Solution Manual : Chapter - 07 Exponential, Logarithmic and Inverse Trigonometric Functions

263
CHAPTER 7
Exponential, Logarithmic, and Inverse
Trigonometric Functions
EXERCISE SET 7.1
1. (a) f(g(x)) = 4(x/4) = x, g(f(x)) = (4x)/4 = x, f and g are inverse functions
(b) f(g(x)) = 3(3x − 1) + 1 = 9x − 2 = x so f and g are not inverse functions
(c) f(g(x)) = 3
(x3 + 2) − 2 = x, g(f(x)) = (x − 2) + 2 = x, f and g are inverse functions
(d) f(g(x)) = (x1/4
)4
= x, g(f(x)) = (x4
)1/4
= |x| = x, f and g are not inverse functions
2. (a) They are inverse functions. 2
-2
-2 2
(b) The graphs are not reﬂections of each other
about the line y = x.
2
-2
-2 2
(c) They are inverse functions provided the domain of
g is restricted to [0, +∞)
5
0
0 5
(d) They are inverse functions provided the domain of f(x)
is restricted to [0, +∞)
2
0
0 2
3. (a) yes; all outputs (the elements of row two) are distinct
(b) no; f(1) = f(6)

264 Chapter 7
4. (a) no; it is easy to conceive of, say, 8 people in line at two diﬀerent times
(b) no; perhaps your weight remains constant for more than a year
(c) yes, since the function is increasing, in the sense that the greater the volume, the greater the
weight
5. (a) yes (b) yes (c) no (d) yes (e) no (f) no
6. (a) no, the horizontal line test fails
6
-2
-3 3
(b) yes, horizontal line test
10
-10
-1 3
7. (a) no, the horizontal line test fails
(b) no, the horizontal line test fails
(c) yes, horizontal line test
8. (d) no, the horizontal line test fails
(e) no, the horizontal line test fails
(f) yes, horizontal line test
9. (a) f has an inverse because the graph passes the horizontal line test. To compute f−1
(2) start
at 2 on the y-axis and go to the curve and then down, so f−1
(2) = 8; similarly, f−1
(−1) = −1
and f−1
(0) = 0.
(b) domain of f−1
is [−2, 2], range is [−8, 8] (c)
-2 1 2
-8
-4
4
8
y
x
10. (a) the horizontal line test fails
(b) −∞ < x ≤ −1; −1 ≤ x ≤ 2; and 2 ≤ x < 4.
11. (a) f (x) = 2x + 8; f < 0 on (−∞, −4) and f > 0 on (−4, +∞); not one-to-one
(b) f (x) = 10x4
+ 3x2
+ 3 ≥ 3 > 0; f (x) is positive for all x, so f is one-to-one
(c) f (x) = 2 + cos x ≥ 1 > 0 for all x, so f is one-to-one
12. (a) f (x) = 3x2
+ 6x = x(3x + 6) changes sign at x = −2, 0, so f is not one-to-one
(b) f (x) = 5x4
+ 24x2
+ 2 ≥ 2 > 0; f is positive for all x, so f is one-to-one
(c) f (x) =
1
(x + 1)2
; f is one-to-one because:
if x1 < x2 < −1 then f > 0 on [x1, x2], so f(x1) = f(x2)
if −1 < x1 < x2 then f > 0 on [x1, x2], so f(x1) = f(x2)
if x1 < −1 < x2 then f(x1) > 1 > f(x2) since f(x) > 1 on (−∞, −1) and f(x) < 1 on
(−1, +∞)
13. y = f−1
(x), x = f(y) = y5
, y = x1/5
= f−1
(x)

Exercise Set 7.1 265
14. y = f−1
(x), x = f(y) = 6y, y =
1
6
x = f−1
(x)
15. y = f−1
(x), x = f(y) = 7y − 6, y =
1
7
(x + 6) = f−1
(x)
16. y = f−1
(x), x = f(y) =
y + 1
y − 1
, xy − x = y + 1, (x − 1)y = x + 1, y =
x + 1
x − 1
= f−1
(x)
17. y = f−1
(x), x = f(y) = 3y3
− 5, y = 3
(x + 5)/3 = f−1
(x)
18. y = f−1
(x), x = f(y) = 5
√
4y + 2, y =
1
4
(x5
− 2) = f−1
(x)
19. y = f−1
(x), x = f(y) = 3
√
2y − 1, y = (x3
+ 1)/2 = f−1
(x)
20. y = f−1
(x), x = f(y) =
5
y2 + 1
, y =
5 − x
x
= f−1
(x)
21. y = f−1
(x), x = f(y) = 3/y2
, y = − 3/x = f−1
(x)
22. y = f−1
(x), x = f(y) =
2y, y ≤ 0
y2
, y > 0
, y = f−1
(x) =
x/2, x ≤ 0
√
x, x > 0
23. y = f−1
(x), x = f(y) =
5/2 − y, y < 2
1/y, y ≥ 2
, y = f−1
(x) =
5/2 − x, x > 1/2
1/x, 0 < x ≤ 1/2
24. y = p−1
(x), x = p(y) = y3
− 3y2
+ 3y − 1 = (y − 1)3
, y = x1/3
+ 1 = p−1
(x)
25. y = f−1
(x), x = f(y) = (y + 2)4
for y ≥ 0, y = f−1
(x) = x1/4
− 2 for x ≥ 16
26. y = f−1
(x), x = f(y) =
√
y + 3 for y ≥ −3, y = f−1
(x) = x2
− 3 for x ≥ 0
27. y = f−1
(x), x = f(y) = −
√
3 − 2y for y ≤ 3/2, y = f−1
(x) = (3 − x2
)/2 for x ≤ 0
28. y = f−1
(x), x = f(y) = 3y2
+ 5y − 2 for y ≥ 0, 3y2
+ 5y − 2 − x = 0 for y ≥ 0,
y = f−1
(x) = (−5 +
√
12x + 49)/6 for x ≥ −2
29. y = f−1
(x), x = f(y) = y − 5y2
for y ≥ 1, 5y2
− y + x = 0 for y ≥ 1,
y = f−1
(x) = (1 +
√
1 − 20x)/10 for x ≤ −4
30. (a) C =
5
9
(F − 32)
(b) how many degrees Celsius given the Fahrenheit temperature
(c) C = −273.15◦
C is equivalent to F = −459.67◦
F, so the domain is F ≥ −459.67, the range
is C ≥ −273.15
31. (a) y = f(x) = (6.214 × 10−4
)x (b) x = f−1
(y) =
104
6.214
y
(c) how many meters in y miles
32. f and f−1
are continuous so f(3) = lim
x→3
f(x) = 7; then f−1
(7) = 3, and
lim
x→7
f−1
(x) = f−1
lim
x→7
x = f−1
(7) = 3

266 Chapter 7
33. (a) f(g(x)) = f(
√
x)
= (
√
x)2
= x, x > 1;
g(f(x)) = g(x2
)
=
√
x2 = x, x > 1
(b)
x
y
y = f(x)
y = g(x)
(c) no, because f(g(x)) = x for every x in the domain of g is not satisﬁed
(the domain of g is x ≥ 0)
34. y = f−1
(x), x = f(y) = ay2
+ by + c, ay2
+ by + c − x = 0, use the quadratic formula to get
y =
−b ± b2 − 4a(c − x)
2a
;
(a) f−1
(x) =
−b + b2 − 4a(c − x)
2a
(b) f−1
(x) =
−b − b2 − 4a(c − x)
2a
35. (a) f(f(x)) =
3 −
3 − x
1 − x
1 −
3 − x
1 − x
=
3 − 3x − 3 + x
1 − x − 3 + x
= x so f = f−1
(b) symmetric about the line y = x
36. y = m(x − x0) is an equation of the line. The graph of the inverse of f(x) = m(x − x0) will be the
reﬂection of this line about y = x. Solve y = m(x − x0) for x to get x = y/m + x0 = f−1
(y) so
y = f−1
(x) = x/m + x0.
37. (a) f(x) = x3
− 3x2
+ 2x = x(x − 1)(x − 2) so f(0) = f(1) = f(2) = 0 thus f is not one-to-one.
(b) f (x) = 3x2
− 6x + 2, f (x) = 0 when x =
6 ±
√
36 − 24
6
= 1 ±
√
3/3. f (x) > 0 (f is
increasing) if x < 1 −
√
3/3, f (x) < 0 (f is decreasing) if 1 −
√
3/3 < x < 1 +
√
3/3, so f(x)
takes on values less than f(1 −
√
3/3) on both sides of 1 −
√
3/3 thus 1 −
√
3/3 is the largest
value of k.
38. (a) f(x) = x3
(x − 2) so f(0) = f(2) = 0 thus f is not one to one.
(b) f (x) = 4x3
− 6x2
= 4x2
(x − 3/2), f (x) = 0 when x = 0 or 3/2; f is decreasing on (−∞, 3/2]
and increasing on [3/2, +∞) so 3/2 is the smallest value of k.
39. if f−1
(x) = 1, then x = f(1) = 2(1)3
+ 5(1) + 3 = 10
40. if f−1
(x) = 2, then x = f(2) = (2)3
/[(2)2
+ 1] = 8/5
41. 6
-2
-2 6
42. 10
-5
-5 10

43. 3
0
0 3
44. 6
0
0 6
45. y = f−1
(x), x = f(y) = 5y3
+ y − 7,
dx
dy
= 15y2
+ 1,
dy
dx
=
1
15y2 + 1
;
check: 1 = 15y2 dy
dx
+
dy
dx
,
dy
dx
=
1
15y2 + 1
46. y = f−1
(x), x = f(y) = 1/y2
,
dx
dy
= −2y−3
,
dy
dx
= −y3
/2;
check: 1 = −2y−3 dy
dx
,
dy
dx
= −y3
/2
47. y = f−1
(x), x = f(y) = 2y5
+ y3
+ 1,
dx
dy
= 10y4
+ 3y2
,
dy
dx
=
1
10y4 + 3y2
;
check: 1 = 10y4 dy
dx
+ 3y2 dy
dx
,
dy
dx
=
1
10y4 + 3y2
48. y = f−1
(x), x = f(y) = 5y − sin 2y,
dx
dy
= 5 − 2 cos 2y,
dy
dx
=
1
5 − 2 cos 2y
;
check: 1 = (5 − 2 cos 2y)
dy
dx
,
dy
dx
=
1
5 − 2 cos 2y
49. f(f(x)) = x thus f = f−1
so the graph is symmetric about y = x.
50. (a) Suppose x1 = x2 where x1 and x2 are in the domain of g and g(x1), g(x2) are in the domain of
f then g(x1) = g(x2) because g is one-to-one so f(g(x1)) = f(g(x2)) because f is one-to-one
thus f ◦ g is one-to-one because (f ◦ g)(x1) = (f ◦ g)(x2) if x1 = x2.
(b) f, g, and f ◦ g all have inverses because they are all one-to-one. Let h = (f ◦ g)−1
then
(f ◦ g)(h(x)) = f[g(h(x))] = x, apply f−1
to both sides to get g(h(x)) = f−1
(x), then apply
g−1
to get h(x) = g−1
(f−1
(x)) = (g−1
◦ f−1
)(x), so h = g−1
◦ f−1
51.
x
y
52. Suppose that g and h are both inverses of f then f(g(x)) = x, h[f(g(x))] = h(x), but
h[f(g(x))] = g(x) because h is an inverse of f so g(x) = h(x).
53. F (x) = 2f (2g(x))g (x) so F (3) = 2f (2g(3))g (3). By inspection f(1) = 3, so g(3) = f−1
(3) = 1
and g (3) = (f−1
) (3) = 1/f (f−1
(3)) = 1/f (1) = 1/7 because f (x) = 4x3
+ 3x2
. Thus
F (3) = 2f (2)(1/7) = 2(44)(1/7) = 88/7.
F(3) = f(2g(3)), g(3) = f−1
(3); by inspection f(1) = 3, so g(3) = f−1
(3) = 1, F(3) = f(2) = 25.

268 Chapter 7
EXERCISE SET 7.2
1. (a) −4 (b) 4 (c) 1/4
2. (a) 1/16 (b) 8 (c) 1/3
3. (a) 2.9690 (b) 0.0341
4. (a) 1.8882 (b) 0.9381
5. (a) log2 16 = log2(24
) = 4 (b) log2
1
32
= log2(2−5
) = −5
(c) log4 4 = 1 (d) log9 3 = log9(91/2
) = 1/2
6. (a) log10(0.001) = log10(10−3
) = −3 (b) log10(104
) = 4
(c) ln(e3
) = 3 (d) ln(
√
e) = ln(e1/2
) = 1/2
7. (a) 1.3655 (b) −0.3011
8. (a) −0.5229 (b) 1.1447
9. (a) 2 ln a +
1
2
ln b +
1
2
ln c = 2r + s/2 + t/2 (b) ln b − 3 ln a − ln c = s − 3r − t
10. (a)
1
3
ln c − ln a − ln b = t/3 − r − s (b)
1
2
(ln a + 3 ln b − 2 ln c) = r/2 + 3s/2 − t
11. (a) 1 + log x +
1
2
log(x − 3) (b) 2 ln |x| + 3 ln sin x −
1
2
ln(x2
+ 1)
12. (a)
1
3
log(x + 2) − log cos 5x (b)
1
2
ln(x2
+ 1) −
1
2
ln(x3
+ 5)
13. log
24
(16)
3
= log(256/3) 14. log
√
x − log(sin3
2x) + log 100 = log
100
√
x
sin3
2x
15. ln
3
√
x(x + 1)2
cos x
16. 1 + x = 103
= 1000, x = 999
17.
√
x = 10−1
= 0.1, x = 0.01 18. x2
= e4
, x = ±e2
19. 1/x = e−2
, x = e2
20. x = 7
21. 2x = 8, x = 4 22. log10 x3
= 30, x3
= 1030
, x = 1010
23. log10 x = 5, x = 105
24. ln 4x − ln x6
= ln 2, ln
4
x5
= ln 2,
4
x5
= 2, x5
= 2, x = 5
√
2
25. ln 2x2
= ln 3, 2x2
= 3, x2
= 3/2, x = 3/2 (we discard − 3/2 because it does not satisfy the
original equation)
26. ln 3x
= ln 2, x ln 3 = ln 2, x =
ln 2
ln 3

27. ln 5−2x
= ln 3, −2x ln 5 = ln 3, x = −
ln 3
2 ln 5
28. e−2x
= 5/3, −2x = ln(5/3), x = −
1
2
ln(5/3)
29. e3x
= 7/2, 3x = ln(7/2), x =
1
3
ln(7/2)
30. ex
(1 − 2x) = 0 so ex
= 0 (impossible) or 1 − 2x = 0, x = 1/2
31. e−x
(x + 2) = 0 so e−x
= 0 (impossible) or x + 2 = 0, x = −2
32. e2x
− ex
− 6 = (ex
− 3)(ex
+ 2) = 0 so ex
= −2 (impossible) or ex
= 3, x = ln 3
33. e−2x
− 3e−x
+ 2 = (e−x
− 2)(e−x
− 1) = 0 so e−x
= 2, x = − ln 2 or e−x
= 1, x = 0
34. (a) y
x
-2
2
2 6
(b) y
x
2
-2 2
35. (a) y
x
2
4
6
-2 4
(b) y
x
-4
2
-4 2
36. (a) y
x
-10
-1
(b) y
x
-1
3
37. log2 7.35 = (log 7.35)/(log 2) = (ln 7.35)/(ln 2) ≈ 2.8777;
log5 0.6 = (log 0.6)/(log 5) = (ln 0.6)/(ln 5) ≈ −0.3174
38. 10
-5
0 2
39. 2
-3
0 3

270 Chapter 7
40. (a) Let X = logb x and Y = loga x. Then bX
= x and aY
= x so aY
= bX
, or aY/X
= b, which
means loga b = Y/X. Substituting for Y and X yields
loga x
logb x
= loga b, logb x =
loga x
loga b
.
(b) Let x = a to get logb a = (loga a)/(loga b) = 1/(loga b) so (loga b)(logb a) = 1.
(log2 81)(log3 32) = (log2[34
])(log3[25
]) = (4 log2 3)(5 log3 2) = 20(log2 3)(log3 2) = 20
41. (a) x = 3.6541, y = 1.2958
2
0.6
2 6
(b) x ≈ 332105.11, y ≈ 12.7132
12.7134
12.7130
332085 332125
42. Since the units are billions, one trillion is 1,000 units. Solve 1000 = 0.051517(1.1306727)x
for x by
taking common logarithms, resulting in 3 = log 0.051517 + x log 1.1306727, which yields x ≈ 77.4,
so the debt ﬁrst reached one trillion dollars around 1977.
43. (a) no, the curve passes through the origin (b) y = 2x/4
(c) y = 2−x
(d) y = (
√
5)x
5
0
-1 2
44. (a) As x → +∞ the function grows very slowly, but it is always increasing and tends to +∞. As
x → 1+
the function tends to −∞.
(b)
1 2
-5
5
x
y
45. log(1/2) < 0 so 3 log(1/2) < 2 log(1/2)
46. Let x = logb a and y = logb c, so a = bx
and c = by
.
First, ac = bx
by
= bx+y
or equivalently, logb(ac) = x + y = logb a + logb c.
Secondly, a/c = bx
/by
= bx−y
or equivalently, logb(a/c) = x − y = logb a − logb c.
Next, ar
= (bx
)r
= brx
or equivalently, logb ar
= rx = r logb a.
Finally, 1/c = 1/by
= b−y
or equivalently, logb(1/c) = −y = − logb c.
47. 75e−t/125
= 15, t = −125 ln(1/5) = 125 ln 5 ≈ 201 days.

48. (a) a If t = 0, then Q = 12 grams
(b) b Q = 12e−0.055(4)
= 12e−0.22
≈ 9.63 grams
(c) 12e−0.055t
= 6, e−0.055t
= 0.5, t = −(ln 0.5)/(0.055) ≈ 12.6 hours
49. (a) 7.4; basic (b) 4.2; acidic (c) 6.4; acidic (d) 5.9; acidic
50. (a) log[H+
] = −2.44, [H+
] = 10−2.44
≈ 3.6 × 10−3
mol/L
(b) log[H+
] = −8.06, [H+
] = 10−8.06
≈ 8.7 × 10−9
mol/L
51. (a) 140 dB; damage (b) 120 dB; damage
(c) 80 dB; no damage (d) 75 dB; no damage
52. Suppose that I1 = 3I2 and β1 = 10 log10 I1/I0, β2 = 10 log10 I2/I0. Then
I1/I0 = 3I2/I0, log10 I1/I0 = log10 3I2/I0 = log10 3 + log10 I2/I0, β1 = 10 log10 3 + β2,
β1 − β2 = 10 log10 3 ≈ 4.8 decibels.
53. Let IA and IB be the intensities of the automobile and blender, respectively. Then
log10 IA/I0 = 7 and log10 IB/I0 = 9.3, IA = 107
I0 and IB = 109.3
I0, so IB/IA = 102.3
≈ 200.
54. The decibel level of the nth echo is 120(2/3)n
;
120(2/3)n
< 10 if (2/3)n
< 1/12, n <
log(1/12)
log(2/3)
=
log 12
log 1.5
≈ 6.13 so 6 echoes can be heard.
55. (a) log E = 4.4 + 1.5(8.2) = 16.7, E = 1016.7
≈ 5 × 1016
J
(b) Let M1 and M2 be the magnitudes of earthquakes with energies of E and 10E,
respectively. Then 1.5(M2 − M1) = log(10E) − log E = log 10 = 1,
M2 − M1 = 1/1.5 = 2/3 ≈ 0.67.
56. Let E1 and E2 be the energies of earthquakes with magnitudes M and M + 1, respectively. Then
log E2 − log E1 = log(E2/E1) = 1.5, E2/E1 = 101.5
≈ 31.6.
57. If t = −2x, then x = −t/2 and lim
x→0
(1 − 2x)1/x
= lim
t→0
(1 + t)−2/t
= lim
t→0
[(1 + t)1/t
]−2
= e−2
.
58. If t = 3/x, then x = 3/t and lim
x→+∞
(1 + 3/x)x
= lim
t→0+
(1 + t)3/t
= lim
t→0+
[(1 + t)1/t
]3
= e3
.
EXERCISE SET 7.3
1.
1
2x
(2) = 1/x 2.
1
x3
(3x2
) = 3/x
3. 2(ln x)
1
x
=
2 ln x
x
4.
1
sin x
(cos x) = cot x
5.
1
tan x
(sec2
x) =
sec2
x
tan x
6.
1
2 +
√
x
1
2
√
x
=
1
2
√
x(2 +
√
x)
7.
1
x/(1 + x2)
(1 + x2
)(1) − x(2x)
(1 + x2)2
=
1 − x2
x(1 + x2)
8.
1
ln x
1
x
=
1
x ln x
9.
3x2
− 14x
x3 − 7x2 − 3

272 Chapter 7
10. x3 1
x
+ (3x2
) ln x = x2
(1 + 3 ln x) 11.
1
2
(ln x)−1/2 1
x
=
1
2x
√
ln x
12.
1
2 2(ln x)(1/x)
1 + ln2
x
=
ln x
x 1 + ln2
x
13. −
1
x
sin(ln x)
14. 2 sin(ln x) cos(ln x)
1
x
=
sin(2 ln x)
x
=
sin(ln x2
)
x
15. 3x2
log2(3 − 2x) +
−2x3
(ln 2)(3 − 2x)
16. log2(x2
− 2x)
3
+ 3x log2(x2
− 2x)
2 2x − 2
(x2 − 2x) ln 2
17.
2x(1 + log x) − x/(ln 10)
(1 + log x)2
18. 1/[x(ln 10)(1 + log x)2
]
19. 7e7x
20. −10xe−5x2
21. x3
ex
+ 3x2
ex
= x2
ex
(x + 3) 22. −
1
x2
e1/x
23.
dy
dx
=
(ex
+ e−x
)(ex
+ e−x
) − (ex
− e−x
)(ex
− e−x
)
(ex + e−x)2
=
(e2x
+ 2 + e−2x
) − (e2x
− 2 + e−2x
)
(ex + e−x)2
= 4/(ex
+ e−x
)2
24. ex
cos(ex
)
25. (x sec2
x + tan x)ex tan x
26.
dy
dx
=
(ln x)ex
− ex
(1/x)
(ln x)2
=
ex
(x ln x − 1)
x(ln x)2
27. (1 − 3e3x
)e(x−e3x
)
28.
15
2
x2
(1 + 5x3
)−1/2
exp( 1 + 5x3)
29.
(x − 1)e−x
1 − xe−x
=
x − 1
ex − x
30.
1
cos(ex)
[− sin(ex
)]ex
= −ex
tan(ex
)
31.
dy
dx
+
1
xy
x
dy
dx
+ y = 0,
dy
dx
= −
y
x(y + 1)
32.
dy
dx
=
1
x tan y
x sec2
y
dy
dx
+ tan y ,
dy
dx
=
tan y
x(tan y − sec2 y)
33.
d
dx
ln cos x −
1
2
ln(4 − 3x2
) = − tan x +
3x
4 − 3x2
34.
d
dx
1
2
[ln(x − 1) − ln(x + 1)] =
1
2
1
x − 1
−
1
x + 1
35. ln |y| = ln |x| +
1
3
ln |1 + x2
|,
dy
dx
= x
3
1 + x2
1
x
+
2x
3(1 + x2)

36. ln |y| =
1
5
[ln |x − 1| − ln |x + 1|],
dy
dx
=
1
5
5 x − 1
x + 1
1
x − 1
−
1
x + 1
37. ln |y| =
1
3
ln |x2
− 8| +
1
2
ln |x3
+ 1| − ln |x6
− 7x + 5|
dy
dx
=
(x2
− 8)1/3
√
x3 + 1
x6 − 7x + 5
2x
3(x2 − 8)
+
3x2
2(x3 + 1)
−
6x5
− 7
x6 − 7x + 5
38. ln |y| = ln | sin x| + ln | cos x| + 3 ln | tan x| −
1
2
ln |x|
dy
dx
=
sin x cos x tan3
x
√
x
cot x − tan x +
3 sec2
x
tan x
−
1
2x
39. f (x) = 2x
ln 2; y = 2x
, ln y = x ln 2,
1
y
y = ln 2, y = y ln 2 = 2x
ln 2
40. f (x) = −3−x
ln 3; y = 3−x
, ln y = −x ln 3,
1
y
y = − ln 3, y = −y ln 3 = −3−x
ln 3
41. f (x) = πsin x
(ln π) cos x;
y = πsin x
, ln y = (sin x) ln π,
1
y
y = (ln π) cos x, y = πsin x
(ln π) cos x
42. f (x) = πx tan x
(ln π)(x sec2
x + tan x);
y = πx tan x
, ln y = (x tan x) ln π,
1
y
y = (ln π)(x sec2
x + tan x)
y = πx tan x
(ln π)(x sec2
x + tan x)
43. ln y = (ln x) ln(x3
− 2x),
1
y
dy
dx
=
3x2
− 2
x3 − 2x
ln x +
1
x
ln(x3
− 2x),
dy
dx
= (x3
− 2x)ln x 3x2
− 2
x3 − 2x
ln x +
1
x
ln(x3
− 2x)
44. ln y = (sin x) ln x,
1
y
dy
dx
=
sin x
x
+ (cos x) ln x,
dy
dx
= xsin x sin x
x
+ (cos x) ln x
45. ln y = (tan x) ln(ln x),
1
y
dy
dx
=
1
x ln x
tan x + (sec2
x) ln(ln x),
dy
dx
= (ln x)tan x tan x
x ln x
+ (sec2
x) ln(ln x)
46. ln y = (ln x) ln(x2
+ 3),
1
y
dy
dx
=
2x
x2 + 3
ln x +
1
x
ln(x2
+ 3),
dy
dx
= (x2
+ 3)ln x 2x
x2 + 3
ln x +
1
x
ln(x2
+ 3)
47. f (x) = exe−1

274 Chapter 7
48. (a) because xx
is not of the form ax
where a is constant
(b) y = xx
, ln y = x ln x,
1
y
y = 1 + ln x, y = xx
(1 + ln x)
49. (a) logx e =
ln e
ln x
=
1
ln x
,
d
dx
[logx e] = −
1
x(ln x)2
(b) logx 2 =
ln 2
ln x
,
d
dx
[logx 2] = −
ln 2
x(ln x)2
50. (a) ex2
(b) ln x
51. (a) f (x) = kekx
, f (x) = k2
ekx
, f (x) = k3
ekx
, . . . , f(n)
(x) = kn
ekx
(b) f (x) = −ke−kx
, f (x) = k2
e−kx
, f (x) = −k3
e−kx
, . . . , f(n)
(x) = (−1)n
kn
e−kx
52.
dy
dt
= e−λt
(ωA cos ωt − ωB sin ωt) + (−λ)e−λt
(A sin ωt + B cos ωt)
= e−λt
[(ωA − λB) cos ωt − (ωB + λA) sin ωt]
53. f (x) =
1
√
2πσ
exp −
1
2
x − µ
σ
2
d
dx
−
1
2
x − µ
σ
2
=
1
√
2πσ
exp −
1
2
x − µ
σ
2
−
x − µ
σ
1
σ
= −
1
√
2πσ3
(x − µ) exp −
1
2
x − µ
σ
2
54. y = Aekt
, dy/dt = kAekt
= k(Aekt
) = ky
55. y = Ae2x
+ Be−4x
, y = 2Ae2x
− 4Be−4x
, y = 4Ae2x
+ 16Be−4x
so
y + 2y − 8y = (4Ae2x
+ 16Be−4x
) + 2(2Ae2x
− 4Be−4x
) − 8(Ae2x
+ Be−4x
) = 0
56. (a) y = −xe−x
+ e−x
= e−x
(1 − x), xy = xe−x
(1 − x) = y(1 − x)
(b) y = −x2
e−x2
/2
+ e−x2
/2
= e−x2
/2
(1 − x2
), xy = xe−x2
/2
(1 − x2
) = y(1 − x2
)
57. (a) f(w) = ln w; f (1) = lim
h→0
ln(1 + h) − ln 1
h
= lim
h→0
ln(1 + h)
h
=
1
w w=1
= 1
(b) f(w) = 10w
; f (0) = lim
h→0
10h
− 1
h
=
d
dw
(10w
)
w=0
= 10w
ln 10
w=0
= ln 10
58. (a) f(x) = ln x; f (e2
) = lim
∆x→0
ln(e2
+ ∆x) − 2
∆x
=
d
dx
(ln x)
x=e2
=
1
x x=e2
= e−2
(b) f(w) = 2w
; f (1) = lim
w→1
2w
− 2
w − 1
=
d
dw
(2w
)
w=1
= 2w
ln 2
w=1
= 2 ln 2
59. 2 ln x + 3ex
+ C
60.
1
2
t−1
−
√
2et
dt =
1
2
ln t −
√
2et
+ C

61. (a)
1
u
du = ln |u| + C = ln | ln x| + C (b) −
1
5
eu
du = −
1
5
eu
+ C = −
1
5
e−5x
+ C
62. (a) −
1
3
1
u
du = −
1
3
ln |u| + C = −
1
3
ln |(1 + cos 3θ)| + C
(b)
du
u
= ln u + C = ln(1 + ex
) + C
63. u = 2x, du = 2dx;
1
2
eu
du =
1
2
eu
+ C =
1
2
e2x
+ C
64. u = 2x, du = 2dx;
1
2
1
u
du =
1
2
ln |u| + C =
1
2
ln |2x| + C
65. u = sin x, du = cos x dx; eu
du = eu
+ C = esin x
+ C
66. u = x4
, du = 4x3
dx;
1
4
eu
du =
1
4
eu
+ C =
1
4
ex4
+ C
67. u = −2x3
, du = −6x2
, −
1
6
eu
du = −
1
6
eu
+ C = −
1
6
e−2x3
+ C
68. u = ex
− e−x
, du = (ex
+ e−x
)dx,
1
u
du = ln |u| + C = ln ex
− e−x
+ C
69. e−x
dx; u = −x, du = −dx; − eu
du = −eu
+ C = −e−x
+ C
70. ex/2
dx; u = x/2, du = dx/2; 2 eu
du = 2eu
+ C = 2ex/2
+ C = 2
√
ex + C
71. u =
√
y + 1, du =
1
2
√
y + 1
dy, 2 eu
du = 2eu
+ C = 2e
√
y+1
+ C
72. u =
√
y, du =
1
2
√
y
dy, 2
1
eu
du = 2 e−u
du = −2e−u
+ C = −2e−
√
y
+ C
73. 1 +
1
t
dt = t + ln |t| + C
74. e2 ln x
= eln x2
= x2
, x > 0, so e2 ln x
dx = x2
dx =
1
3
x3
+ C
75. ln(ex
) + ln(e−x
) = ln(ex
e−x
) = ln 1 = 0 so [ln(ex
) + ln(e−x
)]dx = C
76.
cos x
sin x
dx; u = sin x, du = cos xdx;
1
u
du = ln |u| + C = ln | sin x| + C
77. 5ex
3
ln 2
= 5e3
− 5(2) = 5e3
− 10 78. (ln x)/2
1
1/2
= (ln 2)/2
79. (a)
1
2
1
−1
eu
du =
1
2
eu
1
−1
=
1
2
e − e−1
(b)
2
1
udu =
3
2

276 Chapter 7
80.
6
−6
36 − u2du = π(6)2
/2 = 18π
81. u = ex
+ 4, du = ex
dx, u = e− ln 3
+ 4 =
1
3
+ 4 =
13
3
when x = − ln 3,
u = eln 3
+ 4 = 3 + 4 = 7 when x = ln 3,
7
13/3
1
u
du = ln u
7
13/3
= ln(7) − ln(13/3) = ln(21/13)
82. u = 3 − 4ex
, du = −4ex
dx, u = −1 when x = 0, u = −17 when x = ln 5
−
1
4
−17
−1
u du = −
1
8
u2
−17
−1
= −36
83. ln(x + e)
e
0
= ln(2e) − ln e = ln 2 84. −
1
2
e−x2
√
2
1
= (e−1
− e−2
)/2
85. −
1
3
e−3x
ln 2
0
= −
1
3
(e−3 ln 2
− e0
) = −
1
3
1
8
− 1 = 7/24
86.
0
−1
(1−ex
)dx+
1
0
(ex
−1)dx = (x−ex
)
0
−1
+(ex
−x)
1
0
= −1−(−1−e−1
)+e−1−1 = e+1/e−2
87. (a) y
x
2
4
6
-1 1 2
(b) f(x) = ex
/2 + 1/2
88. y
x
4
-4 4
89. (a) y(t) = 2e−t
dt = −2e−t
+ C, y(1) = −
2
e
+ C = 3 −
2
e
, C = 3; y(t) = −2e−t
+ 3
(b) y(t) = t−1
dt = ln |t| + C, y(−1) = C = 5, C = 5; y(t) = ln |t| + 5
EXERCISE SET 7.4
0
0
- +- +- +
1. (a) critical point x = 0; f :
x = 0: relative minimum
0
ln 2
- +- +- +(b) critical point x = ln 2: f :
x = ln 2: relative minimum
00
1-1
- +- +- -- -+2. (a) critical points x = −1, 1:f :
x = −1: relative minimum;
x = 1: relative maximum
0
1
- --- --(b) x = 1: neither

0.2
0
1 4
3. f (x) = x2
(2x − 3)e−2x
, f (x) = 0 for x in [1, 4] when x = 3/2;
if x = 1, 3/2, 4, then f(x) = e−2
,
27
8
e−3
, 64e−8
;
critical point at x = 3/2; absolute maximum of
27
8
e−3
at x = 3/2,
absolute minimum of 64e−8
at x = 4
0.76
0.64
1 2.7
4. f (x) = (1 − ln 2x)/x2
, f (x) = 0 on [1, e] for x = e/2;
if x = 1, e/2, e then f(x) = ln 2, 2/e, (ln 2 + 1)/e;
absolute minimum of
1 + ln 2
e
at x = e,
absolute maximum of 2/e at x = e/2
5. (a) lim
x→+∞
xex
= +∞, lim
x→−∞
xex
= 0
-1
-3-5
1
x
y
(-2, -0.27)
(-1, -0.37)
(b) y = xex
;
y = (x + 1)ex
;
y = (x + 2)ex
6. (a) lim
x→+∞
xe−2x
= 0, lim
x→−∞
xe−2x
= −∞
-3 31
-0.3
0.3
0.1 x
y
(0.5, 0.18)
(1, 0.14)
(b) y = xe−2x
; y = −2 x −
1
2
e−2x
; y = 4(x − 1)e−2x
7. (a) lim
x→+∞
x2
e2x
= 0, lim
x→−∞
x2
e2x
= +∞
1 2 3
0.3
x
y
(0, 0)
(0.29, 0.05)
(1, 0.14)
(1.71, 0.10)
(b) y = x2
/e2x
= x2
e−2x
;
y = 2x(1 − x)e−2x
;
y = 2(2x2
− 4x + 1)e−2x
;
y = 0 if 2x2
− 4x + 1 = 0, when
x =
4 ±
√
16 − 8
4
= 1 ±
√
2/2 ≈ 0.29, 1.71
8. (a) lim
x→+∞
x2
e2x
= +∞, lim
x→−∞
x2
e2x
= 0.
-3 -2 -1
y
x
0.2
0.3
(-1.71, 0.10)
(-1, 0.14)
(0, 0)
(-0.29, 0.05)
(b) y = x2
e2x
;
y = 2x(x + 1)e2x
;
y = 2(2x2
+ 4x + 1)e2x
;
y = 0 if 2x2
+ 4x + 1 = 0, when
x =
−4 ±
√
16 − 8
4
= −1 ±
√
2/2 ≈ −0.29, −1.71

278 Chapter 7
9. (a) lim
x→+∞
f(x) = +∞, lim
x→−∞
f(x) = −∞ y
x
-100
100
-2 2
(0,0)
(b) y = xex2
;
y = (1 + 2x2
)ex2
;
y = 2x(3 + 2x2
)ex2
no relative extrema, inflection point at (0, 0)
10. (a) lim
x→±∞
f(x) = 1
y
x
0.4
0.8
1
-10 -5 5 10
(0, 0)
(-√2/3, e-3/2) (√2/3, e-3/2)
(b) f (x) = 2x−3
e−1/x2
so f (x) < 0 for x < 0 and f (x) > 0
for x > 0. Set u = x2
and use the given result to find
lim
x→0
f (x) = 0, so (by the First Derivative Test) f(x)
has a minimum at x = 0. f (x) = (−6x−4
+ 4x−6
)e−1/x2
,
so f(x) has points of inflection at x = ± 2/3.
x
y
–4
6
(1, e)
–2 1 3
11. lim
x→+∞
f(x) = +∞, lim
x→−∞
f(x) = 0
f (x) = ex x − 1
x2
, f (x) = ex x2
− 2x + 2
x3
critical point at x = 1;
relative minimum at x = 1
no points of inflection
vertical asymptote x = 0,
horizontal asymptote y = 0 for x → −∞
y
x
-0.8
0.2
1 2
(1, )1
e
(2, )2
e2
12. lim
x→+∞
f(x) = 0, lim
x→−∞
f(x) = −∞
f (x) = (1 − x)e−x
, f (x) = (x − 2)e−x
critical point at x = 1; relative maximum at x = 1
point of inflection at x = 2
horizontal asymptote y = 0 as x → +∞
y
x
0.6
1
1.8
1 2 3 4
(2, )
(3.41, 1.04)
(0.59, 0.52)
(0, 0)
4
e
13. lim
x→+∞
f(x) = 0, lim
x→−∞
f(x) = +∞
f (x) = x(2 − x)e1−x
, f (x) = (x2
− 4x + 2)e1−x
critical points at x = 0, 2;
relative minimum at x = 0,
relative maximum at x = 2
points of inflection at x = 2 ±
√
2 ≈ 0.59, 3.41
horizontal asymptote y = 0 as x → +∞

y
x
-0.4
0.4
0.8
-4 -2
(–4.7, –0.35)
(0, 0)
(–1.27, –0.21)
(–3, –0.49)
1
14. lim
x→+∞
f(x) = +∞, lim
x→−∞
f(x) = 0
f (x) = x2
(3 + x)ex−1
, f (x) = x(x2
+ 6x + 6)ex−1
critical points at x = −3, 0;
relative minimum at x = −3
points of inﬂection at x = 0, −3 ±
√
3 ≈ 0, −4.7, −1.27
horizontal asymptote y = 0 as x → −∞
15. (a) lim
x→0+
y = lim
x→0+
x ln x = lim
x→0+
ln x
1/x
= lim
x→0+
1/x
−1/x2
= 0;
lim
x→+∞
y = +∞
1
x
y
(e-1, -e-1)
(b) y = x ln x,
y = 1 + ln x,
y = 1/x,
y = 0 when x = e−1
1
-0.2
-0.1
0.1
0.2
x
y
(e-1/2, - e-1)1
2
(e-3/2, - e-3)3
2
16. (a) lim
x→0+
y = lim
x→0+
ln x
1/x2
= lim
x→0+
1/x
−2/x3
= 0,
lim
x→+∞
y = +∞
(b) y = x2
ln x, y = x(1 + 2 ln x),
y = 3 + 2 ln x,
y = 0 if x = e−1/2
,
y = 0 if x = e−3/2
,
lim
x→0+
y = 0
1 2 3
-0.4
-0.3
-0.2
-0.1
0.1
0.2
0.3
0.4
x
y
(e1/2, e-1)1
2
(e5/6, e-5/3)5
6
17. (a) lim
x→0+
y = lim
x→0+
ln x
x2
= −∞;
lim
x→+∞
y = lim
x→+∞
ln x
x2
= lim
x→+∞
1/x
2x
= 0
(b) y =
ln x
x2
, y =
1 − 2 ln x
x3
,
y =
6 ln x − 5
x4
,
y = 0 if x = e1/2
,
y = 0 if x = e5/6
18. (a) Let u = 1/x, lim
x→0+
(ln x)/
√
x = lim
u→+∞
−
√
u ln u = −∞ by inspection,
lim
x→+∞
(ln x)/
√
x = 0, by the rule given.
y
x
-1
0.5
2 6 10 14
(e2, 2/e)
(e8/3, e-4/3)8
3
(b) y =
ln x
√
x
, y =
2 − ln x
2x3/2
y =
−8 + 3 ln x
4x5/2
y = 0 if x = e2
,
y = 0 if x = e8/3

280 Chapter 7
19. (a) lim
x→0+
x2
ln x = 0 by the rule given, lim
x→+∞
x2
ln x = +∞ by inspection, and f(x) not defined
for x < 0
y
x
3
8e3
1
2e3/2( ), –
( )1
8e
1
2
, –1
2√e
(b) y = x2
ln 2x, y = 2x ln 2x + x
y = 2 ln 2x + 3
y = 0 if x = 1/(2
√
e),
y = 0 if x = 1/(2e3/2
)
y
x
1
2
–2 2(0, 0)
(1, ln 2)(–1, ln 2)
20. (a) lim
x→+∞
f(x) = +∞; lim
x→0
f(x) = 0
(b) y = ln(x2
+ 1), y = 2x/(x2
+ 1)
y = −2
x2
− 1
(x2 + 1)2
y = 0 if x = 0
y = 0 if x = ±1
21. (a) 0.4
-0.2
-0.5 3
(b) y = (1 − bx)e−bx
, y = b2
(x − 2/b)e−bx
;
relative maximum at x = 1/b, y = 1/be;
point of inflection at x = 2/b, y = 2/be2
.
Increasing b moves the relative maximum
and the point of inflection to the left and
down, i.e. towards the origin.
22. (a) 1
0
-2 2
(b) y = −2bxe−bx2
,
y = 2b(−1 + 2bx2
)e−bx2
;
relative maximum at x = 0, y = 1; points
of inflection at x = ± 1/2b, y = 1/
√
e.
Increasing b moves the points of inflection
towards the y-axis; the relative maximum
doesn’t move.
23. (a) The oscillations of ex
cos x about zero in-
crease as x → +∞ so the limit does not
exist, and lim
x→−∞
ex
cos x = 0.
(b) y
x
-2
4
6
-2 -1 1 2
(0,1)
(1.52, 0.22)

(c) The curve y = eax
cos bx oscillates between y = eax
and y = −eax
. The frequency of
oscillation increases when b increases.
y
x
-5
5
-1 2
a = 1
b = 1
b = 2
b = 3
y
x
5
10
-1 0.5 1
b = 1
a = 1
a = 2
a = 3
24. Let P(x0, y0) be a point on y = e3x
then y0 = e3x0
. dy/dx = 3e3x
so mtan = 3e3x0
at P and an
equation of the tangent line at P is y − y0 = 3e3x0
(x − x0), y − e3x0
= 3e3x0
(x − x0). If the line
passes through the origin then (0, 0) must satisfy the equation so −e3x0
= −3x0e3x0
which gives
x0 = 1/3 and thus y0 = e. The point is (1/3, e).
25. (b) y
x
2
4
6
1 2 3 4
(c)
dy
dx
=
1
2
−
1
x
so
dy
dx
< 0 at x = 1 and
dy
dx
> 0 at x = e
(d) The slope is a continuous function which goes from a negative value to a positive value;
therefore it must take the value zero in between, by the Intermediate Value Theorem.
(e)
dy
dx
= 0 when x = 2
26. (a)
dC
dt
=
K
a − b
ae−at
− be−bt
so
dC
dt
= 0 at t =
ln(a/b)
a − b
. This is the only stationary point and
C(0) = 0, lim
t→+∞
C(t) = 0, C(t) > 0 for 0 < t < +∞, so it is an absolute maximum.
(b) 0.7
0
0 10
27. (a) 100
20
0 8

282 Chapter 7
(b) as t tends to +∞, the population tends to 19
lim
t→+∞
P(t) = lim
t→+∞
95
5 − 4e−t/4
=
95
5 − 4 lim
t→+∞
e−t/4
=
95
5
= 19
(c) the rate of population growth tends to zero
0
-80
0 8
28. (a) 12
0
0 9
(b) P tends to 12 as t gets large; lim
t→+∞
P(t) = lim
t→+∞
60
5 + 7e−t
=
60
5 + 7 lim
t→+∞
e−t =
60
5
= 12
(c) the rate of population growth tends to zero
3.2
0
0 9
29. t = 7.67 1000
0
0 15
30. (a)
dN
dt
= 250(20 − t)e−t/20
= 0 at t = 20, N(0) = 125,000, N(20) ≈ 161,788, and
N(100) ≈ 128,369; the absolute maximum is N = 161,788 at t = 20, the absolute
minimum is N = 125,000 at t = 0.
(b) The absolute minimum of
dN
dt
occurs when
d2
N
dt2
= 12.5(t − 40)e−t/20
= 0, t = 40.

31. (a) y (t) =
LAke−kt
(1 + Ae−kt)2
S, so y (0) =
LAk
(1 + A)2
(b) The rate of growth increases to its maximum, which occurs when y is halfway between 0 and
L, or when t =
1
k
ln A; it then decreases back towards zero.
(c) From (2) one sees that
dy
dt
is maximized when y lies half way between 0 and L, i.e. y = L/2.
This follows since the right side of (2) is a parabola (with y as independent variable) with
y-intercepts y = 0, L. The value y = L/2 corresponds to t =
1
k
ln A, from (4).
32. Since 0 < y < L the right-hand side of (3) can change sign only if the factor L − 2y changes sign,
which it does when y = L/2. From (1) we have
L
2
=
L
1 + Ae−kt
, 1 = Ae−kt
, t =
1
k
ln A.
33.
dk
dT
= k0 exp −
q(T − T0)
2T0T
−
q
2T2
= −
qk0
2T2
exp −
q(T − T0)
2T0T
34. β = 10 log I − 10 log I0,
dβ
dI
=
10
I ln 10
(a)
dβ
dI I=10I0
=
1
I0 ln 10
db/W/m2
(b)
dβ
dI I=100I0
=
1
10I0 ln 10
db/W/m2
(c)
dβ
dI I=100I0
=
1
100I0 ln 10
db/W/m2
35. Solve
dy
dt
= 3
dx
dt
given y = x ln x. Then
dy
dt
=
dy
dx
dx
dt
= (1 + ln x)
dx
dt
, so 1 + ln x = 3, ln x = 2,
x = e2
.
36. v(t) = (1 − t)e−t
, a(t) = (t − 2)e−t
0.4
0
0 10
s(t)
1
-0.2
0 8
v(t)
0.1
-2
0 6
a(t)
(a) v = 0 at t = 1 (b) s = 1/e at t = 1
(c) a changes sign at t = 2, so the particle is speeding up for 1 < t < 2 and slowing down for
0 < t < 1 and 2 < t
37.
3
1
ex
dx = ex
3
1
= e3
− e 38.
5
1
1
x
dx = ln x
5
1
= ln 5 − ln 1 = ln 5

284 Chapter 7
39. A =
ln 2
0
e2x
− ex
dx
=
1
2
e2x
− ex
ln 2
0
= 1/2
2
4
x
y
ln 2
y = e2x
y = ex
40. A =
e
1
dy
y
= ln y
e
1
= 1
1/e 1
1
e
x
y
41. A = A1 + A2 =
0
−1
(1 − ex
)dx +
1
0
(ex
− 1)dx = 1/e + e − 2
42. A = A1 + A2 =
1
1/2
1 − x
x
dx +
2
1
x − 1
x
dx = −
1
2
− ln 2 + (1 − ln 2) = 1/2
43. fave =
1
e − 1
e
1
1
x
dx =
1
e − 1
(ln e − ln 1) =
1
e − 1
44. fave =
1
ln 5 − (−1)
ln 5
−1
ex
dx =
1
ln 5 + 1
(5 − e−1
) =
5 − e−1
1 + ln 5
45. fave =
1
4 − 0
4
0
e−2x
dx = −
1
8
e−2x
4
0
=
1 − e−8
8
46.
k
0
e2x
dx = 3,
1
2
e2x
k
0
= 3,
1
2
(e2k
− 1) = 3, e2k
= 7, k =
1
2
ln 7
47. y(t) = (802.137) e1.528t
dt = 524.959e1.528t
+ C; y(0) = 750 = 524.959 + C, C = 225.041,
y(t) = 524.959e1.528t
+ 225.041, y(12) = 48, 233, 525, 650
48. Vave =
275000
10 − 0
10
0
e−0.17t
dt = −161764.7059e−0.17t
10
0
= $132, 212.96
49. s(t) = (25 + 10e−0.05t
)dt = 25t − 200e−0.05t
+ C
(a) s(10) − s(0) = 250 − 200(e−0.5
− 1) = 450 − 200/
√
e ≈ 328.69 ft
(b) yes; without it the distance would have been 250 ft
50. (a) displacement =
3
0
(et
− 2)dt = e3
− 7
distance =
3
0
|v(t)|dt = −
ln 2
0
v(t)dt +
3
ln 2
v(t)dt = e3
− 9 + 4 ln 2

(b) displacement =
3
1
1
2
−
1
t
dt = 1 − ln 3
distance =
3
1
|v(t)|dt = −
2
1
v(t)dt +
3
2
v(t)dt = 2 ln 2 − ln 3
51. (a) From the graph the velocity is positive, so the displacement
is always increasing and is therefore positive.
2
0.2
0.4
4
x
v
(b) s(t) = t/2 + (t + 1)e−t
v
t
-0.2
-0.1
0.1
0.2 0.6 1
52. (a) If t0 < 1 then the area between the velocity curve and the
t-axis, between t = 0 and t = t0, will always be negative,
so the displacement will be negative.
(b) s(t) =
t2
2
−
1
200
ln(t + 0.1) −
t2
4
+
t
20
−
1
200
ln 10
25
-5
0 c
53. x = 0; also set
f(x) = 1 − ex
cos x, f (x) = ex
(sin x − cos x),
xn+1 = xn −
1 − ex
cos x
ex(sin x − cos x)
x1 = 1, x2 = 1.572512605,
x3 = 1.363631415, x7 = x8 = 1.292695719
1
-4
0 2
54. The graphs of y = e−x
and y = ln x intersect near x = 1.3;
let f(x) = e−x
− ln x, f (x) = −e−x
− 1/x, x1 = 1.3,
xn+1 = xn +
e−xn
− ln xn
e−xn + 1/xn
, x2 = 1.309759929,
x4 = x5 = 1.309799586
55. A graphing utility shows that there are two inﬂection points at x ≈ −0.25, 1.25. These points
are the zeros of f (x) = (x4
− 4x3
+ 8x2
− 4x − 1)
ex
(x2 + 1)3
. It is equivalent to ﬁnd the zeros of
g(x) = x4
−4x3
+8x2
−4x−1. One root is x = 1 by inspection. Since g (x) = 4x3
−12x2
+16x−4,

286 Chapter 7
Newton’s Method becomes
xn = xn−1 −
x4
n−1 − 4x3
n−1 + 8x2
n−1 − 4xn−1 − 1
4x3
n−1 − 12x2
n−1 + 16xn−1 − 4
With x0 = −0.25, x1 = −0.18572695, x2 = −0.179563312, x3 = −0.179509029,
x4 = x5 = −0.179509025. So the points of inﬂection are at x ≈ −0.18, x = 1.
56. (a) Let h(x) = ex
− 1 − x for x ≥ 0. Then h(0) = 0 and h (x) = ex
− 1 ≥ 0 for x ≥ 0, so h(x) is
increasing.
(b) Let h(x) = ex
−1−x− 1
2 x2
. Then h(0) = 0 and h (x) = ex
−1−x. By Part (a), ex
−1−x ≥ 0
for x ≥ 0, so h(x) is increasing.
(c) 6
0
0 2
6
0
0 2
57. V = π
ln 3
0
e2x
dx =
π
2
e2x
ln 3
0
= 4π 58. V = π
1
0
e−4x
dx =
π
4
(1 − e−4
)
1
-1
1
x
y
59. V = 2π
1
0
x
x2 + 1
dx
= π ln(x2
+ 1)
1
0
= π ln 2
-1 1
1
x
y
y = 1
x2 + 1
60. V =
√
3
1
2πxex2
dx = πex2
√
3
1
= π(e3
− e)
-√3 -1 1 √3
10
20
x
y
y = ex2
61. (dx/dt)2
+ (dy/dt)2
= [et
(cos t − sin t)]2
+ [et
(cos t + sin t)]2
= 2e2t
,
L =
π/2
0
√
2et
dt =
√
2(eπ/2
− 1)

62. (dx/dt)2
+ (dy/dt)2
= (2et
cos t)2
+ (−2et
sin t)2
= 4e2t
, L =
4
1
2et
dt = 2(e4
− e)
63. dy/dx =
sec x tan x
sec x
= tan x, 1 + (y )2 =
√
1 + tan2
x = sec x when 0 < x < π/4, so
L =
π/4
0
sec x dx = ln(1 +
√
2)
64. dy/dx =
cos x
sin x
= cot x, 1 + (y )2 =
√
1 + cot2
x = csc x when π/4 < x < π/2, so
L =
π/2
π/4
csc x dx = − ln(
√
2 − 1) = − ln
√
2 − 1
√
2 + 1
(
√
2 + 1) = ln(1 +
√
2)
65. f (x) = ex
, 1 + [f (x)]2
= 1 + e2x
, S =
1
0
2πex
1 + e2x dx ≈ 22.94
66. x = g(y) = ln y, g (y) = 1/y, 1 + [g (y)]2
= 1 + 1/y2
; S =
e
1
2π 1 + 1/y2 ln y dy ≈ 7.05
67. x = et
(cos t − sin t), y = et
(cos t + sin t), (x )2
+ (y )2
= 2e2t
S = 2π
π/2
0
(et
sin t)
√
2e2tdt = 2
√
2π
π/2
0
e2t
sin t dt
= 2
√
2π
1
5
e2t
(2 sin t − cos t)
π/2
0
=
2
√
2
5
π(2eπ
+ 1)
EXERCISE SET 7.5
1. (a) y
t
1
2
3
1 2 3
(b) y
t
1
2
3
0.5 1
(c) y
t
1
2
3
1 e2
2. y
t
1
2
3
12
3
3
2
3. (a) ln t
ac
1
= ln(ac) = ln a + ln c = 7 (b) ln t
1/c
1
= ln(1/c) = −5
(c) ln t
a/c
1
= ln(a/c) = 2 − 5 = −3 (d) ln t
a3
1
= ln a3
= 3 ln a = 6

288 Chapter 7
4. (a) ln t
√
a
1
= ln a1/2
=
1
2
ln a = 9/2 (b) ln t
2a
1
= ln 2 + 9
(c) ln t
2/a
1
= ln 2 − 9 (d) ln t
a
2
= 9 − ln 2
5. ln 5 ≈ 1.603210678; ln 5 = 1.609437912; magnitude of error is < 0.0063
6. ln 3 ≈ 1.098242635; ln 3 = 1.098612289; magnitude of error is < 0.0004
7. (a) x−1
, x > 0 (b) x2
, x = 0
(c) −x2
, −∞ < x < +∞ (d) −x, −∞ < x < +∞
(e) x3
, x > 0 (f) ln x + x, x > 0
(g) x − 3
√
x, −∞ < x < +∞ (h)
ex
x
, x > 0
8. (a) f(ln 3) = e−2 ln 3
= eln(1/9)
= 1/9
(b) f(ln 2) = eln 2
+ 3e− ln 2
= 2 + 3eln(1/2)
= 2 + 3/2 = 7/2
9. (a) 3π
= eπ ln 3
(b) 2
√
2
= e
√
2 ln 2
10. (a) π−x
= e−x ln π
(b) x2x
= e2x ln x
11. (a) lim
x→+∞
1 +
1
x
x 2
= lim
x→+∞
1 +
1
x
x 2
= e2
(b) y = 2x, lim
y→0
(1 + y)
2/y
= lim
y→0
(1 + y)
1/y
2
= e2
12. (a) y = 3x, lim
y→+∞
1 +
1
y
y/3
= lim
y→+∞
1 +
1
y
y 1/3
= lim
y→+∞
1 +
1
y
y 1/3
= e1/3
(b) lim
x→0
(1 + x)
1/3x
= lim
x→0
(1 + x)
1/x
1/3
= e1/3
13. g (x) = x2
− x 14. g (x) = 1 − cos x
15. (a)
1
x3
(3x2
) =
3
x
(b) eln x 1
x
= 1
16. (a) 2x
√
x2 + 1 (b) −
1
x2
sin
1
x
17. F (x) =
cos x
x2 + 3
, F (x) =
−(x2
+ 3) sin x − 2x cos x
(x2 + 3)2
(a) 0 (b) 1/3 (c) 0
18. F (x) =
√
3x2 + 1, F (x) =
3x
√
3x2 + 1
(a) 0 (b)
√
13 (c) 6/
√
13

19. (a)
d
dx
x2
1
t
√
1 + tdt = x2
1 + x2(2x) = 2x3
1 + x2
(b)
x2
1
t
√
1 + tdt = −
2
3
(x2
+ 1)3/2
+
2
5
(x2
+ 1)5/2
−
4
√
2
15
20. (a)
d
dx
a
x
f(t)dt = −
d
dx
x
a
f(t)dt = −f(x)
(b)
d
dx
a
g(x)
f(t)dt = −
d
dx
g(x)
a
f(t)dt = −f(g(x))g (x)
21. (a) − sin x2
(b) −
tan2
x
1 + tan2
x
sec2
x = − tan2
x
22. (a) −(x2
+ 1)40
(b) − cos3 1
x
−
1
x2
=
cos3
(1/x)
x2
23. −3
3x − 1
9x2 + 1
+ 2x
x2
− 1
x4 + 1
24. If f is continuous on an open interval I and g(x), h(x), and a are in I then
g(x)
h(x)
f(t)dt =
a
h(x)
f(t)dt +
g(x)
a
f(t)dt = −
h(x)
a
f(t)dt +
g(x)
a
f(t)dt
so
d
dx
g(x)
h(x)
f(t)dt = −f(h(x))h (x) + f(g(x))g (x)
25. (a) sin2
(x3
)(3x2
) − sin2
(x2
)(2x) = 3x2
sin2
(x3
) − 2x sin2
(x2
)
(b)
1
1 + x
(1) −
1
1 − x
(−1) =
2
1 − x2
26. F (x) =
1
3x
(3)−
1
x
(1) = 0 so F(x) is constant on (0, +∞). F(1) = ln 3 so F(x) = ln 3 for all x > 0.
27. from geometry,
3
0
f(t)dt = 0,
5
3
f(t)dt = 6,
7
5
f(t)dt = 0; and
10
7
f(t)dt
=
10
7
(4t − 37)/3dt = −3
(a) F(0) = 0, F(3) = 0, F(5) = 6, F(7) = 6, F(10) = 3
(b) F is increasing where F = f is positive, so on [3/2, 6] and [37/4, 10], decreasing on [0, 3/2]
and [6, 37/4]
(c) critical points when F (x) = f(x) = 0, so x = 3/2, 6, 37/4; maximum 15/2 at x = 6, minimum
−9/4 at x = 3/2
(d) F(x)
x
-2
2
4
6
2 4 6 8 10

290 Chapter 7
28. fave =
1
10 − 0
10
0
f(t)dt =
1
10
F(10) = 0.3
29. x < 0 : F(x) =
x
−1
(−t)dt = −
1
2
t2
x
−1
=
1
2
(1 − x2
),
x ≥ 0 : F(x) =
0
−1
(−t)dt +
x
0
t dt =
1
2
+
1
2
x2
; F(x) =
(1 − x2
)/2, x < 0
(1 + x2
)/2, x ≥ 0
30. 0 ≤ x ≤ 2 : F(x) =
x
0
t dt =
1
2
x2
,
x > 2 : F(x) =
2
0
t dt +
x
2
2 dt = 2 + 2(x − 2) = 2x − 2; F(x) =
x2
/2, 0 ≤ x ≤ 2
2x − 2, x > 2
31. y(x) = 2 +
x
1
t1/3
dt = 2 +
3
4
t4/3
x
1
=
5
4
+
3
4
x4/3
32. y(x) =
x
1
(t1/2
+ t−1/2
)dt =
2
3
x3/2
−
2
3
+ 2x1/2
− 2 =
2
3
x3/2
+ 2x1/2
−
8
3
33. y(x) = 1 +
x
π/4
(sec2
t − sin t)dt = tan x + cos x −
√
2/2
34. y(x) =
x
0
tet2
dt =
1
2
e−x2
−
1
2
35. P(x) = P0 +
x
0
r(t)dt individuals
36. s(T) = s1 +
T
1
v(t)dt
37. II has a minimum at x = 12, and I has a zero there, so I could be the derivative of II; on the other
hand I has a minimum near x = 1/3, but II is not zero there, so II could not be the derivative of
I, so I is the graph of f(x) and II is the graph of
x
0
f(t) dt.
38. (b) lim
k→0
1
k
(xk
− 1) =
d
dt
xt
t=0
= ln x
39. (a) where f(t) = 0; by the First Derivative Test, at t = 3
(b) where f(t) = 0; by the First Derivative Test, at t = 1, 5
(c) at t = 0, 1 or 5; from the graph it is evident that it is at t = 5
(d) at t = 0, 3 or 5; from the graph it is evident that it is at t = 3
(e) F is concave up when F = f is positive, i.e. where f is increasing, so on (0, 1/2) and (2, 4);
it is concave down on (1/2, 2) and (4, 5)
(f) F(x)
x
-1
-0.5
0.5
1
1 2 3 5

40. (a)
x
-1
1
-4 -2 2 4
erf(x)
(c) erf (x) > 0 for all x, so there are no relative extrema
(e) erf (x) = −4xe−x2
/
√
π changes sign only at x = 0 so that is the only point of inflection
(g) lim
x→+∞
erf(x) = +1, lim
x→−∞
erf(x) = −1
41. C (x) = cos(πx2
/2), C (x) = −πx sin(πx2
/2)
(a) cos t goes from negative to positive at 2kπ − π/2, and from positive to negative at
t = 2kπ + π/2, so C(x) has relative minima when πx2
/2 = 2kπ − π/2, x = ±
√
4k − 1,
k = 1, 2, . . ., and C(x) has relative maxima when πx2
/2 = (4k + 1)π/2, x = ±
√
4k + 1,
k = 0, 1, . . ..
(b) sin t changes sign at t = kπ, so C(x) has inflection points at πx2
/2 = kπ, x = ±
√
2k,
k = 1, 2, . . .; the case k = 0 is distinct due to the factor of x in C (x), but x changes sign at
x = 0 and sin(πx2
/2) does not, so there is also a point of inflection at x = 0
42. Let F(x) =
x
1
ln tdt, F (x) = lim
h→0
F(x + h) − F(x)
h
= lim
h→0
1
h
x+h
x
ln tdt; but F (x) = ln x so
lim
h→0
1
h
x+h
x
ln tdt = ln x
43. Differentiate: f(x) = 3e3x
, so 2 +
x
a
f(t)dt = 2 +
x
a
3e3t
dt = 2 + e3t
x
a
= 2 + e3x
− e3a
= e3x
provided e3a
= 2, a = (ln 2)/3.
44. (a) The area under 1/t for x ≤ t ≤ x + 1 is less than the area of the rectangle with altitude 1/x
and base 1, but greater than the area of the rectangle with altitude 1/(x + 1) and base 1.
(b)
x+1
x
1
t
dt = ln t
x+1
x
= ln(x + 1) − ln x = ln(1 + 1/x), so
1/(x + 1) < ln(1 + 1/x) < 1/x for x > 0.
(c) from Part (b), e1/(x+1)
< eln(1+1/x)
< e1/x
, e1/(x+1)
< 1 + 1/x < e1/x
,
ex/(x+1)
< (1 + 1/x)x
< e; by the Squeezing Theorem, lim
x→+∞
(1 + 1/x)x
= e.
(d) Use the inequality ex/(x+1)
< (1 + 1/x)x
to get e < (1 + 1/x)x+1
so
(1 + 1/x)x
< e < (1 + 1/x)x+1
.
45. From Exercise 44(d) e − 1 +
1
50
50
< y(50),
and from the graph y(50) < 0.06
0.2
0
0 100

292 Chapter 7
46. F (x) = f(x), thus F (x) has a value at each x in I because f is continuous on I so F is continuous
on I because a function that is diﬀerentiable at a point is also continuous at that point
EXERCISE SET 7.6
1. (a) −π/2 (b) π (c) −π/4 (d) 0
2. (a) π/3 (b) π/3 (c) π/4 (d) 2π/3
3. θ = −π/3; cos θ = 1/2, tan θ = −
√
3, cot θ = −1/
√
3, sec θ = 2, csc θ = −2/
√
3
4. θ = π/3; sin θ =
√
3/2, tan θ =
√
3, cot θ = 1/
√
3, sec θ = 2, csc θ = 2/
√
3
␪
3
45
5. tan θ = 4/3, 0 < θ < π/2; use the triangle shown to
get sin θ = 4/5, cos θ = 3/5, cot θ = 3/4, sec θ = 5/3,
csc θ = 5/4
␪
2.6
1
2.4
6. sec θ = 2.6, 0 < θ < π/2; use the triangle shown to get
sin θ = 2.4/2.6 = 12/13, cos θ = 1/2.6 = 5/13,
tan θ = 2.4 = 12/5, cot θ = 5/12, csc θ = 13/12
7. (a) π/7
(b) sin−1
(sin π) = sin−1
(sin 0) = 0
(c) sin−1
(sin(5π/7)) = sin−1
(sin(2π/7)) = 2π/7
(d) Note that π/2 < 630 − 200π < π so
sin(630) = sin(630 − 200π) = sin(π − (630 − 200π)) = sin(201π − 630) where
0 < 201π − 630 < π/2; sin−1
(sin 630) = sin−1
(sin(201π − 630)) = 201π − 630.
8. (a) π/7
(b) π
(c) cos−1
(cos(12π/7)) = cos−1
(cos(2π/7)) = 2π/7
(d) Note that −π/2 < 200 − 64π < 0 so cos(200) = cos(200 − 64π) = cos(64π − 200) where
0 < 64π − 200 < π/2; cos−1
(cos 200) = cos−1
(cos(64π − 200)) = 64π − 200.
9. (a) 0 ≤ x ≤ π (b) −1 ≤ x ≤ 1
(c) −π/2 < x < π/2 (d) −∞ < x < +∞
␪
-3
4
√7
10. Let θ = sin−1
(−3/4) then sin θ = −3/4, −π/2 < θ < 0 and
(see ﬁgure) sec θ = 4/
√
7

11. Let θ = cos−1
(3/5), sin 2θ = 2 sin θ cos θ = 2(4/5)(3/5) = 24/25
␪
3
45
12. (a) sin(cos−1
x) =
√
1 − x2
1
x
√1 - x2
cos-1 x
(b) tan(cos−1
x) =
√
1 − x2
x
1
x
cos-1 x
√1 - x2
(a) csc(tan−1
x) =
√
1 + x2
x
1
x
tan-1 x
√1 + x2
(d) sin(tan−1
x) =
x
√
1 + x2
1
x
tan-1 x
1 + x2
13. (a) cos(tan−1
x) =
1
√
1 + x2
1
x
tan-1 x
1 + x2
(b) tan(cos−1
x) =
√
1 − x2
x
x
1
cos-1 x
1 – x2
(c) sin(sec−1
x) =
√
x2 − 1
x
1
x
sec-1 x
x2 - 1
(d) cot(sec−1
x) =
1
√
x2 − 1
1
x
x2 – 1
sec-1 x

294 Chapter 7
14. (a) x −1.00 −0.80 −0.6 −0.40 −0.20 0.00 0.20 0.40 0.60 0.80 1.00
sin−1
x −1.57 −0.93 −0.64 −0.41 −0.20 0.00 0.20 0.41 0.64 0.93 1.57
cos−1
x 3.14 2.50 2.21 1.98 1.77 1.57 1.37 1.16 0.93 0.64 0.00
(b) y
x
1
1
(c) y
x
-1
1
2
3
0.5 1
15. (a) y
x
-10 10
c/2
c
y
x
c/2
5
(b) The domain of cot−1
x is (−∞, +∞), the range is (0, π); the domain of csc−1
x is
(−∞, −1] ∪ [1, +∞), the range is [−π/2, 0) ∪ (0, π/2].
16. (a) y = cot−1
x; if x > 0 then 0 < y < π/2 and x = cot y, tan y = 1/x, y = tan−1
(1/x);
if x < 0 then π/2 < y < π and x = cot y = cot(y − π), tan(y − π) = 1/x, y = π + tan−1 1
x
(b) y = sec−1
x, x = sec y, cos y = 1/x, y = cos−1
(1/x)
(c) y = csc−1
x, x = csc y, sin y = 1/x, y = sin−1
(1/x)
17. (a) 55.0◦
(b) 33.6◦
(c) 25.8◦
18. (a) Let x = f(y) = cot y, 0 < y < π, −∞ < x < +∞. Then f is diﬀerentiable and one-to-one
and f (f−1
(x)) = cot(cot−1
x) cos(cot−1
x) = −x
√
x2 + 1
x
= − x2 + 1 = 0, and
d
dx
[cot−1
x]
x=0
= lim
x→0
1
f (f−1(x))
= − lim
x→0
x2 + 1 = −1.
(b) If x = 0 then, from Exercise 16(a),
d
dx
cot−1
x =
d
dx
tan−1 1
x
= −
1
x2
1
1 + (1/x)2
= −
1
√
x2 + 1
. For x = 0, Part (a) shows the
same; thus for −∞ < x < +∞,
d
dx
[cot−1
x] = −
1
√
x2 + 1
.
(c) For −∞ < u < +∞, by the chain rule it follows that
d
dx
[cot−1
u] = −
1
√
u2 + 1
du
dx
.

19. (a) By the chain rule,
d
dx
[csc−1
x] = −
1
x2
1
1 − (1/x)2
=
−1
|x|
√
x2 − 1
(b) By the chain rule,
d
dx
[csc−1
u] =
du
dx
d
du
[csc−1
u] =
−1
|u|
√
u2 − 1
du
dx
20. (a) x = π − sin−1
(0.37) ≈ 2.7626 rad (b) θ = 180◦
+ sin−1
(0.61) ≈ 217.6◦
21. (a) x = π + cos−1
(0.85) ≈ 3.6964 rad (b) θ = − cos−1
(0.23) ≈ −76.7◦
22. (a) x = tan−1
(3.16) − π ≈ −1.8773 (b) θ = 180◦
− tan−1
(0.45) ≈ 155.8◦
23. (a)
1
1 − x2/9
(1/3) = 1/ 9 − x2 (b) −2/ 1 − (2x + 1)2
24. (a) 2x/(1 + x4
) (b) −
1
1 + x
1
2
x−1/2
= −
1
2(1 + x)
√
x
25. (a)
1
|x|7
√
x14 − 1
(7x6
) =
7
|x|
√
x14 − 1
(b) −1/
√
e2x − 1
26. (a) y = 1/ tan x = cot x, dy/dx = − csc2
x
(b) y = (tan−1
x)−1
, dy/dx = −(tan−1
x)−2 1
1 + x2
27. (a)
1
1 − 1/x2
(−1/x2
) = −
1
|x|
√
x2 − 1
(b)
sin x
√
1 − cos2 x
=
sin x
| sin x|
=
1, sin x > 0
−1, sin x < 0
28. (a) −
1
(cos−1 x)
√
1 − x2
(b) −
1
2
√
cot−1
x(1 + x2)
29. (a)
ex
|x|
√
x2 − 1
+ ex
sec−1
x (b)
3x2
(sin−1
x)2
√
1 − x2
+ 2x(sin−1
x)3
30. (a) 0 (b) 0
31. x3
+ x tan−1
y = ey
, 3x2
+
x
1 + y2
y + tan−1
y = ey
y , y =
(3x2
+ tan−1
y)(1 + y2
)
(1 + y2)ey − x
32. sin−1
(xy) = cos−1
(x − y),
1
1 − x2y2
(xy + y) = −
1
1 − (x − y)2
(1 − y ),
y =
y 1 − (x − y)2 + 1 − x2y2
1 − x2y2 − x 1 − (x − y)2
33. sin−1
x
1/
√
2
0
= sin−1
(1/
√
2) − sin−1
0 = π/4
34. u = 2x,
1
2
1
√
1 − u2
du =
1
2
sin−1
(2x) + C
35. tan−1
x
1
−1
= tan−1
1 − tan−1
(−1) = π/4 − (−π/4) = π/2

296 Chapter 7
36. u = 4x,
1
4
1
1 + u2
du =
1
4
tan−1
(4x) + C
37. sec−1
x
2
√
2
= sec−1
2 − sec−1
√
2 = π/3 − π/4 = π/12
38. − sec−1
x
−2/
√
3
−
√
2
= − sec−1
(−2/
√
3) + sec−1
(−
√
2) = −5π/6 + 3π/4 = −π/12
39. u = tan x,
1
√
1 − u2
du = sin−1
(tan x) + C
40. u = e−x
, −
√
3/2
1/2
1
√
1 − u2
du = − sin−1
u
√
3/2
1/2
= − sin−1
√
3
2
+ sin−1 1
2
= −
π
3
+
π
6
= −
π
6
41. u = ex
,
1
1 + u2
du = tan−1
(ex
) + C
42. u = t2
,
1
2
1
u2 + 1
du =
1
2
tan−1
(t2
) + C
43. u =
√
x, 2
√
3
1
1
u2 + 1
du = 2 tan−1
u
√
3
1
= 2(tan−1
√
3 − tan−1
1) = 2(π/3 − π/4) = π/6
44. u = cos θ, −
1
u2 + 1
du = − tan−1
(cos θ) + C
45. u = ln x,
1
√
1 − u2
du = sin−1
(ln x) + C
46. u = 3x,
1
u
√
u2 − 1
du = sec−1
(3x) + C if x > 0; − sec−1
(3x) + C if x < 0
47. u = a sin θ, du = a cos θ dθ;
du
√
a2 − u2
= aθ + C = sin−1 u
a
+ C
48. If u > 0 then u = a sec θ, du = a sec θ tan θ dθ,
du
u
√
u2 − a2
=
1
a
θ =
1
a
sec−1 u
a
+ C
49. (a) sin−1
(x/3) + C (b) (1/
√
5) tan−1
(x/
√
5) + C
(c) (1/
√
π) sec−1
(x/
√
π) + C
50. (a) u = ex
,
1
4 + u2
du =
1
2
tan−1
(ex
/2) + C
(b) u = 2x,
1
2
1
√
9 − u2
du =
1
2
sin−1
(2x/3) + C,
(c) u =
√
5y,
1
u
√
u2 − 3
du =
1
√
3
sec−1
(
√
5y/
√
3) + C

51. u =
√
3x2
,
1
2
√
3
√
3
0
1
√
4 − u2
du =
1
2
√
3
sin−1 u
2
√
3
0
=
1
2
√
3
π
3
=
π
6
√
3
52. u =
√
x, 2
√
2
1
1
√
4 − u2
du = 2 sin−1 u
2
√
2
1
= 2(π/4 − π/6) = π/6
53. u = 3x,
1
3
2
√
3
0
1
4 + u2
du =
1
6
tan−1 u
2
2
√
3
0
=
1
6
π
3
=
1
18π
54. u = x2
,
1
2
3
1
1
3 + u2
du =
1
2
√
3
tan−1 u
√
3
3
1
=
1
2
√
3
(π/3 − π/6) =
π
12
√
3
55. (a)
-0.5 0.5
x
y
c/2–
c/2
(b)
x
y
c/2–
c/2
56. (a) sin−1
0.9 > 1, so it is not in the domain of sin−1
x
(b) −1 ≤ sin−1
x ≤ 1 is necessary, or −0.841471 ≤ x ≤ 0.841471
57. (b) θ = sin−1 R
R + h
= sin−1 6378
16, 378
≈ 23◦
58. (a) If γ = 90◦
, then sin γ = 1, 1 − sin2
φ sin2
γ = 1 − sin2
φ = cos φ,
D = tan φ tan λ = (tan 23.45◦
)(tan 65◦
) ≈ 0.93023374 so h ≈ 21.1 hours.
(b) If γ = 270◦
, then sin γ = −1, D = − tan φ tan λ ≈ −0.93023374 so h ≈ 2.9 hours.
59. sin 2θ = gR/v2
= (9.8)(18)/(14)2
= 0.9, 2θ = sin−1
(0.9) or 2θ = 180◦
− sin−1
(0.9) so
θ = 1
2 sin−1
(0.9) ≈ 32◦
or θ = 90◦
− 1
2 sin−1
(0.9) ≈ 58◦
. The ball will have a lower
parabolic trajectory for θ = 32◦
and hence will result in the shorter time of ﬂight.
60. 42
= 22
+ 32
− 2(2)(3) cos θ, cos θ = −1/4, θ = cos−1
(−1/4) ≈ 104◦
61. y = 0 when x2
= 6000v2
/g, x = 10v 60/g = 1000
√
30 for v = 400 and g = 32;
tan θ = 3000/x = 3/
√
30, θ = tan−1
(3/
√
30) ≈ 29◦
.
a
␪
␣
␤
b
x
62. (a) θ = α − β, cot α =
x
a + b
and cot β =
x
b
so
θ = cot−1 x
a + b
− cot−1 x
b
(b)
dθ
dx
= −
1
a + b
1
1 + x2/(a + b)2
−
1
b
1
1 + (x/b)2
= −
a + b
(a + b)2 + x2
−
b
b2 + x2
which is negative for all x. Thus θ is a decreasing function of x, and it has no maximum
since lim
x→0+
θ = +∞.

298 Chapter 7
63. (a) A =
0.8
0
1
√
1 − x2
dx = sin−1
x
0.8
0
= sin−1
(0.8)
(b) The calculator was in degree mode instead of radian mode; the correct answer is 0.93.
64. A =
1/6
0
1
√
1 − 9x2
dx =
1
3
1/2
0
1
√
1 − u2
du =
1
3
sin−1
u
1/2
0
= π/18
65. The area is given by
k
0
(1/ 1 − x2 − x)dx = sin−1
k − k2
/2 = 1; solve for k to get
k = 0.997301.
66. x = sin y, A =
π/2
0
sin y dy = − cos y
π/2
0
= 1
67. The curves intersect at x = a = 0 and x = b = 0.838422 so the area is
b
a
(sin 2x − sin−1
x)dx = 0.174192.
68. The displacement of the particle during the time interval [0, T] is given by
T
0
v(t)dt = 3 tan−1
T − 0.25T2
. The particle is 2 cm from its starting position when
3 tan−1
T − 0.25T2
= 2 or when 3 tan−1
T − 0.25T2
= −2; solve for T to get
T = 0.90, 2.51, and 4.95 sec.
69. V =
2
−2
π
1
4 + x2
dx =
π
2
tan−1
(x/2)
2
−2
= π2
/4
70. (a) V = 2π
b
1
x
1 + x4
dx = π tan−1
(x2
)
b
1
= π tan−1
(b2
) −
π
4
(b) lim
b→+∞
V = π
π
2
−
π
4
=
1
4
π2
71. The area is given by
2
0
1/(1 + kx2
)dx = (1/
√
k) tan−1
(2
√
k) = 0.6; solve for k to get
k = 5.081435.
72. (a) π
1
0
(sin−1
x)2
dx = 1.468384. (b) 2π
π/2
0
y(1 − sin y)dy = 1.468384.
x
y
A(2, 1)
B(5, 4)
P(x, 0)
52
x - 2 5 - x
␣ ␤
␪
73. θ = π − (α + β)
= π − cot−1
(x − 2) − cot−1 5 − x
4
,
dθ
dx
=
1
1 + (x − 2)2
+
−1/4
1 + (5 − x)2/16
= −
3(x2
− 2x − 7)
[1 + (x − 2)2][16 + (5 − x)2]
dθ/dx = 0 when x =
2 ±
√
4 + 28
2
= 1 ± 2
√
2, only 1 + 2
√
2 is in [2, 5]; dθ/dx > 0 for x in
[2, 1 + 2
√
2), dθ/dx < 0 for x in (1 + 2
√
2, 5], θ is maximum when x = 1 + 2
√
2.

74. θ = α − β
= cot−1
(x/12) − cot−1
(x/2)
␪
␣
␤
x
10
2
dθ
dx
= −
12
144 + x2
+
2
4 + x2
=
10(24 − x2
)
(144 + x2)(4 + x2)
dθ/dx = 0 when x =
√
24 = 2
√
6, by the ﬁrst
derivative test θ is maximum there.
75. By the Mean-Value Theorem on the interval [0, x],
tan−1
x − tan−1
0
x − 0
=
tan−1
x
x
=
1
1 + c2
for c in (0, x), but
1
1 + x2
<
1
1 + c2
< 1 for c in (0, x) so
1
1 + x2
<
tan−1
x
x
< 1,
x
1 + x2
< tan−1
x < x.
76.
n
n2 + k2
=
1
1 + k2/n2
1
n
so
n
k=1
n
n2 + k2
=
n
k=1
f(x∗
k)∆x where f(x) =
1
1 + x2
, x∗
k =
k
n
, and ∆x =
1
n
for 0 ≤ x ≤ 1. Thus lim
n→+∞
n
k=1
n
n2 + k2
= lim
n→+∞
n
k=1
f(x∗
k)∆x =
1
0
1
1 + x2
dx =
π
4
.
77. (a) Let θ = sin−1
(−x) then sin θ = −x, −π/2 ≤ θ ≤ π/2. But sin(−θ) = − sin θ and
−π/2 ≤ −θ ≤ π/2 so sin(−θ) = −(−x) = x, −θ = sin−1
x, θ = − sin−1
x.
(b) proof is similar to that in Part (a)
78. (a) Let θ = cos−1
(−x) then cos θ = −x, 0 ≤ θ ≤ π. But cos(π − θ) = − cos θ and
0 ≤ π − θ ≤ π so cos(π − θ) = x, π − θ = cos−1
x, θ = π − cos−1
x
(b) Let θ = sec−1
(−x) for x ≥ 1; then sec θ = −x and π/2 < θ ≤ π. So 0 ≤ π − θ < π/2 and
π − θ = sec−1
sec(π − θ) = sec−1
(− sec θ) = sec−1
x, or sec−1
(−x) = π − sec−1
x.
1 x
sin-1 x
√1 - x2
79. (a) sin−1
x = tan−1 x
√
1 − x2
(see ﬁgure)
(b) sin−1
x + cos−1
x = π/2; cos−1
x = π/2 − sin−1
x = π/2 − tan−1 x
√
1 − x2
80. tan(α + β) =
tan α + tan β
1 − tan α tan β
,
tan(tan−1
x + tan−1
y) =
tan(tan−1
x) + tan(tan−1
y)
1 − tan(tan−1
x) tan(tan−1
y)
=
x + y
1 − xy
so tan−1
x + tan−1
y = tan−1 x + y
1 − xy

300 Chapter 7
81. (a) tan−1 1
2
+ tan−1 1
3
= tan−1 1/2 + 1/3
1 − (1/2) (1/3)
= tan−1
1 = π/4
(b) 2 tan−1 1
3
= tan−1 1
3
+ tan−1 1
3
= tan−1 1/3 + 1/3
1 − (1/3) (1/3)
= tan−1 3
4
,
2 tan−1 1
3
+ tan−1 1
7
= tan−1 3
4
+ tan−1 1
7
= tan−1 3/4 + 1/7
1 − (3/4) (1/7)
= tan−1
1 = π/4
82. sin(sec−1
x) = sin(cos−1
(1/x)) = 1 −
1
x
2
=
√
x2 − 1
|x|
EXERCISE SET 7.7
1. (a) lim
x→2
x2
− 4
x2 + 2x − 8
= lim
x→2
(x − 2)(x + 2)
(x + 4)(x − 2)
= lim
x→2
x + 2
x + 4
=
2
3
(b) lim
x→+∞
2x − 5
3x + 7
=
2 − lim
x→+∞
5
x
3 + lim
x→+∞
7
x
=
2
3
2. (a)
sin x
tan x
= sin x
cos x
sin x
= cos x so lim
x→0
sin x
tan x
= lim
x→0
cos x = 1
(b)
x2
− 1
x3 − 1
=
(x − 1)(x + 1)
(x − 1)(x2 + x + 1)
=
x + 1
x2 + x + 1
so lim
x→1
x2
− 1
x3 − 1
=
2
3
3. lim
x→1
1/x
1
= 1 4. lim
x→0
2 cos 2x
5 cos 5x
= 2/5
5. lim
x→0
ex
cos x
= 1 6. lim
x→3
1
6x − 13
= 1/5
7. lim
θ→0
sec2
θ
1
= 1 8. lim
t→0
tet
+ et
−et
= −1
9. lim
x→π+
cos x
1
= −1 10. lim
x→0+
cos x
2x
= +∞
11. lim
x→+∞
1/x
1
= 0 12. lim
x→+∞
3e3x
2x
= lim
x→+∞
9e3x
2
= +∞
13. lim
x→0+
− csc2
x
1/x
= lim
x→0+
−x
sin2
x
= lim
x→0+
−1
2 sin x cos x
= −∞
14. lim
x→0+
−1/x
(−1/x2)e1/x
= lim
x→0+
x
e1/x
= 0
15. lim
x→+∞
100x99
ex
= lim
x→+∞
(100)(99)x98
ex
= · · · = lim
x→+∞
(100)(99)(98) · · · (1)
ex
= 0

16. lim
x→0+
cos x/ sin x
sec2 x/ tan x
= lim
x→0+
cos2
x = 1 17. lim
x→0
2/
√
1 − 4x2
1
= 2
18. lim
x→0
1 −
1
1 + x2
3x2
= lim
x→0
1
3(1 + x2)
=
1
3
19. lim
x→+∞
xe−x
= lim
x→+∞
x
ex
= lim
x→+∞
1
ex
= 0
20. lim
x→π
(x − π) tan(x/2) = lim
x→π
x − π
cot(x/2)
= lim
x→π
1
−(1/2) csc2(x/2)
= −2
21. lim
x→+∞
x sin(π/x) = lim
x→+∞
sin(π/x)
1/x
= lim
x→+∞
(−π/x2
) cos(π/x)
−1/x2
= lim
x→+∞
π cos(π/x) = π
22. lim
x→0+
tan x ln x = lim
x→0+
ln x
cot x
= lim
x→0+
1/x
− csc2 x
= lim
x→0+
− sin2
x
x
= lim
x→0+
−2 sin x cos x
1
= 0
23. lim
x→(π/2)−
sec 3x cos 5x = lim
x→(π/2)−
cos 5x
cos 3x
= lim
x→(π/2)−
−5 sin 5x
−3 sin 3x
=
−5(+1)
(−3)(−1)
= −
5
3
24. lim
x→π
(x − π) cot x = lim
x→π
x − π
tan x
= lim
x→π
1
sec2 x
= 1
25. y = (1 − 3/x)x
, lim
x→+∞
ln y = lim
x→+∞
ln(1 − 3/x)
1/x
= lim
x→+∞
−3
1 − 3/x
= −3, lim
x→+∞
y = e−3
26. y = (1 + 2x)−3/x
, lim
x→0
ln y = lim
x→0
−
3 ln(1 + 2x)
x
= lim
x→0
−
6
1 + 2x
= −6, lim
x→0
y = e−6
27. y = (ex
+ x)1/x
, lim
x→0
ln y = lim
x→0
ln(ex
+ x)
x
= lim
x→0
ex
+ 1
ex + x
= 2, lim
x→0
y = e2
28. y = (1 + a/x)bx
, lim
x→+∞
ln y = lim
x→+∞
b ln(1 + a/x)
1/x
= lim
x→+∞
ab
1 + a/x
= ab, lim
x→+∞
y = eab
29. y = (2 − x)tan(πx/2)
, lim
x→1
ln y = lim
x→1
ln(2 − x)
cot(πx/2)
= lim
x→1
2 sin2
(πx/2)
π(2 − x)
= 2/π, lim
x→1
y = e2/π
30. y = [cos(2/x)]x2
, lim
x→+∞
ln y = lim
x→+∞
ln cos(2/x)
1/x2
= lim
x→+∞
(−2/x2
)(− tan(2/x))
−2/x3
= lim
x→+∞
− tan(2/x)
1/x
= lim
x→+∞
(2/x2
) sec2
(2/x)
−1/x2
= −2, lim
x→+∞
y = e−2
31. lim
x→0
1
sin x
−
1
x
= lim
x→0
x − sin x
x sin x
= lim
x→0
1 − cos x
x cos x + sin x
= lim
x→0
sin x
2 cos x − x sin x
= 0
32. lim
x→0
1 − cos 3x
x2
= lim
x→0
3 sin 3x
2x
= lim
x→0
9
2
cos 3x =
9
2
33. lim
x→+∞
(x2
+ x) − x2
√
x2 + x + x
= lim
x→+∞
x
√
x2 + x + x
= lim
x→+∞
1
1 + 1/x + 1
= 1/2
34. lim
x→0
ex
− 1 − x
xex − x
= lim
x→0
ex
− 1
xex + ex − 1
= lim
x→0
ex
xex + 2ex
= 1/2

302 Chapter 7
35. lim
x→+∞
[x − ln(x2
+ 1)] = lim
x→+∞
[ln ex
− ln(x2
+ 1)] = lim
x→+∞
ln
ex
x2 + 1
,
lim
x→+∞
ex
x2 + 1
= lim
x→+∞
ex
2x
= lim
x→+∞
ex
2
= +∞ so lim
x→+∞
[x − ln(x2
+ 1)] = +∞
36. lim
x→+∞
ln
x
1 + x
= lim
x→+∞
ln
1
1/x + 1
= ln(1) = 0
38. (a) lim
x→+∞
ln x
xn
= lim
x→+∞
1/x
nxn−1
= lim
x→+∞
1
nxn
= 0
(b) lim
x→+∞
xn
ln x
= lim
x→+∞
nxn−1
1/x
= lim
x→+∞
nxn
= +∞
39. (a) L’Hˆopital’s Rule does not apply to the problem lim
x→1
3x2
− 2x + 1
3x2 − 2x
because it is not a
0
0
form.
(b) lim
x→1
3x2
− 2x + 1
3x2 − 2x
= 2
40. lim
x→1
4x3
− 12x2
+ 12x − 4
4x3 − 9x2 + 6x − 1
= lim
x→1
12x2
− 24x + 12
12x2 − 18x + 6
= lim
x→1
24x − 24
24x − 18
= 0
41. lim
x→+∞
1/(x ln x)
1/(2
√
x)
= lim
x→+∞
2
√
x ln x
= 0 0.15
0
100 10000
42. y = xx
, lim
x→0+
ln y = lim
x→0+
ln x
1/x
= lim
x→0+
−x = 0, lim
x→0+
y = 1 1
0
0 0.5
25
19
0 0.5
43. y = (sin x)3/ ln x
,
lim
x→0+
ln y = lim
x→0+
3 ln sin x
ln x
= lim
x→0+
(3 cos x)
x
sin x
= 3,
lim
x→0+
y = e3
44. lim
x→π/2−
4 sec2
x
sec x tan x
= lim
x→π/2−
4
sin x
= 4

45. ln x − ex
= ln x −
1
e−x
=
e−x
ln x − 1
e−x
;
lim
x→+∞
e−x
ln x = lim
x→+∞
ln x
ex
= lim
x→+∞
1/x
ex
= 0 by L’Hˆopital’s Rule,
so lim
x→+∞
[ln x − ex
] = lim
x→+∞
e−x
ln x − 1
e−x
= −∞
0
-16
0 3
-0.6
-1.2
0 12
46. lim
x→+∞
[ln ex
− ln(1 + 2ex
)] = lim
x→+∞
ln
ex
1 + 2ex
= lim
x→+∞
ln
1
e−x + 2
= ln
1
2
;
horizontal asymptote y = − ln 2
1.02
1
100 10000
47. y = (ln x)1/x
,
lim
x→+∞
ln y = lim
x→+∞
ln(ln x)
x
= lim
x→+∞
1
x ln x
= 0;
lim
x→+∞
y = 1, y = 1 is the horizontal asymptote
1
0
0 50
48. y =
x + 1
x + 2
x
, lim
x→+∞
ln y = lim
x→+∞
ln
x + 1
x + 2
1/x
= lim
x→+∞
−x2
(x + 1)(x + 2)
= −1;
lim
x→+∞
y = e−1
is the horizontal asymptote
49. (a) 0 (b) +∞ (c) 0 (d) −∞ (e) +∞ (f) −∞
50. (a) Type 00
; y = x(ln a)/(1+ln x)
; lim
x→0+
ln y = lim
x→0+
(ln a) ln x
1 + ln x
= lim
x→0+
(ln a)/x
1/x
= lim
x→0+
ln a = ln a,
lim
x→0+
y = eln a
= a
(b) Type ∞0
; same calculation as Part (a) with x → +∞
(c) Type 1∞
; y = (x + 1)(ln a)/x
, lim
x→0
ln y = lim
x→0
(ln a) ln(x + 1)
x
= lim
x→0
ln a
x + 1
= ln a,
lim
x→0
y = eln a
= a

304 Chapter 7
51. lim
x→+∞
1 + 2 cos 2x
1
does not exist, nor is it ±∞; lim
x→+∞
x + sin 2x
x
= lim
x→+∞
1 +
sin 2x
x
= 1
52. lim
x→+∞
2 − cos x
3 + cos x
does not exist, nor is it ±∞; lim
x→+∞
2x − sin x
3x + sin x
= lim
x→+∞
2 − (sin x)/x
3 + (sin x)/x
=
2
3
53. lim
x→+∞
(2 + x cos 2x + sin 2x) does not exist, nor is it ±∞; lim
x→+∞
x(2 + sin 2x)
x + 1
= lim
x→+∞
2 + sin 2x
1 + 1/x
,
which does not exist because sin 2x oscillates between −1 and 1 as x → +∞
54. lim
x→+∞
1
x
+
1
2
cos x +
sin x
2x
does not exist, nor is it ±∞;
lim
x→+∞
x(2 + sin x)
x2 + 1
= lim
x→+∞
2 + sin x
x + 1/x
= 0
55. lim
R→0+
V t
L e−Rt/L
1
=
V t
L
56. (a) lim
x→π/2
(π/2 − x) tan x = lim
x→π/2
π/2 − x
cot x
= lim
x→π/2
−1
− csc2 x
= lim
x→π/2
sin2
x = 1
(b) lim
x→π/2
1
π/2 − x
− tan x = lim
x→π/2
1
π/2 − x
−
sin x
cos x
= lim
x→π/2
cos x − (π/2 − x) sin x
(π/2 − x) cos x
= lim
x→π/2
−(π/2 − x) cos x
−(π/2 − x) sin x − cos x
= lim
x→π/2
(π/2 − x) sin x + cos x
−(π/2 − x) cos x + 2 sin x
= 0
(c) 1/(π/2 − 1.57) ≈ 1255.765849, tan 1.57 ≈ 1255.765592;
1/(π/2 − 1.57) − tan 1.57 ≈ 0.000265
57. (b) lim
x→+∞
x(k1/x
− 1) = lim
t→0+
kt
− 1
t
= lim
t→0+
(ln k)kt
1
= ln k
(c) ln 0.3 = −1.20397, 1024 1024
√
0.3 − 1 = −1.20327;
ln 2 = 0.69315, 1024 1024
√
2 − 1 = 0.69338
58. (a) No; sin(1/x) oscillates as x → 0. (b) 0.05
-0.05
-0.35 0.35
(c) For the limit as x → 0+
use the Squeezing Theorem together with the inequalities
−x2
≤ x2
sin(1/x) ≤ x2
. For x → 0−
do the same; thus lim
x→0
f(x) = 0.
59. If k = −1 then lim
x→0
(k + cos x) = k + 1 = 0, so lim
x→0
k + cos x
x2
= ±∞. Hence k = −1, and by the
rule
lim
x→0
−1 + cos x
x2
= lim
x→0
− sin x
2x
= lim
x→0
− 2
cos x
2
= −
2
2
= −4 if = ±2
√
2.

60. (a) Apply the rule to get lim
x→0
− cos(1/x) + 2x sin(1/x)
cos x
which does not exist (nor is it ±∞).
(b) Rewrite as lim
x→0
x
sin x
[x sin(1/x)], but lim
x→0
x
sin x
= lim
x→0
1
cos x
= 1 and lim
x→0
x sin(1/x) = 0,
thus lim
x→0
x
sin x
[x sin(1/x)] = (1)(0) = 0
61. lim
x→0+
sin(1/x)
(sin x)/x
, lim
x→0+
sin x
x
= 1 but lim
x→0+
sin(1/x) does not exist because sin(1/x) oscillates between
−1 and 1 as x → +∞, so lim
x→0+
x sin(1/x)
sin x
does not exist.
EXERCISE SET 7.8
1. (a) sinh 3 ≈ 10.0179
(b) cosh(−2) ≈ 3.7622
(c) tanh(ln 4) = 15/17 ≈ 0.8824
(d) sinh−1
(−2) ≈ −1.4436
(e) cosh−1
3 ≈ 1.7627
(f) tanh−1 3
4
≈ 0.9730
2. (a) csch(−1) ≈ −0.8509
(b) sech(ln 2) = 0.8
(c) coth 1 ≈ 1.3130
(d) sech−1 1
2
≈ 1.3170
(e) coth−1
3 ≈ 0.3466
(f) csch−1
(−
√
3) ≈ −0.5493
3. (a) sinh(ln 3) =
1
2
(eln 3
− e− ln 3
) =
1
2
3 −
1
3
=
4
3
(b) cosh(− ln 2) =
1
2
(e− ln 2
+ eln 2
) =
1
2
1
2
+ 2 =
5
4
(c) tanh(2 ln 5) =
e2 ln 5
− e−2 ln 5
e2 ln 5 + e−2 ln 5
=
25 − 1/25
25 + 1/25
=
312
313
(d) sinh(−3 ln 2) =
1
2
(e−3 ln 2
− e3 ln 2
) =
1
2
1
8
− 8 = −
63
16
4. (a)
1
2
(eln x
+ e− ln x
) =
1
2
x +
1
x
=
x2
+ 1
2x
, x > 0
(b)
1
2
(eln x
− e− ln x
) =
1
2
x −
1
x
=
x2
− 1
2x
, x > 0
(c)
e2 ln x
− e−2 ln x
e2 ln x + e−2 ln x
=
x2
− 1/x2
x2 + 1/x2
=
x4
− 1
x4 + 1
, x > 0
(d)
1
2
(e− ln x
+ eln x
) =
1
2
1
x
+ x =
1 + x2
2x
, x > 0
5. sinh x0 cosh x0 tanh x0 coth x0 sech x0 csch x0
2 √5 2/√5 √5/2 1/√5 1/2
3/4 5/4 3/5 5/3 4/5 4/3
4/3
(a)
(b)
(c) 5/3 4/5 5/4 3/5 3/4

306 Chapter 7
(a) cosh2
x0 = 1 + sinh2
x0 = 1 + (2)2
= 5, cosh x0 =
√
5
(b) sinh2
x0 = cosh2
x0 − 1 =
25
16
− 1 =
9
16
, sinh x0 =
3
4
(because x0 > 0)
(c) sech2
x0 = 1 − tanh2
x0 = 1 −
4
5
2
= 1 −
16
25
=
9
25
, sech x0 =
3
5
,
cosh x0 =
1
sech x0
=
5
3
, from
sinh x0
cosh x0
= tanh x0 we get sinh x0 =
5
3
4
5
=
4
3
6.
d
dx
cschx =
d
dx
1
sinh x
= −
cosh x
sinh2
x
= − coth x csch x for x = 0
d
dx
sech x =
d
dx
1
cosh x
= −
sinh x
cosh2
x
= − tanh x sech x for all x
d
dx
coth x =
d
dx
cosh x
sinh x
=
sinh2
x − cosh2
x
sinh2
x
= − csch2
x for x = 0
7. (a) y = sinh−1
x if and only if x = sinh y; 1 =
dy
dx
dx
dy
=
dy
dx
cosh y; so
d
dx
[sinh−1
x] =
dy
dx
=
1
cosh y
=
1
1 + sinh2
y
=
1
√
1 + x2
for all x.
(b) Let x ≥ 1. Then y = cosh−1
x if and only if x = cosh y; 1 =
dy
dx
dx
dy
=
dy
dx
sinh y, so
d
dx
[cosh−1
x] =
dy
dx
=
1
sinh y
=
1
cosh2
y − 1
=
1
x2 − 1
for x ≥ 1.
(c) Let −1 < x < 1. Then y = tanh−1
x if and only if x = tanh y; thus
1 =
dy
dx
dx
dy
=
dy
dx
sech2
y =
dy
dx
(1 − tanh2
y) = 1 − x2
, so
d
dx
[tanh−1
x] =
dy
dx
=
1
1 − x2
.
9. 4 cosh(4x − 8) 10. 4x3
sinh(x4
) 11. −
1
x
csch2
(ln x)
12. 2
sech2
2x
tanh 2x
13.
1
x2
csch(1/x) coth(1/x) 14. −2e2x
sech(e2x
) tanh(e2x
)
15.
2 + 5 cosh(5x) sinh(5x)
4x + cosh2
(5x)
16. 6 sinh2
(2x) cosh(2x)
17. x5/2
tanh(
√
x) sech2
(
√
x) + 3x2
tanh2
(
√
x)
18. −3 cosh(cos 3x) sin 3x 19.
1
1 + x2/9
1
3
= 1/ 9 + x2
20.
1
1 + 1/x2
(−1/x2
) = −
1
|x|
√
x2 + 1
21. 1/ (cosh−1
x)
√
x2 − 1
22. 1/ (sinh−1
x)2 − 1
√
1 + x2 23. −(tanh−1
x)−2
/(1 − x2
)
24. 2(coth−1
x)/(1 − x2
) 25.
sinh x
cosh2
x − 1
=
sinh x
| sinh x|
=
1, x > 0
−1, x < 0

26. (sech2
x)/ 1 + tanh2
x 27. −
ex
2x
√
1 − x
+ ex
sech−1
x
28. 10(1 + x csch−1
x)9
−
x
|x|
√
1 + x2
+ csch−1
x
31.
1
7
sinh7
x + C 32.
1
2
sinh(2x − 3) + C 33.
2
3
(tanh x)3/2
+ C
34. −
1
3
coth(3x) + C 35. ln(cosh x) + C 36. −
1
3
coth3
x + C
37. −
1
3
sech3
x
ln 3
ln 2
= 37/375 38. ln(cosh x)
ln 3
0
= ln 5 − ln 3
39. u = 3x,
1
3
1
√
1 + u2
du =
1
3
sinh−1
3x + C
40. x =
√
2u,
√
2
√
2u2 − 2
du =
1
√
u2 − 1
du = cosh−1
(x/
√
2) + C
41. u = ex
,
1
u
√
1 − u2
du = − sech−1
(ex
) + C
42. u = cos θ, −
1
√
1 + u2
du = − sinh−1
(cos θ) + C
43. u = 2x,
du
u
√
1 + u2
= −csch−1
|u| + C = −csch−1
|2x| + C
44. x = 5u/3,
5/3
√
25u2 − 25
du =
1
3
1
√
u2 − 1
du =
1
3
cosh−1
(3x/5) + C
45. tanh−1
x
1/2
0
= tanh−1
(1/2) − tanh−1
(0) =
1
2
ln
1 + 1/2
1 − 1/2
=
1
2
ln 3
46. sinh−1
t
√
3
0
= sinh−1
√
3 − sinh−1
0 = ln(
√
3 + 2)
49. A =
ln 3
0
sinh 2x dx =
1
2
cosh 2x
ln 3
0
=
1
2
[cosh(2 ln 3) − 1],
but cosh(2 ln 3) = cosh(ln 9) =
1
2
(eln 9
+ e− ln 9
) =
1
2
(9 + 1/9) = 41/9 so A =
1
2
[41/9 − 1] = 16/9.
50. V = π
ln 2
0
sech2
x dx = π tanh x
ln 2
0
= π tanh(ln 2) = 3π/5
51. V = π
5
0
(cosh2
2x − sinh2
2x)dx = π
5
0
dx = 5π

308 Chapter 7
52.
1
0
cosh ax dx = 2,
1
a
sinh ax
1
0
= 2,
1
a
sinh a = 2, sinh a = 2a;
let f(a) = sinh a − 2a, then an+1 = an −
sinh an − 2an
cosh an − 2
, a1 = 2.2, . . . , a4 = a5 = 2.177318985.
53. y = sinh x, 1 + (y )2
= 1 + sinh2
x = cosh2
x
L =
ln 2
0
cosh x dx = sinh x
ln 2
0
= sinh(ln 2) =
1
2
(eln 2
− e− ln 2
) =
1
2
2 −
1
2
=
3
4
54. y = sinh(x/a), 1 + (y )2
= 1 + sinh2
(x/a) = cosh2
(x/a)
L =
x1
0
cosh(x/a)dx = a sinh(x/a)
x1
0
= a sinh(x1/a)
55. sinh(−x) =
1
2
(e−x
− ex
) = −
1
2
(ex
− e−x
) = − sinh x
cosh(−x) =
1
2
(e−x
+ ex
) =
1
2
(ex
+ e−x
) = cosh x
56. (a) cosh x + sinh x =
1
2
(ex
+ e−x
) +
1
2
(ex
− e−x
) = ex
(b) cosh x − sinh x =
1
2
(ex
+ e−x
) −
1
2
(ex
− e−x
) = e−x
(c) sinh x cosh y + cosh x sinh y =
1
4
(ex
− e−x
)(ey
+ e−y
) +
1
4
(ex
+ e−x
)(ey
− e−y
)
=
1
4
[(ex+y
− e−x+y
+ ex−y
− e−x−y
) + (ex+y
+ e−x+y
− ex−y
− e−x−y
)]
=
1
2
[e(x+y)
− e−(x+y)
] = sinh(x + y)
(d) Let y = x in Part (c).
(e) The proof is similar to Part (c), or: treat x as variable and y as constant, and diﬀerentiate
the result in Part (c) with respect to x.
(f) Let y = x in Part (e).
(g) Use cosh2
x = 1 + sinh2
x together with Part (f).
(h) Use sinh2
x = cosh2
x − 1 together with Part (f).
57. (a) Divide cosh2
x − sinh2
x = 1 by cosh2
x.
(b) tanh(x + y) =
sinh x cosh y + cosh x sinh y
cosh x cosh y + sinh x sinh y
=
sinh x
cosh x
+
sinh y
cosh y
1 +
sinh x sinh y
cosh x cosh y
=
tanh x + tanh y
1 + tanh x tanh y
(c) Let y = x in Part (b).
58. (a) Let y = cosh−1
x; then x = cosh y =
1
2
(ey
+ e−y
), ey
− 2x + e−y
= 0, e2y
− 2xey
+ 1 = 0,
ey
=
2x ±
√
4x2 − 4
2
= x ± x2 − 1. To determine which sign to take, note that y ≥ 0
so e−y
≤ ey
, x = (ey
+ e−y
)/2 ≤ (ey
+ ey
)/2 = ey
, hence ey
≥ x thus ey
= x +
√
x2 − 1,
y = cosh−1
x = ln(x +
√
x2 − 1).

(b) Let y = tanh−1
x; then x = tanh y =
ey
− e−y
ey + e−y
=
e2y
− 1
e2y + 1
, xe2y
+ x = e2y
− 1,
1 + x = e2y
(1 − x), e2y
= (1 + x)/(1 − x), 2y = ln
1 + x
1 − x
, y =
1
2
ln
1 + x
1 − x
.
59. (a)
d
dx
(cosh−1
x) =
1 + x/
√
x2 − 1
x +
√
x2 − 1
= 1/ x2 − 1
(b)
d
dx
(tanh−1
x) =
d
dx
1
2
(ln(1 + x) − ln(1 − x)) =
1
2
1
1 + x
+
1
1 − x
= 1/(1 − x2
)
60. Let y = sech−1
x then x = sech y = 1/ cosh y, cosh y = 1/x, y = cosh−1
(1/x); the proofs for the
remaining two are similar.
61. If |u| < 1 then, by Theorem 8.8.6,
du
1 − u2
= tanh−1
u + C.
For |u| > 1,
du
1 − u2
= coth−1
u + C = tanh−1
(1/u) + C.
62. (a)
d
dx
(sech−1
|x|) =
d
dx
(sech−1
√
x2) = −
1
√
x2
√
1 − x2
x
√
x2
= −
1
x
√
1 − x2
(b) Similar to solution of Part (a)
63. (a) lim
x→+∞
sinh x = lim
x→+∞
1
2
(ex
− e−x
) = +∞ − 0 = +∞
(b) lim
x→−∞
sinh x = lim
x→−∞
1
2
(ex
− e−x
) = 0 − ∞ = −∞
(c) lim
x→+∞
tanh x = lim
x→+∞
ex
− e−x
ex + e−x
= 1 (d) lim
x→−∞
tanh x = lim
x→−∞
ex
− e−x
ex + e−x
= −1
(e) lim
x→+∞
sinh−1
x = lim
x→+∞
ln(x + x2 + 1) = +∞
(f) lim
x→1−
tanh−1
x = lim
x→1−
1
2
[ln(1 + x) − ln(1 − x)] = +∞
64. (a) lim
x→+∞
(cosh−1
x − ln x) = lim
x→+∞
[ln(x + x2 − 1) − ln x]
= lim
x→+∞
ln
x +
√
x2 − 1
x
= lim
x→+∞
ln(1 + 1 − 1/x2) = ln 2
(b) lim
x→+∞
cosh x
ex
= lim
x→+∞
ex
+ e−x
2ex
= lim
x→+∞
1
2
(1 + e−2x
) = 1/2
65. For |x| < 1, y = tanh−1
x is deﬁned and dy/dx = 1/(1 − x2
) > 0; y = 2x/(1 − x2
)2
changes sign
at x = 0, so there is a point of inﬂection there.
66. Let x = −u/a,
1
√
u2 − a2
du = −
a
a
√
x2 − 1
dx = − cosh−1
x + C = − cosh−1
(−u/a) + C.
− cosh−1
(−u/a) = − ln(−u/a + u2/a2 − 1) = ln
a
−u +
√
u2 − a2
u +
√
u2 − a2
u +
√
u2 − a2
= ln u + u2 − a2 − ln a = ln |u + u2 − a2| + C1
so
1
√
u2 − a2
du = ln u + u2 − a2 + C2.

310 Chapter 7
67. Using sinh x + cosh x = ex
(Exercise 56a), (sinh x + cosh x)n
= (ex
)n
= enx
= sinh nx + cosh nx.
68.
a
−a
etx
dx =
1
t
etx
a
−a
=
1
t
(eat
− e−at
) =
2 sinh at
t
for t = 0.
69. (a) y = sinh(x/a), 1 + (y )2
= 1 + sinh2
(x/a) = cosh2
(x/a)
L = 2
b
0
cosh(x/a) dx = 2a sinh(x/a)
b
0
= 2a sinh(b/a)
(b) The highest point is at x = b, the lowest at x = 0,
so S = a cosh(b/a) − a cosh(0) = a cosh(b/a) − a.
70. From Part (a) of Exercise 69, L = 2a sinh(b/a) so 120 = 2a sinh(50/a), a sinh(50/a) = 60. Let
u = 50/a, then a = 50/u so (50/u) sinh u = 60, sinh u = 1.2u. If f(u) = sinh u − 1.2u, then
un+1 = un −
sinh un − 1.2un
cosh un − 1.2
; u1 = 1, . . . , u5 = u6 = 1.064868548 ≈ 50/a so a ≈ 46.95415231.
From Part (b), S = a cosh(b/a) − a ≈ 46.95415231[cosh(1.064868548) − 1] ≈ 29.2 ft.
71. From Part (b) of Exercise 69, S = a cosh(b/a) − a so 30 = a cosh(200/a) − a. Let u = 200/a,
then a = 200/u so 30 = (200/u)[cosh u − 1], cosh u − 1 = 0.15u. If f(u) = cosh u − 0.15u − 1,
then un+1 = un −
cosh un − 0.15un − 1
sinh un − 0.15
; u1 = 0.3, . . . , u4 = u5 = 0.297792782 ≈ 200/a so
a ≈ 671.6079505. From Part (a), L = 2a sinh(b/a) ≈ 2(671.6079505) sinh(0.297792782) ≈ 405.9 ft.
72. (a) When the bow of the boat is at the point (x, y) and the person has walked a distance D,
then the person is located at the point (0, D), the line segment connecting (0, D) and (x, y)
has length a; thus a2
= x2
+ (D − y)2
, D = y +
√
a2 − x2 = a sech−1
(x/a).
(b) Find D when a = 15, x = 10: D = 15 sech−1
(10/15) = 15 ln
1 + 5/9
2/3
≈ 14.44 m.
(c) dy/dx = −
a2
x
√
a2 − x2
+
x
√
a2 − x2
=
1
√
a2 − x2
−
a2
x
+ x = −
1
x
a2 − x2,
1 + [y ]2
= 1 +
a2
− x2
x2
=
a2
x2
; with a = 15 and x = 5, L =
15
5
225
x2
dx = −
225
x
15
5
= 30 m.
CHAPTER 7 SUPPLEMENTARY EXERCISES
1. (a) f(g(x)) = x for all x in the domain of g, and g(f(x)) = x for all x in the domain of f.
(b) They are reflections of each other through the line y = x.
(c) The domain of one is the range of the other and vice versa.
(d) The equation y = f(x) can always be solved for x as a function of y. Functions with no
inverses include y = x2
, y = sin x.
(e) Yes, g is continuous; this is evident from the statement about the graphs in Part (b) above.
(f) Yes, g must be differentiable (where f = 0); this can be inferred from the graphs. Note that
if f = 0 at a point then g cannot exist (infinite slope).
2. (a) For sin x, −π/2 ≤ x ≤ π/2; for cos x, 0 ≤ x ≤ π; for tan x, −π/2 < x < π/2; for sec x,
0 ≤ x < π/2 or π/2 < x ≤ π.

Chapter 7 Supplementary Exercises 311
(b) y
x
-1
1
c/2
y = sin-1 x
y = sin x
y
x
-1
c
y = cos-1 x
y = cos x
y
x
-2
2
c/2- c/2
y = tan-1 x
y = tan x
y
x
-1
2
c/2
y = sec-1 x y = sec x
y = sec-1 x
y = sec x
3. (a) x = f(y) = 8y3
− 1; y = f−1
(x) =
x + 1
8
1/3
=
1
2
(x + 1)1/3
(b) f(x) = (x − 1)2
; f does not have an inverse because f is not one-to-one, for example
f(0) = f(2) = 1.
(c) x = f(y) = (ey
)2
+ 1; y = f−1
(x) = ln
√
x − 1 = 1
2 ln(x − 1)
(d) x = f(y) =
y + 2
y − 1
; y = f−1
(x) =
x + 2
x − 1
4. f (x) =
ad − bc
(cx + d)2
; if ad − bc = 0 then the function represents a horizontal line, no inverse.
If ad − bc = 0 then f (x) > 0 or f (x) < 0 so f is invertible. If x = f(y) =
ay + b
cy + d
then
y = f−1
(x) =
b − xd
xc − a
.
5. 3 ln e2x
(ex
)3
+ 2 exp(ln 1) = 3 ln e2x
+ 3 ln(ex
)3
+ 2 · 1 = 3(2x) + (3 · 3)x + 2 = 15x + 2
6. Draw equilateral triangles of sides 5, 12, 13, and 3, 4, 5. Then sin[cos−1
(4/5)] = 3/5,
sin[cos−1
(5/13)] = 12/13, cos[sin−1
(4/5)] = 3/5, cos[sin−1
(5/13)] = 12/13
(a) cos[cos−1
(4/5) + sin−1
(5/13)] = cos(cos−1
(4/5)) cos(sin−1
(5/13))
− sin(cos−1
(4/5)) sin(sin−1
(5/13))
=
4
5
12
13
−
3
5
5
13
=
33
65
.
(b) sin[sin−1
(4/5) + cos−1
(5/13)] = sin(sin−1
(4/5)) cos(cos−1
(5/13))
+ cos(sin−1
(4/5)) sin(cos−1
(5/13))
=
4
5
5
13
+
3
5
12
13
=
56
65
.
7. (a) cosh 3x = cosh(2x + x) = cosh 2x cosh x + sinh 2x sinh x
= (2 cosh2
x − 1) cosh x + (2 sinh x cosh x) sinh x
= 2 cosh3
x − cosh x + 2 sinh2
x cosh x
= 2 cosh3
x − cosh x + 2(cosh2
x − 1) cosh x = 4 cosh3
x − 3 cosh x

312 Chapter 7
(b) from Theorem 7.8.2 with x replaced by
x
2
: cosh x = 2 cosh2 x
2
− 1,
2 cosh2 x
2
= cosh x + 1, cosh2 x
2
=
1
2
(cosh x + 1),
cosh
x
2
=
1
2
(cosh x + 1) (because cosh
x
2
> 0)
(c) from Theorem 7.8.2 with x replaced by
x
2
: cosh x = 2 sinh2 x
2
+ 1,
2 sinh2 x
2
= cosh x − 1, sinh2 x
2
=
1
2
(cosh x − 1), sinh
x
2
= ±
1
2
(cosh x − 1)
8. Y = ln(Cekt
) = ln C + ln ekt
= ln C + kt, a line with slope k and Y -intercept ln C
9. (a) y
x
-2
2
4
(b) The curve y = e−x/2
sin 2x has x-intercepts at x = −π/2, 0, π/2, π, 3π/2. It intersects the
curve y = e−x/2
at x = π/4, 5π/4, and it intersects the curve y = −e−x/2
at x = −π/4, 3π/4.
10. (a) y
x
c/2
1
(b) y
x
c/2
1
(c) y
x
-c/2
5
(d) y
x
c/2
1
11. Set a = 68.7672, b = 0.0100333, c = 693.8597, d = 299.2239.
(a) 650
0
-300 300
(b) L = 2
d
0
1 + a2b2 sinh2
bx dx
= 1480.2798 ft

(c) x = 283.6249 ft (d) 82◦
y
x
-2
2
12. (a) If xk
= ex
then k ln x = x, or
ln x
x
=
1
k
. The steps are reversible.
(b) By zooming it is seen that the maximum value of y is
approximately 0.368 (actually, 1/e), so there are two distinct
solutions of xk
= ex
whenever k > 1/0.368 ≈ 2.717.
(c) x ≈ 1.155
13. (a) The function ln x − x0.2
is negative at x = 1 and positive at x = 4, so it must be zero in
between (IVT).
(b) x = 3.654
14. (a)
1
1
2
t
r (b) r = 1 when t ≈ 0.673080 s.
(c) dr/dt = 4.48 m/s.
15. (a) y = x3
+ 1 so y = 3x2
. (b) y =
abe−x
(1 + be−x)2
(c) y =
1
2
ln x +
1
3
ln(x + 1) − ln sin x + ln cos x, so
y =
1
2x
+
1
3(x + 1)
−
cos x
sin x
−
sin x
cos x
=
5x + 3
6x(x + 1)
− cot x − tan x.
(d) ln y =
ln(1 + x)
x
,
y
y
=
x/(1 + x) − ln(1 + x)
x2
=
1
x(1 + x)
−
ln(1 + x)
x2
,
dy
dx
=
1
x
(1 + x)(1/x)−1
−
(1 + x)(1/x)
x2
ln(1 + x)
(e) ln y = ex
ln x,
y
y
= ex 1
x
+ ln x ,
dy
dx
= xex
ex 1
x
+ ln x = ex
xex
−1
+ xex
ln x
(f) y = sinh−1 1
x2
,
dy
dx
=
1
1 + (1/x2)2
−2
x3
= −
2
x
√
x4 + 1
16. y = aeax
sin bx + beax
cos bx and y = (a2
− b2
)eax
sin bx + 2abeax
cos bx, so y − 2ay + (a2
+ b2
)y
= (a2
− b2
)eax
sin bx + 2abeax
cos bx − 2a(aeax
sin bx + beax
cos bx) + (a2
+ b2
)eax
sin bx = 0.
17. sin(tan−1
x) = x/
√
1 + x2 and cos(tan−1
x) = 1/
√
1 + x2, and y =
1
1 + x2
, y =
−2x
(1 + x2)2
, hence
y + 2 sin y cos3
y =
−2x
(1 + x2)2
+ 2
x
√
1 + x2
1
(1 + x2)3/2
= 0.

314 Chapter 7
18. y = a cosh ax, y = a2
sinh ax = a2
y
19. Set y = logb x and solve y = 1: y =
1
x ln b
= 1
so x =
1
ln b
. The curves intersect when (x, x) lies
on the graph of y = logb x, so x = logb x. From
Formula (9), Section 7.2, logb x =
ln x
ln b
from which
ln x = 1, x = e, ln b = 1/e, b = e1/e
≈ 1.4447.
y
x
2
2
20. (a) Find the point of intersection: f(x) =
√
x + k = ln x. The
slopes are equal, so m1 =
1
x
= m2 =
1
2
√
x
,
√
x = 2, x = 4.
Then ln 4 =
√
4 + k, k = ln 4 − 2.
y
x
2
2
(b) Since the slopes are equal m1 =
k
2
√
x
= m2 =
1
x
, so k
√
x = 2.
At the point of intersection k
√
x = ln x, 2 = ln x, x = e2
,
k = 2/e.
y
x
0
2
5
21. lim
x→0+
f(x) = lim
x→+∞
f(x) = +∞ and f (x) =
ex
(x − 2)
x3
, stationary point at x = 2. We know f(x)
has no maximum and an absolute minimum; by Theorem 4.5.5 f(x) has an absolute minimum at
x = 2, and m = e2
/4.
22. f (x) = (1 + ln x)xx
, critical point at x = 1/e; lim
x→0+
f(x) = lim
x→0+
ex ln x
= 1, lim
x→+∞
f(x) = +∞; no
absolute maximum, absolute minimum m = e−1/e
at x = 1/e
23. fave =
1
e − 1
e
1
1
x
dx =
1
e − 1
ln x
e
1
=
1
e − 1
;
1
x∗
=
1
e − 1
, x∗
= e − 1
24. Find t so that N (t) is maximum. The size of the population is increasing most rapidly when
t = 8.4 years.
25. u = ln x, du = (1/x)dx;
2
1
1
u
du = ln u
2
1
= ln 2
26.
1
0
e−x/2
dx = 2(1 − 1/
√
e)
27. u = e−2x
, du = −2e−2x
dx; −
1
2
1/4
1
(1 + cos u)du =
3
8
+
1
2
sin 1 − sin
1
4
28. Divide ex
+ 3 into e2x
to get
e2x
ex + 3
= ex
−
3ex
ex + 3
so
e2x
ex + 3
dx = ex
dx − 3
ex
ex + 3
dx = ex
− 3 ln(ex
+ 3) + C

29. Since y = ex
and y = ln x are inverse functions, their graphs
are symmetric with respect to the line y = x; consequently the
areas A1 and A3 are equal (see ﬁgure). But A1 + A2 = e, so
e
1
ln xdx +
1
0
ex
dx = A2 + A3 = A2 + A1 = e
y
x
1
e
1 e
A3
A1
A2
30.
n
k=1
ek/n
n
=
n
k=1
f(x∗
k)∆x where f(x) = ex
, x∗
k = k/n, and ∆x = 1/n for 0 ≤ x ≤ 1. Thus
lim
n→+∞
n
k=1
ek/n
n
= lim
n→+∞
n
k=1
f(x∗
k)∆x =
1
0
ex
dx = e − 1.
31. Since f(x) =
1
x
is positive and increasing on the interval [1, 2], the left endpoint approximation
overestimates the integral of
1
x
and the right endpoint approximation underestimates it.
(a) For n = 5 this becomes
0.2
1
1.2
+
1
1.4
+
1
1.6
+
1
1.8
+
1
2.0
<
2
1
1
x
dx < 0.2
1
1.0
+
1
1.2
+
1
1.4
+
1
1.6
+
1
1.8
(b) For general n the left endpoint approximation to
2
1
1
x
dx = ln 2 is
1
n
n
k=1
1
1 + (k − 1)/n
=
n
k=1
1
n + k − 1
=
n−1
k=0
1
n + k
and the right endpoint approximation is
n
k=1
1
n + k
. This yields
n
k=1
1
n + k
<
2
1
1
x
dx <
n−1
k=0
1
n + k
which is the desired inequality.
(c) By telescoping, the diﬀerence is
1
n
−
1
2n
=
1
2n
so
1
2n
≤ 0.1, n ≥ 5
(d) n ≥ 1000
32. (a) x∗
k = 0, 1, 2, 3, 4
4
k=1
f(x∗
k)∆x = e0
+ e1
+ e2
+ e3
+ e4
(1) = (1 − e5
)/(1 − e) = 85.791
(b) x∗
k = 1, 2, 3, 4, 5
4
k=1
f(x∗
k)∆x = e1
+ e2
+ e3
+ e4
+ e5
(1) = e(1 − e5
)/(1 − e) = 233.204
(c) x∗
k = 1/2, 3/2, 5/2, 7/2, 9/2
4
k=1
f(x∗
k)∆x = e1/2
+ e3/2
+ e5/2
+ e7/2
+ e9/2
(1) = e1/2
(1 − e5
)/(1 − e) = 141.446
33. 0.351220577, 0.420535296, 0.386502483
34. 1.63379940, 1.805627583, 1.717566087

316 Chapter 7
35. f(x) = ex
, [a, b] = [0, 1], ∆x =
1
n
; lim
n→+∞
n
k=1
f(x∗
k) =
1
0
ex
dx = e − 1
36. In the case +∞ − (−∞) the limit is +∞; in the case −∞ − (+∞) the limit is −∞, because
large positive (negative) quantities are added to large positive (negative) quantities. The cases
+∞ − (+∞) and −∞ − (−∞) are indeterminate; large numbers of opposite sign are subtracted,
and more information about the sizes is needed.
37. (a) when the limit takes the form 0/0 or ∞/∞
(b) Not necessarily; only if lim
x→a
f(x) = 0. Consider g(x) = x; lim
x→0
g(x) = 0. For f(x) choose
cos x, x2
, and |x|1/2
. Then: lim
x→0
cos x
x
does not exist, lim
x→0
x2
x
= 0, and lim
x→0
|x|1/2
x2
= +∞.
38. (a) lim
x→+∞
(ex
− x2
) = lim
x→+∞
x2
(ex
/x2
− 1), but lim
x→+∞
ex
x2
= lim
x→+∞
ex
2x
= lim
x→+∞
ex
2
= +∞
so lim
x→+∞
(ex
/x2
− 1) = +∞ and thus lim
x→+∞
x2
(ex
/x2
− 1) = +∞
(b) lim
x→1
ln x
x4 − 1
= lim
x→1
1/x
4x3
=
1
4
; lim
x→1
ln x
x4 − 1
= lim
x→1
ln x
x4 − 1
=
1
2
(c) lim
x→0
ax
ln a = ln a

Solution Manual : Chapter - 07 Exponential, Logarithmic and Inverse Trigonometric Functions

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Solution Manual : Chapter - 07 Exponential, Logarithmic and Inverse Trigonometric Functions