Lesson 20: Derivatives and the Shapes of Curves (handout)
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This document contains lecture notes on calculus from a Calculus I course. It covers determining the monotonicity of functions using the first derivative test. Key points include using the sign of the derivative to determine if a function is increasing or decreasing over an interval, and using the first derivative test to classify critical points as local maxima, minima, or neither. Examples are provided to demonstrate finding intervals of monotonicity for various functions and applying the first derivative test.
![. V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Deriva ve Test
Concavity
Defini ons
Tes ng for Concavity
The Second Deriva ve Test
.
.
Notes
Recall: The Mean Value Theorem
Theorem (The Mean Value Theorem)
c
Let f be con nuous on [a, b]
and differen able on (a, b).
Then there exists a point c in
(a, b) such that
b
f(b) − f(a) .
= f′ (c). a
b−a
Another way to put this is that there exists a point c such that
f(b) = f(a) + f′ (c)(b − a)
.
.
Why the MVT is the MITC Notes
Most Important Theorem In Calculus!
Theorem
Let f′ = 0 on an interval (a, b). Then f is constant on (a, b).
Proof.
Pick any points x and y in (a, b) with x < y. Then f is con nuous on
[x, y] and differen able on (x, y). By MVT there exists a point z in
(x, y) such that
f(y) = f(x) + f′ (z)(y − x)
So f(y) = f(x). Since this is true for all x and y in (a, b), then f is
constant.
.
.
. 2
.](https://image.slidesharecdn.com/lesson20-derivativesandtheshapeofcurves001handout-110410232658-phpapp01/85/Lesson-20-Derivatives-and-the-Shapes-of-Curves-handout-2-320.jpg)
![. V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Finding intervals of monotonicity I
Example
Find the intervals of monotonicity of f(x) = 2x − 5.
Solu on
f′ (x) = 2 is always posi ve, so f is increasing on (−∞, ∞).
Example
Describe the monotonicity of f(x) = arctan(x).
Solu on
1
Since f′ (x) = is always posi ve, f(x) is always increasing.
. 1 + x2
.
Notes
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
f′ (x) = 2x, which is posi ve when x > 0 and nega ve when x is.
We can draw a number line:
− 0 + f′
.
↘ 0 ↗ f
.
.
Notes
Finding intervals of monotonicity II
Example
Find the intervals of monotonicity of f(x) = x2 − 1.
Solu on
− 0 + f′
.
↘ 0 ↗ f
So f is decreasing on (−∞, 0) and increasing on (0, ∞).
In fact we can say f is decreasing on (−∞, 0] and increasing on
[0, ∞)
.
.
. 4
.](https://image.slidesharecdn.com/lesson20-derivativesandtheshapeofcurves001handout-110410232658-phpapp01/85/Lesson-20-Derivatives-and-the-Shapes-of-Curves-handout-4-320.jpg)
![. V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Finding intervals of monotonicity III
Example
Find the intervals of monotonicity of f(x) = x2/3 (x + 2).
Solu on
.
.
Notes
The First Derivative Test
Theorem (The First Deriva ve Test)
Let f be con nuous on [a, b] and c a cri cal point of f in (a, b).
If f′ changes from posi ve to nega ve at c, then c is a local
maximum.
If f′ changes from nega ve to posi ve at c, then c is a local
minimum.
If f′ (x) has the same sign on either side of c, then c is not a local
extremum.
.
.
Notes
Outline
Recall: The Mean Value Theorem
Monotonicity
The Increasing/Decreasing Test
Finding intervals of monotonicity
The First Deriva ve Test
Concavity
Defini ons
Tes ng for Concavity
The Second Deriva ve Test
.
.
. 5
.](https://image.slidesharecdn.com/lesson20-derivativesandtheshapeofcurves001handout-110410232658-phpapp01/85/Lesson-20-Derivatives-and-the-Shapes-of-Curves-handout-5-320.jpg)
![. V63.0121.001: Calculus I
. Sec on 4.2: The Shapes of .
Curves April 11, 2011
Notes
Finding Intervals of Concavity II
Example
Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2).
Solu on
We have
10 −1/3 4 −4/3
f′′ (x) = x − x
9 9
2 −4/3
= x (5x − 2)
9
.
.
Notes
The Second Derivative Test
Theorem (The Second Deriva ve Test)
Let f, f′ , and f′′ be con nuous on [a, b]. Let c be be a point in (a, b)
with f′ (c) = 0.
If f′′ (c) < 0, then c is a local maximum of f.
If f′′ (c) > 0, then c is a local minimum of f.
Remarks
If f′′ (c) = 0, the second deriva ve test is inconclusive
We look for zeroes of f′ and plug them into f′′ to determine if
their f values are local extreme values.
.
.
Notes
Proof of the Second Derivative Test
Proof.
Suppose f′ (c) = 0 and f′′ (c) > 0.
Since f′′ is con nuous,
f′′ (x) > 0 for all x + + + f′′ = (f′ )′
sufficiently close to c. .
↗ c ↗ f′
Since f′′ = (f′ )′ , we know
0 f′
f′ is increasing near c.
c f
.
.
. 8
.](https://image.slidesharecdn.com/lesson20-derivativesandtheshapeofcurves001handout-110410232658-phpapp01/85/Lesson-20-Derivatives-and-the-Shapes-of-Curves-handout-8-320.jpg)
Lesson 20: Derivatives and the Shapes of Curves (handout)
- 1. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Sec on 4.2 Deriva ves and the Shapes of Curves V63.0121.001: Calculus I Professor Ma hew Leingang New York University April 11, 2011 . . Notes Announcements Quiz 4 on Sec ons 3.3, 3.4, 3.5, and 3.7 this week (April 14/15) Quiz 5 on Sec ons 4.1–4.4 April 28/29 Final Exam Thursday May 12, 2:00–3:50pm . . Notes Objectives Use the deriva ve of a func on to determine the intervals along which the func on is increasing or decreasing (The Increasing/Decreasing Test) Use the First Deriva ve Test to classify cri cal points of a func on as local maxima, local minima, or neither. . . . 1 .
- 2. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Deriva ve Test Concavity Defini ons Tes ng for Concavity The Second Deriva ve Test . . Notes Recall: The Mean Value Theorem Theorem (The Mean Value Theorem) c Let f be con nuous on [a, b] and differen able on (a, b). Then there exists a point c in (a, b) such that b f(b) − f(a) . = f′ (c). a b−a Another way to put this is that there exists a point c such that f(b) = f(a) + f′ (c)(b − a) . . Why the MVT is the MITC Notes Most Important Theorem In Calculus! Theorem Let f′ = 0 on an interval (a, b). Then f is constant on (a, b). Proof. Pick any points x and y in (a, b) with x < y. Then f is con nuous on [x, y] and differen able on (x, y). By MVT there exists a point z in (x, y) such that f(y) = f(x) + f′ (z)(y − x) So f(y) = f(x). Since this is true for all x and y in (a, b), then f is constant. . . . 2 .
- 3. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Deriva ve Test Concavity Defini ons Tes ng for Concavity The Second Deriva ve Test . . Notes Increasing Functions Defini on A func on f is increasing on the interval I if f(x) < f(y) whenever x and y are two points in I with x < y. An increasing func on “preserves order.” I could be bounded or infinite, open, closed, or half-open/half-closed. Write your own defini on (muta s mutandis) of decreasing, nonincreasing, nondecreasing . . Notes The Increasing/Decreasing Test Theorem (The Increasing/Decreasing Test) If f′ > 0 on an interval, then f is increasing on that interval. If f′ < 0 on an interval, then f is decreasing on that interval. Proof. It works the same as the last theorem. Assume f′ (x) > 0 on an interval I. Pick two points x and y in I with x < y. We must show f(x) < f(y). By MVT there exists a point c in (x, y) such that f(y) − f(x) = f′ (c)(y − x) > 0. So f(y) > f(x). . . . 3 .
- 4. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Finding intervals of monotonicity I Example Find the intervals of monotonicity of f(x) = 2x − 5. Solu on f′ (x) = 2 is always posi ve, so f is increasing on (−∞, ∞). Example Describe the monotonicity of f(x) = arctan(x). Solu on 1 Since f′ (x) = is always posi ve, f(x) is always increasing. . 1 + x2 . Notes Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solu on f′ (x) = 2x, which is posi ve when x > 0 and nega ve when x is. We can draw a number line: − 0 + f′ . ↘ 0 ↗ f . . Notes Finding intervals of monotonicity II Example Find the intervals of monotonicity of f(x) = x2 − 1. Solu on − 0 + f′ . ↘ 0 ↗ f So f is decreasing on (−∞, 0) and increasing on (0, ∞). In fact we can say f is decreasing on (−∞, 0] and increasing on [0, ∞) . . . 4 .
- 5. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Finding intervals of monotonicity III Example Find the intervals of monotonicity of f(x) = x2/3 (x + 2). Solu on . . Notes The First Derivative Test Theorem (The First Deriva ve Test) Let f be con nuous on [a, b] and c a cri cal point of f in (a, b). If f′ changes from posi ve to nega ve at c, then c is a local maximum. If f′ changes from nega ve to posi ve at c, then c is a local minimum. If f′ (x) has the same sign on either side of c, then c is not a local extremum. . . Notes Outline Recall: The Mean Value Theorem Monotonicity The Increasing/Decreasing Test Finding intervals of monotonicity The First Deriva ve Test Concavity Defini ons Tes ng for Concavity The Second Deriva ve Test . . . 5 .
- 6. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Concavity Defini on The graph of f is called concave upwards on an interval if it lies above all its tangents on that interval. The graph of f is called concave downwards on an interval if it lies below all its tangents on that interval. . . concave up concave down . We some mes say a concave up graph “holds water” and a concave down graph “spills water”. . Notes Synonyms for concavity Remark “concave up” = “concave upwards” = “convex” “concave down” = “concave downwards” = “concave” . . Notes Inflection points mean change in concavity Defini on A point P on a curve y = f(x) is called an inflec on point if f is con nuous at P and the curve changes from concave upward to concave downward at P (or vice versa). concave up inflec on point . concave down . . . 6 .
- 7. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Testing for Concavity Theorem (Concavity Test) If f′′ (x) > 0 for all x in an interval, then the graph of f is concave upward on that interval. If f′′ (x) < 0 for all x in an interval, then the graph of f is concave downward on that interval. . . Notes Testing for Concavity Proof. Suppose f′′ (x) > 0 on the interval I (which could be infinite). This means f′ is increasing on I. Let a and x be in I. The tangent line through (a, f(a)) is the graph of L(x) = f(a) + f′ (a)(x − a) By MVT, there exists a c between a and x with f(x) = f(a) + f′ (c)(x − a) Since f′ is increasing, f(x) > L(x). . . Notes Finding Intervals of Concavity I Example Find the intervals of concavity for the graph of f(x) = x3 + x2 . Solu on We have f′ (x) = 3x2 + 2x, so f′′ (x) = 6x + 2. This is nega ve when x < −1/3, posi ve when x > −1/3, and 0 when x = −1/3 So f is concave down on the open interval (−∞, −1/3), concave up on the open interval (−1/3, ∞), and has an inflec on point at the point (−1/3, 2/27) . . . 7 .
- 8. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Finding Intervals of Concavity II Example Find the intervals of concavity of the graph of f(x) = x2/3 (x + 2). Solu on We have 10 −1/3 4 −4/3 f′′ (x) = x − x 9 9 2 −4/3 = x (5x − 2) 9 . . Notes The Second Derivative Test Theorem (The Second Deriva ve Test) Let f, f′ , and f′′ be con nuous on [a, b]. Let c be be a point in (a, b) with f′ (c) = 0. If f′′ (c) < 0, then c is a local maximum of f. If f′′ (c) > 0, then c is a local minimum of f. Remarks If f′′ (c) = 0, the second deriva ve test is inconclusive We look for zeroes of f′ and plug them into f′′ to determine if their f values are local extreme values. . . Notes Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′′ is con nuous, f′′ (x) > 0 for all x + + + f′′ = (f′ )′ sufficiently close to c. . ↗ c ↗ f′ Since f′′ = (f′ )′ , we know 0 f′ f′ is increasing near c. c f . . . 8 .
- 9. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. Since f′ (c) = 0 and f′ is increasing, f′ (x) < 0 for x + + + f′′ = (f′ )′ . close to c and less than c, c ↗ ↗ f′ and f′ (x) > 0 for x close − 0 + f′ to c and more than c. c f . . Notes Proof of the Second Derivative Test Proof. Suppose f′ (c) = 0 and f′′ (c) > 0. This means f′ changes sign from nega ve to + + + f′′ = (f′ )′ . posi ve at c, which ↗ c ↗ f′ means (by the First − 0 + f′ Deriva ve Test) that f has a local minimum at c. c ↘ min ↗ f . . Notes Using the Second Derivative Test I Example Find the local extrema of f(x) = x3 + x2 . Solu on f′ (x) = 3x2 + 2x = x(3x + 2) is 0 when x = 0 or x = −2/3. Remember f′′ (x) = 6x + 2 Since f′′ (−2/3) = −2 < 0, −2/3 is a local maximum. Since f′′ (0) = 2 > 0, 0 is a local minimum. . . . 9 .
- 10. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes Using the Second Derivative Test II Example Find the local extrema of f(x) = x2/3 (x + 2) Solu on . . Using the Second Derivative Test II Notes Graph Graph of f(x) = x2/3 (x + 2): y . x . . Notes When the second derivative is zero Remark At inflec on points c, if f′ is differen able at c, then f′′ (c) = 0 If f′′ (c) = 0, must f have an inflec on point at c? Consider these examples: f(x) = x4 g(x) = −x4 h(x) = x3 All of them have cri cal points at zero with a second deriva ve of zero. But the first has a local min at 0, the second has a local max at 0, and the third has an inflec on point at 0. This is why we say 2DT has nothing to say when f′′ (c) = 0. . . . 10 .
- 11. . V63.0121.001: Calculus I . Sec on 4.2: The Shapes of . Curves April 11, 2011 Notes When first and second derivative are zero func on deriva ves graph type ′ 3 ′ f (x) = 4x , f (0) = 0 f(x) = x4 . min f (x) = 12x2 , f′′ (0) = 0 ′′ . g′ (x) = − 4x3 , g′ (0) = 0 g(x) = −x4 max g′′ (x) = − 12x2 , g′′ (0) = 0 h′ (x) = 3x2 , h′ (0) = 0 h(x) = x3 . infl. h′′ (x) = 6x, h′′ (0) = 0 . . Notes Summary Concepts: Mean Value Theorem, monotonicity, concavity Facts: deriva ves can detect monotonicity and concavity Techniques for drawing curves: the Increasing/Decreasing Test and the Concavity Test Techniques for finding extrema: the First Deriva ve Test and the Second Deriva ve Test . . Notes . . . 11 .