# Solution manual calculus 8th edition varberg, purcell, rigdon ch06

Transcendental
Functions

6

CHAPTER

6.1 Concepts Review
x1

1.

∫1 t dt; (0, ∞); (–∞, ∞)

2.

1
x

3.

1
ln(3x – 2)
2
1
1
3
= ⋅
Dx (3 x – 2) =
2 3x – 2
2(3 x – 2)

6. Dx ln 3 x – 2 = Dx

1
; ln x + C
x

7.

dy
1 3
= 3⋅ =
dx
x x

8.

dy
1
= x 2 ⋅ + 2 x ⋅ ln x = x(1 + 2 ln x)
dx
x

4. ln x + ln y; ln x – ln y; r ln x

9. z = x 2 ln x 2 + (ln x)3 = x 2 ⋅ 2 ln x + (ln x)3

Problem Set 6.1
1. a.

b.

⎛3⎞
ln1.5 = ln ⎜ ⎟ = ln 3 − ln 2 = 0.406
⎝2⎠

c.

ln 81 = ln 3 = 4 ln 3 = 4(1.099) = 4.396

d.

ln 2 = ln 21/ 2 =

e.

f.

dz
2
1
= x 2 ⋅ + 2 x ⋅ 2 ln x + 3(ln x) 2 ⋅
dx
x
x
3
= 2 x + 4 x ln x + (ln x)2
x

ln 6 = ln (2 · 3) = ln 2 + ln 3
= 0.693 + 1.099 = 1.792

4

10. r =

1 –2
x – (ln x)3
2
dr –2 –3
1
1 3(ln x) 2
=
x – 3(ln x)2 ⋅ = − −
x
dx 2
x
x3

1
1
ln 2 = (0.693) = 0.3465
2
2

⎛ 1 ⎞
ln ⎜ ⎟ = – ln 36 = – ln(22 ⋅ 32 )
⎝ 36 ⎠
= −2 ln 2 − 2 ln 3 = −3.584

=

11. g ′( x) =

ln 48 = ln(24 ⋅ 3) = 4 ln 2 + ln 3 = 3.871

2. a.

1.792

b. 0.405

c.

4.394

d. 0.3466

e.

–3.584

f. 3.871

=

1
x + 3x + π
2

⋅ Dx ( x 2 + 3 x + π ) =

=

2x + 3
x + 3x + π

=

1
3x + 2 x
3

Dx (3x3 + 2 x)

9 x2 + 2
3 x3 + 2 x

5. Dx ln( x – 4)3 = Dx 3ln( x – 4)
1
3
= 3⋅
Dx ( x – 4) =
x–4
x–4

Instructor’s Resource Manual

x2 + 1

⎡ 1 2

–1/ 2
⋅ 2x⎥
⎢1 + 2 ( x – 1)

x + x –1
1

2

1
x −1
2

2

13.
4. Dx ln(3x3 + 2 x) =

⎡ 1 2

1 + ( x + 1) –1/ 2 ⋅ 2 x ⎥

x + x2 + 1 ⎣ 2
1

1

12. h′( x) =

3. Dx ln( x 2 + 3 x + π)
=

3

ln x
⎛ 1⎞
+ ln
=
+ (– ln x)3
2
2 ⎜⎝ x ⎟⎠
2
x ln x
x ⋅ 2 ln x
ln x

14.

1
f ( x) = ln 3 x = ln x
3
1 1 1
f ′( x) = ⋅ =
3 x 3x
1
1
=
f ′(81) =
3 ⋅ 81 243
1
(– sin x ) = – tan x
cos x
⎛π⎞
⎛π⎞
f ′ ⎜ ⎟ = − tan ⎜ ⎟ = −1 .
⎝4⎠
⎝4⎠

f ′( x) =

Section 6.1

347

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15. Let u = 2x + 1 so du = 2 dx.
1
1 1
∫ 2 x + 1 dx = 2 ∫ u du
1
1
= ln u + C = ln 2 x + 1 + C
2
2

22. Let u = 2t 2 + 4t + 3 so du = (4t + 4)dt .
t +1

1
1
ln u + C = ln 2t 2 + 4t + 3 + C
4
4

=

16. Let u = 1 – 2x so du = –2dx.
1
1 1
∫ 1 – 2 x dx = – 2 ∫ u du
1
1
= – ln u + C = – ln 1 – 2 x + C
2
2
17. Let u = 3v 2 + 9v so du = 6v + 9.
6v + 9
1
∫ 3v2 + 9v dv = ∫ u du = ln u + C

18. Let u = 2 z 2 + 8 so du = 4z dz.
1 1
z
∫ 2 z 2 + 8 dz = 4 ∫ u du
1
1
= ln u + C = ln 2 z 2 + 8 + C
4
4
19. Let u = ln x so du =

)

⎡1

2
∫0 2t 2 + 4t + 3dt = ⎢⎣ 4 ln 2t + 4t + 3 ⎥⎦0
1
1
9
1
ln 9 – ln 3 = ln 4 = ln 4 3 = ln 3
4
4
3
4

=

23. By long division,

x2
1
∫ x − 1 dx = ∫ x dx + ∫ 1 dx + ∫ x − 1 dx
x2
=
+ x + ln x − 1 + C
2

so

–1

∫ x(ln x)2 dx = – ∫ u

1
dx
x

1
dx .
x

–2

21. Let u = 2 x5 + π so du = 10 x 4 dx .
x4

1 1
∫ 2 x5 + π dx = 10 ∫ u du
1
1
= ln u + C = ln 2 x5 + π + C
10
10
3

∫0

x4

3

⎡1

dx = ⎢ ln 2 x5 + π ⎥
5
10

⎦0
2x + π

1
486 + π
= [ln(486 + π) – ln π] = ln 10
≈ 0.5048
10
π

348

Section 6.1

x2 + x x 3
3
= + +
2 x − 1 2 4 4(2 x − 1)

so

3
3
x2 + x
x
∫ 2 x − 1 dx = ∫ 2 dx + ∫ 4 dx + ∫ 4(2 x − 1) dx

25. By long division,
x4
256
= x3 − 4 x 2 + 16 x − 64 +
x+4
x+4

du

1
1
+C =
+C
u
ln x

=

24. By long division,

3
1
x2 3
dx
+ x+ ∫
4 4
4 2x −1
Let u = 2 x − 1 ; then du = 2dx . Hence
1
1 1
1
∫ 2 x − 1 dx = 2 ∫ u du = 2 ln u + C
1
= ln 2 x − 1 + C
2
2
x +x
x2 3
3
+ x + ln 2 x − 1 + C
and ∫
dx =
2x −1
4 4
8

= u 2 + C = (ln x) 2 + C

20. Let u = ln x, so du =

x2
1
= x +1+
x −1
x −1

=

2 ln x
dx = 2∫ udu
x

1

t +1

1

= ln 3v 2 + 9v + C

(

1 1

∫ 2t 2 + 4t + 3 dt = 4 ∫ u du

so

x4
∫ x + 4 dx =

∫x
=

3

dx − ∫ 4 x 2 dx + ∫ 16 xdx − ∫ 64dx + 256∫

x 4 4 x3

+ 8 x 2 − 64 x + 256 ln x + 4 + C
4
3

26. By long division,

1
dx
x+4

x3 + x 2
4
= x2 − x + 2 −
so
x+2
x+2

x3 + x 2
1
dx = ∫ x 2 dx − ∫ xdx + ∫ 2dx − 4∫
dx
x+2
x+2
x3 x 2
=

+ 2 x − 4 ln x + 2 + C
3
2

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27. 2 ln( x + 1) – ln x = ln( x + 1) 2 – ln x = ln
28.

( x + 1)2
x

1
1
ln( x – 9) + ln x = ln x – 9 – ln x
2
2
x–9
x–9
= ln
= ln
x
x

1
31. ln y = ln( x + 11) – ln( x3 – 4)
2
1 dy
1
1
1
=
⋅1 – ⋅
⋅ 3x2
3
y dx x + 11
2 x –4
=

⎡ 1
3x2 ⎤
dy
= y⋅⎢

3
dx
⎣⎢ x + 11 2( x – 4) ⎦⎥

29. ln(x – 2) – ln(x + 2) + 2 ln x
= ln( x – 2) – ln( x + 2) + ln x 2 = ln

1
3x2

x + 11 2( x3 – 4)

x 2 ( x – 2)
x+2

=

30. ln( x 2 – 9) – 2 ln( x – 3) – ln( x + 3)
= ln( x 2 − 9) − ln( x − 3)2 − ln( x + 3)

=–

x2 – 9

1
= ln
= ln
2
x–3
( x – 3) ( x + 3)

x + 11 ⎡ 1
3x2 ⎤

3
x3 – 4 ⎣⎢ x + 11 2( x – 4) ⎦⎥
x3 + 33x 2 + 8
2( x3 – 4)3 / 2

32. ln y = ln( x 2 + 3x ) + ln( x – 2) + ln( x 2 + 1)
1 dy
2x + 3
1
2x
=
+
+
y dx x 2 + 3 x x – 2 x 2 + 1
dy
⎛ 2x + 3
1
2x ⎞
4
3
2
= ( x 2 + 3 x)( x – 2)( x 2 + 1) ⎜
+
+
⎟ = 5 x + 4 x –15 x + 2 x – 6
2
2
dx
x

2
+
+
x
3
x
x
1

1
1
ln( x + 13) – ln( x – 4) – ln(2 x + 1)
2
3
1 dy
1
1
2
=

y dx 2( x + 13) x – 4 3(2 x + 1)

33. ln y =

dy
x + 13
1
1
2
10 x 2 + 219 x – 118
=

=

dx ( x – 4) 3 2 x + 1 ⎢⎣ 2( x + 13) x – 4 3(2 x + 1) ⎥⎦
6( x – 4)2 ( x + 13)1/ 2 (2 x + 1)4 / 3
2
1
ln( x 2 + 3) + 2 ln(3 x + 2) – ln( x + 1)
3
2
1 dy 2 2 x
2⋅3
1
= ⋅
+

y dx 3 x 2 + 3 3x + 2 2( x + 1)

34. ln y =

dy ( x 2 + 3)2 / 3 (3 x + 2)2
=
dx
x +1

⎡ 4x
6
1 ⎤ (3 x + 2)(51x3 + 70 x 2 + 97 x + 90)

+
⎢ 2
⎥=
6( x 2 + 3)1/ 3 ( x + 1)3 / 2
⎢⎣ 3( x + 3) 3 x + 2 2( x + 1) ⎥⎦

35.

36.

y = ln x is reflected across the y-axis.

The y-values of y = ln x are multiplied by
since ln x =

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1
,
2

1
ln x.
2

Section 6.1

349

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37.

y = ln x is reflected across the x-axis since
⎛1⎞
ln ⎜ ⎟ = – ln x.
⎝ x⎠
38.

42. Let r(x) = rate of transmission
1
= kx 2 ln = −kx 2 ln x.
x
⎛1⎞
r ′( x) = −2kx ln x − kx 2 ⎜ ⎟ = − kx(2 ln x + 1)
⎝ x⎠
1
1
r ′( x) = 0 if ln x = − , or − ln x = , so
2
2
1 1
ln = .
x 2
1
1
ln1.65 ≈ , so x ≈
≈ 0.606.
1.65
2
⎛ 1⎞
r ′′( x) = −k (2 ln x + 1) − kx ⎜ 2 ⋅ ⎟ = –k(2 ln x + 3)
⎝ x⎠
′′
r (0.606) ≈ −2k < 0 since k > 0, so
x ≈ 0.606 gives the maximum rate of
transmission.
43. ln 4 > 1

so ln 4m = m ln 4 > m ⋅1 = m
Thus x > 4m ⇒ ln x > m
so lim ln x = ∞
x →∞

y = ln x is shifted two units to the right.
1
so z → ∞ as x → 0+
x
⎛1⎞
Then lim ln x = lim ln ⎜ ⎟ = lim (– ln z )
+
→∞
z
⎝ z ⎠ z →∞
x →0
= – lim ln z = – ∞

44. Let z =

39.

z →∞

45.
y = ln cos x + ln sec x
= ln cos x + ln

1
cos x

⎛ π π⎞
= ln cos x − ln cos x = 0 on ⎜ − , ⎟
⎝ 2 2⎠

40. Since ln is continuous,
sin x
sin x
lim ln
= ln lim
= ln1 = 0
x
x →0
x →0 x
41. The domain is ( 0, ∞ ) .
⎛1⎞
f ′( x) = 4 x ln x + 2 x 2 ⎜ ⎟ − 2 x = 4 x ln x
⎝x⎠
f ' ( x ) = 0 if ln x = 0 , or x = 1 .

f ' ( x ) < 0 for x < 1 and f ' ( x ) > 0 for x > 1
so f(1) = –1 is a minimum.

350

Section 6.1

x

1

x1

1

1

x1

1

1

x1

∫1/ 3 t dt = 2∫1 t dt
x1

∫1/ 3 t dt + ∫1 t dt = 2∫1 t dt
∫1/ 3 t dt = ∫1 t dt
–∫

1/ 3 1

1

t

dt = ∫

x1

1

t

dt

1
– ln = ln x
3
ln 3 = ln x
x=3

46. a.

1 1
<
for t > 1,
t
t
x1
x 1
x
so ln x = ∫ dt < ∫
dt = ∫ t –1/ 2 dt
1 t
1 t
1
x

= ⎡⎣ 2 t ⎤⎦ = 2( x –1)
1

so ln x < 2( x – 1)

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b. If x > 1, 0 < ln x < 2( x – 1) ,

x →∞

and lim

x →∞

∫π 4

ln x
2( x + 1)
≤ lim
=0
x
x
x →∞

1, 000, 000
≈ 72,382
ln1, 000, 000
c

⎛ ax – b ⎞
⎛ ax − b ⎞
f ( x) = ln ⎜
⎟ = c ln ⎜

+
ax
b

⎝ ax + b ⎠
=

a 2 – b2
[ln(ax – b) – ln(ax + b)]
2ab

f ′( x) =
=

f ′(1) =

b.

a 2 – b2 ⎡ a
a ⎤

2ab ⎣ ax – b ax + b ⎦⎥

a 2 − b2
2ab

2ab
a 2 – b2
⎢ (ax − b)(ax + b) ⎥ = 2 2

⎦ a x – b2

a 2 − b2
a 2 − b2

f ′( x) = cos 2 u ⋅

3
4

π

= ⎡⎣ln tan x ⎤⎦π 3 = ln tan π 3 − ln tan π 4
4

cos x

= ln 1 + sin x + C = ln(1 + sin x) + C

(since 1 + sin x ≥ 0 for all x ).
4

1

= 3cos 2 (0) = 3

dx

2πx

4
dx = ⎡⎢ π ln x 2 + 4 ⎤⎥

1
x +4
= π ln 20 − π ln 5 = π ln 4 ≈ 4.355
4

∫1

2

x2
x2 1
– ln x =
– ln x
4
4 2
dy 2 x 1 1 x 1
=
– ⋅ = –
dx
4 2 x 2 2x

54. y =

2

∫1
2

1

x + x –1
2 ⋅1 + 1
f ′(1) = cos [ln(1 + 1 –1)] ⋅
2
1 + 1 –1

x2 + 4

= π ln x 2 + 4 + C

=∫

2

2πx

Let u = x 2 + 4 so du = 2x dx.
2πx
1
∫ x2 + 4 dx = π∫ u du = π ln u + C

du
dx
2x + 1

4

1

53. V = 2π∫ xf ( x)dx = ∫

L=

2

1

∫ 1 + sin x dx = ∫ u du = ln u + C

=1

= cos 2 [ln( x 2 + x –1)] ⋅
2

= ⎡−
⎣ ln cos x + ln sin x ⎤⎦π

52. Let u = 1 + sin x ; then du = cos x dx so that

⎡ 1
1
1 ⎤ 1
⎥⋅
= lim ⎢
+
+ ⋅⋅⋅ +
n⎥ n
n→∞ ⎢1 + 1 1 + 2
+
1
n
n⎦
⎣ n
n ⎛

21
1
1
⎟ ⋅ = ∫ dx = ln 2 ≈ 0.693
= lim ∑ ⎜
i
1
x
n→∞ i =1 ⎜ 1 + ⎟ n
n⎠

49. a.

3 sec x csc x dx

= ln( 3) − ln1 = 0.5493 − 0 = 0.5493

ln x
= 0.
x

1
1⎤
⎡ 1
+
+ ⋅⋅⋅ + ⎥
47. lim ⎢
2n ⎦
n →∞ ⎣ n + 1 n + 2

48.

π

π

ln x 2( x – 1)
<
.
so 0 <
x
x

Hence 0 ≤ lim

51. From Ex 10,

2

2

2
⎛ dy ⎞
⎛x 1 ⎞
1 + ⎜ ⎟ dx = ∫ 1 + ⎜ – ⎟ dx
1
⎝ dx ⎠
⎝ 2 2x ⎠
2

2⎛ x
1 ⎞
⎛x 1 ⎞
⎜ + ⎟ dx = ∫1 ⎜ + ⎟ dx
⎝ 2 2x ⎠
⎝ 2 2x ⎠
2

1 ⎡ x2
1⎡
⎛1
⎞⎤
= ⎢ + ln x ⎥ = ⎢ 2 + ln 2 − ⎜ + ln1⎟ ⎥
2 ⎢⎣ 2
⎝2
⎠⎦
⎥⎦1 2 ⎣
3 1
= + ln 2 ≈ 1.097
4 2

50. From Ex 9,
π

∫0 3 tan x dx = ⎡−
⎣ ln cos x

π

⎤⎦ 0 3

= ln cos 0 − ln cos π 3
⎛ 1 ⎞
= ln(1) − ln(0.5) = ln ⎜

⎝ 0.5 ⎠
= ln 2 ≈ 0.69315

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Section 6.1

351

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55.

b.

c.

1 1
1
+ + ⋅⋅⋅ + = the lower approximate area
2 3
n
1
1
1 + + ⋅⋅⋅ +
= the upper approximate area
2
n –1
ln n = the exact area under the curve

58. a.

Thus,
1 1
1
1 1
1
+ + ⋅⋅⋅ + < ln n < 1 + + + ⋅⋅⋅ +
.
2 3
n
2 3
n −1
y1

56.

ln y – ln x
=
y–x

∫1

t

dt – ∫

x1

1

y–x

= the average value of

t

dt

∫x t dt
y–x

1
on [x, y].
t

1
Since is decreasing on the interval [x, y], the
t
average value is between the minimum value of
1
1
and the maximum value of .
y
x

57. a.

1 + 1.5sin x

(1.5 + sin x)2
On [0,3π ], f ′′( x) = 0 when x ≈ 3.871,
5.553.
Inflection points are (3.871, –0.182),
(5.553, –0.182).

∫0

ln(1.5 + sin x)dx ≈ 4.042

sin(ln x)
x
On [0.1, 20], f ′( x) = 0 when x = 1.
Critical points: 0.1, 1, 20
f(0.1) ≈ –0.668, f(1) = 1, f(20) ≈ –0.989
On [0.1, 20], the maximum value point is
(1, 1) and minimum value point is
(20, –0.989).
f ′( x) = −

b. On [0.01, 0.1], f ′( x) = 0 when x ≈ 0.043.
f(0.01) ≈ –0.107, f(0.043) ≈ –1
On [0.01, 20], the maximum value point is
(1, 1) and the minimum value point is
(0.043, –1).

y1

=

f ′′( x) = −

20

c.

∫0.1 cos(ln x)dx ≈ −8.37

a.

∫0 ⎢⎣ x ln ⎜⎝ x ⎟⎠ − x

59.

1
cos x
⋅ cos x =
1.5 + sin x
1.5 + sin x
f ′( x) = 0 when cos x = 0.

f ′( x) =

π 3π 5π
Critical points: 0, , , , 3π
2 2 2
f(0) ≈ 0.405,
⎛π⎞
⎛ 3π ⎞
f ⎜ ⎟ ≈ 0.916, f ⎜ ⎟ ≈ −0.693,
⎝2⎠
⎝ 2 ⎠
⎛ 5π ⎞
f ⎜ ⎟ ≈ 0.916, f (3π) ≈ 0.405.
⎝ 2 ⎠
On [0,3π ], the maximum value points are
⎛π
⎞ ⎛ 5π

⎜ , 0.916 ⎟ , ⎜ , 0.916 ⎟ and the minimum
2
2

⎠ ⎝

⎛ 3π

value point is ⎜ , −0.693 ⎟ .
2

1⎡

⎛1⎞

2

5
⎛ 1 ⎞⎤
ln ⎜ ⎟ ⎥ dx =
≈ 0.139
x
36
⎝ ⎠⎦

b. Maximum of ≈ 0.260 at x ≈ 0.236
60.

a.

1

∫0 [ x ln x −

x ln x]dx =

7
≈ 0.194
36

b. Maximum of ≈ 0.521 at x ≈ 0.0555

352

Section 6.1

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6.2 Concepts Review
1.

11.

f ′( z ) = 2( z – 1) > 0 for z > 1
f(z) is increasing at z = 1 because f(1) = 0 and
f(z) > 0 for z > 1. Therefore, f(z) is strictly
increasing on z ≥ 1 and so it has an inverse.

12.

f ′( x) = 2 x + 1 > 0 for x ≥ 2 . f(x) is strictly
increasing on x ≥ 2 and so it has an inverse.

13.

f ′( x) = x 4 + x 2 + 10 > 0 for all real x. f(x) is
strictly increasing and so it has an inverse.

f ( x1 ) ≠ f ( x2 )

2. x; f –1 ( y )
3. monotonic; strictly increasing; strictly decreasing
4. ( f –1 )′( y ) =

1
f ′( x)

Problem Set 6.2
14.
1. f(x) is one-to-one, so it has an inverse.

Since f (4) = 2, f −1 (2) = 4 .
2. f(x) is one-to-one, so it has an inverse.

Since f(1) = 2, f −1 (2) = 1 .
3. f(x) is not one-to-one, so it does not have an
inverse.
4. f(x) is not one-to-one, so it does not have an
inverse.
5. f(x) is one-to-one, so it has an inverse.

Since f(–1.3) ≈ 2, f −1 (2) ≈ −1.3 .
6. f(x) is one-to-one, so it has an inverse. Since
1
⎛1⎞
f ⎜ ⎟ = 2, f −1 (2) = .
2
2
⎝ ⎠
7.

8.

9.

f ′( x) = –5 x 4 – 3x 2 = –(5 x 4 + 3 x 2 ) < 0 for all
x ≠ 0. f(x) is strictly decreasing at x = 0 because
f(x) > 0 for x < 0 and f(x) < 0 for x > 0. Therefore
f(x) is strictly decreasing for x and so it has an
inverse.
f ′( x) = 7 x 6 + 5 x 4 > 0 for all x ≠ 0.
f(x) is strictly increasing at x = 0 because f(x) > 0
for x > 0 and f(x) < 0 for x < 0. Therefore f(x) is
strictly increasing for all x and so it has an
inverse.
f ′(θ ) = – sin θ < 0 for 0 < θ < π

f (θ) is decreasing at θ = 0 because f(0) = 1 and
f(θ) < 1 for 0 < θ < π . f(θ) is decreasing at
θ = π because f( π ) = –1 and f(θ) > –1 for
0 < θ < π . Therefore f(θ) is strictly decreasing
on 0 ≤ θ ≤ π and so it has an inverse.
10.

f ′( x) = – csc 2 x < 0 for 0 < x <

f(x) is decreasing on 0 < x <

1

r

r

1

f (r ) = ∫ cos 4 tdt = – ∫ cos 4 tdt
π
f ′(r ) = – cos 4 r < 0 for all r ≠ k π + , k any
2
integer.
π
f(r) is decreasing at r = k π + since f ′(r ) < 0
2
on the deleted neighborhood
π
π

⎜ k π + − ε , k π + + ε ⎟ . Therefore, f(r) is
2
2

strictly decreasing for all r and so it has an
inverse.

15. Step 1:
y=x+1
x=y–1

Step 2: f –1 ( y ) = y – 1
Step 3: f –1 ( x) = x – 1
Check:
f –1 ( f ( x)) = ( x + 1) – 1 = x
f ( f –1 ( x)) = ( x – 1) + 1 = x

16. Step 1:
x
y = – +1
3
x
– = y –1
3
x = –3(y – 1) = 3 – 3y

Step 2: f –1 ( y ) = 3 – 3 y
Step 3: f –1 ( x) = 3 – 3 x
Check:
⎛ x ⎞
f –1 ( f ( x)) = 3 – 3 ⎜ – + 1⎟ = 3 + ( x – 3) = x
⎝ 3 ⎠
–(3 – 3 x)
+ 1 = (–1 + x) + 1 = x
f ( f –1 ( x)) =
3

π
2

π
and so it has an
2

inverse.
Instructor’s Resource Manual

Section 6.2

353

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

17. Step 1:
y = x + 1 (note that y ≥ 0 )
x + 1 = y2

x = 2+

x = y 2 – 1, y ≥ 0

Step 2: f –1 ( y ) = y 2 – 1, y ≥ 0
Step 3: f
Check:
f

–1

–1

x–2=

1
y2

1
y2

,y>0

Step 2: f –1 ( y ) = 2 +

( x) = x – 1, x ≥ 0
2

Step 3: f –1 ( x) = 2 +

( f ( x)) = ( x + 1) – 1 = ( x + 1) – 1 = x
2

f ( f –1 ( x)) = ( x 2 – 1) + 1 = x 2 = x = x

18. Step 1:
y = – 1 – x (note that y ≤ 0 )
1– x = – y

f –1 ( f ( x)) = 2 +

1

(

1
x –2

)

2

= 2+

1

( x 1–2 )

1
⎛2+

1 ⎞–2

x2 ⎠

=

1
= x2
⎛ 1 ⎞
⎜ 2⎟
⎝x ⎠

Step 2: f

( y) = 1 – y , y ≤ 0

Step 3: f
Check:

–1

( x) = 1 – x 2 , x ≤ 0

2

f –1 ( f ( x)) = 1 – (– 1 – x ) 2 = 1 – (1 – x) = x
f ( f –1 ( x)) = – 1 – (1 – x 2 ) = – x 2 = – x

= –(–x) = x

21. Step 1:
y = 4 x 2 , x ≤ 0 (note that y ≥ 0 )
x2 =

y
4

x=–

y
y
=−
, negative since x ≤ 0
4
2

y
2
x
Step 3: f –1 ( x) = −
2
Check:

Step 2: f –1 ( y ) = −

19. Step 1:
1
x–3
1
x–3= –
y
y=–

x = 3–

1
y

Step 2: f

f –1 ( f ( x)) = –
–1

f(f

1
x

( x)) = –

1
– x1–3

1

(3 – )
1
x

x⎞
x
( x)) = 4 ⎜⎜ –
⎟⎟ = 4 ⋅ = x
4
⎝ 2 ⎠

y = ( x – 3)2 , x ≥ 3 (note that y ≥ 0 )
= 3 + ( x – 3) = x

x–3=

y

x = 3+ y

1
=–
=x
1

–3
x

20. Step 1:
1
(note that y > 0)
y=
x–2
1
y2 =
x–2

Section 6.2

–1

22. Step 1:

Check:
f –1 ( f ( x)) = 3 –

4 x2
= – x 2 = – x = –(– x) = x
2
2

1
( y) = 3 –
y

Step 3: f –1 ( x) = 3 –

354

,x>0

x2

= x =x

–1

f(f

1

= 2 + (x – 2) = x

1 – x = (– y ) 2 = y 2

–1

,y>0

y2

Check:

f ( f –1 ( x)) =

x = 1 – y2 , y ≤ 0

1

Step 2: f –1 ( y ) = 3 + y
Step 3: f –1 ( x) = 3 + x
Check:
f –1 ( f ( x)) = 3 + ( x – 3)2 = 3 + x – 3
= 3 + ( x – 3) = x
f ( f –1 ( x)) = [(3 + x ) – 3]2 = ( x )2 = x

Instructor’s Resource Manual

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

23. Step 1:
y = ( x –1)
x –1 = 3 y

Step 3: f –1 ( x) =

x = 1+ 3 y

Step 3: f –1 ( x) = 1 + 3 x

f(f

–1

–1

( f ( x)) = 1 + 3 ( x –1) = 1 + ( x –1) = x
3

( x)) = [(1 + x ) –1] = ( x ) = x
3

3

3

3

24. Step 1:
y = x5 / 2 , x ≥ 0
x= y

Step 3: f –1 ( x) = x 2 / 5
Check:
f ( f –1 ( x)) = ( x 2 / 5 )5 / 2 = x

25. Step 1:
x –1
y=
x +1
xy + y = x –1
x – xy = 1 + y
1+ y
x=
1– y

3 ⎤1/ 3

( )

1/ 3

⎥⎦

=

3⎤

1 – ⎢ xx +–11 ⎥

x +1 + x – 1 2x
=
=
=x
x +1 – x +1 2

1+
1–

x –1
x +1
x –1
x +1

⎛ 2 x1/ 3 ⎞
=⎜
⎟ = ( x1/ 3 )3 = x
⎜ 2 ⎟

27. Step 1:
x3 + 2
y=
x3 + 1
x3 y – x3 = 2 – y
x3 =

2– y
y –1
1/ 3

⎛2– y⎞
x=⎜

⎝ y –1 ⎠

1+ y
( y) =
1− y

1/ 3

1+ x
1– x

⎛2– y⎞
Step 2: f –1 ( y ) = ⎜

⎝ y –1 ⎠

x –1
x +1
x –1
x +1

⎛2– x⎞
Step 3: f –1 ( x) = ⎜

⎝ x –1 ⎠
Check:

Check:

1/ 3

f –1 ( f ( x)) =
( x)) =

1+
1–
1+ x
1– x
1+ x
1– x

=

x + 1 + x –1 2 x
=
=x
x +1 – x +1 2

1 + x –1 + x 2 x
=
=
=x
+1 1+ x +1 – x 2
–1

26. Step 1:
3

⎛ x –1 ⎞
y=⎜

⎝ x +1⎠
x –1
y1/ 3 =
x +1

xy1/ 3 + y1/ 3 = x –1
x – xy1/ 3 = 1 + y1/ 3
x=

( xx+–11 )

x3 y + y = x3 + 2

Step 3: f –1 ( x) =

f(f

1+ ⎢

f –1 ( f ( x)) =

3

f –1 ( f ( x)) = ( x5 / 2 ) 2 / 5 = x

–1

1 – x1/ 3

3

Step 2: f –1 ( y ) = y 2 / 5

Step 2: f

1 + x1/ 3

⎛ 1+ x1/ 3 – 1 ⎞
3
⎛ 1 + x1/ 3 – 1 + x1/ 3 ⎞
⎜ 1– x1/ 3

–1
f ( f ( x)) = ⎜

⎟ = ⎜⎜
1/ 3
1 + x1/ 3 + 1 – x1/ 3 ⎟⎠
⎜⎜ 1+ x + 1 ⎟⎟

⎝ 1– x1/ 3

2/5

–1

1 – y1/ 3

Check:

Step 2: f –1 ( y ) = 1 + 3 y

Check: f

1 + y1/ 3

Step 2: f –1 ( y ) =

3

1 + y1/ 3
1 – y1/ 3

Instructor’s Resource Manual

1/ 3

⎛ 2 – x3 + 2 ⎞

x3 +1 ⎟
f –1 ( f ( x)) = ⎜

3
⎜⎜ x + 2 –1 ⎟⎟
⎝ x3 +1

1/ 3

⎛ 2 x3 + 2 – x3 – 2 ⎞
=⎜

⎜ x3 + 2 – x3 –1 ⎟

1/ 3

⎛ x3 ⎞
=⎜ ⎟
⎜ 1 ⎟
⎝ ⎠

=x

( )

3

⎡ 2– x 1/ 3 ⎤
x +2
⎢⎣ x –1
⎥⎦ + 2 2–
–1
= x –1
f ( f ( x)) =
3
2– x + 1
⎡ 2– x 1/ 3 ⎤
x –1
1
+
⎢⎣ x –1
⎥⎦
2 – x + 2x – 2 x
=
= =x
2 – x + x –1 1

( )

Section 6.2

355

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

28. Step 1:
⎛ x3 + 2 ⎞
y=⎜

⎜ x3 + 1 ⎟

1/ 5

y

29. By similar triangles,

5

This gives

x3 + 2

=

V=

3
V
27
h3 =

x3 + 1

x3 y1/ 5 + y1/ 5 = x3 + 2
x3 y1/ 5 – x3 = 2 – y1/ 5
x3 =

h = 33

2 – y1/ 5
y1/ 5 – 1

⎛2– y
Step 2: f –1 ( y ) = ⎜

⎜ y1/ 5 – 1 ⎟

1/ 3

3

=

4π h3
27

V

v
v2
v2
H = s (v0 / 32) = v0 0 − 16 0 = 0
32
322 64

Check:

v02 = 64 H

1/ 3
5 ⎤1/ 5 ⎫

⎪ 2 – ⎢⎛ x3 + 2 ⎞ ⎥ ⎪
⎜ 3 ⎟
⎪⎪
⎢⎣⎝ x +1 ⎠ ⎥⎦ ⎪⎪
–1
f ( f ( x)) = ⎨

1/ 5
⎪ ⎡ ⎛ 3 ⎞5 ⎤

⎪ ⎢ ⎜ x 3+ 2 ⎟ ⎥ – 1 ⎪
⎪⎩ ⎢⎣⎝ x +1 ⎠ ⎥⎦
⎪⎭

v0 = 8 H

31.

f ′( x) = 4 x + 1; f ′( x) > 0 when x > −

1
and
4

1
f ′( x) < 0 when x < − .
4

1/ 3
1/ 3

⎛ 2 x3 + 2 – x3 – 2 ⎞
=⎜

3
⎜ 3

⎝ x + 2 – x –1 ⎠

1/ 3

⎛ x3 ⎞
=⎜ ⎟
⎜ 1 ⎟
⎝ ⎠

)

s (t ) = v0t − 16t 2 . The ball then reaches a height
of

⎛ 2 – x1/ 5 ⎞
( x) = ⎜

⎜ x1/ 5 – 1 ⎟

⎛ 2 – x3 + 2 ⎞

x3 +1 ⎟
=⎜

3
⎜⎜ x + 2 – 1 ⎟⎟
⎝ x3 +1

(

v
t = 0 . The position function is
32

1/ 5 ⎞1/ 3

Step 3: f

=

π 4h 2 / 9 h

30. v = v0 − 32t
v = 0 when v0 = 32t , that is, when

1/ 3

⎛ 2 – y1/ 5 ⎞
x=⎜

⎜ y1/ 5 – 1 ⎟

–1

π r 2h

r 4
2h
= . Thus, r =
h 6
3

=x
5

⎧⎡

1/ 3 ⎤ 3
⎪ ⎢⎛ 2– x1/ 5 ⎞ ⎥ + 2 ⎪

⎪⎪ ⎢⎝ x1/ 5 –1 ⎠ ⎥
⎪⎪

f ( f –1 ( x)) = ⎨ ⎣

3
⎪ ⎡⎛ 1/ 5 ⎞1/ 3 ⎤

2– x
⎪ ⎢⎜ 1/ 5 ⎟ ⎥ + 1 ⎪
⎩⎪ ⎣⎢⎝ x –1 ⎠ ⎦⎥
⎭⎪

1⎤

The function is decreasing on ⎜ −∞, − ⎥ and
4⎦

⎡ 1 ⎞
increasing on ⎢ − , ∞ ⎟ . Restrict the domain to
⎣ 4 ⎠
1⎤

⎡ 1 ⎞
⎜ −∞, − ⎥ or restrict it to ⎢ − , ∞ ⎟ .
4⎦
⎣ 4 ⎠

Then f −1 ( x) =
f −1 ( x ) =

1
(−1 − 8 x + 33) or
4

1
(−1 + 8 x + 33).
4

5

⎛ 2– x1/ 5

5
⎛ 2 – x1/ 5 + 2 x1/ 5 – 2 ⎞
⎜ x1/ 5 –1 + 2 ⎟
=⎜
=

1/ 5
⎜ 2 – x1/ 5 + x1/ 5 – 1 ⎟
⎜⎜ 2– x + 1 ⎟⎟

1/
5
⎝ x –1

5

⎛ x1/ 5 ⎞
=⎜
⎟ =x
⎜ 1 ⎟

356

Section 6.2

Instructor’s Resource Manual

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

32.

f ′( x) = 2 x − 3; f ′( x) > 0 when x >

3
2

36.

( f −1 )′(3) ≈

1
2

3
and f ′( x) < 0 when x < .
2

3⎤
The function is decreasing on ⎜ −∞, ⎥ and
2⎦

⎡3 ⎞
increasing on ⎢ , ∞ ⎟ . Restrict the domain to
⎣2 ⎠

3⎤
⎡3 ⎞
⎜ −∞, ⎥ or restrict it to ⎢ , ∞ ⎟ . Then
2⎦
⎣2 ⎠

1
(3 − 4 x + 5) or
2
1
f −1 ( x) = (3 + 4 x + 5).
2
f −1 ( x ) =

37.

f ′( x) = 15 x 4 + 1 and y = 2 corresponds to x = 1,

so ( f −1 )′(2) =

33.
38.

f ′( x) = 5 x 4 + 5 and y = 2 corresponds to x = 1,

so ( f −1 )′(2) =

39.

1
1
1
=
= .
f ′(1) 15 + 1 16

1
1
1
=
=
f ′(1) 5 + 5 10

π
,
4
1
1
1
⎛π⎞
so ( f −1 )′(2) =
=
= cos 2 ⎜ ⎟
2
π
π
2
⎝4⎠
f′ 4
2sec 4
f ′( x) = 2sec2 x and y = 2 corresponds to x =

( )

(f

34.

−1

)′(3) ≈

=

1
3

( f −1 )′(3) ≈ −

1
2

40.

( )

1
.
4

f ′( x) =

1
2 x +1

so ( f −1 )′(2) =

and y = 2 corresponds to x = 3,

1
= 2 3 +1 = 4 .
f ′(3)

41. ( g –1 D f –1 )(h( x)) = ( g –1 D f –1 )( f ( g ( x)))
= g –1 D [ f –1 ( f ( g ( x)))] = g –1 D [ g ( x)] = x
Similarly,
h(( g –1 D f –1 )( x)) = f ( g (( g –1 D f –1 )( x)))
= f ( g ( g –1 ( f –1 ( x )))) = f ( f –1 ( x)) = x

Thus h –1 = g –1 D f –1

35.

( f −1 )′(3) ≈ −

1
3

Instructor’s Resource Manual

Section 6.2

357

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

42. Find f −1 ( x) :
y=

44. a.

1
1
, x=
x
y

f −1 ( y ) =

1
y

f −1 ( x ) =

1
x

Find g −1 ( x) :
y = 3x + 2
y−2
x=
3
y−2
g −1 ( y ) =
3
x

2
g −1 ( x) =
3

c.

1
3x + 2

(1)

−2
⎛1⎞
h −1 ( x) = g −1 ( f −1 ( x)) = g −1 ⎜ ⎟ = x
3
⎝ x⎠
⎛ 1 ⎞ (3x + 2) − 2 3x
h −1 (h( x)) = h −1 ⎜
=
=x
⎟=
3
3
⎝ 3x + 2 ⎠

dy − b
cy − a

f −1 ( x) = −

dx − b
cx − a

( )

+(d 2 − bc) x − bd = 0
(ac + dc) x 2 + (d 2 − a 2 ) x + (− ab − bd ) = 0
Setting the coefficients equal to 0 gives three
requirements:
(1) a = –d or c = 0
(2) a = ±d
(3) a = –d or b = 0

( )

( )

If a = d, then f = f −1 requires b = 0 and

43. f has an inverse because it is monotonic
(increasing):

ax
= x . If a = –d, there are
d
no requirements on b and c (other than
c = 0, so f ( x) =

f ′( x) = 1 + cos 2 x > 0

a.

( f −1 )′( A) =

b.

( f −1 )′( B ) =

1

( )

f ′ π2
1

=

( )

f ′ 56π

If f = f −1 , then for all x in the domain we
have:
ax + b dx − b
+
=0
cx + d cx − a
(ax + b)(cx – a) + (dx – b)(cx + d) = 0
acx 2 + (bc − a 2 ) x − ab + dcx 2

⎛ 1 −2⎞
1
1
⎟=
h(h −1 ( x)) = h ⎜ x
=
=x
1
⎜ 3 ⎟ ⎡ 1 − 2⎤ + 2
x

⎠ ⎣ x

c.

f −1 ( y ) = −

b. If bc – ad = 0, then f(x) is either a constant
function or undefined.

h( x ) = f ( g ( x)) = f (3x + 2) =

=

ax + b
cx + d
cxy + dy = ax + b
(cy – a)x = b – dy
b − dy
dy − b
x=
=−
cy − a
cy − a
y=

=

1

( )

1 + cos 2 π2
1

( )

1 + cos 2 56π

bc − ad ≠ 0 ). Therefore, f = f −1 if a = –d
or if f is the identity function.

=1

=

1

45.

7
4

2
7

( f −1 )′(0) =

1
1
1
=
=
2
f ′(0)
2
1 + cos (0)
1 −1

∫0 f

( y ) dy = (Area of region B)

= 1 – (Area of region A)
1
2 3
= 1 − ∫ f ( x) dx = 1 − =
0
5 5

358

Section 6.2

Instructor’s Resource Manual

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

46.

a

∫0

f ( x)dx = the area bounded by y = f(x), y = 0,

47. Given p > 1, q > 1,

and x = a [the area under the curve].
b

f –1 ( y )dy = the area bounded by x = f –1 ( y )

solving

x = 0, and y = b.
ab = the area of the rectangle bounded by x = 0,
x = a, y = 0, and y = b.
Case 1: b > f(a)

1
=
p –1

∫0

1 1
+ = 1, and f ( x) = x p –1 ,
p q

1 1
q
+ = 1 for p gives p =
, so
p q
q –1
1
q
–1
q –1

=

1

=

⎡ q –( q –1) ⎤
⎣⎢ q –1 ⎦⎥

q –1
= q – 1.
1
1

Thus, if y = x p –1 then x = y p –1 = y q –1 , so
f –1 ( y ) = y q –1.

By Problem 44, since f ( x) = x p −1 is strictly
a

b

0

0

increasing for p > 1, ab ≤ ∫ x p –1dx + ∫ y q –1dy
a

The area above the curve is greater than the area
of the part of the rectangle above the curve, so
the total area represented by the sum of the two
integrals is greater than the area ab of the
rectangle.
Case 2: b = f(a)

b

⎡xp ⎤
⎡ yq ⎤
ab ≤ ⎢ ⎥ + ⎢ ⎥
⎣⎢ p ⎦⎥ 0 ⎣⎢ q ⎥⎦ 0
a p bq
ab ≤
+
p
q

6.3 Concepts Review
1. increasing; exp
2. ln e = 1; 2.72
3. x; x

The area represented by the sum of the two
integrals = the area ab of the rectangle.
Case 3: b < f(a)

4. e x ; e x + C

Problem Set 6.3
1. a.

20.086

b. 8.1662
c.

e 2 ≈ e1.41 ≈ 4.1

d.

ecos(ln 4) ≈ e0.18 ≈ 1.20

2. a.

The area below the curve is greater than the area
of the part of the rectangle which is below the
curve, so the total area represented by the sum of
the two integrals is greater than the area ab of the
rectangle.
ab ≤ ∫ f ( x) dx + ∫ f −1 ( y ) dy with equality
a

b

0

0

holding when b = f(a).
Instructor’s Resource Manual

b.

3
e3ln 2 = eln(2 ) = eln 8 = 8

e

ln 64
2

1/ 2 )

= eln(64

= eln 8 = 8

3

3. e3ln x = eln x = x3
4. e –2 ln x = eln x

−2

= x −2 =

1
x2

Section 6.3

359

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

5. ln ecos x = cos x

19.

6. ln e –2 x –3 = –2 x – 3

ex

ex
x

=

eln x

2 – y ln x

=

eln x
e

2

x2

=

y ln x

e

ln x y

x2

=

x

y

= x 2– y

11. Dx e x + 2 = e x + 2 Dx ( x + 2) = e x + 2
12. Dx e
=e

2 x2 – x

2 x2 – x

14. Dx

– 1
2
e x

2 x2 – x

=e

x+ 2

Dx (2 x – x)

=−

Dx x + 2 =

e

x+2

2 x+2

⋅ 2x

ln x 2

x2

–3

=

x2

xe
x

+

−2
2
1 ⎤
⎥ = Dx e x + Dx e − x
2
e x ⎥⎦

−2

Dx x −2 + e− x Dx [− x 2 ]

−2

⋅ (−2 x −3 ) + e− x ⋅ (−2 x)

2

2

x2

2e1

3

2x
ex

2

21. Dx [e xy + xy ] = Dx [2]
e xy ( xDx y + y ) + ( xDx y + y ) = 0
xe xy Dx y + ye xy + xDx y + y = 0
xe xy Dx y + xDx y = – ye xy – y

– 1
2
2e x

x3

= Dx x 2 = 2 x

1
x
x
x (ln x ) ⋅1 – x ⋅
x
x
= e ln x ⋅
16. Dx e ln x = e ln x Dx
2
ln x
(ln x)
=

x2

x

⎝ x ⎠

15. Dx e2 ln x = Dx e

20. Dx ⎢e1
⎢⎣

= ex

2

– 1
⎛ 1
2
= e x Dx ⎜ –
2
– 1
2
=e x

2

= x ex +

= ex

(4 x –1)

x+2

13. Dx e

=e

2

] = Dx (e x )1 2 + Dx e

2
2
1 x2 −1 2
Dx e x + e x Dx x 2
(e )
2
2
2
1 2
x
= (e x )−1 2 e x Dx x 2 + e x ⋅
2
x2
2 x
1 2
= (e x )1 2 2 x + e x ⋅
x
2

2

9. eln 3+ 2 ln x = eln 3 ⋅ e2 ln x = 3 ⋅ eln x = 3 x 2

10. eln x

x2

=

7. ln( x3e –3 x ) = ln x3 + ln e –3 x = 3ln x – 3 x
8. e x –ln x =

2

Dx [ e x + e

x
e ln x (ln x –1)

Dx y =

− ye xy – y
xe

xy

+x

=–

y (e xy + 1)
x (e

xy

+ 1)

=–

y
x

22. Dx [e x + y ] = Dx [4 + x + y ]
e x + y (1 + Dx y ) = 1 + Dx y
e x + y + e x + y Dx y = 1 + Dx y
e x + y Dx y – Dx y = 1 – e x + y

Dx y =

1 – e x+ y
e x + y –1

= –1

23. a.

(ln x) 2

17. Dx ( x3e x ) = x3 Dx e x + e x Dx ( x3 )
= x3e x + e x ⋅ 3x 2 = x 2 e x ( x + 3)

18. Dx e x
= ex
= ex

3 ln x

3 ln x

Dx ( x3 ln x)

3 ln x ⎛ 3

1
2⎞
⎜ x ⋅ + ln x ⋅ 3x ⎟
x

3 ln x

= x2e x
360

= ex

( x 2 + 3x 2 ln x)

3 ln x

The graph of y = e x is reflected across the
x-axis.

(1 + 3ln x)

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portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

b.

26.

f ( x) = e

−x

2

Domain = (−∞, ∞)

1
1 −x
f ′( x) = − e 2 , f ′′( x) = e 2
2
4
Since f ′( x) < 0 for all x, f is decreasing on
(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
−x

The graph of y = e x is reflected across the
y-axis.

y

24. a < b ⇒ – a > – b ⇒ e
increasing function.

–a

>e

–b

x

, since e is an
8

25.

f ( x) = e

Domain = (−∞, ∞)

2x

f ′( x) = 2e , f ′′( x) = 4e2 x
Since f ′( x) > 0 for all x, f is increasing on
(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
2x

4

−5

27.

f ( x) = xe − x

Domain = (−∞, ∞)

f ′( x) = (1 − x)e− x ,

y

(−∞,1)
+

x
f′
f ′′

8

x

5

f ′′( x) = ( x − 2)e − x
1
0

(1, 2)

(2, ∞)

+

2

0

f is increasing on (−∞,1] and decreasing on
[1, ∞) . f has a maximum at (1, 1 )
e
f is concave up on (2, ∞) and concave down on

4

−2

2

x

(−∞, 2) . f has a point of inflection at (2, 2
y

e2

)

5

−3

8

x

−5

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361

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28.

f ( x) = e x + x

Domain = (−∞, ∞)

30.

f ( x) = ln(2 x − 1) . Since 2 x − 1 > 0 if and only if

f ′( x) = e x + 1 , f ′′( x) = e x
Since f ′( x) > 0 for all x, f is increasing on

x>

(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.

(2 x − 1) 2
Since f ′( x) > 0 for all domain values, f is

1
2

1
2

, domain = ( , ∞)

f ′( x) =

2
,
2x −1

f ′′( x) =

−4

1
2

increasing on ( , ∞) .
Since f ′′( x) < 0 for all domain values, f is
1
2

concave downward on ( , ∞) .

y

Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.

5

y

−5

5

5

x

−5

x

8

29.

f ( x) = ln( x 2 + 1) Since x 2 + 1 > 0 for all x,
domain = (−∞, ∞)
f ′( x) =

2x
,
x +1
2

f ′′( x) =

−2( x − 1)
( x 2 + 1) 2

−5

2

x ( −∞ , −1) −1 ( −1,0) 0 (0,1) 1 (1,∞ )
f′
0

+
+
+
f ′′
0
0

+
+
+

f is increasing on (0, ∞) and decreasing on
(−∞, 0) . f has a minimum at (0, 0)
f is concave up on (−1,1) and concave down on
(−∞, −1) ∪ (1, ∞) . f has points of inflection at
(−1, ln 2) and (1, ln 2)
y

31.

f ( x) = ln(1 + e x ) Since 1 + e x > 0 for all x,

domain = (−∞, ∞)
f ′( x) =

ex

,

f ′′( x) =

ex

1 + ex
(1 + e x ) 2
Since f ′( x) > 0 for all x, f is increasing on
(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.
y

5
5

−5

5

x
−5

5

x

−5
−5

362

Section 6.3

Instructor’s Resource Manual

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32.

f ( x) = e1− x

2

f ′( x) = −2 xe1− x ,

f ′′( x) = (4 x 2 − 2)e1− x

2

x

( −∞ , −

f′

+

y

Domain = (−∞, ∞)

2
2
) −
2
2

(−

+

2
2
,0) 0 (0, )
2
2

2
2

+

0

3

2

(

2
,∞ )
2

−1
f ′′

+

0

+

0

f is increasing on (−∞, 0] and decreasing on
[0, ∞) . f has a maximum at (0, e)
f is concave up on (−∞, −
concave down on (−
inflection at (−

2
2

2
2

,

2
2
) ∪ ( , ∞ ) and
2
2
2
).
2

, e ) and (

−3

34.

f ( x) = e x − e− x

f ′( x) = e + e
x

f has points of
2
2

x

4

2

, e)

y

−x

Domain = (−∞, ∞)
,

f ′′( x) = e x − e− x

x (−∞, 0) 0 (0, ∞)
f′
+
+
+
f ′′

+
0

f is increasing on (−∞, ∞) and so has no extreme
values. f is concave up on (0, ∞) and concave
down on (−∞, 0) . f has a point of inflection at
(0, 0)

3

y
−3

3

x
3

−3

33.

f ( x) = e − ( x − 2)

2

Domain = (−∞, ∞)

−3

3

x

f ′( x) = (4 − 2 x)e − ( x − 2) ,
2

f ′′( x) = (4 x 2 − 16 x + 14)e− ( x − 2)

2

−3

Note that 4 x 2 − 16 x + 14 = 0 when
x=

4± 2
≈ 2 ± 0.707
2

x ( −∞ ,1.293) ≈1.293 (1.293,2) 2 (2,2.707) ≈ 2.707 (2.707,∞ )
f′
+
+
+

0
f ′′
+

+
0
0

f is increasing on (−∞, 2] and decreasing on
[2, ∞) . f has a maximum at (2,1)
f is concave up on

(−∞, 4−2 2 ) ∪ ( 4+2 2 , ∞) and

concave down on

(

4− 2 4+ 2
,
) . f has points
2
2

4− 2

of inflection at ( 2

,

1
) and
e

Instructor’s Resource Manual

(

4+ 2
2

,

1
).
e

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363

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35.

f ( x) = ∫0 e − t dt
x

2

f ′( x) = e− x ,
2

Domain = (−∞, ∞)
f ′′( x) = −2 xe− x

2

x (−∞, 0) 0 (0, ∞)
f′
+
+
+
f ′′
+
0

f is increasing on (−∞, ∞) and so has no extreme
values. f is concave up on (−∞, 0) and concave
down on (0, ∞) . f has a point of inflection at
(0, 0)

37. Let u = 3x + 1, so du = 3dx.
1 3 x +1
1 u
1 u
3 x +1
∫ e dx = 3 ∫ e 3dx = 3 ∫ e du = 3 e + C
1
= e3 x +1 + C
3
38. Let u = x 2 − 3, so du = 2x dx.
1 x 2 −3
1 u
x 2 −3
∫ xe dx = 2 ∫ e 2 x dx = 2 ∫ e du
1
1 2
= eu + C = e x −3 + C
2
2

y

39. Let u = x 2 + 6 x , so du = (2x + 6)dx.
1
1
x2 +6 x
dx = ∫ eu du = eu + C
∫ ( x + 3)e
2
2
1 x2 +6 x
= e
+C
2

3

−3

3

x

40. Let u = e x − 1, so du = e x dx .

ex

−3

36.

f ( x) = ∫0 te− t dt
x

f ′( x) = xe

−x

,

( −∞ ,0)

+

x
f′
f ′′

1

∫ e x − 1dx = ∫ u du = ln u + C = ln e
Domain = (−∞, ∞)
f ′′( x) = (1 − x)e
0
0
+

(0,1)
+
+

−x

1
+
0

(1,∞ )
+

f is increasing on [0, ∞) and decreasing on
(−∞, 0] . f has a minimum at (0, 0)
f is concave up on (−∞,1) and concave down on
(1, ∞) . f has a point of inflection at

1
(1, ∫ te−t dt ) .
0

Note: It can be shown with techniques in
2
1
Chapter 7 that ∫0 te− t dt = 1 − ≈ 0.264
e
y

42.

∫e

x +e x

∫e

x

x

43. Let u = 2x + 3, so du = 2dx
1 u
1 u
1 2 x +3
2 x +3
∫ e dx = 2 ∫ e du = 2 e + C = 2 e + C

9

x

⋅ ee dx = ∫ eu du = eu + C = ee + C

x

1

1
1
⎡1

= ⎢ e2 x +3 ⎥ = e5 – e3
2
⎣2
⎦0 2

1 3 2
e (e − 1) ≈ 64.2
2

44. Let u =

−2

x

dx = ∫ e x ⋅ ee dx

Let u = e x , so du = e x dx.

=

−3

−1 + C

1
1
41. Let u = − , so du =
dx .
x
x2
e−1/ x
u
u
−1/ x
∫ x 2 dx = ∫ e du = e + C = e + C

1 2 x +3
e
dx
0

4

x

e3 / x
2

3
3
, so du = − dx.
x
x2

dx = –

1 u
1
e du = – eu + C
3∫
3

x
1
= – e3 / x + C
3
2 e3 / x

2

1 3/ 2 1 3
⎡ 1 3/ x ⎤
∫1 x 2 dx = ⎢⎣ – 3 e ⎥⎦1 = – 3 e + 3 e ≈ 5.2

364

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Instructor’s Resource Manual

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45. V = π∫

ln 3

0

(e x )2 dx = π∫

ln 3 2 x

e dx

0

ln 3

⎡1

= π ⎢ e2 x ⎥
⎣2
⎦0

1 ⎞
⎛1
= π ⎜ e2 ln 3 − e0 ⎟ = 4π ≈ 12.57
2 ⎠
⎝2

2

46. V = ∫ 2πxe− x dx .
1

y = et cos t , so dy = (et cos t − et sin t )dt

Let u = − x 2 , so du = –2x dx.
2

ds = dx 2 + dy 2

2

−x
−x
u
∫ 2πxe dx = −π∫ e (−2 x)dx = −π∫ e du

= et (sin t + cos t )2 + (cos t − sin t )2 dt

2

= −πeu + C = −πe− x + C

∫0 2πxe

− x2

= et 2sin 2 t + 2 cos 2 tdt = 2et dt
The length of the curve is

1

⎡ 2⎤
dx = −π ⎢ e− x ⎥ = – π(e−1 − e0 )

⎦0

π

∫0

−1

= π(1 − e ) ≈ 1.99

1
1− e
1− e
( x − 0);
−1 =
⇒ y −1 =
1− 0 e
e
e
1− e
y=
x +1
e
=

53. a.

⎡1 − e 2

x + 1⎟ − e− x ⎥ dx = ⎢
x + x + e− x ⎥
2
e

⎦0

1− e
1
3−e
=
+1+ −1 =
≈ 0.052
2e
e
2e

48.

f ′( x) =
=
=

(e x –1)2

e x − 1 − xe x
(e x − 1) 2
e x − 1 − xe x
(e x − 1) 2

=−

1
1 − e− x
1
ex −1

1
1 – e– x

(– e

= lim

x →0+

b.

e x − 1 − xe x − (e x − 1)
(e x − 1) 2

Dx [1 + (ln x) 2 ]

ln x

f ′( x) =
=

= lim

x →0+

.

1
x

2 ln x ⋅ 1x

1
=0
2 ln x

x →∞ 1 + (ln x) 2

)(–1)

⎛ 1 ⎞
⎜ x⎟
⎝e ⎠
=

Dx ln x

lim

–x

is of the form

2
x →0+ 1 + (ln x)

x →0+

e

(e x –1)(1) – x(e x )

ln x

lim

= lim

1

1 ⎡⎛ 1 − e

∫0 ⎢⎣⎜⎝

π

2et dt = 2 ⎡ et ⎤ = 2(eπ − 1) ≈ 31.312
⎣ ⎦0

52. Use x = 30, n = 8, and k = 0.25.
(kx) n e− kx (0.25 ⋅ 30)8 e−0.25⋅30
≈ 0.14
Pn ( x) =
=
n!
8!

⎛ 1⎞
47. The line through (0, 1) and ⎜1, ⎟ has slope
⎝ e⎠
1 −1
e

e0.3 ≈ 1.3498588 by direct calculation

51. x = et sin t , so dx = (et sin t + et cos t )dt

0

1

⎧ ⎡⎛ 0.3 ⎞ 0.3 ⎤ 0.3 ⎫
50. e0.3 ≈ ⎨ ⎢⎜
+ 1⎟
+ 1⎥
+ 1⎬ ( 0.3) + 1
⎠ 3
⎦ 2
⎩ ⎣⎝ 4

= 1.3498375

1

= lim

=0

x →∞ 2 ln x

[1 + (ln x) 2 ] ⋅ 1x – ln x ⋅ 2 ln x ⋅ 1x
[1 + (ln x) 2 ]2
1 – (ln x )2
x[1 + (ln x) 2 ]2

f ′( x) = 0 when ln x = ±1 so x = e1 = e

xe x

(e x − 1) 2
When x > 0, f ′( x) < 0, so f(x) is decreasing for
x > 0.

49. a. Exact:
10! = 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1
= 3, 628,800
Approximate:
10

⎛ 10 ⎞
10! ≈ 20π ⎜ ⎟
⎝ e ⎠

⎛ 60 ⎞
b. 60! ≈ 120π ⎜ ⎟
⎝ e ⎠

≈ 3,598, 696

1
e
ln e

or x = e –1 =

f (e) =

1 + (ln e)

=

1
1+1

2

=

1
2

ln 1e

–1
1
⎛1⎞
f ⎜ ⎟=
=
=–
2
2
2
⎝ e ⎠ 1 + ln 1
1 + (–1)
e

( )

Maximum value of
value of −

60

2

1
at x = e; minimum
2

1
at x = e−1.
2

≈ 8.31× 1081

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Section 6.3

365

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F ( x) = ∫

c.

x2

ln t
1 + (ln t )2

1

ln x 2

F ′( x) =

1 + (ln x 2 )2

F ′( e ) =

ln( e )

y-axis so the area is
⎧⎪ 3
2
2
2 ⎨ ∫ 2 [e− x − 2e x (2 x 2 − 1)] dx
0
⎪⎩

dt

⋅ 2x
2

1 + [ln( e ) ]

2 2

⋅2 e =

1
1 + 12

+∫

⋅2 e

= e ≈ 1.65

e x0 – 0
= f ′( x0 ) = e x0 ⇒ e x0 = x0 e x0 ⇒ x0 = 1
x0 – 0

1

a.

b.

2
⎫⎪
(2 x 2 − 1) − e− x ] dx ⎬
⎪⎭

x →∞

f ′( x) = x p e – x (–1) + e – x ⋅ px p –1

b.

= x p –1e – x ( p – x)
f ′( x) = 0 when x = p

so the line is y = e x0 x or y = ex.
59.

ex 2 ⎤
A = ∫ (e – ex)dx = ⎢ e x –

0
2 ⎥⎦
⎣⎢
0
e
e
= e − − (e0 − 0) = –1 ≈ 0.36
2
2

− x2

lim x p e – x = 0

58. a.

x

3 [2e
2

≈ 4.2614

54. Let ( x0 , e x0 ) be the point of tangency. Then

1

3

lim ln( x 2 + e – x ) = ∞ (behaves like − x )

x→ – ∞

lim ln( x 2 + e – x ) = ∞ (behaves like 2 ln x )

x →∞

60.

1

V = π∫ [(e x )2 – (ex) 2 ]dx
0

2 3 ⎤1

⎡1
1
e x
= π∫ (e2 x – e2 x 2 )dx = π ⎢ e2 x –

0
3 ⎥⎦
⎢⎣ 2
0

–1

–1

f ′( x) = –(1 + e x ) –2 ⋅ e x (– x –2 )
=

e1/ x
x 2 (1 + e1/ x )2

⎡1
π
e2 ⎛ 1 ⎞ ⎤
= π ⎢ e2 − − ⎜ e 0 ⎟ ⎥ = (e2 – 3) ≈ 2.30
6
3 ⎝ 2 ⎠ ⎥⎦
⎢⎣ 2

3

1 ⎞

3

1 ⎞

∫−3 exp ⎜⎝ − x2 ⎟⎠ dx = 2∫0 exp ⎜⎝ − x2 ⎟⎠ dx ≈ 3.11

55. a.

8π −0.1x

∫0

b.

e

sin x dx ≈ 0.910

lim (1 + x)1 x = e ≈ 2.72

56. a.

x →0

lim (1 + x)−1 x =

b.

x →0

1
≈ 0.368
e
a.

57.

f ( x) = e

− x2

f ′( x) = −2 xe− x

b.

2

2

2

366

c.

2

e− x = 2e− x (2 x 2 − 1); 1 = 4 x 2 − 2;

d.

3
2
Both graphs are symmetric with respect to the

e.

4 x 2 − 3 = 0, x = ±

Section 6.3

lim f ( x) = 1

x →0 –

2

f ′′( x) = −2e− x + 4 x 2 e− x = 2e− x (2 x 2 − 1)
y = f(x) and y = f ′′( x) intersect when
2

lim f ( x) = 0

x →0 +

lim f ( x) =

x →±∞

1
2

lim f ′( x) = 0

x →0

f has no minimum or maximum values.

Instructor’s Resource Manual

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x + 3 = 5x
3
x=
4

6.4 Concepts Review
3 ln π

1. e

; e x ln a

2. e
3.

9. log5 12 =

ln x
ln a

ln12
≈ 1.544
ln 5

10. log 7 0.11 =

4. ax a −1 ; a x ln a

Problem Set 6.4

ln 0.11
≈ –1.1343
ln 7

11. log11 (8.12)1/ 5 =

1 ln 8.12
≈ 0.1747
5 ln11

12. log10 (8.57)7 = 7

ln 8.57
≈ 6.5309
ln10

1. 2 x = 8 = 23 ; x = 3
2. x = 52 = 25
3. x = 43 / 2 = 8

14. x ln 5 = ln 13
ln13
x=
≈ 1.5937
ln 5

4. x = 64
4

x = 4 64 = 2 2
⎛ x⎞ 1
5. log9 ⎜ ⎟ =
⎝3⎠ 2
x
= 91/ 2 = 3
3

x=

1
2x
1

2⋅4

3

=

1
128

7. log 2 ( x + 3) – log 2 x = 2
log 2

x+3
=2
x

x+3
= 22 = 4
x
x + 3 = 4x
x=1
8. log5 ( x + 3) – log5 x = 1
log5

15. (2s – 3) ln 5 = ln 4
ln 4
2s – 3 =
ln 5
1⎛
ln 4 ⎞
s = ⎜3 +
⎟ ≈ 1.9307
2⎝
ln 5 ⎠
16.

x=9

6. 43 =

13. x ln 2 = ln 17
ln17
x=
≈ 4.08746
ln 2

x+3
=1
x

x+3 1
=5 =5
x

Instructor’s Resource Manual

1

θ –1

ln12 = ln 4

ln12
= θ –1
ln 4
ln12
θ = 1+
≈ 2.7925
ln 4

17. Dx (62 x ) = 62 x ln 6 ⋅ Dx (2 x) = 2 ⋅ 62 x ln 6
18. Dx (32 x

2 –3 x

) = 32 x

= (4 x – 3) ⋅ 32 x

19. Dx log3 e x =

2 –3 x

2 –3 x

ln 3 ⋅ Dx (2 x 2 – 3 x)

ln 3

1

⋅ Dx e x
e ln 3
ex
1
=
=
≈ 0.9102
x
e ln 3 ln 3
Alternate method:
x

Dx log3 e x = Dx ( x log3 e) = log3 e
=

ln e
1
=
≈ 0.9102
ln 3 ln 3

Section 6.4

367

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

20. Dx log10 ( x3 + 9) =
=

21.

3x

( x + 9) ln10
3

⋅ Dx ( x3 + 9)

26.

( x3 + 9) ln10

1
(1) + ln( z + 5) ⋅ 3z ln 3
z +5
⎡ 1

= 3z ⎢
+ ln( z + 5) ln 3⎥
⎣z +5

=

=–

1

1

(θ 2 – θ ) ln 3
ln 3
=
⋅ Dθ θ 2 – θ
ln10
ln10

2θ –1
2 θ 2 –θ

ln 3
ln10

27.

23. Let u = x 2 so du = 2xdx.
2

x
∫ x ⋅ 2 dx =

1 u
1 2u
2
du
=

+C
2∫
2 ln 2

2

24. Let u = 5x – 1, so du = 5 dx.
1
1 10u
u
5 x –1
10
10
dx
=
du
=

+C

5∫
5 ln10
105 x –1
+C
5ln10

5

x

x

=

1
2 x

dx.

5u
+C
ln 5

2⋅5 x
+C
ln 5
4

⎡5 x ⎤
5 ⎞
⎛ 25
∫1 x dx = 2 ⎢⎢ ln 5 ⎥⎥ = 2 ⎜⎝ ln 5 − ln 5 ⎟⎠

⎦1
40
=
≈ 24.85
ln 5

368

Section 6.4

2
2
d ( x2 )
d
= 10( x ) ln10 x 2 = 10( x ) 2 x ln10
10
dx
dx
d 2 10 d 20
(x ) =
x = 20 x19
dx
dx
2
dy d
= [10( x ) + ( x 2 )10 ]
dx dx

d
d
sin 2 x = 2sin x sin x = 2sin x cos x
dx
dx
d sin x
d
2
= 2sin x ln 2 sin x = 2sin x ln 2 cos x
dx
dx
dy d
= (sin 2 x + 2sin x )
dx dx

= 2sin x cos x + 2sin x cos x ln 2

dx = 2∫ 5u du = 2 ⋅

45 x

–3 x

2

28.

25. Let u = x , so du =

3x

= 10( x ) 2 x ln10 + 20 x19

2

2x
2 x –1
=
+C =
+C
2 ln 2
ln 2

=

10 –3 x
+C
3ln10

⎡103 x –10 –3 x ⎤
Thus, ∫ (10 + 10 )dx = ⎢

0
⎢⎣ 3ln10 ⎥⎦ 0
1 ⎛
1 ⎞ 999,999
=
⎜1000 –
⎟=
3ln10 ⎝
1000 ⎠ 3000 ln10
≈ 144.76

) = Dθ (θ 2 – θ ) log10 3

ln 3 1 2
⋅ (θ – θ ) –1/ 2 (2θ –1)
ln10 2

=

1

0

103 x
+C
3ln10
Now let u = –3x, so du = –3dx.
1
1 10u
–3 x
u
10

10

dx
=
du
=

+C

3∫
3 ln10

= 3z ⋅

= Dθ

1

0

+ 10 –3 x )dx = ∫ 103 x dx + ∫ 10 –3 x dx

=

Dz [3z ln( z + 5)]

2 –θ

3x

Let u = 3x, so du = 3dx.
1
1 10u
3x
u
10
10
dx
=
du
=

+C

3∫
3 ln10

2

22. Dθ log10 (3θ

1

∫0 (10

29.

d π+1
x
= (π + 1) x π
dx
d
(π + 1) x = (π + 1) x ln(π + 1)
dx
dy d π+1
= [x
+ (π + 1) x ]
dx dx
= (π + 1) x π + (π + 1) x ln(π + 1)

Instructor’s Resource Manual

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

30.

x
x
d (e x )
d
= 2(e ) ln 2 e x = 2(e ) e x ln 2
2
dx
dx
d e x
(2 ) = (2e ) x ln 2e = (2e ) x e ln 2
dx
dy d (e x )
= [2
+ (2e ) x ]
dx dx

35.

f ′( x) = (− ln 2)2 , f ′′( x) = (ln 2) 2 2− x
Since f ′( x) < 0 for all x, f is decreasing on

(−∞, ∞) .
Since f ′′( x) > 0 for all x, f is concave upward on
(−∞, ∞) .
Since f and f ′ are both monotonic, there are no
extreme values or points of inflection.

x

2 +1)

y

2
dy
d
= e(ln x ) ln( x +1) [(ln x) ln( x 2 + 1)]
dx
dx
2
1
2x ⎤

= e(ln x ) ln( x +1) ⎢ ln( x 2 + 1) + ln x

2
x + 1⎦
⎣x

4

⎛ ln( x 2 + 1) 2 x ln x ⎞
= ( x 2 + 1)ln x ⎜
+

x
x 2 + 1 ⎟⎠

32. y = (ln x 2 ) 2 x +3 = e(2 x +3) ln(ln x
dy
=e
dx

(2 x +3) ln(ln x 2 )

f ( x) = xsin x = esin x ln x
d
(sin x ln x )
dx

⎛1⎞
= esin x ln x ⎢ (sin x) ⎜ ⎟ + (cos x)(ln x ) ⎥
⎝ x⎠

f ′( x) = esin x ln x

⎛ sin x

= xsin x ⎜
+ cos x ln x ⎟
x

sin1

+ cos1ln1⎟ = sin1 ≈ 0.8415
f ′(1) = 1sin1 ⎜
⎝ 1

34.

f (e) = πe ≈ 22.46
g (e) = e π ≈ 23.14
g(e) is larger than f(e).
d
f ′( x) = π x = π x ln π
dx

−3

9

2)

d
[(2 x + 3) ln(ln x 2 )]
dx
2 ⎡
1 1

= e(2 x +3) ln(ln x ) ⎢ 2 ln(ln x 2 ) + (2 x + 3)
(2 x) ⎥
2 2
ln x x

2 x +3 ⎢
2x + 3 ⎥
2
ln
(2
ln
x
)
= (2 ln x)
+




x ln x ⎥

⎥⎦
2
ln x
ln x 2

33.

Domain = (−∞, ∞)

−x

= 2(e ) e x ln 2 + (2e ) x e ln 2

31. y = ( x 2 + 1)ln x = e(ln x ) ln( x

f ( x) = 2− x = e(ln 2)( − x )

x

−2

36.

f ( x) = x 2− x

Domain = (−∞, ∞)

f ′( x) = [1 − (ln 2) x]2− x ,
f ′′( x) = (ln 2)[(ln 2) x − 2]2− x
1
)
ln 2

x

( −∞ ,

f′

+

f ′′

1
ln 2

1 2
,
)
ln 2 ln 2

2
ln 2

0

0

+

(

(

2
,∞ )
ln 2

1 ⎤
f is increasing on ⎜ −∞,
⎥ and decreasing on
ln
2⎦

⎡ 1

1 1
, ∞ ⎟ . f has a maximum at (
,
)

ln
2
ln
2 (e ln 2)

f is concave up on (
(−∞,
(

2
, ∞) and concave down on
ln 2

2
) . f has a point of inflection at
ln 2

2 2
,
)
(e 2 ln 2)
ln 2
y
3

f ′(e) = πe ln π ≈ 25.71
g ′( x) =

d π
x = πx π−1
dx

−2

8

x

g ′(e) = πe π−1 ≈ 26.74
g ′(e) is larger than f ′(e) .
−3

Instructor’s Resource Manual

Section 6.4

369

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

37.

y

ln( x 2 + 1)
. Since
f ( x) = log 2 ( x + 1) =
ln 2
2

5

x 2 + 1 > 0 for all x, domain = (−∞, ∞)
⎛ 2 ⎞⎛ x ⎞
⎛ 2 ⎞ ⎛ 1 − x2
f ′( x) = ⎜
⎟⎜ 2
⎟ ⎜⎜ 2
⎟ , f ′′( x) = ⎜
2
⎝ ln 2 ⎠ ⎝ x + 1 ⎠
⎝ ln 2 ⎠ ⎝ ( x + 1)
x

⎟⎟

−5

5

x

(−∞, −1) −1 (−1, 0) 0 (0,1) 1 (1, ∞)

f′
f ′′

0

+

+

+

0

+

+

+

0

f is increasing on [0, ∞) and decreasing on
(−∞, 0] . f has a minimum at (0, 0)
f is concave up on (−1,1) and concave down on
(−∞, −1) ∪ (1, ∞) . f has points of inflection at
(−1,1) and (1,1)

−5

39.

f ( x) = ∫1 2− t dt
x

2

f ′( x) = 2− x ,

Domain = (−∞, ∞)
f ′′( x) = −2(ln 2) x 2− x

2

2

(−∞, 0) 0 (0, ∞)

x
f′
f ′′

+

+

+

+

0

f is increasing on (−∞, ∞) and so has no extreme
values.
f is concave up on (−∞, 0) and concave down on
(0, ∞) . f has a point of inflection at

y
5

(0, ∫1 2− t dt ) ≈ (0, −0.81)
0

−5

5

2

x
y
5

−5

38.

f ( x) = x log3 ( x 2 + 1) =

x ln( x 2 + 1)
. Since
ln 3

−5

5

x

x 2 + 1 > 0 for all x, domain = (−∞, ∞)
f ′( x ) =

x

1 ⎡ 2 x2
2 ⎡ x3 + 3 x ⎤

+ ln( x 2 +1) ⎥ , f ′′( x) =
ln 3 ⎢ x 2 +1
ln 3 ⎢ x 2 +1 ⎥
⎥⎦

−5

(−∞, 0) 0 (0, ∞)

f′

+

0

+

f ′′

0

+

f is increasing on (−∞, ∞) and so has no extreme
values. f is concave up on (0, ∞) and concave
down on (−∞, 0) . f has a point of inflection at
(0, 0)

370

Section 6.4

Instructor’s Resource Manual

portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

40.

x

f ( x) = ∫0 log10 (t 2 + 1)dt . Since log10 (t 2 + 1) has

domain = (−∞, ∞) , f also has domain = (−∞, ∞)
ln( x 2 + 1)
f ′( x) = log10 ( x + 1) =
,
ln10
2

⎛ 1 ⎞ ⎛ 2x ⎞
f ′′( x) = ⎜
⎟⎜ 2

⎝ ln10 ⎠ ⎝ x + 1 ⎠
x

P=

105.75
≈ 4636 lb/in.2
121.3

45. If r is the ratio between the frequencies of

successive notes, then the frequency of C = r12
(the frequency of C). Since C has twice the

(−∞, 0) 0 (0, ∞)

f′
f ′′

44. 115 = 20 log10 (121.3P )
log10 (121.3P ) = 5.75

+

0

+

frequency of C, r = 21/12 ≈ 1.0595

0

+

Frequency of C = 440(21/12 )3 = 440 4 2 ≈ 523.25

f is increasing on (−∞, ∞) and so has no extreme
values.
f is concave up on (0, ∞) and concave down on
(−∞, 0) . f has a point of inflection at (0, 0)

46. Assume log 2 3 =

p
where p and q are integers,
q

q ≠ 0 . Then 2 p q = 3 or 2 p = 3q. But

2 p = 2 ⋅ 2 … 2 (p times) and has only powers of 2

y

as factors and 3q = 3 ⋅ 3…3 (q times) and has
only powers of 3 as factors.

5

2 p = 3q only for p = q = 0 which contradicts our
assumption, so log 2 3 cannot be rational.
−5

5

x

If y = C ⋅ x d , then ln y = ln C + d ln x, so the
ln y vs. ln x plot will be linear.

−5

41. log1/ 2 x =

47. If y = A ⋅ b x , then ln y = ln A + x ln b, so the
ln y vs. x plot will be linear.

ln x
ln x
=
= − log 2 x
1
ln 2 − ln 2

42.

48. WRONG 1:
y = f ( x) g ( x )
y ′ = g ( x) f ( x) g ( x ) −1 f ′( x)
WRONG 2:
y = f ( x) g ( x )
y ′ = f ( x) g ( x ) (ln f ( x)) ⋅ g ′( x) = f ( x ) g ( x ) g ′( x) ln f ( x)
RIGHT:
y = f ( x) g ( x ) = e g ( x ) ln f ( x )

43. M = 0.67 log10 (0.37 E ) + 1.46

log10 (0.37 E ) =
E=

10

M − 1.46
0.67

M −1.46
0.67

0.37
Evaluating this expression for M = 7 and M = 8

y ′ = e g ( x ) ln f ( x )

d
[ g ( x) ln f ( x)]
dx

1
= f ( x) g ( x ) ⎢ g ′( x) ln f ( x) + g ( x)
f ′( x) ⎥
f ( x)

= f ( x) g ( x ) g ′( x ) ln f ( x) + f ( x) g ( x ) −1 g ( x) f ′( x)
Note that RIGHT = WRONG 2 + WRONG 1.

gives E ≈ 5.017 × 108 kW-h and
E ≈ 1.560 × 1010 kW-h, respectively.

Instructor’s Resource Manual

Section 6.4

371