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

Instructor’s Resource Manual

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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.

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 =

Instructor’s Resource Manual

1

,

2

1

ln x.

2

Section 6.1

349

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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.

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)

Instructor’s Resource Manual

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

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).

3π

∫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

Instructor’s Resource Manual

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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.

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.

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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

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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.

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

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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 =

4π

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

4π

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

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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

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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.

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

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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

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359

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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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Section 6.3

361

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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.

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

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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.

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

Section 6.3

363

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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.

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

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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365

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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.

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.

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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

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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.

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)

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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

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Section 6.4

369

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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.

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

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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.

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.

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Section 6.4

371

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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.

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) =

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

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 =

Instructor’s Resource Manual

1

,

2

1

ln x.

2

Section 6.1

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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.

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

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).

3π

∫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

Instructor’s Resource Manual

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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.

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.

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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

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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.

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

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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.

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 =

4π

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

4π

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

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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.

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

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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.

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

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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.

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

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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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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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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

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

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

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.

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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.

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

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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.

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

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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.

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

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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.

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

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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.

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.

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Section 6.4

371

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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.

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