Problems
5.1 Find the intersection of the lines
9: 3 x - 3 ~ ~ 1 A: 4 x + 2 ~ = 3
3x -3y = 1
4x +2y = 3
We must solvethe system
In order to eliminatey, multiply the first equation by 2 and the second by 3,
6x -6y = 2
12x +6y = 9
Add the two equations,
18x= 11 or x = #
Substitutefifor x in the first equation,
3
(
:
) - 3y = 1
1 = 3y
11
6
--
X
5
6
-= 3y

INTERSECTIONS OF GRAPHS [CHAP. 5
38
5.2
So, the point of intersectionis (M,
&).
x2 y2
Find the intersection of the line 9: y = x +3 and the ellipse-+-= 1.
9 4
To solve the system of two equations, substitute x +3 (as given by the equation of 9
’
)
for y in the
equation of the ellipse,
1
-+-=
9 4
x2 (x +3)2
Multiply by 36 to clear the denominators,
4x2 +9(x +3)2= 36
4x2 +9(x2+6x +9) = 36
4x2 +9x2 +54x +81 = 36
13x2+54x +45 = 0
ALGEBRA The solutions of the quadratic equation ax2 +bx +c = 0 are given by the quadraticformula
By the quadratic formula,
-54 f ,/(54)2 -4(13)(45) -54 f ,
/
- -54 f - -54 f 24
-
-
-
-
-
W3) 26 26 26
x =
Hence,either
15 39 24
and y = x + 3 = - - + - = -
26 26 13 13 13 13
-54+24 30 15
= --= --
X =
-54-24 -78
=-- - - 3 and y = x + 3 = - 3 + 3 = 0
26 26
or x =
The two intersection points are shown in Fig. 5-4.
Notice that we could have solved 13x2+ 54x +45 = 0 by factoringthe left side,
(13x + 15Xx +3) = 0
However, such a factorization is sometimes difficult to discover, whereas the quadratic formula can be
applied automatically.
Fig. 54 Fig. 5-5

CHAP.51 INTERSECTIONS OF GRAPHS 39
53 Find the perpendicular distance from the point P(2,3)to the line 9:3y +x = 3.
X,then the distance
with line A through P perpendicular to 9.
The slope-intercept equation of 9is
Let the perpendicular from P to .5? hit 9at the point X (see Fig. 5-5). If we can find the coordinates
can be computed by the distance formula (2.1). But X is the intersection of line 9
’
1
3
y = - - x + l
which shows that the slope of 9 is -4. Therefore, by Theorem 4.2, the slope of A is -1/ -4 = 3 so that a
point-slope equation of A is
y - 3 = 3(x - 2)
Solvingfor y, we obtain the slope-interceptequation of A,
y - 3 = 3 x - 6 or y = 3 x - 3
Now solve the system
9:3 y + x = 3 a: y = 3 x - 3
By the second equation, substitute 3x - 3 for y in the first equation,
3(3x - 3) +x = 3
9x - 9 +x = 3
1ox = 12
12 6
By the equation for a,when x = 8,
3
3 = 3
18
y = 3 ( 9 - 3 = -
Thus, point X has coordinates (8,3), and
U
s
e the approximation =3.16 (the symbol x means “is approximately equal to”) to obtain
-
PX =0.8(3.16) x 2.53
5.4 Find the intersections of the followingpairs of graphs, and sketch the graphs.
(a) The lines 9:x - 2y = 2 and A: 3x +4y = 6
(b) The lines 9’: 4x +5y = 10and a: 5x +4y = 8
(c) The line x +y = 8 and the circle(x - 2)2+0,- 1)2= 25
(d) The line y = 8x - 6 and the parabola y = 2x2
(e) The parabolas y = x2 and x = y2
(f)The parabola x = y2 and the circle x2 +y2 = 6
(g) The circlesx2 +y2 = 1and (x - 1)2+y2 = 1

40 INTERSECTIONS OF GRAPHS [CHAP. 5
(h) The line y = x - 3 and the hyperbola xy = 4
(i) The circle of radius 3 centered at the origin and the line through the origin with slope 4
(j) The lines 2x - y = 1and 4x - 2y = 3
5.5 Solve Problem 5.4 parts (a)-(j)by using a graphing calculator.
5.6 (a) Using the method employed in Problem 5.3, show that a formula for the distance from the point
P(x,, y , ) to the line 14: Ax +By + C = 0 is
(b) Find the distance from the point (0,3)to the line x - 2y = 2.
5.7 Let x represent the number of million pounds of mutton that farmers offer for sale per week. Let y represent
the number of dollars per pound that buyers are willing to pay for mutton. Assume that the supply equation
for mutton is
y = 0 . 0 2 ~
+0.25
that is, 0.02~
+0.25 is the price per pound at which farmers are willing to sell x million pounds of mutton
per week. Assume also that the demand equation for mutton is
y = -0.025~+2.5
that is, -0.025~+2.5 is the price per pound at which buyers are willing to buy x million pounds of
mutton. Find the intersection of the graphs of the supply and demand equations. This point (x, y ) indicates
the supply x at which the seller’s price is equal to what the buyer is willing to pay.
5.8 Find the center and the radius of the circle passing through the points 4 3 , 0), B(0, 3), and C(6, 0). [Hint:
The center is the intersection of the perpendicular bisectors of any two sides of AABC.]
5.9 Find the equations of the lines through the origin that are tangent to the circle with center at (3, 1) and
radius 3. [Hint: A tangent to a circle is perpendicular to the radius at the point of contact. Therefore, the
Pythagorean theorem may be used to give a second equation for the coordinates of the point of contact.]
5.10 Find the coordinates of the point on the line y = 2x + 1that is equidistant from (0,O)
and (5, -2).

Chapter 6
Symmetry
6.1 SYMMETRY ABOUT A LINE
Two points P and Q are said to be symmetric with respect to a line 9 if P and Q are mirror images
in 9.More precisely, the segment PQ is perpendicular to 9 at a point A such that PA = QA (see
Fig. 6-1).
Fig. 6 1
(i) If Q(x, y)is symmetric to the point P with respect to the y-axis, then P is (-x, y) [see Fig. 6-2(a)].
(ii) If Q(x, y)is symmetric to the point P with respect to the x-axis, then P is (x, -y) [see Fig. 6-2(b)].
Fig. 6 2
A graph is said to be symmetric with respect to a line 9 if, for any point P on the graph, the point Q
that is symmetric to P with respect to 9 is also on the graph. 9 is then called an axis o
f symmetry of
the graph. See Fig. 6-3.
Fig. 6-3
41

[CHAP. 6
42 SYMMETRY
Consider the graph of an equation f ( x , y) = 0. Then, by (i) above, the graph is symmetric with
respect to the y-axis if and only if f ( x , y) = 0 impliesf(-x, y) = 0. And, by (ii) above, the graph is
symmetric with respect to the x-axis if and only iff(x, y) = 0 impliesf(x, -y) = 0.
EXAMPLES
(a) The y-axis is an axis of symmetry of the parabola y = x2 [see Fig. 6-4(u)]. For if y = x2, then y = (- x ) ~ .
The
x-axis is not an axis of symmetry of this parabola. Although (1, 1) is on the parabola, (1, -1) is not on the
parabola.
X L
4
(b) The ellipse -+y2 = 1 [see Fig. 6-4(b)] has both the y-axis and the x-axis as axes of symmetry. For if
X2
-+y2 = 1, then
4
(_x)2+y2= 1 and -+(-y)’=
X2 1
4 4
6.2 SYMMETRY ABOUT A POINT
Two points P and Q are said to be symmetric with respect to U point A if A is the midpoint of the
line segment PQ [see Fig. 6-5(u)].
The point Q symmetric to the point P(x,y) with respect to the origin has coordinates (-x, -y). [In
Fig. 6-5(b),APOR is congruent to AQOS. Hence,
Symmetry of a graph about a point is defined in the expected manner. In particular, a graph Y is
said to be symmetric with respect to the origin if, whenever a point P lies on Y, the point Q symmetricto
P with respect to the origin also lies on 9.The graph of an equation f ( x , y) = 0 is symmetric with
respect to the origin if and only iff(x, y) = 0 impliesf( -x, -y) = 0.
= 0sand =a.]
0
t’

CHAP. 61 SYMMETRY 43
EXAMPLES
(a) The ellipsegraphed in Fig. 6-4(b)is symmetric with respect to the origin because
X2 (-XI2
-+y2 = 1 implies -+( - Y ) ~= 1
4 4
(b) The hyperbola xy = 1(see Fig. 6-6)is symmetric with respect to the origin,for if xy = 1, then (-x)(- y) = 1.
(c) If y = ax, then - y = a(-x). Hence, any straight line through the origin is symmetric with respect to the
origin.
Fig. 6-6
Solved Problems
6.1 Determine whether the line y = --x (see Fig. 6-7) is symmetric with respect to: (a) the x-axis;
(b) the y-axis; (c) the origin.
(U) The line is not symmetric with respect to the x-axis, since (-1, 1) is on the line, but (-1, -l), the
reflection of (- 1, 1)in the x-axis, is not on the line.
(b) The line is not symmetricwith respect to the y-axis, since (- 1, 1)is on the line, but (1, l), the reflection
of (- 1, 1) in the y-axis, is not on the line.
t'
,
U
I 2 3
Fig. 6-7 Fig. 6 8

44 SYMMETRY [CHAP. 6
(c) The line is symmetricwith respect to the origin, by example (c)above.
6.2 Determine whether the parabola x = y2 (see Fig. 6-8) is symmetric with respect to: (a) the
x-axis;(b) the y-axis;(c) the origin.
(a) The parabola is symmetricwith respect to the x-axis, since x = y2 implies x = (- Y ) ~ .
(6) It is not symmetricwith respect to the y-axis, since (1, 1)is on the parabola, but (- 1, 1)is not.
(c) It is not symmetricwith respect to the origin, since(1, 1)is on the parabola, but (- 1, -1) is not.
6.3 Show that if the graph of an equationf(x, y) = 0 is symmetric with respect to both the x-axis
and the y-axis, then it is symmetricwith respect to the origin. (However, the converseis false, as
is shown by Problem 6.1.)
Assume that f(x, y) = 0; we must prove that f(-x, -y) = 0. Sincef(x, y) = 0 and the graph is sym-
metric with respect to the x-axis,f(x, -y) = 0. Then, sincef(x, -y) = 0 and the graph is symmetric with
respect to the y-axis,f( -x, -y) = 0.
6.4 Let points P and Q be symmetric with respect to the line 9: y = x. If P has coordinates (a, b),
show that Q has coordinates (6,a).
Let Q have coordinates (U,U), and let B be the intersection point of 9 and the line PQ (see Fig. 6-9).
B bisects PQ; hence, its coordinates are given by (2.2) as
from which, since B lies on 9,
b + v a + u
2 2
-=-
b + u = a + u
U - u = a - b
Furthermore, the perpendicular lines P
Q and 9 have respective slopes
b - t ,
and 1
a - u
so that, by Theorem 4.2,
b - u = u - a
b + a = v + u
v + u = a + b
tY
I
Fig. 6 9

CHAP.61 SYMMETRY 45
To solve (I)and (2) simultaneouslyfor U and U, first add the two equations, yielding 20= 24 or U = a Then
subtract (I)from (2),yielding 2u = 2b,or U = b. Thus, Q has coordinates (b, U).
6.5 If the graph of 3x2 +xy = 5 is reflected in the y-axis (that is, each point on the graph is replaced
by the point symmetric to it with respect to the y-axis),find an equationof the new graph.
A point (x, y) lies on the new graph if and only if (- x, y) is on the original graph; that is, if and only if
3( -x)’ +(- x)y = 5. This is equivalent to 3x2 -xy = 5. Note that the new equation is obtained from the
old equation by replacing x by -x.
.
6.6 Determine whether the graphs of the following equations are symmetric with respect to the x-axis, the
y-axis, the origin, or none of these:
(U) y = -x2 (b) y = x3 1
x2 y2
(c) ---=
9 4
(d) x2 + y 2 = 5 (e) (x - +y2 = 9 ( f ) y = (x -
(g) 3x2 -xy + y 2 = 4 (h) y=- (i) y = (x2 + 1)2 -4
0) y = x 4 - 3 x 2 + 5 (k) y = -x5 +7x (0 y = (x - 213 + 1
x + l
X
6.7 Find an equation of the new curve when:
(a) The graph of x2 - xy +y2 = 1 is reflected in the x-axis.
(b) The graph of y3 - xy2 +x3 = 8 is reflected in the y-axis.
(c) The graph of x2 - 12x +3y = 1 is reflected in the origin.
(d) The line y = 3x +1is reflected in the y-axis.
(e) The graph of (x - 1)2+y2 = 1 is reflected in the x-axis.

Chapter 7
Functionsand Their Graphs
7.1 THE NOTION OF A FUNCTION
To say that a quantity y is afunction of some other quantity x means, in ordinary language, that the
value of y depends on the value of x. For example, the volume V of a cube is a function of the length s
of a side. In fact, the dependence of V on s can be made precise through the formula V = s3.Such a
specific association of a number s3 with a given number s is what mathematicians usually mean by a
function.
In Fig. 7-1, we picture a functionfas some sort of process which, from a number x, produces a
numberf(x); the number x is called an argument off and the numberf(x) is called the value off for the
argument x.
EXAMPLES
(a) The square-root function associates with each nonnegative real number x the value &;that is, the unique
nonnegative real number y such that y2 = x.
(b) The doubling function g associates with each real number x the value 2x. Thus, g(3) = 6,
g(-1) = -2, g(&) = 2
a
.
The graph of a functionfconsists of all points (x, y) such that y =f(x). Thus, the graph off is the
graph of the equation y =f(x).
EXAMPLES
(a) Consider the functionfsuch thatf(x) = x + 1 for all x. The graph off is the set of all points (x, y) such that
y = x + 1. This is a straight line, with slope 1 and y-intercept 1(see Fig. 7-2).
(b) The graph of the functionfsuch thatf(x) = x2 for all x consists of all points (x, y) such that y = x2. This is the
parabola of Fig. 3-2.
(c) Consider the function f such that f ( x ) = x3 for all x. In Fig. 7-3(a), we have indicated a few points on the
graph, which is sketched in Fig. 7-3(b).
The numbers x for which a functionfproduces a valuef(x) form a collection of numbers, called the
domain off. For example, the domain of the square-root function consists of all nonnegative real
OUTPUT
INPUT N N C n O N
(argument) (value)
f( x )
-
Fig. 7-1 Fig. 7-2
46

CHAP. 71 FUNCTIONS AND THEIR GRAPHS
2718
A Y
(4
Fig. 7-3
47
numbers; the function is not defined for negative arguments. On the other hand, the domain of the
doubling function consists of all real numbers.
The numbers that are the values of a function form the range of the function.
The domain and the range of a functionfoften can be determined easily by looking at the graph
offi The domain consists of all x-coordinates of points of the graph, and the range consists of all
y-coordinates of points of the graph.
EXAMPLES
(U) The range of the square-root function consists of all nonnegative real numbers. Indeed, for every nonnegative
real number y there is some number x such that &= y ; namely, the number y2.
(b) The range of the doubling function consists of all real numbers. Indeed, for every real number y there exists a
real number x such that 2x = y ; namely, the number y/2.
(c) Consider the absolute-value function h, defined by h(x)= Ix I. The domain consists of all real numbers, but the
range is made up of all nonnegative real numbers. The graph is shown in Fig. 7-4.When x 2 0, y = Ixl is
equivalent to y = x, the equation of the straight line through the origin with slope 1. When x < 0, y = Ix I is
equivalent to y = -x, the equation of the straight line through the origin with slope -1. The perpendicular
projection of all points of the graph onto the y-axis shows that the range consists of all y such that y 2 0.
b Y
Fig. 7-4
(d) Consider the function g defined by the formula g(x) = ,
/
=
,
whenever this formula makes sense. Here the
value ,/=
is defined only when 1 - x 2 0; that is, only when x _< 1. So the domain of g consists of all real
numbers x such that x I 1. It is a little harder to find the range of g. Consider a real number y ; it will belong
to the range of g if we can find some number x such that y = ,
/
=
.
Since ,
/
=
cannot be negative
(by definition of the square-root function), we must restrict our attention to nonnegative y. Now if

48 FUNCTIONS AND THEIR GRAPHS
1
0
- 1
- 2
-3
[CHAP. 7
0
1
fi 5 1.4
fi z 1.7
2
y = ,
/
=
,
then y2 = 1 - x, and, therefore,x = 1 - y2. Indeed, when x = 1 -y2, then
The range of g therefore consists of all nonnegative real numbers. This is clear from the graph of g [see Fig.
7-5(b)].
The graph is the upper half of the parabola x = 1 -y2.
tY
-1
-’ O I
-3 -2 1 X
(e) A function can be defined “by cases.” For instance,
x2 i f x < o
l + x i f 0 s x s l
Here the valuef(x) is determined by two different formulas.The first formula applies when x is negative, and
the second when 0 5 x s 1. The domain consists of all numbers x such that x 5 1. The range turns out to be
all positive real numbers. This can be seen from Fig. 7-6, in which projection of the graph onto the y-axis
produces all y such that y > 0. .
1 I 1 )
X
Fig. 7-6
Note: In many treatments of the foundations of mathematics, a function is identified with its graph. In
other words, a function is defined to be a setfof ordered pairs such thatfdoes not contain two pairs
(a, b) and (a, c) with b # c. Then “y =j(x))’ is defined to mean the same thing as “(x, y) belongs to 5’’
However, this approach obscures the intuitive idea of a function.
7.2 INTERVALS
convenient to introduce special notation and terminology for them.
In dealing with the domains and ranges of functions, intervals of numbers occur so often that it is

CHAP. 73 FUNCTIONS AND THEIR GRAPHS 49
Closedintend: [a,b] consists ofall numbers x such that a 5 x 5 b.
The solid dots on the line at a and b means that a and b are included in the closed interval [a, b].
a b
Openintend: (a,b)consists ofall numbers x such that a < x < b.
The open dots on the line at a and b indicate that the endpoints a and b are not included in the open
interval (a,b).
a b
n n
W
Hdf+pen intervals: [a,b)consists ofall numbers x such that a 5 x < b.
a b
(a,b] consists of all numbers x such that a < x s b.
EXAMPLE Consider the functionfsuch thatf(x) = ,
/
=
,
whenever this formula makes sense. The domain off
consists of all numbers x such that
1 - x 2 2 0 or x’11 or - I S X S ~
Thus, the domain off is the closed interval [-1, 13. To determine the range off; notice that as x varies from -1 to
0, x2varies from 1 to 0, 1 - x2 varies from 0 to 1, and ,
/
=
also varies from 0 to 1. Similarly,as x varies from 0
to 1, ,
/
=
varies from 1 to 0. Hence, the range offis the closed interval [0, 13. This is confirmed by looking at
the graph of the equation y = ,
/
=
in Fig. 7-7. This is a semicircle. In fact, the circle x2 +y2 = 1 is equivalent
to
y2 = 1 - x2 or y = +JC7
The value of the functionfcorresponds to the choice of the + sign and gives the upper half of the circle.
- 1
O I
Fig. 7-7

50 FUNCTIONS AND THEIR GRAPHS [CHAP. 7
Sometimeswe deal with intervalsthat are unbounded on one side.
[a, CO) is made up of all x such that U 5 x.
a
(a, 00) is made up of all x such that a <x.
a
n
V
b
(- CO, b] is made up of all x such that x 5 b.
b
(-CO, b)is made up of all x such that x < b.
n
V
EXAMPLE The range of the function graphed in Fig. 7-6 is (0, CO). The domain of the function graphed in Fig.
7-5(b)is (- CO, 13.
7.3 EVEN AND ODD FUNCTIONS
A functionf is called even if, for any x in the domain off, - x is also in the domain off and
f(-4 =f(x).
EXAMPLES
(a) Letf(x) = x2for all x. Then
f(-x) = (-Xy = x2 =f(x)
and sof is even.
(b) Letf(x) = 3x4 - 5x2 + 2 for all x. Then
f(-x) = 3(-x)4 - 5 ( - X y +2 = 3x4 - 5x2 +2 = f ( x )
Thus,f is even. More generally, a function that is defined for all x and involves only even powers of x is an
even function.
(c) Letf(x) = x3 + 1 for all x. Then
j-(-X) = ( 4 + 1 = 4 + 1
Now -x3 + 1 is not equal to x3+ 1except when x = 0. Hence,fis not even.
A function f is even i
f and only if the graph off is symmetric with respect to the y-axis. For the
symmetry means that for any point (x,f(x)) on the graph, the image point (-x, f (x)) also lies on the
graph; in other words,f(- x) =f ( x )(see Fig. 7-8).

CHAP. 71
t’
FUNCTIONS AND THEIR GRAPHS 51
Fig. 7-8 Fig. 7-9
A function f is called odd if, for any x in the domain off, --x is also in the domain off and
f(--x)= -f(-x).
EXAMPLES
(U) Letf(x) = x3for all x. Then
f(-x) = (-x)3 = -x3 = -f(x)
Thus,fis odd.
(b) Letf(x) = 4x for all x. Then
f(-x) = q-x) = -4x = -f(x)
Thus,fis odd.
(c) Letf(x) = 3x5- 2x3 +x for all x. Then
f(-X) = 3(-x)5 - 2(-x)3 +(-x) = 3(-x5) - 2+3) -x
= - 3 2 +2x3 - = 4 3 2 - 2x3 +x) = -f(x)
Thus,fis odd. More generally, iff(x) is defined as a polynomial that involves only odd powers of x (and lacks
a constant term), thenf(x) is an odd function.
(d) The functionf(x) = x3 + 1,which was shown above to be not even,is not odd either. In fact,
but
f(1) = (1)3 + 1 = 1 + 1 = 2 -f(-1) = -[(-1)3 + 13 = -[-1 + 13 = 0
A function f is odd i
f and only if the graph off is symmetric with respect to the origin. For the
symmetry means that for any point (x,~(x))
on the graph, the image point (-x, -f(x)) is also on the
graph; that is,f( -x) = -f(x) (see Fig. 7-9).
7.4 ALGEBRA REVIEW: ZEROS OF POLYNOMIALS
Functions defined by polynomials are so important in calculus that it is essential to review certain
basic facts concerningpolynomials.
Definition: For any functionf, a zero or root offis a number r such thatf(r) = 0.

EXAMPLES
(a) 3 is a root of the polynomial2x3 -4x2- 8x +6, since
2(3)3 -4(3)2- 8(3)+6 = 2(27) -4(9) - 24 +6 = 54 - 36 -24 +6 = 0
X' - 2~ - 15
x + 2
(6) 5 is a zero of because
(5)2 - 2(5)- 15 25 - 10- 15 0
5 + 2 7 7
-
- - 0
- - =
Theorem 7.1: Iff ( x )= a, x" +a,- lx"- + +a,x +a, is a polynomial with integral coefficients
(that is, the numbers a,,, a,,-,, ...,a,, a, are integers), and if r is an integer that is a root
off, then r must be a divisor of the constant term a,.
EXAMPLE By Theorem 7.1, any integral root of x3 - 2x2 - 5x +6 must be among the divisors of 6, which are
f1, f2, f3, and f6. By actual substitution,it is found that 1, -2, and 3 are roots.
Theorem 7.2: A number r is a root of the polynomial
f ( x )= a,x" +a,-,xn-~+ * * * +a,x +a0
if and only iff(x) is divisible by the polynomial x -r.
EXAMPLES
Consider the polynomialf(x) = x3 - 4x2+ 14x - 20. By Theorem 7.1, any integral root must be a divisor of
20; that is, f1, f2
, f4, f5, f10, or 20. Calculation reveals that 2 is a root. Hence (Theorem 7.2), x - 2
must dividef(x); we carry out the division in Fig. 7-1qa). Sincef ( x )= (x - 2)(x2 - 2x + lO), its remaining
roots are found by solving
x2 - 2x + 10 = 0
Applyingthe quadratic formula(see Problem 5.2),
Thus, the other two roots off(x) are the complexnumbers 1 +3
- and 1- 3 n .
Let us find the roots of f(x) = x3 - 5x2 +3x +9. The integral roots (if any) must be divisors of 9: f1, +,3,
f9. 1is not a root, but -1 is a root. Hencef(x) is divisibleby x + 1[see Fig. 7-1qb)l. The other roots off(x)
must be the roots of x2-6x +9. But x2 -6x +9 = (x - 3)2.Thus, -1and 3 are the roots off(x); 3 is called
a repeated root because(x - 3)' is a factor of x3 - 5x2+3x +9.
xz - 2x + 10
x3 - 2x2
x - 2 I x3 - 4x2 + 14x - 20
- 2x2 + 1 4 ~
- 2x2 + 4x
1ox - 20
1ox - 20
x2 - 6 ~
+ 9
x3 + x2
x + 1 ) x 3 - 5 x * + 3 x + 9
- 6x2 + 3~
- 6x2 - 6~
9x +9
9x +9
Theorem 7.3: (Fundamental Theorem of Algebra): If repeated roots are counted multiply, then every
polynomial of degree n has exactly n roots.

CHAP. 71 FUNCTIONS AND THEIR GRAPHS 53
EXAMPLE In example (b) above, the polynomial x3 - 5x2 +3x +9 of degree 3 has two roots, -1 and 3,but 3 is
a repeated root of multiplicity 2and, therefore, is counted twice.
Since the complex roots of a polynomial with real coefficients occur in pairs, a f b n , the poly-
nomial can have only an even number (possibly zero) of complex roots. Hence, a polynomial of odd
degree must have at least one real root.
Solved Problems
7.1 Find the domain and the range of the functionfsuch thatf(x) = -x2.
Since -x2 is defined for every real number x, the domain offconsists of all real numbers. To find the
range, notice that x2 2 0 for all x and, therefore, -x2 I
0 for all x. Every non ositive number y appears as
a value -xz for a suitable argument x; namely, for the argument x = &(and also for the argument
x = -A).
Thus the range off is (- 00, 01.This can be seen more easily by looking at the graph of
y = -x2 [see Fig. 7-11(u)].
tY A Y
-I 0 1 X
Fig. 7-11
7.2 Find the domain and the range of the functionfdefined by
The domain off consists of all x such that either -1 < x <0 or 0 Ix < 1. This makes up the open
interval (- 1, 1
)
. The range off is easily found from the graph in Fig. 7-1l
(
b
)
,whose projection onto the
y-axis is the half-open interval [0, 1
)
.
7.3 Definef(x) as the greatest integer less than or equal to x; this value is usually denoted by [x].
Find the domain and the range, and draw the graph off:
Since [x] is defined for all x, the domain is the set of all real numbers. The range offconsists of all
integers. Part of the graph is shown in Fig. 7-12.It consists of a sequence of horizontal, half-open unit
intervals.(A function whose graph consists of horizontal segments is called a stepfunction.)
7.4 Consider the functionfdefined by the formula
x2 - 1
f(x) = -
x - 1
whenever this formula makes sense. Find the domain and the range, and draw the graph off:

54
1 1 1
-2 - 1 0
- I
0
cllo -2
- -3
4 r
3 - -
2 - -
I - cllo
. A 1 1 I I
.. * *
1 2 3 4 5 X
0
-
-
FUNCTIONS AND THEIR GRAPHS [CHAP. 7
Fig. 7-12
1
I
f I 2 X
Fig. 7-13
The formula makes sense whenever the denominator x - 1 is not 0. Hence, the domain offis the set of
all real numbers different from 1. Now x2 - 1 = (x - 1Xx + 1) and so
x2 - 1
x - 1
-- - x + l
Hence, the graph off(x) is the same as the graph of y = x + 1, except that the point corresponding to x = 1
is missing(see Fig. 7-13). Thus, the range consists of all real numbers except 2.
7.5 (a) Show that a set of points in the xy-plane is the graph of some function of x if and only if the
set intersectsevery vertical line in at most one point (vertical line test).
(6) Determine whether the sets of points indicated in Fig. 7-14 are graphs of functions.
(a) If a set of points is the graph of a functionf, the set consists of all points (x, y), such that y =f(x). If
(xo, U) and (xo, U) are points of intersection of the graph with the vertical line x = xo,then U =f(x,)
and U =f(x,). Hence, U = U, and the points are identical. Conversely, if a set d of points intersects
each vertical line in at most one point, define a functionfas follows. If at intersects the line x = xo at
some point (xo,w),letf(x,) = w. Then at is the graph of$
(b) (i)and (iv)are not graphs of functions,since they intersect certain vertical lines in more than one point.
In each case, we substitute the specified argument for all occurrencesof x in the formula forf(x).
(a) f(2) = (2)2+2(2)- 1 = 4 +4 - 1 = 7
(b) f(-2) = (-2)2 +2(-2) - 1 = 4 - 4 - 1 = - 1
(c) f(-X) = (-x)2 +2(-x) - 1 = x2 - 2x - 1
(6) f ( x + 1) = (x + 1)2 +2(x + 1) - 1 = (x2 +2x + 1) +2x +2 - 1 = x2 +4x +2

CHAP. 71
I X
FUNCTIONS AND THEIR GRAPHS 55
t Y
(iii)
Fig. 7-14
(e) f ( x - 1) = (x - +2(x - 1) - 1 = (x2 - 2x + 1) +2x - 2 - 1 = x2 - 2
(f)f(x +h) = (X +h)2+2(x +h) - 1 = (x’ +2hx +h2)+2~ +2h - 1
= X
’ +2hx +2~ +h2 +2h - 1
(9) Using the result of part (f),
f ( x +h) - f ( ~ ) = (x2 +2hx +2~ +hZ+2h - 1) - (x’ +2~ - 1)
= x2 +2hx +2~ +h2 +2h - 1 -X’ - 2~ + 1
= 2hx +h2 +2h
(h) Using the result of part (g),
f ( x +h)-f(x) 2hx +h2 +2h -h(2x +h +2) = 2x + +
-
-
-
h h h
7.7 Find all rootsof the polynomialf(x) = x3 - 8x2 +21x - 20.
4 is a root.Hence, x -4 must be a factor off(x) (see Fig. 7-15).
The integral roots(ifany)must be divisorsof 20: f1, f2, f4, f5, f10, f20. Calculationshows that
To find the roots of x2 -4x +5, use the quadratic formula
= 2 f / z
- ( - 4 ) f , / - - 4 * 0 4*$- - 4&2-
-
-
- -
2 2 2 2
X =
Hence, there are one real root,4, and two complex roots, 2 +&
i
and 2 - n.

x2-4x + 5
x3 - 4x2
x - 4 Ix3 - 8x2+ 21x - 20
-4x2 +21x
- 4x2 + 1 6 ~
5x - 20
5x -20
Fig. 7-15
SupplementaryProblems
7.8 Find the domain and the range, and draw the graphs of the functionsdetermined by the followingformulas
(for all arguments x for which the formulasmake sense):
(a) h(x) = 4 - x2 (b) G(x) = -2J;; (c) H(x) = J7.7
J(x)= -Jc7
(f) f ( 4= C2xI
1
(i) F(x) = -
x - 1
3 - x f o r x s l
5x- 3 for x > 1
x i f x s 2
4 i f x > 2
(s) Z(x) = x - [ X I (t) f ( x )= 6
7
.
9 Check your answers to Problem 7.8, parts (a)-(j), (n),@), (
s
)
, (t), by using a graphing calculator.
7.10 In Fig. 7-16, determinewhich sets of points are graphs of functions.
7.11 Find a formula for the function f whose graph consists of all points (x, y) such that (a) x3y - 2 = 0;
(b) x = -
+ ;(c)x2 - 2xy +y2 = 0. In each case, specify the domain off:
1-Y
7.12 For each of the following functions, specify the domain and the range, using interval notation wherever
possible. [Hint: In parts (a),(b), and (e), use a graphing calculator to suggestthe solution.]

CHAP. 71
-2 - 1 0
FUNCTIONS AND THEIR GRAPHS
w
1 2 X
t Y
I1
X
T'
tY
(4
Fig. 7-16
x2 - 16
i f x # -4
i f x = -4
7.13 (a) Letf(x) = x - 4 and let g(x) = .Determine k so that f ( x ) = g(x) for all x.
X L - x
(b) Let f(4= 7 g(x) = x - 1
Why is it wrong to assert thatfand g are the same function?
7.14 In each of the following cases, define a function having the given set 9 as its domain and the given set 9 as
its range: (a) 9 = (0, 1) and 9 = (0, 2); (b) 9 = [0, 1) and 41 = [- 1
, 4); (c) 9 = [0, 00) and 9 = (0, 1);
(6)9 = (-CO, 1) U (1,2) [that is, (-CO, 1)together with (1,2)] and 41 = (1, CO).
7.15 For each of the functions in Problem 7.8, determine whether the graph of the function is symmetric with
respect to the x-axis, the y-axis, the origin, or none of these.
7.16 For each of the functions in Problem 7.8, determine whether the function is even, odd, or neither even nor
odd.
7.17 (a) Iffis an even function andf(0) is defined, mustf(0) = O?
(b) Iffis an odd function andf(0) is defined, mustf(0) = O?
(c) Iff(x) = x2 +kx + 1for all x and iffis an even function,find k.
(d) Iff(x) = x3 - (k- 2)x2 +2x for all x and iffis an odd function,find k.
(e) Is there a functionfwhich is both even and odd?
7.18 Evaluate the expression f ( x + h,
h
for the following functions$

7.19 Find all real roots of the followingpolynomials:
(a) x4 - l0x2 +9 (6) X’ +2x2 - 1 6 ~
- 32 (c) x4 - X’ - l0x2 +4~ +24
(d) x3 - 2x2 +x - 2 (e) x3 +9x2 +26x +24 (f) x3 - 5x - 2 (g) x3 - 4x2 - 2x +8
7.20 How many real roots can the polynomial ax3 +6x2+cx +d have if the coefficients a, 6, c, d are real
numbers and Q # O?
7.21 (a) Iff(x) = (x + 3Xx +k) and the remainder is 16whenf(x) is divided by x - 1,find k.
(6) Iff(x) = (x +5)(x - k) and the remainder is 28 whenf(x) is divided by x - 2, find k.
ALGEBRA The division of a polynomialf(x) by another polynomial g(x)yields the equation
f(x) = s(x)q(x)+ fix)
in which q(x) (the quotient) and r(x) (the remainder) are polynomials, with r(x) either 0 or of lower degree
than g(x). In particular, for g(x) = x - a, we have
f(x) = (x - a)q(x)+ r = (x -aMx) +f(4
that is, the remainder whenf(x) is dioided by x -a isjustf(a).
7.22 If the zeros of a functionf(x) are 3 and -4, what are the zeros of the function g(x) =f(x/3)?
7.23 If f(x) = 2x3+Kx2 +J x - 5, and if f(2) = 3 and f(-2) = -37, which of the following is the value of
K +J?
(i) 0 (ii) 1 (iii) -1 (iv) 2 (v) indeterminate
7.24 Express the set of solutions of each inequality below in terms of the notation for intervals:
(a) 2x + 3 < 9 (6) 5x + 1 2 6 (c) 3x + 4 5 5 (d) 7 ~ - 2 > 8
(e) 3 < 4 x - 5 < 7 (f) - 1 _ < 2 x + 5 < 9 (9) 1 x + 1 ) < 2 (h) 1 3 ~ - 4 1 5 5
(9 yj-
< 1 ( j ) x2 s 6 (k) (X - 3 ) ( ~
+ 1)< 0
2x - 5
7.25 For what values of x are the graphs of (a)f(x) = (x - l)(x +2) and (6)f(x) = x(x - 1)(x+2) above the
x-axis? Check your answers by means of a graphing calculator.
7.26 Prove Theorem 7.1. [Hint: Solvef(r) = 0 for a, ,]
7.27 Prove Theorem 7.2. [Hint: Make use of the ALGEBRA following Problem 721.1

Chapter 8
Limits
8.1 INTRODUCTION
To a great extent, calculus is the study of the rates at which quantities change. It will be necessary
to see how the valuef(x) of a functionfbehaves as the argument x approaches a given number. This
leads to the idea of limit.
EXAMPLE Consider the functionfsuch that
whenever this formula makes sense.Thus,fis defined for all x for which the denominator x - 3 is not 0; that is, for
x # 3. What happens to the value f ( x ) as x approaches 3? Well, x2 approaches 9, and so x2 - 9 approaches 0;
moreover, x - 3 approaches 0. Since the numerator and the denominator both approach 0, it is not clear what
x2 - 9
happens to -
x - 3 ’
However, upon factoring the numerator, we observe that
x2 -9
x - 3 x - 3
(X -3)(x +3)
= x + 3
-- -
Since x +3 unquestionably approaches 6 as x approaches 3, we now know that our function approaches 6 as x
approaches 3. The traditional mathematical way of expressingthis fact is
x2-9
lim -
-
- 6
x+3 x - 3
x2 -9
x - 3
This is read: “The limit of -
as x approaches 3 is 6.”
approaches 4, x2 - 9 approaches 7 and x - 3 approaches 1. Hence,
Notice that there is no problem when x approaches any number other than 3. For instance, when x
x 2 - 9 7
lim -- - = 7
-
x+4 x - 3 1
8.2 PROPERTIES OF LIMITS
In the foregoing example we assumed without explicit mention certain obvious properties of the
notion of limit.Let us write them down explicitly.
PROPERTY I.
l i m x = a
x-ra
This followsdirectlyfrom the meaning of the limit concept.
PROPERTY 11. If c is a constant,
lim c = c
x+a
As x approaches a, the value of c remains c.
59

60 LIMITS [CHAP. 8
PROPERTY 111. If c is a constant andfis a function,
lim c . f ( x ) = c limf(x)
x+a x+a
EXAMPLE
lim 5x = 5 lim x = 5 3 = 15
x 4 3 x+ 3
lim - x = lim (- 1)x = (- 1) lim x = (-1) 3 = -3
X'3 x-3 x-3
PROPERTY IV. Iffand g are functions,
lim [ f ( x ) g(x)] = limf ( x ) lim g(x)
x+a x+a x+a
The limit o
f a product is the product of the limits.
EXAMPLE
lim x2 = lim x lim x = a a = u2
X'O X'O X'O
More generally, for any positive integer n, lim x" = U".
x+o
PROPERTY V. Iffand g are functions,
lim [ f ( x ) g(x)] = limf(x) & lim g(x)
x+a x+a x+a
The limit o
f a sum (diflerence)is the sum (diflerence)of the limits.
EXAMPLES
(4 Iim (3x2+ 5x) = lim 3x2+ lim 5x
x-2 X'2 x-2
= 3 lim x2 + 5 lim x = 3(2)2+ 5(2) = 22
x-2 x-2
(b) More generally, if f ( x ) = a,P +a,- lxn-l + - - - +u0 is any polynomial function and k is any real number,
then
limf(x) = a,k" +u,-lk"-l + +a, = f ( k )
x+k
PROPERTY VI. Iff and g are functions and lim g(x) # 0, then
x+a
x +a
The limit o
f a quotient is the quotient of the limits.
EXAMPLE
lim (2x3- 5)
2(4)3- 5 123
=-=-
2 2 - 5 x+4
lim -
=
x+4 3x +2 lim (3x +2) 3(4) + 2 14
x+4
PROPERTY VII.
lim ,
,
&
j = ,/m
x+a x -+a
The limit o
f a square root is the square root o
f the limit.

CHAP. 81 LIMITS 61
EXAMPLE
lim ,
/
- = d m= fi = 3
x-42 x+2
Properties IV-VII have a common structure. Each tells us that, providedfand/or g has a limit as x
approaches a (see Section 8.3), another, related function also has a limit as x approaches a, and this
limit is as given by the indicated formula.
83 EXISTENCE OR NONEXISTENCE OF THE LIMIT
In certain cases, a functionf(x) will not approach a limit as x approaches a particular number.
EXAMPLES
(a) Figure 8-l(a)indicates that
1
lim -
x-ro x
does not exist. As x approaches 0,the magnitude of l/x becomes larger and larger. (If x > 0, l/x is positive and
very large when x is close to 0. If x <0, l/xis negative and very “small”when x is close to 0.)
(b) Figure 8-l(b)indicates that
1x1
lim -
x-ro x
does not exist. When x>O, 1x1= x and IxI/x= 1; when x<O, 1x1 = -x and IxI/x= -1. Thus, as x
approaches 0, IxI/x approaches two different values, 1 and -1, depending on whether x nears 0 through
positive or through negative values. Since there is no unique limit as x approaches 0, we say that
1x1
lim -
x-ro x
does not exist.
(4 Let
Then [see Fig. 8-l(c)], limf(x) does not exist. As x approaches 1from the left (that is, through values of x < l
)
,
f ( x )approaches 1.But as x approaches 1 from the right (that is, through values of x > l),f(x) approaches 2
.
x-r 1
(4
Fig. 8-1

62 LIMITS [CHAP. 8
Notice that the existence or nonexistence of a limit forf ( x ) as x 4a does not depend on the value
f(a), nor is it even required that f be defined at a. If lirnf ( x ) = L, then L is a number to which f ( x ) can
be made arbitrarily close by letting x be sufliciently close to a. The value of L - o r the very existence of
L-is determined by the behavior offnear a, not by its value at a (if such a value even exists).
x+a
Solved Problems
8.1 Find the following limits (if they exist):
(a) Both y2 and l / y have limits as y +2. So, by Property V,
(b) Here it is necessary to proceed indirectly. The function x2 has a limit as x +0. Hence, supposing the
indicated limit to exist, Property V implies that
lim [x' - (x2 - 91= lim -
1
x-0 x+o x
also exists. But that is not the case. [See example (a)in Section 8.3.1 Hence,
lim (x2 -i)
x-0
does not exist.
u2 - 25 (U - 5XU +5)
lim -
-
- lim = lim (U + 5) = 10
"-5 U - 5 "+5 U - 5 U 4 5
(d) As x approaches 2 from the right (that is, with x > 2), [x] remains equal to 2 (see Fig. 7-12). However,
as x approaches 2 from the left (that is, with x < 2), [x] remains equal to 1. Hence, there is no unique
number that is approached by [x] as x approaches 2. Therefore, lirn [x] does not exist.
x + 2
8.2 Find lim f ( x + h, -'@) for each of the following functions. (This limit will be important in the
study of differentialcalculus.)
h - 0 h
1
(a) f ( x )= 3x - 1 (b) f ( x )= 4x2 - x (c) f ( x )=
;
(U) f ( x +h) 3(x +h) - 1 = 3~ +3h - 1
f ( x )= 3x - 1
f ( x +h) - f ( x ) = ( 3 ~
+ 3h - 1) - ( 3 ~
- 1) = 3~ +3h - 1 - 3~ + 1 = 3h
f ( x + h) -fW = - = 3
3h
h h
f ( x +h) - f ( x ) = lim
= 3
h h-0
Hence, lim
h+O

CHAP. 83 LIMITS 63
(b) f ( x +h) = 4(x +h)’ - (X +h) = 4(x2 +2hx +h2)- x - h
= 4x2 +8hx +4h2 - x - h
f ( x )= 4x2 - x
f(x +h) - f ( x ) = (4x2+8hx +4h2 -x -h) - (4x2- X)
= 4x2 +8hx +4h2 - x - h - 4x2 +x
= 8hx +4h2 - h = h ( 8 ~
+4h - 1)
-
- = 8 ~ + 4 h -1
f(x +h)-f(x) h(8x +4h - 1)
h h
= lim (8x - 1)+lim 4h = 8x - 1 +0 = 8x - 1
h - 0 h+O
1
x + h
(c) f(x +h) = -
1 1
f ( x +h) -f(x) = ---
x + h x
ALGEBRA
a c ad - bc
b d bd
---=-
Hence,
and
x - ( x + ~ ) x - x - ~ -h
--
-
-
-
-
c
(x +h)x (x +h)x (x +h)x
1 1 1 1
.-=--.- = - -
1
= -
lim (x +h) x x x X2
h+O
x3 - 1
8.3 Find lim -
x-1 x - 1 -
Both the numerator and the denominator approach 0 as x approaches 1. However, since 1 is a root of
x3 - 1, x - 1 is a factor of x3 - 1 (Theorem 7.2). Division of x3 - 1 by x - 1 produces the factorization
x3 - 1 = (x - 1Xx2+x + 1).Hence,
x3 - 1 (x - lXX2 +x + 1)
lim -= lim = l i m ( ~ ~ + x + 1 ) = 1 ~ + 1 + 1 = 3
x+l x - 1 x-1 x - 1 X+ 1
[See example (b)following Property V in Section 8.2.1
8.4 (a) Give a precise definition of the limit concept; limf ( x )= L.
x+a
(b) Using the definition in part (a)prove Property V of limits:
limf(x) = L and lirn g(x) = K imply lim [ f ( x )+g(x)] = L +K
x+a x-a x-a

64 LIMITS
tY
[CHAP. 8
-t------
I I
I I
I I
I 1 1 -
0 - 6 a a + 6 x
Fig. 8-2
(a) Intuitively, limf(x) = L means that f ( x )can be made as close as we wish to L if x is taken suficiently
close to a. A mathematically precise version of this assertion is: For any real number E > 0, there exists
a real number 6 > 0 such that
0 < Ix - a1 < 6 implies If(x) - L
I c E
for any x in the domain off. Remember that I x - a I < 6 means that the distance between x and a is
smaller than 6, and that If(x) - L I < E means that the distance betweenf ( x )and L is smaller than E.
We assume that the domain off is such that it contains at least one argument x within distance 6 of a
for arbitrary 6 > 0. Observe also that the condition 0 < Ix - a Iexcludesconsideration of x = a, in line
with the requirement that the value (if any) off@) has nothing to do with limf(x).
A pictorial version of this definition is shown in Fig. 8-2. No matter how thin a horizontal strip (of
width 2 4 that may be taken symmetricallyabove and below the line y = L, there is a thin vertical strip
(of width 26) around the line x = a such that, for any x-coordinate of a point in this strip except x = a,
the corresponding point (x,f(x)) lies in the @venhorizontal strip. The number 6 depends on the given
number E; if is made smaller, then 6 may also have to be chosen smaller.
The precise version just given for the limit concept is called an epsilon-delta definition because of
the traditional use of the Greek letters 6 and 6.
(b) Let E > 0 be given. Then 4 2 > 0 and, sincelimf(x) = L, there is a real number 6 , > 0 such that
x-a
x-+a
x-a
€
0 < [ x - a I < 6 , implies If(x) - L I <-
2
for all x in the domain ofJ: Likewise,since lim g(x) = K, there is a number 6, > 0 such that
x-w
for all x in the domain of g. Let 6 be the minimum of 6, and 6,. Assume that 0 < Ix - a I< 6 and that
x is in the domains offand g. By the triangle inequality (ZJO),
(3)
For the number x under consideration, ( I ) and (2)show that the two terms on the right-hand side of (3)
are each less than 42. Hence,
I(fM+dx)) - ( L+ WI = I(f(x)-4 +(sr(x)- K)l5 If(4- Ll +Id4 - KI
E €
I(f(x) +g(x)) - (L+K ) (<- +- = E
2 2
and we have thereby established that lim ( f ( x )+g(x)) = L +K.
x-a

CHAP. 8) LIMITS 65
8.5 Find the followinglimits(if they exist):
(a) lim 7
x+2
(4 lim Cxl
~ 4 3 1 2
5u2-4
(b) lim -
u+o U + 1
(e) lim 1x1
x+o
.(h) lim (x - [x])
x-2
4 - w2
(c) lim -
w 4 - 2 w + 2
(f) lim (7x3 - 5x2 +2x -4)
x-*4 x - 4
x 4 2
x2 - x - 12
(i) lim
x3 - x2 - x - 15 2x4 - 7x2 +x -6 x4 +3x3 - 13x2- 2 7 ~
+36
(k) lirn (I) lirn
(n) lim
x - 3 x-2 x - 2 x- 1 x2 +3x - 4
(/I lim
(m) lim
x 4 3
J Z - 2 4
$ x z 3 -fi
x 4 0 X x-1 x - 1 x 4 2
f(x’+ h, and then lirn f ( x -+ h,
h h+O h
(if the latter exists) for each of the following
8.6 Compute
functions:
8.7 Give rigorous proofs of the followingproperties of the limit concept:
(a) lirn x = a (b) lim c = c (c) limf(x) = L for at most one number L
x+a x+a x+a
8.8 Assuming that limf(x) = L and lim g(x) = K,prove rigorously:
x+a x+a
(a) lirn c * f ( x )= c L, where c is any real number,
(b) lim ( f ( x ) g(x)) = L K.
x-a
x+a
x+a
(e) If lim ( f ( x )- L) = 0, then limf(x) = L.
(f) If limf(x) = L = lirn h(x)and iff(x) 5 g(x) 5 h(x) for all x near a, then lim g(x) = L.
x+a x+a
x+a x+a x-a
[Hints:
In part (4,for L > 0,
In part (f),
iff(x) and h(x) lie within the interval (L -E
, L +E), so must g(x).J
8.9 In an epsilon-delta proof of the fact that lirn (2 +5x) = 17, which of the following values of 6 is the
largest that can be used, given E?
x+3

66
8.10 (a)
(b)
a
1
1 (a)
(b)
LIMITS [CHAP. 8
x4 - 1
Find lim -
x-1 x - 1
Solve part (a)with a graphing calculator by graphing y =-
x4 - 1 and tracing along the curve as x
x - 1
approaches 1.
Jx+21-5
Find lim
x-64 x - 4
Solve part (a) with a graphing calculator by graphing y = and tracing along the
4 G - 5 - 5
x - 4
curve as x approaches 4.
8.12 Find the following limits:
x2 +3x +2 x3 - 3x2
X2
(c) lim
(a) lim (b) l
a
-
x-.3 x2 -4x +3 x+o
x-ro X x + - 2 3x2 - x - 14

Chapter 9
Special Limits
9.1 ONE-SIDED LIMITS
It is often useful to consider the limit of a functionf(x) as x approachesa given number either from
the right or from the left.
EXAMPLES
x - 1 i f x s O
x + 2 ifx>O'
(a) Consider the functionf(x) =
Its graph is given in Fig. 9-l(a).As x approaches 0 from the left,f ( x )approaches -1. As x approaches0
from the right,f(x) approaches2. We will denote these facts as follows:
lim f ( x ) = -1 and lim f ( x )= 2
x+o- x+o+
(b) Consider the function g(x) = I;' y;.
Its graph is given in Fig. 9-l(b).Then lim g(x) = 1 and lim g(x) = 2.
x-rl- X + l +
.y
I
I I I I )
-2 - I 0 I 2 x
-3 -
( a)
Fig. 9-1
It is clear that limf(x) exists if and only if both lirn f ( x ) and lim f ( x ) exist and
x+a x+a+ x+a-
lim f(x) = lim f(x).
x+a+ x+a-
67

68 SPECIAL LIMITS [CHAP. 9
9.2 INFINITE LIMITS: VERTICAL ASYMPTOTES
If the valuef(x) of a function gets larger without bound as x approaches a, then limf(x) does not
X-Cl
exist. However, we shall write
limf(x) = +GO
x-a
to indicate thatf(x) does get larger without bound.
EXAMPLE Letf ( x )= 1/x2 for all x # 0. The graph off is shown in Fig. 9-2. As x approaches0 from either side,
l/x2grows without bound. Hence,
1
lim-= +a
x-ro x2
The notation lim f ( x )= -CO shall mean that f ( x ) gets smaller without bound as x approaches a;
x-a
that is,
limf ( x )= -C
O if and only if lim(-f(x)) = +CO
x-a x-a
Y
Fig. 9-2
EXAMPLE
* - 0
l i m ( - $ ) = - a
on the basis of the preceding example.
Sometimes the valuef(x) will get larger without bound or smaller without bound as x approaches U
from one side (x +a- or x -+ a').
EXAMPLES
(a) Letf(x) = l/x for all x # 0. Then we write
1
lim - = +a
x+o+ x

CHAP. 91 SPECIAL LIMITS 69
to indicate thatf(x) gets larger without bound as x approaches0 from the right [see Fig. 8-l(u)]. Similarly,we
write
1
lim - = -00
x-ro- x
to express the fact thatf(x) decreases without bound as x approaches0 from the left.
- for x > O
(b) Letf ( x )=
[x for x 50
Then (see Fig. 9-3), lim f ( x )= +00 and lim f ( x ) = 0
X’O+ x+O-
t
Fig. 9-3
Whenf(x) has an infinite limit as x approaches a from the right and/or from the left, the graph of
the function gets closer and closer to the vertical line x = a as x approaches U. In such a case, the line
x = a is called a oertical asymptote of the graph. In Fig. 9-4, the lines x = a and x = 6 are vertical
asymptotes(approached on one side only).
If a function is expressed as a quotient, F(x)/G(x),the existence of a vertical asymptote x = U is
usually signaled by the fact that Gfa)= 0 [except when F(a)= 0 also holds].
Fig. 9-4 Fig. 9-5

x - 2
x - 3
EXAMPLE Letf(x) = -for x # 3. Then x = 3 is a vertical asymptote of the graph off, because
x - 2 x - 2
3 - +a and lim --
lirn --
x-b3+ x - x+3-
x - 3 - - c o
In this case, the asymptote x = 3 is approached from both the right and the left (seeFig. 9-5).
[Notice that, by division, -- 1
- +- Thus, the graph off@) is obtained by shifting the hyperbola y = l/x
x - 2
x - 3 x - 3 '
three units to the right and one unit up.]
9.3 LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES
As x gets larger without bound, the valuef ( x )of a functionfmay approach a fixed real number c.
In that case, we shall write
lim f ( x )= c
In such a case, the graph off gets closer and closer to the horizontal line y = c as x gets larger and
larger. Then the line y = c is called a horizontal asymptote of the graph-more exactly, a horizontal
asymptote to the right.
X++aD
x - 2
EXAMPLE Consider the functionf(x) = -whose graph is shown in Fig. 9-5. Then lirn f ( x ) = 1 and the line
x - 3 X-.+aD
y = 1is a horizontal asymptote to the right.
Iff(x) approaches a fixed real number c as x gets smaller without bound,' we shall write
lim f ( x ) = c
X ' - W
In such a case, the graph offgets closer and closer to the horizontal line y = c as x gets smaller and
smaller. Then the line y = c is called a horizontal asymptote to the left.
EXAMPLES For the function graphed in Fig. 9-5, the line y = 1 is a horizontal asymptote both to the left and to
the right. For the function graphed in Fig. 9-2, the line y = 0 (the x-axis) is a horizontal asymptote both to the left
and to the right.
If a function f becomes larger without bound as its argument x increases without bound, we shall
write lim f ( x ) = +00.
X + + W
EXAMPLES
lirn (2x + 1)= +oo and lirn x3 = +a
X + + Q ) X - r + a D
If a functionf becomes larger without bound as its argument x decreases without bound, we shall
write lim f ( x ) = +00.
x+--a3
To say that x getssmaller without bound means that x eventuallybecomes smaller than any negative number. Of course, i
n that
case, the absolute value Ix Ibecomes larger without bound.

EXAMPLES
lirn x2 = +a and lim --x= +a
.X+-OO x-+-UJ
If a functionfdecreases without bound as its argument x increases without bound, we shall write
lim f ( x )= -CQ.
X++CO
EXAMPLES
lim -2x = -CO and lirn (1 -x2) = --CO
X ' + Q x + + w
If a function f decreases without bound as its argument x decreases without bound, we shall write
lim f ( x )= - W .
x+-CO
EXAMPLES
EXAMPLE Consider the functionfsuch thatf(x) = x - [x] for all x. For each integer n, as x increases from n up
to but not including n + 1,the value off@)increases from 0 up to but not including 1.Thus, the graph consists of a
sequenceof line segments, as shown in Fig. 9-6. Then lirn f ( x ) is undefined, since the valuef(x) neither approaches
a fixed limit nor does it become larger or smaller without bound. Similarly, lim f ( x )is undefined.
x + + m
x-+-OO
-5 -4 -3 -2 -1 1 2 3 4 s x
Fig. 9-6
Finding Limits at Infinity of Rational Functions
3x2 - 5x +2
x + 7
A rational function is a quotient f(x)/g(x)of polynomials f ( x ) and g(x). For example,
x2 - 5
4x' +3x
and are rational functions.
GENERALRULE. Tofind lirn f O a n d lirn -
f ( x ) ,divide the numerator and the denominator by
the highest power of x in the denominator, and then use the fact that
x-++oo g(x) x-'-Q) dx)
C C
lim - = 0 and lim - = 0
x + + m x' x-t-a, x'
for any positive real number r and any constant c.

EXAMPLE A
1 2 5
-(2x +5) -+-
2x + 5 X2 x x2
7 3
-(x2-7x+3) x + + w 1 - - + -
X2 x x2
2 5
lim -+ lim -
lirn 1 - lirn -+ lim -
lim = lim = lirn
.++m x2 -7 x+3 x + + m 1
o + o 0
-
- =-=o
7 3 1-o+o 1
x + + m X x - . + w X 2
x + + w x - + w x x + + w X 2
-
-
Thus, y = 0 is a horizontal asymptoteto the right for the graph of this rational function.
Notice that exactly the same calculation applies when +0
0 is replaced by -00. Hence, y = 0 is also
a horizontal asymptote to the left for the graph of this function.
GENERAL RULE A. Iff(x) and g(x) are polynomials and the degree of g is greater than the degree of
- 0.
f,then lim -= 0 and lim --
f(XI f(4
X’ - a
, g(x)
x+ + a
, g(x)
EXAMPLE B
1 4 2
-(3x3 - 4x +2) 3 - - + -
3x3- 4x +2 x3 x2 x3
= lim
5
x3 x3
lim = lim
7x3+5 x + + w 1
-(7x3+ 5) x-.+w 7 + -
X + + W
4 2
lirn 3 - lirn -+
x2 x-.+wx3
lim - 3 - o + o 3
5 7 + 0 7
--
-
-
-
x + + w x + + w
-
-
lim 7 + lim -
x + + m X’+aO x3
Notice that exactly the same calculation works when +0
0 is replaced by -00.
GENERAL RULE B. Iff(x) and g(x) are polynomials and the degree of g is the same as the degree of
f, then lim -and lim fO are both equal to the quotient of the leading coefficients off and g.
f(XI
x + + w g(x) x - - w g(x)
EXAMPLE C
1 . . ?
1
-(4x” - 1) 4x2 - A
4x5- 1 x3 x3
= lim -
1 X - r + a O 7
x * + m I
(3x3 + 7)
lim -- - lim
x + + m 3x3+ 7
x3 3 + 2
4
lim 4x2 - lirn -
1
lim 3 + lim -T
lim 4x2 - O
= - Iim x 2 = +a
x + + m x3 - x * + m
-
X-.+CO
-
-
7 3 + 0 3 x + + m
GENERAL RULE C. If f(x) and g(x) are polynomials and the degree of g is smaller than the degree
off, then lim -= +_ 00. The result is +0
0 when the leading coeficients off and g have the same
sign, and the result is -0
0 when the leading coefficients offand g have opposite signs.
f(x)
x + + w g(x)

EXAMPLE D
7
3
3x -L
3x2- 7 X
2 + -
(a) lim -
-
- lim -= lim - x = - 0 0
x + - Q 2 x + 7 x'-m 7 x'-m2
X
7
3x2 --
3
X
- lim -= lim j x 2 = +00
(b) lim -
-
7
-3x--
2 + -
3x3 - 7
x + - m 2 x + 7 x + - m 7 x'-m
2+;
(c) lim = lim -= lim -x=++oo
-3x2 - 7 -3
x + - Q 2 x + 7 x+--Q) 7 x 4 - m 2
X
GENERAL RULE D. Iff(x) and g(x) are polynomials and the degree of g is smaller than the degree
SolvedProblems
9.1 Evaluate the following limits:
(a)
(c)
lirn ( 5 2 - 20x2 +2x - 14)
lim (7x4 - 12x3+4x - 3)
(b)
(d)
lim (-2x3 +70x2 + 50x + 5)
lim (72+ 10x2+3x +5)
X - r + W x + + w
x-r-03 X + - w
Iff(x) is a polynomial,
f(x) = a,x" +a,,-,xn-l +an-2xn-2+ ...+a,x +u0
with a,, # 0, then
It follows from (9.2) that as x 3 foo,f(x)/a,,xn becomes arbitrarily close to 1. Therefore,f(x) must become
unboundedexactlyas does U,, xn;that is,
l
i
m f ( x )= lim a,x"
X ' f Q X + f Q
Applying this rule to the givenpolynomials,we find:
(a) lim 5x3 = +00 (b) lim (-22) = -00
(c) lirn 7x4 = +00 (d) lim 7x3 = -00
x++a, X ' + Q
x + - w X - - W
The general rule for determining whether +a or -a holds i
s complex. If a, and bk arc the leading cocfiuents off and g,
respectively,then the limit i
s equal to lim !
!
? x"-' and the correctsign is the Ggn of a,, bk(-l)"-'.
x+-w

GENERAL RULE P. Iff(x) = Q,X" +an-l ~ n - + * * * +alx +ao,then:
(i)
(ii)
lim f(x) = f00 and the correct sign is the sign of U,.
lim f(x) = fa and the correct sign is the sign of an(-1)".
X + + c D
X + - m
x + 3 x + 3
9.2 Evaluate: (a) lim -,
- (6) lim -
x+1+ x - 1 x+l-
x - 1'
As x approaches 1, the denominator x - 1 approaches 0, whereas x + 3 approaches 4. Thus,
increaseswithout bound.
(a) As x approaches 1 from the right, x - 1is positive. Sincex +3 is positive when x is close to 1,
x + 3
X - x + 1 + x - 1
' + : > O and lim -- - +a
(b) As x approaches 1 from the left, x - 1is negative, while x +3 is positive. Therefore,
x + 3
x - 1 x-1- x - 1
- -a
-
x + 3 < 0 and lim --
The line x = 1 is a vertical asymptote of the graph of the rational function (see Fig. 9-7).
ifx>O
9.3 Givenf(x) = {'" find: (a) lim f ( x ) ;(6) lim f ( x ) ;(c) lim f ( x ) ;(d) lim f(x).
3 x + 2 ifx<O x+o+ x-ro- X + + W x+-OO
1
lim f(x) = lirn - = +a
x+o+ x+o+ x
Hence, the line x = 0 (the y-axis)is a vertical asymptote of the graph off(Fig. 9-8).
(4
5-3
Y I
I
I
I
I
I
I
1
I
1
I
I
I 1 1 b
X
Y
X
Fig. 9-7 Fig. 9-8

lim f(x) = lim (3x + 2) = 2
x-+o- x-ro-
1
X-++CIJ X ' + W x
lim f(x) = lim - = 0
Hence, the line y = 0 (the x-axis) is a horizontal asymptote (to the right) of the graph off:
lim f ( x ) = lim (3x +2) = --a
x-r-CIJ x-+-CIJ
3x2 - 5 2x - 7 5x +2 5x +2
9.4 Evaluate: (a) lirn -
,
* (b) lim -
;(c) lim ;(d) lim
x + - m 4x2 +5 x + + a x2 - 8 X'+Q) J;cT-3x+l x--03 J z x T i '
(a) Apply Rule B of Section 9.3:
3x2 - 5 3
lim -=-
x - + - W 4x2+5 4
(b) Apply Rule A of Section 9.3:
2x - 7
lim -- - 0
x - + + W x2 - 8
(c) By Rule P, developed in Problem 9.1, the denominator behaves like ,
/
? as x 3 +00. But ,
/
? = x
when x > 0. Thus, by Rule B of Section 9.3,
5 x + 2 5
- lim - - _ -
5x +2
lim - - 5
(d) By Rule P, developed in Problem 9.1, the denominator behaves like ,
/
? as x + -00. But @= -x
when x <0. Thus, by Rule B of Section 9.3,
X-*+mJm-x-
x 1
5 x + 2 5
= lim -=--
5x +2
lim - - 5
x - + - W p = z T x + - m -x -1
9.5 Find the vertical and horizontal asymptotes of the graph of the rational function
3x2- 5x +2
f ( x )= 6x2 - 5x + 1
The vertical asymptotes are determined by the roots of the denominator:
6x2 - 5~ + 1 = 0
(3x - 1X2x - 1) = 0
3 x - 1 = 0 or 2 x - 1 = 0
3x= 1 or 2x = 1
x = $ or x = i
Because the numerator is not 0 at x = 4 or x = 4, If(x) Iapproaches +00 as x approaches4 or 3 from one
side or the other. Thus, the vertical asymptotes are x = 3 and x = 3.
3x2- 5 x + 2 3 1
lim - -
x , + , 6 ~ 2 - 5 ~ + 1 6 2
Thus, y = 4 is a horizontal asymptote to the right. A similar procedure shows that lirn f(x) = 9, and so
y = $ is also a horizontal asymptote to the left.
The horizontal asymptotes may be found by Rule B of Section 9.3,
- - - -
x-+-aI

9.6
9.7
9.8
9.9
9.10
Evaluate the following limits:
(a)
(c)
(e)
lirn (2x" - 5x6 +3x2 + 1)
lim (2x4 - 12x2+x - 7)
lim (-x8 +2x7 - 3x3 +x)
(6)
(d)
(f)
lim (-4x7 +23x3 + 5x2 + 1)
lim (2x3 - 12x2 +x - 7)
lim (-2x5 + 3x4 - 2x3 +x2 - 4)
x * + m X * + W
x * - a , x * - w
X * - W x - - w
Evaluate the following limits (if they exist):
7x-2 i f x 2 2
3 x + 5 i f x < 2
limf(x) and limf(x), iff(x) =
x * 2 + x-2-
lim (t- i) hint:;
1 -7
I = -3
x - 1
x * o + X2
x2 - 5x +6 X' - 5~ +6
lim and lim
x * 3 + x - 3 x*3- x - 3
x * + m 3x - 5
x * + w 3x+ 1 x+--a, 3x + 1
1
and lim -
lim -
3x - 5
1
x2 +4x - 5 x2 +4x - 5
lim and lim
lim -
and lim -
4x - 1 4x - 1
X * + W Jm X * - Q d m
7x - 4 7x - 4
lim -
and lim -
2x +5
x - + w J7TT x - - w J m
x * + w d77-4 * - - m 4 5 4
2x + 5
lim -
and lirn -
x2 - 5
x++a, 3 + 5x - 2x2
x4 - 7x3 +4
x * - - a x 2 - 3
lim
lim
x + 3 x + 3
(6) lim -andlim -
**3+ x2 - 9 x*3- x2 - 9
1 1
($1 lim and lim
x+3+ x2 - 7x + 12 x*3- x2 - 7x + 12
3x3 +x2
x'+a, 5x3- 1
3x3 +x2
(h) lim -
and lim -
x * - w 5x3 - 1
(j) lim and lim
3x3 +2 3x3 +2
(0 lim ~ and lim -
2x3 + - 5 2x3 + - 5
3 x 4 + 4 x * - m 3 x 4 + 4
X * + W
x* + w J
G
'x+- 00 JZ
J2T3
(n) lim @and lim -
x + + a ) 3~ - 2 x-.-w 3x2-2
4x - 3 4x - 3
(p) lim -
and lim ~
x * + w 4
- **-a2 d?T
4 x 2 - 9
(r) lim -
x 4 3 t x - 3
Find any vertical and horizontal asymptotes of the graphs of the following functions:
, / x + 4 - 2
X
(h) f ( x )= Jz-i
- J
;
; (i) f(x)=
2x + 3
J
-
(9) f(x) =
Assume thatfis a function defined for all x. Assume also that, for any real number c, there exists 6 > 0 such
that 0 c. Which of the following holds?
lim f(x) = +CO (6) limf(x) = c (c) lim f(x) = +00 (6) lim f(x) = 0
X * + C f J X+d x-0 x+o
Assume that f(x) 2 0 for arguments to the right of and near a [that is, there exists some positive
number 6 such that a < x < a +6impliesf(x) 2 01. Prove:
lim f(x) = L implies L 2 0
x*o +

(b) Assume thatf(x) s 0 for argumentsto the right of and near U. Prove:
lim f(x) = M implies M s0
(c) Formulate and prove results similar to parts (a)and (b)for lim f(x).
x+a+
x+a-
9.11 Find the vertical and horizontal asymptotes of the graphs of the followingfunctionswith the help of a
graphingcalculator, and then use analyticmethods to verify your answers.
5 - 3x3 J K i 1 4 - 4
4x3 + x - 1 (b) 5 - 2x (4 4-x
JZZ -4 (e) (4x4 +3x3+2x2 +x + 1)l’2
X x 2 + x + 1
(4

Chapter 10
Contlnuity
10.1 DEFIMON AND PROPERTIES
can be made precise in the following way.
A function is intuitively thought of as being continuous when its graph has no gaps or juilips. This
D
Definition: A functionfis said t
c
(i) lirn f ( x )exists.
(ii) f(a) is defined.
(iii) lirnf ( x ) =f(a)<
x+a
x-a
EXAMPLES
be continuous at a if the following three conditions hold:
0 i f x = O
1 ifx+O'
The function is discontinuous (that is, not continuous) at 0. Condition (i)is satisfied:
Letf(x) =
lirn f(x) = 1. Condition (ii) is satisfied:f(0) = 0. However, condition (iii) fails: 1 # 0. There is a gap in the
graph off(see Fig. 10-1)at the point (0, 1).The function is continuous at every point different from 0.
Let f(x) = x2 for all x. This function is continuous at every a, since lim f ( ~ )
= lim x2 = a2 =f(a). Notice that
there are no gaps or jumps in the graph off(see Fig. 10-2).
The function f such that f(x) = [x] for all x is discontinuous at each integer, because condition (i) is not
satisfied (see Fig. 7-12). The discontinuities show up asjumps in the graph of the function.
The functionfsuch thatf(x) = Ix I/x for all x # 0 is discontinuous at 0 [see Fig. 8-1(b)]. lirn f(x) does not exist
andf(0) is not defined. Notice that there is a jump in the graph at x = 0.
x-0
x+a x ' U
x-ro
If a functionfis not continuous at a, thenfis said to have a remooable discontinuity at a if a suitable
change in the definition offat Q can make the resulting function continuous at a.
I I
Fig. 10-1 Fig. 10-2
EXAMPLES In example (a)above, the discontinuity at x = 0 is removable, since if we redefinedf so that f(0) = 1,
then the resulting function would be continuous at x = 0. The discontinuities of the functions in examples (c) and
(d)above are not removable.
A discontinuity of a functionfat a is removable if and only if lim f ( x )exists. In that case, the value
x-a
of the function at a can be changed to limf(x).
x+a
78

CHAP. 101 CONTINUITY 79
A function f i s said to be continuous on a set A iffis continuous at each element of A. Iffis
continuous at every number of its domain, then we simply say thatfis continuous or thatfis a contin-
uousfunction.
EXAMPLES
(a) Every polynomialfunction is a continuousfunction.This followsfrom example (b) of Property V in Section 8.2.
(b) Every rationalfunction h(x) = fo,wherefand g are polynomials, is continuous at eoery real number a except
g(x)
the real roots (ifany)ofg(x).This follows from Property VI in Section 8.2.
There are certain properties of continuity that follow directly from the standard properties of limits
8.2). Assumefand g are continuous at a. Then,
The sumf+ g and the differencef- g are continuous at a.
NOTATION f+ g is a function, such that (f+ gxx) = f ( x ) +g(x)for every x, that is in both the domain of
fand the domain of g . Similarly,(f- gxx) = f ( x ) - g(x) for all x common to the domains offand g.
If c is a constant, the function cfis continuous at a.
~ ~~ ~~~~~~~~~~~~~~~~~~~~
NOTATION cfis a function such that (cf)(x)= c - f ( x )for every x in the domain off:
The product fg is continuous at a, and the quotient f l s is continuous at a provided that
s(4 +0-
NOTATION fg is a function, such that (fg)(x)= f ( x ) g(x)for every x, that is in both the domain off and
the domain of g. Similarly,(f/gXx)= -for all x in both domains such that g(x) # 0.
f(4
d x )
is continuous at a iff(a) > 0.
Note: Sincefis continuous at a, the restriction thatf(a) > 0 guarantees that, for x close to a,
S(x) > 0, and therefore that is defined.
10.2 ONE-SIDED CONTINUITY
A functionfis said to be continuous on the right at a if it satisfies conditions(i)-(iii) for continuity at
a, with lirn replaced by lim; that is, (i) lirn f ( x ) exists; (ii)f(a) is defined; (iii) lirn f ( x )=f(a). Simi-
larly,fis continuous on the Zeft at a'if it satisfies the conditions for continuity at a with lirn replaced by
lim .Note that fis continuous at a if and only iffis continuous both on the right and on the left at a,
sincelimf ( x )exists if and only if both lim f ( x )and lim f ( x )exist and lim f ( x )= lirn f(x).
x-a x+a+ x-a+ x-a+
x-a
x-a-
x+a x+a+ x-a- x-a+ x-a-
EXAMPLES
is continuous on the right at 1 (see Fig. 10-3). Note that lirn f ( x ) =
i f x < l x+1+
(a) The function f ( x ) =
lim (x + 1) = 2 =f(l). On the other hand,fis not continuous on the left at 1, since lirn f(x) = lim x = 1 #
x+1+ x-1- x-1-
f(1). Consequently,fis not continuous at 1,as is evidenced by thejump in its graph at x = 1.

80 CONTINUITY [CHAP. 10
(b) The function of example(c) of Section 8.3 [see Fig. 8-l(c)] is continuouson the left,but not on the right at 1.
(c) The function of example(b) of Section 9.1 [see Fig. 9-l(b)] is continuouson the right, but not on the left at 1.
(d) The functionof example (b) of Section 9.2 (seeFig. 9-3) is continuous on the left, but not on the right at 0.
Fig. 10-3
10.3 CONTINUITY OVER A CLOSED INTERVAL
ignoring the function’s behavior at any other points at which it may be defined.
Definition: A functionfis continuous ouer [a, b] if:
(i) fis continuous at each point of the open interval (a,b).
(ii) fis continuous on the right at U.
(iii) fis continuous on the left at b.
We shall often want to restrict our attention to a closed interval [a, b] of the domain of a function,
EXAMPLES
(a) Figure 10-4(a)shows the graph of a functionthat is continuousover [a, b].
2x i f 0 l ; x l ; l
1 otherwise
is continuous over [0, 13 [see Fig. 10-4(b)].
(b) The function f ( x ) =
Note thatfis not continuousat the points x = 0 and x = 1.Observealso that if we redefinedfso thatf(1) = 1,
then the new function would not be continuous over CO, 13, since it would not be continuous on the left at
x = 1.
6 X
(4
Fig. 10-4
ty
A Y
I X

Solved Problems
x2 - 1
10.1 Find the points at which the functionf(x) = /=if x # -1
is continuous.
For x # -1,fis continuous, sincefis the quotient of two continuous functions with nonzero denomi-
nator. Moreover, for x # -1,
x2 - 1
x + l x + l
(x - 1XX + 1)
f(x) = -
-
- = x - 1
whence lirn f ( x )= lirn (x - 1) = -2 =f( -1).Thus,fis also continuous at x = -1.
x + - 1 x - r - 1
10.2 Consider the function f such that f ( x ) = x - [x] for all x. (See the graph off in Fig. 9-6.) Find
the points at whichf is discontinuous. At those points, determine whether f is continuous on the
right or continuous on the left (or neither).
For each integer n,f(n)= n - [n] = n - n = 0. For n < x < n + l,f(x) = x - [x] = x - n. Hence,
lim f ( x )= lim (x - n) = 0 = f ( n )
x+n+ x+n+
Thus,fis continuous on the right at n.On the other hand,
lim f(x) = lim [x - (n- l)] = n - (n- 1) = 1 # 0 =f(n)
so that f is not continuous on the left at n. It follows thatfis discontinuous at each integer. On each open
interval (n,n + l),fcoincides with the continuous function x - n. Therefore, there are no points of discon-
tinuity other than the integers.
x+n- x-rn-
10.3 For each function graphed in Fig. 10-5,find the points of discontinuity (if any). At each point of
discontinuity, determine whether the function is continuous on the right or on the left (or
neither).
(a) There are no points of discontinuity (no breaks in the graph).
(b) 0 is the only point of discontinuity. Continuity on the left holds at 0, since the value at 0 is the number
approached by the values assumed to the left of 0.
(c) 1 is the only point of discontinuity. At 1 the function is continuous neither on the left nor on the right,
since neither the limit on the left nor the limit on the right equalsf(1). (In fact, neither limit exists.)
(d) No points of discontinuity.
(e) 0 and 1 are points of discontinuity. Continuity on the left holds at 0, but neither continuity on the left
nor on the right holds at 1.
f o r O s x s 1
2 x - 2 for k x s 2
.(See Fig. 10-6.)Isfcontinuous over:
10.4 Definef such that f ( x )=
(a) Yes, sincefis continuous on the right at 0 and on the left at 1.
(b) No, sincefis not continuous on the right at 1. In fact,
lim f(x) = lim (2x - 2) = 0 # 1 =f(l)
x + l + % + I +
(c) No, sincefis not continuous at x = 1,which is inside (0,2).

82
0
I
I
IY
I X 0 1 2 3 X
CONTINUITY
I’
[CHAP. 10
Fig. 10-6
10.5 For each of the following functions,determine the points of discontinuity (if any).For each point
of discontinuity,determinewhether it is removable.

CHAP. 101 CONTINUITY a3
(a) There are no points of discontinuity [see Fig. 10-7(a)J.At x = O,f(O) = 0 and lim f(x) = 0.
(b) The only discontinuity is at x = 1, since g(1) is not defined [see Fig. 10-7(b)]. This discontinuity is
removable. Since-= (x - ’)(’ -t = x + 1, lim g(x)= lim (x + 1)= 2. So, if we define the func-
tion value at x = 1to be 2, the extended function is continuous at x = 1.
x-0
x2 - 1
x - 1 x - 1 x-. 1 x-. 1
Fig. 10-7
10.6 Determine the points at which each of the following functions is continuous. (Draw the graphs of the
functions.) Determine whether the discontinuities are removable.
x + l i f x r 2
x - 1 i f x s l
i f l < x < 2
x2 - 4
i f x # -2 i f x z -2
i f x = -2
10.7 Find the points of discontinuity (if any) of the functions whose graphs are shown in Fig. 10-8.
10.8 Give simpleexamples of functions such that:
(a) fis defined on [-2,23, continuous over [-1,13, but not continuous over [-2,23.
(b) g is defined on [0, 13,continuous on the open interval (0, l), but not continuous over [0, 1).
(c) h is continuous at all points except x = 0, where it is continuous on the right but not on the left.
10.9 For each discontinuity of the following functions,determine whether it is a removable discontinuity.
(a) The functionfof Problem 7.4 (seeFig. 7-13).
(b) The functionfof example (c) in Section 8.3 [see Fig. 8-l(c)].
(c) The functionfof example (a)in Section 9.1 [see Fig. 9-l(a)].
(d) The functionfin example (a)in Section 9.2 (see Fig. 9-3).
(e) The examples in Problems 10.3and 10.4.

84
-3 -2 - 1 0
Y
1 2 3 %
T'
CONTINUITY
0
[CHAP. 10
Y
Fig. 10-8
3x +3
x2 - 3x -4'
10.10 Letfbe defined by the formulaf(x) =
(a) Find the argumentsx at whichfis discontinuous.
(b) For each number a at whichf is discontinuous, determine whether limf ( x ) exists. If it exists, find its
value.
(c) Write an equation for each vertical and horizontal asymptote of the graph off.
x-0
10.11 Letj'(x) = x +(l/x)for x # 0.
(a) Find the points of discontinuity off:
(b) Determineall vertical and horizontal asymptotes of the graph off:

10.12
10.13
10.14
10.15
10.16
10.17
For each of the followingfunctions determine whether it is continuous over the given interval:
i f x > O
0 ifx=O
(4f(x) = {: over CO, 13
2x i f O < x < l
x - 1 i f x > l
over CO, 11 (d) f as in part (c) over Cl, 21
X’ - 16
i f x # 4
If the functionf(x) = [
T is continuous, what is the value of c?
cc i f x = 4
x2 - b2
Let b # 0 and let g be the function such that g(x) = 1- i f x # b
lo ifx = b
(a) Does g(b)exist? (b) Does lim g(x) exist? (c) Is g continuous at b?
x-+b
(a) Show that the followingf is continuous:
c7b i f x = 6
(b) For what value of k is the followinga continuous function?
U i f x = 2
Determine the points of discontinuity of the following functionf:
1 if x is rational
0 if x is irrational
[Hint: A rational number is an ordinary fraction p/q, where p and q are integers. Recall Euclid’s proof that
fi cannot be expressed in this form; it is an irrational number, as must be &I, for any integer n. It
follows that any fixed rational number r can be approached arbitrarily closely through irrational numbers
of the form r +&/n. Conversely, any fixed irrational number can be approached arbitrarily closely
through rational numbers.]
U
s
e a graphing calculator to find the discontinuities (if any) of the followingfunctions

Chapter 11
The Slope of a Tangent Line
The slope of a tangent line to a curve is familiar in the case of circles [see Fig. 11-l(u)]. At each
point P of a circle, there is a line 9 such that the circle touches the line at P and lies on one side of the
line (entirely on one side in the case of a circle). For the curve of Fig. 11-l(b), shown in dashed lines, Y 1
is the tangent line at P,, 9, the tangent line at P,, and 9,
the tangent line at P,. Let us develop a
definition that corresponds to these intuitive ideas about tangent lines.
Figure ll-2(u) shows the graph (in dashed lines) of a continuous functionf: Remember that the
graph consists of all points (x, y ) such that y =f(x). Let P be a point of the graph having abscissa x.
Then the coordinates of P are (x,~(x)).
Take a point Q on the graph having abscissa x +h. Q will be
close to P if and only if h is close to 0 (becausefis a continuous function). Since the x-coordinate of Q is
x +h, the y-coordinate of Q must bef(x +h). By the definition of slope, the line PQ will have slope
f ( x +h) -fW -
- f ( x +h) -m
(X +h) -x h
Observe in Fig. ll-2(6)what happens to the line PQ as Q moves along the graph toward P. Some of
the positions of Q have been designated as Q1,
Q2,Q 3 , ..., and the corresponding lines as A,,A2,
Fig. 11-1
4 y
I I b b
X x + h X X
( a) ( 6 )
Fig. 11-2
86

CHAP. 11) THE SLOPE OF A TANGENT LINE 87
d 3 ,
....These lines are getting closer and closer to what we think of as the tangent line 5 to the graph
at P.Hence, the slope of the line PQ will approach the slope of the tangent line at P; that is, the slope of
the tangent line at P will be given by
What we havejust said about tangent lines leads to the followingprecise definition.
Definition: Let a functionf be continuous at x. By the tangent line to the graph off at P(x,f(x))is
meant that line which passes through P and has slope
Solved Problems
11.1 Consider the graph of the functionfsuch thatf(x) = x2(the parabola in Fig. 11-3). For a point P
on the parabola having abscissa x, perform the calculations needed to find
We have:
f ( x +h) = (x +h)2 = x2 +2xh +h2
f ( x )= x2
f ( x +h) - f ( x ) = (x2+2xh +h2)-x2 = 2xh +h2 = h(2x +h)
f ( x +h) - f ( x ) h(2x +h)
= - = 2 x + h
h h
lim f ( x + h, -
'
(
'
) = lim (2x +h) = l
i
m 2x +lirn h = 2x +0 = 2x
h-0 h h+O h-rO h+O
and the slope of the tangent line at P is 2x. For example, at the point (2, 4), x = 2, and the slope of the
tangent line is 2x = 2(2)= 4.
tY
Fig. 11-3 Fig. 11-4

88 THE SLOPE OF A TANGENT LINE [CHAP. 11
11.2 Consider the graph of the functionf such that f ( x )= x3 (see Fig. 11-4).For a point P on the
graph having coordinates (x, x3),compute the value of
f(x + h) -fW
h
lim
h+O
We havef(x +h) = (x +h)3= x3 +3x2h +3xh2 +h3andf(x) = x3.
ALGEBRA For any x and h,
( X +h)3= [(x+h)(x +h)](x+h) = (xz +2xh +hZXx+h)
= (x3+2x2h +h2x)+(x2h+2xh2 +h3)
= x3 +3x2h +3xhZ+h3
Hence,
and
f ( x +h) -f(x) = (x3+3x2h +3xh2 +h3)- x3
= 3x2h +3xh2 +h3 = h(3x2+ 3xh +h2)
f ( x +h) -f(x) h(3x2 +3xh +h2)
= 3x2 + 3xh +h2
-
-
h h
lim f ( x + h, - f ( x ) = lim (3x2+ 3xh +h2)
h - 0 h h-0
= 3x2 +3 ~ ( 0 )
+O2 = 3x2
This shows that the slope of the tangent line at P is 3x2. For example, the slope of the tangent line at (2, 8)
is 3x2 = 3(2)2= 3(4)= 12.
11.3 (a) Find a formula for the slope of the tangent line at any point of the graph of the functionf
such thatf ( x )= l/x(the hyperbola in Fig. 11-5).
(b) Find the slope-interceptequation of the tangent line to the graph offat the point (2, i).
1 1
f ( x +h) = - and f ( x ) =
;
x + h
h
= --
1 1 x - ( x + ~ )
- x - x - ~
f ( x +h) -f(x) = -- -= -
x +h x (x +h)x (x +h)x (x +h)x
'f:
I
Fig. 11-5

CHAP. 113 THE SLOPE OF A TANGENT LINE
Thus, the slope of the tangent line at (x,f(x))is 6x - 6.
89
ALGEBRA
a c ad bc ad-bc
b d bd bd bd
---=---=-
and
h-rO
1 1
x x x2
= --= _ -
1
lim (x +h) lim x
- -
-
h-0 h+O
Thus,the slope of the tangent line at (x, l/x)is -1/x2.
From part (a), the slope of the tangent line at (2,$) is
1 1 1
x2 22 4
- - = - - = - -
Hence, the slope-interceptequation of the tangent line 9has the form
1
y = - - x + b
4
Since 9 passes through (2,i),
substitution of 2 for x and 4 for y yields
- -
1 1 1 1
2 - - 4 ( 2 ) + b or - = - - + b
2 2 or b = l
Hence, the equation of 9
’
is
1
y = - - x + l
4
Find a formula for the slope of the tangent line at any point of the graph of the functionf
such thatf(x) = 3x2 - 6x +4.
Find the slope-interceptequation of the tangent line at the point (0,4) of the graph.
Draw the graph offand show the tangent line at (0,4).
Computef(x +h) by replacing all occurrences of x in the formula forf(x) by x +h:
f ( x+h) = 3(x +h)2 - 6(x +h) +4
= 3(x2+2xh +h2)- 6x - 6h +4
= 3x2 +6xh +3h2 -6~ - 6h +4
f(x) = 3x2 - 6~ +4
f ( x +h) -f(x) = (3x2+6xh +3h2 -6x - 6h +4) - (3x2 - 6x +4)
= 3x2 +6xh +3h2 - 6x - 6h +4 - 3x2 +6x - 4
= 6xh +3h2 - 6h = h(6x +3h - 6)
f(x +h) - f ( x ) h(6x +3h - 6)
= 6x +3h - 6
-
-
h h

90 THE SLOPE OF A TANGENT LINE [CHAP. 11
(b) From part (a), the slope of the tangent line at (0, 4) is 6x - 6 = 6
(
0
)- 6 = 0 - 6 = -6. Hence, the
slope-intercept equation has the form y = -6x +6. Since the line passes through (0,4), the y-intercept
b is 4. Thus, the equation is y = -6x +4.
(c) We want the graph of y = 3x2 - 6x +4. Complete the square:
y = 3 x 2 - 2 x + - = 3 ( x - l ) 2 + - =3(x-1)2+1
( 3 ( :>
The graph (see Fig. 11-6)is obtained by moving the graph of y = 3x2 one unit to the right [obtaining
the graph of y = 3(x - l)'] and then raising that graph one unit upward.
-
I 2 x
"1 Fig. 11-6
11.5 The normal line to a curve at a point P is defined to be the line through P perpendicular to the
tangent line at P. Find the slope-intercept equation of the normal line to the parabola y = x2 at
the point (4,
a). I
By Problem 11.1, the tangent line has slope 2(f) = 1. Therefore, by Theorem 4.2, the slope of the
normal line is - I, and the slope-intercept equation of the normal line will have the form y = - x +b.
When x = f,y = x2 = (4)' = 4, whence,
3
4 4
-
' = - ( ' ) + b or b = -
Thus, the equation is
3
y = - x + -
4
11.6 For each function f and argument x = a below, (i) find a formula for the slope of the tangent line at an
arbitrary point P(x,f ( x ) )of the graph off; (ii) find the slope-intercept equation of the tangent line corre-
sponding to the given argument a; (iii)draw the graph offand show the tangent line found in (ii).
(a) f ( x )= 2x2 +x; a = - (b) f ( x )= - x3 + 1; a = 2
1
(c) f ( x ) = x2 - 2x; a = 1 (d) f ( x )= 4x2 +3; a = -
2
1 1
4 3

CHAP. 113 THE SLOPE OF A TANGENT LINE 91
11.7 Find the point(s) on the graph of y = x2 at which the tangent line is parallel to the line y = 6x - 1. [Hint:
Use Theorem 4.1.)
11.8 Find the point(s) on the graph of y = x3 at which the tangent line is perpendicular to the line 3x +9y = 4.
[Hint: Use Theorem 4.2.1
11.9 Find the slope-interceptequation of the line normal to the graph of y = x3at the point at which x = 4.
11.10 At what point(s)does the line normal to the curve y = x2 - 3x +5 at the point (3, 5) intersect the curve?
11.11 At any point (x, y) of the straight line having the slope-intercept equation y = mx +b, show that the
tangent line is the straight line itself.
11.12 Find the point(s) on the graph of y = x2 at which the tangent line is a line passing through the point
(2, -12). [Hint: Find an equation of the tangent line at any point (xo, xi) and determine the value@)of xo
for which the line contains the point (2, -12).]
11.13 Find the slope-intercept equation of the tangent line to the graph of y = fi at the point (4, 2). [Hint: See
ALGEBRA in Problem 10.15.1

Chapter 12
The Derivative
The expression for the slope of the tangent line
f ( x +h) -f(4
h
lirn
h-+O
determines a number which depends on x. Thus, the expression defines a function, called the derivative
Definition: The derivativef’ offis the function defined.by the formula
off.
NOTATION There are other notations traditionallyused for the derivative:
dY
D,f(x) and -
dx
dY
When a variable y representsf(x), the derivative is denoted by y’, D,y, or - We shall use whichever notation is
dx *
most convenient or customary in a given case.
The derivative is so important in all parts of pure and applied mathematics that we must devote a
great deal of effort to finding formulas for the derivatives of various kinds of functions. If the limit in the
above definition exists, the functionfis said to be dz@erentiabZe at x, and the process of calculatingf‘ is
called diferentiation off.
EXAMPLES
(a) Letf(x) = 3x +5 for all x. Then,
Hence,
f(x +h) = 3(x +h) +5 = 3x +3h +5
f(x +h) -f(x) = ( 3 ~
+ 3h +5) - ( 3 ~
+ 5) = 3x + 3h + 5 - 3x - 5 = 3h
S(X + h) -I@)-
- - = 3
3h
h h
f(x + h) -f(x) = lim =
h h+O
f’(x)= lim
h+O
or, in another notation, D,(3x +5) = 3. In this case, the derivativeis independent of x.
(b) Let us generalizeto the case of the functionf(x) = Ax +B, where A and B are constants.Then,
f(x +h) - f ( x ) [A(x +h) +B] - (Ax +B) AX +Ah +B - AX - B Ah
-
- -
- = - - = A
h h h h
Thus, we have proved:
Theorem12.1: D,(Ax +B) = A
By letting A = 0 in Theorem 12.1, we obtain:
92

CHAP. 121 THE DERIVATIVE 93
CoroUary 12.2: Dx(B)= 0; that is, the derivative o
f a constantfunction is 0.
Letting A = 1 and B = 0 in Theorem 12.1,we obtain:
Corollary12.3: Dx(x)= 1
By the computations in Problems 11.1, 11.2,and 11.3(a),we have:
Theorem 12.4: (i) Dx(xz)= 2x
(ii) Dx(x3)
= 3x2
1
X2
(iii) D
x
(
:
) = - -
We shall need to know how to differentiate functions built up by arithmetic operations on simpler
functions. For this purpose, several rules of differentiationwill be proved.
RULE 1. (0 D X ( f ( 4 +g(4) = Dxf(4+D xg(x)
Dx(f(4- g(xN = Dxf(4- D xg(x)
The derivative o
f a sum is the sum o
f the derivatives.
The derivative o
f a diflerence is the diflerence o
f the derivatives.
For proofs of (i)and (ii),see Problem 12.1(a).
EXAMPLES
(a) Dx(x3+x2)= D,(x3) +D,(x2) = 3x2 +2x
(b) D,(x2 -1)= DX(x2)
- Dx(-!) = 2x - (--$)= 2~ +-
1
X X2
RULE 2. D,(c *f(x))= c D,f(x)
where c is a constant.
For a proof, see Problem 12.1(b).
EXAMPLES
DX(7x2)
= 7 D,(x2) = 7 2x = 14x
D,(12x3) = 12 D,(x3) = 12(3x2)= 36x2
D,(3x3 +5x2 +2x +4) = D,(3x3) +Dx(5x2)+DA2x) +Dx(4)
= 3 Dx(x3)+5 D,(x2) +2 Dx(x)+0
= 3(3x2)+ 5(2x) +2(1) = 9x2 + 1Ox +2
RULE 3 (Product Rule). Dx(f(x) g(x)) =f(x) Dxg(x)+g(x) Dxf(x)

T
EXAMPLES
THE DERIVATIVE [CHAP. 12
D,(x4) = 44x3 x)
ALGEBRA
= x3 D,(x) +x D,(x3) [by the product rule]
= x3(1) +x(3x2) = x3 +3x3 = 4x3
D,(xS) = D,(x4 x)
= x4 D,(x) +x D,(x4)
= x4(1) +x(4x3)
= x4 +4x4 = 5x4
[by the product rule]
[by example (a)]
DX((x3+x)(x2- x +2)) = (x3 +X) D,(x' - x +2) +(x' - x +2) D,(x3 +X)
= (x3+x)(~x
- 1) +(x2 - x +2 x 3 ~ '+ 1)
The reader may have noticed a pattern in the derivativesof the powers of x:
D,(x) = 1 = 1 xo D,(x2) = 2x D,(x3) = 3x2 Dx(x4)= 4x3 Dx(x5)= 5x4
This pattern does in fact hold for all powers of x.
RULE 4. D,(x") = nx"-'
where n is any positive integer.
EXAMPLES
(U) D,(x9) = 9x8
(b) D,(~x") = 5 D,(x") = 5(11x1O)= 55x"
Using Rules 1,2, and 4, we have an easy method for differentiatingany polynomial.
EXAMPLE
Dx(; x3- 4x2 +2x - ;
) = D X ( i 2)- DX(4X2)+Dx(2x)- D,(i) [by Rule 1
3
3
5
3
5
= - D,(x3) -4 D,(x2) +2 D,(x) - 0 [by Rule 2 and Corollary 12.2)
[by Rule 41
= - (3x7 - 4 (24 +2 (1)
9
5
-
- - x2 - 8~ +2
More concisely,we have:
RULE 5. To differentiatea polynomial,change each nonconstant term akxk to kakxk-' and drop the
constant term (if any).
EXAMPLES
(a) D,(8x5 - 2x4 +3x2+5x +7) = 40x4-8x3 +6x +5
8
3
4
3
3x7 +8 x 5 -- x2 +9x - It = 21x6 + 5 8 x 4 -- x +9

SolvedProblems
12.1 Prove: (a)Rule l(i, ii);(b) Rule 2. Assume that D,f(x) and Ox&) are defined.
cCf(x + h) -fWI
h
= lim
c *f(x+h) - c *f(x)
h h+O
(b) D,(c *f(x))= lim
h+O
f ( x + h, -
'
(
'
)
h
[by Section 8.2, Property 1111
= c lim
h-rO
= c D,f(x)
12.2 Prove Rule 4, D,(x") = nx"- ',for any positive integer n.
We already know that Rule 4 holds when n = 1,
D,(X') = D,(x) = 1 = 1 xo
(Remember that xo = 1.)We can prove the rule by mathematical induction. This involves showing that if
the rule holdsfor any particular positive integer k, then the rule also must holdfor the next integer k + 1. Since
we know that the rule holds for n = 1,it would then follow that it holds for all positive integers.
Assume, then, that D&ck) = kx'-'. We have
D,(xk+ = Dx(xk x) [since xk+1= x k x1 = xk x]
= Xk D,(x) +x D,(xk)
= xk 1 +x(kx"-')
= xk +kx'
= (1 + k)xk= (k + l)x(k+l)-l
[by the assumption that Dx(x")= kxk-']
[since x 2 - 1 = x1 2 - 1 = xkl
and the proof by induction is complete.
12.3 Find the derivative of the polynomial 5x9 - 12x6+4x5 - 3x2 +x - 2.
By Rule 5,
D,(5x9 - 12x6+4x5 - 3x2 +x -2) = 4 5 ~ '
- 72x5+20x4 - 6x + 1
124 Find the slope-interceptequations of the tangent lines to the graphs of the followingfunctions at
the given points:
(a) f ( x )= 3x2 - 5x + 1, at x = 2
(a) Forf(x) = 3x2 - 5x + 1,Rule 5 givesf'(x) = 6x - 5. Then,
(b) f ( x )= x7 - 12x4+2x, at x = 1
f'(2) = 6(2)-5 = 12 - 5 = 7
f(2) = 3(2)2- 5(2)+ 1
and 3(4) - 10 + 1 = 12 - 9 = 3
Thus, the slope of the tangent line to the graph at (2, f(2)) = (2, 3) is f'(2) = 7, and we have as a
point-slope equation of the tangent line y - 3 = 7(x -2), from which we get
JJ- 3 = 7~ - 14
y = 7 x - 11

96 THE DERIVATIVE [CHAP. 12
(b) Forf(x) = x7 - 12x4+2x, Rule 5 yieldsf’(x) = 7x6-4 8 2 + 2. Now
and
f(i) = (117 - 12(14) +2(1) = i- 12 +2 = -9
f’(1) = 7(1)6 -48(l)j +2 = 7 - 48 +2 = -39
Thus, the slope of the tangent line at x = 1 is -39, and a point-slope equation of the tangent line is
y - (-9) = -39(x - l), whence
y + 9 = -39x+39
y = - 3 9 ~+30
12.5 At what point@)of the graph of y = x5 +4x - 3 does the tangent line to the graph also pass
through the point (0, l)?
The slope of the tangent line at a point (xo, yo) = (xo, x: +4x0 - 3) of the graph is the value of the
derivativedy/dx at x = xo.By Rule 5,
= 5x; +4
dY
dx dx X‘XO
-=5 x4 +4 andso dyI
NOTATION The value of a function g at an argument x = 6 is sometimesdenoted by g(x)IxXb. For example,
x2jx=3= (3)2= 9.
The tangent line F to the graph at (xo, yo) goes through (0, 1) if and only if F is the line 9that
connects (xo, yo) and (0, 1). But that is true if and only if the slope m y of .
T is the same as the slope my of
9.Now m y = 5xi +4 and my =
(
X
: +4x0 - 3) - 1 X i +4x0 - 4
-
- .Thus, we must solve
xo - 0 x0
X
: +4x0 - 4
5x2 +4 =
X O
5 X i +4x0 = X i -t4x0 - 4
4 4 = -4
x;= -1
xo= -1
Thus, the required point of the graph is
(-1,(-l)’+4(-1)-3)=(-1, -1 -4-3)=(-1, -8).
12.6 Use the basic definitionoff‘(x) as a limit to calculate the derivativesof the followingfunctions:
12.7 Use Rule 5 to find the derivatives of the followingpolynomials:
(a) -8x5 +f i x 3 +2nx2 - 12
3x3- 4x2 + 5x - 2 (b)
(cl 3xi3 - 5xi0 + i o x 2 (4 2 2 1 +3x12 - 14~2
+fix
+fi

12.8
12.9
12.10
12.11
12.12
12.13
12.14
12.15
12.16
12.17
12.18
12.19
d(3x2 - 5x + 1)
(a) Find D, 3x7-- x5
( ). (b) Find
dx
d(3t7- 12t2)
dt *
(d) Find (e) If U = f i x 5 -x3, find D,u.
1 d
Y
(c) If y = - x4 +5x, find -
2 dx *
Find slope-intercept equations of the tangent lines to the graphs of the following functionsfat the specified
points:
(a) f ( x ) = x2 - 5x +2, at x = -1
(c) f ( x ) = -x4 +2x2 +3, at x = O
Specify all straight lines that satisfy the followingconditions:
(a) Through the point (0,2)and tangent to the curve y = x4 - 12x + 50.
(b) Through the point (1,5) and tangent to the curve y = 3x3 +x +4.
(b) f ( x ) = 4x3 -7x2, at x = 3
Find the slope-interceptequation of the normal line to the graph of y = x3 - x2at the point where x = 1.
Find the point(s)on the graph of y = *x2 at which the normal line passes through the point (4, 1).
Recalling the definition of the derivative, evaluate
(3 +h)' - 3' 5(3 +h)4 - 5(3)4
h
(a) lim (b) lim
h-0 h-0
A function f, defined for all real numbers, is such that: (i)f(1) = 2; (ii)f ( 2 )= 8; (iii)f(w +U) -f(u) =
kuo - 2u2for all U and U, where k is some constant. Findf'(x) for arbitrary x.
Letf(x) = 2x2 +Jsx for all x.
(a) Find the nonnegative value(s)of x for which the tangent line to the graph offat (x,f(x))is perpendicu-
lar to the tangent line to the graph at ( - x , f ( -x)).
(b) Find the point of intersection of each pair of perpendicular lines found in part (a).
If the line 4x - 9y = 0 is tangent in the first quadrant to the graph of y = 3x3 +c, what is the value of c?
For what nonnegative value of b is the line y = -&x +b normal to the graph of y = x3 + $?
Letfbe differentiable(that is,f' exists).Define a functionf " by the equation
f ( x + h) - f ( x - h)
h
f " ( x ) = lim
h+O
(a) Findf"(x) iff(x) = x2 - x. (b) Find the relationship betweenf" and the derivativef'.
1
f ( x + h, - f ( x - h, - f ( x + h, - f ( x ) + f ( x + k, - f ( x ) where k = -h.
-
h h k
Hint:
Letf(x) = x3 +x2 - 9x - 9.
(a) Find the zeros off:
ALGEBRA Iff(X) = Q,X" +a,-,X"-' +
root k off@)must be a divisor of the constant term a, .l
+a,x +a,, where the a,'~ are integers, then any integer

98 THE DERIVATIVE [CHAP.12
(b) Find the slope-interceptequation of the tangent line to the graph offat the point where x = 1.
(c) Find a point (xo,yo) on the graph off such that the tangent line to the graph at (xo,yo) passes through
the point (4, -1).
12.20 Letf(x) = 3x3- llx’ - 15x +63.
(a) Find the zeros off:
(b) Write an equation of the line normal to tue graph offat x = 0.
(c) Find all points on the graph offwhere the tangent line to the graph is horizontal.
f ( x + h, -f(x), andf’-(x), the left-hand deriv-
Definef’+(x),the right-hand derivative off at x, to be lim
ative off and x, to be lim f(x + h, -f(x). Show that the derivativef’(x) exists if and only if bothf‘+(x)
andf’-(x) exist and are equal.
h
12.21
h+O+
h-0- h
12.22 Determine whether the following functions are differentiable at the given argument. [Hint: Use Problem
(a) The functionfof Problem 10.5(a)at x = 0.
12.21.1
i f x > 3
{Tx- 1 i f x s 3 ‘
12.23 Letf(x) =
(a) Find the value of A for whichfis continuousat x = 3.
(b) For the value of A found in (a), isfdifferentiable at x = 3?
1
12.24 Use the originaldefinitionto find the derivativeof the functionfsuch thatf(x) = -.
& - 1

Chapter 13
More on the Derivative
13.1 DIFFERENTIABILITY AND CONTINUITY
In the formula lim + h,
h-0 h
for the derivativef’(a), we can let x = a +h and rewritef’(a) as
lim j
(
’
) Iffis differentiable at a, then
x-0
limf ( x )= lim [ f ( x )-f(a) +f(a)]
x+a x+a
= lim [ f ( x )-f(a)] +limf(a)
x+a x+a
=f’(a) 0 +f(a) =f(a)
Thus,fis continuous at a. This proves:
Tkorem 13.1: Iffis differentiable at a, thenfis continuous at a.
EXAMPLE Differentiability is a stronger condition than continuity. In other words, the converse of Theorem 13.1
is not true. To see this, consider the absolute-value functionf(x) = 1x1 (see Fig. 13-1).fis obviously continuous at
x = 0; but it is not differentiableat x = 0. In fact,
h - 0
= lim -- - l i m 1 = 1
f(0+ h) -f(O)
h h-+O+ h-O+
lirn
h+O+
- h - 0
lim f(O + h, -f(o) = lim -
= lim - 1 = -1
h-0- h h-+O- h - 0 -
and so the two-sided limit needed to definef’(0) does not exist. (The sharp corner in the graph is a tip-off. Where
there is no unique tangent line, there can be no derivative.)
X
Fig. 13-1
99

100 MORE ON THE DERIVATIVE [CHAP. 13
13.2 FURTHER RULES FOR DERIVATIVES
Theorem 13.1 will enable us to justify Rule 3, the product rule, of Chapter 12 and to establish two
additional rules.
RULE 6 (Quotient Rule). Iffand g are differentiableat x and if g(x) # 0, then
EXAMPLES
x + 1 (x2 - 2) Dx(x + 1)- (x + 1) Dx(x2- 2)
(x2- 2K1) - (x + 1K2x) - x2 - 2 - 2x2 - 2x
(4 Dx(=)= (x2- 2)2
-
-
-
(x2- 2)2 (x2-2)2
-x2 -2x - 2
(x2- 2)2
x2 +2x +2
(x2 -2)2
x2 Dx(l)- 1 Dx(x2) X2(O)- l(2x) 2x 2
(x2)2 x4 x4 x3
- -
-
-
-
= - - = - -
a
-
The quotient rule allows us to extend Rule 4 of Chapter 12:
RULE 7. D,(2) = kxk- for any integer k (positive,zero, or negative).
EXAMPLES
D x ( ~ ) = D x ( x - ~ ) = ( - l ) x - ~ 3 ( - l ) ; i =
1 --
1
X2
2
(b) Dx(-$) = DAx-~)= - Z X - ~ = - -
x3
Solved Problems
In the last step, lim f(x +h) =f(x) follows from the fact thatfis continuous at x, by Theorem 13.1.
h - 0

CHAP. 131 MORE ON THE DERIVATIVE 101
1
3
.
2 Prove Rule 6, the quotient rule: Iffand g are differentiable at x and if g(x) # 0, then
If g(x) # 0, then l/g(x) is defined. Moreover, since g is continuous at x (by Theorem 13.1), g(x +h) # 0
for all sufficientlysmall values of h. Hence, l/g(x +h) is defined for those same values of h. We may then
calculate
1 1
g(x) -g(x +
- h, [by algebra: multiply top and bottom by g(x)g(x +h)]
9(x +h) g(x)= lim
h h-0 hg(xk(x + h
,
lim
h-rO
lim c-l/S(X)l
lim g(x +h) D,g(x) [by Property VI of limits and differentiability of g]
h+O
-
-
h-rO
[by Property I1 of limits and continuity of g]
-l/gW .Dxg(x)
=-
g(x)
-1
=-
[g(x)I2 Dx g(x)
Having thus proved that
we may substitute in the product rule (proved in Problem 13.1) to obtain
which is the desired quotient rule.
13.3 Prove Rule 7:Dx(xk)= kxL-' for any integer k.
When k is positive, this is just Rule 4 (Chapter 12).When k = 0,
DJxk) = Dx(xo)= Dx(l)= 0 = 0 x-' = kxk-'
Now assume k is negative; k = -n, where n is positive.
~ ~~
ALGEBRA By definition,
By (1)of Problem 13.2,
-1

102 MORE ON THE DERIVATIVE [CHAP. 13
But (x")' = x2"and, by Rule 4, DAY) = nx"- l
.
Therefore,
= k 2 - l
-1 nx-"- 1
Dx(2)=- ny-1 = -nx"-"-2" = -
X2"
ALGEBRA We have used the law of exponents,
x 2 + x - 2
x 3 + 4 *
13.4 Find the derivative of the functionfsuch thatf(x) =
Use the quotient rule and then Rule 5 (Chapter 12),
(x3 +4)oX(x2+ - 2) -(x2 +x -2)~,(~3
+4)
(x3+4)2
- (x' +4 x 2 ~
+ 1) -(x' +x - 2 x 3 ~ ~ )
-
(x3 +4)2
- (2x4+x3 +8x +4) -(3x4+3x3- 6x2)
-
(x3 +4)2
-x4 - 2x3 +6x2 +8x +4
-
-
(x3 +4)2
13.5 Find the slope-intercept equation of the tangent line to the graph of y = l/x3when x = $.
The slope of the tangent line is the derivative
3
dx x4
When x = 3,
So, the tangent line has slope-intercept equation y = -48x +b. When x = 4, the y-coordinate of the point
on the graph is
1 1
Substituting 8 for y and 4 for x in y = -48x +b, we have
8 = -48($) +b or 8 = -24 +b or b = 32
Thus, the equation is y = -48x +32.
13.6 Find the derivativesof the functions defined by the following formulas:
x2 - 3
(b) x+4
(a) (x1O0+2xS0- 3x7~'+20x + 5)
x s - x + 2
x3 + 7
3x7 +x5 -2x4 + -3
x4
(f1
2 4
(e) 8x3 -x2 +5 - - +-
x x3

CHAP. 13) MORE ON THE DERIVATIVE 103
13.7 Find the slope-interceptequation of the tangent line to the graph of the function at the indicated point:
x + 2
a t x = - 1
1
(4 f(x) = 2’
at x = 2 (b) f(x) =
13.8 Letf(x) = -
+ for all x # 2. Findf’( -2).
x - 2
13.9 Determine the points at which the functionf(x) = I x - 3I is differentiable.
13.10 The parabola in Fig. 13-2is the graph of the functionf(x) = xz - 4x.
(a) Draw the graph of y = If(x) I. (b) Where does the derivative of I f(x)I fail to exist?
Fig. 13-2
13.11 Use a graphing calculator to find the discontinuities of the derivativesof the followingfunctions:
(a) f(x) = x2’3 (b) f ( x )= 4Ix - 2I +3 (c) f(x) = 2 J X

Chapter 14
Maximum and Minimum Problems
14.1 RELATIVE EXTREMA
A functionf is said to have a relative maximum at x = c if
f ( x )s f ( c )
for all x near c. More precisely, f achieves a relative maximum at c if there exists 6 > 0 such that
Ix - c I -c6 impliesf(x) <f(c).
EXAMPLE For the function f whose graph is shown in Fig. 14-1, relative maxima occur at x = c1 and x = c 2 .
This is obvious, since the point A is higher than nearby points on the graph, and the point B is higher than nearby
points on the graph.
The word “relative” is used to modify “maximum” because the value of a function at a relative maximum is
not necessarily the greatest value of the function. Thus, in Fig. 14-1, the valuef(c,) at c1 is smaller than many other
values off@);in particular,f(c,) <f(c2).In this example, the valuef(c,) is the greatest value of the function.
A functionfis said to have a relative minimum at x = c if
f(42f(4
for x near c. In Fig. 14-1,fachieves a relative minimum at x = d, since point D is lower than nearby
points on the graph. The value at a relative minimum need not be the smallest value of the function; for
example, in Fig. 14-1, the valuef(e) is smaller thanf(d).
By a relative extrernum is meant either a relative maximum or a relative minimum. Points at which
a relative extremum exists possess the followingcharacteristic property.
ty B
Fig. 14-1
Theorem14.1: Iffhas a relative extremum at x = c and iff’(c) exists, thenf’(c) = 0.
The theorem is intuitively obvious. Iff’(c) exists, then there is a well-defined tangent line at the
point on the graph offwhere x = c. But at a relative maximum or relative minimum, the tangent line is
horizontal (see Fig. 14-2),and so its slopef’(c) is zero. For a rigorous proof, see Problem 14.28.
The converse of Theorem 14.1 does not hold. Iff’(c) = 0, thenfneed not have a relative extremum
at x = c.
104

CHAP. 14) MAXIMUM AND MINIMUM PROBLEMS 105
A Y
2 -
1 -
I *
I 2 X
Fig. 14-2 Fig. 14-3
EXAMPLE Consider the functionf(x) = x3. Becausef’(x) = 3x2,f’(x) = 0 if and only if x = 0. But from the graph
offin Fig. 14-3it is clear thatfhas neither a relative maximum nor a relative minimum at x = 0.
In Chapter 23, a method will be given that often will enable us to determine whether a relative
extremum actually exists whenf’(c) = 0.
14.2 ABSOLUTE EXTREMA
Practical applications usually call for finding the absolute maximum or absolute minimum of a
function on a given set. Letfbe a function defined on a set d (and possibly at other points, too), and let
c belong to b. Then f is said to achieve an absolute maximum on Q at c if f ( x )<f(c) for all x in b.
Similarly,fis said to achieve an absolute minimum on E at d iff(x) >f(d) for all x in 8.
If the set d is a closed interval [a, b], and if the functionfis continuous over [a, b] (see Section
1
0
.
3
)
,
then we have a very important existence theorem (which cannot be proved in an elementary way).
Theorem14.2 (Extreme-Value Theorem): Any continuous function f over a closed interval [a, b] has
an absolute maximum and an absolute minimum on [a, b].
EXAMPLES
(a) Letf(x) = x + 1 for all x in the closed interval [0, 21. The graph offis shown in Fig. 14-*a). Thenfachieves
an absolute maximum on CO, 2) at x = 2; this absolute maximum value is 3. In additionJachieves an absolute
minimum at x = 0; this absolute minimum value is 1.
(b) Let f(x) = l/x for all x in the open interval (0, 1). The graph off is shown in Fig. 14-4(b).f has neither an
absolute maximum nor an absolute minimum on (0, 1). If we extendedf to the half-open interval (0, 11, then
there is an absolute minimum at x = 1,but still no absolute maximum.
x + l i f - l < x < O
ifx=O
Iox - 1 i f O < x < l
(4 k t f ( x ) =
See Fig. 14-q~)
for the graph off: f has neither an absolute maximum nor an absolute minimum on the
closed interval [- 1, 13.Theorem 14.2does not apply, becausefis discontinuous at 0.
Critical Numbers
To actually locate the absolute extrema guaranteed by Theorem 14.2,it is useful to have the follow-
ing notion.

106 MAXIMUM AND MINIMUM PROBLEMS [CHAP. 14
(6)
Fig. 14-4
Definition: A critical number of a functionfis a number c in the domain off for which eitherf’(c) = 0
orf’(c) is not defined.
EXAMPLES
(a) Let f ( x ) = 3x2-2x +4. Then f‘(x) = 6x - 2. Since 6x - 2 is defined for all x, the only critical numbers are
given by
6 ~ - 2 = 0
6x = 2
X = $ = i
Thus, the only critical number is 3.
(b) Letf(x) = x3 -x2 -5x +3. Thenf’(x) = 3x2 -2x -5, and since 3x2 - 2x -5 is defined for all x, the only
critical numbers are the solutions of
3x2 -2x - 5 = 0
(3x - 5)(x + 1) = 0
3 x - 5 = 0 or x + 1 = 0
3x = 5 or x = - 1
x = 3 or x - -1
Hence, there are two critical numbers, -1 and 3.
We already know from the example in Section 13,l thatf’(0)is not defined. Hence, 0 is a critical number.
SinceD,(x) = 1and DA-x) = -1, there are no other critical numbers.
Method for FindingAbsolute Extrema
Let f be a continuous function on a closed interval [a, b]. Assume that there are only a finite
number of critical numbers c1, c2, ...,ck off inside [a, b]; that is, in (a, b). (This assumption holds for
most functions encountered in calculus.) Tabulate the values off at these critical numbers and at the
endpoints U and b, as in Table 14-1. Then the largest tabulated value is the absolute maximum off on
[a, b], and the smallest tabulated value is the absolute minimum off on [a, b].(This result is proved in
Problem 14.1.)

CHAP. 14) MAXIMUM AND MINIMUM PROBLEMS
Table 14-1
107
~
EXAMPLE Find the absolute maximum and minimum values of
f(x) = x3 - 5 2 + 3x + 1
on CO,13 and find the arguments at which these values are achieved.
The function is continuous everywhere; in particular, on CO,13. Sincef '(x) = 3x2 - 1Ox +3 is defined for all x,
the only critical numbers are the solutions of
3x2- 1ox + 3 = 0
(3x - 1)(x - 3) = 0
3 x - 1 = 0 or x - 3 = 0
3x = 1 or x = 3
x = 4 or x = 3
Hence, the only critical number in the open interval (0, 1)is 4. Now construct Tat-: 1#
1 5
f($) = ($ - 5(;y +3
(
;
) + 1 = 27- + 1 + 1
1 15 14 40
27 27
=---
27 2 7 + 2 = 2 - - = -
f(0) = O3 - 5(0)' +3(0) + 1 = 1
f(1) = l3
- 5(1)2+3(1) + 1 = 1- 5 +3 + 1 = 0
The absolute maximum is the largest value in the second column, 3,
and it is achieved at x = 3. The absolute
minimum is the smallest value, 0, which is achieved at x = 1.
Table 14-2
Solved Problems
14.1 Justify the tabular method for locating the absolute maximum and minimum of a function on a
closed interval.

2/3
1
-1
2
By the extreme-value theorem (Theorem 14.2), a functionfcontinuous on [a, b] must have an absolute
maximum and an absolute minimum on [a, b]. Let p be an argument at which the absolute maximum is
achieved.
CaseZ: p is one of the endpoints, a or b. Thenf(p) will be one of the values in our table. In fact, it will be
the largest value in the table, sincef(p) is the absolute maximum offon [a, b].
Case 2: p is not an endpoint andf’(p) is not defined. Then p is a critical number and will be one of the
numbers c1, c2, ...,ck in our list. Hence,f(p) will appear as a tabulated value, and it will be the largest of
the tabulated values.
Case 3.- p is not an endpoint and f‘b)
is defined. Since p is the absolute maximum off on [a, b],
f(p) 2f(x) for all x near p. Thus,fhas a relative maximum at p, and Theorem 14.1 givesf’b) = 0. But then
p is a critical number, and the conclusion follows as in Case 2.
A completely analogous argument shows that the method yields the absolute minimum.
1/27
0
-12 min
3 max
14.2 Find the absolute maximum and minimum of each function on the given interval:
x2 +3
(a) f ( x )= 2x3 - 5x2 +4x - 1 on [-I, 2) (6) f ( x ) = x+l on CO,31
(a) Sincef’(x) = 6x2 - 10x +4, the critical numbers are the solutions of:
6x2 - 10x +4 = 0
3x2 - 5x +2 = 0
( 3 ~
- 2)(x - 1)= 0
3 x - 2 = 0 or x - 1 = 0
3x = 2 or x = l
x = 3 or x = l
Thus, the critical numbers are 3 and 1, both of which are in (- 1,2). Now construct Table 14-3:
f
(
’
) = 2
(
‘
)
3 - ( ‘ ) l + 4
(
;
) - = -
16 - -
20+8
- - = -
16 - -
60 + -
72 - -
27 = -
1
27 9 3 27 27 27 27 27
f(i) = 2(1)3 - 5(i)2 +qi)- 1 = 2 - 5 +4 - 1 = o
f(- 1) = 2(- 1)3- 5(- 1)2+4(- 1) - 1 = -2 - 5 - 4 - 1 = - 12
f(2) = 2(2)3- 5(2)2+q2) - 1 = 16 - 20 +8 - 1 = 3
Thus, the absolute maximum is 3, achieved at x = 2, and the absolute minimum is - 12, achieved at
x = -1.
Table 14-3
(X + l)DX(x2+3) - (x’ +3)DX(x+ 1) - (X + 1x2~)
- (x’ +3x1)
2x2 +2x - x2 - 3
(x + 1)2
-
(x + 1)2 (x + 1)2
(4 f ’ ( 4 =
x2 +2x - 3
(x + 1)2
-
-
-
-
f’(x) is not defined when (x + 1)2= 0; that is, when x = -1. But since - 1 is not in (0, 3), the only
critical numbers that need to be considered are the zeros of x2 +2x - 3 in (0, 3):
x2 + 2x - 3 = 0
(X+ 3 ) ( ~
- 1) = 0
x + 3 = 0 or x - 1 = 0
x = -3 or x = l

CHAP. 141 MAXIMUM AND MINIMUM PROBLEMS 109
Thus, 1 is the only critical number in (0,3).
Now construct Table 14-4:
- 2
(1)2+3 1 + 3 4
1 + 1 2 2
f(1) = -
= -= - -
(o)2+3 3
f(0) = -
= - = 3
0 + 1 1
(3)2 + 3 9 + 3 12
f(3) = -
= -= -=
Thus, the absolute maximum, achieved at 0 and 3, is 3; and the absolute minimum, achieved at 1,is 2.
3 + 1 4 4
Table 14-4
14.3
14.4
Among all pairs of positive real numbers U and U whose sum is 10, which gives the greatest
product UV?
Let P = uu. Since U +U = 10, U = 10 - u, and so
P = u(10 - U) = 1ou - u2
Here, 0 < u < 10. But since P would take the value 0 at U = 0 and U = 10, and 0is clearly not the absolute
maximum of P,we can extend the domain of P to the closed interval CO, 101. Thus, we must find the
absolute maximum of P = 1Ou - u2 on the closed interval [0, 101.The derivative dP/du = 10 - 214vanishes
only at U = 5, and this critical point must yield the maximum. Thus, the absolute maximum is
P(5)= 5(10 - 5) = 5(5) = 25, which is attained for u = 5. When u = 5,u = 10 - U = 10 - 5 = 5.
(U + - (U - 102- (U -
-
-
4 4
ALGEBRA Calculus was not really needed in this problem, for P = 9
which is largest when U - u = 0, that is, when U = U. Then 10 = u +u = 2u and, therefore, U = 5.
An open box is to be made from a rectangular piece of cardboard that is 8 feet by 3 feet by
cutting out four equal squares from the corners and then folding up the flaps (see Fig. 14-5).
What length of the side of a square will yield the box with the largest volume?
X x
I I I I 3
X X
Fig. 14-5

Let x be the side of the square that is removed from each corner. The volume V = Iwh, where I, w, and
h are the length, width, and height of the box.Now 1 = 8 - 2x, w = 3 - 2x, and h = x, giving
V(X)= (8 - 2xK3 -2x)x = (4x2 - 2 2 ~
+2 4 ) ~
= 4 2 - 22x2 +2 4 ~
The width w must be positive. Hence,
3 - 2 x > O or 3 > 2 x or 3 > x
Furthermore, x > 0. But we also can admit the values x = 0 and x = 3, which make V = 0 and which,
therefore, cannot yield the maximum volume. Thus, we have to maximize V(x)on the interval CO, 31.Since
-=
dV 12x2 -4 4 ~
+24
dx
the critical numbers are the solutions of
12x2- 4 4 ~
+24 = 0
3x2 - llx +6 = 0
(3x - 2xx - 3) = 0
3 x - 2 = 0 or x - 3 = 0
3x = 2 or x = 3
x = j or x = 3
The only critical number in (0,j)is 3. Hence, the volume is greatest when x = 3.
14.5 A manufacturer sells each of his TV sets for $85. The cost C (in dollars) of manufacturing and
sellingx TV sets per week is
C = 1500 + 10x +0 . 0 0 5 ~ ~
If at most loo00 sets can be produced per week, how many sets should be made and sold to
maximizethe weekly profit?
For x sets per week, the total income is 85x. The profit is the income minus the cost,
P = 8 5 ~
- (1500+ 1 0 ~
+0 . 0 0 5 ~ ~ )
= 7 5 ~
- 1500 -0 . 0 0 5 ~ ~
We wish to maximize P on the interval CO, lOOOO], since the output is at most 1OOOO.
--
- 75 -0.01x
dP
dx
and the critical number is the solution of
75 - 0.01x = 0
0.01x= 75
x=-- 75 - 7500
0.01
We now construct Table 14-5:
P(7500) = 75(7500) - 1500 - 0.0005(7500)2
= 562500 - 1500-0.0005(56250000)
= 5
6
1OOO - 281250 = 279750
P(0)= 75(0) - 1500-0.0005(0)2= -1500
P(1OOOO)= 75(10000) - 1500 - 0.0005(10000)2
750OOO - 1500 -O.OOOS( 100OOOOOO)
= 748 500 - 500000 = 248 500
Thus, the maximum profit is achieved when 7500 TV sets are produced and sold per week.

Table 14-5
279750
248 500
-1500
14.6 An orchard has an average yield of 25 bushels per tree when there are at most 40 trees per acre.
When there are more than 40 trees per acre, the average yield decreases by 9 bushel per tree for
every tree over 40. Find the number of trees per acre that will give the greatest yield per acre.
Let x be the number of trees per acre, and let f(x) be the total yield in bushels per acre. When
0 sx 5 40,f(x) = 25x. If x > 40, the number of bushels produced by each tree becomes 25 - i(x - 40).
[Here x -40 is the number of trees over 40, and i(x -40) is the corresponding decrease in bushels per
tree.] Hence, for x > 40,f(x) is given by
1 X
(X -40)). = (25 --
2 x +2 0 ) ~= (45 - f X)X = (90 - X)
Thus,
X
f ( x ) is continuous everywhere, since 25x =-(90 - x) when x = 40. Clearly,f(x) < 0 when x > 90.Hence,
we may restrict attention to the interval CO,901.
For 0 < x < 40,f ( x )= 25x, and f’(x)= 25. Thus, there are no critical numbers in the open interval
(0,40). For 40 < x < 90,
2
X X2
f(x) = -.(90 -X) = 4 5 ~
--
2 2
and f’(x)= 45 - x
Thus, x = 45 is a critical number. In addition, 40 is also a critical number sincef‘(40) happens not to exist.
We do not have to verify this fact, since there is no harm in adding 40 [or any other number in (0, 9O)J to
the list for which we computef(x).
We now construct Table 14-6:
f(45) = -
45 (90 -45) = -
45 (45) = -
2025 = 1012.5
2 2 2
f(40) = 25(40) = loo0
f(0)= 25(0) = 0
90 90
f(90) = -(90 -90) = -(0) = 0
2 2
The maximum yield per acre is realized when there are 45 trees per acre.
Table 14-6

14.7 Find the absolute maxima and minima of the followingfunctions on the indicated intervals:
(a) f(x)= -4x +5 on [-2, 3) (b) f(x) = 2x2 - 7x - 10 on [-1, 3)
(c) f ( x )= x3 +2x2 +x - I on [-I,
13 (d) f ( x )= 4x3 - 8x2 + 1 on [-1, 11
(e) f ( x ) = x4 - 2x3 - x2 - 4x +3 on CO, 41
2x + 5
(f)f ( x )= on [-5, -31
14.8 A farmer wishes to fence in a rectangular field. If north-south fencing costs $3 per yard, and east-west
fencing costs $2 per yard, what are the dimensions of the field of maximum area that can be fenced in for
$600?
14.9 A farmer has to fence in a rectangular field alongside a straight-running stream. If the farmer has 120 yards
of fencing, and the side of the field alongside the stream does not have to be fenced, what dimensions of the
field will yield the largest area?
14.10 The distance by bus from New York to Boston is 225 miles. The bus driver gets paid $12.50 per hour, while
the other costs of running the bus at a steady speed of x miles per hour amount to 90 +0.5x cents per mile.
The minimum and maximum legal speeds on the bus route are 40 and 55 miles per hour. At what steady
speed should the bus be driven to minimize the total cost?
14.11 A charter airline is planning a flight for which it is considering a price of between $150 and $300per person.
The airline estimates that the number of passengers taking the flight will be 200 - OSx, depending on the
price of x dollars that will be set. What price will maximize the income?
14.12 Suppose that a company can sell x radios per week if it charges 100 - 0.1 x dollars per radio. Its pro-
duction cost is 30x +5000 dollars when x radios are produced per week. How many radios should be
produced to maximize the profit, and what will be the selling price per radio?
14.13 A box with square base and vertical sides is to be made from 150 square feet of cardboard. What dimen-
sions will provide the greatest volume if: (a) the box has a top surface; (b) the box has an open top?
14.14 A farmer wishes to fence in a rectangular field, and also to divide the field in half by another fence (AB in
Fig. 14-6). The outside fence costs $2 per. foot, and the fence in the middle costs $3 per foot. If the farmer
has $840 to spend, what dimensions will maximize the total area?
14.15 On a charter flight the price per passenger is $250for any number of passengers up to 100. The flight will be
canceled if there are fewer than 50 passengers. However, for every passenger over 100, the price per pas-
A
B
Fig. 14-6 Fig. 14-7

senger will be decreased by $1. The maximum number of passengers that can be flown is 225. What number
of passengers will yield the maximum income?
14.16 Among all pairs x, y of nonnegative numbers whose sum is 100, find those pairs: (a) the sum of whose
squares x2 +y2 is a minimum; (b) the sum of whose squares x2 +y2 is a maximum; (c) the sum of whose
cubes x3 +y3 is a minimum.
14.17 A sports complex is to be built in the form of a rectangular field with two equal semicircular areas at each
end (see Fig. 14-7). If the border of the entire complex is to be a running track 1256 meters long, what
should be the dimensions of the complex so that the area of the rectangular field is a maximum?
14.18 A wire of length L is cut into two pieces. The first piece is bent into a circle and the second piece into a
square. Where should the wire be cut so that the total area of the circle plus the square is: (a) a maximum;
(b) a minimum?
GEOMETRY The area of a circle of radius r is nr', and the circumference is 2nr.
14.19 A wire of length L is cut into two pieces. The first piece is bent into a square and the second into an
equilateral triangle. Where should the wire be cut so that the total area of the square and the triangle is
greatest?
S2
GEOMETRY The area of an equilateral triangle of side s is fi -.
4
14.20 A company earns a profit of $40 on every TV set it makes when it produces at most loo0 sets. If the profit
per item decreases by 5 cents for every TV set over 1O00,what production level maximizes the total profit?
14.21 Find the radius and the height of the right circular cylinder of greatest volume that can be inscribed in a
right circular cone having a radius of 3 feet and a height of 5 feet (see Fig. 14-8).
~ ~ ~~
GEOMETRY The volume of a right circular cylinder of radius r and height h is nr2h.By the proportionality
of the sides of similar triangles (in Fig. 14-8),-= -
r 3 '
5 - h 5
-3-
Fig. 14-8 Fig. 14-9
X
Fig. 14-10

14.22 Find the height h and the radius r of the right circular cylinder of greatest volume that can be inscribed in a
sphere of radius U. [Hint: See Fig. 14-9. The Pythagorean theorem relates h/2 and r, and provides bounds
on each.]
14.23 Among all isosceles triangles with a fixed perimeter p, which has the largest area? [Hint: See Fig. 14-10 and
solve the equivalent problem of maximizing the square of the area.]
x2 y2
14.24 Find the point(s) on the ellipse -+-= 1 that is (are): (a)closest to the point (1,O);(6)farthest from the
25 9
point (1, 0
)
.[Hint: It is easier to find the extrema of the square of the distance from (1, 0) to (x, y). Notice
that - 5 I
x I
5 for points (x, y) on the ellipse.]
14.25 A rectangular swimming pool is to be built with 6-foot borders at the north and south ends, and 10-foot
borders at the east and west ends. If the total area available is 6OOO square feet, what are the dimensions of
the largest possible water area?
14.26 A farmer has to enclose two fields. One is to be a rectangle with the length twice the width, and the other is
to be a square. The rectangle is required to contain at least 882 square meters, and the square has to
contain at least 400 square meters. There are 680 meters of fencing available.
(a) If x is the width of the rectangular field, what are the maximum and minimum possible values of x?
(b) What is the maximum possible total area?
14.27 It costs a company 0 . 1 ~ ~
+4x +3 dollars to produce x tons of gold. If more than 10 tons is produced, the
need for additional labor raises the cost by 2(x - 10)dollars. If the price per ton is $9, regardless of the
production level,and if the maximum production capacity is 20 tons, what output maximizes the profit?
f(c+ h, -f(c) 0 for h
h
14.28 Prove Theorem 14.1. [Hint: Take the case of a relative minimum at x = c. Then
f(c + h,
h
sufficiently small and positive, and
Sincef’(c) exists,f’(c) = lirn f(c + h, -
’
(
‘
) 2 0 andf’(c) = lim
I
0 for h negative and sufficiently small in magnitude.
f(c + h) --f(c) ~ 0*1
h
h+O+ h h - 0 -
14.29 A rectangle is inscribed in an isosceles triangle with base 9 inches and height 6 inches. Find the dimensions
of the rectangle of maximum area if one side of the rectangle lies inside the base of the triangle.
14.30 A rectangular yard is to enclose an area of 200 square meters. Fencing is required on only three sides, since
one side will lie along the wall of a building. The length and width of the yard are each required to measure
at least 5 meters.
(a) What dimensions will minimize the total fencing required? What will be the minimum fencing?
(b) What dimensions will maximize the total fencing, and what will be the maximum fencing?
2x - 3
14.31 Letf(x) = -
x2 *
(a) Find the maximum and minimum values offon [l, 101.
(b) Does the extreme-value theorem apply tofon [-10, lO]? Why?
14.32 Find the point@)on the curve y = ,/6x4 +8x3 + 1lx2+9 that is (are) closest to the origin.
14.33 Find the shortest distance between points of the curve y = ,/x2 + 3x +2 and the origin.
14.34 Find the dimensions of the rectangle of maximum area that can be inscribed in a right triangle whose sides
are 3, 4, and 5, if one side of the rectangle lies on the side of the triangle of length 3 and the other side lies
on the side of the triangle of length 4.

CHAP. 141 MAXIMUM AND MINIMUM PROBLEMS
14.35 Find the relative extrema off(x) = x4 +2x2 - 5x - 2 in two ways.
(a) Trace the graph offto find the relative maximum and relative minimum directly.
(b) Trace the graph off’ to find the critical numbers off.
115
14.36 Find the relative extrema off(x) = x3 - 3x2 +4x - 1on (- 5,3).

Chapter 15
The Chain Rule
lS.l COMPOSITE FUNCTIONS
There are still many functions whose derivatives we do not know how to calculate; for example,
(i) J
- (ii) $Z7 (iii) (x2 +3x - 1)23
In case (iii), we could, of course, multiply x2 +3x - 1 by itself 22 times and then differentiate the
resulting polynomial. But without a computer, this would be extremely arduous.
The above three functions have the common feature that they are combinations of simpler func-
tions:
(i) d m ’ is the result of starting with the functionf(x) = x3 - x +2 and then applying the
function g(x) = fi to the result. Thus,
Jx’ - x +2 = g(f(x))
dZ4= G(F(x))
(x2 +3x - 1)23 = K ( H ( ~ ) )
(ii) 1 5is the result of starting with the function F(x)= x +4 and then applying the function
G(x) = fi.Thus,
(iii) (x2 + 3x - 1)23 is the result of beginning with the function H(x)= x2 +3x - 1 and then applying
the function K(x)= x23.Thus,
Functions that are put together this way out of simpler functions are called compositefunctions.
Definition: Iff and g are any functions, then the composition g 0 f off and g is the function such that
(9O f K 4 = g(f(x))
The “process” of composition is diagrammed in Fig. 15-1.
Fig. 15-1
EXAMPLES
(a) Letf(x) = x - 1 and g(x) = x2.Then,
(9 0 fXx) = g(f(x))= g(x - 1) = (X -
(fO sKx)=f(g(4) = f ( x 2 )= x2 - 1
On the other hand,
Thus,fo g and g ofare not necessarily the same function(and usually they are not the same).
116

CHAP. 151 THE CHAIN RULE 117
(6) Letf(x) = x2 +2x and g(x) = &.Then,
Again,g 0 fandf 0 g are different.
A composite function g 0 fis defined only for those x for whichf(x) is defined and g(f(x))is defined.
In other words, the domain of g ofconsists of those x in the domain offfor whichf(x) is in the domain
Theorem15.1: The composition of continuous functions is a continuous function. Iff is continuous at
a, and g is continuous atf(a), then g 0 fis continuous at a.
of 9.
15.2 DIFFERENTIATION OF COMPOSITE FUNCTIONS
First, let us treat an important special case. The function [f(x)]”is the composition g 0 f off and
the function g(x) = x”. We have:
Theorem15.2: (Power Chain Rule): Letfbe differentiable and let n be any integer. Then,
W f(4)”)
= n(f(x))”
- ‘D,(f(xN (25.2)
EXAMPLES
(a) Dx((x2- 5)3) = 3(x2 - 5)’DX(x2 - 5) = 3(x2 - 5)2(2~)
= 6 4 ~ ’
- 5)’
(b) oX((x3- 2x2 +3x - 1)7) = 7(x3 - 2x2 +3x - 1)6~x(x3
- 2x2 + 3x - 1)
= 7(x3 - 2x2 +3~ - 1)6(3~2
- 4~ + 3)
1
(4 Dx(-) = Dx((3x - 5 r 4 ) = -q3x - 5)-sDx(3x - 5)
4 12
= --(3)= --
(3x - 5)5
(3x - 5)5
Theorem15.3 (Chain Rule): Assume that f is differentiable at x and that g is differentiable at f(x).
Then the composition g 0 fis differentiable at x, and its derivative (g 0 f)’is given by
(25.2)
that is,
W g ( f ( x ) )= g’(f(x))Dxf(4
The proof of Theorem 15.3 is tricky; see Problem 15.27. The power chain rule (Theorem 15.2)
follows from the chain rule (Theorem 15.3) when g(x) = x”.
Applications of the general chain rule will be deferred until later chapters. Before leaving it,
however, we shall point out a suggestive notation. If one writes y = g(f(x))and u = f ( x ) , then y = g(u),
and (25.2) may be expressed in the form
dy dy du
dx - du dx
---- (25.3)
just as though derivatives were fractions (which they are not) and as though the chain rule were an
identity obtained by the cancellation of the du’s on the right-hand side. While this “identity” makes for
an easy way to remember the chain rule, it must be borne in mind that y on the left-hand side of (25.3)
stands for a certain function of x [namely (g of)(x)], whereas on the right-hand side it stands for a
different funcion of u [namely, g(u)].

118 THE CHAIN RULE [CHAP. 15
EXAMPLE Using (25.3)to rework the precedingexample (c), we write
y = (3x - 5)-4 = U-4
where U = 3x - 5. Then,
12 12
(-4u-5)(3) = --= - -
dx du dx us (3x - 9 5
dy dy du
-=--=
Differentiation of Rational Powers
We want to be able to differentiate the functionf(x) = x', where r is a rational number. The special
case of r an integer is already covered by Rule 7 of Chapter 13.
ALGEBRA A rational number r is one that can be represented in the form r = n/k,where n and k are integers,with
k positive. By definition,
anlk = ( f i y
except when a is negative and k is even (in which case the kth root of a is undefined).For instance,
(8)2/3
= (
,
$
)
' = (2)2 = 4
1 1
(32)-2/5 = (@)-2 = (2)-2 = -
22 =
;
(-27)4/3 = (m)'
= (-3)4 = 81
(-4)'/* is not defined
Observethat
(&)n = .tJ;;;;
whenever both sides are defined.In fact,
which shows that (fi)"
is the kth root of a". In calculations we are free to choose whichever expression for a"Ikis
the more convenient.Thus: (i)642/3is easier to compute as
than as
but (ii)(J8)2/3
is easier to compute as
(,$Z)2 = (4)2= 16
m
=
@
i
E
The usual laws of exponentshold for rational exponents:
(1) d a" = ar+s
d
d
(2) -= d-s
(3) (a")' = a''
(4) (ab)' = a'b'
where r and s are any rational numbers.
Theorem 15.4: For any rational number r, D,(x? = rxr?

EXAMPLES
Theorem 15.4, together with the chain rule (Theorem 15.3), allows us to extend the power chain rule
(Theorem 15.2)to rational exponents.
Curuffizry
15.5: Iffis differentiableand r is a rational number,
4r((fW)= r ( f W- 4f(x)
EXAMPLES
1
2
(a) Dx(J-) = D,((x2 - 3x + l p 2 ) = -(x2- 3x + 1)- l’2Dx(x2- 3x + 1)
1 2x - 3
(2x - 3) =
1
= -
2 (x2 - 3x + 1 y 2 2
J
-
7
(7) = -
1 1
= --
3 (7x +2)4/3 3(,’J7xr?)4
Solved Problems
15.1 For each pair of functionsfand g, find formulas for g 0 fandf 0 g, and determine the domains of
g ofandfo g.
(a) g(x) = 6andf(x) = x + 1
(4
(b) g(x) = x2 andf(x) = x - 1
(g 0 fXx) = df(4)
= B(x + 1) = d x
Because Jx + 1is defined if and only if x 2 -1
, the domain of g 0 fis [-1, 00).
(fosK4=f@W
=f(J;) = J;+ 1
Because J
;
;+ 1is defined if and only if x 2 0, the domain off 0 g is CO, 00).
(b) (9 O f X 4 = g ( f ( 4 ) )= d x - 1) = (x -
(f0 gXx) =f(g(x)) =f ( X 2 ) = x2 - 1
Both composite functionsare polynomials,and so the domain of each is the set of all real numbers.
15.2 Calculatethe derivativesof:
1
(cl(5x2 +4)3
(U) (x4 - 3x2 +5x - 2)3 (b) J7x3 - 2x2 + 5

The power chain rule is used in each case.
(a) ~,((x4- 3x2 + 5x - 2)3) = 3(x4 - 3x2 +5x - 2)2~,(~4
- 3x2 + sX -2)
= 3(x4- 3x2+SX - 2)2(4~3
-6~ +5)
1
2
(b) D,(,/7x3 - 2x2+ 5) = D,((7x3 - 2x2+ 5)'12) =- (7x3- 2x2+5)-"20,(7~3 - 2x2+5)
1 x(21x -4)
(21x2- 4x) =
1
2 (7x3- 2x2+5 y 2
= -
2,/7x3 - 2x2+5
1
(4 D,((5x2 + 4)1) = D,((5x2 +4)-3) = -3(5x2 +4)-4Dx(5x2 +4)
-3 30x
-
- (lOx)= -
(5x2+4)4 (5x2+4)4
153 Find the derivative of the functionf(x) = ,
/
-
) = [l +(x + 1)1/2]1/2.
By the power chain rule, used twice,
1
2
f'(x) = - (1 +(x + 1p2)-"2D,(l +(x + 1)1/2)
1
2
= - (1 +(x + q1l2)- (x + 1)- 1/2Dx(x
+ 1))
= -
1(1 +(x + l)1/2)-112(x
+ l)-lf2(l)
4
1
4
1 1 1 1
= - ((1 +(X+ 1)"')(~ +
-
= -
4 ((1 +(x + W ~ N X + U ) ~ / ~
-4~ ( ( i
+, / m j N x + 1)
15.4 Find the absolute extrema of f ( x )= xJ1 - x2 on CO, 13.
D,(xJi=7) = X D A J c 2 ) +J m D x ( x )
= xD,(( 1- X2y2)+JcT
= x - (1 - x2)-1/2Dx(1- x2) +Jm-
G )
m y the power chain rule]
x 1 -X2
2 (1 -x 2 p 2
- -
- (-2x) +J
c
7
= -
Ji-J+J-
-
- -x2 +(1 - x2) --1 - 2x2
(by;+b=*)
a
J
G
? -Jm C
The right-hand side is not defined when the denominator is 0; that is, when x2 = 1. Hence, 1and -1 are
critical numbers. The right-hand side is 0 when the numerator is 0; that is, when
2x2=1 or x2=+ or x = kJI
Thus, fi and -fiare also critical numbers. The only critical number in (0, 1) is 4,
ALGEBRA $= 8= <x 0.707

At the endpoints, f(0)=f(l) = 0. Hence, 9 is the absolute maximum (achieved at x = A)and 0 is the
absolute minimum (achieved at x = 0 and x = 1).
15.5 A spy on a submarine S, 6 kilometers off a straight shore, has to reach a point B, which is 9
kilometers down the shore from the point A opposite S (see Fig. 15-2).The spy must row a boat
to some point C on the shore and then walk the rest of the way to B. If he rows at 4 kilometers
per hour and walks at 5 kilometers per hour, at what point C should he land in order to reach B
as soon as possible?
6
A C B
Fig. 15-2
Let x = x;
then = 9 - x. By the Pythagorean theorem,
-
SC2=(6)2 + x 2 or E
=
,
/
-
The time spent rowing is, therefore,
[hours]
4
Tl =
and the time spent walking will be T,= (9 - x)/5 [hours]. The total time T(x)is given by the formula
,/m
~ 9 - x
4 5
T(x)= Tl + T, =
We have to minimize T(x)on the interval [0,9], since x can vary from 0 (at A) to 9 (at B),
(36 +x2)lI2 9 x
1 1
-
--.- (36 +x ~ ) - ” ~
D,(36 +x’) --
4 2 5
[by the power chain rule]
1 X 1
(2X) -- =
1 1
= -
8(36 +x ~ ) ~ / ~
5 4J3-3
The only critical numbers are the solutions of
X 1
-0
4J36T-;i-j-
X 1
4J%7 =j
5x = 4,/%7 [cross multip~y~
25x2 = 16(36 +x2) [square both sides]
25x2 = 576 + 16x2
9x2= 576
x2 = 64
X = f8

X
8
The only critical number in (49)is 8. Computingthe valuesfor the tabular method,
T(x)
27
--in
10
33
0 -
10
~ ~ + 9 - O - f l i 9
- = -
6 + 9
- = - + - = -
3 9 33
4 5 4 5 4 5 2 5 1 0
T(0)=
f i
J r n 9 - 9 J r n
+-= +o=-=-=-
J i i i f i f i 3
4 5 4 4 4 4
T(9)=
we generate Table 15-1. The absolute minimum is achieved at x = 8 ; the spy should land 8 kilometers
down the shorefrom A.
Table 15-1
ALGEBRA T(9)> T(8);for, assumingthe contrary,
27
4
[multiply by $1
[by squaring]
36
1296
100
13 5 -= 12.96
which is false.
15.6 Prove Theorem 15.4: DJx') = rx'- for any rational number r.
Let r = n/k,where n is an integer and k is a positive integer. That x"lk is differentiable is not easy to
prove; see Problem 15.26.Assuming this, let us now derive the formulafor the derivative.Let
f(x) = x"/k = *
Then, since(f(x))' = x",
D,((f(x))') = D,(x") = n3-l
But, by Theorem 15.2,D,((f(x))') = k ( f ( ~ ) ) ~ -
'f'(x). Hence,
k(f(x)r- '
f '(x) = nx"-
and solvingforf'(x), we obtain
nx"-' n Yk-'
k(f(x)F-'= 5; i?j=
f'(4=
,rk - 1

CHAP. 15) THE CHAIN RULE 123
15.7
15.8
15.9
15.10
15.11
15.12
15.13
15.14
For each pair of functionsf(x) and g(x), find formulas for (f0 gxx) and (g 0 f)(x).
For each pair of functionsf and g, find the set of solutions of the equation (f0 gxx) = (g 0 fxx).
2
(a) f(x) = x3,g(x) = x2 (b) f(x) = x+l’
g(x) = 3x
Express each of the following functions as the composition (g of)(x) of two simpler functions. [The func-
tionsf(x) and g(x) obviously will not be unique.]
1
(a) (x3- x2 +2)’ (b) (8 - x)* (c) ,
/
- (d)
Find the derivativesof the followingfunctions:
(a) (x3- 2x2+7x - 3)4 (b) (7 +3xy (c) (2x - 3)-2
(d) (3x2+5)-3 (e) (4x2- 3)2(x+5)3
4
(h) 3x2- x +5
Find the derivativesof the followingfunctions:
Jx-1
Find the slope-intercept equation of the tangent line to the graph of y =-
at the point (2, $).
Find the slope-interceptequation of the normal line to the curve y = ,
/
=
at the point (3,5).
x2 + 1
x + 2
Let g(x) = x2- 4 andf(x) = -
x - 2 ‘
(a) Find a formula for (g 0 fxx) and then compute (g 0 f)’(x).
(b) Show that the chain rule gives the same answer for (g 0 f)’(x) as was found in part (a).

15.15 Find the absolute extrema of the followingfunctions on the given intervals:
X
(4 f(x)= -
(c) f(x) = ,/CG on [- 1,13
(b) f ( x )= (x - 2)2(x+3)3 on [-4, 31
(d) f ( x )= - x - x213on [0, 8)
2
3
1
(e) f ( x )= x215 --
9 x7/5 on [-1, 13
15.16 Two towns, P and Q, are located 2 miles and 3 miles, respectively, from a railroad line, as shown in
Fig. 15-3. What point R on the line should be chosen for a new station in order to minimize the sum of the
distances from P and Q to the station, if the distance between A and B is 4 miles?
-
4
-
Fig. 15-3
15.17 Assume that F and G are differentiable functions such that F’(x) = -G(x) and G’(x) = -F(x). If
H(x)= ( F ( x ) ) ~
- (G(x))’, find a formula for H‘(x).
15.18 If y = x3 - 2 and z = 3x + 5, then y can be considered a function of z. Express dyjdz in terms of x.
15.19 Let F be a differentiablefunction, and let G(x) = F‘(x).ExpressD,(F(x3))in terms of G and x.
15.20 If g(x) = x1I5(x- 1)315,
find the domain of g’(x).
15.21 Letfbe a differentiableodd function (Section7.3). Find the relationship betweenf’( -x) andf‘(x).
15.22 Let F and G be differentiablefunctions such that
F(3)= 5 F’(3) = 13 F‘(7) = 2
G(3) = 7 G‘(3) = 6 G(7) = 0
If H(x)= F(G(x)),find H’(3).
15.23 Let F(x)= ,
/
-
.
(a) Find the domain and the range of F.
(b) Find the slope-intercept equation of the tangent line to the graph of F at x = 5.
(c) Find the coordinates of the point(s) on the graph of F such that the normal line there is parallel to the
line 4x +3y = 1.
15.24 Find the dimensions of the rectangle of largest area that can be inscribed in a semicircleof radius 1 if a side
of the rectangle is on the diameter.

15.25 Prove Theorem 15.1: Iffis continuous at a and g is continuous atf(a), prove that g ofis continuous at a.
[Hint: For arbitrary E > 0, let 6, >0 be such that Jg(u)- g(f(a))l < whenever lu -f(a)I < 6,. Then
choose 6 > 0 such that J f ( x )-!(a) 1 < 6, whenever I x - a I < 6.)
15.26 Prove that x"lh is differentiable. [Hint: It is enough to show that f ( x )= x1Ik (k > 1) is differentiable.]
Proceed as follows:
15.27 Prove the chain rule (Theorem 15.3): (g 0 f)'(x)= g'(f(x))f'(x),wherefis differentiableat x and g is differ-
9cv + t) - dY)
entiable at f(x). [Hint: Let H = g of: Let y = f ( x ) and K = f ( x +h) -f(x). Also let G(t)=
t
- g'(y) for t # 0. Since lim G(t)= 0, let G(0)= 0. Then g(y +t) - g(y) = t(G(t)+g'(y)) holds for all t. When
t+O
t = K ,

Chapter 16
Implicit Differentiation
A function is usually defined explicitly by means of a formula.
EXAMPLES
(a) f(x) = x 2 - x +2 (b) f ( x ) = J
;
; (c) f(x)= ("- 1
i f x r l
1 - x i f x < l
However, sometimes the value y = f ( x ) is not given by such a direct formula.
EXAMPLES
(a) The equation y 3 - x = 0 implicitly determines y as a function of x. In this case, we can solve for y explicitly,
y 3 = x or y = f i
(b) The equation y3 + 12y2+48y - 8x +64= 0 is satisfied when y = 2 6 - 4, but it is not easy to find this
solution. In more complicated cases, it will be impossible to find a formula for y in terms of x.
(c) The equation x2 +y2 = 1implicitly determines two functions of x,
y = , / ' D and y = - , / ' D -
The question of how many functions an equation determines and of the properties of these func-
tions is too complex to be considered here. We shall content ourselves with learning a method for
finding the derivatives of functions determined implicitlyby equations.
EXAMPLE Let us find the derivative of a function y determined by the equation x2 +y2 = 4. Since y is assumed
to be some function of x, the two sides of the equation represent the same function of x, and so must have the same
derivative,
D,(x2 +y') = D,(4)
2x +2yD,y = 0 [by the power chain rule]
2yD,y = -2x
2x x
D,y = --- - -
-
2
Y Y
Thus, D,y has been found in terms of x and y. Sometimes this is all the information we may need. For example, if
we want to know the slope of the tangent line to the graph of x2 +y2 = 4 at the point (fi,
l),then this slope is the
derivative
D,y = --
x f i
= --=-fi
Y 1
The process by which D, y has been found, without first solvingexplicitly for y, is called implicit diflerentiation.
Note that the given equation could, in this case, have been solved explicitly for y,
and from this, using the power chain rule,
D, y = DJk(4 -x2)1/2) = f3(4 - x2)- '120,(4 -2)
- X X
(-2x) = +-
=
1
= +
- 2 J G i - J G j T J C - 7
126

CHAP. 161 IMPLICIT DIFFERENTIATION
Solved Problems
16.1 Consider the curve 3x2 -xy +4y2 = 141.
127
Find a formula in x and y for the slope of the tangent line at any point (x, y) of the curve.
Write the slope-intercept equation of the line tangent to the curve at the point (1,6).
Find the coordinates of all other points on the curve where the slope of the tangent line is
the same as the slope of the tangent line at (1,6).
We may assume that y is some function of x such that 3x2- xy +4y2 = 141. Hence,
D,(3x2 - XY +4y2)= D,(141)
6~ - D,(xY) +D,(4y2) = 0
d
Y
6x - (x 2+ye 1) +8y -
dx = 0
d
Y d
Y
-X -+8y -= y - 6~
dx dx
d
Y
(-X +8y) -= y - 6~
dx
which is the slope of the tangent line at (x, y).
The slope of the tangent line at (1,6) is obtained by substituting 1for x and 6 for y in the result of part
(a).Thus, the slope is
6-6(1) 6 - 6 0
---- --=o
8(6) - 1 - 48 - 1 47
and the slope-interceptequation is y = b = 6.
If (x, y) is a point on the curve where the tangent line has slope 0, then,
Substitute 6x for y in the equation of the curve,
3x2- 4 6 ~ )
+4 ( 6 ~ ) ~
= 141
3x2-6x2 + 1 4 4 ~ ~
= 141
141x2= 141
x2 = 1
x = +1
Hence, (- 1, -6) is another point for which the slope of the tangent line is zero.
16.2 If y = f ( x ) is a function satisfying the equation x3y2 -2x +y3 = 36, find a formula for the
derivative dy/dx.
D,(x3y2 -2x +y3)= DJ36)
Dx(x3y2)- 2Dx(x)+DJy3) = 0
x3D,(y2) +y2Dx(x3)- 2(1) +3y29= 0
dx
d
Y
+y2(3x2)- 2 +3y2-= 0
dx
d
Y
(2x3y +3y2) -= 2 - 3x52
dx
dy 2 - 3x2y2
--
dx - 2x3y+3y2

[CHAP. 16
128 IMPLICIT DIFFERENTIATION
16.3 If y = f ( x ) is a differentiable function satisfying the equation x2y3 - 5xy2 -4y = 4 and if
f(3) = 2, find the slope of the tangent line to the graph offat the point (3,2).
DJx2y3 - 5xy2-4y) = 0,(4)
x2(3y2y‘) +y3(2x) - 5(x(2yy’) +y 2 ) - 4y‘ = o
3x2y2y’ +2xy3 - ioxyY’- 5y2 - 4y‘ = o
Substitute 3 for x and 2 for y,
108~’
+48 - 6 0 ~ ’
- 20 - 4y’ = 0
My‘ +28 = 0
28 7
y‘= ---
44--11
Hence, the slope of the tangent line at (3,2) is -A.
16.4 (a)Find a formula for the slope of the tangent line to the curve x2 - xy +y2 = 12 at any point (x, y). Also,
find the coordinates of all points on the curve where the tangent line is: (b) horizontal; (c)vertical.
165 Consider the hyperbola 5x2- 2y2= 130.
(a) Find a formula for the slope of the tangent line to this hyperbola at (x, y).
(b) For what value(s) of k will the line x - 3y +k = 0 be normal to the hyperbola at a point of intersec-
tion?
16.6 Find y’ by implicit differentiation.
2x +y 1 1
2x -y * Y
(a) x2 +y2= 25 (b) X’ =- (c) - + - = 1
xz yz X + Y
(9) - + - = 1 (h) y +xy3 = 2x (i) x2 =-
9 4 x - Y
16.7 Use implicit differentiation to find the slope-intercept equation of the tangent line at the indicated point.
(a) y3- xy = 2 at (3, 2)
X2
16
(b) -+ y2 = 1 at
(c) (y - x ) ~
+y3 = xy +7 at (1, 2)
(e) 4xy2 +98 = 2x4 -y4 at (3, 2)
(d) x3 - y3 = 7xy at (4, 2)
(f)4x3 - xy - 2y3 = 1 at (1, 1)
x3 -
(9) -
-
- x at (1, -1)
1-Y
(h) Zy = xy3 +Zx3 -3 at (1, -I)
16.8 Use implicit differentiation to find the slope-intercept equation of the normal line at the indicated point.
(a) y3x +2y = x2 at (2, 1)
(c) y f i - x& = 12 at (9, 16)
(b) 2x3y +2y4-x4 = 2 at (2, 1)
(d) x2 +y2 = 25 at (3, 4) .
16.9 Use implicit differentiation to find the slope of the tangent line to the graph of y = ,/E
at x = 6.
[Hint: Eliminate the radicals by squaring twice.]

Chapter 17
The Mean-Value Theorem and
the Sign of the Derivative
17.1 ROLLE’S THEOREM AND THE MEAN-VALUE THEOREM
Let us consider a functionf that is continuous over a closed interval [a, b] and differentiable at
every point of the open interval (a, b). We also suppose thatf(a) =f (b) = 0.Graphs of some examples of
such a function are shown in Fig. 17-1. It seems clear that there must always be some point between
x = a and x = b at which the tangent line is horizontal and, therefore, at which the derivativeoffis 0.
Y
a b X
Y
(6)
Fig. 17-1
Theorem17.1 (Rolle’s Theorem): Iffis continuous over a closed interval [a, b], differentiable on the
open interval (a, b), and if f(a) =f (b) = 0, then there is at least one number c in (a, b)
such thatf’(c) = 0.
See Problem 17.6for the proof.
Rolle’s theorem enables us to prove the followingbasic theorem (which is also referred to as the law
of the meanfor derivatives).
Theorem17.2 (Mean-Value Theorem): Letfbe continuous over the closed interval [a, b] and differen-
tiable on the open interval (a,b). Then there is a number c in the open interval (a,b) such
that
EXAMPLE In graphic terms, the mean-value theorem states that at some point along an arc of a curve, the
tangent line is parallel to the line connecting the initial and the terminal points of the arc. This can be seen
in Fig. 17-2, where there are three numbers (c1, c2, and cj) between a and b for which the slope of the
tangent line to the graph f ’(c) is equal to the slope of the line AB, f(b)-f(4
b - a *
129

130 THE MEAN-VALUE THEOREM AND THE SIGN OF THE DERIVATIVE [CHAP. 17
4 Y
0
17.2 THE SIGN OF THE DERIVATIVE
Similarly,fis decreasing on a set a
t if, for any U and t, in d ,U < t, impliesf(u) >f(t,).
A function f is said to be increasing on a set d if, for any U and U in d,U 0 for all x in the open interval (a,b), thenfis increasing on (a,b). Iff’(x) < 0
for all x in (a,b), thenfis decreasing on (a,b).
For the proof, see Problem 17.8. The converse of Theorem 17.3 does not hold. In fact, the function
f ( x ) = x3 is differentiable and increasing on (- 1, 1
)
-
a
n
d everywhere else-but f’(x)= 3x2 is zero for
x = 0 [see Fig. 7-3(b)].
Theorem 17.4 (Intermediate-Value Theorem): Let f be a continuous function over a closed interval
[a, 61, withf(a)#f(6).Then any number betweenf(a) and f(b)is assumed as the value
offfor some argument between a and 6.
While Theorem 17.4 is not elementary, its content is intuitively obvious. The function could not
“skip” an intermediate value unless there were a break in the graph; that is, unless the function were
discontinuous. As illustrated in Fig. 17-4, a functionfsatisfying Theorem 17.4 may also take on values
that are not betweenf(a) andf(b). In Problem 17.9, we prove:
The following important property of continuous functions will often be useful.

CHAP. 17) THE MEAN-VALUE THEOREM AND THE SIGN OF THE DERIVATIVE 131
A Y
‘ P
1 I
*
a b X
Fig. 17-4
Corollary17.5: Iffis a continuous function with domain [a, b], then the range offis either a closed
interval or a point.
SolvedProblems
17.1 Verify Rolle’s theorem forf(x) = x3 - 3x2 - x +3 on the interval [l, 31.
fis differentiableeverywhere and, therefore,also continuous. Furthermore,
f(i) = ( 9 3 - 3(1)2 - 1 +3 = 1 - 3 - 1 +3 = o
f(3) = (3)3- 3(3)2- 3 +3 = 27 - 27 - 3 +3 = 0
so that all the hypotheses of Rolle’s theorem are valid. There must then be some c in (1, 3) for which
f ‘(c) = 0
.
Now, by the quadratic formula,the roots off’(x) = 3x2 - 6x - 1 = 0 are
Consider the root c = 1 + 4fi. Since fi < 3,
1 < 1 + @
L 1 +3(3)=
L 1 + 2 = 3
Thus, c is in (1, 3)andf’(c) = 0.
17.2 Verify the mean-value theorem forf(x) = x3 - 6x2 - 4x +30 on the interval [4,6].
f is differentiable and, therefore,continuous for all x.
whence,
f(6) = (6)3- 6(6)2- 416) +30 = 216 - 216 - 24 +30 = 6
f(4) = (4)3- 6(4)2-414) + 30 = 64 -96 - 16 +30 = -18
f(6) -f(4) 6 -(-18) 24
-
- -
--=I2
6 - 4 6 - 4 2

We must therefore find some c in (4,6) such thatf’(c) = 12.
Now,f‘(x) = 3x2 - 12x - 4, so that c will be a solution of
3x2 - 12x - 4 = 12 or 3x2- 12x - 16 = 0
By the quadratic formula,
12 +_ J144 -4(3)(- 16) -
- 12 & d m-
- 12
6 6
x =
Choose c = 2 +j f l . Since4 < < 5,
4 < 2 +-
8 = 2 +j
2 (4) < 2 +j
2 J21<2 +j
2 (5) < 2 +4 = 6
3
Thus, c is in (4, 6) andf’(c) = 12.
17.3 Determine when the functionf ( x )= x3 - 6x2 +9x +2 is increasing and when it is decreasing,
and sketch its graph.
We havef’(x) = 3x2 - 12x +9 = 3(x2 - 4x +3) = 3(x - 1Xx - 3). The crucial points are 1and 3 [see
Fig. 17-5(a)].
(i) When x < 1, both (x - 1)and (x - 3) are negative and sof’(x) >0 in (- 00, 1).
(ii) When x moves from (-CO, 1) into (1, 3), the factor (x - 1) changes from negative to positive, but
(x - 3) remains negative. Hence,f’(x) < 0 in (1, 3).
(iii) When x moves from (1, 3) into (3, CO), (x - 3) changes from negative to positive, but (x - 1) remains
positive. Hence,f’(x) > 0 in (3, CO).
Thus, by Theorem 17.3, f is increasing for x < 1, decreasing for 1 < x < 3, and increasing for x > 3.
Note that f(1) = 6, f(3) = 2, lim f ( x ) = +CO,
and lim f(x) = -CO. A rough sketch of the graph is
shown in Fig. 17-5(b).
X + + a , x---O3
4Y
Fig. 17-5
17.4 Verify Rolle’s theorem forf(x) = 2x6 - 8x5 +6x4 -x3 +6x2 - llx +6 on [l, 33.
f is differentiableeverywhere,and
f(1) = 2 - 8 +6 - 1 +6 - 11 +6 = 0
f(3) = 1458 - 1944 +486 -27 +54 - 33 +6 = 0
.
It is difficult to compute a value of x in (1,3) for which
f‘(x) = 12x5- 40x4 +24x3- 3x2 + 12x - 11 = 0

CHAP. 173 THE MEAN-VALUE THEOREM AND THE SIGN OF THE DERIVATIVE 133
However,f’(x) is itselfa continuous function such that
f’(1) = 12 - 40 +24 - 3 + 12 - 11 = -6 <0
f’(3) = 2916 - 3240 +648 - 27 +36 - 11 = 322 > 0
Hence, the intermediate-value theorem assures us that there must be some number c between 1 and 3 for
whichf’(c) = 0.
17.5
17.6
17.7
Show thatf(x) = 2x3 +x - 4 = 0 has exactly one real solution.
zero between 0 and 2; call it xo.
f ( x ) >0; and when x < xo, f ( x )< 0. In other words, there is no zero other than xo.
Sincef(0) = -4 and f(2) = 16 +2 - 4 = 14, the intermediate-value theorem guarantees that f has a
Because f’(x)= 6x2 + 1 > 0, f ( x ) is increasing everywhere (Theorem 17.3). Therefore, when x > xo,
Prove Rolle’s theorem (Theorem 17.1).
Cuse I: f ( x ) = 0 for all x in [a, b]. Then f’(x)= 0 for all x in (a, b), since the derivative of a constant
function is 0.
Case 2
: f ( x ) > 0 for some x in (a, b). Then, by the extreme-value theorem (Theorem 14.2), an absolute
maximum off on [a, b] exists, and must be positive [since f ( x )> 0 for some x in (a, b)]. Because
f(a) =f(b)= 0, the maximum is achieved at some point c in the open interval (a, b). Thus, the absolute
maximum is also a relative maximum and, by Theorem 14.l,f’(c) = 0.
Cuse 3: f ( x )<0 for some x in (a, b). Let g(x)= -f(x). Then, by Case 2, g’(c) = 0 for some c in (a, b).
Consequently,f’(c) = -g’(c) = 0.
Prove the mean-value theorem (Theorem 17.2).
Let
Then g is continuous over [a, b] and differentiableon (a,b). Moreover,
db) =f@)
- f(b)
b - (6 - a) -f(a) =f(b)- (f(b)-f(a)) -f(a)
=f(b) -f(b) +f(4-f(4= 0
By Rolle’s theorem, applied to g, there exists c in (a,b) for which g’(c) = 0. But,
whence,
Assume that f‘(x)> 0 for all x in (a, b) and that a 0 and U - U > 0; hence,f(u) -f(u) > O,f(u) <f(u).
The casef’(x) < 0 is handled similarly.

17.9 Prove Corollary 17.5.
By the extreme-value theorem,fhas an absolute maximum valuef(d) at some argument d in [a, b], and
an absolute minimum valuef(c) at some argument c in [a, b]. Iff(c) =f(d) = k, thenfis constant on [a, b],
and its range is the single point k. Iff(c) #f(d), then the intermediate-value theorem, applied to the closed
subinterval bounded by d and c, ensures that fassumes every value betweenf(c)andf(d). The range off is
then the closed interval [f(c),f(d)] (which includes the values assumed on that part of [a, b] that lies
outside the subinterval).
17.10 Determine whether the hypotheses of Rolle's theorem hold for each functionf; and if they do, verify the
conclusion of the theorem.
(a) f ( x )= x2 - 2x - 3 on [-1, 3)
(c) f ( x ) = 9x3- 4x on [-3, 31
(b) f ( x )= x3 -x on [0, 13
(4 f(x) = x3 - 3x2 +x + 1 on 11, I +$
1
17.11 Verify that the hypotheses of the mean-value theorem hold for each function f on the given interval, and
find a value c satisfyingthe conclusion of the theorem.
(a) f ( x ) = 2x + 3 on [1, 41 (b) f ( x ) = 3x2- 5x + 1 on [2, 51
(c) f ( x )= x314on [0, 16)
17.12 Determine where the functionfis increasing and where it is decreasing.Then sketch the graph off:
(a) f ( x ) = 3x + 1 (b) f ( x )= -2x +2 (c) f ( x )= x2 - 4x +7
1
(9) f ( x ) x3 - 9x2 + 15x - 3 (h) f(x) = x +-
X
(i) f ( x ) = x3 - 12x +20
17.13 Letfbe a differentiable function such thatf'(x) # 0 for all x in the open interval (a, b). Prove that there is at
most one zero off@) in (a, b). [Hint:
Assume, for the sake of contradiction, that c and d are two zeros off,
with a < c < d < b, and apply Rolle's theorem on the interval [c, 4
.
3
17.14 Consider the polynomialf(x) = 5x3 - 2x2 +3x - 4.
(a) Show thatfhas a zero between 0 and 1.
(b) Show thatfhas only one real zero. [Hint:
Use Problem 17.13.1
17.15 Assumefcontinuous over CO, 13 and assume thatf(0) =f(l). Which one@)of the following assertions must
be true?
(a) Iffhas an absolute maximum at c in (0, l), thenf'(c) = 0.
(b) f ' exists on (0, 1).

CHAP. 171 THE MEAN-VALUE THEOREM AND THE SIGN OF THE DERIVATIVE 135
17.16
17.17
17.18
17.19
17.20
17.21
17.22
17.23
17.24
17.25
17.26
(c) f’(c) = 0 for some c in (0, 1).
(d) lim f ( x )=f(c) for all c in (0, 1).
(e) fhas an absolute maximum at some point c in (0, 1).
x-c
Letfand g be differentiablefunctions.
(a) Iff(a) = g(a) andf(b) = g(b),where a < b, show thatf’(c) = g’(c) for some c in (a, b).
(b) Iff(a) 2 g(a) andf’(x) > g‘(x) for all x, show thatf(x) > g(x) for all x > a.
(c) Iff’(x)>g’(x) for all x, show that the graphs off and g intersect at most once. [Hint:
In each part,
apply the appropriate theorem to the function h(x)= f ( x ) -g(x).]
Letfbe a differentiablefunction on an open interval (a,b).
(a) Iffis increasingon (a,b), prove that f’(x) 2 0 for every x in (a, b).
[Hint :f’(x)= lim f ( x + h, - f ( x ) and Problem 9.1qa) applies.
(b) Iffis decreasingon (a, b), prove thatf’(x) 5 0 for every x in (a, b).
1
h-O+ h
The mean-value theorem predicts the existence of what point on the graph of y = fi between (27, 3) and
(125, 5)?
(Generalized Rolle’s Theorem) Assumefis continuous on [a, b) and differentiable on (a, b). Iff(a) =f(b),
prove that there is a point c in (a,b) such thatf’(c) = 0. [Hint:
Apply Rolle’s theorem to g(x) = f ( x ) -f(a).]
Letf(x) = x3 - 4x2 +4x and g(x) = 1for all x.
(a) Find the intersection of the graphs offand g.
(b) Find the zeros o f t
(c) If the domain offis restricted to the closed interval [0, 31, what would be the range off?
Prove that 8x3 - 6x2 - 2x + 1 has a zero between 0 and 1. [Hint: Apply Rolle’s theorem to the function
2x4 - 2x3 - x2 +x . ~
Show that x3 +2x - 5 = 0 has exactly one real root.
Prove that the equation x4 +x = 1has at least one solution in the interval CO, 13.
Find a point on the graph of y = x2 +x + 3, between x = 1 and x = 2, where the tangent line is parallel to
the line connecting (1, 5) and (2,9).
(a) Show thatf(x) = x5 +x - 1has exactly one real zero.
(b) Locate the real zero of x5 +x - 1correct to the first decimal place.
(a)
is increasing and the intervals in which it is decreasing.
(b) As in part (a),but for the functionf(x) = x3 - 2x2 +x -2.
Use a graphing calculator to estimate the intervals in which the functionf(x) = x4 - 3x2 +x - 4

Chapter 18
Rectilinear Motion and InstantaneousVelocity
Rectilinear motion is motion along a straight line. Consider, for instance, an automobile moving
along a straight road. We can imagine a coordinate system imposed on the line containing the road (see
Fig. 18-1).(On many highways there actually is such a coordinate system, with markers along the side
of the road indicating the distance from one end of the highway.)Ifs designates the coordinate of the
automobile and t denotes the time, then the motion of the automobile is specified by expressing s, its
position, as a function oft: s =f(t).
1 L 1 I 1 I 1 1 &
- 3 - 2 - 1 0 I 2 3 4 S
Fig. 18-1
The speedometerindicates how fast the automobile is moving. Since the speedometer reading often
varies continuously, it is obvious that the speedometer indicates how fast the car is moving at the
moment when it is read. Let us analyze this notion in order to find the mathematical concept that lies
behind it.
If the automobile moves according to the equation s =f(t), its position at time t isf(t), and at time
t +h, very close to time t, its position isf(t +h). The distance’ between its position at time t and its
position at time t +h isf(t +h) - f ( t ) (which can be negative). The time elapsed between t and t +h is
h. Hence, the average velocity2 during this time interval is
f ( t +h) -m
h
(Averagevelocity = displacement- time.) Now as the elapsed time h gets closer to 0, the average veloc-
ity approaches what we intuitively think of as the instantaneous velocity U at time t. Thus,
v = lim f ( t +h) -f(O
h
h-0
In other words, the instantaneous velocity v is the derivativef’(t).
EXAMPLES
(a) The height s of a water column is observed to follow the law s =f(t) = 3t +2. Thus, the instantaneous veloc-
ity U of the top surfaceisf’(t) = 3.
(6) The position s of an automobile along a highway is given by s =f(t) = t2 - 2t. Hence, its instantaneous
velocity is U =f‘(t) = 2t - 2. At time t = 3, its velocity U is 2(3) - 2 = 4.
The sign of the instantaneous velocity v indicates the direction in which the object is moving. If
U = ds/dt > 0 over a time interval, Theorem 17.3 tells us that s is increasingin that interval. Thus, if the
s-axis is horizontal and directed to the right, as in Fig. 18-2(a),then the object is moving to the right;
but if the s-axis is vertical and directed upward, as in Fig. 18-2(b),then the object is moving upward. On
More precisely, the displacement,since it can be positive,negative,or zero.
’We use the term velocity rather than speed because the quantity referred to can be negative.Speed is defined as the magnitude of
the velocityand is nevernegative.
136

CHAP. 181 RECTILINEAR MOTION AND INSTANTANEOUS
the other hand, if U = ds/dt -c0 over a time interval, then s must be
137
VELOCITY
decreasing in that interval. In
Fig. 18-2(a),the object would be moving to the left (in the direction of decreasing s);in Fig. 18-2(b),the
object would be moving downward.
A consequence of these facts is that at an instant t when a continuously mooing object reverses
direction, its instantaneous oelocity o must be 0. For if U were, say, positive at t, it would be positive in a
small interval of time surrounding t ; the object would therefore be moving in the same direction just
before and just after t. Or, to say the same thing in a slightly different way, a reversal in direction means
a relative extremum of s, which in turn implies (Theorem 14.1)ds/dt = 0.

EXAMPLE An object moves along a straight line as indicated in Fig. 18-3(a).In functional form,
s = f ( t ) = (t +2)2 [s in meters, t in seconds]
as graphed in Fig. 18-3(b).The object’s instantaneous velocity is
t, =f’(t) = 2(t +2) [meters per second]
For t +2 < 0, or t < -2, o is negative and the object is moving to the left; for t +2 > 0, or t > -2, t, is positive
and the object is moving to the right. The object reverses direction at t = -2, and at that instant U = 0. [Note that
f ( t )has a relative minimum at t = -2.1
t = -3 t = - 4
(--------
+
0 S
-*--- +-a----
I = -0.5 I = 0 I = I
Free Fall
(b)
Fig. 18-3
Consider an object that has been thrown straight up or down, or has been dropped from rest, and
which is acted upon solely by the gravitational pull of the earth. The ensuing rectilinear motion is called
jieefall.
Let us put a coordinate system on the vertical line along which the object moves, such that the
s-axis is directed upward, away from the earth, with s = 0 located at the surface of the earth (Fig. 18-4).

138 RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY [CHAP. 18
Fig. 18-4
Then the equation of free fall is
s SO +VO t - 16t2 (18.1)
where s is measured in feet and t in seconds? Here so and tro are, respectively,the position (height)and
the velocity of the object at time t = 0. The instantaneous velocity U is obtained by differentiating(18.1),
( I8.2)
EXAMPLES
At t = 0, a rock is dropped from rest from the top of a building 256 feet high. When, and with what velocity,
does it strike the ground?
With so = 256 and uo = 0, (18.1) becomes s = 256 - 16t2 and the time of striking the ground is given by
the solution of
0 = 256 - 16t2
16t2 = 256
t2 = 16
t = +,4 seconds
Since we are assuming that the motion takes place when t 2 0, the only solution is t = 4 seconds.
The velocity equation (18.2)is U = -32t, and so, for t = 4,
U = -32(4)= -128 feet per second
the minus sign indicating that the rock is moving downward when it hits the ground.
ALGEBRA x feet per second = 6Ox feet per minute
= 60(60x) feet per hour
=-3600x miles per hour
5280
15
22
= -x miles per hour
For example, 128feet per second = 8(128)= 8 7 6 miles per hour.
(28.3)
A rocket is shot vertically from the ground with an initial velocity of 96 feet per second. When does the rocket
reach its maximum height, and what is its maximum height?
With so = 0 and uo = 96, (18.1)and (28.2)become
d S
dt
s = 96t - 16t2 and U =-= 96 - 32t
If the position is measuredi
n meters,the equationreads s = so +uo t -4.9t2.

CHAP. 18) RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY 139
At a maximum value, qr turning point, U = 0. Hence,
0 = 96 - 32t
32t = 96
t = 3
Thus, it takes 3 secondsfor the rocket to reach its maximum height, which is
s = 96(3) - 16(3)2= 288 - 16(9) = 288 - 144 = 144 feet
(c) When does the rocket of part (b) hit the ground?
It sufficesto set s = 0 in the free-fall equation (18J),
0 = 96t - 16t2
0 = 6t - t2 [divide by 163
0 = t(6 - t)
from which t = 0 or t = 6. Hence, the rocket hits the ground again after 6 seconds.
Notice that the rocket rose for 3 seconds to its maximum height, and then took 3 more seconds to fall
back to the ground. In general, the upward flight from point P to point Q will take exactly the same time as
the downward flight from Q to P. In addition, the rocket will return to a given height with the same speed
(magnitude of the velocity)that it had upon leaving that height.
Solved Problems
18.1 A stone is thrown straight down from the top of an 80-foot tower. If the initial speed is 64 feet
per second, how long does it take to hit the ground, and with what speed does it hit the ground?
Here so = 80 and uo = -64. (The speed is the magnitude of the velocity. The minus sign for uo indi-
cates that the object is moving downward.)Hence,
ds
dt
s = 80 - 64t - 16t2 and U = -= -64 - 322
The stone hits the ground when s = 0,
0 = 80 - 64t - 16t2
0 = t2 +4t - 5
t + 5 = 0 or t - 1 = 0
t = - 5 or t = l
[divide by -161
0 = (t +5Xt - 1)
Sincethe time of fall must be positive, t = 1 second.The velocity U when the stone hits the ground is
o(1) = -64 - 32(1) = -64 - 32 = -96 feet per second
15 5
22 11
By (18.3),96 feet per second = -(96) = 65 -miles per hour.
18.2 A rocket, shot straight up from the ground, reaches a height of 256 feet after 2 seconds. What
was its initial velocity, what will be its maximum height, and when does it reach its maximum
height?
Since so = 0,
ds
dt
s = u,t - 16t2 and U = - = UO - 32t

140 RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY [CHAP. 18
When t = 2, s = 256,
256 = 00(2)- 16(2)2
256 = 200 - 64
320 = 2
0
,
160= 00
The initial velocity was 160feet per second, so that
s = 160t - 16t2 and U = 160- 32t
To find the time when the maximum height is reached, set D = 0,
0 = 160- 32t
32t = 160
t = 5 seconds
To find the maximum height, substitute t = 5 in the formula for s,
s = 160(5)- 16(5)2= 800 - 16(25)= 800 -400 = 400feet
18.3 A car is moving along a straight road according to the equation
= f ( t ) = 2t3 - 3 t 2 - 12t
Describe its motion by indicating when and where the car is moving to the right, and when and
where it is moving to the left. When is the car at rest?
We have U =f'(t) = 6t2-6t - 12 = 6(t2 -t -2) = 6(t -2)(t + 1). The key points are t = 2 and
t = -1(seeFig. 18-5).
-I 2 t
Fig. 18-5
(i) When t > 2, both t -2 and t + 1 are positive. So, U > 0 and the car is moving to the right.
(ii) As t moves from (2, CO) through t = 2 into (-1, 2), the sign of t - 2 changes, but the sign of t + 1
remains the same. Hence, U changes from positive to negative. Thus, for -1 < t < 2, the car is moving
to the left.
(iii) As t moves through t = -1 from (- 1, 2) into (- a,-l), the sign of t + 1 changes but the sign of
t -2 remains the same. Hence, U changes from negative to positive. So, the car is moving to the right
when t < -1.
When t = -1,
s = 2(- 1)3-3(- 1)2- 12(- 1)= -2 - 3 + 12 = 7
When t = 2
s = 2(2)3- 3(2)2- 12(2) = 16 - 12- 24 = -20
Thus, the car moves to the right until, at t = -1,it reaches s = 7,where it reverses direction and moves left
until, at t = 2, it reaches s = -20, where it reverses direction again and keeps moving to the right thereafter
(see Fig. 18-6).
Fig. 18-6

CHAP. 181 RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY 141
The car is never at rest. It makes sense to talk about the car being at rest only when the position of the
car is constant over an interval of time (not at just a single point). In such a case, the velocity would be zero
on an entire interval.
18.4 (a) If an object is released from rest at any given height, show that, after t seconds, it has dropped 16t2feet
(assumingthat it has not yet struck the ground).
(b) How many seconds does it take the object in part (a)to fall: (i) 1 foot; (ii) 16feet; (iii)64 feet; (iv) 100
feet?
185 A rock is dropped down a well that is 256 feet deep. When will it hit the bottom of the well?
18.6 Assuming that one story of a building is 10feet, with what speed, in miles per hour, does an object dropped
from the top of a 40-story building hit the ground?
A rocket is shot straight up into the air with an initial velocity of 128feet per second.
(a)How far has it traveled in 1 second? in 2 seconds? (b) When does it reach its maximum height? (c)What
is its maximum height? (d) When does it hit the ground again? (e) What is its speed when it hits the
ground ?
18.7
18.8 A rock is thrown straight down from a height of 480 feet with an initial velocity of 16feet per second.
(a)How long does it take to hit the ground? (b) With what speed does it hit the ground? (c) How long does
it take before the rock is moving at a speed of 112 feet per second? (d)When has the rock traveled a
distance of 60feet?
3
An automobile moves along a straight highway, with its position given by s = 12t3- 18t2+9t --
2
[s in miles, t in hours].
(a) Describe the motion of the car: when it is moving to the right, when to the left, where and when it
changes direction.
(b) What distance has it traveled in 1 hour from t = 0 to t = l ?
18.9
18.10 The position of a moving object on a line is given by the formula s = (t - l)3(t - 5).
(a) When is the object moving to the right? (b) When is it moving to the left? (c) When is it changing
direction? (d) When is it at rest? (e)What is the farthest to the left of the origin that it moves?
18.11 A particle moves on a straight line so that its position s (miles) at time t (hours) is given by
(a) When is the particle moving to the right? (b)'When is the particle moving to the left? (c) When does it
change direction? (d)When the particle is moving to the left, what is the maximum speed that it achieves?
(The speed is the absolute value of the velocity.)
s = (4t - 1Kt - 1)2.
18.12 A particle moves along the x-axis according to the equation x = 10t - 2t2. What is the total distance
covered by the particle between t = 0 and t = 3?
18.13 A rocket was shot straight up from the ground. What must have been its initial velocity if it returned to
earth in 20 seconds?

[CHAP. 18
142 RECTILINEAR MOTION AND INSTANTANEOUS VELOCITY
18.14 Two particles move along the x-axis. Their positions f ( t ) and g(t) are given by f ( t ) = 6t - t2 and
(a) When do they have the same position? (b) When do they have the same velocity? (c) When they have the
same position, are they moving in the same direction?
g(t)= t2 - 4t.
18.15 A rock is dropped and strikes the ground with a velocity of -49 meters per second. (a) How long did it
fall?(b) Find the height from which it was dropped.
18.16 A ball is thrown vertically upward from the top of a 96-foot tower. Two seconds later, the velocity of the
ball is 16 feet per second. Find: (a) the maximum height that the ball reaches; (6) the speed of the ball
when it hits the ground.

Chapter 19
Instantaneous Rate of Change
One quantity, y, may be related to another quantity, x, by a functionf: y =f(x). A change in the
value of x usually induces a corresponding change in the value of y.
EXAMPLE Let x be the length of the side of a cube, and let y be the volume of the cube. Then y = x3.In the case
where the side has length x = 2 units, consider a small change Ax in the length.
NOTATION Ax (read “delta-ex”) is the traditional symbol in calculus for a small change in x. Ax is considered a
single symbol, not a product of A and x. In earlier chapters, the role of Ax often was taken by the symbol h.
The new volume will be (2 +Ax)~,
and so the change in the value of the volume y is (2 +A x ) ~
- Z3. This change in
y is denoted traditionally by Ay,
Ay = (2 +A x ) ~
- 23
Now the natural way to compare the change Ay in y to the change Ax in x is to calculate the ratio Ay/Ax. This
ratio depends of course on Ax, but if we let Ax approach 0, then the limit of Ay/Ax will define the instantaneousrate
o
f change of y compared to x, when x = 2. We have (ALGEBRA, Problem 11.2)
Ay = (2 +A x ) ~
-23 = [(2)3+3(2)2(A~)1
+3(2)’(A~)~
+(Ax)~]- 23
= 1 2 6 ~
+~ ( A x ) ~
+(Ax)~= (AxX12 +6Ax +(Ax)~)
Hence, --
Ay - 12 +6Ax +(Ax)’
Ax
and AY
lim -= lim (12 + 6Ax +(AX)~)
= 12
Ax-0 Ax Ax+O
Therefore, when the side is 2, the rate of change of the volume with respect to the side is 12. This means that, for
sides close to 2, the change Ay in the volume is approximately 12 times the change Ax in the side (since Ay/Ax is
close to 12).Let us look at a few numerical cases.
If Ax = 0.1, then the new side x +Ax is 2.1, and the new volume is (2.1)3= 9.261. So, Ay = 9.261 - 8 = 1.261,
and
Ay 1.261
Ax 0.1
----= 12.61
If Ax = 0.01, then the new side x +Ax is 2.01, and the new volume is (2.01)3= 8.120601. So,
Ay = 8.120601 -8 = 0.120601, and
12.0601
Ay 0.120601
Ax 0.01
-=-=
If Ax = 0.001, a similar computation yields
--
Ay - 12.006001
Ax
Let us extend the result of the above example from y = x3 to an arbitrary differentiable function
y =f(x). Consider a small change Ax in the value of the argument x. The new value of the argument is
143

[CHAP. 19
144 INSTANTANEOUS RATE OF CHANGE
then x +Ax, and the new value of y will bef(x +Ax).Hence, the change Ay in the value of the function
is
Ay =f(x +Ax) -f ( x )
The ratio of the change in the function value to the change in the argument is
The instantaneousrate o
f change of y with respect to x is defined to be
AY
lim - = lim
A x - O A x Ax-0 Ax
The instantaneous rate o
f change is evaluated by the derivative. It follows that, for Ax close to 0, Ay/Ax
will be close tof’(x), so that
Ay x f’(x)Ax (19.1)
SolvedProblems
19.1 The weekly profit P, in dollars, of a corporation is determined by the number x of radios pro-
duced per week, according to the formula
P = 7 5 ~
- 0 . 0 3 ~ ~
- 15000
(a)Find the rate at which the profit is changing when the production level x is loo0 radios per
week. (6) Find the change in weekly profit when the production level x is increased to 1001
radios per week.
(a) The rate of change of the profit P with respect to the production level x is dP/dx = 75 - 0.06~.
When
x = 1000,
- =
dP 75 - O.O6(1ooO)= 75 -60 = 15 dollars per radio
dx
(b) In economics, the rate of change of profit with respect to the production level is called the marginal
pro& According to (29.2), the marginal profit is an approximate measure of how much the profit will
change when the production level is increased by one unit. In the present case, we have
P(1000)= 75(1OOO)- 0.03(1000)2- 15000
= 75000 - 30000 - 15000 = 30000
P(1001) = 75(1001)- 0.03(1001)2- 15000
= 75075 - 30060.03 - 15000 30014.97
A P = P(1001) - P(1OOO) = 14.97 dollars per week
which is very closely approximated by the marginal profit, 15dollars, as computed in part (a).
19.2 The volume V of a sphere of radius r is given by the formula V = 4nr3/3.(a) How fast is the
volume changing relative to the radius when the radius is 10millimeters? (b)What is the change
in volume when the radius changesfrom 10to 10.1millimeters?

CHAP. 191
When r = 10,
INSTANTANEOUS RATE OF CHANGE
-=
dV 4n(10)’ = 4OOn w 400(3.14) = 1256
dr
4 4000n
3 3
4 4 4121.204~
3 3
~ ( 1 0 )
= - 410)3 = -
3
V(lO.1) = -~(10.1)~
= - lt(1030.301) =
4121.204~ 4 0 0 0 ~
--
3 3
AV = V(10.1) - V(10) =
n R 3.14
3 3 3
= - (4121.204 - 4ooo) = - (121.204) w -(121.204)
= 126.86 cubic millimeters
The change predicted from (29.2)and part (a)is
145
dV
dr
AV x -Ar = 125qO.l) = 125.6cubic millimeters
19.3 An oil tank is being filled. The oil volume V ,in gallons, after t minutes is given by
V = 1.W +2t
How fast is the volume increasing when there is 10 gallons of oil in the tank? [Hint: To answer
the question, “how fast?,” you must always find the derivative with respect to time.]
When there are 10gallons in the tank,
1 3 ’ +2t = 10 or 1.9’ +2t - 10 = 0
Solvingby the quadratic formula,
-2 f ,/4 - 4(1.5K-10) -2 k , / 4 3 -2 f@ -2 +8 10
-
- -
- =
:
=
2 or --
3 3 3 3
t =
2(1.5)
Since t must be positive, t = 2 minutes. The rate at which the oil volume is growing is
dV
dt
-= D,(1.5t2 +2t) = 3t +2
Hence, at the instant t = 2 minutes when V = 10gallons,
--- 3(2) +2 = 6 + 2 = 8 gallons per minute
dV
dt
19.4 The cost C, in dollars per day, of producing x TV sets per day is given by the formula
C = 7000 +5 0 ~
- 0 . 0 5 ~ ~
Find the rate of change of C with respect to x (called the marginal cost) when 200 sets are being produced
each day.

146 INSTANTANEOUS RATE OF CHANGE [CHAP. 19
19.5 The profit P, in dollars per day, resulting from making x units per day of an antibiotic, is
P = 5x +0.02x2- 120
Find the marginal profit when the production level x is 50 units per day.
19.6 Find the rate at which the surfacearea of a cube of side x is changing with respect to x, when x = 2 feet.
19.7 The number of kilometers a rocket ship is from earth is given by the formula
E = 30t +0.005t2
where t is measured in seconds. How fast is the distance changing when the rocket ship is 35000 kilometers
from earth?
19.8 As a gasoline tank is being emptied, the number G of gallons left after t seconds is given by G = 3(15 - t)2.
(a) How fast is gasoline being emptied after 12seconds? (b) What was the average rate at which the gasoline
was being drained from the tank over the first 12 seconds? [Hint: The average rate is the total amount
emptied divided by the time during which it was emptied.]
19.9 If y = 3x2-2, find: (a) the average rate at which y changes with respect to x over the interval [1, 21; (b)
the instantaneous rate of change of y with respect to x when x = 1.
19.10 If y =f(x) is a function such thatf’(x) # 0 for any x, find those values of y for which the rate of increase of
y4 with respect to x is 32 times that of y with respect to x.

Chapter 20
Most quantities encountered in science or in everyday life vary with time. If two such quantities are
related by an equation, and if we know the rate at which one of them changes, then, by differentiating
the equation with respect to time, we can find the rate at which the other quantity changes.
EXAMPLES
(a) A 6-foot man is running away from the base of a streetlight that is 15 feet high (see Fig. 20-1). If he moves at
the rate of 18feet per second,how fast is the length of his shadow changing?
Let x be the distance of the man from the base A of the streetlight, and let y be the length of the man's
shadow.
GEOMETRY Two triangles,
X
are similar if their angles are equal in pairs: 3c A = %X, 3c B = % Y, % C = 3cZ. (For this condition to hold,
it suffices that two angles of the one triangle be equal to two angles of the other.) Similar triangles have
corresponding sides in fixed ratio:
147

148 RELATED RATES [CHAP. 20
In Fig. 20-1, ASMN and ASAL are similar, whence
Y 6
S M N M
__ - or ---
SA LA y + x 15
-
-=-
which is the desired relation between x and y. In this case, it is convenient to solve (1)for y in terms of x,
--
Y 2
--
y + x 5
5y = 2y +2x
3y = 2x
2
3
y = - x
Differentiation of (2)with respect to t gives
dy 2 dx
- = - -
dt 3 dt
(3)
Now because the man is running away from A at the rate of 18 feet per second, x is increasing at that rate.
Hence,
dx
-= 18 feet per second
dt dt 3
and -
dy = -(18) = 12 feet per second
that is, the shadow is lengthening at the rate of 12 feet per second.
(b) A cube of ice is melting. The side s of the cube is decreasing at the constant rate of 2 inches per minute. How
fast is the volume V decreasing?
Since V = s3,
ds
dt dt dt
d V 4s’) - 3s2 -
-=--
The fact that s is decreasing at the rate of 2 inches per minute translates into the mathematical statement
ds
dt
--
- -2
d V
dt
-= 3s2(-2) = -6s2
Hence,
Thus, although s is decreasing at a constant rate, V is decreasing at a rate proportional to the square of s. For
instance, when s = 3 inches, V is decreasing at a rate of 54 cubic inches per minute.
(c) Two small airplanes start from a common point A at the same time. One flies east at the rate of 300 kilometers
per hour and the other flies south at the rate of 400 kilometers per hour. After 2 hours, how fast is the distance
between them changing?
Refer to Fig. 20-2. We are given that dx/dt = 300 and dy/dt = 400 and wish to find the value of du/dt at
t = 2 hours. The necessary relation between U,x, and y is furnished by the Pythagorean theorem,
U2 = x2 +y2
Therefore,
d(u2) d(x2 +y2)
-- -
dt dt
du d(x2) d(y2)
2u-=-+-
dt dt dt
du dx
2u -= 2x -+2y -
dy
dt dt dt
du dx dy
dt dt dt
U -= x -+y -= 300x +400y

CHAP. 20) RELATED RATES 149
Fig. 20-2
Now we must find x, y, and U after 2 hours. Since x is increasing at the constant rate of 300 kilometers per
hour and t is measured from the beginning of the flight, x = 300t (distance = speed x time, when speed is
constant). Similarly,y = 400t. Hence, at t = 2,
x = 300(2) = 600 y = 400(2) = 800
and U’ = (600)2+(800)2= 360000 +64OOOO = 1OOOOOO
U = 1OOO
Substituting in (4),
du
dt
loo0- 300(600) +400(800) = 180000 +320000 = 500000
--
-
du
dt
500000
= 500 kilometers per hour
Solved Problems
20.1 Air is leaking out of a sphericalballoon at the rate of 3 cubic inchesper minute. When the radius
is 5 inches,how fast is the radius decreasing?
Since air is leaking out at the rate of 3 cubic inches per minute, the volume V of the balloon is
decreasing at the rate of dV/dt = -3. But the volume of a sphere of radius r is V = 3m3.Hence,
Substituting r = 5,
3
dr
dt l00n 314
- - - % - - % -0.00955
--
Thus, when the radius is 5 inches, the radius is decreasing at about 0.01 inch per minute.
20.2 A 13-foot ladder leans against a vertical wall (see Fig. 20-3). If the bottom of the ladder is
slipping away from the base of the wall at the rate of 2 feet per second, how fast is the top of the
ladder moving down the wall when the bottom of the ladder is 5 feet from the base?

- X
Fig. 20-3
Let x be the distance of the bottom of the ladder from the base of the wall, and let y be the distance of
the top of the ladder from the base of the wall. Since the bottom of the ladder is moving away from the base
of the wall at 2 feet per second, dx/dt = 2. We wish to compute dy/dt when x = 5 feet. Now, by the
Pythagorean theorem,
(1)
(13)2 = x2 +y2
Differentiation of this, as in example (c), give
dx dY
dt
0 = x dt+y $= 2x +y -
But when x = 5, (1)gives
y = J
- = Jlas-25= Jr;r;T= 12
so that (2)becomes
dY
0=2(5)+ 12-
dt
Hence, the top of the ladder is moving down the wall (dy/dt <0) at 3 feet per second when the bottom of
the ladder is 5 feet from the wall.
20.3 A cone-shaped paper cup (see Fig. 20-4) is being filled with water at the rate of 3 cubic centi-
meters per second. The height of the cup is 10 centimeters and the radius of the base is 5
centimeters. How fast is the water level rising when the level is 4 centimeters?
At time t (seconds), when the water depth is h, the volume of water in the cup is given by the cone
formula V = 3nr2hwhere r is the radius of the top surface. But by similar triangles in Fig. 20-4,
r h 5h h
5 10 Or r = l o = 2
- = -
(Only h is of interest, so we are eliminating r.) Thus,
1 1 h 2 n
V = - R (i)h = 5n(7)h = -h3
3 12
and, by the power chain rule,

CHAP. 203 RELATED RATES
SubstitutingdV/dt -3 and h = 4, wc obtain
3 4 3 ;
dlr 3 3
---e- 0.24 antimtccr per aeand
L 4s q3.14)
Hmot.at the momeat when t k water kvel i
s4 centimetax,the level is rismgat about 0.24 alrmccer per
sccand.
A ship B is moving westward lowaid a fixed point A at a .speed 0112 knots (nautical mik, per
hour).At tht moment when ship B is 72 nautical miks from A. ship C pa.. through A. heading
due south at 1
0 knots. How fast is the distance between the ships changing 2 hours &er ship C
has passedthrough A?
Figure XL5 showsthe dtuatioaat time t >6.At t -0. ship C w8s at A. Since
U* = xa + (1)
*re ohrin. .sin exampk(c},
since x is deacuing at 12 knots and y is incnuing at 10 knols. At r=2 we have (since
distanu =s p e d x time)
y = 1 0 x 2 - M
72- x = 12 x 2 = 24
x = 72 -24 -48
Fii265

152 RELATED RATES
0
0
0
/
0
0
0
0
;0‘
0
0
/
/
0
0
0
0
0
0
0
r
,
[CHAP. 20
S
and, by (0
U = JM
= JZGTZZi = ,/Z@= 52
Substitution in (2)yields
du
dt
52 = -12(48)+ lO(20) = -576 +200 = -376
x -7.23
du 376
dt 52
--
- --
which shows that, after 2 hours, the distance between ships B and C is decreasing at the rate of 7.23 knots.
20.5 The top of a 25-foot ladder, leaning against a vertical wall, is slipping down the wall at the rate of 1foot per
minute. How fast is the bottom of the ladder slipping along the ground when the bottom of the ladder is 7
feet away from the base of the wall?
20.6 A cylindrical tank of radius 10feet is being f
i
l
l
e
d with wheat at the rate of 314 cubic feet per minute. How
fast is the depth of the wheat increasing? [Hint: The volume of a cylinder is nr2h,where r is its radius and h
its height.]
20.7 A 5-fOOt girl is walking toward a 20-foot lamppost at the rate of 6 feet per second. How fast is the tip of her
shadow (cast by the lamp) moving?
20.8 A rocket is shot vertically upward with an initial velocity of 400 feet per second. Its height s after t seconds
is s = 400t - 16t2. How fast is the distance between the rocket and an observer on the ground 1800 feet
away from the launching site changing when the rocket is still rising and is 2400 feet above the ground (see
Fig. 20-6)?
Fig. 20-6
20.9 A small funnel in the shape of a cone is being emptied of fluid at the rate of 12cubic millimetersper second.
The height of the funnel is 20 millimeters and the radius of the base is 4 millimeters. How fast is the fluid
level dropping when the level stands 5 millimeters above the vertex of the cone? [Remember that the
volume of a cone is 3nr’h.I

CHAP. 201 RELATED RATES 153
20.10
20.11
20.12
20.13
20.14
20.15
20.16
20.17
20.18
20.19
20.20
A balloon is being inflated by pumping in air at the rate of 2 cubic inches per second. How fast is the
diameter of the balloon increasingwhen the radius is one-half inch?
Oil from an uncapped oil well in the ocean is radiating outward in the form of a circular film on the surface
of the water. If the radius of the circle is increasing at the rate of 2 meters per minute, how fast is the area of
the oil film growingwhen the radius reaches 100meters?
The length of a rectangle having a constant area of 800 square millimeters is increasing at the rate of 4
millimeters per second.(a)What is the width of the rectangle at the moment when the width is decreasing at
the rate of 0.5 millimeter per second? (b) How fast is the diagonal of the rectangle changing when the width
is 20 millimeters?
A particle moves on the hyperbola x2 - 18y2= 9 in such a way that its y-coordinate increases at a constant
rate of nine units per second. How fast is its x-coordinate changing when x = 9?
An object moves along the graph of y =f(x). At a certain point, the slope of the curve (that is, the slope of
the tangent line to the curve) is 3 and the abscissa (x-coordinate) of the object is decreasing at the rate of
three units per second. At that point, how fast is the ordinate (y-coordinate)of the object changing? [Hint:
y =f(x), and x is a function oft. So y is a composite function oft, which may be differentiated by the chain
rule.]
If the radius of a sphere is increasingat the constant rate of 3 millimeters per second, how fast is the volume
changing when the surface area (4ar2)is 10square millimeters?
What is the radius of an expanding circle at a moment when the rate of change of its area is numerically
twice as large as the rate of change of its radius?
A particle moves along the curve y = 2x3 - 3x2 +4. At a certain moment, when x = 2, the particle’s
x-coordinate is increasing at the rate of 0.5 unit per second. How fast is its y-coordinate changing at
that moment?
A plane flying parallel to the ground at a height of 4 kilometers passes over a radar station R (see Fig. 20-7).
A short time later, the radar equipment reveals that the plane is 5 kilometers away and that the distance
between the plane and the station is increasing at a rate of 300 kilometers per hour. At that moment, how
fast is the plane moving horizontally?
A boat passes a fixed buoy at 9 A.M.,heading due west at 3 miles per hour. Another boat passes the same
buoy at 10A.M.,heading due north at 5 miles per hour. How fast is the distance between the boats changing
at 11:30A.M.?
Water is pouring into an inverted cone at the rate of 3.14 cubic meters per minute. The height of the cone
is 10 meters and the radius of its base is 5 meters. How fast is the water level rising when the water stands
7.5 meters in the cone?
R
Fig. 20-7 Fig. 20-8

20.21 A particle moves along the curve y = i x 2+2x. At what point(s) on the curve are the abscissa and the
ordinate of the particle changingat the same rate?
20.22 A boat is being pulled into a dock by a rope that passes through a ring on the bow of the boat (see
Fig. 20-8).The dock is 8 feet higher than the bow ring. How fast is the boat approachingthe dock when the
length of rope between the dock and the boat is 10feet, if the rope is being pulled in at the rate of 3 feet per
second?
20.23 A girl is flying a kite, which is at a height of 120 feet. The wind is carrying the kite horizontally away from
the girl at a speed of 10 feet per second. How fast must the string be let out when the kite is 150 feet away
from the girl?
20.24 The bottom of a 17-footladder is on the ground and the top rests against a vertical wall. If the bottom is
pushed toward the wall at the rate of 3 feet per second, how fast is the top moving up the wall when the
bottom of the ladder is 15 feet from the base of the wall?
20.25 At a given moment, one person is 5 miles north of an intersection and is walking straight toward the
intersection at a constant rate of 3 miles per hour. At the same moment, a second person is 1 mile east of
the intersectionand is walking away from the intersection at the constant rate of 2 miles per hour. How fast
is the distance between the two people changing 1 hour later? Interpret your answer.
20.26 An object is moving along the graph of y = 3x -x3, and its x-coordinate is changing at the rate of two
units per second. How fast is the slope of the tangent line to the graph changingwhen x = -l?

Chapter 21
Approximation by Differentials; Newton's Method
In Chapter 19, we obtained an approximate relation between the change
Ay =f(x +Ax)-f(x)
in a differentiable functionfand the change Ax in the argument off. For convenience,we repeat (19.1)
here and name it the approximation principle,
Ay e f ' ( x ) Ax (21.1)
21.1 ESTIMATING THE VALUE OF A FUNCTION
Many practical problems involve finding a valuef(c) of some functionf: A direct calculation off(c)
may be difficult or, quite often, impossible. However, let us assume that an argument x close to c, the
closer the better, can be found such that f ( x ) andf'(x) can be computed exactly. If we set Ax = c - x,
then c = x +Ax and the approximation principle (21.1)yields
(21.2)
EXAMPLE Let us estimate m.Here f is the square root function and c = 9.2. If we choose x = 9, then
Ax = 9.2 - 9 = 0.2. Bothf(x) andf'(x) can be computedeasily,
f(x) = fi = 3
and (21.2)yields
1
,
/
% =f(9.2) x 3 +6 (0.2)= 3 +0.0333 ...= 3.0333 ...
The actual value of m,correct to four decimal places,is 3.0331.
21.2 THE DIFFERENTIAL
The product on the right side of (21.l)is called the dzflerential off and is denoted by df:
Definition: Let f be a differentiable function. Then, for a given argument x and increment Ax, the
diflerential dfoffis defined by
df=f'(~) AX
Notice that dfdepends on two quantities, x and Ax.Although Ax is usually taken to be small, this
is not explicitly required in the definition. However, if Ax is small, then the content of the approx-
imation principle is that
f(x +Ax)-f(x) x df (21.3)
155

156 APPROXIMATION BY DIFFERENTIALS; NEWTONS METHOD [CHAP. 21
EXAMPLE A graphic picture of this last form of the approximation principle is given in Fig. 21-1. Line 9 is
tangent to the graph offat P;its slope is thereforef‘(x).But then
-
or RT =f’(x) Ax =df
R T R T
f’(x) = == -
PR Ax
- -
Now it is clear that, for Ax very small, RT x RQ; that is,
d f ~ f ( x+Ax) -f(x)
If the value of a function f is given by a formula, say, f ( x )= x2 +2 ~ - ~ ,
let us agree that the
differential dfmay also be written
d(x2 + = df=f’(x) Ax = (2x - 6 ~ - ~ )
Ax
In particular, iff(x) = x, we shall write
dx = df=f’(x) AX = 1 AX = AX
Since dx = Ax,the definition of the differentialdfcan be rewritten as
df=f‘(x) dx
Assuming that dx = Ax # 0, we may divide both sides by dx, obtaining the result
df
f’(x)= -
dx
If we let y = f ( x ) , this may explain the traditional notation dy/dx for the derivative.
A
O(x+ bx,f(x + Ax))
Y
I I m
X x + Ax
Fig. 21-1
2
1
.
3 NEWTON’S METHOD
Let us assume that we are trying to find a solution of the equation
f(x)= 0 (214)

CHAP. 211 APPROXIMATION BY DIFFERENTIALS; NEWTONS METHOD
0
157
and let us also assume that we know that xo is close to a solution. If we draw the tangent line 3 to the
graph off at the point with abscissa xo, then F will usually intersect the x-axis at a point whose
abscissa x1is closer to the solution of (21.4)than xo(see Fig. 21-2).
A point-slope equation of the tangent line Fis
If Fintersects the x-axis at the point (xl, 0),then
0 -f(xo) =f’(xoXxi - xo)
Iff l x o ) z 0,
If we repeat the same procedure, but now starting with x1 instead of xo, we obtain a value x2 that
should be stillcloser to the solution of ( 2 1 4 ,
If we keep applying this procedure, the resulting sequence of numbers xo, xl, x2, ...,x,, ... is deter-
mined by the formula
(21.5)
This process for finding better and better approximations to a solution of the equation f ( x )= 0 is
known as Newton’s method. It is not always guaranteed that the numbers generated by Newton’s
method will approach a solution off(x) = 0. Some difficulties that may arise will be discussed in the
problems below.
EXAMPLE If we wish to approximate ,/?,we should try to find an approximate solution of x2 - 2 = 0.Here
f ( x ) = x2 - 2. Thenf’(x) = 2x, and (22.5)becomes
(22.6)
If we take the first approximation xo to be 1 (since we know that ,/? is between 1 and 2), then we obtain by
successively substituting n = 0, 1, 2, ...in (21.6),’
The computations required by Newton’s method are usually too tedious to be done by hand. A calculator,preferably a pro-
grammablecalculator,shouldbe used.

158 APPROXIMATION BY DIFFERENTIALS; NEWTONS METHOD [CHAP. 21
- 1.5
x1=--
1 + 2
2
(1.5)2+2 2.25 +2 4.25
2(1.5)
= -
= -x 1.416666667
3 3
xq =
(1’416666667)2+ 1.414215686
XJ X
x4 X
xg x
2(1.416666667)
2(1.414215686)
(1.414213562)2+2
2(1.414213562)
(1’414215686)2+ 1.414213562
x 1.414213562
Since our approximationsfor x4 and x5 are equal, all future values will be the same, and we have obtained the best
approximationto nine decimal places within the limits of our calculator. Thus, fi w 1.414213562.
Solved Problems
2
1
.
1 Using the approximation principle, estimate the value of @.
Lettingf(x) = fi and c = 62, choosex = 64(theperfect squareclosest to 62).Then,
A x = c - ~ = 6 2 - 6 4 = -2
f(x) = J64= 8
and (21.2)yields
1 1 7
16 8 8
,/Gw 8 +-(-2) = 8 -- = 7 - = 7.875
Actually,,/G= 7.8740 ....
21.2 Use the approximation principle to estimate the value of @3.
Letf(x) = fi,c = 33, and x = 32. Then Ax = c -x = l,f(x) = @= 2, and
Hence, by (22.2),
1
80
.
@=f(c) x 2 +-(1) = 2.0125
Since the actual value is 2.0123 ...,the approximationis correct to three decimal places.
213 Measurement of the side of a square room yields the result of 18.5 feet. Hence, the area is
A = (18.5)2 = 342.25 square feet. If the measuring device has a possible error of at most 0.05 foot,
estimate the maximum possible error in the area.

CHAP. 211 APPROXIMATION BY DIFFERENTIALS ; NEWTONS METHOD 159
The formula for the area is A = x2, where x is the side of the room. Hence, dA/dx = 2x. Let x = 18.5,
and let 18.5 +Ax be the true length of the side of the room. By assumption, IAxl I 0.05. The approx-
imation principle yields
dA
dx
A(x +Ax) -A(x) x -AX
A(18.5 +Ax) - 342.25 x q18.5) AX
IA(18.5 +Ax) - 342.25 I x I37 Ax I I
37(0.05) = 1.85
Hence, the error in the area should be at most 1.85 square feet, putting the actual area in the range of
(342.25 +_ 1.85) square feet. See Problem 21.13.
21.4 Use Newton’s method to find the positive solutions of
x 4 + x - 3 = 0
Let f(x) = x4 +x - 3. Then f’(x) = 4x3 +1. Since f(1) = -1 and f(2) = 15, the intermediate-value
theorem tells us that there is a solution between 1and 2. [The interval (1, 2) is suggested by drawing the
graph off with a graphing calculator.] Sincef’(x) > 0 for x 2 0,f is increasing for x >0, and, therefore,
there is exactly one positive real solution. Start with xo = 1.Equation (22.5) becomes
x
: +x
, - 3 4x: +x, - (x: +x
, - 3) 3x: +3
4 4 + 1 4x,3 + 1 4x: + 1
--
-
-
-
= X n -
Successive calculations yield x1 = 1.2, x2 = 1.165419616, x3 = 1.164037269, xq = 1.164035 141, and
x5 = 1.164035 141.Thus, the approximate solution is x = 1.164035 141.
21.5 Show that if Newton’s method is applied to the equation x113= 0 with xo = 1, the result is a
divergentsequence of values (which certainlydoes not converge to the root x = 0).
1
3X2l3
Letf(x) = x’/~.
So,f’(x)= -and (22.5) becomes
x,+1 = x
, - -
= x, - 3x, = -2x,
1/(3x,213)
Hence, x1 = -2, x2 = 4, x3 = -8, and, in general, x, = (-2)”.
Note: If we are seeking a solution r of an equation f(x) = 0, then it can be shown that a sufficient
condition that Newton’s method yields a sequence of values tht converges to r is that
for all x in an interval around r that includesxo.However, this is not a necessary condition.
21.6 Use the approximation principle to estimate the followingquantities:
(a) f i (b) ,/% (c) .
@ (d) (8.35)2/3 (e) (33)-’/’
(f) & (g) diE5 (h) Jsos (i) f l

160
21.7
21.8
21.9
21.10
21.11
21.12
21.13
21.14
21.15
21.16
21.17
21.18
21.19
21.20
APPROXIMATION BY DIFFERENTIALS; NEWTONS METHOD [CHAP. 21
The measurement of the side of a cubical container yields 8.14 centimeters, with a possible error of at most
0.005 centimeter. Give an estimate of the maximum possible error in the value of V = (8.14)3= 539.353 14
cubic centimeters for the volume of the container.
It is desired to give a spherical tank 20 feet (240 inches) in diameter a coat of paint 0.1 inch thick. Use the
approximation principle to estimate how many gallons of paint will be required. (V = 41cr3, and 1 gallon is
about 231 cubic inches.)
A solid steel cylinder has a radius of 2.5 centimeters and a height of 10 centimeters. A tight-fitting sleeve is
to be made that will extend the radius to 2.6 centimeters. Find the amount of steel needed for the
sleeve: (a)by the approximation principle; (b) by an exact calculation.
If the side of a cube is measured with a percentage error of at most 3%, estimate the maximum percentage
error in the volume of the cube. (If AQ is the error in measurement of a quantity Q, then IAQ/Q I x 100% is
the percentage error.)
Assume, contrary to fact, that the Earth is a perfect sphere, with a radius of 4000 miles. The volume of ice at
the North and South Poles is estimated to be about 8000000 cubic miles. If this ice were melted and if the
resulting water were distributed uniformly over the globe, approximately what would be the depth of the
added water at any point of the Earth?
(a) Let y = x3/2.When x = 4 and dx = 2, find the value of dy.
(b) Let y = 2 x , / m ’ . When x = 0 and dx = 3, find the value of dy.
For Problem 21.3, calculate exactly the largest possible error in the area.
Establish the very useful approximation formula (1 +U
)
’ x 1 +ru, where r is any rational exponent and Iu I
is small compared to 1. [Hint: Apply the approximation principle tof(x) = x
‘
, letting x = 1and Ax = u.]
Use Newton’s method to approximate the followingquantities:
Show that Newton’s method for finding 4yields the equation
for the sequence of approximating values.
Use part (a)to approximate fi by Newton’s method.
Use Newton’s method to approximate solutions of the following equations:
x3 - x - 1 = 0 (b) x3 +x - 1 = 0 (c) x4 - 2 ~ ’
+0.5 = 0 (d) x3 +2~ - 4 = 0
x 3 - 3 x 2 + 3 = o (f) x 3 - x + 3 = 0 (9) x 3 - 2 x - 1 = 0
Show that x3 +x2 - 10 = 0 has a unique root in (1, 2) and approximate this root by Newton’s method,
with xo = 2.
Show that xs +5x - 7 = 0 has a unique solution in (1,2) and approximate this root by Newton’s method.
Explain why Newton’s method does not work in the following cases:
(a) Solvex3 - 6x2+ 12x - 7 with xo = 2.
(b) Solve x3 - 3x2 +x - 1 = 0 with xo = 1.
Jx-1 f o r x 2 1
(c) Solvef(x) = 0, wheref(x) = and xo > 1 (say, x = 1 +b, b > 0
)
.

Chapter 22
Higher-Order Derivatives
The derivativef’ of a functionfis itself a function,which may be differentiable.Iff’ is differentiable,
its derivative will be denoted byf”. The derivativeoff”, if it exists,is denoted byf”’, and so on.
Definition: f”(x)= D,(f’(x))
f”’W
= DX(f”(4)
f‘*’(x)= D,(f”’(x))
We callf‘ thefirst derivative off;f”the second derivative off; andf ’” the third derioatioe of
f:If the order n exceeds3, we writef‘”)for the nth derivative of$
EXAMPLES
(a) Iff(x) = 3x4- 7x3+5x2+2x - 1
, then,
fyx) = 12x3 - 21x2 + iox +2
f”(x) = 36x2-4 2 ~
+ 10
f’”(x) = 7 2 ~
- 42
f‘4‘(~)= 72
f(”)
= O for all n 2 5
(b) Iff(x) = 3x3- 5x2+x +4
, then,
3
f’(x) = - x2- 1ox + 1
2
f”(x) = 3x - 10
f”’(x) = 3
f(”)(x)= O for all n 2 4
It is clear that iffis a polynomial of degree k,then the nth derivativef‘”)will be 0 for all n > k.
(c) Iff(x) = - = x-l, then,
1
X
In this case, the nth derivative will never be the constant function 0.
Alternative Notation
First derivative: D X Y = Y’
f’(x)= D x f ( x )= -= -=
df dY
dx dx
- p y = y”
d2f d2y
dx2 -dx2
Second derivative: f”(x)= 0
: f ( x )= ----
161

162 HIGHER-ORDER DERIVATIVES [CHAP. 22
d3f d3y
f’”(X) = 0,”
f(x) = -- -= D,”y y”t
f (“)(x)= 0
; f(x) = -
d”f -
--
‘“Y -
- D;y = y(n)
dx3 -dx3
dx” dx”
Third derivative:
nth derivative:
Higher-Order ImplicitDifferentiation
EXAMPLE Let y = f ( x ) be a differentiablefunctionsatisfyingthe equation
x2 +yz = 9
(we know that y = ,
/
=
or y = -
,
/
-
; their graphs are shown in Fig. 22-1.)
Let us find a formula for the second derivativey”, where y stands for either of the two functions.
DJx2 +y2)= Dx(9)
x +yy’ = 0
2x +2yy’ = 0 [D,y2 = 2yy’ by the power chain rule]
Next, differentiateboth sidesof (I)with respect to x,
Dx@+YY’) = DX(0)
1 +yD#) +y’Dxy = 0
1 +yy” +y‘ y’ = 0
1 +yy” +(y’y = 0
Solve(I)for y’ in terms of x and y,
Substitute -(x/y) for y’ in (2),
X*
Y
1 +yy” +2= 0
y2 +y3y” +x2 = 0 [multiplying by y2]
Solvefor y”,
t’
(a) y = d9- x! (b) y = -v9- x!
Fig. 22-1

CHAP. 221 HIGHER-ORDER DERIVATIVES
From (0), we may substitute9 for x2 +y2,
9
y” - -
Y3
163
Acceleration
Let an object move along a coordinate axis according to the equation s =f(t), where s is the
coordinate of the object and t is the time. From Chapter 18,the object’s velocity is given by
ds
dt
U = - - - -f ‘@)
The rate at which the velocity changesis called the acceleration a.
dv d2s
Definition: a = -
dt = -
dt2 -f’’(t)
-
EXAMPLES
(a) For an object in free fall, s = so +oo t - 16t2,where s,measured in feet, is positive in the upward direction and
t is measured in seconds. Recall that so and oo denote the initial position and velocity; that is, the values of s
and o when t = 0. Hence,
ds
dt
U = - - - UO - 32t
do
dt
a = - = -32
Thus, the velocity decreases by 32 feet per second every second. This is sometimesexpressed by saying that the
(downward)accelerationdue to gravity is 32 feet per second per second,which is abbreviated as 32 ft/sec2.
(b) An object moves along a straight line accordingto the equation s = 2t3 - 3t2 +t - 1.Then,
do
dt
a = - = l 2 t - 6
In this case, the acceleration is not constant. Notice that a > 0 when 12t - 6 >0, or t > 4. This implies (by
Theorem 17.3)that the velocity is increasingfor t > 4.
Solved Problems
22.1 Describe all the derivatives(first,second, etc.)of the followingfunctions:
1 Y
A
(a) f ( x ) = ;
x4 - 5x - a (b) f ( x )= x+l

( x + l)D,X - xD,(x + 1)
( x + 1)2
(4 f”x) = [by the quotient rule]
1
-
--
= ( x + 1)-2
( X + 1x1)- ~ ( 1 )
- x + 1 - x
-
-
-
( x + 1)2 (x + 1)2 ( x +
f y x ) = -2(x + 1 ) - 3 ~ , ( ~
+ 1) [by the power chain rule]
~~
ALGEBRA (-1y-l will be 1 when n is odd and -1 when n is even. n! stands for the product
1 x 2 x 3 x - - - x n of the first n positive integers.
22.2 Find y” if
y3 - xy = 1 (0)
Differentiation of(0), using the power chain rule for D,y3 and the product rule for D,(xy), gives
3y2y’ - (xy’ +y) = 0
3y2y’ - xy‘ - y = 0
(3y2 - x)y’ - y = 0
Next, differentiate(Z),
(3y2 - x)D,y’ +y’D,(3y2 - x) - y’ = 0
(3y2 - x)y” +y’(6yy’ - 1)- y’ = 0
(3y2- x)y” +y’((6yy’ - 1)- 1) = 0
[factor y’]
(3y2- X)Y” +y’(6yy’ - 2) = 0
Now solve(1)for y‘,
Y
y’ = -
3y2 - x
Finally, substitute into (2)and solve for y”,
(3y2 - x)y” + +
L (K
- 2) = 0
3y - x 3y2 - x
(3y2 - X)~Y’’ +(3y2 - x ) 7
Y (3y2 - x)(- 6Y - 2 ) = 0 [multiply by (3y2 - x ) ~ ]
3y - x 3y2 - x
(3y2- X)~Y’’
+y (6y2 - 2(3y2 - x)) = 0 [U(: - c ) = b - cu]
(3y2 - x ) ~ Y ”
+v(6v2- 6y2 +2 ~ )0
(3y2 - x)3y” +2xy = 0
-2xy
(3y2 - x)3
y” =

CHAP. 221 HIGHER-ORDER DERIVATIVES 165
2
2
.
3 If y is a function of x such that
x3 - 2xy +y3 = 8 (0)
and such that y = 2 when x = 2 [note that these values satisfy (O)], find the values of y’ and y”
when x = 2.
Proceed as in Problem 22.2.
x3 - 2xy +y3 = 8
D,(x3 - 2xy +y3)= D,(8)
3x2 - 2(xy’ +y) +3y2y’ = 0
3x2 -2xy’ - 2y +3y2y’ = 0
DX(3X2- 2xy’ - 2y +3y2y’) = D,(O)
6~ - ~(XY’’
+y’) - 2y’ +3dy’y’‘ +y’(2yy’)) = 0
6~ - 2xy” - 2y’ - 2y’ +3y2y” +6 ~ 4 ~ ’ ) ~
= 0
Substitute 2 for x and 2 for y in (I),
Substitute 2 for x, 2 for y, and -1for y’ in (2),
28 7
y”= -T= - -
2
12 - 4y” +2 +2 + 12y” + 12 = 0 or 8y” +28 = 0 or
22.4 Let s = t3 - 9t2 +24t describe the position s at time t of an object moving on a straight line.
(a)Find the velocity and acceleration. (b) Determine when the velocity is positive and when it is
negative. (c) Determine when the acceleration is positive and when it is negative. (6)Describe the
motion of the object.
ds
dt
U = - = 3t2 - 18t +24 = 3(t2- 6t +8) = 3(t -2Xt - 4)
dv
dt
a = - = 6t - 18 = 6(t - 3)
U is positive when t - 2 > 0 and t - 4 > 0 or when t - 2 < 0 and t - 4 < 0; that is, when
t > 2 and t > 4 or t < 2 and t <4
which is equivalent to t > 4 or t < 2. v = 0 if and only if t = 2 or t = 4. Hence, U <0 when 2 < t < 4.
a > Owhen t > 3, and a <Owhen t < 3.
Assuming that s increases to the right, positive velocity indicates movement to the right, and negative
velocity movement to the left. The object moves right until, at t = 2, it is at s = 20, where it reverses
direction. It then moves left until, at t = 4, it is at s = 16, where it reverses direction again. Thereafter,
it continues to move right (see Fig. 22-2).
Fig. 22-2

225 Find the second derivativeOf y of the followingfunctionsy:
(a) y = x --
1 (b) y = & -7x (c) y = J x 7 3 (d) y = & i
X
22.6 Use implicit differentiationto find the second derivativey” in the followingcases:
(U) x2 + y 2 = 1 (b) X’ - y2 = 1 (c) x3 -y3 = 1 (d) X Y + y2 = 1
22.7 If &+&= 1, calculate y”: (a)by explicitly solving for y and then differentiatingtwice; (b) by implicit
differentiation.(c) Which of methods (a) and (b) is the simpler one?
22.8 Find all derivatives(first,second,etc.)of y:
1
X
(a) y=4x4 -2x2 + 1 (b) y = 2x2+ x - 1 +- (c) y = &
x + l
(d) y = X _ l
1
(4 Y = 3+x
1
(f)Y = 2
229 Find the velocity the first time that the accelerationis 0 if the equation of motion is:
(a) s = tZ - 5t +7 (b) s = t3 - 3t +2 (c) s = t4 - 4t3 +6t2 - 4t +3
22.10 At the point (1,2) of the curve x2 - xy +y2 = 3, find the rate of changewith respect to x of the slope of the
tangent line to the curve.
22.11 If xz +2xy +3y2 = 2, find the values of dy/dx and d2y/dx2when y = 1.
1 +3K(x - 2) +(x - 2)2 if x 5 2
Lx +K i f x > 2
where L and K are constants.
22.12 Letf ( x ) =
(a) Iff(x) is differentiable at x = 2, find L and K.
continuousfor all x?
(b) With L and K as found in part (a), isf”(x)
22.13 Let h(x) =f(x)g(x) and assume that f and g have derivativesof all orders. (a) Find formulas for h”(x),h‘“(x),
and h(4)(x).
(b) Find a generalformula for h(”)(x).
22.14 Let H(x) =f(x)/g(x),wherefand g have first and second derivatives. Find a formulafor H”(x).
22.15 The height s of an object in free fall on the moon is approximately given by s = so +uo t - fit2, where s is
measured in feet and t in seconds. (a) What is the acceleration due to “gravity” on the moon? (b) If an
object is thrown upward from the surface of the moon with an initial velocity of 54 feet per second,what is
the maximumheight it will reach,and when will it reach that height?

Chapter 23
Applications of the Second Derivative and
Graph Sketching
2
3
.
1 CONCAVITY
If a curve has the shape of a cup or part of a cup (like the curves in Fig. 23-l), then we say that it is
concuue upward. (To remember the sense of “concave upward,” notice that the letters c and up form the
word cup.) A mathematical description of this notion can be given. A curve is concave upward if the
curve lies above the tangent line at any given point of the curve. Thus, in Fig. 23-l(a) the curve lies
above all three tangent lines.
/
/
/
/
Fig. 23-1 Concavity *ward
A curve is said to be concavedownwardif it has the shape of a cap or part of a cap (see Fig. 23-2).In
mathematical terms, a curve is concave downward if it lies below the tangent line at an arbitrary point
of the curve [see Fig. 23-2(a)].

(4 (b)
Fig. 23-2 Concavity downward
A curve may, of course, be composed of parts of different concavity. The curve in Fig. 23-3 is
concave downward from A to B, concave upward from B to C, concave downward from C to D, and
concave upward from D to E. A point on the curve at which the concavitychanges is called an inflection
point. B, C, and D are inflectionpoints in Fig. 23-3.
167

168 THE SECOND DERIVATIVE AND GRAPH SKETCHING [CHAP. 23
Fig. 23-3
From Fig. 23-1 we see that if we move from left to right along a curve that is concave upward, the
slope of the tangent line increases. The slope either becomes less negative or more positive. Conversely,
if the tangent line has this property, the curve must be concave upward. Now for a curve y =f(x), the
tangent line will certainly have this property iff”(x) > 0 since, in that case, Theorem 17.3 implies that
the slopef’(x) of the tangent line will be an increasing function. By a similar argument, we see that if
f”(x)< 0, the slope of the tangent line is decreasing, and from Fig. 23-2 we see that the curve y = f ( x ) is
concavedownward. This yields:
Theorem 2
3
.
l
: Iff”(x) > 0 for all x in (a, b), then the graph off is concave upward between x = a and
x = b. Iff”(x) <0 for all x in (a, b), then the graph offis concave downward between
x = a and x = b.
For a rigorous proof of Theorem 23.1, see Problem 23.17.
Corollary2
3
.
2
: If the graph off has an inflection point at x = c, and f”exists and is continuous at
x = c, then$”@)= 0.
In fact, if f”(c) # 0, the f”(c) > 0 or f”(c) < 0. If f”(c) > 0, then f”(x)> 0 for all x in some open
interval containing c, and the graph would be concave upward in that interval, contradicting the
assumption that there is an inflection point at x = c. We get a similar contradiction iff”(c) <0, for in
that case, the graph would be concavedownward in an open interval containing c.
EXAMPLES
(a) Consider the graph of y = x3 [see Fig. 23-4(a)]. Here y’ = 3x2 and y” = 6x. Since y” > 0 when x > 0, and
y” < 0 when x < 0, the curve is concave upward when x > 0, and concave downwardwhen x <0. There is an
inflection point at the origin, where the concavity changes. This is the only possible inflection point, for if
y” = 6x = 0, then x must be 0.
(b) Iff”(c) = 0, the graph offneed not have an inflection point at x = c. For instance, the graph off($ = x4 [see
Fig. 23-4(b)] has a relative minimum, not an inflection point, at x = 0, wheref”(x) = 12x2= 0.
Fig. 23-4

CHAP. 231 THE SECOND DERIVATIVE AND GRAPH SKETCHING 169
23.2 TEST FOR RELATIVE EXTREMA
We already know, from Chapter 14, that the conditionf'(c) = 0 is necessary, but not sufficient, for a
differentiable function f to have a relative maximum or minimum at x = c. We need some additional
information that will tell us whether a function actually has a relative extremum at a point where its
derivative is zero.
Theorem 2
3
.
3 (Second-Derivative Test for Relative Extrema): If f'(c) = 0 and f"(c) < 0, then f has a
relative maximum at c. Iff'(c) = 0 andf"(c) > 0, thenfhas a relative minimum at c.
Proof: Iff'(c) = 0, the tangent line to the graph off is horizontal at x = c. If, in addition,f"(c) -c0,
then, by Theorem 23.1,' the graph off is concave downward near x = c. Hence, near x = c, the graph
off must lie below the horizontal line through (c,f(c));f thus has a relative maximum at x = c [see
Fig. 23-5(a)].A similar argument leads to a relative minimum whenf"(c) > 0 [see Fig. 23-5(b)].
I
I I
I I
C C
I 1
EXAMPLE Consider the functionf(x) = 2x3 +x2 - 4x +2. Then,
f'(x)=
r 6x2 +2x - 4 = 2(3x2 +x -2) = 2(3x -2Xx + 1)
Hence, iff'(x) = 0, then 3x - 2 = 0 or x + 1 = 0;that is, x = 3 or x = -1. Nowf"(x) = 12x +2. Hence,
f"(-1) = 12(-1) +2 = -12 +2 = -10 < 0
f''(4) = 12(3) +2 = 8 +2 = 10 > 0
Fig. 23-6
In order to use Theorem 23.1, we must assume thatf" i
s continuous at c and exists in an open interval around c. However,a
more complicatedargumentcan avoid that assumption.

Sincef"(- 1) <0,fhas a relative maximum at x = -1, with
f ( 4= 2(-u3 +(-q2- q-1)+2 = -2 + 1 +4 + 2 = 5
Sincef"(3) > 0,fhas a relative minimum at x = 3,with
16 4 8 16 12 72 54 1
0
The graph offis shown in Fig. 23-6. Now because
f"(x) = 12x +2 = 12 x +- = 12 x -
( 3 [ (-81
f"(x) > 0 when x > -4, andf"(x) <0 when x < -4. Hence, the curve is concave upward for x > -4 and concave
downward for x < -&.So there must be an inflection point I , where x = -&.
From Problem 9.1we know that
lim f ( x )= lirn 2x3 = +a and lirn f ( x )= lirn 2x3= -a
x - r + m x + + w x - - w x * - w
Thus, the curve moves upward without bound toward the right, and downward without bound toward the left.
The second-derivativetest tells us nothing whenf'(c) = 0 andf"(c) = 0. This is shown by the exam-
ples in Fig. 23-7, where, in each case,f'(O) =f"(O) = 0.
To distinguish among the four cases shown in Fig. 23-7, consider the sign of the derivativef' just to
the left and just to the right of the critical point. Recalling that the sign of the derivative is the sign of
the slope of the tangent line, we have the four combinations shown in Fig. 23-8. These lead directly to
Theorem 23-4.
(a) Inflection point at 0 (6) Relative minimum at 0
(c) Inflection point at 0 (d) Relative maximum at 0
Fig. 23-7

CHAP. 231 THE SECOND DERIVATIVE AND GRAPH SKETCHING

171
Fig. 23-8
Theorem 23.4 (First-Derivative Testfor Relative Extrema): Assumef’(c) = 0.
{ -, +} If f‘is negative to the left of c and positive to the right of c, then f has a
relative minimum at c.
{ +, -} If f’is positive to the left of c and negative to the right of c, then f has a
relative maximum at c.
{ +, +1 If f’ has the same signs to the left and to the right of c, then f has an
{ -, -} inJlectionpoint at c.
2
3
.
3 GRAPH SKETCHING
We are now equipped to sketch the graphs of a great variety of functions. The most important
features of such graphs are:
(i) Relative extrema (if any)
(ii) Inflection points (if any)
(iii) Concavity
(iv) Vertical and horizontal asymptotes (if any)
(v) Behavior as x approaches +00 and -00
tional example follows.
The procedure was illustrated for the functionf(x) = 2x3 +x2 - 4x +2 in Section 23.2. An addi-

EXAMPLE Sketch the graph of the rational function
X
f(x) = -
x2 + 1
First of all, the function is odd [that is,f(-x) = -f(x)], so that it need be graphed only for positive x. The
Compute the first two derivativesoff,
graph is then completed by reflection in the origin (seeSection 7.3).
(x’ + ~)D,(x)- xD,(x2 + 1) x2 + 1 - x(~x) 1 - X’
=-
-
-
f‘(4= (x2+ 1)2 (x2+ 1)2 (x2+ 1)2
(x2+ 1)2Dx(1
- x2)- (1 - x2)D,((x2 + 1)2)
(x2 + 1)4
f”(x) = Dxf’(x) =
- (x’ + 1)2(-2~)
- (1 - x2)[2(x2+ 1)(2~)]
-
(x2 + 114
-2x(x2 + 1)[x2 + 1 +2(1 - x2)] -2x(3 - x2)- 2x(x - &)(x +&)
-
-
-
-
-
(x2 + 114 (x2 + 1)3 (xz + 113
Since 1 -x2 = (1 - xX1 +x), f’(x) has a single positive root, x = 1, at which f”(1) = [-2(2)]/(2)3 = -4.
If we examine the formula forf”(x),
Hence, by the second-derivativetest,fhas a relative maximum at x = 1.The maximum value isf(1) = #.
~ X ( X
- J3Xx +J3)
f“(x)=
(x2 + 113
we see that f”(x) > 0 when x > fi and that f”(x) -c0 when 0 < x < fi.By Theorem 23.1, the graph off
is concave upward for x > & and concave downward for 0 < x < fi.Thus, there is an inflection point I at
x = &,where the concavity changes.
Now calculate
-0
0
llX - -
X
lim f(x) = lim --
- lim
x + + w X’+a) x2 + 1 X‘+W 1 +(1/x2) 1 +0
which shows that the positive x-axis is a horizontal asymptote to the right.
The graph, with its extension to negative x (dashed),is sketched in Fig. 23-9. Note how concavity of one kind
reflects into concavity of the other kind. Thus, there is an inflectionpoint at x = -&and another inflection point
at x = 0. The valuef(1) = # is the absolute maximum off, andf( -1)= -4 is the absolute minimum.
Fig. 23-9

CHAP. 231 THE SECOND DERIVATIVE AND GRAPH SKETCHING 173
Solved Problems
2
3
.
1 Sketch the graph off@) = x - l/x.
The function is odd. Hence, we can first sketch the graph for x > 0 and, later, reflect in the origin to
The first and second derivativesare
obtain the graph for x < 0.
I
1
X2
f ' ( x ) = D , ( x - X - ' ) = l - ( - l ) x - 2 = 1 + -
2
fryx) = ~ ~ ( 1
+ x-2) = -2x-3 = - -
x3
Sincef'(x) = 1+(l/xz) > 0,fis an increasing function. Moreover, for x > 0, the graph off is concave
downward, sincef"(x) = -(2/x3) < 0 when x > 0. The line y = x turns out to be an asymptote, because
1
X - + m x + + m x
lim [x -f(x)] = lim - = 0
Since
the graph offhas the negative y-axis as a vertical asymptote.
Notice that x = 0, at whichfis undefined,is the only critical number. The graph is sketched, for all x,
in Fig. 23-10. Although the concavity changes at x = 0, there is no inflection point there becausef(0) is not
defined.
6 C
I L X
(-1. -i)
Fig. 23-10 Fig. 23-11

2
3
.
2 Sketchthe graph off(x) = 4x4 -x3 +4x +2.
The first derivative isf’(x) = x3 - 3x2 +4. We can determine that -1is a root of x3 - 3x2 +4.
ALGEBRA When looking for roots of a polynomial, first test the integral factors of the constant. In this
case, the factors of 4 are f1, f2, +4.
So (Theorem 7.2),f’(x) is divisibleby x + 1.The division yields
X’ - 3x2+4 = (X + l)(x2 - 4~ +4) = (X+ 1Xx - 2)2
Hence, the critical numbers are x = -1and x = 2. Now
f”(x) = 3x2 - 6~ = ~ X ( X
- 2)
Thus, f”(-1)= 3(-1X- 1 -2) = 9. Hence, by the second-derivative test, f has a relative minimum at
x = -1.
Sincef”(2) = 3(2)(2 - 2) = 0, we do the first-derivative test at x = 2.
f‘(x) = (X+ 1Xx -2)2
On both sides of x = 2, f’(x) > 0, since x + 1 > 0 and (x - 2)’ > 0. This is the case { +, +}. There is an
inflection point at (2,6). Furthermore,f”(x) changes sign at x = 0, so that there is also an inflection point at
(0,2). Because
1
lim f(x)= lim - x4 = +CO
X+fCXJ x + * w 4
the graph moves upward without bound on the left and the right. The graph is shown in Fig. 23-11.
2
3
.
3 Sketch the graph off(x) = x4 - 8x2.
As the function is even, we restrict attention to x 2 0.
f’(x) = 4 ~ ’
- 1 6 ~
= 4x(x2- 4) = ~ X ( X
- 2Xx +2)
f”(x) = 12x2- 16 = 4(3x2 - 4) = 12(X’ - -
:>= 12(x +-
$
(
x - 5)
The nonnegative critical numbers are x = 0 and x = 2. Testing,
f“(x)
-16 rel. min.

Checking the sign of ”(x),we see that the graph will be concave downward for 0 < x < 2/* and concave
upward for x >2/ 3. Because lim f(x) = +00, the graph moves upward without bound on the right.
The graph is sketched in Fig. 23-12. Observe that, on the set of all real numbers, f has an absolute
minimum value of -16,assumed at x = _+2,but no absolute maximum value.
J X‘+W

CHAP. 231
23.4
23.5
23.6
THE SECOND DERIVATIVE AND GRAPH SKETCHING
t’
I
(-2, -16)
Fig. 2312
175
Determine the intervals where the graphs of the following functions are concave upward and the intervals
where they are concave downward. Find all inflection points. aCheck your solutions with a graphing
calculator.
(a) f(x) = X2 -x + 12 (b) f(x) = x4 + 18x3+ 1 2 0 ~ ~
+x + 1
X
(C) f(x) = x3 + 15x2+6~ + 1 (a) f(x) = (e) f ( x )= 5x4-x5
Find the critical numbers of the following functions and determine whether they yield relative maxima,
relative minima, inflection points, or none of these. Check your solutions with a graphing calculator.
(U) f(x) = 8 - 3~ +X’ (b) f(x) = x4 - 18x2+9
X2 X2
(c) f ( x ) = x3 - 5x2- 8~ + 3 (a) f(x) = x-l (4 f ( x ) =
Sketch the graphs of the following functions, showing extrema (relative or absolute), inflection points,
asymptotes, and behavior at infinity. Check your solutions with a graphing calculator.
(a) f(x) = (xz - 1
1
3 (b) f(x) = x3 - 2x2 - 4x +3 iC) f(x) = x(x - 212
(d) f(x) = x4 +4x3 (e) f(x) = 3x5- 2oX3 (f)f(x) = d=

23.7 If,for all x,f’(x) > 0 andf”(x) <0, which of the curves in Fig. 23-13 could be part of the graph off?
(4
Fig. 23-13
23.8 At which of the five indicated points on the graph in Fig. 23-14 do y’ and y” have the same sign?
t Y
I I I I
1 1 1 1 1 +
X
Fig. 2S14
23.9 Letf ( x ) = ux2 +bx +c, with U # 0. (a) How many relative extrema doesf have? (6) How many points of
inflectiondoes the graph offhave? (c) What kind of curve is the graph off?
23.10 Letfbe continuousfor all x, with a relative maximum at (- 1,4) and a relative minimum at (3, -2). Which
of the following must be true? (a)The graph off has a point of inflection for some x in (- 1, 3). (b) The
graph off has a vertical asymptote. (c) The graph off has a horizontal asymptote. (d)f’(3) = 0. (e) The
graph off has a horizontal tangent line at x = -1. (f) The graph offintersects both the x-axis and the
y-axis. (g)fhas an absolute maximum on the set of all real numbers.

CHAP. 23) THE SECOND DERIVATIVE AND GRAPH SKETCHING 1I7
23.11
23.12
23.13
23.14
23.15
23.16
23.17
23.18
23.19
Iff(x) = x3 +3x2 +k has three distinct real roots, what are the bounds on k? [Hint: Sketch the graph off;
usingf’ andf”. At how many points does the graph cross the x-axis?]
Sketch the graph of a continuous functionfsuch that:
(a) f(1) = -2,f‘( 1)= O,f”(x) > 0 for all x
(6) f(2) = 3,f’(2) = O,f”(x) <0 for all x
(c) f(1) = l,f”(x) < 0 for x > l,f”(x) >0 for x < 1, lim f(x) = +00, lim f(x) = -00
(d) f(0) = O,f”(x) < 0 for x > O,f”(x)> 0 for x < 0, lim f(x) = 1, lim f ( x )= -1
(e) f(0) = l,f”(x) < 0for x # 0, lim f’(x) = +00, lirn f‘(x) = -CO
(f) f(0) = O,f”(x) > 0 for x < O,f”(x)< 0 for x > 0, lim f‘(x)= +00, lim f’(x) = +CO
(9) f(0) = l,f”(x) < 0 if x # 0, litp f‘(x) = 0, lirn f’(x) = -CO
X’+aO x + - w
X‘+a, x--CD
x+o+ x-ro-
x+o- x-o+
x+o+ x-ro-
Let f(x) = x I x - 1I for x in [-1, 21. (a) At what values of x isf continuous? (6) At what values of x isf
differentiable? Calculate f’(x). [Hint: Distinguish the cases x > 1 and x < 1.3 (c) Where is f a n increasing
function? (d) Calculate f”(x). (e) Where is the graph off concave upward, and where concave downward?
(f) Sketch the graph of5
Given functionsf and g such that, for all x, (i) (g(x))2-( f ( ~ ) ) ~
= 1; (ii)f’(x) = (g(x))2;(iii)f”(x) and g”(x)
exist; (iv)g(x) < 0; (v)f(O)= 0. Show that: (a)g‘(x) =f(x)g(x); (b) g has a relative maximum at x = 0; (c)f
has a point of inflection at x = 0.
For what value of k will x - kx- have a relative maximum at x = -2?
Let f(x) = x4 +Ax3 +Bx2 +Cx +D. Assume that the graph of y =f(x) is symmetric with respect to the
y-axis, has a relative maximum at (0, l), and has an absolute minimum at (k,-3). Find A, B, C, and D, as
well as the possible value@)of k.
Prove Theorem 23.1. [Hint: Assume that f”(x) > 0 on (a, b), and let c be in (a, 6). The equation of the
tangent line at x = c is y =f’(c)(x - c) +f(c). It must be shown that f(x) >f’(c)(x - c) +f(c). But the
mean-value theorem gives
f(x) =f’(x*Xx - 4+f(c)
where x* is between x and c, and sincef”(x) > 0 on (a, b),f’ is increasing.]
Give a rigorous proof of the second-derivative test (Theorem 23.3). [Hint: Assumef’(c) = 0 and f”(c) c 0.
f’(c + h) -f’(c) < o,
h
Sincef”(c) <0, lirn ”(‘ + h, -”(‘) <0. So, there exists 6 >0 such that, for Ih I < 6,
h - 0 h
and since f’(c) = 0, f‘(c +h) < 0 for h > 0 and f‘(c +h) > 0 for h < 0. By the mean-value theorem, if
f ( c + h, -
’
(
‘
) =f’(c +h,) for some c +h, between c and c +h. So,I h, I < IhI, and whether h > 0
lhl< 6,
or h < 0, we can deduce thatf(c +h) -f(c) < 0; that is,f(c +h) <f(c). Thus,fhas a relative maximum at
c. The case whenf“(c) > 0 is reduced to the first case by considering -5.3
h
3(x2 - 1)
Considerf(x) = -
x 2 + 3 *
(a) Find all open intervals wherefis increasing. (b) Find all critical points and determine whether they
correspond to relative maxima, relative minima, or neither. (c) Describe the concavity of the graph offand
find all inflectionpoints (if any). (d)Sketch the graph off. Show any horizontal or vertical asymptotes.

23.20 In the graph of y = f ( x ) in Fig. 23-15: (a)find all x such thatf’(x) > 0;(b)find all x such thatf”(x) > 0.
2
f
-I 0
Y
I
I
(3, 1.5)
I
1 I 1 I 1 I 1 L
I
I
I
I
I
I
I
I
I
I
I
I
I
I
1 3 4
-
X
Fig. 23-15

Chapter 24
More Maximum and Minimum Problems
Until now we have been able to find the absolute maxima and minima of differentiable functions
only on closed intervals (see Section 14.2).The following result often enables us to handle cases where
the function is defined on a half-open interval, open interval, infinite interval, or the set of all real
numbers. Remember that, in general, there is no guarantee that a function has an absolute maximum or
an absolute minimum on such domains.
Theorem 24.1: Letfbe a continuous function on an interval 9,
with a single relative extremum within
9.
Then this relative extremum is also an absolute extremum on 9.
Intuitive Argument: Refer to Fig. 24-1. Suppose that f has a relative maximum at c and no other
relative extremum inside 3.Take any other number d in 9.
The curve moves down-
ward on both sides of c. Hence, if the valuef(d) were greater thanf(c), then, at some
point U between c and d, the curve would have to change direction and start moving
upward. But thenfwould have a relative minimum at U,contradicting our assumption.
The result for a relative minimum follows by applying to -f the result just obtained
for a relative maximum.
For a rigorous proof, see Problem 24.20.
1 1 1
*
C U d
Fig. 24-1
EXAMPLES
(a) Find the salortest4 istancefrom the point P(1,O) to the parabola x = y2 [see Fig. 24-2(a)].
The distance from an arbitrary point Q(x, y) on the parabola to the point P(1,O) is, by (24,
U = J -
= d
- [y2 = x at Q]
=Jx2-2X+l+x=J-
But minimizing U is equivalent to minimizing u2 =F(x) = x2 -x + 1 on the interval CO, +co)(the value of x
is restricted by the fact that x = y2 2 0).
F(x)= 2x - 1 F”(x) = 2
or x = 3
The only critical number is the solution of
F(x) = 2x - 1 = 0
Now F”(3)= 2 > 0. So by the second-derivative test, the function F has a relative minimum at x = 3. Theorem
24.1 implies that this is an absolute minimum. When x = 3,
y 2 = x = - 1 and y = + - = 1 .4 f-
4
2
-$ +JzJz= 2
Thus, the points on the parabola closest to (1,O) are (4, &2) and (4, - 4 / 2 ) .
179

180 MORE MAXIMUM AND MINIMUM PROBLEMS [CHAP. 24
(b) An open box (that is, a box without a top) is to be constructed with a square base [see Fig. 24-2(b)] and is
required to have a volume of 48 cubic inches. The bottom of the box costs 3 cents per square inch, whereas the
sides cost 2 cents per square inch. Find the dimensions that will minimize the cost of the box.
Let x be the side of the square bottom, and let h be the height. Then the cost of the bottom is 3x2and the
cost of each of the four sides is 2xh, giving a total cost of
C = 3x2+4(2xh) = 3x2 +8xh
The volume is V = 48 = x2h.Hence, h = 48/x2and
384
c = 3x2 +fix@= 3x2 +-
X = 3x2 + 3fj4x-l
which is to be minimized on (0, +00). Now
and so the critical numbers are the solutions of
384
6~ ---
x2 - O
384
6~ = -
X2
x3 = 64
x = 4
Now
768
dZC
dx2 x3
- 6 - ( - 2 ) 3 8 4 ~ - ~
= 6 +-> 0
--
for all positive x; in particular, for x = 4. By the second-derivative test, C has a relative minimum at x = 4.
But since 4 is the only positive critical number and C is continuous on (0, +oo),Theorem 24.1 tells us that C
has an absolute minimum at x = 4. When x = 4,
48 48
x2 16
h = - = - = 3
So, the side of the base should be 4 inches and the height 3 inches.
tY
Fig. 24-2

CHAP. 241 MORE MAXIMUM AND MINIMUM PROBLEMS 181
Solved Problems
24.1 A farmer must fence in a rectangular field with one side along a stream; no fence is needed on
that side. If the area must be 1800square meters and the fencing cost $2 per meter, what dimen-
sions will minimize the cost?
Let x and y be the lengths of the sides parallel and perpendicular to the stream, respectively. Then the
cost C is
c = 2(x +2y) = 2x +4y
But 1800 = xy, or x = 1800/y, so that
+4y = 3600y-1 +4y
3600
Y
c = *(y)
+4y = -
and
3600
- - 3 6 0 0 ~ - ~
+ 4 = - -+ 4
dC
--
d
Y Y2
X
We wish to minimize C(y)for y > 0. So we look for positive critical numbers,
+ 4 = 0
3600
--
Y2
3600
4=--
Y2
+-= 3600 900
4
y = +30
,which is positive at y = +30. Thus, by the second-
derivative test, C has a relative minimum at y = 30. Since y = 30 is the only positive critical number, there
can be no other relative extremum in the interval (0, +CO). Therefore, C has an absolute minimum at
y = 30, by Theorem 24.1.When y = 30 meters,
d2C d 7200
NOW
-= -(-3 6 0 0 ~ - ~
+4) = 72OOy3= -
dY2 d
Y Y3
24.2 If c,, c2,...,c, are the results of n measurements of an unknown quantity, a method for estimat-
ing the value ofthat quantity is to find the number x that minimizes the function
f ( x ) = (x - C
l
)
, +(x - c,), + ' * +(x - c,),
This method is called the least-squares principle. Find the value of x determined by the least-
squares principle.
f'(x) = 2(x -c,) +2(x - c,) + * * * +2(x - C")

To find the critical numbers,
2(x - c1) +2(x - Cq) + * * +2(x -c,) = 0
(x -c1) +(x -c2) +* * * +(x -c,) = 0
nx -(c1 +c2 + *..+c& = 0
nx = c1 +c2 + . * .+c,
x = c1 +c2 +.*.+c,
n
Asf"(x) = 2 +2 + - .- +2 = 2n >0, we have, by the second-derivativetest, a relative minimum off at the
unique critical number. By Theorem 24.1, this relative minimum is also an absolute minimum on the set of
all real x. Thus, the least-squaresprinciple prescribes the average ofthe n measurements.
4X2 - 3
x - 1
2
4
3 Letf(x) = -
for 0 I x < 1. Find the absolute extrema,if any, offon CO,l),
(X - 1)D,(4x2 - 3) - (4x2 - ~)D,(x - 1) -(X - 1 x 8 ~ )
-(4x2 - 3x1)
-
(x - 1)2 (x - 1)2
f'W =
8x2 - 8~ - 4x2 +3 4x2 - 8~ +3 ( 2 ~
- 3 x 2 ~
- 1)
-
-
-
-
-
-
(x - 1)2 (x - 1)2 (x - 1)2
To find the critical numbers,setf'(x) = 0,
( 2 ~
- 3 x 2 ~
- 1)
= O
( 2 ~
- 3 x 2 ~
- 1) = 0
(x - 1)2
2 x - 3 = 0 or 2 x - 1 = 0
x = 3 or x = 4
So the only critical numberin CO, 1) is x = 4.
Let us apply the first-derivativetest (Theorem23.4),
For x immediately to the left of 4, x - 4 < 0 and x - 3 < 0, and so, f'(x) > 0. For x immediately
to the right of 4, x - 4 > 0 and x -4< 0, and so, f'(x) < 0. Thus, we have the case { +, -}, and f
has a relative maximum at x = 3. (The second-derivativetest could have been used instead.) The function
f has no absolute minimum on CO, 1). Its graph has the line x = 1 as a vertical asymptote, since
4x2 -3
lim f ( x )= lim -
-
- -a(see Fig. 24-3).'
x-1- x 4 1 - x - 1
t'
X
Fig. 24-3
4x2-3 1
Notcthat-=4(x+ 1)+-+ -coasx+l-.
x - 1 x - 1

CHAP. 241
24.4
245
24.6
24.7
24.8
24.9
24.10
24.11
24.12
MORE MAXIMUM AND MINIMUM PROBLEMS
183
A rectangular field is to be fenced in so that the resulting area is 100 square meters. Find the dimensions of
that field (if any) for which the perimeter is: (a) a maximum; (b)a minimum.
Find the point(s)on the parabola 2x = y2 closest to the point (1,O).
Find the point(s)on the hyperbola x2 - y2 = 2 closest to the point (0, 1).
A closed box with a square base is to contain 252 cubic feet. The bottom costs $5 per square foot, the top
costs $2 per square foot, and the sides cost $3 per square foot. Find the dimensions
cost.
that will minimize the
x2 + 4
x - 2
Find the absolute maxima and minima (if any) of f(x) = -
on the interval CO,2).
A printed page must contain 60 square centimeters of printed material. There are to be margins of
5 centimeters on either side, and margins of 3 centimeters each on the top and the bottom. How long
should the printed lines be in order to minimize the amount of paper used?
A farmer wishes to fence in a rectangular field of loo00 square feet. The north-south fences will cost $1.50
per foot, whereas the east-west fences will cost $6.00 per foot. Find the dimensions of the field that will
minimize the cost.
1
Sketch the graph of y = -
1 +x2'
Find the point on the graph where the tangent line has the greatest slope.
Find the dimensions of the closed cylindrical can [see Fig. 24-4(a)] that will have a capacity of k
volume units and used the minimum amount of material. Find the ratio of the height h to the radius r
of the top and bottom. (The volume is I/ =-nr2h,and the lateral surfacearea is S = 2nrh.)
If the bottom and the top of the can have to be cut from square pieces of metal and the rest of these
squares is wasted [see Fig. 24-4(b)], find the dimensions that will minimize the amount of material
used, and find the ratio of the height to the radius.

t Y
Fig. 24-5 Fig. 24-6
24.13
24.14
24.15
24.16
24.17
24.18
24.19
24.20
24.21
24.22
A thin-walled cone-shaped cup is to hold 36n cubic inches of water when full. What dimensions will mini-
mize the amount of material needed for the cup? (The volume is V = *m2h and the surface area is A = nrs;
see Fig. 24-5.)
X
(a) Find the absolute extrema on CO, +CO) (if they exist)of f ( x )= (b) Sketch the graph off:
(x2 + 1)3/2*
A rectangular bin, open at the top, is required to contain a volume of 128 cubic meters. If the bottom is to
be a square, at a cost of $2 per square meter, whereas the sides cost $0.50 per square meter, what dimen-
sions will minimize the cost?
The selling price P of an item is 100- 0.02~
dollars, where x is the number of items produced per day. If
the cost C of producing and selling x items is 4Ox + 15OOO dollars per day, how many items should be
produced and sold every day in order to maximize the profit?
Consider all lines through the point (1,3) and intersecting the positive x-axis at A(x, 0) and the positive
y-axis at B(0,y) (see Fig. 24-6).
Find the line that makes the area of ABOA a minimum.
1 k
2 x
Conisder the functionf ( x )= - x2 +-.(a) For what value of k willf have a relative minimum at x = -2?
(b) For the value of k found in part (a), sketch the graph of 1: (c) For what value(s) of k will f have an
absolute minimum?
Find the point@)on the graph of 3x2 + lOxy + 3y2= 9 closest to the origin. [Hint: Minimize x2 +y2,
making use of implicit differentiation.]
Fill in the gaps in the followingproof of Theorem 24.1. Assume thatfis continuous on an interval 9.
Letf
have a relative maximum at c in 9,
but no other relative extremum in 9.
We must show that f has an
absolute maximum on 9 at c. Assume, to the contrary, that d # c is a point in 9 withf(c) <f(d). On the
closed interval 9 with endpoints c and d,f has an absolute minimum at some point U. Sincef has a relative
maximum at c, U is different from c and, therefore,f(u) <f(c). Hence, U # d. So, U is in the interior of 9,
whencefhas a relatioe minimum at U # c.
Prove the following theorem, similar to Theorem 24.1: If the graph offis concave upward (downward)
over an interval 9,
then any relative minimum (maximum) offin 9is an absolute minimum (maximum)on
9.
[Hint: Consider the relationship of the graph offto the tangent line at the relative extremum.)
Find the absolute extrema (if any) off(x) = x2I5- 3x7/’ on (- 1, 1
3
.

Chapter 25
Angle Measure
25.1 ARC LENGTH AND RADIAN MEASURE
Figure 25-l(a) illustrates the traditional system of angle measure. A complete rotation is divided
into 360 equal parts, and the measure assigned to each part is called a degree. In modern mathematics
and science, it is useful to define a different unit of angle measure.
Definition: Consider a circle with a radius of one unit [see Fig. 25-l(b)]. Let the center be C, and let
CA and CB be two radii for which the intercepted arc A
% of the circle has length 1. Then
the central angle ACB is taken to be the unit of measure, one radian.
Let X be the number of degrees in 3;ACB of radian measure 1. Then the ratio of X to 360" (a
complete rotation) is the same as the ratio of A
% to the entire circumference,211. Since A
% = 1,
360 180
or X = - = -
x 1
---
360 - 2
1
1 2
1
1 It
(25.1)
180
Thus, 1 radian = -degrees
11
If we approximate 11 as 3.14, then 1 radian is about 57.3 degrees. If we multiply (25.1) by n/180, we
obtain
11
1 degree = -radians (25.2)
180
EXAMPLE Let us find the radian measures of some "common" angles given in degrees. Clearly, the null angle is 0
in either measure. For an angle of 30", (25.2)gives
30"= 30(& radians) = radians
for an angle of 45",
45" = 45(& radians) = 4
K radians
185

186 ANGLE MEASURE [CHAP. 25
and so on, generatingTable 25-1. T
h
i
stable should be memorized by the student, who will often be going back and
forth between degrees and radians.
Table 25-1
Degrees
0
30
45
60
90
180
270
360
Radians
0
n
-
6
a
-
4
a
-
3
n
-
2
f
3n
2
2n
-
Consider now a circle of radius r with center 0(see Fig. 25-2).Let <DOE contain 8radians and let
s be the length of arc m.The ratio of 8 to the number 2n of radians in a complete rotation is the same
as the ratio of s to the entire circumference2713, 8/2n = s/2nr. Hence,
s = r8 (25.3)
gives the basic relationship between the arc length, the radius, and the radian measure of the central
angle.
Fig. 25-2
25.2 DIRECTED ANGLES
Angles can be classified as positive or negative, according to the direction of the rotation that
generates them. In Fig. 25-3(a),we shall agree that the directed angle AOB is taken to be positive when
it is obtained by rotating the arrow OA counterclockwise toward arrow OB.On the other hand, the
directed angle AOB in Fig. 25-3(b) is taken to be negative if it is obtained by rotating arrow OA
clockwise toward arrow OB. Some examples of directed angles and their radian measures are shown in
Fig. 25-4.

ANGLE MEASURE 187
CHAP. 251
I
0LA o*T*A
B
n
n radians
7radians
n
-radians
4
(90") (180")
n I n
(-45") (-900)
- - radians
2
- - radians
4
Fig. 25-4
2n radians
(360")
1 (270")
radians
(3-(3-
-- radians -2n radians
(- 270") (-360")
Some directed angles corresponding to more than one complete rotation are shown in Fig. 25-5. It
is apparent that directed angles whose radian measures differ by an integral multiple of 2n (e.g., the first
and the last angles in Fig. 25-5) represent identical configurations of the two arrows. We shall say that
such angles "have the same sides."
+I: turns
o
r
+ 7radians
+ lf turns -la turns
or or
+3n radians - - radians
Iln
+2: turns
Of
+ radians
Fig. 25-5

188 ANGLE MEASURE [CHAP. 25
SolvedProblems
25.1 Express in radians an angle of: (a)72"; (b)150".
Use (25.2).
2 36 2
7
2
5
(a) 72" = 72(& radians) = 5
.
3
6(nradians) = -radians
150 5(30) 5n
(b) 150" = -n = -R = -radians
180 q30) 6
25.2 Express in degrees an angle of: (a)5n/12 radians; (6)0 . 3 ~
radians; (c) 3 radians.
Use (25.1).
0 . 3 ~
(b) 0.3n radians = -x 180" = 54"
n
(c) 3 radians = -
3 x 180" = ('3"
- x 172"
n
Since n e 3.14.
25.3 (a) In a circle of radius 5 centimeters,what arc length along the circumferenceis intercepted by
a central angle of n/3 radians?
(b) In a circle of radius 12 feet, what arc length along the circumference is intercepted by a
central angle of 30°?
Use (25.3):s = re.
n 5n
(a) s = 5 x - = - centimeters
3 3
(b) The central angle must be changed to radian measure. By Table 25-1,
n A
30" = - radians and so
6 6
s = 12 x - = 2n feet
25.4 The minute hand of an ancient tower clock is 5 feet long. How much time has elapsed when the
tip has traveled through an arc of 188.4 inches?
In the formula 8 = s/r, s and r must be expressed in the same length unit. Choosing feet, we have
s = 188.4/12 = 15.7feet and r = 5 feet. Hence,
15.7
5
(?=-= 3.14 radians
This is very nearly IC radians, which is a half-revolution, or 30 minutes of time.
25.5 What (positive)angles between 0 and 2n radians have the same sides as angles with the following
measures?
9 R n
4 2
(a) -radians (6) 390" (c) - - radians (d) -3n radians
= (2 +$c = 2n +-
n
4 4
Hence, 9744 radians determines a counterclockwise complete rotation (2n radians) plus a counter-
clockwise rotation of n/4 radians (45") [see Fig. 25-qa)I. The "reduced angle"; that is, the angle with
measure in CO, 2n) and having the same sides as the given angle, therefore is n/4 radians.

CHAP. 251 ANGLE MEASURE 189
(b) 390" = 360" +30" = (271 radians) +( 4 6 radians) [see Fig. 25-6(b)]. The reduced angle is 4 6 radians
(or 30").
(c) A clockwise rotation of 4 2 radians (90")
is equivalent to a counterclockwise rotation of 271 - n/2
= 3n/2 radians [see Fig. 25-6(c)].Thus, the reduced angle is 3n/2 radians.
(d) Adding a suitable multiple of 2
7
1 to the given angle, we have -3a +4
7
1 = + A radians; that is,
a clockwise rotation of 371 radians is equivalent to a counterclockwise rotation of R radians [see
Fig. 25-6(d)].The reduced angle is II radians.
$
25.6 Convert the followingdegree measuresof angles into radian measures: (a)36"; (b) 15";(c)2";(d)(90/n)O;
(e) 144".
25.7 Convert the following radian measuresof angles into degreemeasures: (a)2 radians; (b)4 5 radians;
(c) 7z/12 radians; (6)571/4 radians; (e)7n/6 radians.
25.8 If a bug moves a distance of 371 centimeters along a circular arc and if this arc subtends a central angle of
45", what is the radius of the circle?
25.9 In each of the following cases, from the information about two of the quantities s (intercepted arc), r
(radius),and 8 (central angle),find the third quantity. (If only a number is given for 8, assume that it is the
number of radians.)(a)r = 10, 8 = 71/5;(b)8 = 60°, s = 11/21; (c)r = 1, s = w/4; (d) r = 2, s = 3 ; (e)r = 3,
e = 900;(f)
e = 1800, = 6.283 18;(g) t = 10, e = 1200.
25.10 If a central angle of a circle of radius r measures 8 radians, find the area A of the sector of the circle
determined by the central angle (see Fig. 25-7). [Hint:
The area of the entire circle is 71r2.]

Fig. 25-7
25.11 Draw pictures of the rotations determininganglesthat measure: (a)405";(b) 11n/4radians;
(c)7n/2 radians; (d)-60"; (e)-71/6 radians; (f) -5a/2 radians.
25.12 Reduce each angle in Problem 25.11to the range of 0 to 271 radians.

Chapter 26
/ * - '
/'
/
/
Sine and Cosine Functions
-

26.1 GENERAL DEFINITION
1 0
The fundamental trigonometric functions, the sine and the cosine, will play an important role in
calculus. These functionswill now be defined for all real numbers.
Definition: Place an arrow 03 of unit length so that its initial point 0 is the origin of a coordinate
system and its endpoint A is the point (1, 0) of the x-axis (see Fig. 26-1). For any given
number 8, rotate about the point 0 through an angle with radian measure 8. Let 0
be the final position of the arrow after the rotation. Then: (i) the x-coordinate of B is
defined to be the cosine of 8, denoted by cos 8; (ii) the y-coordinate of B is defined to be
the sine of 8, denoted by sin 8. Thus, B = (cos 8, sin 8).
-
1 X
f'
, I
Fig. 26-1
EXAMPLES
(a) Let 8 = n/2.If we rotate OA x/2 radians in the counterclockwise direction, the final position B is (0, 1) [see
Fig. 26-2(u)].Hence,
11 x
sin - = 1
2
cos-=o
2
(b) Let 8 = x. If we rotate FA through K radians in the counterclockwisedirection, the final position B is (- 1,O)
[see Fig. 26-2(b)].So,
cos Ic = -1 sin 7c = 0
f Y
(6)
Fig. 26-2
190

CHAP. 261 SINE AND COSINE FUNCTIONS 191
(c) Let 6 = 3z/2. Then the final position B, after a rotation through 3z/2 radians in the counterclockwise direc-
tion, is (0, -1)[see Fig. 26-2(c)]. Hence,
3n 3n
2 2
cos -= 0 sin -= -1
(6) Let 8 = 0. If is rotated through 0 radians, the finalposition is still (1,O). Therefore,
cos 0 = 1 sin 0 = 0
(e) Let 8 be an acute angle (0 < 6 < n/2)of right triangle DEF, and let AOBG be a similar triangle with hypot-
enuse 1 (see Fig. 26-3). By proportionality, = b/c and = a/c. So, by definition, cos 6 = a/c,
sin 8 = b/c.This agreeswith the traditional definitions:
adjacent side
hypotenuse
cos e =
E
opposite
side,b
opposite side
hypotenuse
sin 8 =
Y
B(COS
8, sin 8)
D adjacent F O E G X
Fig. 26-3
side, a C
Consequently,we can appropriate the values of the functions for 8 = n/6, 44, n/3 from high-school trigonom-
etry. The results are collectedin Table 26-1, which ought to be memorized.
Table 26-1
6

192 SINE AND COSINE FUNCTIONS [CHAP. 26
The above definition implies that the signs of cos 8 and sin 8 are determined by the quadrant in
which point B lies. In the first quadrant, cos 8 > 0 and sin 8 > 0. In the second quadrant, cos 8 <0 and
sin8 > 0. In the third quadrant, cos8 < 0 and sin 8 < 0. In the fourth quadrant, cos8 > 0 and
sin 8 < 0 (see Fig. 26-4).
+ +
t (cos 0,sin 0)
- +
(cos 0,sin 6)
(cos 0,sin 0 )
(COS 8, sin e) / I + -
- -
Fig. 26-4
26.2 PROPERTIES
We list the following theorems, which give the most important properties of the sine and cosine
functions.
Theorem 26.1: sin’ 8 +cos’ 8 = 1.
Proof: In Fig. 26-1, the length of 03is given by (2.1),
1 = ,/(COS e -0)’ +(sin e -o ) ~
= ,/(COS el2 +(sin 8)’
Squaring both sides, and using the conventional notations (sin 8)’ E sin28 and (cos 8)’ =cos28 gives
Theorem 26.1.
Corollary: 1 - sin2 8 = cos2 8and 1 - cos2 8 = sin’ 8.
EXAMPLE Let 8 be the radian measureof an acute angle such that sin 8 = 3.By Theorem 26.1,
9 16
cos28= 1 --=-
25 25
Since 8 is an acute angle,cos 8 is positive;cos 8 = 4.
We have already seen (Chapter 25) that two angles that differ by a multiple of 2
1
1 radians (360”)
have the same sides. T
h
i
s establishes:

/
Theorem26.2: The cosine and sine functions are periodic,of period 211;that is, for all 8,
cos (e +2 4 = cos e sin (8 +2n)= sin 8
(Moreover, 211 is the smallest positive number with this property.)
In view of Theorem 26.2,it is sufficient to know the values of cos 8 and sin 8for 0 I 8 < 2n.
EXAMPLES
n f i
(a) sin -= sin 2n +- = sin -= -
77c
3 ( I) 3 2
(b) cos 5n = cos (3n +274 = cos 3n = cos (n+2n)= cos n = - 1
J
3
(c) COS 390"= COS (30"+360")= COS 30" =-
2
fi
(4 sin 405" = sin (45" +360") = sin 45" =-
2
Theorem26.3; cos 8 is an even function, and sin 8 is an odd function.
The proof is obvious from Fig. 26-5. Points B
'and B have the same x-coordinates,
but their y-coordinates differ in sign,
sin (- 8) = -sin 8
Because of Theorem 26.3, we now need to know the values of cos 8 and sin 8 only for 0 5 8 5 n.
Consider any point A(x, y) different from the origin 0,
as in Fig. 26-6. Let r be its distance from the
origin and let 8 be the radian measure of the angle that line OA makes with the positive x-axis. We call
r and 8 polar coordinatesof point A.
t'
Fig. 26-5 Fig. 266

Theorem 26.4: The polar coordinates of a point and its x- and y-coordinates are related by
x = t cos e y = r sin 8
For the proof, see Problem 26.2.
Theorem 26.5 (Law ofcosines): In any triangle ABC (see Fig. 26-7),
c2 = a2+b2 - 2ab cos 8
For a proof, see Problem 26.3. Note that the Pythagorean theorem is obtained as the special case
e = 4 2 .
-----
C a B c 5 B
Fig. 26-7 Fig. 26-8
EXAMPLE Find angle 8 in the triangle of Fig. 26-8.
Solving the law of cosines for cos 8,
U
' +b2 - c2 (5)2 +(212-(@)' 25 +4- 19 1
--
-
-
-
-
-
2ab 2(5M2) 20 2
cos e =
Then,from Table 26-1,8= x/3.
Theorem 26.6 (Sumand Diflerence Formulas):
(a) cos (U +v) = cos U cos U -sin U sin v
(6) cos (U - U) = cos U cos U +sin U sin U
(c) sin (U +U) = sin U cos U +cos U sin U
(d) sin (U - U) = sin U cos U - cos U sin U
Once any one of these four formulas has been proved, the other three can be derived as corollaries.
A proof of the differenceformula for the cosine is sketched in Problem 26.14.
EXAMPLES
II R R R I I
(a) cos -= cos (
: - i)
= cos - cos - +sin - sin -
12 3 4 3 4
(b) cos 135"= cos (900+45")= cos 90"cos 45"- sin 90" sin 45"

CHAP. 26) SINE AND COSINE FUNCTIONS 195
In 7 1 7 1 7 1 7 1
12 2 12 2 12
(c) sin -= sin (2 +5)= sin 5cos 5+cos - sin -
= (l)( Jz+& )+(O)(sin :
) [by example (a)]
SubstitutingU = a/2 and U = 8in the differenceformulasof Theorem 26.6yields
a 1c
cos (i- 8 ) = cos 5cos 8 +sin - sin 8 = 0 cos 8 + 1 sin 8 = sin 8
sin (F- 8) = sin 5cos 8 - cos - sin 8 = 1 cos 8 - 0 sin 8 = cos 8
2
2
a II
Thus we have:
Theorem26.7: cos (i- 8) = sin 8 and sin (;,- 8) = cos 8; that is, the sine of an angle is the cosine
of its complement.
We also have the useful double-angleformulas :
Theorem26.8: cos 28 = cos28 -sin28 = 2 cos28 - 1 = 1 - 2 sin28 and sin 28 = 2 sin 8 cos 8.
For cos 28, note that:
(i) cos 28 = cos (8 +8)= cos 8 cos 8 - sin 8 sin 8
(ii)
[by Theorem 26.6(a)]
p y Theorem 26.11
[by (ii) and Theorem 26.11
= cos2 8 - sin28
= cos28 - (1 - cos28)
= 2 cos2 8 - 1
= (1.- sin28) - sin28
= 1 - 2 sin28
For sin 28,
sin 28 = sin (8 +8) = sin 8 cos 8 +cos 8 sin 8 [by Theorem 26.6(c)]
= 2 sin 8 cos 8
From Theorem 26.8,we deduce:
Theorem26.9 (Half-AngleFormulas):
8 1 +cos 8
2
8 1 -cos 8
2
2
2
(a) cos2- =
(b) sin2- =
In fact,replace 8 by 812 in Theorem 26.8:
8 1 +COS e
2 '
2
8 8 1 -cos 8
2' 2 2 .
(a) cos 8 = cos - 1. so, cos2 - =
(6) COS 8 = COS - So, sin2- =
EXAMPLES
(a) sin 120"= sin (2 x 60")
= 2 sin 60"cos 60"

2n n
4 4 2
J
3
1+cos- 1 + -
2
=-
6
2
n
(c) cos2-= cos2
12
=- + J 3 [multiply numerator and denominator by 2)
4
~~~ ~
CHECK Earlier, we deduced from Theorem 26.6 that
n Jz+&
4
cos -=
12
Hence, in view of the identity (U +u ) ~
= u2 +2uu +u2,
1-cos-n J z
1--
.Hence, since n/8 is in the first quadrant,
2 2-$
=-=-
1 n 4
8 2 4 2 2 4
(d) sin2 = sin2 (-x -)=
n &m
2
sin - = +
8
SolvedProblems
'26.1 Evaluate: (a)cos ( 4 8 ) ;(6) cos 210";(c) sin 135"; (d) cos 17".
(a) By Theorem 26.9,
Hence, since n/8 is an acute angle,
J
m
m
2
cos -=
8
(b) Writing 210" = 180" + 30",we have, by Theorem 26.6,
cos 210" = cos (180" + 30") = cos 180" cos 30" - sin 180" sin 30"
= -I(+) J- - (O)(i) = - 2
J5
(c) Using the previously derived value cos 135" = -&2, we have, from Theorem 26.1,

CHAP. 26) SINE AND COSINE FUNCTIONS 197
Hence, since 135"is in the second quadrant,
sin 135"= + 8= $
(d) 17"cannot be expressed in terms of more familiar angles (such as 30", 45", 60")in such a way as to
allow the application of any of our formulas. We must then use the cosine table in Appendix D, which
gives 0.9563as the value of cos 17". This is an approximation, correct to four decimal places.
Refer to Fig. 26-9. Let D be the foot of the perpendicular from A to the x-axis. Let F be the point on
the ray OA at a unit distance from the origin. Then, F = (cos8, sin 8). If E is the foot of the perpendicular
from F to the x-axis, we have
- -
OE = cos e FE = sin 8
Now, AADO is similar to AFEO (by the AA criterion),whence
Therefore, x = r cos 8 and y = r sin 8. When A(x, y) is in one of the other quadrants, the proof can be
reduced to the case where A(x, y) is in the first quadrant. The proof is easy when A(x, y) is on the x-axis or
on the y-axis.
E D X
Fig. 26-9
In Fig. 26-7, set up a coordinate system with C as origin and B on the positive x-axis. Then B has
coordinates (a,0).Let (x,y) be the coordinates of A. By Theorem 26.4,
y = b sin 8
= b cos e
By the distance formula (24,
c = J(x -a)' +(y -0)' = & q T j 7
Hence,
c2 = (x - a)2+y2 = (b cos 8 - a)' +(b sin q2
ALGEBRA (P - Q)' = P
' - 2PQ +Q'
= b2 cos2 8 - h b COS e +a2 +b2 sin' 8
= U
' +b2(cos28 +sin' 8) - 2ab COS 8
= a2 +b2 - 2ab cos e

26.4 Derive the followingidentities:
sin 8 1 -COS 28
(b) = sin 28
(a) (sin 8 -cos 8)2= 1 - sin 28
(a) By the ALGEBRA of Problem 26.3,
(sin 8 - COS q2= sin28 -2 sin 8 COS 8 +cos2 8
= 1 - 2 sin 8 cos 8
= 1 - sin 28
[by Theorem 26.1J
[by Theorem 26.8)
sin2 8 +sin2 8
2 sin 8 COS 8
(2 sin @)(sin8) sin 8
(2 sin excos e)-COS 8
-
-
- --
-
[by Theorem 26.1J
26.5 Given that 8is in the fourth quadrant and that cos 8 = 3,
(c)COS (8/2); (d)sin (8/2).
find: (a)cos 28; (b) sin 8;
/? 2
(b) By Theorem 26.1,
4 5
sin2 8 = 1 - cos2 8 = 1 -
Then, sincesin 8 is negative for 8 in the fourth quadrant, sin 8 = -&3.
(c) By Theorem 26.9,
e 1+COS e 1 +p/q 3 +2 5
=-=--- -
2 2 6 6
cos2 - =
2
Since 8is in the fourth quadrant, 3z/2 < 8 < 2z; therefore,
31L e
- - < - < a
4 2
Thus, 8/2 is in the second quadrant, and cos (8/2) is negative. Hence,
e
cos - = - $
2
(4 By Theorem 26.9,
e i - c o s e 1 - 3 3 - 2 1
sin2 - = =-=-=-
2 2 2 6 6
From part (c), 8/2 is in the second quadrant. Hence, sin (8/2) is positive. Thus,sin 8/2 = l / f i = $/6.
26.6 Let AABC be any triangle.Using the notation in Fig. 26-10, prove
sin A sin B sin C
a b C
law o
f sines
-----
- -
(Here,sin A is the sine of 3= CAB, and similarly for sin B and sin C.)

Let D be the foot of the perpendicularfrom A to side BC. Let h = AD. Then,
-
or h = c sin B
AD h
sin B === -
AB c
and so
1 1 1
2 2 2
area of AABC = -(base x height) = - ah = - ac sin B
(Check that this is also correct when 9:B is obtuse.) Similarly,
1
2
area =- ab sin C
1
area = - bc sin A
2
Hence,
1 1 1
- ac sin B = - bc sin A = - ab sin C
2 2 2
and division by 4abc gives the law of sines.
A
B D a C
Fig. 26-10
26.7 Evaluate: (a) sin (4n/3); (b) cos (lln/6); (c) sin 240"; (6)cos 315";(e) sin 75"; (f)sin 15";(9) sin (11n/12);
(h) cos 71";(i) sin 12".
26.8 If 8 is acute and cos 8 = 4, evaluate: (a) sin 8;(b) sin 28; (c)cos 28;(d) cos (8/2);(e) sin (8/2).
26.9 If 8 is in the third quadrant and sin 8 = -3,evaluate: (a) cos 8;(b) sin 28; (c)cos 28;(d) cos (8/2);
(e)sin (8/2).
26.10 In AABC, AB = 5, AC = 7, and = 6. Find cos B.
26.11 In Fig. 26-11, D is a point on side BC of AABC such that AB = 2, AD = 5,
[Hint:
U
s
e the law of cosines twice.]
= 4, and = 3. Find x.
26.12 Prove the followingidentities:
1 +cos 28 cos e
--
(a) cos258 = (1 - sin 5e)(i +sin 58) (b) -
sin 28 sin 8
(c) (sin 8 +cos 8)2= 1 +sin 28 (6) cos4 8 - sin48 = cos2 8 - sin28

Fig. 26-11 Fig. 26-12
26.13 For each of the following, either prove that it is an identity or give an examplein which it is false:
sin 8 cos 8 1
+-=-
(b) I +cos e sin e sin e
(a) sin 48 = 4 sin 8 cos 8
cos 8 sin 8 2 3 3
2 2
(e) -+-=-
sin 8 COS 8 sin 28
(f)2 sin - 8 cos - 8 = sin 38
26.14 Fill in all details of the followingproof of the identity
cos (U - U) = cos U cos U +sin U sin U
Consider the case where 0 5 U < U < U +n (see Fig. 26-12).By the law of cosines,
-
BC2 = l2 + 1' -2(1)(1)COS #BOC
(cos U - cos U
)
' +(sin U -sin U
)
' = 2 -2 cos (U - U)
cos2U - 2 cos U cos U +cos2U +sin' U -2 sin U sin U +sin2U = 2 - 2 cos (U - U)
(cos2U +sin' U) +(cos2U +sin2U) - COS U cos U +sin U sin U) = 2 - 2 cos (U -U)
1 + 1 - CCOS U cos U +sin U sin U) = 2 - 2 cos (U - U)
cos U cos U +sin U sin U = cos (U - U)
Verify that all other casescan be derived from the casejust considered.
26.15 Prove the followingidentities:
= -sin 8 (b) sin
(c) COS (n+8) = - COS 8 (d) sin (n+8) = -sin 8
n n 1 7c n J 5
n *
26.16 Show: (a) sin - = cos - =- (b) sin - = cos -=-
4 4 2 6 3 2 3 6 2
(c) sin - = cos -= -
n
[Hint: (a)Take an isosceles right triangle AOGB with right angle at G and let b = OG = BG and c = OB.
L
1 n 1
f = (:r= = - Hence, sin - = -= g.
(b) Take an equilateral
Then, c
' = b2 +b2 = 2b2. So, sin' -
c2 2' 4 J z 2

1
= -
2'
triangle AABC of side 1 and let AD be the altitude from A to the midpoint D of BC. Then
-
x x B D 4 1 x a
By Theorem 26.7, sin-=cos
6
Since iKABD contains - radians, cos - === - = -
3 3 AB 1 2'
a x 1 3
(c) sin2 - = 1 - cos2 - = 1 - - =-
3 3 4 4' 3 2 6 3 1
a a
Hence,sin = and cos - = sin -by Theorem 26.7.
26.17 Derive the other parts of Theorem 26.6from part (a),which was proved in Problem 26.14. Hint:First note
r
that cos (i- 8 ) = sin 8 follows from part (a). The formula for cos (U +U) follows by applying part (a)to
cos (U - (-U)) and using the identitiescos (-U) = cos U and sin (-U) = - sin 0.The formula for sin (U +v)
follows by applying part (a) to cos ((; - U
) - v), and then the formula for sin (U - U) follows from the
formulafor sin (u +v) by replacing U by -U.
3

Chapter 27
Graphs and Derivatives of Sine and
Cosine Functions
27.1 GRAPHS
Let us first observe that cos x and sin x are continuous functions. This means that, for any fixed
x = e,
lim cos (0 +h) = cos 8 lim sin (0 +h) = sin 8
h - 0 h - 0
as is obvious from Fig. 27-1. Indeed, as h approaches 0, point C approaches point B. Therefore, the
x-coordinate (the cosine) of C approaches the x-coordinate of B, and the y-coordinate (the sine) of C
approachesthe y-coordinate of B.
Table 27-1
Now we can sketch the graphs of y = cos x and y = sin x. Table 27-1 contains the values of cos x
and sin x for the standard values of x between 0 and n/2;these values are taken from Table 26-1. Also
listed are the values for 2n/3 (120"),3n/4 (135"),and 5n/6(150"). These are obtained from the formulas
202

CHAP. 271 GRAPHS AND DERIVATIVES OF SINE AND COSINE FUNCTIONS 203
(COS (8+h), sin (0 +
-(COS e, sin 8)
Fig. 27-1
(see Problem 26.15)
= -sin 8 and sin
The graph of y = cos x is sketched in Fig. 27-2(a). For arguments between --a and 0, we have used
the identity cos (-x) = cos x (Theorem 26.3). Outside the interval [- -a, 4, the curve repeats itself in
accordance with Theorem 26.2.
t'
(a) y = cosx
IY
(b) y=sinx
Fig. 27-2

204 GRAPHS AND DERIVATIVES OF SINE AND COSINE FUNCTIONS [CHAP. 27
The graph of y = sin x [see Fig. 27-2(6)] is obtained in the same way. Notice that this graph is the
result of moving the graph of y = cos x to the right by 4 2 units. This can be verified by observingthat
cos (x -;
) = sin x
The graphs of y = cos x and y = sin x have the shape of repeated waves, with each complete wave
extending over an interval of length 2n (the period). The length and height of the waves can be changed
by multiplying the argument and the functional value, respectively, by constants.
EXAMPLES
(a) y = cos 3x. The graph of this function is sketchedin Fig. 27-3. Because
cos 3 x +- = cos (3x +2n) = cos 3x
( 3
the function is of period p = 2n/3. Hence, the length of each wave is 2n/3. The number of waves over an
interval of length 2n (corresponding to one complete revolution of the ray determining angle x) is 3. This
number is called thefrequencyfof cos 3x. In genera),
pf= (length of each wave) x (number of waves in an interval of length 2n) = 2%
and so
2R
P
f = -
For an arbitrary constant b > 0, the functioncos bx (or sin bx) has frequencyb and period 2n/b.
Fig. 27-3
(b) y = 2 sin x. The graph of this function (see Fig. 27-4) is obtained from the graph of y = sin x (see Fig. 27-2) by
multiplying each ordinate by 2. The period (wavelength)and the frequencyof the function 2 sin x are the same
as those of the function sin x: p = 27c,f= 1. But the amplitude, the maximum height of each wave, of 2 sin x
is 2, or twice the amplitude of sin x.
NOTE The total oscillation of a sine or cosine function, which is the vertical distance from crest to trough, is
twice the amplitude.

CHAP. 271 GRAPHS AND DERIVATIVES OF SINE AND COSINE FUNCTIONS
Y
205
Fig. 27-4
For an arbitrary constant A, the function A sin x (or A cos x) has amplitude I A I.
(c) Putting together examples (a) and (b) above, we see that the functions A sin bx and A cos bx (b > 0) have
period 27& frequencyb, and amplitude IA I.Figure 27-5 gives the graph of y = 1.5 sin 4x.
t'
Fig. 27-5
27.2 DERIVATIVES
sin 8 1-cos e
Lemma 27.1: (a) lim -= 1; (6) lim = 0.
8-0 8 8+0 6
(a) For the proof, see Problem 27.4.

1 -cos e 1 +cos e 1 - cos2 e
= lim
i - m s e
(b) lim = lim
e-ro 6 0-0 8 1+cos 0 e+o e(l +cos 0)
sin 8 sin 8
Xlim-0
sin2 8
= lim
= lim - lim
Theorem 2
7
.
2
; D,(sin x) = cos x and D,(cos x) = -sin x.
0-0 e(l +cos 8) e+o 0 1 +cos 8
0 4 0 8 e-ro 1 +cos8 1 + 1
= I * - = O O
sin 8 sin 8
For the proof, see Problem 27.5.
EXAMPLES
(a) Find DJsin 2x). The function sin 2x is a composite function. If f ( x )= 2x and g(x) = sin x, then
sin 2x = g(f(x)).The chain rule (15.2)gives
or
(b) Find Dx(c0s4x). The function cos4x is a composite function. Iff(x) = cos x and g(x) = x4, then
cos4 x = (COS
x ) ~
= g(f(x))
Therefore, D,(c0s4 x) = D,((cos x)~)
= *cos x ) ~D,(cos x)
= qcos x)’ (-sin x)
= -4 cos3 x sin x
Solved Problems
27.1 Find the period, frequency,and amplitude of:
X 1
2 2
(a) 4 sin - (b) -cos 3x
and sketch their graphs.
2n 2n
(a) For 4 sin i x A sin bx, the frequency isf= 6 = 4, the period is p = -= -= 4n, and the amplitude
f 3
is A = 4. The graph is sketched in Fig. 27-qa).
2n 2n
(b) For 4 cos 3x E A cos bx, the frequencyf is b = 3, the period is p = -= - and the amplitude is
b 3 ’
A = 4. The graph is sketched in Fig. 27-6(b).
Fig. 27-6

27.2 Find all solutions of the equations:
1
(a) cos x = 0 (b) sin x = -
2
(a) For -a 5 x < a, inspection of Fig 27-2(a)shows that the only solutions of cos x = 0 are
a
x = -- and x=lf
2 2
Since 211 is the period of cos x, the solutions of cos x = 0 are obtained by adding arbitrary integral
multiples of 271 to -n/2 and a/2,
a ‘ a a a
2 2 2 2
- - +2zn = - (4n - 1) and - +2an = - (4n - 1)
Together, 4n - 1 and 4n + 1 range over all odd integers, and so the solutions of cos x = 0 are all odd
multiples of 4 2 ,
n
x = (2k + 1)5 [k = 0, f1, f 2 , ...I
(b) Figure 27-2(b)shows that, for -a/2 5 x < 37c/2, the only solutions are x = a/6 and x = 5a/6. Hence,
all the solutions are
a 5n
-+2an and -++an
6 6
wheren=O,+l,f2 ,....
27.3 Find the derivatives of:
X X
(a) 3 sin Sx (b) 4 cos - (c) sin’ x (6) cos3-
2 2
(a) D,(3 sin 5x) = 3D,(sin 5x)
= COS 5x)DX(5x)
= COS 5 ~ x 5 )
= 15 COS 5x
[by the chain rule and Theorem 27.2)
(b) Dx(4 cos ;
) = 4D,(cos ;
)
= 4( -sin ;)Dx( ;
) [by the chain rule and Theorem 27.2)
= -(sin ;)(:) = -2 sin -
X
2
(c) DJsin’ x) = DJsin x)’)
= (2 sin x)D,(sin x)
= 2 sin x cos x
= sin 2x
b y the power chain rule]
[by Theorem 27.23
[by Theorem 26.83
= 3( COS COS ;
) [by the power chain rule]
= (3 cos’ ;)( - (sin i)D,(;))
3 x x
2 2 2
= cos2 :)( - sin :
) = -- cos2 - sin -

27.4 Prove Lemma 27.1.
Consider the case when 8 > 0. Using the notation of Fig. 27-7,
area (AOBC) < area (sector OBA)< area (AODA)
Now
t'
I
Fig. 27-7
1 - - 1
2 2
27t
area (AOBC)
= -(OC)(BC)= -cos 8 sin 8
e
area (sector OBA) = - x (area of circle)
-
e x (7t12) =-
e
27t 2
Furthermore,since AOBC and AODA are similar (by the AA-criterion),
-
- sin 8
or DA =-
DA 1
or -=-
D A O A
-=-
B C O C sin 8 COS 8 cos e
area (AODA)= - (DAXOA)= - -
2 -- 2 tin
cos 9
and ( I ) becomes, after dividing through by the positive quantity 4sin 8,
e 1
cos e <-<-
sin 8 COS 8
which is equivalentto
ALGEBRA If a and b are positive numbers,then
a < b ifandonlyif
a b

In (
2
)
,let 8 approach 0 from the right. Both cos 8 and l/(cos 8) approach 1. Therefore, by Problem
8 4 f1,
sin 8
lim -- - 1
e+o+ 8
(3)
sin (-8) sin 8 sin 8
--. Hence, (3)implies lim -= 1,
e B+O- 8
Now, by Theorem 26.3,sin (- 8)= -sin 8 and, therefore, -
-
-8
sin 8
which, together with (3),yields lirn -= 1.
0-0 8
(a) Let us first prove D,(sin x) = cos x. By Theorem 26.6(c),sin (x +h) = sin x cos h +cos x sin h, and so
sin (x +h) -sin x sin x cos h +cos x sin h - sin x
cos x sin h +sin x (cos h - 1)
h
-
-
h h
-
-
sin h 1 -COS h
h
= cos x --sin x
h
Therefore,
sin (x +h) -sin x
h
D,(sin x) = lim
1-0
sin h l - c o s h
h
= lim cos x -- lim sin x
= cos x lim -- sin x lim
= cos x 1-sin x 0
h+O h+O
sin h 1 -COS h
h - 0 h-rO
[by Lemma 27.11
= cos x
(b) By Theorem 26.7,cos x = sin (i- x) .Hence, by the chain rule,
27.6 Sketch the graph of the functionf(x) = sin x - sin2 x.
Along with sin x,f(x) has 2n as a period. Therefore,we need only sketch the graph for 0 5 x < 2n.
f'(x) = cos x -2 sin x cos x = cos x -sin 2x
f"(x) = -sin x -(cos 2x)Dx(2x)= -sin x -2 cos 2x
To find the critical numbers, solvef'(x) = 0.
cos x -2 sin x cos x = 0
(cos xX1 - 2 sin x) = 0
cos x = 0 or 1 - 2 s i n x = O
1
sin x =-
2
cos x = 0 or
n 3n n 5A
6' 6
x = - - or
2' 2
x = - -
where Problem 27.2@) has been used. Application of the second-derivative test (Theorem 23.3) at each
critical number gives the results shown in Table 27-2.The graph is sketched in Fig. 27-8.

x f(x>
1
6 4
0
2
1
5n
6 4
3n
- 2
2
-
n
-
n
-
-
-
- 2
-'I
f"(x)
rel. max.
3
2
1 rel. min.
1
2
3 re1 min.
--
rel. max.
--
Fig. 27-8
Table 27-2
27.7 Find the absolute extrema of the functionf(x) = 2 sin x -cos 2x on the closed interval [0, lt].
For this simple function it is unnecessary to resort to the tabular method of Section 14.2. It suffices to
observe that the sine and cosine functions have maximum absolute value 1. Thus, if there is an argument x
in [0, n] such that
f(x) = 2(+1) -(-1) = 3
that argument must represent an absolute maximum (on any interval whatever). Clearly, there is one and
only one such argument, x = n/2.
On the other hand, since sin x 2 0 on [0, n], the arguments x = 0 and x = n minimize sin x and, at the
same time, maximize cos 2x. Hence,
f(0) =f(n)= 2(0) -(+I) = -1
is the absolute minimum value on [0, n].
27.8 Find the period,frequency, and amplitude of each function, and sketch its graph.
4x 1
2
3
(6) - sin 3x
(a) cos -

27.9 Find the period, frequency,and amplitude of:
X 5x
3 2
(a) 2 sin - (b) -cos 2x (c) 5 sin - (d) cos (-4x)
[Hint: In part (d), the function is even.]
27.10 Find all solutions of the followingequations:
(a) sin x = 0 (b) cos x = 1 (c) sin x = 1 (d) cos x = -1
(e) sin x = -1 (j)cos x = -
1 (9) sin x = Jz (h) cos x = -
J5
2 2 2
[Hint: By Theorem 26.1, only two of parts (a),(b), and (d)need be solved.]
27.11 Find the derivativesof the followingfunctions:
(a) 4 sin3 x (b) sin x +2 cos x (c) x sin x (d) x2 cos 2x
sin x 1- cos x
(4 x ( f ) x2 (9) 5 sin 3x cos x (h) cos2 2x
(i) cos (2x2- 3) 0
1 sin3 (5x +4) (k) JZZG (0 sin3 (sin2x)
27.12 Calculate the followinglimits:
sin x
(a) lirn -
x+o 3x
cos x - 1 sin 2x x + l
(b) lirn (c) lirn - (d) lirn -
x+o 2x x-ro sin 3x x-ro cos x
sin 4x
(e) lim -
x-ro 5x
cos 3x
(h) lirn -
x-rn/2 cosx
sin3 x sin x2
(f) lim -
im x cos -
0 1'
(9) lim x
x-ro 4x3 x+o
2 -x cos x
(I) lim
1 (k) sin2(x - 4)
x-ro X2 x-ro X2 x+4 (x -412 x-r+Q)x2 +x + 1
x sin 4x -sin2x
(i) lirn
27.13 Sketch the graphs of the followingfunctions:
(a) f(x) = sin2x (b) g(x) = sin x +cos x (c) f(x) = 3 sin x -sin3 x
(d) h(x)= cos x -cos2x (e) g(x) = Isin x I
(9) f(x) = sin (x - 1)
(h)
(f)f ( x )= sin x +x
Check your answers to parts (a)-(g) on a graphing calculator.
27.14 Find the absolute extrema of each function on the given interval:
sin x +x on [0, 2
1
1
1
2 sin x +sin 2x on CO, 2
7
1
1
3 sin x -4 cos x on CO, 2
1
1
1
(b) sin x -cos x on CO, n]
(e) lcos x - on CO, 2nl
(c) cos2 x +sin x on [0, A]
1
(f)- x - sin x on CO, 2nl
2
Show that f(x) = A cos x +B sin x has pdriod 2n and amplitude ,
/
-
. Hint: Clearly,
[
B
S 1. Hence, there exists cp such that
A
IJAand IJml
f(x +2n) =f(x). Note that
sin cp = Jm
and cos cp =
1
.Show thatf(x) = ,/mi
sin (x +cp).
A B
Jm
Find the amplitude and period of: (i)3 cos x -4 sin x; (ii)5 sin 2x + 12cos 2x;
(iii)2 cos x +,/5 sin x.

212
2
7
.
1
6
27.17
27.18
27.19
27.2~
2
7
.
2
1
27.22
27.23
27.24
27.25
27.26
GRAPHS AND DERIVATIVES OF SINE AND COSINE FUNCTIONS [CHAP. 27
COS (
: +h) - 1
2
h
(b) lim
h b+O
Evaluate: (a) lim
[Hint: Recall the definitionof the derivative.]
h-0
For what value of A does 2 sin Ax have period 2?
Find an equation of the tangent line to the graph of y = sin2x at the point where x = n/3.
Find an equation of the normal line to the curvey = 1 +cos x at the point (n/3,3/2).
Find the smallest positive integer n such that D:(cos x) = cos x.
Letj(x) = sin x +sin Ix I.(a)Sketch the graph off: (b) Isfcontinuous at O? (c) Isfdifferentiable at x = O?
Find the slope of the tangent line to the graph of:
(a) y = x +cos (xy) at (0, 1); (b) sin (xy) = y at (n/2, 1).
[Hint:Use implicit differentiation.]
Use implicit differentiationto find y':
cos y = x (b) sin (xy) = y2
A ladder 26 feet long is leaning against a vertical wall (see Fig. 27-9). If the bottom of the ladder A is
slippingaway from the base of the wall at the rate of 3 feet per second,how fast is the angle 8 between
the ladder and the ground changingwhen the bottom of the ladder is 10feet from the base of the wall?
An airplane is ascending at a speed of 400 kilometers per hour along a line making an angle of 60"
with the ground. How fast is the altitude of the plane changing?
A man at a point P on the shore of a circular lake of radius 1 mile (see Fig. 27-10) wants to reach the
point Q on the shore diametrically opposite P.He can row 1.5 miles per hour and walk 3 miles per
hour. At what angle 8 (0 5 8 5 4 2 ) to the diameter PQ should he row in order to minimize the time
required to reach Q? (When 8 = 0, he rows all the way;when 8 = n/2,he walks all the way.)
Rework part (a) if, instead of rowing, the man can paddle a canoe at 4 miles per hour.
Fig. 27-9 Fig. 27-10
Find the absoluteextrema off@) = x - sin x on CO, 421.
From part (a),infer that sin x < x holds for all positive x.
sin x
Verifyon a graphing calculator that cos x 5 -5 1for x # 0 and -(n/2) I x 5 (n/2).
X

CHAP. 27) GRAPHS AND DERIVATIVES OF SINE AND COSINE FUNCTIONS 213
sin x for x 5 7t/4
cos x for x > 4 4
is continuous at x = 7t/4. Is the function differ-
27.27 Determine whether the functionf(x) =
entiable at x = n/4?
27.28 (a) The hypotenuse of a right triangle is known to be exactly 20 inches, and one of the acute angles is
measured to be 30°, with a possible error of 2". Use differentials to estimate the error in the computa-
tion of the side adjacent to the measured angle.
(b) Use differentialsto approximate cos 31".
27.29 (a) Find the first four derivativesof sin x. (b) Find the 70th derivativeof sin x.
27.30 (a) Show that there is a unique solution of x3 - cos x = 0 in the interval CO, 13. (b)
method to approximate the solution in part (a).
Use Newton's
27.31 Approximate7t by applyingNewton's method to find a solution of 1 +cos x = 0.
2732 U
s
e Newton's method to find the unique positive solution of sin x = 4 2 .

Chapter 28
X
0
It
6
-
a
-
4
3
The Tangent and
Other Trigonometric Functions
tan x
0
x 0.58
-
3
1
fi x 1.73
Besides the sine and cosine functions, there are four other important trigonometric functions, all of
them expressiblein terms of sin x and cos x.
sin x
Definitions: Tangentfunction tan x = -
cos x
cos x 1
cot x = --
--
sin x tan x
Cotangentfinction
1
1
Cosecantfiurction csc x = -
sin x
Secantfirnction secx=-
cos x
EXAMPLE Let us calculate, and collect in Table 28-1, some of the values of tan x.
sin 0 0
cos0 1
tan 0= -= - = 0
a sin (n/6) 112 1 1 9 Js
a sin (n/4) J z / 2
a sin (a/3) J5/2
3 cos (a/3) 1/2
tan -= -
= - -= --= -
6 cos (46) J3/2 J3 J3Js 3
4 cos (d4) fi/2
= 1
tan -= -
=
=--
-8
tan - = -
Notice that tan ( 4 2 )is not defined, since sin ( 4 2 )= 1and cos (n/2) = 0. Moreover,
sin x sin x
x-qn/2)- x--r(x/2)+ cos x x + n + / 2 x-+n+/2 cos x
lim tan x = lim -- - +a and lim tanx = lim -- - -00
because cos x > 0 for x immediately to the left of n/Z and cos x <0 for x immediately to the right of
n/2.
Table 28-1
214

CHAP. 281 THE TANGENT AND OTHER TRIGONOMETRIC FUNCTIONS 215
Tkurem 28.1: The tangent and cotangent functionsare odd functionsthat are periodic,of period A.
That they are odd followsfrom Theorem 26.3,
sin (- x) -sin x
tan (-x) = - - -tan x
cos (-x) cos x
- -cot x
tan (-x) - -tan x
---
1
---
1
cot (-x) =
The periodicity of period ~t
followsfrom Problem 26.15(c) and (4,
sin (x +z) -sin x
---
tan (x +A) = - tan x
cos (x +It) - -cos x
Theorem28.2 (Deriuatiues): D,(tan x) = sec2x
&(sec x) = tan x sec x
D,(cot x) = -csc2 x
D,(csc x) = -cot x csc x
For the proofs, see Problem 28.1.
EXAMPLE From Theorem 28.2 and the power chain rule,
D,2(tan x) = D,((sec x)’) = 2(sec x)D,(sec x)
= 2(sec x)(tan x sec x) = 2 tan x sec’ x
Now in (0, n/2), tan x > 0 (since both sin x and cos x are positive), making D:(tan x) > 0. Thus (Theorem 23.1), the
graph of y = tan x is concave upward on (0, 42). Knowing this, we can easily sketch the graph on (0, n/2), and
hence everywhere(see Fig. 28-1).
I
I
I
I
I
I
I
I
I
I
I
1
‘ I
I
I
Fig. 28-1

216 THE TANGENT AND OTHER TRIGONOMETRIC FUNCTIONS [CHAP. 28
Tkorem 283 (Identities): tan’ x + 1 = sec’ x and cot’ x + 1= csc’ x.
Proof: Divide sin2x +cos2x = 1 by cos’ x or sin2x.
Traditional Definitions
As was the case with sin 8 and cos 8, the supplementary trigonometric functions were originally
defined only for an acute angle of a right triangle. Referringto Fig. 26-3, we have
opposite side
adjacent side
adjacent side
opposite side
hypotenuse
adjacent side
hypotenuse
opposite side
tan 8 =
cot 8 =
sec 8 =
csc e =
SolvedProblems
DAtan x) = D,(-)sin x
cos x
(cos x)D,(sin x) -(sin x)D,(cos x)
-
- [by the quotient rule]
(cos x)2
(cos xxcos x) -(sin x)(-sin x)
cos2x
-
- [by Theorem 27.21
cos2x +sin’ x 1
- -
cos2x cos’ x
[by Theorem 26.11
= sec2x
-1
(tan x ) ~
D,(cot x) = D,((tan x)- l) = -
DAtan x) [by the power chain rule]
-1 1 -1
-1 sin x
=
-
-
=
(tan x ) ~
(cos x ) ~
(sin x)’
= -csc2 x
(tan x cos x)’
I
=- [tan x =
Differentiatingthe first identity of Theorem 28.3,
2(tan x)(sec2 x) = 2(sec x)D,(sec x)
and dividing through by 2(sec x), which is never zero, gives
D,(sec x) = tan x sec x
Similarly,differentiationof the second identity of Theorem 28.3 gives
D,(csc x) = -cot x csc x

CHAP. 281 THE TANGENT AND OTHER TRIGONOMETRIC FUNCTIONS
I
I
1 I 1
-3 -? -? -:()
217
I
I
1 1 1
I 5 & ; +
4 1 x
28.2
2
8
.
3
28.4
Draw the graph of csc x.
only find the graph for 0 c x < a. Now
Because csc x = l/(sin x), it is, along with sin x, periodic, of period 2n, and odd. Therefore, we need
- 1
n 1 1
2 sin (42) 1
csc - = -
= - -
As x decreases from n/2 toward 0, sin x decreases from 1 toward 0. Therefore, csc x increases from 1 to
+CO. Likewise, as x increases from z/2 toward n,sin x decreases from 1toward 0, and csc x increases from
1 to +a.
In fact, since sin ((a/2)+U) = sin ((42) - U), the graph will be symmetric about the line x = n/2
(see Fig. 28-2).
Fig. 28-2
tan U - tan U
1 +tan u tan U ’
Prove the identity tan (U - U) =
sin (U - U) sin U cos U -cos U sin U
cos (U - U) - cos U cos U +sin U sin U
tan (U - U) = - [by Theorem 26.63
sin U sin U
---
cos U cos U
-
- .[divide numerator and denominator by cos U cos U]
sin U sin U
1+--
tan U - tan U
1 +tan U tan U
cos U cos U
-
-
Calculate: (a)0,(3 tan2x); (b)D,(sec x tan x).
(a) D,(3 tan’ x) = 3D,(tan2 x) = 3(2 tan x)D,(tan x) [by the power chain rule]
= 6 tan x sec’ x
(b) D,(sec x tan x) = (sec x)D,(tan x) +(tan x)DJsec x) [by the product rule]
[by Theorem 28.21
[by Theorem 28.31
= (sec x)(sec2x) +(tan x)(tan x sec x)
= (sec x)(seczx +tan’ x)
= (sec x)(sec2x +tan’ x)
= (sec x)(tan2x + 1+tan’ x)
= (sec x)(2 tan’ x + 1)

28.5 A lighthouse H, 1mile from a straight shore, has a beam that rotates counterclockwise(as seen
from above) at the rate of six revolutions per minute. How fast is point P, where the beam of
light hits the shore, moving?
Let A be the pont on shore opposite H, and let x be the distance AP (see Fig. 28-3). We must find
dx/dt. Let 0 be the measure of 3:PHA. Since the beam makes six revolutions (of 2n radians) per minute,
--
- 12n radians per minute
d8
dt
Now
Hence,
I
opposite side
adjacent side
dx dt?
- D,(tan 0) = &(tan 0) -
dt dt
- - [by the chain rule]
1
hypotenuse
adjacent side
= (sec2 0x127c) = ( ~ ~ ~ i ) 2 ( 1 2 n )
[since sec e =
= 12n(x2+ 1)
For instance, when P is 1 mile from A, the light is moving along the shore at 24n miles per minute (about
4522 miles per hour).
28.6 The angle o
f inclination of a nonvertical line 2' is defined to be the smaller counterclockwise
angle a from the positive x-axis to the line (see Fig. 28-4). Show that tan a = m,where m is the
slope of 9.
By taking a parallel line, we may always assume that 9 goes through the origin (see Fig. 28-5). 9
intersects the unit circle with center at the origin at the point P (cos a, sin a). By definition of slope,
sin a - 0 sin a
cosa-0 cosa
=--
m = - tan a

I
I
A X P
'

,
Fig. 28-3 Fig. 28-4
28.7 Find the angle at which the lines S1
: y = 2x + 1 and s2:y = -3x +5 intersect.
Let a1 and a2 be the angles of inclination of LZ1and S2(see Fig. 28-6),and let ml and m2 be the slopes
of Y1
and g2.
By Problem 28.6, tan a, = rn, = 2 and tan a2 = m2 = -3. a2 -a1 is the angle of intersec-
tion. Now
tan a2 - tan a1
1 +tan a, tan a2
m2 - m1
1 +mlm2
tan (a2- a l ) =
-
-
[by Problem 28.33
[by Problem 28.63
1
-5 -5
1+(-3x2) 1 - 6 -5
=-=-=
-3 - 2
-
-

CHAP. 281 THE TANGENT AND OTHER TRIGONOMETRIC FUNCTIONS 219
9 7 t'
Fig. 28-5
Since tan (a2- al) = 1,
Fig. 28-6
II
a2 - al= - radians = 45"
4
In general, given tan (a2 - al), the value of a2 - alcan be estimated from the table in Appendix D.
It should be noted that, in certain cases,the above method will yield the larger angle of intersection.
28.8 Sketch the graphs of: (a)sec x; (b)cot x.
tan U +tan U
1 -tan U tan U'
28.9 Prove the identity tan (U +U ) =
28.11 Find y' by implicit differentiation:
(4 tan ( X Y ) = Y (b) sec' y +csc2x = 3
(c) tan2 (y + 1) = 3 sin x (4 y = tan2 (x +y)
28.12 Find an equation of the tangent line to the curve y = tan x at the point ( 4 3 ,fi).
28.13 Find an equation of the normal line to the curve y = 3 sec2x at the point (z/6,4).
tan x tan3 2x sin 3x
28.14 Evaluate: (a) lim - (b) lim - (c) lim -
x-ro x x-ro x3 x-ro tan 4x

28.15 A rocket is rising straight up from the ground at a rate of loo0 kilometers per hour. An observer 2
kilometers from the launchingsite is photographing the rocket (see Fig. 28-7). How fast is the angle 8of the
camera with the ground changingwhen the rocket is 1.5 kilometers above the ground?
0
’
0
’
0
’
0
0
’
Y
0
0
’
2
Fig. 28-7
/-
A/?@
2
Fig. 28-7
28.16 Find the angle of intersectionof the lines Yl: y = x - 3 and Y 2 : y = -5x +4.
28.17 Find the angle of intersection of the tangent lines to the curves xy = 1 and 9 = x3 at the common point
(1, 1).
28.18 Find the angle of intersection of the tangent lines to the curves y = cos x and y = sin 2x at the common
point (46, &2).
28.19 Find an equation of the tangent line to the curve 1 + 16x2y = tan (x -2y) at the point (n/4,0).
28.20 Find the relative maxima and minima, inflection points, and vertical asymptotes of the graphs of the follow-
ing functions,on [O,a] :
(a) f ( x ) = 2x -tan x (6) f ( x )= tan x - 4x
2.8.21 Find the intervalswhere the functionf(x) = tzn x -sin x is increasing.
28.22 Use Newton’s method to approximatesolutions of the followingequations: (a)sec x = 4 on CO, a/2];
(b) tan x - x = 0 on [n,3n/2]; (c) tan x = l/x on (0,a).

Chapter 29
Antiderivatives
29.1 DEFINITION AND NOTATION
Definition: An antideriuatiue of a functionf is a function whose derivative is$
EXAMPLES
(a) x2 is an antiderivativeof 2x, since D,(x2) = 2x.
(b) x4/4 is an antiderivativeof x3, since D,(x4/4) = x3.
(c) 3x3 - 4x2 +5 is an antiderivativeof 9x2 - 8x, since D,(3x3 - 4x2 +5) = 9x2 - 8x.
(d) x2 +3 is an antiderivativeof 2x, since D,(x2 +3) = 2x.
(e) sin x is an antiderivativeof cos x, since D,(sin x) = cos x.
Examples (a)and (d) show that a function can have more than one antiderivative. This is true for all
functions. If g(x) is an antiderivative off(x), then g(x) +C is also an antiderivative off(x), where C is
any constant. The reason is that Dx(C)= 0, whence
Let us find the relationship between any two antiderivatives of a function.
Theorem29.1: If F'(x) = 0 for all x in an interval Y, then F(x)is a constant on 9.
The assumption F'(x) = 0 tells us that the graph of F always has a horizontal tangent. It is then
obvious that the graph of F must be a horizontal straight line; that is, F(x) is constant. For a rigorous
proof, see Problem 29.4.
Corollary 29.2: If g'(x) = h'(x) for all x in an interval Y, then there is a constant C such that
g(x) = h(x)+C for all x in Y.
Indeed,
D,(g(x) - h(x))= g'(x) - h'(x) = 0
whence, by Theorem 29.1,g(x) - h(x)= C, or g(x) = h(x) +C.
Thus, if we know one antiderivative of a function, we know them all.
According to Corollary 29.2, any two antiderivatives of a given function differ only by a constant.
NOTATION If(.) dx stands for any antiderivativeofJ Thus,
OTHER TERMINOLOGY Sometimes the term indefinite integral is used instead of antiderivative, and the process of
finding antiderivativesis termed integration. In the expression I f ( x )dx,f(x)is called the integrand.The motive for
this nomendature will become clear in Chapter 31.

222 ANTIDERIVATIVES [CHAP. 29
EXAMPLES
x3
(a) Ix2 dx = 7+C. Since D,(x3/3) = x2, we know that x3/3 is an antiderivativeof x2. By Corollary 29.2, any
other antiderivativeofx2 is of the form (x3/3) +C, where C is a constant.
(b) !cos x dx = sin x +C
(c) Isin x dx = -cos x +C
(d) jsec2 x dx =tan x +C
(e) I O d x = C
(f) [ l d x = ~ + C
29.2 RULES FOR ANTIDERNATIVES
spondingrulesfor antiderivatives.
The rules for derivatives-in particular,the sum-or-differencerule and the chain rule-yield corre-
RULE 1. [ a dx = ax +cfor any constant a.
EXAMPLE
3 dx = 3x + c
s
A
RULE 2. J xr dx = -+C for any rational numberr other than r = -1.
r + l
NOTE The antiderivativeof x- will be dealt with in Chapter 34.
Rule 2 followsfrom Theorem 15.4,accordingto which
EXAMPLES
1 1
2
X-2
-2
I-$dx = J x - ~dx = -+ c -- x - 2 + c = + C
2x2
This followsfrom Dx( a [f ( x )d x ) = a Dx( 1f ( x )dx) = af(x).

CHAP. 291 ANTIDERIVATIVES 223
EXAMPLE
Notice that we finda specific antiderivative,x3/3 +x4/4, and then add the “arbitrary”constant C.
Rules 1through 4 enable us to compute the antiderivative of any polynomial.
1 1 xs
EXAMPLE J(3x5 -2 x4 +7x2 +x - 3) dx = 3 ( 3 - -
2 5
(-) + 7 ( 3 +2-
2 3x + c
x6 x5 7 x2
2 10 3 2
+- x3 +-- 3x +c
-
_ - _ -
The next rule will prove to be extremelyuseful.
(g(x))’+ + c
RULE 5 (Quick Formula I). (g(x))’g’(x)dx = r-fl
s
The power chain rule implies that
which yields quick formula I.
EXAMPLES
J(; x2 +5)7x dx =; (; xz + 5)’ +c
1 (2x - 5)3’2 1
2 3
+c = - (2x - 5)3’z +c
1
3
{J- dx = 5I ( 2 x - 5)’12(2)dx = -
RULE 6 (SubstitutionMethod). Deferring the general formulation and justification to Problem 29.18,
we illustrate the method by three examples.
(i) Find x2 cos x3 dx. Let x3 = U. Then, by Section21.3, the differentialof u is given by
1
3
du = Dx(x3)dx = 3x2 dx or x2 dx = - du
Now substitute U for x3 and idu for x2 dx,
1 1
cos u du = - sin u + c = - sin x3 +C
x2 cos x3 dx = - cos u du =
s s: 3 3

(ii) Find j(x2+3x - 5)3(2x+3) dx. Let U = x2 +3x - 5, du = (2x +3) dx. Then
u4 1
(x2+3x - 5)3(2~
+3) dx = u3 du = -+C = - (x2+3x - 5)4 +C
s s 4 4
(iii) Find sin2x cos x dx. Let u = sin x. Then du = cos x dx, and
u3 sin3x
sin2x cos x dx = u2 du = -+ c = -+ C
s s 3 3
Notice that quick formula I (Rule 5) is a special case of Rule 6, corresponding to the substitution
U = g(x). The beauty of quick formula I is that, when it is applicable, it allows us to avoid the bother of
going through the substitution process.
SolvedProblems
29.1 Find the following antiderivatives:
(a) /(fi
- 5x2)dx = (X1l3 - 5x2)dx
sx4/3
s
-
--- 5(;) +C [by Rules 2 and 41
(b) +p-2) dx = 5(4. + - 2) dx = 4 ( 3 +T-
2x +c
z
2
7
= 2x2 +-X1l2 - 2x +c
(c) /(x2 - sec2 x) dx = tan x +C
3
dx = [(?+3x) dx = 2 !x-'/' dx +3 s x dx [by Rules 1 and 41
x1/2 x 2 3
2 2
= 2 9
+ 3 -+c = 4J;; +- x2 +c

CHAP. 291 ANTIDERIVATIVES 225
(a) Notice that DJ2x3 -x) = 6x2- 1. So, by quick formula I,
1
(2x3- ~ ) ~ ( 6 x ~
- 1) dx = - (2x3- x)' +C
I 5
(b) Observe that Dx(5x2- 1) = 1Ox. Then, by Rule 1,
J ~ G Z dx = ( 5 2 - 1 p 3 dx = -J ( 5 2 - 11113 iox dx
10
1 ( 5 2 - 1)4/3 +
10 4
p y quick formula I]
I
=Ad-+c
40
=-
3 3
40 40
= -pX2
- 11413 +c = -(
4
-
1
~+c
(For manipulations of rational powers, review Section 15.2.)
293 Use the substitution method to evaluate:
(a) Let u = &.Then,
1
du = DJ&) dx = Dx(X"2) dx = - x-1/2 dx = -
2 2> dx
d x = 2 sinudu= - ~ C O S U + C = - ~ C O S & + C
s
Hence,
(b) Let U = 3x2 - 1. Then du = 6x dx, and
x sec2(3x2 - 1) dx =-
1 1
s 6
'
I 6 6
sec2u du = - tan u + C = - tan (3x2 - 1) +C
(c) Let u = x +2. Then du = dx and x = u - 2. Hence,
Ix 2 , / z dx = I(U - 2)2& du = I(u2 -4u +4)u1I2du
= { ( u ~ / ~
- 4u3I2+4u1/') du
2 8 8 2 8 8
-
-- u7i2 -- u5i2 + - u ~ i z
7 5 3
[by du" = ur+q
+c = (x +2)7/2- (x +2)5/2+ (x +2)3/2+c
The substitution U = ,
/
=
would also work.
Let a and b be any two numbers in 9.
By the mean-value theorem (Theorem 17.2), there is a number c
between a and b, and therefore in 9,
such that
But by hypothesis, F'(c)= 0; hence, F(b)-F(a) = 0, or F(b)= F(a).

29.5 A rocket is shot straight up into the air with an initial velocity of 256 feet per second. (a)When
does it reach its maximum height? (6) What is its maximum height? (c) When does it hit the
ground again? (d)What is its speed when it hits the ground?
In free-fall problems, U = I a dt and s = 5 U dt because, by definition, a = du/dt and U = ds/dt. Since
a = -32 feet per second per second (when up is positive),
U = 1 - 3 2 dt = -32t +C,
s = ji-32t +C,) dt = (-32) +C,t +C, = -16t2+C,t +C,
t2
in which the values of C1 and C
, are determined by the initial conditions of the problem. In the present
case, it is given that 40)= 256 and 40)= 0. Hence, 256 = 0 +C, and 0 = 0 +0+C, ,so that
U = -32t+256 (1)
s = -16t2+256t (2)
(a) For maximum height, ds/dt = o = -32t +256 = 0. So,
- 8 seconds
256
t = - -
32
when the maximum height is reached.
(b) Substituting t = 8 in (2),
(c) Setting s = 0 in (2),
-16t2+256t = 0
-16t(t - 16) = 0
t = O or t = 1 6
The rocket leaves the ground at t = 0 and returns at t = 16.
(d) Substituting t = 16 in (Z), 416) = -32(16) +256 = -256 feet per second. The speed is the magnitude
of the velocity,256 feet per second.
29.6 Find an equation of the curve passing through the point (2, 3) and having slope 3x3 - 2x +5 at
each point (x, y).
The slope is given by the derivative.So,
dY
-
dx = 3x3-2x + 5
3
4
Y = l ( 3 2 - 2x + 5) dx = -x4 - x2 + 5x +C
Hence,
Since(2,3) is on the curve,
3
4
3 = - (2)4 -(2), +5(2) + C = 12 - 4 + 10 + C = 18 +C
Thus, C = -15, and

CHAP. 29) ANTIDERIVATIVES 221
(U) ](2x3 - 5x2 +3~ + 1)dx (c) 5 2 f i dx
(6) 1 5 P d x (4 j$dx (f) j ( x ' - 1)fidx
(0) I x ( x 4 +2)2dx
s
(9) I($-+
U) s ( 7 sec2 x - sec x tan x) dx
(m)I
' dx
(h) s"'-;+ 1 dx
(k) (csc2 x +3x2) dx
(n) !tan2 x dx
(i) I ( 3 sin x +5 cos x) dx
(0 1x ~ 3 x
dx
sec x
[Hint:
Use Theorem 28.3 in (n).)
29.8 Evaluate the followingantiderivatives by using Rule 5 or Rule 6. [In (m),a # 0.1
X
(d) Isin (3x - 1) dx (e) /sec2 5dx
(9) I ( 4 - 2t2)7t dt (h) sx2,./;'Ti dx (i) X dx
x + l
(f) l v d x
U) 1d x m , dx (k) (x* + 1)ll3x7dx
s
(m)I.,/=
dx
(p) J(3x - 5)l2x dx
(s) 1ssec2 2dx
(n) I & X
cos 3x
(4) "4 - 7t2)7tdt
(0) s,/G
x2 dx
sin (l/x) cos (l/x)
dx
3
29.9 A rocket is shot vertically upward from a tower 240 feet above the ground, with an initial velocity of 224
feet per second. (a)When will it attain its maximum height? (b) What will be its maximum height? (c)When
will it strike the ground? (d) With what speed will it hit the ground?
29.10 (Rectilinear Motion, Chapter 18) A particle moves along the x-axis with acceleration a = 2t - 3 feet per
second per second. At time t = 0, it is at the origin and moving with a speed of 4 feet per second in the
positive direction. (a) Find a formula for its velocity U in terms of t. (b) Find a formula for its position x in
terms of t. (c) When and where does the particle change direction? (d) At what times is the particle moving
toward the left?
29.11 Rework Problem 29.10 if a = t2 - 9feet per second per second.
29.12 A rocket shot straight up from ground level hits the ground 10 seconds later. (a) What was its initial
velocity? (b) How high did it go?

29.13 A motorist applies the brakes on a car moving at 45 miles per hour on a straight road, and the brakes cause
a constant deceleration of 22 feet per second per second. (a) In how many seconds will the car stop? (b)
How many feet will the car have traveled after the time the brakes were applied? [Hint: Put the origin at
the point where the brakes were initially applied, and let t = 0 at that time. Note that speed and deceler-
ation involvedifferent units of distance and time; changethe speed to feet per second.]
29.14 A particle moving on a straight line has acceleration a = 5 - 3t, and its velocity is 7 at time t = 2. If s(t) is
the distancefrom the origin at time t, find 42)-$1).
29.15 (a) Find the equation of a curve passing through the point (3, 2) and having slope 2x2 - 5 at point (x, y),
(b) Find the equation of a curve passing through the point (0, 1)and having slope 12x + 1 at point (x, y
)
.
29.16 A ball rolls in a straight line, with an initial velocity of 10 feet per second. Friction causes the velocity to
decrease at a constant rate of 4 feet per second per second until the ball stops. How far will the ball roll?
[Hint: U = -4 and uo = 10.1
29.17 A particle moves on the x-axis with acceleration u(t) = 2t - 2 for 0 I; t I; 3. The initial velocity uo at t = 0
is 0. (a) Find the velocity Nt). (6)When is u(t) c O? (c) When does the particle change direction?(d)Find the
displacement between t = 0 and t = 3. (Displacement is the net change in position.) (e) Find the total dis-
tance traveled from t = 0 to t = 3.
29.18 Justify the followingform of the substitutionmethod (Rule6):
If(g(x))c?’(x)
dx = 1f(4du
/
where U is replaced by g(x) after integration on the right. The “substitution” would be applied to the
left-hand side by letting U = g(x) and du = g’(x) dx. [Hint: By the chain rule,

Chapter 30
The Definite Integral
30.1 SIGMA NOTATION
The Greek capital letter Z is used in mathematics to indicate repeated addition.
EXAMPLES
99
X i = 1 +2 +3 +-..+99
that is, the sum of the first 99 positive integers.
c ( 2 i - 1)= 1 + 3 + 5 + 7 + 9 + 11
that is, the sum of the first six odd positive integers.
i = 1
6
i=
1
5 5
c 3 i = 6 +9 + 12 + 15 = 3(2 +3 +4 + 5) = 3
c
i
i = 2 i = 2
j2 = l2+22 +32 + * . * + 152 = 1 +4 +9 +
15
j = 1
5
+ 225
C sinjn = sin n +sin 2n +sin 311 +sin 4n +sin 5n
j = 1
In general, given a functionfdefined on the integers,and given integers k and n 2 k,
n
f(i) = f ( k ) + f ( k + 1) + -..+f(n)
i = k
30.2 AREA UNDER A CURVE
Letfbe a function such thatf(x) 2 0 for all x in the closed interval [a, b]. Then its graph is a curve
lying on or above the x-axis (see Fig. 30-1). We have an intuitive idea of the area A of the region 9 lying
under the curve, above the x-axis, and between the vertical lines x = a and x = 6. Let us set up a
procedure for finding the value of the area A.
Select points xl, x2,...,x,- inside [a, 61 (see Fig. 30-2). Let xo = a and x, = 6,
a = xo <x1 < x2 < * ' < x,-1 <x, = b
These divide [a, b] into the n subintervals [xo, xl], [xl, x2], ..., [ x ~ - ~ ,
x,]. Let the lengths of these
subintervalsbe Alx, A2x, ...,A,,x, where
Aix E xi - x1-1
Draw vertical lines x = xi from the x-axis up to the graph, thereby dividing the region 9 into n strips. If
AiA is the area of the ith strip, then
n
A = C A i A
I=1
229

230 THE DEFINITE INTEGRAL [CHAP. 30
1
-
Y
a
Fig. 30-1
Y
/
/
x = b
b X
/
-
I X
b
&-1 a
...
Fig. 30-2
Approximate the area AiA as follows. Choose any point x,* in the ith subinterval [xi-1, xi] and draw
the vertical line segment from the point x,* up to the graph (see the dashed lines in Fig. 30-3); the length
of this segment is f(x,*).The rectangle with base Aix and height f ( x f ) has area f(x,*)Aix, which is
approximately the area AiA of the ith strip. So, the total area A under the curve is approximately the
sum
(30.1)
The approximation becomes better and better as we divide the interval [a, b] into more and more
subintervals and as we make the lengths of these subintervals smaller and smaller. If successive approx-

CHAP. 3
0
1 THE DEFINITE INTEGRAL 231
imations get as close as one wishes to a specificnumber, then this number is denoted by
[f(X) dx
and is called the deJinite integral o
f ffrom a to b. Such a number does not exist for all functions5 but it
does exist,for example,when the functionfis continuous on [a, 61.
Y
a x? x i
I
-
x i x i ... x : b X
Fig. 30-3
EXAMPLE Approximating the definite integral by a small number n of rectangular areas does not usually give
good numerical results. To see this, consider the functionf(x) = x2 on CO, 13.Then x2 dx is the area under the
parabola y = x2, above the x-axis, between x = 0 and x = 1. Divide CO, 13 into n = 1
0 equal subintervals by the
points 0
.
1
,0
.
2
,..., 0
.
9 (see Fig. 30-4).Thus, each Aix equals 1/10.In the ith subinterval, choose xi* to be the
left-hand endpoint (i - l)/lO.Then,
6’
1 10
= 2
- (i - 1
)
2 [by example (c)above]
loo0 t=1
1 1
loo0 loo0
= -(0+ 1 +4 + * * - +81)= -(285)
= 0.285
As will be shown in Problem 3
0
.
2
,the exact value is
1
s,’x2 dx = 3 = 0
.
3
3
3...
So the above approximation is not too good. In terms of Fig. 30-4,there is too much unfilled space between the
curve and the tops of the rectangles.
Now, for an arbitrary (not necessarily nonnegative)functionfon [a, b], a sum of the form (30.1)can
be defined, without any reference to the graph off or to the notion of area. The precise epsilon-delta
procedure of Problem 8.4(a) can be used to determine whether this sum approaches a limiting value as n
approaches c
o and as the maximum of the lengths Aix approaches 0. If it does, the functionfis said to

In the followingsection, we shall slateseveral prapertksof the definite integral, omittins any proof that
depends on the ptadse definition in favor o
f the intuitive picture o
f the definite integral as an area
[whmf(x) 2 01.
3
0
.
3 PROPERTIES OF THE DEFINITE ~ , G R A L
Thrmam3U.f: Jffis continuouson [ab].thenfir integrrbk on [ab].
TlcIbnn163: lgwdx =c l / ( x l dx for any constant c.
Obviously, since the rqxctivt approximatingsumscnjoy this relationship[cxampk (c) above]. the
h i t s enjoy it as well.
EXAMPLE Supoor that f(x) 5 0 lor all x m [ab] Thc graph d/4lang with its mirror image, he graph o
f
-j-ts shownin Fig 3 M . Siou -f(x) 2 0,
[-f(x) dx -area 2)
sb

CHAP. 301 THE DEFINITE INTEGRAL 233
But by symmetry,area B = area A; and by Theorem 30.2, with c = -1,
[-fW dx = -I'm dx
It followsthat
[j(x) dx = -(area A)
In other words, the definite integral of a nonpositive function is the negatioe of the area aboue the graph of the
function and below the x-axis.
Theorem 30.3: Iffand g are integrable on [a, b],then so aref+ g andf- g, and
I"(f(x, _+ g(x))dx = l'fW dx fI"(.,dx
Again, this property is implied by the corresponding property of the approximating sums,
CW)_+ Q(91= P(i)_+ Q(i)
i= 1 I=1 i = 1
Theorem 30.4: If a < c < b and iffis integrable on [U, c] and on [c, b], thenfis integrable on [a, b],
and
[f(x) dx = p ( x )dx +I"f(4dx
Forf(x) 2 0, the theorem is obvious: the area under the graph from a to b must be the sum of the areas
from a to c and from c to b.
EXAMPLE Theorem 30.4 yields a geometric interpretation for the definite integral when the graph off has the
appearance shown in Fig. 30-6. Here,
p x ) dx = I"m dx +p x ) dx + p x , dx +p x ) dx +I:f(X) dx
= A , - A2 +A, - A, +A,
That is, the definiteintegral may be considered a total area, in which areas aboue the x-axis are counted as positioe,
and areas below the x-axis are counted as negatioe.Thus, we can infer from Fig. 27-2(b)that
['sin x dx = 0
because the positive area from 0 to ?I isjust canceled by the negaive area from n to 2n.
t'
a
Fig. 30-6
Arbitrary Limitsof Integration
Extend the definition as follows:
In defining C f ( x )dx, we have assumed that the limits o
f integration a and b are such that a < b.

234 THE DEFINlTE INTEGRAL (CHAP.
[I) E
/
(
.
) dx = 0.
(2) If a > b, let g j ( x ) dx = -kf(x) dx (with the definite integral on the right falling under the
originaldefinition).
Under this extendaddefinition. iruerchgitq rhr limits ofintegrution in any definite ktcgrul rewTscs
rhc ulg4b*Qicsign o f r h integral. Moreover, the equationsof Thaorrms 302.30.3, and 30.4 now hold for
arbitrarylimits of integrationU,b,and E.
solved Prokm
30.1 Show that 1dx = b -a.
I”
Forany subdivisiona =xoc x1 c x2 < --- c x,- ,< x, = bo
f [a,bj, the approximatingsum (M.t)is
f(x:) A,X - A+ [since/(x) = 1 for all XJ
1-1 I - I
-(x, -x*] +(x, -x,)+(xg -x,) + - * - +(x.-x”-,) -x, - x
, =b -4
Sinceevery approximatingsum isequal 10b -a.
[l dx = b -11
As an dtmatiw, intuitive proof, note that Q 1 dx is equal to the 8
m o
f a rectarde with bart o
f length
h-rr a d height 1, since the graph of the conslant function 1 is the line y = 1. This area is
(b-a)(l]-b -a (see Fig. W
7
)
.
U
I b r
,which
@
I + 1K2n+ 1)
6
303 Calculate r x 2dx. [
.
.
U may assume the formula Is +22 + +r ~ *
=
1
J
isestablishedin Problem30.12(a.it).
Divide the iaterval [O. 1J into m cqlul pmtr, as indicated in Fig. 30-8. makinga c h A,x = I/n. In the
Hh subinterval [(i -1 ) hi/n], let xf be the right mdpoint i/n. Then (30.1)bccomcs

CHAP. 30) THE DEFINITE INTEGRAL 235
t Y
Fig. 30-8
We can make the subdivisionfiner and finer by letting n approach infinity. Then,
This kind of direct calculation of a definite integral is possible only for the very simplest functionsf(x).
A much more powerful method will be explained in Chapter 31.
30.3 Letf(x) and g(x) be integrable on [a, 6).
Iff(x) 2 0 on [a, 61,show that
I ' f ( x ) dx 2 0
I
'
f
(
.
)
dx [e(x) dx
l
Iff(x) I g(x) on [a, 61, show that
If rn < f ( x ) 5 M on [a, 61, show that
m(b - U ) I f(x) dx IM(6 - U )
The definite integral, being the area under the graph off, cannot be negative. More fundamentally,
every approximating sum (30.1) is nonnegative, since f ( x f )2 0 and A,x > 0. Hence (as shown in
Problem 9.10),the limiting value of the approximating sums is also nonnegative.
Because g(x) - f ( x ) z 0 on [a, b],
[ M X ) -f(x)) dx 2 0 C
b
Y (41
[g(x) dx - [ f ( x ) dx 2 0 [by Theorem 30.3)
[s(x) dx 2 [
f
(
. dx
m dx I f ( x )dx < M dx [by(b)]
l I ' l
m [l dx I [ f ( x ) dx < M 1
1dx [by Theorem 30.2)
f ( x )dx 5 M(b -a) [by Problem 30.1)

30.4 Evaluate:
(a) r 8 dx (6) [5x2 dx (c) 6
’
(
.
.+4) dx
2
[Hint: Use Problems 30.1 and 30.2.1
3
0
.
5 For the functionfgraphed in Fig. 30-9,express f ( x )dx in terms of the areas A,, A,, A,.
l
tY
Fig. 30-9
30.6 (a) Show that You may assume the formula 1 +2 +.-- +n =-
n(n + ’) proved in Problem
30.12(a).Check your result by using the standard formula for the area of a triangle. [Hint: Divide the
interval CO,b] into n equal subintervals,and choose xi* = ib/n,the right endpoint of the ith subinter-
2
val.]
b2 -a2
2
(b) Show that [x dx = -
.[Hint: Use (a)and Theorem 30.4.)
5x dx. [Hint: Use Theorem 30.2 and (b).]
30.7 Show that the equation of Theorem 30.4,
I’f(x) dx + p x ) dx = [ f ( x ) dx
holds for any numbers a, b, c, such that the two definiteintegrals on the left can be defined in the extended
sense. [Hint: Consider all six arrangements of distinct a, b, c: a < b < c, a <c < b, b c a c c, b c c < a,
c < a c 6, c < b <a. Also considerthe caseswhere two of the numbers are equal or all three are equal.]
30.8 Show that 1 5 x3 dx 5 8. [Hint: Use Problem 30.3(c).]
I,’
30.9 (a) F i n d j J D dx by using a formula of geometry. [Hint: What curve is the graph of
y = 4-x2?]
(b) From part (a)infer that 0 s A 5 4. (Muchcloser estimatesof A are obtainablethis way.)
30.10 Evaluate:
3 4
(a) C(3i- 1) (b) c(3k2 +4)
i=1 k = O

CHAP. 301 THE DEFINITE INTEGRAL 237
30.11 Iffis continuous on [a, b],f(x) 2 0 on [a, b],andf(x) > 0 for some x in [a, b], show that
I’f (x) dx ’0
[Hint:By continuity, f ( x )> K > 0 on some closed interval inside [a, b]. Use Theorem 30.4 and
Problem 30.3(c).]
30.12 (a) Use mathematical induction (see Problem 12.2)to prove:
n(n + 1X2n + 1)
6
(ii) l2+22 + ...+n2 =
n(n + 1)
(i) 1 +2 + ...+n = -
2
(b) By looking at the cases when n = 1,2, 3,4,5, guess a formula for l3+23+- -- +n3 and then prove it
by mathematical induction. [Hint:
Compare the values of formula (i) in part (a)for n = 1,2, 3,4,5.)
30.13 If the graph off between x = 1and x = 5 is as shown in Fig. 30-10, evaluate 1
;f ( x )dx.
t’
Fig. 30-10
30.14 Letf(x) = 3x + 1 for 0 I
;x I1. If the interval [0, 13 is divided into five subintervals of equal length, what
is the smallest corresponding Riemann sum (30.1)?

Chapter 31
The FundamentalTheorem of Calculus
31.1 CALCULATION OF THE DEFINIT~INTEGRAL
We shall developa simple method for calculating
lf(x)dx
a method based on a profound and surprising connection between differentiation and integration. This
connection, discovered by Isaac Newton and Gottfried von Leibniz, the co-inventors of calculus, is
expressedin the following:
Theorem31J: Letfbe continuous on [a, b). Then, for x in [a, b],
is a function of x such that
A proof may be found in Problem 31.5.
Now for the computation of the definite integral, let F(x)= f ( x )dx denote some known anti-
derivative off(x) (for x in [a, b)). According to Theorem 31.1, the function f(t) dt is also an anti-
derivativeoff(x). Hence, by Corollary 29.2,
p ( t )dt = F(x) +c
for some constant C. When x = a,
Thus, when x = b,
p t ) dt = F(b)- F(a)
and we have proved:
Theorem 31.2 (Fundamental Theorem of Calculus): Let f be continuous on [a, b), and let
F(x) = j f ( x )dx. Then,
[ f ( x ) dx = F(b)- F(a)
NOTATION The difference F(b)- F(a)will often be denoted by F(x)],b,and the fundamental theorem notated as
238

CHAP. 31) THE FUNDAMENTAL THEOREM OF CALCULUS 239
EXAMPLES
(a) Recall the complicated evaluation x2 dx = in Problem 30.2. If, instead, we choose the antiderivative x3/3
6’
and apply the fundamental theorem,
(b) Let us find the area A under one arch of the curve y = sin x; say, the arch from x = 0 to x = R. With
I sin x dx = -cos x +fi the fundamental theorem gives
A = [“ sin x dx = (-cos x +fi)T = (-cos n +fi)-(-cos o +fi)
Jo J o
= [ - ( - l ) + f i ] - ( - l + f i ) = l + l + J S - J S = 2
Observe that the &terms canceled out in the calculation of A. Ordinarily, we pick the “simplest” anti-
derivative (here, -cos x) for use in the fundamental theorem.
31.2 AVERAGE VALUE OF A FUNCTION
The aoerage or mean of two numbers a, and a, is
a1 + a
2
2
For n numbers a,, a 2 , ...,a,, the average is
a, +a, + *..+ a ,
n
Now consider the functionfdefined on an interval [a,b]. Sincefmay assume infinitely many values, we
cannot directly use the above definition to talk about the average of all the values of&However, let us
divide the interval [a, b] into n equal subintervals, each of length
6 - a
n
AX = -
Choose an arbitrary point x
; in the ith subinterval. Then the average of the n numbersf(xf),f(x;), ...,
f(4
is
If n is large, this value should be a good estimate of the intuitive idea of the “average value off on
[a, 61.’’But,
f ( x f ) Ax
1if(xi*)=- [since - = -
n i = 1 b - a i = l n 6 - a
1
As n approaches infinity, the sum on the right approaches f ( x )dx (by definition of the definite
integral), and we are led to:
l b
Definition: The average value offon [a, b] is -
b - a If ( x )dx*

240 THE FUNDAMENTAL THEOREM OF CALCULUS [CHAP. 31
EXAMPLES
(a) The average value V of sin x on CO, n] is
1 1
V=- - [sin x dx =- (2) [by example (b)above]
R
2
= - % O h 4
R
(6) The average value Y of x3on CO, 13 is
1
v = 1
-
01 x 3 dx = l x 3 dx
Now x3 dx = x4/4. Hence, by the fundamental theorem,
With the mean value of a function defined in this fashion,we have the followinguseful
Tkorem 313 (Mean-Value Theoremfor Integrals): If a functionfis continuous on [a, b], it assumes
its mean value in [a, b);that is,
l b
b-a f ( x )dx =f(4
for somec such that a s c b.
For the proof, see Problem 31.4. Note that, by contrast, the average of a finite set of numbers
a,, u2,...,a, in general does not coincidewith any of the a,.
313 CHANGE OF VARIABLE I
N A DEFINITE INTEGRAL
To evaluate a definite integral by the fundamental theorem, an antiderivative j f ( x )dx is required.
It was seen in Chapter 29 that the substitution of a new variable U may be useful in finding If(.) dx.
When the substitution is made in the definite integral too, the limits of integration a and b must be
replaced by the correspondingvalues of U .
EXAMPLE Let us compute
*
Let U = 5x +4; then du = 5 dx. Consider the
Therefore,
f 9
limits of integration: when x = 0, U = 4; when x = 1, U = 9.
38
- 15 15 15
15
--
2 (33 - 23) = -
2 (27 - 8) = -
2 (19) = -
See Problem 31.6 for ajustification of this procedure.
SolvedProblems
31.1 Calculate the area A under the parabola y = x2 +2x and above the x-axis, between x = 0 and
x = 1.

CHAP. 311 THE FUNDAMENTAL THEOREM OF CALCULUS 241
Since x2 +2x 2 0 for x 2 0, we know that the graph of y = x2 +2x is on or above the x-axis between
x = 0 and x = 1. Hence, the area A is given by the definiteintegral
[(x2 +2x) dx
Evaluating by the fundamental theorem,
sin x dx. (Compare the example followingTheorem 30.4, where U = 0.)
I"""
31.2 Compute
By the fundamental theorem,
sin x dx = -cos x3:"" = 0
sincethe cosine function has period 2n.
31.3 Compute the mean value V of & on [0, 4). For what x in CO, 4) does the value occur (as
guaranteed by Theorem 31.3)?
V = I&dx = f [x'12 dx = f (ix312)y
0
This average value, 3, is the value of &when x = (3)' = 9.Note that 0 < 4f < 4.
31.4 Prove the mean-value theorem for integrals (Theorem 31.3).
Write
Let m and M be the minimum and maximum values offon [a, b].(The existenceof m and M is guaranteed
by Theorem 14.2.)Thus, rn S f ( x ) M for all x in [a, b], so that Problem 30.3(c)gives
m(b -a) s f f ( x ) dx S M(b -a) or rn s V s M
a
But then, by the intermediate-value theorem (Theorem 17.4), the value V is assumed by f somewhere in
Ca,bl.
Write
Then, g(x +h) -g(x) = [+>(t) dt - [f(t)dt
= I'f(t) dt +["f(t) dt - l'f(t) dt
= [ " f ( t ) dt
p y Theorem 30.41


By the mean-value theorem for integrals, the last integral is equal to hf(x*) for some x* between x and
x +h. Hence,
and
Now as h -+ 0, x +h +x, and so x* + x (sincex* lies between x and x +h).Sincefis continuous,
lim f ( x * )= f ( x )
h+O
and the proof is complete.
31.6 (Change o
f Variables) Consider f ( x )dx. Let x = g(u), where, as x varies from a to b, U
increases or decreases from c to d. [See Fig. 31-1; in effect, we rule out g'(u) = 0 in [c, 4
.
3 Show
that
r
[The right-hand side is obtained by substituting g(u) for x, g'(u) du for dx, and changing the
limits of integration from a and b to c and d.]
Fig. 31-1
Let F(x) = 1f ( x )dx or F'(x)=f(x)
The chain rule gives
Hence,
By the fundamental theorem,
31.7 Calculate s,"Tix dx.

Let us find the antiderivative of , / G ’ x by making the substitution U = x2 + 1. Then, du = 2x dx,
and
Hence, by the fundamental theorem,
= -
1 ( ( f i ) 3 - (Ji)3j = 3
1( 2 f i - 1)
3
ALGEBRA ($)’ = (fi)2
$= 2 d and (fi)3
= l 3 = 1
Alternate Method: Make the same substitution as above, but directly in the definite integral, changing the
limits of integration accordingly. When x = 0, U = O2 + 1 = 1; when x = 1, u = l2+ 1 = 2. Thus, the first
line of the computation above yields
1 1
3
= - (($)3 - (Ji)3) = 5( 2 f i - 1)
31.8 (a) Iffis an even function (Section 7.3), show that, for any a > 0,
dx = 2 p x )
0 dx
(b) Iffis an odd function (Section 7.3), show that, for any a > 0,
dx = 0
If U = -x, then du = -dx. Hence, for any integrable functionf(x),
NOTATION Renaming the variable in a definite integral does not affect the value of the integral:
Thus, changing U to x,
and so
f(x) dx = f( -x) dx
J-: l
= f f ( - x ) dx + [ f ( X ) dx
f ( x ) dx = f(x) dx +[ f ( x ) dx [by Theorem 30.41
C
b
Y (01
-a -a
[by Theorem 30.31

244 THE FUNDAMENTAL THEOREM OF CALCULUS [CHAP.31
(U) For a n e w runction,J(x) + j ( - x ) = Y(xA whcna.
(b) For an odd fun&a.f(x) +f
(-x) = 0,whence,
31.9 (a) LetJ(x) 2 0 on [
a
,b],and let [cr. b] be divided into II equal parts, of length AJC
= (b -u)/n,
by meansof points x,+ x,, ...,x,, [see Fig 31-2(u)].Show that
(b) U
s
ethe trapczoicial rule, with n = 10,to approximate
l2dx (- 0.333...)
(U) fhe area in the strip over the interval [x,-~,x,] is approximately tbe area o
f trapezoid ABCD id
Fig. 31.2@), which i
s
WRY Theareaof a trapezoid ofhdght Ib and hesesh, and 6
, is
1
5w,4- b3

(b) By the trapezoidal rule, with n = 10, a = 0, b = 1, Ax = 1/10, xi = i/lO,
[by arithmetic or Problem 30.12(a, ii)]
285 1
loo0 20
= -+-= 0.285 +0.050 = 0.335
whereasthe exact value is 0.333 ....l
31.10 Use the fundamentaltheorem to computethe followingdefiniteintegrals:
(a) (3x2 - 2x +1)dx (b) cos x dx (c) JU" sec2x dx
- 1 0
(d) l 6 x 3 l 2
dx
31.11 Calculate the areas under the graphs of the following functions, above the x-axis and between the two
indicated valuesa and b of x. [In part (g), the area below the x-axisis counted negative.]
II
(a) f(x) = sin x (a = 6,b =:
) (b) f ( x ) = x2 +4x (a = 0, b = 3)
(d) f ( x ) = $ z T (a = 0, b = 2)
1
(c) f(x) = - (U = 1, b = 8)
fi
(e) f ( x )= x2 -3x (U = 3, b = 5)
(g) f(x) = x2(x3- 2) (U = 1, b = 2) (h) f(x) = 4~ - x2 (U = 0, b = 3)
(f)f ( x ) = sin2x cos x
31.12 Compute the followingdefiniteintegrals:
cos x sin x dx tan x sec2x dx (c) J
-
l
1
J
- (3x - 1) dx
(d) J"',/-
0 cos x dx (e) I-21,/=x2 dx
0
9 s'Ix-11dx
- 1
(m) l , / z x 5 dx sec22x tan3 2x dx
[Hint: Apply Theorem 30.4 to part U).]
Whenfhas a continuoussecond derivative,it can be shown that the error in approximating j ( x ) dx by the traptzoidal rule is
l
at most ((b-u)/12n2)M,where M is the maximumof If"(x)Ion [U, b] and n is the number of subintervals.

246 THE FUNDAMENTAL THEOREM O F CALCULUS [CHAP. 31
31.13 Compute the average value of each of the followingfunctions on the given interval:
(b) f ( x ) = sec2x on 0
,-
(4f ( x ) = sin x +cos x on CO,n]
[ :I
(a) f ( x ) = fi on CO, 11
(c) f ( x ) = x2 -2x - 1on [-2, 31
31.14 Verify the mean-value theorem for integrals in the followingcases:
(a) f(x) = x +2 on [l, 23 (b) f ( x )= x3 on [0, 13 (c) f ( x ) = x2 +5 on CO, 3)
31.15 Evaluate by the change-of-variabletechnique:
(a) dbJS;rS.. dx (b) r2sin’ x cos x dx
31.16 Using only geometric reasoning, calculate the average value of f ( x )= ,
/
- on [0, 2
)
. [Hint: If
y =f(x), then (x - 1)2 +y2 = 1.Draw the graph.]
31.17 If, in a period of time T, an object moves along the x-axis from x1 to x2, calculate its average velocity.
[Hint: 1v dr = x.]
31.18 Find:
[Hint: In part (c),use Problem 31.8(a).]
31.19 Evaluate sin x dx.
31.20 (a) Find D,(rx2
,/mdt). [,int: With U = 3x2, the chain rule yields D,(6’,
/
- dr) =
I
D
u
(
[ ,/m
dt) g,and Theorem 31.1 applies on the right side.
Mx)
(b) Find a formula for DA[ f(r) dt).
(c) Evaluate Dx( lx
4dr) and Dx( (
i
+ 1) dt).
31.21 Solvefor b:
n
31.22 If f ( x - k)dx = 1,compute
1
[Hint:
Let x = U - k
.
]

3
1
.
2
4 Given that 2x2 - 8 = f ( t )dt, find: (a)a formulaforf(x); (b) the value of a.
I^
31.25 Define H(x)
2
(a) Find H(1) (b) Find H'(1) (c) Show that H(4)-H(2) <-
5
3
1
.
2
6 If the average value off(x) = x3 +bx - 2 on CO,2) is 4, find b.
3
1
.
2
7 Find lim (i
r + h d m
dx) .
b+O 2
3
1
.
2
8 If g is continuous,which of the followingintegrals are equal?
g(x - 1) dx (c) J y l ( x +a) dx
0
3
1
.
2
9 The region above the x-axis and under the curve y = sin x, between x = 0 and x = a,is divided into two
parts by the line x = c. The area of the left part is 3 the area of the right part. Find c.
3130 Find the value@ of k for which
f2dx = [(2 -x r dx
3131 The velocity U of an object moving on the x-axis is cos 3t. It is at the origin at t = 0.(a) Find a formula for
the position x at any time t. (b) Find the average value of the position x over the interval 0 5 t 5 a/3.
(c) For what values of t in CO, n/3] is the object moving to the right? (d) What are the maximum and
minimum x-coordinates of the object?
3132 An object moves on a straight line with velocity U = 3t - 1, where U is measured in meters per second. How
far does the object move in the period of 0 5 t 5 2 seconds?[Hint: Apply the fundamental theorem.]
31.33 Evaluate:
+sin E)
(a) lim - (sin - +sin -+6 --
(b) lim {sec2 (5)
4n +sec2 (z5)
4n + .- +sec2 ((n - 1) i
)
+2)
1 n 2a
n4+a,n n n n
n + + m
n
31.34 (Midpoint Rule) In a Riemann sum (304, f ( x f ) Aix, if we choose xf to be the midpoint of the ith
subinterval, then the resulting sum is said to be obtained by the midpoint rule. Use the midpoint rule to
approximate x2 dx, using a division into five equal subintervals, and compare with the exact result
obtained by the fundamental theorem.
i = 1
6'

31.35 (Simpson’s Rule) If we divide [a, b] into n equal subintervals by means of the points a = xo,xl, x2, ...,
x, = b, and n is even, then the approximation to f ( x ) dx given by
I’
is said to be obtained by Simpson’s rule. Aside from the first and last terms, the coefficients consist of
alternating 4‘s and 2’s. (The underlying idea is to use parabolas as approximating arcs instead of line
segments as in the trapezoidal rule.)2 Apply Simpson’s rule to approximate sin x dx, with n = 4, and
compare the result with the exact answer obtained by the fundamental theorem.
l-
31.36 Consider the integral x3 dx.
f,’
(a) Use the trapezoidal rule [Problem 31.9(a)], with n = 10,to approximate the integral, and compare the
result with the exact answer obtained by the fundamental theorem. [Hint: You may assume the
formula l3+Z3 + - -- +n3= (n(n + 1)/2)2.]
(b)
(c)
Approximate the integral by the midpoint rule, with n = 10.
Approximate the integral by Simpson’s rule, with n = 10.
Simpson’s rule is usually much more accurate than the midpoint rule or the trapezoidal rule. Iff has a continuous fourth
f ( x ) dx by Simpson’srule is at most ((b-a)’/180n4)M4,where M, is the
derivative on [a, b], then the error in approximating
maximumof If(*)(x)Ion [a, b] and n isthe number of subintervals.
l

Chapter 32
AppClcatlonsof IntegrationI:Area and Arc Length
321 AREA R h m E N A CURVE A N D THE FAXIS
what happenswhen x andy are intcrchanbd.
We have leamad how to And the area o
fa region like that shown in Fig.32-1. Now let us consider
EXAMPLES
(U) The graph ofx = y
' + 1 is 8 parabola. 4 t h its 'nod at (I, 0) and the positive x-axis ,
a
s its axis o
f symmetry
(seeFig 32-21. Consider the regionacoruistingofall points to the left of this graph. to thc right of chc y-axis,
a d between y --1 and p = 2. If me apply the rcamning urad 10 calculate the n r a d a region like that
shown in Fig. 32.1, but with x and y interchar@, we must integrate "along the y-axis." Thus,the area d i
s
givenby the definite integral
Thefundamental theorem g i w
(b) Find the area afthe region above tbc line y = x -3 in the f
i
a qwdnnt and b
e
l
m the line y = 4 (the shaded
regimof Fig. 32-31,Thinkingofx asa function o
fy. namely, x = y + 3. we can e x p m the a m as
C
l
m
k result by computing the a m oftnpezaid OBCD by the gamctricrl formulagiwiin Probkm31.9.
249

250 APPLICATlONS OF INTEGRATION I: AREA AND ARC LENGTH [CHAP. 32
3
2
.
2 AREA BETWEEN TWO CURVES
Assum that 0 S dx) SI(x) for x in [u,b]. La us find the area A of the region 9
P consisting o
fall
points between the graphs of y = g(x) and y =f(x), and between x =a and x = 6
.A
s may be 8een from
Fig. 32-4,A is the area under the upper curve y =f(x) minus the a m under the lower curve y = dx);
that is,
(32.1)
figurn 32-5 shows the region i# under the line y = fx +2.8boVC the parabola y =xa, and berwern
EXAMPLE
thc y-axis and x -1. Its U#L i
s
- +30)-E)
-(
;
+ a,,--- 3
3
I 1 9 1 27-4 23
=-+2------=-=-
4 3 4 3 12 12
Formula(32.1) i
sstill valid when the condition on the two functionsis relaxed 10
dx) 5f(XI
that is, when the curvesarc allowed to lie partly or totally below the x-axis, as in Fig. 32-6.
Sec Problem
32.3 for a proof of thisstatemeat
Another application of(32.1) is in findingthe a m o
f a regionencloscdby two curves.

CHAP. 321 APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH 251
EXAMPLE Find the area d the region bounded by the paraholay =x* and the tine y =x +2 (sec Fig 32-7).
respectively.These arc found by solvingsimultamously the equations ofthe c u m y =xz and y = x +2
. Thus,
The limits of in1egralic.w U and h in (32.1)musi be rhe xcoordinatcs o
f the inkmc1ion points P and Q
.
x * = x + 2 or x z - x - 2 - 0 or ( ~ - 2 N x + l ) = O
whm%x D - 1 WVJX = b 2. Th&,
4.v
3
2
.
3 ARC LENGTH
Considera differentiable (notjust continuous)functionfon a closed interval [U, b]. The graph off
isa curve running from (o.f(u))
to (b,f(b)). We shall find3 formula for the length L of this curve
Dividc [U, h] into n equal parts, each of lmgrh Ax.To each xt in this subdivision corresponds the
point PAX,,/(xi)) on the curve (see Fig. 32-8). For large n, the sum POP,+ v,
+ .-.+Pm- s
E- P,-rYi ofthe lengths of the line segments P,- ,P,
is an approximation to the length of the curve.
Now,by the distance formula (2.1).
-
-
p
4 -1 = J(xi - xi-a)* +(/(xi) - j ( x i -
But x4-xi-l = Ax;also. by the mean-valuetheorem (Theorem 17.2).
I(x3-f(x4- 1) = Ixr -xi- a).f’(xF) = 4W.f’(%’)
for some xr in(x,- xi). Hence.
f,-
*pi= J(W’ + (Wmm2
= J{1 +~Y(XW)(WZ
= J
- rn= J
- Ax

252 APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH [CHAP. 32
The right-handsum approximatesthe definite integral
Therefore, letting n + GO, we obtain
L = [,/m
dx arc-lengthformula
EXAMPLE Find the arc length of the graph of y = x312from (1, 1) to (4,8).
We have
Hence,by the arc-lengthformula,
4
9
dx = - du
9
4
du = - dx
9
Let u = 1 + 4 x
When x = 1, U = 9;
when x = 4, U = 10.Thus,
where, in the next-to-laststep, we have used the identity(J)3 = (&)2(J) = CA.
SolvedProblems
(32.2)
32.1 Find the area A of the region to the left of the parabola x = -y2 +4 and to the right of the
y-axis.

CHAP.321 APPLICATIONS OF INTEGRATION 1: AREA AND ARC LENGTH 253
The rqbn is shown in Fu.32-9.Nolice thai the parabola cuts the p
a
x
b a1 y = f2
. (Sec x =0 in the
equation ofthe curvc.)Henoe.
A-[-'I(-Y-t4)dy=2 [byPrabkm31.8yo)]
3
2
.
2 Find the area of thc region bctween the c u m s y = x3 and y = 2x,between x = 0 and x = 1 (sec
Fig. 32-10).
For0 s x IS 1,
2x- x' -1 2 -2 1 -&/i+
xxfi -x) 2 0
siaoC all three faaOr5 arc nonnegative Thus, y -x' is the lower curve, and y =2x b the upper curve.By
(32.1),
3
2
.
3 Provethat the formula I
' the a m A = (f(x) -A%))
dx holds whenever e(x) s f ( x )on [u,b].
I'
A = [ M X l + ImI) -Mx) -F 1mI11d
-
K
-[VI.) -dx) +0)dx =
Lel m <0 bc the abdoklk minimum of g on [a, b] [see Fig. 32-1I(u)].(If m 2Q both curves lie above
o
r on the x-axis. and this case balready known.) "Raise" both curyes by I mI units; the new graphsshown
in Fig. 32-1I(& arc om o
r above the x-axis and inclrlde the OUM arm A ac the original graphs. Thus, by
(32.1).

254 APPLICATIONS OF INTEGRATION I: AREA A N D ARC LENGTH [CHAP. 32
I
I
I'
1
!
L
b .
Y
I
-2 -I 0
tr
I
2 X
I
0
y = - ( 2 - I)
I
I I 'b)+I d
a
(h)
F i i 32-11
b x
3 U Find the area A betweenthe parabolasy = x* -1and y = -(x2 -1)
From the symmetry of Fw 32-12 it bclear thal A will bc equal to four times the area o
f the shaded
resion.
325 Find the area between the parabla x =y2 and the liney = 3x -2 (seeFig. 32-13).
Find the intersectionpoints. x = J? and y = 3x -2 imply
y = 3)J -2
3 f - y - 2 - 0
(3y +2)Q -1) -0
3 y + 2 = 0 or y-1-0
or Y ' 1
2
y = --3
A Y
t9
F
e
.3212 Fit.3t13

CHAP. 321 APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH 255
Notice that we cannot find the area by integrating “along the x-axis” (unless we break the region into two
parts). Integration along the y-axis is called for (which requires only the ordinates of the intersection
points),
A = Cl3
(F
-Y.) dy = I’
-213 (
l
+
2- p) dy
Here the “upper” curve is the line y = 3x - 2. We had to solve this equation for x in terms of y, obtaining
x = (y +2)/3. Evaluating by the fundamental theorem,
1 1
1 22 81 +44 125
--
- 2 + 8 1 = 1 6 2 = 1 6 2
x3 1
6 2x
3
2
.
6 Find the length of the curve y = -+-from x = 1to x = 2.
x2 1 x2 1 ’
y’ = --- x-2 = ---
x3 1
y = -+- x-1
6 2 2 2 2 2x2
Then,
x4 1 1
(y’)2 = -- -
4 2 + 7
1 +01’)2 =4
x4 +T1 +2
1 = (
; +&y
Hence, the arc-length formula gives
L = f 6
’
(
2
+x-2) dx =
32.7 Sketch and find the area of: (U) the region to the left of the parabola x = 2y2, to the right of the y-axis,
and between y = 1 and y = 3; (b) the region above the line y = 3x - 2, in the first quadrant, and below the
line y = 4; (c) the region between the curve y = x3and the lines y = --x and y = 1.
32.8 Sketch the followingregions and find their areas:
(a) The region between the curves y = x2and y = x3.
(b) The region between the parabola y = 4x2 and the line y = 6x - 2.
(c) The region between the curves y = &,y = 1, and x = 4.
(d) The region under the curve &+&= 1and in the first quadrant.
(e) The region between the curves y = sin x, y = cos x, x = 0, and x = 44.
(f) The region between the parabola x = -y2 and the line y = x +6.
(9) The region between the parabola y = x2-x - 6 and the line y = -4.

256 APPLICATIONS OF INTEGRATION I: AREA AND ARC LENGTH [CHAP. 32
(h) The region between the curvesy = fi and y = x3.
(i) The region in the first quadrant between the curves4y +3x = 7 and y = x - ~ .
(j) The region bounded by the parabolas y = x2 and y = -x2 +6x.
(k) The region bounded by the parabola x = y2 +2 and the line y = x - 8.
(Z) The region bounded by the parabolas y = x2 - x and y = x - x2.
(m) The region in the first quadrant bounded by the curves y = x2and y = x4.
(n) The region between the curve y = x3and the lines y = -x and y = 1.
32.9 Find the lengthsof the followingcurves:
x4 1
8 4x2
(a) y = -+-from x = 1to x = 2.
(b) y = 3x - 2 from x = 0 to x = 1.
(c) y = x2I3from x = 1 to x = 8.
(d) x2/3+y2l3= 4 from x = 1to x = 8.
(4 Y = 15+7
4x from x = 1to x = 2.
(f) y = 3 &(3 - x) from x = o to x = 3.
(9)
(h)
x5 1
1
24xy = x4 +48 from x = 2 to x = 4.
y = 3 (1 +x2)3/2
from x = O to x = 3.
2
32.10 Use Simpson’s rule with n = 10to approximatethe arc length of the curve y =f(x) on the given interval.
(a) y = x2 on [O,13 (b) y = sin x on CO, n] (c) y = x3on CO,5)

Chapter 33
Applications of Integration II:Volume
The volumesofcertainkinds of solidscan kcalculated by means of definite integrals.
33.1 SOLIDS OF REVOLUTION
Diskand Ring Mcrhods
Letfbc a continuous hnclion such thatf(x) 2 0 for a s x s b. Consider the region Sr under the
graph o
fp -f(x}, above the x-axis, and between x =U and x = b (sacFik 33-11, If 9is revolved about
the x-axis, the resultingsolid is called a d i d ofredution.The generating regions i# T
o
r some familiar
solids of revolution are shownin Fig 33-2.
I
" I
-
i:
t
i
h
Tkorem 33.1: The volume Y of the solid of revolution obtained by revolvingthc region of Fig. 33-1
about the x-axis is givenby
h
v = 11 (
I
@
)
)
'dx = %I
J dx diskfnrnrub
l
An argument for the disk formula is skedlcdin Problem 33.4.
If we interchange the roles of x and p and rcvolw ~ h c
area 'under" the grapb of x =g(y) about the
pnxis, then the same rraroaingleads to the disk formula
257

258 APPLlCATlONS OF INTEGRATION 11: VOLUME [CHAP.33
EXAMPLE Applying the disk formulato Fig. 33-34, wc obtain
which htht standard formula for the ~ J u m
ora conewith height h and radiusof base r.
Now let land g be two runctions such that 0 sg(x) sJ(x) for U g x <b, and revolve the region S
between tlle curves y -f(x) and y =g(x) about the x-axis (see Fig. 33-35 The resulting solid of rcvolu-
tion has a volumc V which i
s the differencebctwecn thc volume of the d i d of revolutiongatcrated by
the @on under y = j ( x ) and the volume of the solid of revolution generated by the region under
y = dx). Hence, by Theorem33.1.
V = R I{[~(X))~
-(g(x))’) dx washerformukd
EXAMPLE Cornider tbe region fl bounded by the cunm y = dy = x (rse Fig 33-41, Thecurve obviously
intersect in the points (0. 01 and (1. 1). The bowl-shaped solid of revolution generated by revolving R about thc
x-axis has volume
h X
Fig. 3S3 Fe.33-4
Cjlibdrial shcll M e d d
Lelfbe a conlinuous function such thatfCx) 2 0 fur a s r 5 h. where U 2 0. As usual, k t abe the
region under the curve y =f(x), above the x-axis, and between x =U and x = 6(sec Fig. 33-51. Mow,
howwer, revolve 4
)about the y-axis.The resultingsolid of revolution has volume
Y = 24 r
/
(
x
) dx = 2r xy dx cylindricul shellfiwlu
l I‘
For tbe basic idea behind this formula and its namc.sec Problem 33.5.

CHAP.331 APPLICATIONS OF INTEGRATION 11: VOLUME 259
EXAMPLE Consider the hctiOn /(XI
-,
,
&
- for 0 sx 5 r. Thc graph o
f f is the part or the circle
Y' +y2 =r' that lies in the first quadrrnL Revolution about ihc y-axis of the region a under the graph orjlsee
Fb.3%) prodica a solid heinisphm of radiusr. By the cylindricalshell formula,
To evaluate V substirutc U = rz -xz. T
h
e
n du = -2x dx, and the limits or integration x = 0 and x =r
koomeU = rzand U =0. respectively,
vhir result is moreeasilyobtainedby thedisk rormula V = A x2 dy. Try it.)
r
3
3
.
2 VOLUME BASED ON CROSS SECTIONS
Assume that a salid (not nacxssarily a solid of wolution) l
i
a entirely between the plane perpen-
dicularto the x-axiu at x -U and the plane perpendicular to the x-axisat x = b. For a 5 x 5 b. let the
pllne perpendicular to the x-axis at that value or x intersect the solid in a region of area A(x), as
indicated in Fig. 33-7. Then thc volume V of the solidis given by
dx cross-sectionformdo
For i
lderivation, see Problem 33.6.

260 APPLICATIONS OF INTEGRATION 11: VOLUME [CHAP.33
F i i 33%
EXAM?LCS
F
i
g
. 33-9
Assume that hrlrola salami or length h is such that a cros!%
mtisn perpendicularto the axis of the salami at a
distancex from the end 0 isa circlco
fradiusfi(see Fig. 33-8).Thus,
A(x)= x(fi,’ = nkx
and the crors-section hmula @ves
Note thri for thissolid of revolutionthe disk lbrmulawould give the same expression for V.
Amme that a solid has a b e which i
s a cide of radius r. Assume that there is a diameter D such that all
pkm sections o
f the wlid perpendicularto diameter D arcsquares (
s
e
e
Fie, 33-9).Findthe mlume.
Let the origin he the enter of the circle and kt the x-axis be the sped01 diameter D.For a @irm
value of
I, 4 t h -r 5 x s F
, the side 4x1ofthe square cross scction is obtained by applying the Pythagorean theorem
to rhc right triangle with sidesx. $
2
. and r (secFik 33-9),
+J‘=8
,
’
4
$2 -q? -x’) = A(x)
mCn,by the crclsaaec(ionformula,
Solved Problems
33.1 Find the volume o
fthe solid gmerated by revolvingthe a
v
e
nregionaboutthe given axis.
(a) The region under the parabola y = x’. above the x-axis, between x =0 and x = I; about
the x-axis.
(b) fhc same ~ ~ g i o n
as in part (a), but about the y-axis. The region is shown in Fig. 33-10.

CHAP. 331 APPLICATIONS OF INfEGRATlON 11: VOLUME 261
Fig. -10 Fig. 33-11
{a) Use the disk f6nnuh.
(h) Use the cylindricalshell formula,
33.2 Let 9
l be the region between y =x2 and y =x (see Fig. 33-11). Find the volume of the sdid
obtainedby revolving around: (a)the x-axis; (h)the y-axis.
(0.0)and (1. I).
The curves intersect
(a) By the w d m rormulo,
{b) (Method 1) Use the m h c r lormula along the y-axis,
(Mnhod 2
) We can integrate alongthe x-axis and use the difference or two cylindricalshell knndy.
The formula u
s
c
d in mthod 2a n be lormulatcdas follows:
Y = 2x x(g(x) -JM) dx diflcrmce tfcylindrid shells
w b c V is rhc volume or the solid obtaid by evolving about the paxk the region bounded above
by y -dx). below by y =f(x), and lyingbetween x =o and x = b,wirh 0 5 a <h.
1

262 APPLICATIONS OF INTEGRATION 11: VOLUME [CHAP. 33
33.3 Find the volume of the salid whose bax i
s a circle of radius I and such that every cross section
perpendicular to a particular fixed diameter D is an equilateral triangle.
Let the center of the circular base bc rhe origin. and let the x-axis he the diamcrer D. The a m of the
cross saclion at x isA(x)= hs/2{secFig 33-12). Now. in the horizontal right triangle,
and in the vertical right triungle,
h = Jsf -J5J-
Hcncc.A(x) = v/s(ra - x’ban cvcn function-and the cross-section formula gives
Fig, 3 sI2
33.4 Establish the disk formula V = R (((x))~ dx.
s:
We assume as valid the expression ir% fbr rhc volume of a cylinder d radius r and height h. Divide
the interval [a. b] into n qual subintemla each o
f length Ax -[h-o)/n(we Fig. 33-13). Consider the
volume V
, obtained by revolving the region d,above the ith subinterval about rhe x-axis. U mi and h4,
dcnotc the absolute minimum and the absolutc maximum dJon the Fbsutiritensl. it is plain that b: must
lie heirnen the rolumc of a cylinder or radiusm, and height Ax and the volumc of a cylinder or radius M,
and height Ax,

CHAP. 331 APPLICATIONS OF INTEGRATION XI: VOLUME 263
Hence,
I
Sincethis relation holds(for suitable numbcnx
:
) for arbitraryA. it must holdin the limL as n +m
,
which is the disk formula. The name derives from the ux of cylindrical disks [ofthicknca Ax) to approx-
imate the V;.
Fig. 33-13 Fig. 33-14
3
3
5 Establish the cylindrical shell formula V = 2n xf(x) dx.
I'
Divide the interval [a, b] into n equal subintervals, each of length AV.
Ler 9,be the region abwc the
P" subinterval (secFig. 33-14). Letx
: be the midpoint ofthc Fbsubinterval. x
: -(.x,-, +n#2.
Now the solid obtained by revolving the region Yt
about the paxit is approximately the solid oblainud
by revolvingthc mangle with basc AYand height y
: =/(x:). The latter solid is a cylidrical Ml;that is.
i t is lhe difference between the cylinders obtained by rotating the rectangles with the same heigbtf[x:) and
with bares [O.x,- end [O. XJ H m .it has volume
A X ~ J ( X ~ )
- XX:- f(.~:)rt/(x:n~: -x
:
- 1) = M(x:Mx,+ x,- ~Mx, -x,- k)
= nf(x:W2x:HAx) = Znx:f(xf) Ax
Thusthe tau V is approximated by
2nEx:f(xf) Ax
x/(x) dx.
1 - 1
which in turn approximatmthe definiteintegral 2r
I'
33A Esmblish the cross-section formula V = A(x) dx.
I"
Divide the interval [a, h] into n equal wbmrerrab[x,_,,xi].each of length Ax. Choose a point x? in
[x,- ,,xJ. If n is larp, making Ax mall, the piox of the d i d between x, -,and x, will bc very nearly a

264 APPLICATIONS OF INTEGRATION 11: VOLUME [CHAP. 33
(noncircular) disk, of thickness Ax and base area A(xF) (see Fig. 33-15). This disk has volume A(x:) Ax.
Thus,
Fig. 33-15
3
3
.
7 (Solids of Revolution about Lines Parallel to a Coordinate Axis) If a region is revolved about a
line parallel to a coordinate axis, we translate the line (and the region along with it) so that it
goes over into the coordinate axis. The functionsdefining the boundary of the region have to be
recalculated. The volume obtained by revolving the new region around the new line is equal to
the desired volume.
(a) Consider the region 9 bounded above by the parabola y = x2, below by the x-axis, and
lying between x = 0 and x = 1 [see Fig. 33-16(a)].Find the volume obtained by revolvingW
around the horizontal line y = -1
.
(b) Find the volume obtained by revolving the region W of part (a) about the vertical line
x = -2.
(a) Move W vertically upward by one unit to form a new region W*.
The line y = -1 moves up to become
the x-axis. W*is bounded above by y =x' +1 and below by the line y = 1. The volume we want is
obtained by revolvingW*about the x-axis. The-washerformula applies,
V = II ((x' + I)' - 1
'
) dx =II
I'
(b) Move W two units to the right to form a new region 9
2
' [see Fig. 33-16(b)]. The line x = -2 moves
over to become the y-axis. W x is bounded above by y = (x -2)' and below by the x-axis and lies
Y
I X
1
-
a
Y
1 2 3 x
Fig. 33-16

CHAP. 33) APPLICATIONS OF INTEGRATION 11: VOLUME 265
hetwee11 x -2 aid x -3. the volume wc want is obtained by revolving9
. about tbe y-axis The
cylindrial shell formula applies,
.
S
m
#
m
t In calculating the volume of a solid of revolution we usually apply either the disk formula (or the
washer brmula) or the cylindrical shells formula (or the diffcrmcc of cylindrical shells formula). To decide which
formula to use:
(I) Decide along which axis you are going to inteerate. This depends on the shape 8nd posifion o
f the regiona
that U rcvolvod.
(2) (I) U
s
e the disk formula (or the washer rormulr) if the region a is revolved pwpmdimlar to tbe axis of
integration.
(ii) Usc the cylindrical shells formula (or the difference of cytindriwl sMls formula) if the region is
revolvcdparullcl to the axis o
f integration.
3
3
.
8 Find thc volume of thesolid gmcnld by m k n g the givm regkmaboulthe @wnaxis.
The qion above the curve y -x’. under the line y -1. and h t m x = 0 and x = 1:about the
x-axis.
The rcgionof par1(a);about f he)-axis.
The region below the line y = Lr,abow the x-axis. and bctwecnx =0and x -I ;about rite puk
Thc region betweenthe parabolasJ -x2 and x -ya;abut ntbcr thex-axisor the paxis.
The region (secFig. 33-17) inride the cirdc xa +y3 =r2,with 0 5 x 5U <r; about the y-axis (This
giver the volume CUI from a sphere of radius r by 8 pipc d radius a whose u h b 8 diameter of the
sphere.)
The region (ice Fig. 33-18) inside the circk x3 +y’ = 9,with x 2 0 and y z 0. and above the lii
y =a,wherc 0 $ a < r; about the y-axis. {Thisgives the volume o
fa polar cap ofa sphere.)
The region bounded by p -1 +xzand y -5; about the x-axis.
Tht region (sec Fig. 33-191 inside the citde xz +(y -h)’ -U*, with 0 <a c b, about the x-axis.
[Hint: When you obtain an integral d the ronn ,
/
- dx notice that this is the a m o
f 8
micirck o
f radius0.3 This problem givts the volume of a doaghnut-shapcd soli.
The region boundedby x2=4y and y =x/2; about the y-axis.
t’
0 ’
I
F
3
g
. 35.17 F k 33.18 F
i
& 33-19

266 APPLICATIONS OF INTEGRATION I1: VOLUME [CHAP,33
0
1 The region bounded by y =4/x and y -(x -3)z;about the x-axis. poticc that the curwca intersect
w h x = l a n d x = 4 . ~ 1 i s ~ l a b o u l 1 h e i n t e r s a c d o n a t x = l ? )
(k) The region of part 0;aboul he y-axis.
(0 The region bounded by xy = I. x = 1, x = 3. y =0: about the x-axis.
(m) The region ofp
a
n(0;about the y-axis.
UK the crmsiectionformula 10 Rad the volumc o
f the followingsolick
(U) The Wlid har 8 base which b a a r c k o
f radius r, Each m
s
)ssction perpendicularto a fixed diametet
of thecircle is an Lsorcelestrianglewith altitude equal tooae-haUofits base.
(b) Tbc solid is a wedge. cut from a pcriaCtly round trac o
f radius r by two planes, one perpendicular to
the axis ofthe I t e t m d the other inlensecting the first plane at an angle of W along a diameter (see
(4 A square pyramid with 8 heishtof h unitsa d a bortof Side r unira [Hint: Locate the x-axb 8s in F
i
g
.
33-21. By similarriglit ~riangks.
3
M
Pig, 33-20).
d h - x
-
I
-
e h
and
which &ermines A(xl.1
(4The tetrrhedron (see Fig. 33-22) formed by Ihroe mutually perpendicular CaccS and thm m u l d y
perpendicularedges o
flengthsn, b,c. [Hh:Another pyrsmid;proceed 85 in part (cl-]
I
TX
t X

CHAP. 33) APPLICATIONS OF INTEGRATION 11: VOLUME 267
33.10 (a)Let 41 be the region between x = 0 and x = 1 and bounded by the curves y = x3 and y = 2x. Find the
volume of the solid obtained by revolving 9 about the y-axis. (b) Let W be the region between the curves
y = 2x - x2 and y = ix. Find the volume of the solid obtained by revolving9 about the y-axis.
33.11 Let 9 be the region in the first quadrant bounded by y = x3 +x, x = 2, and the x-axis. (a)Find the volume
of the solid obtained by revolving 9 about the line y = -3. (b) Find the volume of the solid obtained by
revolving41about the line x = -1.
33.12 Let 41 be the region in the first quadrant bounded by x = 4 - y2 and y2 = 4 - 2x. (a) Sketch 41. (b) Find
the volume of the solid obtained by revolvingW about: (i)the x-axis; (ii)the y-axis.
33.13 Let 41 be the region in the second quadrant bounded by y = 2x2,y = x2 +x +2, and the y-axis. (a) Sketch
9.(b) Find the volume of the solid obtained by revolving 9 about the y-axis.
33.14 Let 41 be the region in the second quadrant bounded by y = 1 +x2and y = 10.(a)Sketch 41. Then find the
volume of the solid obtained by revolving W about: (b) the x-axis; (c) the y-axis; (d)the line y = -1;
(e) the line x = 1.

Chapter 34
The Natural Logarithm
34.1 DEFlMTlON
We already know the formula
There mains the problem of finding an anridcrivativeotx- '.
definiteintegral
Figure 34-1 shows the graph of y = 1/r tor r >0; it is MK branch or a hyperbola Far x z 1, the
II'+
representsthe area underthe curve y = l/r and above tbe r-axis, betwecn r = 1 and c = x.
DcRnitkn:
I 3 4 r r
Fig. 341
In x = -br [x >01
J1':
Thefunction In x is called the natural Iogmirhm.By Thcorcm 31.1.
1
D,(ln x) =- cx '01 (34.I)
X
and 50 tbe natural logarithm is the desired antiderivativeo
f x", h
r mdy on (0. oo). An antiderivative
forall x # 0 will heconstructedin the followingsection.
3
4
.
2 PROPERTlES
PROPERTY 1. In 1 =O.
This Collows from
PROPERTY 2. If x > 1, In x > 0.
268

CHAP. 341 THE NATURAL LOGARITHM 269
This is apparent from the area interpretation (Fig. 34-1)or, more rigorously, from Problem 30.11.
PROPERTY 3. If 0 c x < 1, In x < 0.
" 1 ' 1
In fact, In x = 7tit = - [- tit [reversing limits of integration]
t
and, for 0 c x < 1,
by Problem 30.11.
PROPERTY 4. - dx = In 1x1 +C [x # 0).
1:
In other words, In 1x1 is an antiderivative of x - l for all nonzero x. The proof is simple. When
x > 0, then 1x1= x, and so
When x <0, then 1x1 = -x, and so
D,(ln I x I) = D,(ln (-x)) = D,(ln u)D,(u) [by the chain rule; U = - x > O]
= (t>(-1)
PROPERTY 5. In uv = In U +In v.
U
V
PROPERTY 6. In - = In U - In v.
U
V
Proof: In Property 5, replace U by -.
1
PROPERTY 7. In - = -In v.
Proof:
PROPERTY 8. For any rational number r, In x' = r In x.
V
Let U = 1 in Property 6 and use Property 1.
See Problem 34.3 for a proof.
PROPERTY 9.
Proof: Since D,(ln x) = - >0, In x must be an increasing function.
PROPERTY 10. In U = In D implies U = v.
In x is an increasing function.
1
X
This followsfrom Property 9. Since In x is increasing, it can never repeat a value.
PROPERTY 11. c In 2 c 1.

270 THE NATURAL LOGARITHM [CHAP. 34
Proof: The maximum of l/t on [l, 2) is 1, and the minimum is 4. Hence, by Problem 30.3(b),
i(2 - 1) < (l/t) dt < l(2 - 1); that is, 4 <In 2 < 1. The strict inequalities follow from Problem 30.11.
A more intuitive proof would use the area interpretation of
6' (l/t) dt.
6'
We shall see later that In 2 is 0.693...,and we shall assume this value in what follows.
PROPERTY 12.
Proof: By Property 9, we need only show that In x eventually exceeds any given positive integer k. For
lim In x = +W.
X - + + a ,
x > 22k,
In x >In 22k= 2k In 2 [by Property 8)
so In x >2k(l) = k [by Property 11)
PROPERTY 13. lim In x = -00.
x+o+
1
Proof: Let U = -. As x +O+, U + +W. So,
X
1
lim In x = lim In - = lim -In U [by Property 71
x+o+ u-r+(o U u-++a3
= - lim l n u = - W [by Property 121
U-.+CU
SolvedProblems
34.1 Sketch the graph of y = In x.
We know that In x is increasing (Property 9
)
, that In 1 = 0 (Property l
)
, and that 4 < In 2 < 1
(Property 1
1
)
.From the value y = In 2 = 0.693...we can estimate the y-values at x = 4,8, 16,... and at
x = +,a, 6, ...by Property 8,
In 4 = 2 In 2
ln-= -In2 ln-= -2ln2 ln-= - 3 l n 2 . . .
In 8 = 3 In 2 In 1
6 = 4 In 2 ...
1 1 1
2 4 8
D:(ln x) = D,(x-') = - x - ~= -1/x2 <0and, therefore,the graph is concave downward.There is no hori-
zontal asymptote (by Property 12), but the negative y-axis is a vertical asymptote (by Property 1
3
)
.The
graph is sketched in Fig. 34-2.Notice that In x ussumes all real numbers us oalues.
3
4
.
2 Prove: In uu = In U +In U.
In
l n u = L" 1
; d t
make the change of variable w = ut (U fixed). Then dw = U dt, and the limits of integration,t = 1 and t = o,
go over into w = U and w = uu, respectively,
yu U 1
In = -- dw = ["dw = 1dt
w u t

CHAP. 343 THE NATURAL LOGARITHM 271
f'
Fig. 34-2
Then, by Theorem 30.4,
3
4
3 Prove: In x' = r In x for rational r.
By the chain rule,
Then, by Corollary 29.2, In x
'= r In x +C, for some constant C. Substituting x = 1, we find that C = 0,
and the proof is complete.
34.4 Evaluate:
(a) D,(h (x3 - 2x)) (b) D,(ln (sin x)) (c) Dices (In x))
Use the chain rule.
3x2-2
(3x2- 2) =-
1
Dx(x3- 2x) = - x3 - 2x
x3 - 2x
1
(a) D,(h (x3- 2x)) = -
x3 - 2x
(cos x) =-= cot x
(b) D,(ln (sin x)) = -D,(sin x) = -
1 sin (ln x)
(c) D,(cos (In x)) = -(sin (In x)) DJln x) = -(sin (In x)) - = - -
cos x
1 1
sin x sin x sin x
X X
34.5 Quick Formula 11: 1- dx = In If(x)l+ c.
*f'(x) D y the chain rule]
Proof: D,(ln If (x) I)= -
f(4
1
3
4
.
6 Find the following antiderivatives:

272 THE NATURAL LOGARITHM [CHAP. 34
(a) Since DJcos x) = - sin x,
= - In I cos x I +C
= In lcos XI-' + C
=In !sec xl +C
[by quick formula 111
[by Property 7)
[since sec x = (cos x)- '1
(b) Since DJsin x) = cos x,
cot x dx = -dx = In Isin x I+ C [by quick formula 111
I 1:::1
dx = - In I3x - 1 I +C [by quick formula I13
3
'I3 x - 1 3
(c) [ - d x = -
1 -
3x - 1
1
dx = - -dx = - In Ix2 - 5I +C [by quick formula I13
2
'Ix2-5
2x 2
X
34.7 Find sec x dx.
s
The solution depends on a clever trick,
sec x +tan x dx = /sec2 x +sec x tan x
sec x dx = (sec x) dx
I s sec x +tan x sec x +tan x
Here we have applied quick formula 11, using the fact that DJsec x +tan x)= sec x tan x +sec2 x.
=In !sec x +tan xl + C
34.8 (LogarithmicDiflerentiation) Find the derivative of
Instead of using the
the absolute values,'
In l Y
and then to differentiate,
product and quotient rule for differentiation, it is easier to find the logarithm of
. .
= In ix',/ii~~) - In I2x - 113 CbY Property 61
[by Property 51
= In (x2)+In (8x +5)'12 - In I2x - 1l3
1
2
= 2 In x +- In (8x +5) - 3 In I2x - 1I [by Property 8)
1 1 2 4 6
- D,y = 2(5) +f (A8
) - 3
(
- 2
) = - +-
--
Y 2x - 1 x 8 ~ + 5 2 ~ - 1
Therefore,
This procedure of first taking the logarithm and then differentiating is called logarithmic differentiation.
We take the absolute values to make sure that the logarithm is defined. In practice, we shall omit the absolute values when it is
clear that the functionsare nonnegative.

CHAP. 341 THE I~ATURALLOGARITHM 273
1
34.9 Show that 1-- 5 In x 5 x - 1for x > 0
.
X
For x 2 1, note that l/t is a decreasing function on [l,
x] and, therefore, its minimum on [l,
x] is l/x
and its maximum is 1.Then, by Problem 30.3(c),
1
- (x - 1) I
In x =
X
1
l - - s l n x ~ x - l
X
Note that 1- l/x < In x < x - 1 for x > 1 by Problem 30.11. For 0 < x c 1
, - l/t is an increasing func-
tion on [x, 13. By Problem 30.3(c),
--(1
1 - x ) I l n x = [ S d t = [ (-:)dts -1(1 -x)
X
1
Hence,l - - l ; l n x < x - 1.
X
Supplementary
34.10 Find the derivatives of the following functions:
(a) In (4x - 1) (b) (In x ) ~ (c) 6
x - 1
(e)* x2 In x (f)In x+l (9) In I5x - 2I
Problems
34.11 Find the following antiderivatives. Use quick formula I1 whenever possible.
34.12 Use logarithmic differentiation to find y':
,/xL - 1 sin x
(c) =
(2x +3)4
34.13 (a) Show that In x < x. [Hint: Use Problem 34.9.1
In x 2
(b) Show that -<-. [Hint: Replace x by fiin part (a).]
x &
In x
(c) Prove: lim --
-0. [Hint: Use part (b).]
x + + m x

274
34.14
34.15
34.16
34.17
3
4
.
1
8
34.19
34.20
34.21
34.22
3
4
.
2
3
34.24
34.25
34.26
-~
THE NATURAL LOGARITHM
1
1
(d) Prove: lim (x In x) = 0. [Hint: Replacex by ;
in part (c). I
X-.O+ L
(e) Show that lim (x - In x) = +a.
X + + c O
Calculate in terms of In 2 and In 3:
2
(a) In (212) (b) In -
9
Calculate in terms of In 2 and In 5:
1
(c) In -
5
1
(a) In 10 ~ (b) In -
2
(d) In 25
(h) In 2’
[CHAP. 34
Find an equation of the tangent-lineto the curve y = In x at the point (1’0).
Find the area of the region bounded by the curves y = x2,y = l/x, and x = 4.
Find the average value of l/x on [l, 41. .~
Find the volume of the solid obtained by revolving about the x-axis the region in the first quadrant under
y = x-lI2 between x = 3 and x = 1.
Sketch the graphs d:
1u x (d) y =In (cos x)
(a) y = In (X-+ 1) (b) y =In - (c) y = x -1-
1
X
6
t
An object moves along the x-axis w
i
t
h acceleration a = t - 1 +-. (a) Find a formula for the velocity u(t)if
v(1) = 1.5. (b) What is the maximum value of U in the interval [l, 931
Use implicit differentiationto find y’:
(a) y2 = In (x2 +-y2) (b) In xy +2x - y = 1 (c) !I?
(x +y2) = y3
1 3 + h
Find lim (
i
In T)
.
h+O
Derive ihe formula csc x dx = In l-&cx -cot x I +C. [Hint: Similarto Problem 34.7.)
(b) s,’*
f 2 dx
Find: (a) -
.Io 4 + x 1 - 2 x 2 ’
34.27
ApproximateIn 2 = (l/x) dx in the following ways:
Ey the trapezoidal rule [Problem 31.9(a)J,
with n = 10.
By the midpoint ruie (Problem 31.34)’with n = 10.
By Simpson’s rule (Problem31.35)’ with n = 10.
6’
Use Nzwton’s method to approximatea solution of: (a) In x +x = 0; (b) In x = l/x.

Chapter 35
35.1 INTRODUCTION
x is rational. For example, we want to obtain the results
Let a be any positive real number. We wish to define a function axthat has the usual meaning when
1 1
5-2 =- =-
52 25
43 = 4 4 4 = 64
In addition, an expression such as 5J5, which does not yet
reasonable value by the new definition.
Definition: aXis the unique positive real number such that
In ax= x In a
8213 = (fi)2 = 22 = 4
have any meaning, will be assigned a
(35.2)
To see that this definition makes sense, observe that the equation In y = x In a must have exactly
one positive solution y for each real number x. This follows from the fact that In y is an increasing
function with domain (0, +00) and range the set of all real numbers. (See the graph of the In function in
Fig. 34-2.)
35.2 PROPERTIES OF ax
It is shown in Problem 35.4 that the function axpossesses all the standard properties of powers:
(I) ao = I.
(11) a' = a.
(111) a"+' = aUaV.
1
(V) a - v = -
av*
(VI) (ab)" = aXbX.
aX
(VII) (
;
) =b".
(VIII) (a")"= a"'.
35.3 THE FUNCTION ex
For a particular choice of the positive real number a, the function ax becomes the inverse of the
function In x.
TERMINOLOGY Two functionsfand g are inoerses of each other iffundoes the effect of 9, and g undoes the effect of
f.In terms of compositions (Section 15.1), this means:
f(g(x))= x and B(f(4)
= x
275

276 EXPONENTIAL FUNCTIONS [CHAP. 35
Definition: Let e denote the unique positive real number such that
l n e = 1 (35.2)
Since the range of In x is the set of all real numbers, there must exist such a number e. It can be
shown that e = 2.718 ....[SeeProblem 35.3o(c).]
Tkorem 35.1: The functions e" and In x are inverses of each other:
In ex= x and elnx- x
-
Indeed, substituting e for U in (354,
In ex = x In e = x 1 = x
If we replace x in this result by In x,
In elnx= In x
Hence, ,1nx -
- x [since In u = In U implies u = U by Property 10, Section 34.21
Theorem 35.1 shows that the natural logarithm In x is what one would call the "logarithm to the
base e"; that is, In x is the power to which e has to be raised to obtain x: ,Inx -
- x.
Theorem35.2: ax= e"In '.
Thus, every exponential function a" is definable in terms of the particular exponential function ex,
which for this reason is often referred to as the exponential function. To see why Theorem 35.2 is true,
notice that, by Theorem 35.1,
In ex'"' = x In a
But y = uxis the unique solution of the equation In y = x In a. Therefore,
,xlna -
- ax
In Problem 35.9 it is shown that exis differentiable and that:
Theorem35.3: Dx(e")= ex.
Thus, ex has the property of being its own derivative. All constant multiples Ce"share this property,
since Dx(Cex)
= C Dx(ex)
= Ce". Problem 35.28 shows that these are the only functions with this
property.
From Theorem 35.3, we have
exdx = ex+ C (35.3)
s
Knowing the derivative of ex,we obtain from Theorem 35.2
D,(aX) = DX(ex'"')
= e"In' D,(x In a) [by the chain rule]
In a = axIn a
-
- ex In a .
This proves:
Theorem35.4:
or, in terms of antiderivatives, / a . dx = + c
We know that Dx(Y)= r$-' for any rational number T. Now the formula can be extended to
arbitrary exponents. If, in Theorem 35.2, we replace x by T and U by x (thereby making x positive), we
Dx(ax)
= (In a)ax
(35.4)
ax

CHAP. 351 EXPONENTIAL FUNCTIONS 277
get x' = e'In ".Hence,
D,(xr) = D,(e' In = e
' In "D,(r In x) [by the chain rule]
Thus, we have the following:
Theorem 355: For any real number r and all positive x,
D,(x? = rxr-
Solved Problems
35.1 Evaluate: (a)e2In"; (b) In e2; (c)e('"")-';(d) 1".
(a) e2InX
= (e'"?' [by Property (VIII)]
(b) In e2= 2 [by Theorem 35.1)
(c) e('nU)- 1 = - [by Property (IV)]
= x2 [by Theorem 35.1)
U
e1
U
= -
e
[by Theorem 35.1 and Property (II)]
(d) By definition,In 1" = x(ln 1)= x(0)= 0 = In 1.Hence, 1" = 1.
35.2 Find the derivatives of: (a)e6; (b) 32";(c)xe"; (d) 3x3.
(a) Dx(e3= eJ;D,(&) [by the chain rule]
(b) D,(32? = (In 3)32x0d2x)
(c) D,(xe? = xD,(e? +DJx)ex [by the product rule]
= (ln 3)32x(2)= (2 In 3)32x
= xe" +( l w = ex(x+ 1)

(d) D"(3Xfi) = 3Dx(xJi) = 3 ( d x f i - I ) = 3JzxJi-1
353 Find the following antiderivatives (ex' stands for e("')): (a) 10" dx; (b) xe"' dx.
s I
I10" dx = 1w+c
Ixr"dx =; I e udu =?
1 e" +C = -
1 e2
+C
(b) Let U = x2.Then du = 2x dx, and
2

278
35.4
35.5
35.6
EXPONENTIAL FUNCTIONS
Prove Properties(I)-(VIII) of aX(Section35.2).
By definition,h a * = x In a. We shall use the fact that II? U = In U implies U = U.
-0
U = l
In 1 = 0 = 0 (In a) = In a'. Hence, 1 = a'.
d = u
In a = 1 *!n a = In a'
In (a"a")= In a" +In a' = U In c +U In a
= (U i
U) In a = In a"+'
so, a"+"= a"a".
Let U = 0 in (IV), and use (I).
(abr = aXbX
In ju"b? = In a" +In b" = x In a +x In b
= x(ln a +In b) = x In (ab)= In (ab)"
So, (abj" = axbx.
($ =
;
By (VI), (i y b x= (
. b)" = a".Now divide by b".
b 1
(
a
'
)
"
= a""
In (a")v = U In a" = u(u In a) = uu(1n a) = In a""
so,(0")" = a"".
r
Show that Ig'(x)8(x)dx = 8'")
+C.
By Theorem 35.3 and the chain rule,
J
Thus, egtx)is a particular antiderivative of g'cx)e@("),and so k(")
+C is the general antiderivative.
dY ,/x
P
Use logarithmic differentiationto-find -: (a) y = xx; (b)y = -
.
dx 2JJr
[CHAP. 35

(a) In y = In xx = x In x. Now differentiate,
In 2 1 In 2
+ 5 --x - w = -
1dy 1 1
ydx 2 x + 1 2
--=--
2(x + 1) +
-
1
dx = y ( m + 5 - 2J;;
35.7 Prove the followingfacts about ex:
ex >0 (b) ex is increasing (c) ex > x (d) lim ex = +a (e) lim ex = 0
ux> 0, by definition (Section 35.1).
Dx(e3= ex > 0, and a function with positive derivative is increasing. More generally, the definition of
a", together with the fact that In y is increasing, implies that axis increasingif U > 1.
We know (Problem 34.8) that In U < U. Hence, x = In ex < ex.
This is a direct consequenceof part (c).
Let U = -x; then, by Property (V),
X++OO x+-OO
As x + -00, U --* +00, and the denominator on the right becomes arbitrarily large [by part (d)].
Hence, the fraction becomes arbitrarily small.
35.8 Sketch the graph of y = ex, and show that it is the reflection in the diagonal line y = x of the
graph of y = In x.
From Problem 35.7 we know that ex is positive and increasing, and that it approaches + m on the
right and approaches 0 on the left. Moreover, since D;(e? = ex > 0, the graph will be concave upward for
all x. The graph is shown in Fig. 35-l(a).
By Theorem 35.1, y = ex is equivalent to x = In y. So, a point (a, b) is on the graph of y = ex if and
only if (b, a) is on the graph of y = In x. But the points (a, b) and (b, a) are symmetricwith respect to the line
y = x, by Problem 6.4.
In general, the graphs of any pair of inverse functions are mirror images of each other in the line y = x.
35.9 Prove: (a) ex is continuous (6) ex is differentiable,and &(ex) = ex
Let E > 0. To prove continuity at x, we must show that there exists S > 0 such that IU - x I < 6 implies
Ie" - exI < E. Let E, be the minimum of ~ / 2
and 8/2. SinceIn is increasingand continuous, the range of
In on (ex - E,, ex +E ~ )
is an interval (c, d) containing x. Let 6 > 0 be such that (x - 6, x +6) is
included within (c, d). Then, for any U, if IU - x I < 6, it follows that I e" - exI < E, < E.
The proof will consist in showing that
e X + h - 8
lim -
= ex
h+O

L 1 I I
-2 -1 0 I 2 3 X
m
(b)
Fig. 35-1
e X + h - ex ex$ - ex eh- 1 @ - 1
,it will suffce to show that lim --
- 1.
Because-
= -
= ex -
h h h h+O
Let k.= eh - 1. Then t
? = 1 +k and, therefore, h = In (1 +k), by Theorem 35.1. Since e" is contin-
uous and eo = 1,k +0 as h 0. Hence,
eh- 1 k
lim -= lim
= lim
h - 0 k-oln(1 + k )
1
[since In 1 = 0)
k-ro In (1 +k) - In 1
k
In (1 +k)-In 1
k
1
-
-
lim
k-rO
In (1 +k)-In 1 1
k X
is the derivative DJln x) = - at x = 1; that is, it is equal to 1. Hence,
But lirn
k - 0
eh- 1
lim -= 1.
h+O
35.10 (a) Evaluate: (i)e2In 3 ; (ii)In e2,
(b) Solve for x: (i) In x2 = 5; (ii)In (In x + 1) = 3; (iii)ex - 6e-" = 5.
(a)
(b) (i) In x2 = 5
2 1 n x = 5
I n x = j
(i) By Property (VIII)and Theorem 35.1, e2 In = (e'"3)2 = 32 = 9. (ii) By Theorem 35.1, In e2 = 2.
x = e5/2 [since In U = b implies U = 81
(ii) In (In x + 1)= 3
h x + 1 = e 3
In x = e3 - I
= ee3-1

(iii) ex - 6e-" = 5
e2x - 6 = Sex [multiply by 81
e2x- 5e" - 6 = 0
(ex -6Xex + 1) = 0
e x - 6 = 0 or e x + l = O
ex = 6
x = I n 6
[ex + 1 > 0, since ex > 01
[e" = b implies u = In b]
35.11 Evaluate the followingexpressions:
(a) ,-Inx (b) In e-x (c) (e4)lnx (d) (3eYnX
(e) e l n ( X - l ) (f) In (f) (9) eln(2x) (h) In fi
35.12 Calculate the derivativesof the followingfunctions:
ex
(a) e-x (b) e"" (c) ecosx (d) tan ex (e) ;
(f)ex In x (9) x" (h) nx (i) In e2x U) ex - e-x
35.13 Evaluate the following antiderivatives:
r r r
(a) J e3x dx
(d) Jecosxsin x dx
(b) J eexdx
(e) J32x dx
(c) J e x d F Z dx
e2x
(h) dx (i) dx
(
J 1x22" dx (k) 1 x 3e-x4 dx
35.14 Use implicit differentiation to find y':
(a) ey = y +In x (b) tan ey-x = x2 (c) elly +ey = 2x
(d) x2 +exY +y2 = 1 (e) sin x = e
Y
35.15 Use logarithmic differentiation to find y':
(d) y = (In x)'" (e) y2 = (x + 1Xx +2)
(a) y = 3 S i " X (4 Y = (J2)C' (c) y = XInx
35.16 Solve the followingequations for x:
(a) e3x = 2 (b) In x3 = - 1 (c) ex - 2e-" = 1 (d) In (In x) = 1 (e) In (x - 1) = O
35.17 Consider the region 9 under the curve y = ex, above the x-axis, and between x = 0 and x = 1. Find:
(a) the area of 9;(b) the volume of the solid generated by revolving 9 about the x-axis.
35.18 Consider the region B bounded by the curve y = e"'2, the y-axis, and the line y = e. Find: (a) the area of
9 ;(b) the volume of the solid generated by revolving 3about the x-axis.

35.19 Let 9 be the region bounded by the curve y = ex', the x-axis, the y-axis, and the line x = 1. Find the
volume of the solid generated by revolving 9 about the y-axis.
35.20 Find the absolute extrema of y = 8'"
"on the interval [-n, n]. [Hint: euis an increasingfunction of u.]
35.21 If y = enx,where n is a positive integer,find the nth derivativey(").
35.22 Let y = 2 esin .(a)Find y' and y". (b) Assume that x and y vary with time and that y increases at a constant
rate of four units per second. How fast is x changing when x = n?
35.23 The acceleration of an object moving on the x-axis is 9e3'. (a)If the velocity at time t = 0 is four units per
second, find a formula for the velocity dt).(b) How far does the object move while its velocity increases
from four to ten units per second? (c) If the object is at the origin when t = 0, find a formula for its position
x(t)-
35.24 Find an equation of the tangent line to the curve y = 2e" at the point (0,2).
35.25 Sketch the graphs of the followingfunctions,indicating relative extrema, inflection points, and asymptotes:
(a) y = e-"' (b) y = x In x
In x
(4 Y = x
1
(6) y = e - " (e) y = ( I - l n ~ ) ~ (f)y = ; + l n x
[Hint: For parts (b)and (c) you will need the results of Problem 34.13(d)and (c).]
35.26 Sketch the graphs of y = 2" and y = 2-". [Hint: a" = exIn ".]
In x
-. This function is called the logarithm ofx to the base a (log,, x is
In a
35.27 For a > 0 and a # 1, define log,x
called the cornon logarithm ofx). Prove the following properties:
(b) alofi" = x
1
x In a
(a) D,(lOg, x) = - (c) log, a" = x
U
U
(d) log, x = In x (e) log,, (uu) = log, U +log, U (f) log, - = log, U -log,, U
log, x
(9) log, U
' = r log, U (h) In x = -
log, e
35.28 Show that only the functionsf(x) such thatf'(x) =f(x) are the functions Ce",
where C is a constant. [Hint:
Let F(x) =f(x)/ex and find F'(x).]
35.29 Find the absolute extrema off(x) = (In x)'/x on [l, e].
35.30 (a) Prove ex = lim (1 +E)". [Hint: Let y = (1 +f)'. Then,
Y * + W
[by the mean-value theorem]
1
= u ( l n ( u + x ) - l n u ) = u - x * -
U*
U* X
where U < U* 0, and U +x < U* 0 or
1 + <-< 1 if x < 0. In either case, lim -= 1. So, lim In y = x and, therefore, lim y =
U U
x U* U*
U Y * + W U U++Q) U * + c O
1
lim eInY = ex.
U * + W

(b) Prove e = lim .[Hint: Use part (a).]
n + + w
(c) rnApproximate e by finding (1 +(t)rfor large values of n (say,n 2 10OOo).
X"
3531 Show that, for any positive n, lim -= 0.
x-, +w ex
1
1 1
Hint: -= -- -
=
ex -
In ex'l- Inxlx). Now apply Problems 34.13(c)and 35.7(d).
-
35.32 Evaluate the following definiteintegrals:
1
35.33 Evaluate lim - (ell"+elln + - +enln).
n + + w n
35.34 Show that e" > ne. [Hint: Prove generally that U' > U" when U > U 2 e. (u/e)"-"2 1 since u/e 2 1 and
U > U. so,
u'e" 2 e'u" (1)
U u u U ue ue
By Problem 34.9, - - 1 > In -, - > In - + 1 = In -, e'/" > -, and so,
U u u U U U
By (1) and (2),u'e" > e"u", U' > U".]
3535 If interest is paid at r percent per year and is compounded n times per year, then P dollars become
P(1+-lknrdollars after 1 year. If we let n -+ 00, then the resulting interest is said to be continuously
compounded.
(a) Show that, if continuously compounded at r per cent per year, P dollars become Pe' dollars after 1
year. [Hint: Use Problem 35.30.1 Then, after t years, P dollars would become Per' dollars.
(b) Continuously compounded at 6 percent per year, how long would it take to double a given amount of
money?
(c) rnWhen r = 5 percent, compare the result of continuous compounding with that obtained by com-
pounding once a year and with that obtained by compounding monthly.
35.36 Use Newton's method to approximate a solution of: (a)ex = l/x; (b) e-x = In x.
35.37 Use Simpson's rule with n = 4 to approximate e-xz/2dx.
Jr,'

Chapter 36
L’HGpital’s Rule;ExponentialGrowth and Decay
36.1 L’HOPITAL’S RULE
The followingtheorem allows us to compute limits of the form
in the “indeterminate case” when the numeratorf(x) and the denominator g(x) both approach 0 or both
approach fCO.
Theorem 36.1 (L‘H6pital‘sRule): Iff(x) and g(x) either both approach 0 or both approach & 00, then
In L’H6pital’s rule, the operator “lirn” stands for any of
lim lim lim lim lim
See Problem 36.21 for a sketch of the proof. It is assumed in Theorem 36.1 that g(x) # 0 and g’(x) # 0
for x sufficientlyclose to a.
X++CCl x+-CCl x-a x-a+ x-+a-
EXAMPLES Let us verify L‘Hbpital’s rule for four limits found earlier by other means.
(a) See Problem 9 4 4 ,
6x 6 6 3
,,-,4x2+5 x-r-,8x x+-,8 8 4
- ]im - = lim - = - = -
3x2 - 5
lirn --
(b) See Problem 27.4,
sin 8 cos e
lim -= lim --
- cos 0 = 1
e-ro 8 e-ro 1
(c) See Problem 34.13(4. To apply Theorem 36.1, we must rewrite the function as a fraction that becomes indeter-
minate,
In x X - ’
lim (x In x) = lim -
1
= lim - [Dx(ln x) = x-’]
X‘O+ X‘O+ x x-ro+ -x-2
= lim - x
X‘O+
= o
(d) See Problem 35.31, where now n is supposed to be an integer. By L‘H6pital’s rule,
X” nx”-
x++, ex X’+Co ex
lim - = lim -
As the right side is indeterminate, L’H6pital’s rule can be applied again,
nx”- n(n - I)x”-~
lirn -- - lim
e”
x + + w e“ X’+al
Continuing in this fashion, we reach, after n steps,
n(n - 1) ..-(2x1) 1
lim = n! lirn - = n ! ( 0 ) = 0
X - r + a , ex X’+, ex
284

CHAP. 361 L'HOPITAL'S RULE; EXPONENTIAL GROWTH AND DECAY 285
This example illustrates the importance of having the function expressed as a fraction "in the right way."
Suppose we had chosen "the wrong way," and tried to evaluate the same limit as lirn F .Then repeated
application of L'Hbpital's rule would have given
e-x
X + + m
- ...
e-x (- l)e-. (- 1)2e-x
= lim
(-n)x-'"+') (- 1)2(n)(n+ 1 ) ~ - ( " + ~ )
-
lim 7
= lim
X
and we should never have arrived at a definite value.
(e) See Problem 34.13(c).By L'H8pital's rule,
FURTHER EXAMPLES
(1) Find lirn (In x)lIX.
X++O3
Since In x --* +00 and l/x 40,it is not clear what the limit is. Let y = (In x)'/~.Then In y =
1
- In (In x). Hence, by L'HGpital's rule,
X
1 1
-.-
In (In x) l n x x 1
lim In y = lim -
-
- lim -- - lim -=0
X + + m x + + m x X + + a ) 1 ,++,xInx
Therefore,
(2) Find lirn x'/~.
X + + W
1
Since x -P +00 and l/x +O, the limit is not obvious. Let y = x1IX. Then In y = - In x, and
X
In x
lim In y = lim --
- 0 by (e)above. Hence, lirn y = lim e'" = e0= 1.
Warning: When the conditions for L'HBpital's rule do not hold, use of the rule usually leads to false
results.
EXAMPLE
x 2 + 1 2 2 + 1 5
x+2 x2- 1 22- 1 3
- -=-
lim --
If we used L'Hbpital's rule, we would conclude mistakenly that
2x
lim -
x2 + 1 -
- lim -= lim 1 = 1
x+2 x2 - 1 x+2 2x x-+2
36.2 EXPONENTIAL GROWTH AND DECAY
Example (d) above shows that ex grows much faster than any power of x. There are many natural
processes, such as bacterial growth or radioactive decay, in which quantities increase or decrease at an
"exponential rate."

286 L'HOPITAL'S RULE; EXPONENTIAL GROWTH AND DECAY [CHAP. 36
Definition: Assume that a quantity y varies with time t. Then y is said to grow or decay exponentially if
its instantaneous rate of change (Chapter 19) is proportional to its instantaneous value;
that is,
* = K y
dt
(36.2)
where K is a constant.
Suppose that y satisfies(36.1).Let us make the change of variable U = Kt. Then, by the chain rule,
dy dydu - d ~ K or - = y
dY
du
K Y = - = - - - -
dt du dt du
and so, by Problem 35.28,
y = Ce" = CeKt (36.2)
where C is another constant.
We can now see why the process y is called exponential. If K > 0, K is called a growth constant, and
y increases exponentially with time. If K < 0, K is called a decay constant, and y decreases exponentially
with time.
Let y, be the value of y at t = 0. Substituting 0 for t in (36.2),we obtain
y, = Ceo = C(1) = C
so that (36.2)can be rewritten as
y = y,eK' (36.3)
EXAMPLES
(a) Assume that a culture consisting of loo0 bacteria increases exponentially with a growth constant K = 0.02,
where time is measured in hours. Let us find a formula for the number y of bacteria present after t hours, and
let us compute how long it will take until 1
O
O
O
O
Obacteria are present in the culture.'
Since y, = 10o0, the desired formula for y is given by (36.3),
y = 1O()OeO.O2*
Now set y = 1
O
O
O
O
Oand solve for t,
1OOOOO = 1000e0*02'
100 = eo.02*
In 1
O
O = In eo.OZ'
2 In 10 = 0.02t pn 10' = 2 In 10; In e" = U]
t = 100 In 10
Appendix E gives the approximate value of 2.3026for In 10.Thus,
t x 230.26 hours
Note: Sometimes, instead of specifying the growth constant K,say, K = 0.02, one says that the quantity is
increasing at the rate of 2 percent per unit time. This is not quite accurate. A rate of increase of r percent per
unit time is approximately the same as a value of K = O.Or when r is relatively small (say, r 5 3). In fact, with
Although populationsare measured in positiveintegers, (36.3)seems to be applicable, even though that formula was derivedfor
a quantity measuredi
n real numbers.

CHAP. 363 L'H~PITAL'SRULE; EXPONENTIAL GROWTH' AND DECAY 287
an r percent growth rate, y = yo(l +O.Or) after one unit of time. Since y = yoeK when t = 1, eR= 1 +O.Or,
and, therefore, K = In (1 +0.Or). This is close to O.Or, since In (1 +x) x x for small x. [For example,
In (1.02) x 0.0198 and In (1.03) x 0.02956.1 Thus, many textbooks will automatically interpret a rate of
increase of r percent per unit time to mean that K = O.Or.
If the decay constant of a given radioactive element is K < 0, compute the time T after which only half of any
original quantity remains.
At t = T, (36.3)gives
- I n 2 = K T I n - = - l n x
[
I 1
In 2
K
--- - T (36.4)
The number T is called the half-life of the given element. Knowing either the half-life or the decay constant, we
can find the other from (36.4).
Solved Problems
(In x)"
36.1 Show that lim --
- 0 for any positive integer n.
Y
The proof will proceed by mathematical induction (see Problem 12.2).The assertion is true for n = 1,
by Problem 34.13(4. Assuming that it is true for n = k, we have, for n = k + 1,
(In x ) ~
+ (k + lMln x)~x-
lim -- - lim I [by L'H8pital's rule]
Thus, the assertion is also true for n = k + 1,and the proof is complete.
ex - 1
x+o sin x
3
6
.
2 Find lim -.
The numerator and the denominator are continuous functions, each 0 at x = 0. Therefore, L'Hbpital's
rule applies,
e0 1
- --- - = 1
ex - 1 ex
lim --
- lirn --
x+o sin x x+o cos x cos0 1

288 L'H~PITAL'SRULE; EXPONENTIAL GROWTH AND DECAY [CHAP. 36
3
6
3 Find lim
X-.+oO
Since n/x -
+ 0, it follows that sin (R/x) -
+ sin 0 = 0 and there is no obvious way of solving the problem.
However, L'HBpital's rule turns out to be applicable,
II
II
X
sin -
= II lim cos -
x + + m x
-x-2
lim (x sin E)= lim --
- lim
X + + W x - + m x-l x + + m
= n c o s O = n * l = n
36.4 Find lim xx.
x+o+
By example (c)in Section 36.1, lim x In x = 0. Hence,
x+o+
36.5 Sketch the graph of y = xex.
By the product rule, y' = xe" +ex = ex(x + 1). Since ex is always positive, the only critical number
occurs when x + 1 = 0; that is, when x = -1. When x = -1,
Again by the product rule, y" = ex(l) +eX(x + 1) = ex(x +2). When x = -1,
1
e
y" = e- '[( -1) +21 = - > O
So, by the second-derivative test there is a relative minimum at the point P(-1, -l/e). From the expression
for y", the curve is concave upward (that is, y" > 0) when x > -2, and concave downward (y" < 0) when
x < -2. Thus, the point I(-2, -2/e2) is an inflection point.
It is clear that the curve rises without bound as x + +00. To see what happens as x + -00, let
U = -x. Then,
lim xex= lim - ue-u= - lim E= O
x+-CO U + + m u + + m eu
by Problem 35.31. Hence, the negative x-axis is an asymptote. This enables us to sketch the graph in
Fig. 36-1.
36.6 If y(t) defines an exponential growth or decay process, find a formula for the average value of y
over the time interval [O, 23.
By definition (seeSection 31.2),the desired average value is given by
But (36.1)states that an antiderivative of Ky is y itself. Hence, by the fundamental theorem of calculus,

-2
X
P
-l t
Fig. 36-1
Note that this relation was obtained without referenceto the explicit form of y(t). Rewritten as
Y.", = -
Ay or A y = Ky,,, At
K At
it provides a useful description of exponential processes: the change in a quantity over any time intervul is
proportional to the size of the interval and to the average value of the quantity over that interval.
36.7 If the bacteria in a culture grow exponentially and if their number y doubles in 1 hour, how long
will it take before 10times the original number is present?
We know that y = y, eKt.Sincethe number after 1 hour is 2y,,
2y, = y, 8')
= y, eR or 2 = eK or K = In 2
So,y = y, e(ln')'; and when y is lOy, ,
IOy, = yoe('"')'
In 10 = (In 2)t
10 = e(ln 2)r
In 10 2.3026
In 2 0.6931
t = - X - N -3.32
Thus, it takes a little less than 33 hours for the number of bacteria to increase tenfold.
36.8 Given that the half-life T of radium is 1690 years, how much will remain of 1 gram of radium
after 10OOO years?
In 2 In 2
T 1690
By (36.4),K = --= - -and the quantity y of radium is given by y = y, eRr= e-(In2)2/1690.
Appendix F was used to evaluate e-4.1. Hence, about 16.6 milligrams is left after 100oO years.

290 L'HQPITAL'S RULE; EXPONENTIAL GROWTH AND DECAY [CHAP. 36
36.9 Find the indicated limits:
5x3- 4x +3 In (1 +ex) 1 -cos x
(a) lim (b) lim (c) lim
X + + Q ' + ' x+o x
1 - ex
(e) lim -
(4 lim (
: -J-)
x-ro sin x x+o x
X'
(f) lim -
x+ +a0 (In XI3
tan x
(h) lim ( - !) (i) lim -
x3 - x2 +x - 1
x+o In (x + 1) x x-0 x
(9) lim
x + l x + l n x - 1
1 - sin x
(m) lim -
x-u/2 x -4 2
3" - 2"
(k) lim -
x+o x
(n) lim xsin
x-o+
ex
(I) lim -
x+o x2
x3 - 1
(0) lim -
x-1 x + 1
sin x
(r) lim -
x-o+ J
x
In x
(s) lim -
x-rl tan nx
sin 3x e3x- I
(t) lim - (U) lim -
x+o sin 7x x-+o tanx
1 - cos2 2x tan x - sin x
(U) lim (w) lim
x-0 x2 x-0 x3
36.10 Sketch the graphs of the following functions, indicating relative extrema, concavity, inflection points, and
asymptotes:
X
(4 Y = 2 (b) y = x2e-x (c) (d) y = x 2 1 n x (e) y = e - x s i n x
36.11 A bacteria culture grows exponentially so that the initial number has doubled in 3 hours. How many times
the initial number will be present after 9 hours?
36.12 The half-life of radium is 1690years. If 10per cent of an original quantity of radium remains, how long ago
was the radium created?
36.13 If radioactive carbon-14 has a half-life of 5750years, what will remain of 1gram after 3000 years?
36.14 If 20 per cent of a radioactive element disappears in 1year, compute its half-life.
36.15 Fruit flies are being bred in an enclosure that can hold a maximum of 640 flies. If the flies grow exponen-
tially with growth constant K = 0.05 per day, how long will it take an initial population of 20 to fill the
enclosure? (Recallthat In 2 2 0.6931.)
36.16 A certain chemical decomposes exponentially. Assume that 200 grams become 50 grams in 1 hour. How
much will remain after 3 hours?
36.17 If 100 bacteria in a colony reproduce exponentially and in 12 hours there are 300 bacteria, how many
bacteria are in the colony after 72 hours?
36.18 If the world population in 1990 was 4.5 billion and it is growing exponentially with growth constant
K = 4 In 2 when time is measured in years, find the population in the year 2020.
36.19 Bacteria in a culture growing exponentially increase from 100 to 400 in the first hour. (a)What will be the
population in 1.5 hours? (b) What is the average number present in the first 2 hours?

3
6
.
2
0 Prove Cauchy's extended mean-value theorem: If f and g are differentiable in (a, b) and continuous on
[a, b],and ifg'(x) # 0 for all x in (a, b),then there exists a point c in (a, b)such that
[Hint:
Apply the generalized Rolle's theorem (Problem 17.19)to
h(x) = (f(b) -f(a))g(x) - (g(b)- s(a))f(x)
Note that g(b) # g(a) since,otherwise, the generalized Rolle's theorem would imply that g'(x) = 0 for some x
in (a, b).)
36.21 Prove L'H6pital's rule. [Hint:
Consider the case lim,,,+ (f(x)/g(x)),
where lim,,,, f ( x ) = lirn,,,, g(x)= 0.
We may assume that f(a) = g(a) = 0. Then, by Problem 36.20, f(x)/g(x)= ( f ( x )-f(a))/
@(x) -g(a))=f'(x*)/g'(x*) for some x* between a and x. Therefore, as x + U + , x* -11'. Hence, if
(f'(x)/g'(x))= L, then lim,,,, (f(x)/g(x))= L. The other cases can be handled in the same way or
can be reduced to this case.]

Chapter 37
Inverse Trigonometric Functions
37.1 ONEONE FUNCTIONS
In Section 35.3, we introduced the notion of the inverse of a function, and we showed that the
inverse of In x is ex,and vice versa. Not all functions, however, have inverses.
EXAMPLES
(a) Consider the functionf such that f ( x )= x2 for all x. Thenf(1)= 1 and f(-1)= 1. If there were an inverse g
of f, then g(f(x)) = x. Therefore, g(1) = g(f(1)) = 1 and g(1) = g(f(- 1
)
) = -1, implying that 1 = -1,
which is impossible.
(b) Letfbe any periodic function,f(x +p) =f(x), for all x (see Section 26.2). The argument of example (a),for two
points xo and xo +p, shows that f cannot have an inverse. Now all the trigonometric functions are periodic
(with either p = 2n or p = n).Hence, the trigonometric functions do not have inverses!
The functions that have inverses turn out to be the one-one functions.
Definition: A functionfis one-one if, whenever U # v,
f(U)# f (0)
Thus, a one-one function takes different numbers into different numbers. A function is one-one if
and only if its graph intersects any horizontal line in at most one point. Figure 37-l(a) is the graph of a
one-one function; Fig. 37-l(b) graphs a function that is not one-one becausef(U)=f(t))= c.
-7
c
-
J
*
X
Y
1 1 *
U U
Fig. 37-1
X
*
~~~~~~~~ ~ ~~~ ~ ~ ~ ~ ~ ~ ~
NOTATION The inverse of a one-one functionf will be denoted byf -
f
Warning: Do not confuse the inversef-' with the reciprocal 1
6
292

CHAP. 371 INVERSE TRIGONOMETRIC FUNCTIONS 293
If a one-one function f is defined by means of a formula, y =f(x), we can sometimes solve this
equation for x in terms of y. This solution constitutes the formula for the inverse function, x =f-'(y).
EXAMPLES
(a) Letf(x) = 3x + 1(a one-one function). Let y stand forf(x); then y = 3x + 1. Solve this equation for x in terms
of Y,
y - l = 3 x
--
Y - L x
3
Therefore,the inversef-' is given by the formula
Y-1
f-'(y) = -
3
(6) Consider the one-one functionf(x) = 2e" - 5,
y = 2e" - 5
y +5 = 2e"
Y + 5
2
-- - ex
Y + 5
In -= In ex = x
2
Thus,
(c) Let f(x) = xs +x. Sincef'(x) = 5x4+ 1 > 0,f is an increasing function, and therefore one-one (see Problem
37.12). But if we write y = xs +x, we have no obvious way of solving the equation for x in terms of y.
37.2 INVERSES OF RESTRICTED TRIGONOMETRIC FUNCTIONS
restricted to some subset of one period.
For a periodic function to become one-one-and so to have an inverse-its domain has to be
Inverse Sine
The domain off(x) = sin x is restricted to [-4 2 , 421, on which the function is one-one [in fact,
increasing; see Fig. 37-2(a)]. The inverse function f - '(x) = sin- 'x is called the inverse sine of x (or,
sometimes, the arc sine of x, written arc sin x or arcsin x). Its domain is [-1, 13.Thus,
tY
( a ) y = sin x
Fig. 37-2
(b) y = sin-'x

294 INVERSE TRIGONOMETRIC FUNCTIONS [CHAP. 37
The graph of sin-' x is given in Fig. 37-2(6). By a general theorem, the inverse of an increasing
function is itself an increasing function. You will find it helpful to think of sin- ' x as the angle (between
-n/2 and 4 2 ) whose sine is x.
EXAMPLES
1 a
2 6
fi a (4 sin-'-=-
(c) sin-' -= -
2 4
a J s a
(a) sin-' 1 = - (b) sin-' -= -
2 2 3
n n
(e) sin-' 0 = 0 (f)sin-l( - i)= - - (g) sin-'( - 1) = - -
6 2
The derivative of sin-' x may be found by the general method of Problem 35.9(6).In this case, a
shortcut evaluation is possible, using implicit differentiation. Let y = sin- ' x. Then,
sin y = x
d d
-(sin y) = -(x)
dx dx
dY
(cos y) -= 1 [by the chain rule]
dx t
dy 1
-1 1
-
_ -
-
dx cos Y + J i T F y = +JS
where the positive square root is implied because cos y > 0 for -n/2 < y < 4 2 . Thus we have found
dx = sin-' x +C
1 1
DJsin - x) =
Jc7 (37.1)
The importance of the inverse sine function in calculus lies mainly in the integration formula (37.1).
The procedure for the other trigonometric functions is exactly analogous to that for the sine func-
tion. In every case, a restriction of the domain is chosen that will lead to a simple formula for the
derivative of the inverse function.
Inverse Cosine (Arc Cosine)
(a) y = COS% (b) y =cos-'x
Fig. 37-3

CHAP.371 INVERSE TRIGONOMETRIC FUNCTIONS 295
y=cosx [ O I x 1 a ] e> x=cos-'y [ - 1 I y I 1 ]
Interpret cos-' x as the angle (between0 and II)
whose cosine is x. Let y = cos-' x. Then,
cos y = x
&(cos y) = D,(x) = 1
* d
Y
- m y - = 1
dx
Note that sin y = +,/1 - cos2y since sin y > 0 when 0 < y < a. Thus,
dx = -COS-' x +C
1 1
D,(cos-' x) = -
Ji=7
Observe that - cos-' x and sin-' x have the same derivative;see Problem 37.17.
Inverse Tangent (Arc Tangent)
(a) y = tan x
iII
(b) y = tan-'x
Fig. 37-4
(37.2)
Interpret tan-' x as the angle (between -a/2 and z/2) whose tangent is x. Let y = tan-' x. Then,
tan y = x
D,(tan y) = 1
d
Y
sec2y -= 1
dx
1
--
-
1
-
d
Y 1
----
-
dx sec2y 1+tan2 y 1 +x2
Thus, dx = tan-' x +C
1 and 1-1
&(tan-' x) = -
1 + x 2 1 + x 2
(37.3)

296 INVERSE TRIGONOMETRIC FUNCTIONS
- 1
Inverse Cotangent (Arc Cotangent)
I
Y
(a) y = c o t x (6) y = cot-'x
Fig. 37-5
y = c o t x [ O < x < n ] - x=cot-'y [ally]
Interpret cot-' x as the angle (between0 and n)whose cotangent is x. Then,
dx = -cot-' x +c
1
DJcot-' x) = --
1 +x2 Or
For the relation between cot-' x and tan-' x, see Problem 37.17.
Inverse Secant(Arc Secant)
I
I
-
I - I
I
I
I
I
I
I
I
I
I
Y
)I
I
I
I
I
I 1
I
w 3
x
;;
-X
[CHAP. 37
(37.4)
W
f
I
(6) y = sec-'x
X
( a ) y =secx
Fig. 37-6

The secant function is restricted to two separate subintervals of the fundamental period
(-71/233 7 m ,
0I x < 71/2
y=secx [or ] e> x=sec-'y [Iy( 2 11
71 I x < 37112
Interpret sec-' x as the angle (in the first or third quadrant) whose secant is x. Then,
dx = sec-' x +C
1 1
Dx(sec-' x) =
X J Z T
See Problem 37.7.
Inverse Cosecant (Arc Cosecant)
(37.5)
( U ) y = cscx
Fig. 37-7
1 I 1 L
(6)y = csc-' x
0< x I71/2
y=cscx [or ] * x=csc-' y [IyI 2 13
?I < x I 37112
Interpret csc-' x as the angle (in the first or third quadrant) whose cosecant is x. Then,
dx = -csc-' x +c
1 1
Dx(csc-'x) = -
X J Z i
For the relation between csc-' x and sec-' x, see Problem 37.17.
(37.6)

SolvedProblems
37.1 A one-one function has an inverse. Prove, conversely, that a function with an inverse is one-one.
If the domain off consists of a single point, there is nothing to prove. Let f have the inverse g, but
suppose, contrary to hypothesis, that the domain off contains two distinct numbers, U and U, such that
f(u) =f(v).Then,
= g(f(u)) = g(f(d) = "
a contradiction. We must then admit thatf(u) #f(o); that is, thatfis one-one.
37.2 Determine whether each of the following functionsfis one-one. If it is one-one, find a formula for
its inversef - '.
(a) f ( x ) = x2 +6x - 7 (6) f ( x ) = In (x2 + 1) (c) f ( x )= 5x3 -2
b)f ( x )= x2 +6x +7 = (x +7)(x - 1).Becausef( -7) = 0 =f(l),fis not one-one.
(4Obviously,f( +1)=f(-1);fis not one-one.
(c) If y = 5x3 - 2, then,
y +2 = 5x3
--
y + * - x 3
5
Thus,f(x) = 5x3 - 2 has an inverse,
f-'(y) = 3 Y
J ;*
and so, by Problem 37.1,fis one-one.
37.3 Find formulas for the inverses of the following one-one functions:
(a) f ( x )= 1Ox - 4 (6) f ( x )= 3e" + 1
(a) Let y = 1Ox - 4. Then,
- x
Y + 4
y+4=1Ox or --
10
Y + 4
f-'(J) = -
10
Hence,
(b) Let y = 3e" + 1.Then,
y - 1 =3ex or -- - e x or In --
Y - L
3
Y-1
3
Y-1
f-'(y) =In -
3
Hence,
37.4 Find the following values:
(a) sin-' (-k) (6) cos-' (-k) (c) tan-' (-1)
a n 1 A
(a) sin-' (-5)= the angle in [- 5,T] whose sine is -- = - -
2 6 '

1 a 5n
(b) cos- '(-i)= the angle in [0, a] whose cosine is - - -
(c) tan- '(- 1)= the angle in (-;
, ;
) whose tangent is -1 = --
4 '
2 - -6 = 6'
a
37.5 Evaluate: (a)cot-' fi;(6)sec-' 2; (c)csc-' (2/fi).
(a) cot-' fi is the angle 8between 0 and a such that cot 8 = fi.This angle is a/6 (see Fig. 37-8). Note
that an inverse trigonometric function must be represented by an angle in radians.
(b) The angle 8 whose secant is + 2 must lie in the first quadrant. The secant function is negative in the
third quadrant. From Fig. 37-8,8 = a/3.
(c) The angle 8 whose cosecant is + 2 / f i must lie in the first quadrant. The cosecant function is negative
in the third quadrant. From Fig. 37-8,8 = a/3.
Fig. 37-8
37.6 Find the derivatives of:
X
(a) sin-' 2x (6) cos-' x2 (c) tan-' - (d) sec-' x3
2
Dx(2x) [by the chain rule]
1
J
W
Ji=G JGG
J
-
(a) D,(sin-' 2x) =
1 2
(2) =
-
-
- 1
(b) DJCOS- 'x2) = Dx(x2) [by the chain rule]
- 1 -2x
(2x) = -
=
-
Ji=7 JD
1
D(;) [by the chain rule]
4 1
1+- 1+-
4
Dx(x3) [by the chain rule]
1
(d) DJsec - 'x3)=
x3J-
3
(3x2)=
1
-
-
x 3 J G X p T

1
3
7
.
7 Assuming that sec-' x is differentiable,verify that Dx(sec-' x) =
xJ.T-1'
Let y = sec-' x. Then,
s e c y = x
&(sec v)= 4 ( x )
dY
dx
dY
dx sec y tan y x tan y
(sec y tan y ) -= 1 [by the chain rule]
1
--
-
1
-
--
But the identity 1 +tan' y = sec2y gives
1 +tan2 y = x2 or tan2 y = x' - I or tan y = +JZi
By definition of sec-' x, the angle y lies in the first or third quadrant, where the tangent is positive. Hence,
1
-
dY
tan y = +
,
/
- and --
dx xJ;;i_l
3
7
.
8 Prove the followingformulas for antiderivatives:
1 X
a' +x2 a a
(a) J'L
= - tan-' - + c
NOTATION It is common to write
In each case, we employ the chain rule to show that the derivative of the function on the right is the
integrand on the left.
1 1 1
a
' x2 a2 +x'
=--=-
l + T
a

(a) By Problem 37.8(a), with a = 2,
(b) By Problem 37.8(b), with a = 3,
dx X
3
1 = sin- 1 -+c
(c) Complete the square: x2 +4x +5 = (x +2)2+ 1. Thus,
dx
Let U = x +2. Then du = dx, and
(d) Complete the square: x2 - 2x +7 = (x - 1)' +6. Let U = x - 1, du = dx, 2x = 2u +2. So,
f x 2 T t + 7 u 2 + 6
= s-2u +2 du = 1
m
2u du +1%
= In (
U
' +6) +2 -
2 x - 1
=In (x2 - 2x +7) +-tan-' -
U
[by quick formula I1 and Problem 37.8(a)]'
1
fi
37.10 Evaluate sin (2 cos- (- 4)).
Let 8 = cos-' (-4). Then, by Theorems 26.8and 26.1,
sin 28 = 2 sin e COS e = 2(fJi -COS2 excos e)
By definition of the function cos-', angle 8 is in the second quadrant (its cosine being negative), and so its
sine is positive. Therefore, the plus sign must be taken in the above formula,
We have written In (u2 +6) instead of In Iu2 +6 Ibecauseu2 +6 >0.

37.11 Show that sin-' x and tan-' x are odd functions.
More generally, we can prove that if a one-one function is odd (as are the restricted sin x and the
restricted tan x, but not the restricted cot x), then its inverse function is odd. In fact, iffis an odd one-one
function and g is its inverse, then
g(-f(x)) = s(f(
-x)) [sincef(4= -f(-41
= -x
= -g(f(x))
[since g is inverse off]
[since g is inverse off]
which shows that g is odd. (Could a one-one function be euen?)
37.12 For each of the following functionsf, determine whether it is one-one; and if it is, find a formula for the
inversefunctionf - '.
37.13 Evaluate :
(6) tan-' -
3
3
37.14 (a) Let 8 = cos- '(4). Find the values of sin 8, tan 8, cot 8, sec 8, and csc 8.
(b) Let 8 = sin- ' (- 4). Find cos 8, tan 8, cot 8, sec 8, and csc 8.
37.15 Compute the following values:
(a) sin cos-' -
( :> (e) sin-' (sin n)
[Hints: Part (b)52 + 122= 132;part (c) cos (U +U ) = cos U cos U - sin U sin U ; part (e)is a trick question.]
37.16 Find the domain and the range of the functionf(x) = cos (tan- 'x).
37.17 Differentiate:
1
(a) sin-' x +cos-' x (b) tan-' x +cot-' x (c) sec-' x +csc-' x (6) tan-' x +tan-' -
(e) Explain the significance of your answers.
X

37.18 Differentiate:
(a) x tan-' x (b) sin-' J;; (c) tan-' (cos x) (6) In (cot-' 3x)
(e) ex cos-' x (f) In (tan-' x)
X
(h) x , / n +U
' sin-' -
a
1
(9) csc-' -
X
1 2
U) sin-' - +sec-' x (k) tan-' -
X X
37.19 What identity is implied by the result of Problem 37.18(i)?
37.20 Find the followingantiderivatives:
x dx
(q)
[Hints:(b) let U = 3x; (d) let U = 4x; (e) let U = x - 3; (0 let U
' = x4 +9; (m)complete the square in
x2 - 6x; (n)D,(6x - x') = 6 - 2x; (p)divide x3by x2 - 2x +4.1
37.21 Find an equation of the tangent line to the graph of y = sin-' (x/3) at the origin.
37.22 A ladder which is 13 feet long leans against a wall. The bottom of the ladder is sliding away from the base
of the wall at the rate of 5 feet per second. How fast is the radian measure of the angle between the ladder
and the ground changing at the moment when the bottom of the ladder is 12feet from the base of the wall?
37.23 The beam from a lighthouse 3 miles from a straight coastline turns at the rate of 5 revolutions per minute.
How fast is the point P at which the beam hits the shore moving when that point is 4 miles from the point
A on the shore directly opposite the lighthouse (see Fig. 37-9)?
L
A P
Fig. 37-9

1
37.24 Find the area under the curve y = -
above the x-axis, and between the lines x = 0 and x = 1.
1 + x 2
above the x-axis,and between the lines x = 0 and x = 1/2.
37.25 Find the area under the curve y = -
1
Ji=7
1
37.26 Tine region 9 under the curve y = ~ above the x-axis, and between the linesx = 2 / f i and x = 2,
X2. f i
is revolved around the y-axis. Find th; volume of the resultingsolid.
37.27 Use the arc-length formula (32.2) to find the circumferenceof a circle of radius r. [Hint: Find the arc length
of the part of the circle x2 +y2 = r2 in the first quadrant and multiply by 4.1
37.28 A person is viewing a painting hung high on a wall. The vertical dimensionof the painting is 2 feet and the
bottom of the painting is 2 feet above the eye level of the viewer. Fiild the distance x that the viewer should
stand from the wall in order to maximizethe angle 8 subtended by the painting. [Hint: Express 8 as the
differenceof two inverse cotangents;see Fig. 37-10.]
X
Fig. 37-10
37.29 For which values of x is each of the followingequationstrue?
(a! sin-' (sin x) = x (b) cos-' (cos x) = x (c) sin-' (-x) = -sin-' x
(6) sin (sin-' x) = x (e) cos (cos-' x) = x
(J^)
Part (4
Use a graphing calculator to draw the graph of y = sin-' (sin x) and review your answer to
3730 Find y' by implicit differentiation:
(a) x2-- x tan-' y =In y (b) cos-' xy = e2Y
3731 Sketch the gaph of y = tan-' x - In J1 +x2.
3732 Assuming that cot-' x and csc-' x are differentiable,use implicit differentiationto derive the formulas for
their derivatives.
1
3733 Find the averagevaluesoff(x) = -
+ x2 on [-2,21.
37.34 (a) Show that IC = (b) ApproximateIC by applyingSimpson's rule to the definite integral in part
(a),with n = 8.

Chapter 38
Integration by Parts
In this chapter, we shall learn one of the most useful techniques for finding antiderivatives. Letland
g be differentiablefunctions. The product rule tells us that
d
dx
-(f(x)g(x))=f(x)g’(x)+g(x)f’(x)
or, in terms of antiderivatives,
The substitutions U = f ( x ) and v = g(x) transform this into’
uv = [ U do +[ V du
from which we obtain
U dv = uv - v du integration by parts
s s
The idea behind integration by parts is to replace a “difficult” integration 1U do by an “easy”
integration 1v du.
EXAMPLES
(a) Find xex dx. This will have the form U du if we choose
I Iu = x and du=e“dx
Since d t ~
= u’(x) dx and do = ex dx, we must have U’@) = ex. Hence,
u = I e x dx = ex +C
-
and we take the simplest case, C = 0, making U = ex.
Sincedu = dx, the integrationby parts procedure assumesthe followingform:
U d v = u ~ - v du
I I
s
xex dx = xe” - ex dx
= xex -ex +C = ex(x - 1) +C
For example, 1f(x)g’(x) dx = U do, where, in the result of the integrationon the right, we replace U byf(x) and U by g(x).In
fact,by the chain rule,
Hence,j U do = f(x)g’(x)dx. Similarly, o du = g(x)f’(x) dx.
305

306 INTEGRATION BY PARTS [CHAP. 38
Integration by parts can be made easier to apply by setting up a box such as the following one for
example (a) above:
u = x do = ex dx
~ d u = d x U = ex
In the first row, we put U and do. In the second row, we write du and U. The result UD - U du is
obtained from the box by first multiplying the upper left corner U by the lower right corner o and then
subtracting the integral { D du of the two entries in the second row.
Notice that everything depends on a wise choice of U and U. If we had instead picked U = ex and
du = x dx, then U = j x dx = x2/2 and we would have obtained
I
which is true enough, but of little use in evaluating
integration 5 xe“ dx by the even more “difficult”integration 1(x2/2)ex dx.
(b) Find
xex dx. We would have replaced the “difficult”
x In x dx. Let us try U = In x and dv = x dx:
I I u = l n x d v = x d x
Then, v = x dx = x2/2.Thus,
I U dV=uv - v du
I I
J
X2
2 2
=-In x -1{x dx
X2 1 x2
- l n x - - -
2 2 2 + c
--
(c) Find sIn x Let us try U = In x and L- = d x :
Then, v = dx = x. Thus,
I
r U = In x dv = dx I
U dv=ut, - v du
I I
In x dx = x In x - x - dx
I I:
= x l n x - dx
= x In x - x +C
= x(ln x - 1) +C
s

CHAP. 381 INTEGRATION BY PARTS 307
(d) Sometimes two integrations by parts are necessary.Consider excos x dx. Let U = ex and du = cos x dx:
I
U = ex du = cos x dx
I
du = exdx U = sin x
Thus, ex cos x dx = e" sin x - ex sin x dx
s I
Let us try to find ex sin x dx by another integration by parts. Let U = e"and du = sin x dx:
I
du = ex dx U = - cos x
Thus, ex sin x dx = -ex cos x -
= -ex cos x +
(-ex cos x) dx
ex cos x dx
I
I
I
Substitute this expression for exsin x dx in (1) and solve the resultingequation for the desired antiderivative,
I
sex cos x dx = 8 sin x - (-ex cos x +s e x cos x dx)
Iex cos x dx = 8 sin x +8 cos x - 1ex cos x dx
J
a
J
2 ex cos x dx = e" sin x +ex cos x = e"(sin x +cos x)
e" cos x dx =
e"(sin x +cos x)
2
+ C
J
Solved Problems
38.1 Find xe-x dx.
s
Let
Integration by parts gives
xe-x dx = -xe-x - (-e-x) dx = -xe-* + e-x dx
I s
= -xe-x -e-x + C = -eex(x + 1)+ C
s
Another method would consist in making the change of variable x = --t and using example (a)of this
chapter.

308 INTEGRATION BY PARTS
38.2 (a) Establish the reductionformula
x"exdx = x"e" -n xn-'eX dx
s s
[CHAP. 38
(2)
for xnexdx (n = 1,2, 3, ...).
s
(4 Let
U = x" du = ex dx
and integrate by parts,
x " 8 dx = Yex - ex(nx"-')dx = x"8 - n x"-'ex dx
s s s
s s
s
(b) For n = 1, (2)gives
xe" dx = xe" - ex dx = xex - ex = (x - 1)ex
as in example (a).We omit the arbitrary constant C until the end of the calculation.Now let n = 2 in (2),
x2ex dx = x2ex- 2 xex dx = x2e"- 2((x - 1)eX)
= (x2 - 2(x - l)P= (x2 - 2x +2)e" +c
3
8
.
3 Find tan'' x dx.
Let
s
I
I u=tan" x d u = d x
du = -dx u = x
I 1 +x2
Hence,
X
tan-' x dx = x tan-' x - -
J 1 +x2 dx
=xtan-l x - - 1-
2x dx
J
2 1 + x 2
1
= x tan-' x -- In II +x21+ C
2
1
= x tan-' x -- In (1 +x2) + C
2
38.4 Find cos2 x dx.
Let
s
[by quick formula 11, Problem 34.53
[since 1 +x2 > O]
du = - sin x dx U = sin x

CHAP. 38)
Then,
INTEGRATION BY PARTS
cos’ x dx = cos x sin x -
= cos x sin x +
= cos x sin x +
= cos x sin x +
sin2x dx
(1 - cos’ x) dx
1 dx -
I
s
1 5cos’ x dx
Solving this equation for cos’ x dx,
s2 cos’ x dx = cos x sin x + 1dx = cos x sin x +x
s I
1
cos’ x dx = - (cos x sin x +x) +C
I 2
This result is more easily obtained by use of Theorem 26.8,
s
2
38.5 Find x tan-’ x dx.
s
Let
Then,
But
and so
Hence,
I u=tan-’x dt,=xdx 1
I
I 1 + x 2 2
1 X 2
du = -dx U = -
2
‘s1+ X 2 d x
x2
X’
2
x tan-’ x dx = -tan-’ x -- -
x2 (1 +x’)- 1 1+x2 1 1
1 + x 2 - 1+x2 1 + x 2 l + x z 1 +x2
1--
-- =---=
X’ 1
x tan-’ x dx =-tan-’ x -- (x - tan-’ x) + C1
s 2 2
1
2
1
2
= - (x’ tan-’ x -(x -tan-’ x)) +C,
= -(x2tan-’ x - x +tan-’ x) + C,
= -
1 ((tan-’ XXX’ + 1) -x) +C,
2

310
3
8
.
6
38.7
38.8
38.9
38.10
38.11
38.12
38.13
Compute:
(a) dx
(e) I x cos x dx
(i) Ieaxcos bx dx
(m) dx
(4) I x sin 2x dx
(U) I x 2 tan-' x dx
INTEGRATION BY PARTS
(b) I e x sin x dx
(f) I x 2 sin x dx
0
9 I s i n ' x d x
(n) {xse!c2 x dx
(r) { x sin x2 dx
(U) 11. (x2 + 1)dx
(c) s x 3 e xdx
(g) 1.0.(In x ) dx
(k) 1 ~ 0 s '
x dx
(0) I x cos2x dx
(s) {%dx
[CHAP. 38
(d) [ s i n - ' x d x
J
P
(h) Ix cos (5x - 1)dx
J
*
J
fl
[Hint: Integration by parts is not a good method for part (r).]
Let 41 be the region bounded by the curve y = In x, the x-axis, and the line x = e. (a) Find the area of 41.
(b) Find the volume generated by revolvingW about (i)the x-axis;(ii) the y-axis.
Let 9 be the region bounded by the curve y = x-' In x, the x-axis, and the line x = e. Find: (a)the area
of 9;(b) the volume of the solid generated by revolving 41 about the y-axis; (c) the volume of the solid
generated by revolvingW about the x-axis. [Hint:
In part (c), the volume integral,let U = (In x ) ~ ,
Y = -l/x,
and use Problem 38.6(s).]
Derive from Problem 38.8(c)the (good)bounds: 2.5 s e I; 2.823. [Hint: By Problem 30.3(c),
5 e - 1
0 I; 2 - - 5 -
or
0I; [(yydx I; 5
e2 e e2
The left-hand inequality gives e 2 3;the right-hand inequalitygives e 5 (3 +&2.]
Let 9t be the region under one arch of the curve y = sin x, above the x-axis, and between x = 0 and x = n.
Find the volume of the solid generated by revolving9 about: (a)the x-axis;(b) the y-axis.
If n is a positive integer,find:
(a) I x x cos nx dx (b) J'bx sin nx dx
For n = 2, 3,4, ...,find reduction formulasfor:
(a) !cos" x dx
(c)Use these formulasto check the answers to Problems 38.4,38.6(k), 38.6(1),and 38.60).
(b) Isin" x dx
(a) Find a reduction formula for I sec" x dx (n = 2, 3, 4, ...). (b) Use this formula, together with Problem
34.7, to compute: (i)I sec3x dx; (ii) sec4 x dx.

Chapter 39
Trigonometric lntegrandsand
Trigonometric Substitutions
39.1 INTEGRATION OF TRIGONOMETRIC FUNCTIONS
We already know the antiderivativesof some simple combinations of the basic trigonometric func-
tions. In particular, we have derived all the formulas given in the second column of Appendix B.Let us
now look at more complicatedcases.
EXAMPLES
(a) Consider I sinkx cos" x dx, where the nonnegative integers k and n are not both even. If, say, k is odd
(k= 2j + l), rewrite the integral as
sin2'
s x cos" x sin x dx = (sin2x)' cos" x sin x dx
= J(1- cos2 x)' cos" x sin x dx
s
Now the change of variable u = cos x, du = -sin x dx, produces a polynomial integrand. (For n odd, the
substitutionu = sin x would be made instead.)For instance,
I- I-
cos2x sin5x dx = cos2 x sin* x sin x dx
J J
= J cos2 x(1- cos2 x)' sin x dx
= Iu'(1 - u2)2(-1) du = +(1 - 2u2 +U*) du
u3 us
= -
= -- cos3 x +- cos5 x --COS' x +c
- 2#4 +u6) du = - (
s- 2 +:
) +c
1 2 1
3 5 7
(b) Consider the same antiderivative as in part (a), but with k and n both eoen; say, k = 2p and n = 2q. Then, in
view of the half-angleidentities
1+cos2x 1 -cos 2x
2
and sin2x =
2
cos2x =
we can write
[sink x cos" x dx = [(sin2 xy(cos2 xp dx
J
1
2P+Q
= -[(l - cos 2x)P(l +cos 2x)4 dx
311

312 TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS [CHAP. 39
When the binomials are multiplied out, the integrand will appear as a polynomialin cos 2x,
1 +(q - pxcos 2x) + .* * &
- (cos 2x)P+Q
and so
1dx +(q -p) (cos 2x) dx + - - * (cos 2X)P+4dx
I I
sin' x cos" x dx = 2p+4
I
On the right-hand side of ( I ) are antiderivatives of odd powers of cos 2x, which may be evaluated by the
method of example (U), and antiderivatives of etfenpowers of cos 2x, to which the half-angle formula may be
applied again. Thus, if the sixth power were present, we would write
and expand the polynomial in cos 4x, and so forth. Eventually the process must end in a final answer, as is
shown in the followingspecific case:
cos2x sin4x dx = (cos' xxsin' x)' dx
I I
1 +cos 2x 1- 2 cos 2x +cos22x
=I( 2 >( 4
= 1J(l(1 - 2 cos 2x +cos22x) +(cos 2xK1 - 2 cos 2x +cos22x)) dx
8
1
8
8
= - I ( 1 - 2 COS 2x +cos22x +COS 2x - 2 cos22x +cos32x) dx
= 1J
'
(
i - cos 2x - cos2 2x +cos3 2x1dx
=! (I1 dx -Ices2x dx - Icos2 2x dx + cos32x dx
8 I )
dx + (cos 2 ~ x 1
- sin22x) dx
I
sin 2x sin 4x
=I(
8 x - --I
2 2 (x +7)
+Jcos 2x dx - A
2 1 . 2 du) uetting u = sin 2x1
(c) From Problem 3444, we know how to integratethe first power of tan x,
tan x dx = In lsec xl +C
I
Higher powers are handled by means of a reduction formula. We have, for n = 2
, 3, ...,
Itan" x dx = tan"-' x(tan2x) dx = tan"-' x(sec2 x - 1) dx
s I
I
= J'tann-2 x sec2 x dx - tan"-2 x dx
= Sum-. du - !tan"-2 x dx [let u = tan x3
tan"-' x
n - 1
-
--
- Jtann-2 x dx (39.2)

CHAP. 391 TRIGONOMETRIC INTEGRANDS AND TRIGONOMETRIC SUBSTITUTIONS 313
Similarly,from
sec x dx =In lsec x +tan xl +C
I
and the reduction formulaof Problem 38.13(a), we can integrate all powers of sec x.
Antiderivatives of the forms
using the identities
sin Ax cos Bx dx, 5 sin Ax sin Bx dx, cos Ax cos Bx dx can be computed by
1
2
1
2
1
2
sin Ax cos Bx = - (sin (A +B)x +sin (A -B)x)
sin Ax sin Bx = - (cos (A -B)x - cos (A +B)x)
cos Ax cos Bx = - (cos (A -B)x +cos (A +B)x)
For instance,
11
sin 8x sin 3x dx = - (cos 5x - cos 1lx) dx =-
s I:
39.2 TRIGONOMETRIC SUBSTITUTIONS
,
/
=
,
it is often helpful to substitute a trigonometric function for x.
To find the antiderivative of a function involving such expressions as ,
/
-
’ or ,
/
G
2
or
EXAMPLES
(a) Evaluate ,
/
- dx.
I
None of the methods already available is of any use here. Let us make the substitution x = fi tan 8,
where -n/2 -c8 < n/2. Equivalently, 8= tan-’ (x/&). Figure 39-1 illustrates the relationship between x
and 8,with 8interpreted as an angle. We have dx = fi sec’ 8d8 and, from Fig. 39-1,
--
m - e c e or , / ~ = a = e
a
where sec 8> 0 (since -n/2 < 8 -cn/2).Thus,
Id mdx = I($ sec 8 X f i sec2 8)d8 = 2
= sec 8 tan 8 +In lsec 8 +tan 81+C [by Problem 38.13(b)]
J2 J2 I J2 J2l
Note how the constant -In was absorbed in the constant term in the last step. The absolutevalue signs in
the logarithm may be dropped, since Jm
+x > 0 for all x. This follows from the fact that d m>
p=
1x1 2 -x.
This example illustrates the following general rule: If ,
/
G
2
occurs in an integrand, try the
substitution x = U tan 8 with -(n/2) < 8 -=(42).

Fig. 39-1 Fig. 39-2
(b) Evaluate dx.
Make the substitution x = 2 sin 8, where - ~ / 25 8 s n/2; that is, 8 = sin-’ (x/2). The angle interpreta-
tion of 8is shown in Fig. 39-2. Now dx = 2 cos 8 d8 and
JC7-J- 2 J C Z Z j COS^
-- -
- =-= cot e
X 2 sin 8 2 sin 8 sin 8
Note that ,
/
- = ,
/
a
= ICOS 8I = COS 8, since COS 8 2 0 when -~ / 2
5 8 5 4 2 . Thus,
1 -sin’ 8
I sin 8
= 2 I* do = 2
sin 8
T
h
i
s example illustrates the following general rule: 1
’ ,
/
G
2
occurs in an integrand, try the
substitution x = a sin 8 with -n/2 I 8 I; a/2.
(c) Evaluate dx.
x’
Let x = 2 sec 8,where 0 5 8 < n/2 or x s 8 < 3 ~ / 2 ;
that is, 8 = sec- (x/2). Then dx = 2 sec 8 tan 8 d8 and
Jn
= J G Z i Z = 2 J G Z j Z = 2 J G Z = 21tan el = 2 tan e
Note that tan 8 > 0, since 8is in the first or third quadrant. Then
I F d x = I 2 t a n 8 (2 sec 8 tan 8) d8
X 8 sec’ 8
sin2 8
x x
1
4
= - (e - sin 8 COS 8) +c =
The general rule illustrated by this example is: Ifd- occurs in an integrand, try the substitu-
tion x = a sec 8, with 0 I; 8 <( 4 2 )or ~t
I; 8 < 31112.

Solved Problems
3
9
.
1 Find sin3x cos’ x dx.
s
The exponent ofsin x is odd. So,let U = cos x. Then, du = -sin x dx, and
sin3x cos2x dx = sin2x cos2 x sin x dx
= / ( I - cos2 x) cos2 x sin x dx
I
I
L L
3
9
.
2 Find cos4 x sin4 x dx.
I
The exponents are both even; in addition, they are equal. This allows an improvement on the method
of Section 39.1, example(b).
j’cos4 x sin4 x dx = I(yr
dx = Isin4 2x dx
dx
16
1
64
= -I ( 1 -2 cos 4x +cos24x) dx
=A(Jl
64 d x - ~ I c o s u d u + ~ j ’ c o s 2 u d u )
4 nettingu=4xI
= -(x - - sin u +- (u +sin u cos u)
1 1 1
64 2 8
1 x sin 4x cos4x
64 2 8
= -(x - 5sin 4x +- +
+-)+c
16
sin 8x
3x - sin 4x +-
8
2u3
C O S ~
x dx = (1 - 2u2 +U ~ )
d~ = U --+-
I I 3 5
2 sin3x sin’ x
3 5
= sin x --
+-+ C

(b) This antiderivative was essentiallyobtained in Problem 39.2,
sin4 x dx = 2 sin4 (2u)du
pet u = i]
I I sin 8u
= 2 -16
(3u -sin 4
u +-
sin 4x
= f (F-sin 2x +-
8
128 8
>,,
39.4 Find tan' x dx.
s
From the reduction formula (39.2),
tan2 x
tan x dx = -- In Isec x I
I 2 I 2
tan2 x
tan3 x dx = --
tan4 x tan2 x
tan3 x dx = -- -+In lsec x l +C
I 4 I 4 2
tan4 x
tan' x dx = -
-
39.5 Show how to find { tanPx secqx dx: (a) when 4 is even; (6) when p is odd. (c) Illustrate both
techniques with 1tan3 x sec4x dx and show that the two answers are equivalent.
(a) Let q = 2r (r = 1,2,3, ...).Then
a a
J J
a
= J tanp x ( l + tan2 xy-l(sec2 x) dx
since 1 +tan2 x = sec2 x. Now the substitution u = tan x, du = sec2x dx, produces a polynomial
integrand.
(b) Let p = 2s + 1(s = 0, 1, 2,. ..). Then,
tan2s+'x sec4 x dx = tan2' x secq- x(sec x tan x) dx
I
s
= J' (sec2 x - ly sec4- 1 x(sec x tan x) dx
I
I
since tan2 x = sec2x - 1.Now let U = sec x, du = sec x tan x dx, to obtain a polynomial integrand.
(4 BY Part (4
tan3 x sec4x dx = tan3 x(1 +tan2 x)(sec2 x) dx
= 1u3(1 +u2) du = I ( u 3 + du
u4 u6 tan4 x tan6 x
4 6 4 6
=-+-+c=-+- + C
tan3 x sec4 x dx = !(sec2 x - 1) sec3 x(sec x tan x) dx
= [(U. - l)u3 du = (us - u3) du
+ C
u6 u4 sec6 x sec4 x
= - - - + c = - - -
6 4 6 4
I
I

Since 1 +u2 = u2,
u6 u4 4u6- 6u4 4(1 +u ~ ) ~
-6(1 +u ~ ) ~
-
-
-_-- -
6 4 24 24
ALGEBRA (1 +t)3 = 1 +3t +3t2 +t3
q1 +3u2 +3u4 +u6) - q1 +2u2+u4)
-
24
4 + 12u2+ 12u4+4u6 -6 - 12u2- 6u4
-
24
6u4 +4u6 - 2 u4 u6 1
-
- =-+---
24 4 6 12
and so the two expressions for tan3 x sec4 x dx are equivalent. (The -Ais soaked up by the arbi-
trary constant C.)
39.6 Find tan' x sec x dx.
s
Problem 39.5 is of no help here.
tan2 x sec x dx = (sec2x - 1) sec x dx = (sec3x - sec x) dx
I s I
1
2
1 1
2 2
= - (sec x tan x +In Isec x +tan x I)- In Isec x +tan x I+ C
=-sec x tan x --In /secx +tan xl +C
[by Problem 38.13(b)]
39.7 Prove the trigonometric identity sin Ax cos Bx = *(sin (A +B)x +sin (A - B)x).
The sum and differenceformulas of Theorem 26.6give
sin ( A +B)x = sin (Ax +Bx) = sin Ax cos Bx +cos Ax sin Bx
sin ( A - B)x = sin (Ax - Bx) = sin Ax cos Bx -cos Ax sin Bx
and so, by addition, sin ( A +B)x +sin ( A - B)x = 2 sin Ax cos Bx.
39.8 Compute the value of sin nx cos kx dx for positive integers n and k.
1
'
Case I : n # k. By Problem 39.7, with A = n and B = k,
Jr:^sin nx cos kx dx = - [=(sin (n +k)x +sin (n - k)x)dx
2 n + k 0
1 cos (n +k)x + cos; ,t*>12*
--( - = O
because cos px is, for p an integer, a periodic function of period 2n.
Case 2: n = k. Then, by the double-angle formula for the sine function,
1 cos 2nx 2*
sin nx cos nx dx =- sin 2nx dx = -5(
T
)
]
= 0
r ;

dx
J;m*
39.9 Find
The presenceof ,
/
- suggestsletting x = 3 tan 8. Then dx = 3 sec28 do, and
,
/
=
= ,
/
- = J
- = 3 , / S = 3 sec 8
NOTE
and the constant -In 3 can be absorbed in the arbitrary constant K.Furthermore,
Jm-
+x > 0
dx
39.10 Find
and
The presenceof ,
/
- suggeststhe substitution x = J?sin 8.Then, dx = J?cos 8 do.
,
/
=
= ,/KG23 = J
- = & J c X = JScos e
dx
1
3
= - - c o t e + c
But
Hence,
dx
3x
39.11 Find
The occurrenceof ,
/
- suggeststhe substitutionx = 2 sec 8. Then dx = 2 sec 8 tan 8 do.
JZ'=
JGZiFi = ,
/
- = 2,/- = 2 tan e
d8 = 4 sec38 d8
5
X2 sec28x2 sec 8 tan 8)
and 1- dx = 5'" 2 tan 8
= 2(sec 8 tan 8 +In Isec 8 +tan 81) +C [by Problem 38.13(b)]
+ C
-
- xF-
x - 4 + 2 In Ix +J
=
2 2
+2 In ~x+,/Z
- xJx2 -4
-
2
where K = C - 2 In 2 (compareProblem 39.9).
+ C
+ K

(a) J'sin x cos2 x dx (b) !cos2 3x dx (c) ]sin4 x cos' x dx (d) [cos6 x dx
(e) x sin2x dx (f) J'tan2 5dx (g) J'tan6 x dx (h) [sec' x dx
(i) [tan2 x sec4 x dx
(m) Isin nx cos 3nx dx
6
9 [tan3 x sec3x dx
(n) J'sin 5x sin 7x dx
(k) [tan4 x sec x dx
(0) l c o s 4x cos 9x dx
(0 [sin 2x cos 2x dx
39.13
39.14
39.15
39.16
39.17
39.18
39.19
Prove the followingidentities:
1
2
1
2
(a) sin Ax sin Bx = -(cos (A -B)x -cos (A +B)x)
(b) COS Ax COS Bx =- (COS (A -B)x +COS (A +B)x)
Calculate the followingdefiniteintegrals,where the positive integers n and k are distinct:
(a) sin nx sin kx dx (b) 1%
sin2nx dx
0
Evaluate:
dx
(i) [ x 2 , / E i d x 0
9 [ e 3 x , / s dx
(k) I(x' - 6x + 13)2
[Hint: In part (k),completethe square.]
Find the arc length of the parabola y = x2from (0,O) to (2,4).
Find the arc length of the curve y = In x from(1,O) to (e, 1).
Find the arc length of the curve y = 8 from (0, 1)to (1,e).
Find the arc length of the curve y = In cos x from (0,O) to (n/3, -In 2).
39.20
x2 y2
Find the a
r
e
a enclosed by the ellipse-+-= 1.
9 4

Chapter 40
Integration of Rational Functions;
The Methodof Partial Fractions
This chapter w
i
l
lgive a general method for evaluating indefinite integrals of the type
where N(x)and D(x)are polynomials. That is to say, we shall show how to find the antiderivative of any
rational functionf(x) = N(x)/D(x)(see Section 9.3). Two assumptions will be made, neither of which is
really restrictive: (i) the leading coeficient (the coefficient of the highest power of x) in D(x) is +l;
(ii)N(x)is of lower degree than D(x)[that is,j'(x) is a proper rational function].
EXAMPLES
(a) - 4 ~ ' ~
+3x - 11
(b) Considerthe improper rationalfunctionf(x) = -
.Long division(see Fig.40-1) yields
-56x4
-
-
-7 8x4
-7 -jx'O + 3x - 11
--
-
8x4
x10 - 2 1 ~
+77
x4 +7x
x2 - 1
7x + 1
f(x) = x2 + 1 +-
x2 - 1
Consequently,
I f ( x ) dx = 1 ( x 2 + 1) dx +I-7x + 1 dx = -+
x3 x + 7x + 1
x2 - 1 3
and the problem reduces to finding the antiderivativeof a proper rational function.
x 2 + 1
x2 - 1 I x4 + 7x
x4 - x2
x2 + 7x
x2 - 1
7x + 1
Fig. 40-1
The theorems that follow hold for polynomials with arbitrary real coefficients. However, for sim-
plicity we shall illustrate them only with polynomials whose coefficientsare integers.
Theorem 40.1: Any polynomial D(x) with leading coefficient 1 can be expressed as the product of
linearfactors, of the form x - a, and irreducible quadraticfactors (that cannot be fac-
tored further),of the form x2 +bx +c, repetition of factors being allowed.
As explained in Section 7.4, the real roots of D(x)determineits linear factors.
320

CHAP. 4
0
1
EXAMPLES
THE METHOD OF PARTIAL FRACTIONS 321
x2 - 1 = (x - 1)(x + 1)
Here, the polynomial has two real roots (k1)and, therefore, is a product of two linear factors.
The root x = 3, which
D(x) by x - 3 yielded
formula,its roots are
x3 +2x2- 8x - 21 = (x - 3XX2 +5x +7)
generates the linear factor x - 3, was found by testing the divisors of 21. Division of
the polynomial x2 +5x +7. This polynomial is irreducible, since, by the quadratic
which are not real numbers.
Theorem 40.2 (Partial Fractions Representation): Any (proper) rational functionf (x) = N(x)/D(x)may
be written as a sum of simpler, proper rational functions. Each summand has as denomi-
nator one of the linear or quadratic factors of D(x),raised to some power.
By Theorem 40.2, f ( x )dx is given as a sum of simpler antiderivatives-antiderivatives which, in
It will now be shown how to construct the partial fractions representation and to integrate it term
fact, can be found by the techniques already known to us.
by term.
Case I : D(x)is a product of nonrepeated linearfactors.
The partial fractions representation off (x) is
A n
+-
+-+ ...
x - a,,
A2
x - a,
AI
x - a,
--
-
N(x)
(x - a,& - a,) * . . (x -a,)
The constant numerators A,, ...,A, are evaluated as in the followingexample.
A, A2
+-x - 1
--
2x + 1
(x + 1)(x - 1) - x + 1
EXAMPLE
Clear the denominatorsby multiplying both sidesby (x + 1)(x - I),
A2
+(x + 1XX - 1) -
(x + 1XX - 1) x + l x - 1
2x + 1 A1
(X + 1Xx - 1) -
(x + 1Nx -
2x + 1 A ~ ( x
- 1) +A2(x + 1)
In (I), substituteindividually the roots of D(x).With x = - 1,
1
-1 = A1(-2) +0 or A , = -
2
and with x = 1
,
3
3 = 0 + A2(2) or A2 = -
2
With all constantsknown, the antiderivativeoff@) w
i
l
lbe the sum of terms of the form
dx = A In I x -a1
Case 2: D(x)is a product of linearfactors, at least one of which i
s repeated.

322 THE METHOD OF PARTIAL FRACTIONS [CHAP. 40
This is treated in the same manner as in Case 1,except that a repeated factor (x - a)’ gives rise to a
sum of the form
A2 A&
x - a ( x - a ) (x -
+
y
+* * * +-
A1
EXAMPLE A2 A3
+-+-
A ,
(x - 1)2(x -2) x - 1 (x - 1)2 x -2
=-
3x + 1
Multiply by (x - 1)’(x - 2),
3x + 1 = A,(x - 1)(x - 2) +A,(x -
Letting x = 1,
4 = O + A 2 ( - 1 ) + 0 or
Letting x = 2,
7 = 0 +0 +A3(l) or
The remaining numerator, A,, is determined by the condition thal
zero (sinceit is zeroon the left side). Thus,
2) +A,(x - 1)2
A2 = -4
A, = 7
the coefficient of x2 on the right side of (2) be
A , +A3 =0 or A , = -A3 = -7
[More generally, we u
s
e all the roots of D(x) to determine some of the A’s, and then compare coefficients-of as
many powers of x as necessary-to find the remainingA’s.]
Now the antiderivatives off (x) will consist of terms of the form In Ix - a Iplus at least one term of
the form
C a e 3: D(x)has irreduciblequadraticfactors, but none is repeated.
In this case, each quadratic factor x2 +bx +c contributes a term
A x + B
x2 +bx +c
to the partial fractions representation.
EXAMPLE
Multiply by (x2 + 1)(x +2),
-t
x2 - 1 A1
=-
(x2 + 1)(x +2) x +2
x2 - 1 = A1(x2 + 1) +( A ~ x
+A3Xx +2)
Letx= -2,
3
3 = Al(5) +O or A, = -
5
Comparingcoefficients of xo(theconstant terms),
1 4
- 1 = A , + 2 A 3 or A , = - - ( ~ + A , ) = - -
2 5
Comparing coefficientsof x2,
CI
1 = A , + A 2 or A 2 = 1 - A , = 4
5

CHAP. 4
0
1 THE METHOD OF PARTIAL FRACTIONS 323
The sum for f ( x )dx will now include, besides terms arising from any linear factors, at least one
term of the form
Ax +B
sx2 +bx +c
l e t u = x + -
[ ;I
C U
2 6 6
= A In (U2 +62) +-tan-' -
(For a guarantee that 6 is a real number, see Problem 40.7.)
Case 4: D(x)has at least one repeated irreducible quadraticfactor.
A repeated quadratic factor (x2 +bx +c ) ~
contributes to the partial fractions representation the
expression
Alx +A2 ASx +A, + ... A 2 k - G + A 2 k
ax2 +bx +c + (ax2+bx +c ) ~ + (ax2+bx +c)'
The computationsin this case may be long and tedious.
EXAMPLE
x 3 + 1 A , ~ + A , A , ~ + A ,
(x2 + 1), - x2 + 1 (x2 + l),
+
--
Multiply by (x2+ l),,
x3 +1 = (A,x +A2)(x2+1) +A 3 x +A,
Compare coefficients of x3,
l = A ,
Compare coefficientsof x2,
0= A,
Compare coefficients of x,
O = A , + A 3 or A , = - A , = - l
Compare coefficients of xo,
1 = A 2 + A , or A , = 1 - A 2 = 1
The new contribution to I f ( x ) dx will consist of one or more terms of the form
du
[as in Case 3)
8 do net U = S tan 81
E
-
-
(U2 +a2)1-'
and we know how to evaluate the trigonometric integral [see Problem 38.12(a) or example (b)of Section
39.11.

SolvedProblems
2x3 +x2 - 6x +7
dx.
s x 2 + x - 6
40.1 Evaluate
The numerator has greater degree than the denominator. Therefore, divide the numerator by the
denominator,
2x - 1
x2 +x -6I 2x3 +x2 - 6x +7
2x3 +2x2 - 12x
-x’+ 6 ~ + 7
- x2 - x + 6
7x + 1
2x3+x2 -6x +7 7x + 1
Thus, = 2 x - 1 +
x 2 + x - 6 x 2 + x - 6
Next, factor the denominator, x2 +x - 6 = (x + 3Kx - 2). The partial fractions decomposition has the
form (Case 1)
7x + 1 =-+-
A1 A2
(x+3)(x-2) x + 3 x - 2
Multiply by the denominator(x +3)(x - 2),
7~ + 1 = Al(x - 2) +A ~ ( x
+3)
Let x = 2,
Letx=-3,
Thus,
and
15=O+5A2 or A 2 = 3
-20= -5A1 + O or A, = 4
4 3
-
--+-
( ~ + 3 X x - 2 ) x + 3 X - 2
7x + 1
2x3+x2 - 6x +7 dx = I(2. - 1) dx +s”dx +I
’
d
x
s x 2 + x - 6 x + 3 x - 2
= x 2 -x+41n ( x + 31 +3 In I X - 21 +C
x2 dx
x3 - 3x2 - 9x +27’
40.2 Find 1
Testing the factors of 27, we find that 3 is a root of D(x). Dividing D(x) by x - 3 yields
X’ - 3x2 - 9~ +27 = (X - 3Hx2-9) = (X- 3)(x - 3Xx +3) = (X - 3)2(~
+3)
and so the partial fractions representation is (Case2)
A2 A3
+-+-
A,
(x - 3)2(x+3) x -3 (x - 3)2 x +3
=-
X2
Multiply by (x - 3)2(x +3),
x2 = Al(x - 3Xx +3) +A,(x +3) +A3(x - 3)2
Letx=3,
3
9 = O + 6 A 2 + 0 or A,=-
2

CHAP. 4Q] THE METHOD OF PARTIAL FRACTIONS 325
1
Let x = -3, 9 = 0 +0 +A3(-6)2 or A, = 4
3
l =A ,+A , or A,=l-A,=-
4
Thus,
and
3 1 3 1 1 1
=-- +--+--
x3-3x2-9x+27 4 ~ - 3 2 ( ~ - 3 ) ~4 ~ + 3
X2
x2 dx 3 3 1 1
2 x - 3 4
= - In I x - 31 - - -+- In I x + 31 +C
Ix3 - 3x2- 9~ +27 4
4
0
.
3 F i n d 1 dx.
x(x2 +2)
This is Case 3,
x + l A , A , x + A ,
-=-+
x(x2 +2) x x2 +2
Multiply by x(x2 +2),
x + 1 = A,(x’ +2) +x(A,x +A,)
Letx=o,
1
1 =2A, + O or A, = -
2
1
O = A , + A 2 or A , = - A , = - -
2
Compare coefficientsof x,
1=A,
Thus,
and
-
x + l = 1(A) + (-4). +-1
x(x2 +2) 2 x x2 +2
Because the quadratic factor x2 +2 is a complete square, we can perform the integrations on the right
without a change of variable,
x + l 1 1 X
1x1--In (x2+2) +-tan-’ -+C
4 & f i
P .
1
dx.
J 1- sin x +cos x
40.4 Evaluate
Observe that the integrand is a rationalfunction a. sin x and cos x. Any rational function c. the six
trigonometric functions reduces to a function of this type, and the method we shall use to solve this
particular problem will work for any such function.
Make the change of variable z = tan (x/2); that is, x = 2 tan- z. Then,
dx = -
dz
1 + z 2

and, by Theorem 26.8,
x x tan (42)
sin x = 2 sin - cos - = 2 -
2 2 sec2(x/2)
22
1+z2
=-
tan (x/2)
1+tan2 (x/2)
= 2
X tan2 (x/2)
2 sec2 (x/2)
cos x = 1-2 sin2- = 1 - 2
2z2 1- 22
- I - - = -
tan2 (@)
= 1 - 2
1+tan2 (x/2) - 1+ z 2 1+ 2 2
When these substitutions are made, the resulting integrand will be a rational function of z (because com-
positions and products of rational functions are rational functions).The method of partial fractions can
then be applied,
-sinx+cosx)-ldx= dz
= I('1
= I ~ d r =
1
+z2) -22 +(1 -z2)
1+ z 2
dz
2-22
2-22 1+ z 2
-In 11 - 2 1 + C
= -In 1-tan + C
I fl
x dx
4
0
.
5 Find [(x + 1Hx2+2x +2)2'
This is Case 4 for D(x),and so
X A, A 2 x + A 3 A 4 x + A 5
+ +
=-
(x + 1XX2 +2x +2)2 x + 1 x2 +2x +2 (x2+2x +2)2
Multiply by (x + 1Xx2+2x +2)2,
x = A1(x2 +2x +2)2 +(A2 x +A3Xx + 1Xx2 +2x +2) +(A4x +A& + 1)
or, partially expandingthe right-hand side,
x = A1(x4 +4x3+8x2 +8x +4) +(A, x +A,Xx3 +3x2 +4x +2) +(A4x +A,Xx + 1) ( I )
In (I),let x = -1, -1 = AI(1) = AI
O = A l + A 2 or
0 = 4A1 +3A2 +A, or
0= 8A1+ 4A2 +3A3 +A4 or
Compare coefficientsof xo,
0 = 4A, +2 4 +A, or

CHAP. 403 THE METHOD OF PARTIAL FRACTIONS 327
Therefore,
x dx dx (x + 1)dx (x +2) dx
/(x + 1)(x2+2x +2)2= -1 + Jx2 +2x +2 + J(x2 +2x +2)2
U du U du du
[by quick formulas I1 and I]
= -InIx+II+-ln(u2+1)--
1 '(-
)+Jms2ede
2 2 u 2 + 1
[Case 4: let U = tan 01
1 1 8 sin 28
2
= --InIx+ ll+-In(x2+2x+2)-- + C
Now
and (see Problem 40.4)
8 = tan-' U = tan-' (x + 1)
2 tan 9 - 2(x+ 1)
1+tan28 - x2+2x +2
sin 28 =
so that we have, finally,
x dx 1
2
= -lnIx+ ll+-1n(x2+2x+2)--
J(x + lXX2 +2x +2)2
+-
1tan-' (x + 1) +-
1( x + l ) + c
2 2 x2+2x+2
X
=1(In (x2 +2x +2) + +tan-' (x + 1)) - In Ix + 1I +c
2 x2+2x +2
x dx
x2- 9
2x2+ 1 x2- 4
x dx x - 5
dx
(
' /(x - 1Xx - 2)(x - 3)
(
'
) /x4 - 13x2+36
x4 dx
x2dx dx
(n)
(
'
) I ( x - l)(x2 +4)2
+ 1xx2+4)
x2dx
x - 1 x2+2
dx
/x(x2 +5x +6)
dx
x3 +2x2-x - 2
(c) 5"-4x2+x + 1
I-
dx
x2 -4
x3+ 1
dx
(f) Jx(x +3Kx +2)(x - 1)
(i) J( 2x dx
x - 2)2(x +2)
dx
x4 + 1
(r)
x3 +
dx
x(x2+x +1)2

40.7 Show that p(x) = x2 +bx +c is irreducible if and only if c -(b2/4)
> 0
.[Hint: A quadratic polynomial is
irreducibleif and only if it has no linear factor; that is (byTheorem 7.2),if and only if it has no real root.]
4
0
.
8 (a) Find the area of the region in the first quadrant under the curve y = l/(x3 +27) and to the left of the
line x = 3.
(b) Find the volume of the solid generated by revolving the region of part (a)around the y-axis.
dx
1 -sin x
4
0
.
9 Find I
- . [Hint: See Problem 4
0
.
4
.
1
cos x dx
sin x - 1*
answers are equivalent.
40.10 Find 1
- (a) Use the method of Problem 40.4.(b) Use the quick formula 11. (c) Verify that your
40.11 Evaluate the followingintegralsinvolvingfractionalpowers:
[Hints: Part (a)let x = z3;part (b)let x - 1 = z4; part (c) let 1 +3x = z2; part (d)let x = P
.
3

sin (8+2n)= sin 8
cos (- 8)= cos 8
Appendix A
Trigonometric Formulas
sin (- 8)= -sin 8
cos (U +U) = cos U cos U -sin U sin U
cos (U - U) = cos U cos U +sin U sin U
sin (U +U) = sin U cos U +cos U sin U
sin x 1
tanx=--- -
cos x cot x
cos x 1
cot x=--- -
sin x tan x
1
sec x = -
cos x
1
sin x
csc x = -
sin (U - U) = sin U cos U - cos U sin U tan (-x) = -tan x
sin 28 = 2 sin 8 cos 8
cos 28 = cos2 8 - sin2 8
tan (x +a) = tan x
1+tan2x = sec2 x
= 2 cos2 8 - 1= 1- 2 sin2 8 1+cot2 x = csc2x
8 1+cos 8
2 2
cos2 - =
8 1-cos 8
2 2
sin2 - =
tan U +tan U
1- tan U tan U
tan (U +U) =
tan U - tan U
1 +tan U tan U
tan (U - U) =
cos (
; - 8) = sin 8; sin (a- 8) = sin 8; sin (8 +z) = -sin 8
sin (
; - 8) = cos 8; cos (a- 8) = -cos 8; cos (8 +z) = -cos 8
B
Law of cosines: c2 = a2 +b2 - 2ab cos 8
sin A sin B sin C
Law of sines:
a b C
~ -
- --
- -
C
a
329

Appendix B
Basic Integration Formulas
a d x = a x + C
s
X
tan-' - +C
a2 + x 2 a a
dx 1 X
a
--sec-'-+C [a>O]
ex dx = e" +C
s
xe" dx = eyx - 1) +c
In x dx = x In x - x +C
s
s
s
s
sinxdx= - c o s x + C
cos x dx = sin x +C
tan x dx =In lsec x l +C
sec x dx = In /sec x +tan xl +C
cot x dx =In (sinxl +C
s
s
s
s
s
s
s
s
csc x dx = In lcsc x - cot xl +C
sec2 x dx = tan x + C
csc2 x dx = -cot x + c
sec x tan x dx = sec x +C
csc x cot x dx = -csc x +c
x sin 2x
2 4
x sin 2x
sin2x dx = - --+ C
cos2 x dx = - +-+ C
s
s 2 4
tan2 x dx = tan x - x +C
s
330

Appendix C
Geometric Formulas
( A = area, C = circumference, V = volume, S = lateral surface area)
Triangle Trapezoid Parallelogram Circle
b bi
1
2
A = - (b, +b2)h
1
A = - b h
2
Sphere
0
---3--
4
V = - nr3
3
S = 4zr2
Cylinder
V = ar2h
S = 27rh
A = bh
Cone
1
V = - nr2h
3
A = nr2
C = 2nr
331

0"
1"
2"
3"
4"
5"
6"
7"
8"
9"
10"
11"
12"
13"
14"
15"
16"
17"
18"
19"
20"
21"
22"
23"
24"
25"
26"
27"
28"
29"
30"
31"
32"
33"
34"
35"
36"
37"
38"
39"
40"
41"
42"
43"
44"
45"
-
Appendix D
Trigonometric Functions
sin
O.oo00
0.0175
0.0349
0.0523
0.0698
0.0872
0.1045
0.1219
0.1392
0.1564
0.1736
0.1908
0.2079
0.2250
0.2419
0.2588
0.2756
0.2924
0.3090
0.3256
0.3420
0.3584
0.3746
0.3907
0.4067
0.4226
0.4384
0.4540
0.4695
0.4848
0.5000
0.5150
0.5299
0.5446
0.5592
0.5736
0.5878
0.6018
0.6157
0.6293
0.6428
0.6561
0.6691
0.6820
0.6947
0.7071
~~
cos
cos
1
.m
0.9998
0.9994
0.9986
0.9976
0.9962
0.9945
0.9925
0.9903
0.9877
0.9848
0.98 16
0.9781
0.9744
0.9703
0.9659
0.9613
0.9563
0.95 11
0.9455
0.9397
0.9336
0.9272
0.9205
0.9135
0.9063
0.8988
0.8910
0.8829
0.8746
0.8660
0.8572
0.8480
0.8387
0.8290
0.8192
0.8090
0.7986
0.7880
0.777 1
0.7660
0.7547
0.7431
0.73 14
0.7193
0.7071
~~
sin
tan
O.oo00
0.0175
0.0349
0.0524
0.0699
0.0875
0.1051
0.1228
0.1405
0.1584
0.1763
0.1944
0.2126
0.2309
0.2493
0.2679
0.2867
0.3057
0.3249
0.3443
0.3640
0.3839
0.4040
0.4245
0.4452
0.4663
0.4877
0.5095
0.5317
0.5543
0.5774
0.6009
0.6249
0.6494
0.6745
0.7002
0.7265
0.7536
0.78 13
0.8098
0.8391
0.8693
0.9004
0.9325
0.9657
1.Ooo
~
cot
COt
...
57.29
28.64
19.08
14.30
11.43
9.514
8.144
7.115
6.3 14
5.671
5.145
4.705
4.331
4.011
3.732
3.487
3.271
3.078
2.904
2.747
2.605
2.475
2.356
2.246
2.145
2.050
1.963
1.881
1.804
1.732
1.664
1.600
1.540
1.483
1.428
1.376
1.327
1.280
1.235
1.192
1.150
1.111
1.072
1.036
1.Ooo
sec
1.Ooo
1.Ooo
1.001
1.001
1.002
1.004
1.006
1.008
1.010
1.012
1.015
1.019
1.022
1.026
1.031
1.035
1.040
1.046
1.051
1.058
1.064
1.071
1.079
1.086
1.095
1.103
1.113
1.122
1.133
1.143
1.155
1.167
1.179
1.192
1.206
1.221
1.236
1.252
1.269
1.287
1.305
1.325
1.346
1.367
1.390
1.414
~~
csc
csc
...
57.30
28.65
19.11
14.34
11.47
9.567
8.206
7.185
6.392
5.759
5.241
4.8 10
4.445
4.134
3.864
3.628
3.420
3.236
3.072
2.924
2.790
2.669
2.559
2.459
2.366
2.28 1
2.203
2.130
2.063
2.000
1.942
1.887
1.836
1.788
1.743
1.701
1.662
1.624
1.589
1.556
1.524
1.494
1.466
1.440
1.414
~~
sec
~~~
90"
89"
88"
87"
86"
85"
84"
83"
82"
81"
80"
79"
78"
77"
76"
75"
74"
73"
72"
71"
70"
69"
68"
67"
66"
65"
64"
63"
62"
61"
60"
59"
58"
57"
56"
55"
54"
53"
52"
51"
50"
49"
48"
47"
46"
45"
-
332

n
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.o
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
Appendix E
Natural Logarithms
In n
-
-2.3026
-1.6094
-1.2040
-0.9 163
-0.6931
-0.5108
-0.3567
-0.2231
-0.1054
O.oo00
0.0953
0.1823
0.2624
0.3365
0.4055
0.4700
0.5306
0.5878
0.6419
0.6931
0.7419
0.7885
0.8329
0.8755
0.9163
0.9555
0.9933
1.0296
1.0647
1.0986
1.1314
1.1632
1.1939
1.2238
1.2528
1.2809
1.3083
1.3350
1.3610
1.3863
1.4110
1.4351
1.4586
1.4816
-
n
4.5
4.6
4.7
4.8
4.9
5.0
5.1
5.2
5.3
5.4
5.5
5.6
5.7
5.8
5.9
6.0
6.1
6.2
6.3
6.4
6.5
6.6
6.7
6.8
6.9
7.0
7.1
7.2
7.3
7.4
7.5
7.6
7.7
7.8
7.9
8.0
8.1
8.2
8.3
8.4
8.5
8.6
8.7
8.8
8.9
-
I
n n
1.5041
1.5261
1.5476
1.5686
1.5892
1.6094
1.6292
1.6487
1.6677
1.6864
1.7047
1.7228
1.7405
1.7579
1.7750
1.7918
1.8083
1.8245
1.8405
1.8563
1.8718
1.8871
1.9021
1.9169
1.9315
1.9459
1.9601
1.9741
1.9879
2.0015
2.0149
2.028 1
2.0142
2.0541
2.0669
2.0794
2.0919
2.1041
2.1163
2.1282
2.1401
2.1518
2.1633
2.1748
2.1861
n
9.0
9.1
9.2
9.3
9.4
9.5
9.6
9.7
9.8
9.9
10
11
12
13
14
15
16
17
18
19
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
200
300
400
500
600
700
800
900
I
n n
2.1972
2.2083
2.2192
2.2300
2.2407
2.25 13
2.2618
2.2721
2.2824
2.2925
2.3026
2.3979
2.4849
2.5649
2.6391
2.708 1
2.7726
2.8332
2.8904
2.9444
2.9957
3.2189
3.4012
3.5553
3.6889
3.8067
3.9120
4.0073
4.0943
4.1744
4.2485
4.3 175
4.3820
4.4427
4.4998
4.5539
4.6052
5.2983
5.7038
5.9915
6.2146
6.3069
6.5511
6.6846
6.8024
333

-
X
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.95
1.o
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
-
Appendix F
ExponentialFunctions
ex
1.oooo
1.0513
1.1052
1.1618
1.2214
1.2840
1.3499
1.4191
1.4918
1.5683
1.6487
1.7333
1.8221
1.9155
2.0138
2.1 170
2.2255
2.3396
2.4596
2.5857
2.7183
3.0042
3.3201
3.6693
4.0552
4.48 17
4.9530
5.4739
6.0496
6.6859
7.3891
8.1662
9.0250
9.9742
11.023
e- x
1.m
0.95 12
0.9048
0.8607
0.8187
0.7788
0.7408
0.7047
0.6703
0.6376
0.6065
0.5769
0.5488
0.5220
0.4966
0.4724
0.4493
0.4274
0.4066
0.3867
0.3679
0.3329
0.3012
0.2725
0.2466
0.2231
0.2019
0.1827
0.1653
0.1496
0.1353
0.1225
0.1108
0.1003
0.0907
X
2.5
2.6
2.7
2.8
2.9
3.0
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
3.9
4.0
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5
6
7
8
9
10
-
ex
12.182
13.464
14.880
16.445
18.174
20.086
22.198
24.533
27.113
29.964
33.115
36.598
40.447
44.701
49.402
54.598
60.340
66.686
73.700
81.451
90.017
99.484
109.95
121.51
134.29
148.41
403.43
1096.6
2981.O
8103.1
22026
e-x
0.0821
0.0743
0.0672
0.0608
0.0550
0.0498
0.0450
0.0408
0.0369
0.0334
0.0302
0.0273
0.0247
0.0224
0.0202
0.0183
0.0166
0.0150
0.0136
0.0123
0.0111
0.0101
0
.
0
0
91
0.0082
0.0074
0.0067
0.0025
0.0009
0.0003
o.Oo01
O.ooOo5
334

Answers to Supplementary Problems
CHAPTER 1
1.10 use (1.3): (a)x = 4 or x = -3; (b)x = 6 or x = S.
1.11 (a)O<x<2;(b) - 3 s ~ ~
-$;(c)x< - 6 o r x > -2;(d)x5 1 o r x 2 4 ;
(e)2I;x54or - 4 5 x 5 -2;(f) - 8 < x < -4.
1.12 ( a ) x > -5;(b)-13<x< -3;(c)x< - - f o r x > * ; ( d ) - l < x < 3 ; ( e ) - l < x < l ; ( f ) l S x < 3 .
1.13 (a)x >0or x < -2; (b) -4 < x < 1;(c) x < 1or x > 5 ; (d) -8 < x < 1;(e) -1 < x <4;
( f ) - l < x < O o r x > l ; ( g ) x < - 7 o r - $ < x < 3 .
1.15 (a)By (2.5).(b)By (2.5)and (a),Ia3I = Ia2aI = Ia2II a I = Ia l2IU I = Ia 13.
(c) Ia"I = Ia I"for all positive integers n.
1.16 Use (2.3): (a)x = 5 or x = 4; (b)x = 3 or x = &;(c) x = 8.
1.17 (a)4 < x < 5 ; (b)4 5 x 5 3.
1.19 Yes: ( u ~ ) ~
= u4and a2 2 0.
1.20 No: @<@implies Ia I < Ib I, but a < b does not hold when, for example,a = 1 and b = -2.
1.21
-1 0 1 2 3 4
B D O I C A
-
-
IA = 3,AI = 3,OC = 3,BC = 4,iB+BD = 2 +4 = $ , E
= $,iB+B C = 2 +4=l,IC=
3.
1.22 (a)b = 10;(b) b = - 5 ; (c) b = -5.
CHAPTER 2
2.4 A(O,41, B(2,2), C(430
1
,o(-3, I),E(O, -4), F(2, -3).
2.6
2.7
(a)5 ; (6)5 ; (c) 2; (6)8.
Area (righttriangle)= -f(x)(m)
= &(5)(10)= 25.
335

336 ANSWERS. TO SUPPLEMENTARY PROBLEMS
0
c
/
//
0
/
/
2.8 (3,4). 2
.
9 (- 1, 1) and (3,O). 2.10 (0,2),(6,2),(4, -4).
2.11 (2,y)for some real numbery.
0
’
0
0
/
/
*0
-X
J389
2.12 (a)p 4 ; (b)3Jz; (c) .
2.13 (a)Isoscelesonly; (b)right only, area = 10;(c)isoscelesright, area = 17.
2.14 k = 5. 2.15 (a)No; (b)yes.
2.16 (4(4, 2);(b)(i,
2); (4(
7
,
2).
5 + $
2-17 (5,8).
CHAPTER 3
3.9 See Fig. A-1.
( h )
Fig. A-1
3~10 See Fig. A-2. 3.11 SeeFig. A-3. 3.12 See Fig. A-4. 3.13 x2 = 4py (parabola).
3.14 (U) (X -4)2+0) - 3)2 = 1; (b)( X + +0) - 5)2 2;(c)x2 +(U - 2)2= 16;(a)( X - 3)2 +0) - 3)2 = 18;
(e) (x -412+(y + 112= 20; (f)
(x - +0) -2J2= 5.
3.15 (a)Circle [center (6, -lO), radius 1I]; (b)circle [center (0, -15), radius 141;(c)null set;
(a)circle [center (a,0),radius 41;(e) point (- 1, 1); (f)
circle [center (-3, -2), radius 7).

ANSWERS TO SUPPLEMENTARY PROBLEMS
I
I
I
I
I
I
I
I
I

I
I
Fig. A-2
X
Fig. A-3
3.16 (b)4F < 0’ +E‘.
3.17 (U) (X - 3)’ +(y +2)2 = 100; (b)(X - 4)’ +y’ = 26.
337

338

I -

- I I
/
/
1
/ - 1
/
1
/
A Y
/
/
/
/
/
I
I

*
I X
-

3.18 k = 2and k = -4.
t’
Fig. A 4
CHAPTER 4
4.8 (a) y - 5 = 3(x - 2
)or y - 4 = $(x +1
)
;(b)y = 4x or y -4 =4(x - 1
)
;
(c)y+1 = - ( x - 7 ) o r y - 7 = -(x+ 1
)
.
4.10 (a)m = 5, b = 4
,(1,
9
)
;(b)m= $, b = -2, (4,
5
)
; (c)m= -4,b = 2
,(1,-2);(a)m = 0
,b = 2
,(1,2);%
(e)rn = -4, b = 4, (3,O).

ANSWERS TO SUPPLEMEWARY PROBLEMS 339
4.11 k -9. 4.12 NO.
4.13 (a)Yes; (c) in all cua;
(a)k = -4.
414 (a] Yvrlkl:(h)&her; (c)paralkl;(d)perpendimtar;(c) neither.
115 (a)y -fx +32;(h) -40".
116 (a)10; (6) -15; (c) -f ;(d)9.
117 ( 4 y = -+X +#;(b)y - 4 ~
+3 2 : ( ~ ) p
;
X -4.
118 (12.9) is not an the linc;(d, 3)is on tbc line.
4
.
2
2 4x +3y -9 >Oondx > I;= F
i
g A-5.
4
.
2
3 x<200/3. 4J4 SacFig.A-6.
166 (U) AN nonvcrtid lines throughtbe point (0.2); (b)all Lineswith slope3.
4.27 (4Horizmlal lina;(b)(4
2; (194; (If04;(h13; (n)none.
a (a)y- -$x +3 ; ( b ) ~ -- X +9 ; ( ~ ) y m 4 ~ + # .
f'

340 ANSWERS TO SUPPLEMENTARY PROBLEMS
(f)
t'

CHAPTER 5
341
(i) (Aa,
Af i )
and (-Afi,- @);(j)null set. See Fig. A-7.
8
5.6 (b)5fi z 3.58. 5.7 x = 50. 5.8 Center
4
5.9 x = 0and y = -- X. 5.10 (33/2,34).
3
CHAPTER 6
6.6 (a)y-axis; (b)origin; (c)x-axis, y-axis,origin; (d)x-axis, y-axis, origin; (e)x-axis; (f) none;
(9)origin; (h) none; (i) y-axis;(j)y-axis;(k) origin; (0none.
6.7 (a)x2 +xy +y2 = 1;(b) y3 +xy2 -x3 = 8; (c)x2 + 12x - 3y = 1;(d) y = -3x + 1; (e) no change.
CHAPTER 7
7.8 (Let R denote the set of real numbers. In each answer, the first set is the domain, the second set the
range. The graphs are sketched in Fig. A-8.) (a)R, (- a,41; (b) CO,a
)
,
(- a , O ] ; (c) [-2,2], range of H
is CO,21, range of J is [-2,Ol; (d)
(- a,-21 U [2, a)
(union or “sum” of the two intervals),CO, a
)
;
(e)R, CO,a
)
;
(f) R, set of all integers;(9)R, set of all integers; (h) R - (0) (set of all real numbers except 0),
R - (0);
(0R - {I},R - (0);
U)R, R;(4
R, R;(0R,c2, 4;(m)
(1,2,4), { -43); (4
R -{--2}, R - { -4};
(0) R, (-a,
21 U (4); (PI R - (O), {-1, 1);(4)
R, c2, 4;(r)R, R; (4R, CO, 1); (t)R,R.
7.10 (c)and (d).
2 x - 1
x3’ x + l
7.11 (a)f(x) = - domain is R - (0); (b)f(x) = -
,domain is R - { -1);(c)f(x) = x, domain is R.
7.12 (a) Domain is R - (2, 3}, range is (0,00) U (- a,-41; (b) (- 1, l),(1, a
)
;
(c)(- 1, a
)
,
(0,2];
(4CO,4), c- 1921;(4R, CO, 4.
7.13 (a) k = -8; (b)fis not defined when x = 0, but g is.
7.14 Of the infinitely many correct answers, some examples are: (a)f(x) = 2x for 0 < x < 1;
(b)f(x) = 5x - 1for 0 Ix < 1;(c)f(x) =
0 ifx=O
1 ifx>O’* (d)f(x) = (x - + 1for x < 1 or 1 < x < 2.
7.15 (U) y-axis; (b) none; (c)y-axis for both; (d) y-axis; (e) none; (f)
none; (g) none; (h)origin; (4 none;
( j )origin; (k) origin; (0none; (m) none; (n)none; (0)
none; (p) origin; (4)
none; (r)none; (s) none;
(t) origin.
7.16 (a)Even; (b) neither; (c) both even; (d) even; (e)neither; (f) neither; (g) neither; (h)odd; (i) neither;
(
1
1odd; (k) odd; (I) neither; (m) neither; (n)neither; (0)
neither; Cp) odd; (4)
neither; (r)neither;
(s) neither; (t) odd.
7.17 (a)No; (b) yes; (c) k = 0;(d) k = 2; (e)yes;f(x) = 0for all x.

.ty
- 2 - 1 0 I 2 3 4 x
t(4
4F
3 -
t-I
ty
- I
+--2
t’
.
7
+
!
f
G
-2

ANSWERS TO SUPPLEMENTARY PROBLEMS 343
7.18
7.19
7.20
7.24
7.25
1 1 Ix +h l - 1x1
- 5 ; w h
JxT-i;+&;(e)J Z i +J;
(U) 2x +h - 2; (b) 1;(c) 3x2+3hx +h2;(d)
(U) 1, -1,3, -3; (b) -2,4, -4; (c)2, -2,3; (42; (e) -2, -3, -4; (f)-2, 1+fi,1 - fi;
(B)4,49 -fi.
One, two (oneof them repeated),or three. 7.21 (a) k = 3; (b) k = -2. 7.22 9 and -12. 7.23 (iii).
(a)x< - 2 o r x > l;(b) - 2 < x < O o r x > 1.
CHAPTER 8
(c)lim 7 = 7;
(f)
lim (1Ox +5h - 2) = 1Ox - 2.
1
1
- -
-
'
-1
8.6 (a) lim (6x +3h) = 6x; (b) lim
(d) lim (3x2+3xh +h2)= 3x2;(e)lim
h+O h+O (X +h + 1)(X+1)- (X + h+O
=-.
1
h-rO h-0 J F i +& 2&' h+O
8.9 (4. 8.10 (a) lim (x3+x2 +x + 1)= 4. 8.11 (a) &. 8.12 (a) 0; (b) A;(c)8;(d) no limit.
x+ 1
CHAPTER 9
9.6
9.7
9.8
9.9
9.11
(a)12andll;(b)land -l;(c) -oo;(d) +wand -co;(e)landl;(f) -wand +m;(g)OandO;
(h)3 and 3;(i) +00 and -00; (I]0 and 0; (k)4 and -4; (Z) +00 and -00; (m)0 and undefined
(denominatorundefined when x3 < -5);(n)0 and 0; (0)2 and 2; (p) +00 and -00 ;(4)-3;(r) +00;
(s) +0O.
(a)No asymptotes;(b) vertical asymptote:x = -3; horizontalasymptote: y = 3(to the left and the right);
(c)vertical asymptotes:x = -3, x = 2; horizontal asymptote:y = 0 (to the left and the right);
(a)vertical asymptotes: x = -4, x = 1;horizontal asymptote:y = 0 (to the left and the right);
(e) vertical asymptotes: x = -2; horizontal asymptote:y = 0(to the left and the right);
(f)
vertical asymptotes:x = -4, x = 2; horizontal asymptotes:y = 1(to the right)and y = -1(to the left);
(8)horizontal asymptotes:y = 2 (to the right)and y = -2 (to the left);(h)horizontal asymptote:y = 0 (to
the right);(i) horizontal asymptote:y =0 (to the right).
(a) Vertical asymptote: x = 4; horizontal asymptote:y = -3; (b)vertical asymptote:x = 4; horizontal
asymptotes:y = -1(to the right) and y = 1(to the left);(c) horizontalasymptotes:y = -1(to the right),
y = 1(to the left);(4horizontal asymptote:y = 0 (to the right);(e)horizontal asymptote:y = 2.
CHAPTER 10
10.6 (a) Continuouseverywhere;(b) continuous for x # 0;(c)continuouseverywhere;
(6)continuousfor x # -2; (e)continuouseverywhere;(f)
continuousexcept at x = 1 and x = 2.

10.7 (a) Continuouson the right but not on the left at x = 0; (b) discontinuous on both the left and the right at
any nonnegativeintegerx; (c) no points of discontinuity;(6)continuouson the left but not on the right at
x = -3andatx=2.
1 i f x r o
O ifx<O'
0 i f - 2 r ; x < - 1
0 if 1 < x < 2
10.9 (a) Yes; (b)-(d)no; (e) Problem lO.?)--none, Problem 10.4-no.
10.10 (a) x = 4, x = -1;(b) lim f ( x )does not exist, lim f ( x )= -3;
(c) x = 4 (verticalasymptote),y = 0(horizontalasymptote).
x+4 x + - 1
10.11 (a) x = 0;(b)x = 0 (verticalasymptote),no horizontal asymptote.
10.12 (a) No; (b) no; (c) yes; (d)no. 10.13 c = 8.
10.14 (a) Yes; (b)yes; (c) no. 10.15 (b) 6. 10.16 Discontinuousfor all x.
CHAPTER 11
11.6
11.7
11.10
(a) (i)slope = 4x + 1;(ii)y = 2x - 6;(iii) see Fig. A-9(a); (b)(i)slope = x2;(ii)y = 4x - 9,
(iii)see Fig. A-9(b); (c) (i)slope = 2x - 2; (ii) y = -1;(iii)see Fig. A-9(c); (6)(i)slope = 8x; (ii)y = 4x +2;
(iii)see Fig. A-qd).
28
27
(3,9). 11.8 (1, 1)and (- 1, -1). 11.9 y = -3x +-
1 55 1
4
(3, 5) and (-j,7)
. 11.12 (6, 36) and (-2,4). 11.13 y = -x + 1.
CHAPTER 12
126
127
128
129
12.10
1211
1213
1214
(a) 9x2 - 8x + 5 ; (b) -40x4 +3 8 x 2 +4nx; (c) 3 9 ~ ' ~
- 50x9 +20x; (d)lO2XS0 +36x" - 28x +fi.
(a) 21x6-x4; (b) 6x - 5; (c) 2x3+5; (d)21t6 -24t; (e) 5 8 x 4 - 3x2.
(u)Y= -7x+ l;(b)y=66x- 1 5 3 ; ( ~ ) ~ = 3 .
85 65
(a)y = 20x +2andy = -44x +2;(b)y = x +4andy =- x - -
4 4 '
y = --x + 1. 1212 (2,2).
20
(a) ox(xS)
IX=j = 405; (b)0,t(5x4)I x t 113 = 27-
f'(x) = 8x [(iii) givesf'(u) = ku;then choose U = U = 1in (iii)].

t Y
(c)
Fig. A-9
16 17
12.17 b = -
2 ’
81’
12.15 (U) x = 4; (b)(0, -4). 12.16 c = -
12.18 (U) 4~ - 2; (b)f”(x)= 2f’(x).
11 1105
12.19 (a) -1, 3, -3; (b)y = -4x - 12; (c)(2, -151, (-2, 51, Of (798).
12.20 (a)3, -J ;
7 (b)y= 15
1 x +63;(c)(3,0)and(-:,$)= (--
5 -)
16384
9’ 243 ’
12.22 (a)Yes; (b)yes; (c)no;(4no.
10 1
12.23 (a)-; (b)no. 12.24 -
3 2&(& - 1)2’
CHAPTER 13
13.6 (U) (xlo0+2x50- 3x56~’+20) +(7x8 +20x +5)(10Oxg9+ ~OOX*~);

(X +4x2~)
-(x’ - 3) X’ +8x +3 .
-
-
(x +4)2 (x +4)2 ’
(x3 +7 x 5 ~ ~
- 1) -(x5- x +2 x 3 ~ ~ )
2x7 +35x4+3x3 - 6x2 -7
-
-
(x3 +7)2 (x3 +7)2
(4 Y
2 12 3 12
(d)-5;(e) 24x2- 2x +7 --;(f)
9x2 + 1 --+-
x x4 x4 x5’
1 3 5 7 1 13.9 At all points except x = 3.
4 4
13.8 --
4’
13.7 (u)Y= - - x + - ; ( b ) Y = - 4 X - z -
13.10 (b) At x = 0 and x = 4. 13.11 (a)x = 0; (b)x = 2; (c)x = 1. 13.12 -8.
CHAPTER 14
14.7
14.8
14.9
14.10
14.13
14.14
14.16
14.17
14.18
14.19
14.22
14.24
(a) max = 13 (at x = -2), min = -7 (at x = 3); (b) max = -1(at x = -I), min = --
129 (at x = :);
(c)max=3(atx=l),min= --(atx=
27 -~);(d)rnax=l(atx=O),min= - l l ( a t x = -1);
(e) max = 99 (at x = 4), min = -9 (at x = 2); (f) max = --(at x = -3), min = --(at = -4);
(9)max = -(at x = 4), min = -(at x = 2);(h)max = -(at x = I),
min = -1(at x = -1);
8
31
1 1
5 4
5 3 1
4 4 3
(i) max = -
14(at x = 2),min = --(at x = i).
3 27
75 yd east-west by 50 yd north-south.
60yd parallel to stream, 30 yd perpendicularto stream.
50 d/h. 14.11 $200. 14.12 x = 350, at $65 per radio.
(a)Side of base = 5 ft, height = 5 ft; (b) side of base = 5 f i ft, height = 5 / f i ft.
Length = 105 ft, width (parallelto divider)= 60ft.
(a)x = 50, y = 50;(b) x = 100,y = 0 or x = 0, y = 100;(c)x = 50, y = 50.
1 = 314 m, w = 628/n =200 m.
14.15 175.
nL 4L
n + 4 n + 4
(a)The entire wire for the circle;(b)-for the circle,-for the square.
5
3
The whole wire for the square. 14.20 loo0 sets. 14.21 r = 2 ft, h = -ft.
14.23 The equilateral triangle of side p/3.
2a
h = -
Jf=7

14.25 East-west dimension = 80 ft, north-south dimension = 48 ft.
14.26 (a) 21 s x s 100m; (b) 20400 m2(when x = 100m). 14.27 15tons.
CHAPTER 15
15.7
15.8
15.9
15.10
15.11
15.12
15.14
15.15
15.16
15.20
15.21
15.23
15.24
6
x + 1’
,(Q of)@) = -- (b) x6 +2x3 - 5, (x2 +2x - 5)3; (c) 443; (d)x6,x6; (e)x, x;
2
(4(fO gXx) =
(f)
x2 -4, x2 -4.
(a) ~ l l
x; (b)x = - ;(c) x = +;(4x = 0;(e)x = *fi.
(a)f(x)= x3 -x2 +2, g(x) = x’; (b)f(x)= 8 -X, g(x) = x4;(c)f(x) = 1 +x2,dx) = J;;;
(d)f(x)= x2-4, g(x) = x- l.
(U) 4(x3 -2x2 +7x - 3)3(3~2
- 4x +7);(b) 15(7+3 ~ ) ~ ;
(c)-4(2x -3)-3; (d)-18x(3x2+5)-4;
4(1 - 6 ~ ) 3 x2
;(i) --
(x +2)2 20x(x2-2).
(e)(4x2-3)(x +5)2(28x2+80x - 9);(f)-15 -
*
(x - 3)43 (Q)
(2x2 + 1)3 (h)(3x2 - x +5)2 2 J
-
*
(a) max = $/2 at x = 1,min = -&2 at x = -1; (b) max = 216 at x = 3, min = -36 at x = -4;
(c)max = 3at x = -1, min = 1at x = l;(d)max = 3 at x = 8, min = -$at x = l;(e)max = !#at
x = -l,min=Oatx=O.
R is 9 mifrom A. 15.17 H’(x) = 0. 15.18 x2. 15.19 3x2G(x3).
All real numbers except 0 and 1.
f’(-x) =f’(x)(the derivative is an even function). 1522 12.
(a) Domain [- $, a
)
,
range CO, 00); (b)y = i x +q;
(c)(1,2).
Base = $
,height = */2.

5x
2
Y
165 ( a ) - ; ( b ) k = f21.
11 4 1 1
7 7 2 2
(f) y = -x --;(g) y = -- x --;(h) y = -5x +4.
CHAPTER 17
17.10 (a)f’(l)= 0;(b)f’($) =0;(c)f’($) = 0;(a)f’(l +$)= 0;(e)fnot continuous at x = 1;
(f)f’(;) = 0; (g)f‘(l) = 0; (h)fnot differentiableat x = 1.
7 81
2’ 16’
17.11 (a) 1 < c <4; (b)c = -. (c)c = -* (d)c = 4 - fi;(e) c = & $;(f)
c = 4 - 2fi.
17.12 The functionis increasingin the first-listed region, decreasingin the second: (a) everywhere,nowhere;
(b) nowhere,everywhere;(c) x > 2, x < 2; (d) x < -2, x > -2; (e) -1 < x < 0,O < x < 1;
(f)-3 < x <O,O<x < 3;(g)x < 1o r x > 5,l < x < 5 ; ( h ) x< -1 orx > 1, -1 < x < O o r O < x < 1;
( i ) x < - 2 0 r x > 2 , - 2 c x < 2 .
343 343J5
3 f i
9 -
17.15 (6). 17.18 -=-
17.20 (a) The points (1, l), (3
-
+2fi, l), and (v,
1); (b) 0, 2 (double root); (c) CO, 31.
17.24 (:,:). 17.25 (b)0.8.
17.26 (a) Increasingon (- 1.3,0.17)and (1.13, +00); (b) increasingon (- 00,0.333)and (1, +00).
CHAP’ITR 18
18.4
185 4 seclater. 18.6 108.8 mi/h.
18.7
(a) so -s = 16t2;(b) (i)0.25; (ii)1; (iii)2; (iv)2.5.
(a) 112ft, 192ft; (b) 4 sec;(c) 256 ft; (d) 8 sec;(e) 128ft/sec = oo.
18.8 (a) 5 sec;(b) 476 ft/sec;(c) 3 sec;(d)1.5sec.

18.9 (a)It is always moving to the right; (b) 3 mi.
18.10 (a)t > 4; (b) t c 4; (c) t = 4; (6)
never; (e) 27 units.
18.11 (a)t < 3h and t > 1h; (b) 4 < t < 1h; (c) t = 4 h and t = 1h; (d) 0.75 mi/h.
18.12 13units. 18.13 320 ft/sec. 18.14 (a)t = 0 and t = 5; (b) t = 2.5; (c)no.
18.15 (a)5 sec; (b) 122.5 m.
18.16 (a)196ft; (b) 112ft/sec.
CHAPTER 19
19.4 $30per set. 19.5 $7 per unit. 19.6 24 ft2/ft.
19.7 40km/sec (at t = 10oOsec). 19.8 (a)18gal/sec (i.e., G = -18);(b) 54 gal/sec (i.e., AG/At = -54).
19.9 (U) 9; (b) 6. 19.10 y = 2.
CHAPTER 20
205
20.9
20.12
20.14
20.17
20.20
20.24
24/7 ft/min. 20.6 3.14/n x 1ft/min. 20.7 8 ft/sec. 20.8 64 ft/sec.
12/x w 3.82 mm/sec. 20.10 4/nx 1.27in/sec. 20.11 400zx 1256m2/min.
(a)10mm; (b)increasing at 6/fi x 2.69 mm/sec.
Decreasing at 4 units/sec. 20.15 30 mm3/sec. 20.16 r = l/n.
Increasing at 6 units/sec. 20.18 500 km/h. 20.19 4fi x 5.64 mi/h.
20.13 36 units/sec.
E(*) x 0.64 m/min.
25 n
20.21 (-1, -3). 20.22 5 ft/sec. 20.23 6 ft/sec.
45
-ft/sec. 20.25 0 mi/h. 20.26 12units/sec.
8
CHAPTER 21
21.6
21.7
21.10
647
(g)480 36 108
193x 0.4021; (h) 323x 8.9722; (i) -x 5.9907.
x 78.3 gal.
1920n
0.994cm3. 21.8 -
77
21.9 (a) 5n w 15.7 cm3;(b) 5.ln x 16.0cm3.
1
8n
9%. 21.11 -mi w 210 ft. 21.12 (U) 6; (b) 6. 21.13 max (1.8725, 1.8475) = 1.8725ft2.

21.15 (a) 1.189207115;(6) 1.587401052;(c) 1.872171231;(d)1.817120593.
21.17 (a) 1.324717957; (b)0.6823278038; (c) f1.306562965 and f0.5411961001; (d)1.179509025;
(e) -0.8793852416,1.347296355, and 2.532088886; ( f ) -1.671699882; (8) 1.618033989,
-0.6180339888, and -1.
21.18 1.867460025. 21.19 1.090942496.
CHAPTER 22
22.5
22.6
227
22.0
22.9
22.10
22.13
2214
22.15
1 4x +5
(e) - (2 - x2)(x2+ 1)-7/4;(f)
4(x - 3 x 5 ~ ~
- 6x - 3); (9)-
4 (1- x y '
1 1 2x 2
(a) --* (6)--* (c)--;(d)-
Y3 ' Y3 ' y5 (x +2y)3*
(a) y' = 16x3-4x, y" = 48x2 -4, y" = 96x, y(*) = 96, y(") = 0for n > 4;
(b)yf=4x+ 1 - x - ~ , Y " = 4 + 2 ~ - ~ , y ' " =-( 2 * 3 ) ~ - ~ ,
...,y(")=(-l~n!x-'"+";
(d)y' = -2(x - 1)-2, y" = 4(x - 1)-3, y " = -12(x - 1)-4, ...,y'") = (- 1)"2(n!)(x - l)-("+
'I;
(e)y'= - ( 3 + ~ ) - ~ , y "=2(3 + ~ ) - ~ , y " =
- 6 ( 3 ; ~ ) - ~ , . . . , y ( ~ ) = ( - l ) " n ! ( 3 + ~ ) - ( " + ~ ) ;
(f)
yf = -zx-3, yff = 3 .~ ~ - 4 ,
yf" = -4 3 .2x- ,...,y(n) = (- - I)! x-("+2).
1 3
dy d2y 22.12 (a) K = - L =-;@)no.
22.11 -= 0
,-= --
dx dx2 2' 7' 7
2
y" = -
(a) h"(x)=f(x)g"(x) +2f'(x)g'(x) +f"(x)g(x), h"@)=f(x)g'"(x) +3ff(x)g"(x) +3f"(x)g'(x) +f"(x)g(x),
h(4)(~)
=f(x)g'"(x) +4ff(x)g"(x) +6f"(x)g"(x) +4ff"(x)g'(x) +f(*)(x)g(x);
(6)h(")(x)= $ (3f'.)(x)g("-')(x), where = andf(O)(x)=f(x).
n!
k 0 (3' k!(n- k)!
(a) 5.4 ft/sec2;(6) 270 ft at t = 10 sec.
CHAPTER 23
23.4 (a)Concave upward for all x. (b)Upward for x > -5, downwardfor x < -5; I( -5,221). (c) Upwardfor
x < -5 or x > -4, downwardfor -5 <x < -4; I( -5,1371), I( -4,1021). (6)Upward for x > 3,
downwardfor x < 3; no inflection points.(e)Upward for x < 3, downwardfor x > 3; 43, 162).

t Y
351
f(-l.O) X(1.0)
w
X
I
- 1
f(-2, -16) I
(4
Y
X(1.0)
6
Y
4
1 2 X
I’
Fig. A-I0

23.5 (a) 1.5(min);(b)0 (max),3 (min), -3 (min);(c)4 (min), -4 (max);(d)0 (max),2 (min);(e)0 (min).
23.6 SeeFig. A-10. 23.7 (a). 23.8 Band E.
23.9 (a)One;(b)none; (c)parabola. 23.10 (f). 23.11 -4 < k < 0
. 23.12 See Fig. A-11.
23.13 (a)Allx in [-1,2]; @)allx in [-1,2] but x = 1 [f'(x) = 2x - 1 for x > 1, andf'(x) = 1 - 2x for x < 1);
(c) 1 < x < 2 or -1 < x < 4; (d)f"(x)= 2 for x > 1, andf"(x) = -2 for x < 1;(e)concaveupward for
x > 1, concavedownwardfor x < 1; (I)
see Fig. A-12.
23.15 k = -4. 23.16 A = C = 0
,D = 1, B = -4, k = -+d.
t' t'
I I'
-'I
X
Fig. A-11

t Y
Fig. A-12
353
23.19 (U) (0, +00); (b)x = 0, rel. min.;(c) concave upward for Ix I <fi,I( Itfi,1);
(6)see Fig. A-13, horizontal asymptotey = 3.
23.20 (U) -1 <x < 2or 2 < x < 3;(b)O <x < 2 or x > 4.
CHAPTER 24
24.4 (a)No maximum;(b)10m by 10m. 245 (0’0). 24.6 (3,i),
(-3,i).
24.7
24.9 10cm. 24.10 200ft north-southby 50 ft east-west.
24.11 (U) See Fig. A-14; (b)(- 1, 9).
Height = 7 ft, side of base = 6 ft. 24.8 Absolute max = -2 (at x = 0),no absolute min.
24.13 h = 6 in, t = 3 f i in.
24.14 (U) Absolute max =-
2J5 (at x = $),
absolute min = 0 (at x = 0);(b)see Fig. A-15.
9
4 Y
3
-I
Fig. A-13
t Y
- 1 1 X
Fig. A-14

I rI 4
2
1
X -2
-2
Y
Fig. A-15 Fig. A-16
2
4
.
1
5 Height = 8 m
, side of base = 4 m. 2
4
.
1
6 1500per day. 2
4
.
1
7 x = 2, y = 6.
24.18 (a) k = -8; (b)see Fig. A-16;(c) k = 0
.
2
4
.
2
2 Absolute min = 0 at x = 0, no absolutemax.
2
4
.
1
9 (3, $), (-3, -3).
CHAPTER 25
360
2
5
.
7 (a)-;(b)36; (c)105;(6)225;(e)210. 25.8 12 cm.
n
Y28
2 '
25.10 A =- 2
5
.
1
1 See Fig. A-17.
(e)
Fig. A-17

CHAPTER 26
(h)0.3256 (to four decimal places);(i) 0.2079(to four decimalplaces).
26.8 ( u ) @ ; ( b ) $ ; ( c )
9 -;;(a)$@)$.
It
26.13 (a)and (c)are false for 8 = -
4‘
1
26.10 - 2411 -
5 ’ 2 .
CHAPTER 27
2n 1
7,f= 3, A = 5,see Fig. A-l8(b).
--9f =3’ A = 1, see Fig. A-l8(a);(b) p =
27.8 (a)p-3n 4
2
t Y A Y
1
- 2
-
- 1 -
(4
Fig. A-18
It
27.10 (a) nn (forall integersn);(b) 2nn (for all integers n);(c)- +2nn (forall integers n);
(d)kn (fora
l
l odd integersk);(e)-+2nn (for all integersn);(f)
(g)- +2nn and -+2nn (forall integers n);(h) f - +2nn (forall integers n),
2
3n n
2 3
- +2nn (for all integersn);
n 3n n
4 4 6
x cos x - sin x .
X2
27.11 (a) 12sin2x cos x; (b)cos x -2 sin x; (c)x cos x +sin x; (a) 2 4 ~ 0 s
2x -x sin 2x);(e) 9
x sin x +2 cos x -2
x3
;(g) 5(3 cos x cos 3x -sin 3x sin x); (h) -4 cos 2x sin 2x = -2 sin 4x;
(f)
(0 -4x sin (2x2- 3);( j ) 15 sin2(5x +4)cos (5x +4);(k)--
(06(sin2(sin2x)Xcos(sin2x)) sin x cos x.
sin 2x .
&&Tx’

27.13 Figure A-19 shows the graphs for one period of each function,except in the case of the aperiodic
function(f).
1' t Y
-c/i-
f Y
.F
I
!
A Y
-4w - 3 r - Z n -
I'
(8)
Fig. A-19

27.14
27.15
27.16
27.19
27.22
27.24
27.25
27.27
27.29
2730
(a)max = 2n (at 2n),min = 0 (at 0);(b) max = 2 at - ,min = -1 (at 0);
J( 7)
(c) max = (at and ;
)
y min = 1 (at 0,f ,and n); (d) max = -
3 fi (at :), min = --
3 fi (at F);
2 2
6
(f) max =
3 1 1
5 2
at xo,where sin xo =-and- < xo c n min = - 5 (at xo +n).
(b)(i) Amplitude = 5, period = 2n; (ii)amplitude = 13, period = n;(iii) amplitude = fi,period = 2n.
$ ; ( b ) -$. 27.17 A = % . 27.18 y = & x +9 -12
2 4 .
2
(42
. 27.20 n = 4. 27.21 (a)See Fig. A-20; (b) yes; (c)no.
Fig. A-20
1 1 Y cos ( X Y )
(U) 1;(b)0. 27.23 (U) y’ = - --
- *-
; (b)Y’ = - cos (xy) - 2y’
sin y ,
/
-
1
8
(a) - - rad/sec; (b) 200& km/h.
n
(U) 8 = -;(b)8 = 0
. 27.26 (U) max =
2
Continuous, but not differentiable. 27.28 (a)!
!x 0.349;(b)0.857.
9
(a)cos x, -sin x, -cos x,sin x;(b) -sin x.
(b) 0.8654740331. 2731 3.141592523. 2732 1.895494267.
CHAPTER 28
28.8 See Fig. A-21.
X
28.10 (a)sec2- ;(b)(sec xxsec x - tan x); (c) -2 cot x csc2x;(d) 3 -4 csc24x; (e)6 sec’ 3x tan 3x;
2
cot& csc & sec2x
;(4
3 3 J Z -
(f) -3 cot (3x - 5) csc (3x - 5); (g) -
2
J
;
;

I
'I I
I
I
1
I I
I
- 2 I
I
I I
w '
I )
0 ; I W X
I I
I
I
-"I -;
I
I
I
li
I I
I
I
I
I
I
I
I
I
0
b
0 I[ 2 n l
I
I X
III
I
I
I
I
I I'I
I'
I
I
csc2x cot x 3 cos x
;(c)y' = -
;(4
y' = sec2
tan 2 tan (y + 1) sec2 (y + 1)'
Y(Y2 + 1)
x(y2 + 1) - 1
= -
Y sec2( X Y )
x sec2(xy)- 1
28.11 (a)y' = -
2 tan (x +y) sec2 (x +y) - 2(1 +y) tan (x +y)
('~'=1-2ttan(x+y)sec2(x+y)-1-2(1 +y)tan(x+y)*
2
8
.
1
2 y = 4(. -;
) +J5. 2
8
.
1
3 y = - q X - ; ) + 4 .
8 2
8
.
1
4
4
45
28.15 320 rad/h =-rad/sec x 5" per second. 2
8
.
1
6 tan (a2-al)= 1.5, a2 - al w 56".
4x -II
2
8
.
1
9 y = -
8(n2+ 1)'
2
8
.
1
7 tan (a2- al)= 2, a2 - a1 x 63". 2
8
.
1
8 tan (a2 -al)= 3, a2 - a1 w 71".
2
8
.
2
0 (a)Rel. max. at 4 4 , rel. min.at 3n/4, vertical asymptote at n/2, inflectionpoints at 0 and n;(b) rel. max. at
2n/3, rel. min. at 43, vertical asymptote at n/2,inflection points at 0 and n.
(-;, ;), (2,
n T),
3n etc.
2
8
.
2
1 On all intervalswhere it is defined:
2
8
.
2
2 (a)1.318116072;(b)4.493409579;(c) 0.860333589.
CHAPTER 29
x4 5 3 8 1
2 3 2 5 x3
2 2 1 1 1
2
2
9
.
7 (U) ---X' +- x2 +x +C;(b)5x -2& +C;(c) -xSI4+C;(d)3 ~ " ~
+C;(e) --+C;
+-+c = -(1 -2x2)+c;
+c = -x3'2(3x2- 7) +c;(g) --
(f)
-i x7/2 --x3/2
3 21 2x2 4x4 4x4

6 4 2
3 15
(h);
x512 -- X3/2 + 2x112 +C = -&9x2 - 1Ox + 15) +C;(i)
-3 cos x +5 sin x +C;
1 2
10 3
(m) sin x +C; (n)tan x -x +C;(0)-x10 +- x6 +2x2+C.
2 1 1
21 39 3
298 (a)-(7x +4)3/2+c;(b)2(x - 1p2+c;(c) -(3x - 5)13 +c;(6)- -cos (3x - 1) +c;
(e) 2 tan - +C; (f)
2 sin J;;+C;(g) --(4 - 22)' +C;(h)i(x3 + 5)4/3 +C;
X 1 1
2 32
2 2 3
3 3 5
3 1 1 3 1
(i) -(x + 11312- 2 ( ~
+ iy2
+c = - ,/iTT(x -2) +c;( j )-(x - 1)5/3+ c;
(k) [?
(x4 + 1)7/3
-- (x4 + 1)4/3] +c= (x4 + 1)4/3(4~4
- 3) +C;(0 Js
+C;
4
2(ax +b)3/2 1 1
+ c =--csc3x+c;
(3ax - 2b) +C;(n)- -
(m) 15a2 3 sin 3x
(0) -2(1 - x)~/~[-
-- (1 -x) + (1 - x)I] +C = - -(1 -x)312(15~2
+ 12x +8) +C;
3
1 2 1 2
3 5 105
1 1 1 1 1 3
112 2 x 12 x4
(p)5(3x - 5)13 +c;(4)--(4 - 7t2)' +C;(r) -- sin2- +C;(s) --tan -+ C.
29.9 (a) t = 7 sec;(b)1024ft;(c) t = 15 sec;(6) 256 ft/sec.
40 35
+4t;(c)at x = -when t = 4,at x = --when t = -1;
3 6
t3 3t2
(U) U = t2 -3t +4; (b)x =-- -
3 2
(6) -1 < t <4.
29.10
t4 13t2 23 3
12 4
12 6
+4t; (c)at x =-when t = 1, at x = - -when t = 3, and at
t +4; (b)x = -- -
t3 13
29.11 (a) U = ---
3 3
when t = -4; (6)t < -4 and 1 < t < 3.
88
3
X = - -
29.12 (a) 160ft/s;(b)400 ft. 29.13 (a)3 s; (b)99 ft. 29.14 7 units.
2x3
29.15 (U) y = -- 5~ - 1; (b)y = 6x2+x + 1. 29.16 12.5 ft.
3
CHAPTER 30
5 13
30.4 (a) 24; (b);3;(c) - 30.5 A2 -A1 - A , . 30.6 (~)20. 30.9 (a)n.
3 '
3+fi,(6) 15. 30.12 ( b ) ( T ) .
n(n +1) 30.13 4 - 1 +2=5. 30.14 -
11
5 '
30.10 (U) 15; (b)110; (c)7
*

CHAPTER 31
1 1 8 2 2 113226
31.12 (U) ;(b) ;(c) (1 - 2 a ) ; (d)
3(
0
- 1);(e) - +p;(f)
45 c(11)3(383)- 2561 = -
'
5 '
3 4 1 2
31.13 (U) 4;(b);
;;(c) -;(d)
;
.
3
3 1 857 1 n
31.16 -
35 4'
31.14 (U) c = - ;(b)c = -;(c) c = Js. 31.15 (a) -;(b)6.
2 fi
31.17 U
.31.18 (a) ,/m
;(b) -sin3 x; (c) 2 d m , 31.19 0.
T
25x
31.20 (a) 6x,/- ;(b)f(h(x)) h'(x);(c) 3 f i and - 125xj+ 1.
31.21 b = fl. 31.22 1. 31.23 0.
1
2'
31.24 ( ~ ) 4 x ;
(b) f2. 31.25 (U) 0; (b)- 31.26 b = 4. 31.27 fi.
n
31.28 All three are equal. 31.29 c = -
3'
3130 All values such that xk is integrableon CO,21; these include all positive values.
sin 3t 2 n 1 1
3 3n 6 3 3
3131 (a)x=-;(b)-;(c)O<t<-;(d)-and --.
13 2 1
3132 -m. 3133 (a) - ;(b) 1. 31.34 0.33 as compared to -
2 n 3'
31.35 (2& + 1)x 2.004559755 as compared to 2. 31.36 (a)0.2525, as compared to the exact answer 0.25;
6
(b)0.24875; (c)0.25.
CHAPTER 32
52 16 5
3 3 4
32.7 (a) -[Fig. A-22(a)]; (b) -[Fig. A-22(b)]; (c)- [Fig. A-22(c)].

ANSWERS TO SUPPLEMENTARY PROBLEMS 36I
CHAPTER 33
148 27n 7312n 452x 64rr
15 32 105 15 5
=I0 (a)-;(h)-~ 33.1 1 (01-;(b] -. 33.12 (a)See Fig A-23; (h)4n;(c)-.
A-23
llS2n 81x 1332~ 1%
;(4
-
;(a)-
.
5 5 2
33.11 (4
SOC Fib A-25; (b)-
;[c)

34.14 (a) 12 In 2; (b) I
n 2 -2 I
n 3.
34.15 (a)I
n 2 +In 5 ; (b) -In 2; (c) -In 5 ; (42 In 5 ; (e) 4 In 2; (f)
4 In 5 ; (g) -(2 I
n 2 +In 5); (h)7 1x1 2.
7
34.16 y = x - 1. 34.17 In 2--
24*
2 In 2
3 .
34.18 - 34.19 271I
n 2.
X
Y
34.20 SeeFig. A-26.
-
I
)
f
l
yI I I
I
I
I
I
Fig. A-26
1 67
2 2
34.21 (a)t, =- tZ- t +6 In I t I +2; (b)-+ 12 I
n 3.
1
34.23 -
3’
X Y ( h + 1) 1
;(b) y’ = -
;(c) y’=
xdv- 1)
3432 (a)y’ =
Y(x2+yz - 1) y(3xy +3y3 -2)-
1
4
34.25 (a)I
n 3 -In 2; (b) --In 7.
34.26 (a)0.6937714032; (b)0.6928353604; (c)0.6931473747 (to 10decimal places, the correct answer is
0.6931471806).
34.27 (a)0.5671432904; (b) 1.763222834.
CHAPTER 35
35.11 (U) -
1 ;(b) -x; (c) x4; (4xl+ln
3; (e) x - 1;(f)
x -In x; (g) 2x; (h)-
1
X 3‘

t Y
O I
+
I
)
.
X
ty
*
0 e e2 X
A Y
1 2
O I
X
(f)
Fig. A-27
ellX ex(x - 1)
X2 X
35.12 (a)-e-X; (b)--; (c)(-sin x)ewSx;
(d)ex sec2 8;( e ) ~ ;
(f)
ex(: +In x
(h)(In1c)RX; (i) 2 ; ( j )8 +e-x.
35.13 (~);e~~+C;(b)
-e-’+C;(~)7(,/=)~+C;(d)
2 -ecmx+C;(e)-32X+C;($)2ex~2+C;
1
2 In 3
1
+c;(k)--e-++C.
4
+C;(h)- eJX+C; (i) ex - In (e“ + 1) +C ;( j )-
@) a+l 3 3 In 2
1 2x3
xff+1
2x +yexy
2Y2 ;(e)cot x.
1 2x 2x
35.14 (U) -
;(b)1 +eY-x sec’ d-” = + ey-x(l + x4) ; y2ep -ef/9;(d)- 2y + xp‘
x(e‘ - 1)
I
n 2 1
3
5
.
1
5 (a)(In 3)(cos x)3”” ”;(b)2 eJ2°.5cx.
,(c)(2 I
n x)x(lnX)-’;
(d) - [I +I
n (Inx)](ln x)””’;
X
(e)
Y (-1 +-)1 =
2x +3
2 x + l x + 2 2 J r n ) ‘

In 2 1
35.16 (a)-;(b)-* (c) In 2; (d)eC;(e)2.
3 ~'
n
35.17 (a) e - 1;(b)5(e2 - 1). 35.18 (a)2; (b) n(e2+ 1). 35.19 z(e - 1).
35.20 max = e (at x = i),min =f (at x = - ;
) . 35.21 nnenx.
35.22 (a) y' = (2 cos x)Lin
x, y" = 2esin
x(cos2x - sin x); (b) -2 rad/sec.
In 3
3
35.23 (a) o = 3e3' + 1;(b)2 +-; (c)x = e3' +t - 1.
35.24 y = 2x +2. 35.25 See Fig. A-27.
35.26 In Fig. 35-l(a) and Fig. A-27(d), multiply the horizontal scale by l/ln 2.
35.29 max = e- '(at x = e), min = 0 (at x = 1).
35.30 (c) 2.718145927. (The correct approximation to 10significantfigures is 2.718281828.)
3532 (a)-;(b)ln(e+
3 1)-1112;(c)j(l--$).
1 3533 e - 1.
2
3535 (b) Approximately 11.55 years; (c) after 1year, 1dollar yields: (i) 1.05 dollars when compounded once a
year; (ii) about 1.05116dollars when compounded monthly; (iii)about 1.0512dollars when compounded
continuously.
35.36 (a) 0.5671432904;(b) 1.309799586. 35.37 0.8556260464.
CHAPTER 36
36.10 See Fig. A-28. 36.11 8 times.
In 10
In 2
36.12 1690-x 169q3.32) x 5611 years. 36.13 e-(I2'*2)/23x 0.7 g.
In 2 0.6931
36.14 T = x 3.1 years. 36.15 69.31 days. 36.16 3.125 g.
In 10 - 3 In 2 2.3026 - 2.0793
375
In 2
36.17 72900. 36.18 288 billion. 36.19 (a) 800;(b)-x 541.
CHAPTER 37

IY
ty
ty
I (2. $)
2
1 2-L/i
X
( 3 . 2 )

I
I I 1
b T r _
- 4 - 2 - 4
- n
Y
(e) Damped sine wave
Fig. A-28
3 J 2
4 ’
37.14 (a) sin 6 = Q,tan 8 = 2&, cot 6 =-, sec 6 = 3, csc 8 = --
,csc 6 = -4.
(b)COS 6 = -
&,tan6=
4 - q , c o t 8 = -&,secB=- 15
J 2
3 4
4 J l j

3 12 2
5 5
3
7
.
1
5 (a)- ;(b)-;(c) 15(2 - 3$); (d) (fi- 1); (e)0
.
3
7
.
1
6 Domain = (- 00, 00)' range = (0'13.
R 5a/2 for x 5 -1.
2 ' for x 2 1 '
37.17 (a) sin-' x +cos-' x = - (b) tan-' x +cot-' x =- * (c) sec-' x +csc- 'x = {
(d)tan-' x +tan-' - = - or cot-' x = tan-' -
1). 2 '
x 2 X
X 1 sin x . -3
( 2 f i ) J G ; (c) - 1 +cos2 x ' (1 +9x2)cot-' 3x;
37.18 (U) -
+tan-' x;
1+x2
1
1 p Jc7 ifx>O
i f x c o
(e) ,.(cos-1 x - ;
) (tan-' xX1 +x2);
(h)[ 2 J ; n
2x2
i f a > o
i f a < O
'
1
(i) -
-
1+ x 2 '
u + v
3
7
.
1
9 tan-' u +tan-' v = tan-' -
1-uv'
1 1 3x X 1 4x
2 2 6 2 5 4 5
3
7
.
2
0 (a)- tan-' +C;(b)- tan-' -+ C;(c)sin-' - +C;(d)- sin-' -+C;(e)sec-' (x - 3) +C;
1 X 1 3x
2 2 4 4
(qx)
+C; (h) - sec-' - +C; (i) -sec-' -+C;
1 sin-' (T)
d x +c;(g) -tan-' -
Jr;i
1 x + 4
(n)2(3 sin-' 9
- ,/-) +C; (o)zln(x2+8x +20) - 2 tan-' -+C;
2
1 X2 1 ex 1
8 J stan-' - + C; (4)
- sin-' -+C; (r)- tan-' -+C; (s)-tan-'
X2
2 2 fi
Jj 2 2
(p) 2 +2x --
3
(t)-
1tan-' -
x + l + C .
3 3
mi/min = 5000~
mi/h FZ 15708 mi/h.
1 25On
3
7
.
2
1 y = - x. 3
7
.
2
2 -1 rad/sec. 3
7
.
2
3 -
3 3
R R R2
4' 6' 3 '
3
7
.
2
4 - 3
7
.
2
5 - 3
7
.
2
6 - 3
7
.
2
8 2
,
/
i ft.
3
7
.
2
9 (a)[-;,;];(b)[O,~];(c)[-l,l];(d) -1 5 x 5 l;(e) -1 5 x 5 1.
y(1 +y2)(2x-tan-' Y
3
7
.
3
0 (U)
1+y2+xy

A
3731 See Fig. A-29. 3733 -
16'
t'
Fig. A-29
3734 (b) 3.141592651. (Thecorrect approximationto 10 significantfiguresis 3.141592654.)
CHAPTER 38
e"(sin x -cos x)
2
+C; (c) ex(x3- 3x2+6x - 6) +C;
3
8
.
6 (a) -eex(x2 +2x +2) +C; (b)
(d)x sin-' x +,
/
- +C; (e) x sin x +cos x +C; (I)
2x sin x +(cos x)(2 - x') +C;
(g)- [cos (In x) +sin (In x)] +C; (h) -[5x sin (5x - 1) +cos (Sx - l)]
+C;
X 1
2 25
eu 1
(b sin bx +U cos bx) +C; ( j )- (x - sin x cos x) +C;
(0 2
e3*
9
sin3x sin x
+c = -(2 +cos2 x) +C; (01 x +sin 2x + +c;(rn)-(3x - 1) + C;
3 4 (I
(k) sin x --
3
1
8
(n) x tan x - In Isec x I+C;(0)-(2x2+2x sin 2x - cos 2x) +C;
(p) xOn x ) ~
- 2x(h x - 1) +C;
1 1 1 e3x
2 X 27
(q)4(sin 2x -2x cos 2x) +C;(r) --cos x2 +C;(s) -- (1 +In x) +C;(t)-(9x2 - 6x +2) +C;
1
6
(U)- [2x3 tan- 'x -x2 +In (1 +x2)] +C; (U)x In (x2+ 1) - 2(x - tan-' x) +C;
x3
9
- 2) + C;(x) -(3 In x - 1) +C.
A
3
8
.
7 (a) 1;(b) (i) n(e -2); (ii)-(1 +e2).
2
A2 2
A
2 n
1
3
8
.
8 (U) j;(b) 2 ~ ;
(c)A . 3
8
.
1
0 (U)- ;(b) 2z2. 3
8
.
1
1 (U) 0;(b) --.
cosn-' x sin x ~ n - 1Jcosn-2
38.12 (a) Icos" x dx = x dx.
n n
sinn-' x cos x + n - 1
(b) I s i d x dx = - x dx.
n n
sec"-2 x tan x n - 2
I n - 1 n - 1
3
8
.
1
3 (a) sec"xdx= +-I s e C 2x dx;
sec2x tan x 2
3 3
+- tan x +C = (tan x)
1
2
(b) (i)-(sec x tan x +In Isec x +tan x I)+C; (ii)

CHAPTER 39
369
1 sin 6x 1 2 1
1
3 2 12 5 7 9
3
8
+C;(c) - sin5 x - - sin' x +- sing x +C;
39.12 (a) -- cos3 x +C; (b)- x +-
sin 4x sin3 2x sin 8x
(4f (F+2 sin 2x +- sin 4x -- sin3 2x +c;(e) - - x --
+-
--
)+c;
6 ) :
6
(
: 8 3 64
X 1 1
2 5 3
(f)2 tan--x + C;(g)- tan5 x -- tan3 x +tan x - x +C;
1 3 3 1 1
8 8 5 3
(h)4sec3x tan x +- sec x tan x +- In Isec x +tan x I +C;(i) -tan5 x +- tan3 x +C;
1 1 1 5 3
3 4 8 8
( j ) sec5x -- sec3 x + C;(k)- sec3x tan x --sec x tan x +- In Isec x +tan x I +C;
1 1 1
8 8a 24
(Z)--cos 4x +C; (m) -(2 cos 2nx - cos 4ax) +C;(n)-(6 sin 2x -sin 12x) +C;
13
39.14 (a)0; (b)n.
J l + x 2 - 1
39.15 (a),
/
=
- sec-' x + C; (b) 2 sin-' !
!- ,
/
=
+C ;(c) ,
/
- +In +c;
2 2 1x1
-
1 , / ~ ' - 9
(4 -,/E7
+C;(e) - -
X
+C;(9)54 tan-' - +-
3x )+c;
+ c;(f)
4 J Z l ( 3 x 2 + 9
9 x
J Z T - 1
$ - 1 '
39.16 f i + In (fl+
4). 39.17 ,
/
=
-
-(1 +fi)+In
39.18 Same answer as to Problem 39.17 (becausethe two arcs are mirror imagesin the line y = x).
39.19 In (2 +fi). 39.20 a f i & = 6n.
CHAPTER 40
1 x - 3 x3 1 3
+C;(b)3 In I x +3 ) - 2 In I x +2) +C;(c)- +-In I x +21 +-In I x -
3 4 4
40.6 (a)- In -
6 Jxf31
3 19 3 3 5
( d ) ~
In Ix - 1I- 9 In Ix - 21 +-In Ix - 31 + C; (e)- In I x - 1I -8 In Ix + 1I +- In
2 4 8
(f)- 6 In 1x1+6In I x +3 I - - In Ix +2I +- In Ix - 1I +c;( g ) z In IAI+c;
6 6
1 13 7 1 1 x2-
+c;
( h ) 6 l n l ~ l + - + C ; ( i ) ~ l n
5 1 1
x + l x
25 x + l
266 In Ix - 41 +9 In Ix + 1I +-
21 +c;
x - 3 1 + c ;
1 9 13 1 1 X
x - 6 tan-' 5+C;
(m) 10In Ix - 1I +-In (x' +4x + 5) --tan-' (x +2) + C;(n)- tan-'
20 10 3

x2 1 1 1 1
2 9 2 2 x + 1
(0)- +- Dn 1x1-41 In (x2 +9)] +C;(p)In 1x1--In (x2 + 1)+
-
,
+ C;
1 1 x 1 x - 4
(4)
25In Ix - 1I --I
n (x2 +
50
x + l
+c;(s) ln 1-1x + 2 +c ;
1
2
(r)InIx(--In(x2+x+1)-
1 11
3
(t)? In Ixl +-In Ix +3I - 3 In Ix +21 +C;(U) x - In (1 +e? + C.
+c.
1 1
40.8 (a)81(3 In 2 +x d ) ; (6)5( n f i-3 In 2). 40.9
X
1 -tan -
2
40.10 (a)In(l -sinx)+C;(b)ln(sinx- l ( + C ; ( c ) s i n x < l .

Index
A
Abscissa, 8
Absolute extremum (maximum, minimum), 105
tabular method, 106
Absolute value, 2
Absolute value function, 47
Acceleration, 163
of gravity, 163
Amplitude, 204
Angle, directed, 185
Angle of inclination, 218
Angle measure, 185
Antiderivative, 221
Approximation Principle, 155
Arc cosecant, 297
Arc cosine, 293
Arc cotangent, 296
Arc length, 251
Arc length formula, 252
Arc secant, 295
Arc sine, 292
Arc tangent, 295
Area:
between curves, 250
circle, 113
under a curve, 229, 249
equilateral triangle, 113
trapezoid, 244
Argument, 46
Asymptote, horizontal, 70
vertical, 69
Average, 239
Average speed (velocity), 136
Average value of a function, 239
Axis of symmetry, 41
B
b (y-intercept), 28
Base (of logarithm), 282
C
Chain rule, 117
power, 117
Change of variable in an integral, 240
Circle, 14
standard equation of, 14
area, 113
circumference, 13
Closed interval, 49

Collinear points, 30
Common logarithm, 282
Completing the square, 17
Composite function, 116
Composition, 116
Concave:
downward, 167
upward, 167
Concavity, 167
Continuity, 78
over a closed interval, 80
on the left, 79
on the right, 79
one-sided, 79
Continuous function, 78, 79
Continuously compounded interest, 283
Coordinate systems, on a line, 1, 2
in a plane, 8
Coordinates, on a line, 1
polar, 193
Cosecant, 214
Cosine, 190
Cotangent, 214
Critical number, 106
Cross-section formula, 259
Cylindrical shell formula, 258, 261
D
Decay constant, 286
Decreasing function, 130
Definite integral, 231, 232
Degree, 185
Demand equation, 40
Derivative, 92
first, 161
higher-order, 161
second, 161
Difference of cylindrical shells, 261
Differentiable function, 92
Differential, 155
Differentiation, 92
higher-order implicit, 162
implicit, 126
logarithmic, 272
Directed angle, 186
Discontinuity, 78
Displacement, 136
Disk formula, 257
Distance formula, 9
Division of polynomials, 58
Domain, 46
Double-angle formulas, 195

E
e, 276
Ellipse, 14
Epsilon-delta definition, 64
Even function, 50
Exponential decay, 285
Exponential functions, 275
tables, 334
Exponential growth, 284
Exponents, 118
Extreme-value theorem, 105
Extremum, absolute, 105
relative, 104
F
First-derivative test, 171
Free fall, 137
Frequency, 204
Function, 46
composite, 116
continuous, 78, 79
even, 50
odd, 51
one-one, 292
periodic, 193
Fundamental theorem of algebra, 52
Fundamental theorem of calculus, 238
G
Gap, 78
Generalized Rolle's theorem, 135
Geometric formulas, 331
Graphs:
of equations, 14
of functions, 46
intersections of, 36
sketching, 171
Greatest-integer function, 53
Growth constant, 286
H
Half-angle formulas, 195
Half-life, 287
Half-open interval, 49
Higher-order:
derivative, 161
implicit differentiation, 162
Horizontal asymptote, 70
Hyperbola, 14
I
Implicit differentiation, 126
higher-order, 162
Increasing function, 130
Indefinite integral, 221
Inequalities, 3

Infinite limits, 68
Inflection point, 167
Instantaneous rate of change, 144
Instantaneous velocity, 136
Integrable function, 231-232
Integral, definite, 231, 232
indefinite, 221
Riemann, 232
Integrand, 221
Integration, 221
by parts, 305
Integration formulas, 330
Interest, continuously compounded, 283
Intermediate value theorem, 130
Intersections of graphs, 36
Intervals, 48
Inverse:
cosecant, 297
cosine, 293
cotangent, 296
secant, 296
sine, 292
tangent, 295
Inverse function, 275, 292
trigonometric functions, 293
Irrational number, 85
Irreducible factors, 320
J
Jump, 78
L
Law of cosines, 194
Law of the mean, 129
Law of sines, 198
Laws of exponents, 118
Least-squares principle, 181
Leibniz, Gottfried von, 238
L'Hôpital's rule, 284
Limit, 59, 67
infinite, 68
at infinity, 70
one-sided, 67
Limits of integration, 233
Line (see Straight line)
Linear factors, 320
and roots of polynomials, 52
Logarithm:
to the base a, 282
common, 282
natural, 268, 333
Logarithmic differentiation, 272
M
m (slope), 24

Marginal cost, 145
Marginal profit, 144
Mathematical induction, 95
Maximum, absolute, 105
relative, 104
Mean, 239
Mean-value theorem:
for derivatives, 129
for integrals, 240
Midpoint formulas, 10
Midpoint rule, 247
Minimum, absolute, 105
relative, 104
N
Natural logarithm, 268
tables, 333
Newton, Isaac, 238
Newton's method, 156
Normal line, 90
Null set, 18
O
Odd function, 51
One-one function, 292
One-sided continuity, 79
One-sided limit, 67
Open interval, 49
Ordinate, 8
Origin, 1
P
Parabola, 14
Parallel lines, 28
Partial fractions, 321
Percentage error, 160
Period, 193
Perpendicular lines, 29
Point of inflection, 167
Point-slope equation, 27
Polar coordinates, 193
Polynomials, 51
differentiation of, 94
Power chain rule, 117
Product rule, 93
Proper rational function, 320
Pyramid, 266
Q
Quadrants, 8
Quadratic factors, 320
Quadratic formula, 38
Quick formula I, 223
Quick formula II, 271
Quotient of a division, 58
Quotient rule, 100

R
Radian, 185
Radicals, 118
Range, 47
Rational function, 71
proper, 320
Rational number, 85
Rational power, 118
Rectilinear motion, 136
Reduced angle, 188
Reduction formulas, 308, 310, 312
Related rates, 147
Relative extremum, 104
maximum, 104
minimum, 104
Remainder of a division, 58
Removable discontinuity, 78
Repeated root, 52
Rhombus, 30
Riemann integral, 232
Riemann sums, 232
Rolle's theorem, 129
Roots of a number, 118
Roots of a polynomial, 51
S
Secant, 214
Second-derivative test, 169
Sigma notation, 229
Simpson's rule, 247
Sine, 190
Slope, 24
of a tangent line, 86
Slope-intercept equation, 28
Solid of revolution, 257
Speed, 136
Standard equation of a circle, 14
Step function, 53
Straight line, 24
equations of, 26
Substitution method for antiderivatives, 223
Supply equation, 40
Symmetry:
with respect to a line, 41
with respect to a point, 42
with respect to the origin, 42
T
Tabular method, 106
Tangent function, 214
Tangent line, 86, 87
Total oscillation, 204
Trapezoid, area of, 244
Trapezoidal rule, 244
Triangle inequality, 3

Trigonometric:
formulas, 329
functions, 190, 214
function tables, 332
integration of, 311
Trigonometric integrands, 311
Trigonometric substitutions, 313
V
Value of a function, 46
Velocity, 136
Vertical asymptote, 69
Vertical line test, 54
Volume, 257
of a cylinder, 113
of a sphere, 144
W
Washer formula, 258
X
x-axis, 8
x-coordinate, 8
x-intercept, 35
Y
y-axis, 8
y-coordinate, 8
y-intercept, 28
Z
Zero of a polynomial, 51

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