23. y

Not to scale

• • = 1'-. <' i-x-3 '.I 1 3

II

24. y = In (x3) = 3 In x

The coefficient of 3 "amplifies" the graph ofy = lnx.

The correct choice is A.

25. Since f is a continuous function, it must cross thex-axis in going from y = -2 to y = 3.

The correct choice is B.

PROBLEM SET 791. aCt) = 2 cos t

vet) = 2 sin t + C

v( ~) = 2 sin ~ + C

-4 = 2 + CC = -6

vet) = 2 sin t - 6

x(t) = -2 cos t - 6t + Cx(O) = -2 cos 0 - 6(0) + C

8 = -2 + CC=lO

x(t) = -2 cos t - 6t + 10

2. aCt) = -6t

vet) = -3t2 + C

v(l) = -3(12) + C

-1 = -3 + C

C=2

vet) = -3t2 + 2

v(3) = -25

x(t) = _t3 + 2t + C

x(2) = _(23) + 2(2) + C

-3 = -4 + C

C = 1

x(t) = _t3 + 2t + 1

x(3) = -20

Calculus, Second Edition

Problem Set 79

3.

2~1 k' .••• , ."

1 2 3 4

w = f depth . weight density . volume

= f (2 - y)(9800)(4)(l0) dy

= 392,000 f (2 - y) dy

= 392,000[ 2y - ~y2 I= 392,000[ 4 - 2 - (2 - ~)]

= 392,000( ~) = 196,000 joules

4. W = f depth . weight density . volume

= f~(1 - y)(1000)(2y)(6) dy

= 12,000 f~(y - y2) dy

[1 1 ]1= 12,000 _y2 - _y3 = 2,000 joules2 3 0

5. y

6-y

-------6 Iii5

43

2

1

~x

F = f weight density . depth . area

= f~2000(6 - y)(l) dy

= 2000 f~(6 - y) dy

= 2000 [ 6y - ~y21

= 2000 ( 6 - ~) = 11,000 newtons

6. lim sin x =x~o X

lim cos x = 1x~o 1

181

Problem Set 79

7. lim 2 - 2 cos xx~o sin x

= lim 2sinx =x~o cos x

14. f (log x + 43X) dx

= f (~ + 43X) dx

In 10

= _1_(x In x - x) + _1_43x + CIn 10 In 43

x 43x

= x log x - -- + -- + CIn 10 In 43

o = 01

8. lim __ x_X~~ (In x)2

= lim _x_x~~ 2 In x

1. 1= Im-

x~~ 215. (a) f(x) = Ix2

- 81 = 0 when x2 - 8 = 0x2 = 8x = ±2../2x

= lim'::'X~~ 2 (b) Y

= 00

9.x + sin xlim -----,,-----

x~~ x2lim 1 + cos x = 0

X~~ 2x

-~r-r-~+--- x10. 1

. e" - xIm--- =

x~o sin x

-4 -2 2 41 - 0o =

1o

{2x when Ixl > 2...fi

(c) f'(x) = -2x when Ixl < 2...fie" - xlim ---

x~o+ sin x= +00

116. u = 2x- x = -(u + 1)

2

1du = 2 dx dx= - du

2

= -00

Therefore, lim eX - x does not exist.x~o sin x

13 1 1-(u + l)ull2- du

122

= f3 .!..(u3l2 + ull2) duJ1 4

r x-hx - 1 dx =11. Iirn

x_M

g(x) = 1 by the sandwich theorem becauseIimX---7" -cos x = 1, lirnX---7" (2 + cos x) = 1, andf(x) ~ g(x) ~ h(x) for values of x near n:

12. f sirr' x dx = f sin x (1 - cos2 x) dxThe correct choice is E.

= f sin x dx - f cos2 x sin x dx

17. h(x) = f(x)g(x) = 3 tan x sin x

h(-x) = 3 tan (-x) sin (-x)

= 3(-tan x)(-sin x)

= 3 tan x sin xu = cos x du = -sin x dx

f sin x dx - f cos? x sin x dx

= -f du + f u2 du

Since h(-x) = h(x), the graph of h is even andtherefore symmetric about the y-axis.

= .!. cos ' X - cos x + C3

13. J cos x sirr' x dx = ~ sin" x + C

182 Calculus, Second Edition

18. y = x3 + 6x2 + 1 21. (a)

y' = 3x2 + 12x

y" = 6x + 12

o = 6x + 12

x = -2

y = (_2)3 + 6(-2)2 + 1 = 17

y'(-2) = 12 - 24 = -12

y - 17 = -12(x + 2)

y = -12x - 24 + 17

Y = -12x - 7

19. f(x) = a sin x + b cos xf(O) = a sin 0 + b cos 0

2 = b

f(x) = a sin x + 2 cos xf(x) = a cos x - 2 sin x

f(O) = a cos 0 - 2 sin 0 (b)2 = a

a + b = 4

20. (a) y

I"t, I I •• 1 •. x

-3

A = f2 Ix 2 + X - 21 dx-3

(b) A = fn I nt. (abs Co::::: +::'::-2) , ::.::,-3;t 2)

"" 8.1667 units2

Calculus, Second Edition

Problem Set 79

n f(n)(x) f(n)(o)

0 sin x 0

1 cosx 1

2 -sin x 0

3 -cosx -1

4 sin x 0

5 cosx 1

6 -sin x 0

sin x =5 7

x3

X _ ~ + ...x - 3! + 5! 7!

(2n-l)~, lX~ __L (-I)n- (2n - I)!n=l

sinx =

n f(n)(x) f(n)(o)

0 sin x2 0

1 2x cos x? 0

2 _4x2 sin x2 + 2 cos x2 2

3 -12x sin x2 - 8x3 cos x2 0

4 (16x4 - 12) sin x2 - 48x2 cos x2 0

160x3 sin x25

+ (32x5 - 120x) cos x2 0

6(-64x6 + nOx2) sin x2

-120+ (480X4 - 120) cos x2

sin x2 = 2x2 120x6

2! - -6-! + ...

. 2 6smx = x2 _ x-+ ...3!

(c) sin x2

2 (x2)3 (x2)5 (x2)1= x ---+-----+ ...

3! 5! 7!x6 xlO x14

sin x2 = x2 - - + - - - + ...3! 5! 7!

They are the same.

183

Problem Set 80

22. x + sin x + arctan (x2) + eX csc (2x)y = cos x

dydx

cos x(l + cos x) - (x + sin x)(-sin x)cos? x

dy =dx

+ 2 ~ + eX[-2 csc (2x) cot (2x)](x) + 1

+ e" csc (2x)

l+cosx sinx(x + sin x)---+cos? xcos x

+ ~ + e' csc (2x) [-2 cot (2x) + 1]x4 + 1

dy = see x + 1 + see x tan x (x + sin x)dx

+ ~ + eX csc (2x) [-2 cot (2x) + 1]X4 + 1

dy = see x + 1 + x see x tan x + tarr' xdx

2x [ ]+ -4-- + e' csc (2x) -2 cot (2x) + 1x + 1

dy 2 2x= see x + x see x tan x + see x + -4--dx x +1

+ eX csc (2x) [-2 cot (2x) + 1]d 2xJx = see x (1 + x tan x + see x) + x4 + 1

+ eX ese (2x) [-2 eot (2x) + 1]

23. hex) = f(g(x)) = f(sin x) = sin2 xh'(x) = 2 sin x cos xh'(x) = sin (2x)

24. lim f(a + h) - f(a) = f'(a)h-40 h

= lim f(x) - f(a)X-4a X - a

The correct choice is B.

25. (a)

r J

\J(b) f(x) = 2x3 - 3x2 - 12x + 20

f'(x) = 6x2 - 6x - 12o = 6(x2 - X - 2)o = (x - 2)(x + 1)x = -1,2

f(-I) = 27, f(2) = 0

(-1,27), (2, 0)

184

PROBLEM SET 801. aCt) = 2t

vet) = t2 + C

v(3) = 32 + C

1O=9+C

C = 1

v(t) = t2 + 1

17 = t2 + 1

t2 = 16

t = -4,4

2. y

4

dy

w = f depth . weight density . volume

= f: (4 - y)(2000)(I)(3) dy

= 6000 f: (4 - y) dy

= 6000 [ 4y - ~y2 J:= 6000(8 - 2) = 36,000 joules

3. (a) aCt) = -9.8

vet) = -9.8t + C

vet) = -9.8t + 50

x(t) = -4.9t2 + SOt + C

x(t) = -4.9t2 + SOt + 160

(b) vet) = -9.8t + 50

o = -9.8t + 50

t '" 5.1020 s

(c) x(t) = -4.9t2 + SOt + 160

o = -4.9t2 + SOt + 160

-50 ± ~2500 - 4(-4.9)(160)t = -9.8

t '" 12.7626 s (t "f; -2.5585 s)

Calculus, Second Edition