g(x) = 3x2 - x3

251) Chapter 3 Applications of Differentiation

108. First observe that

tanx+cotx+secx+cscx-sin x cos x 1 1

----+ +cos x sin x cos x sin x

sin2x + cos2x + sin x + cos xsin x cos x

l+sinx+cosx(sinx+cosx-~) .sin x cos x ~ x + cos x

(sin x + cos x)~ - 1sin x cos x[sin x + cos x - 1)

2 sin x cos xsin x cos x(sin x + cos x - 1)

2sin x + cosx - 1

Let t = sin x + cos x - 1. The expression inside the absolute value sign is

2sin x + cos x +

sin x + cos x - 12

(sin x + cos x - 13 + 1 +\ sin x + cos x - 1

2=t+l+-

t

-- + cosxsln--Because sin x + = sin x cos 4 4

- ~(sin x + cos x),2

t= sin x + cos x- 1 ~ [-1-w/~,-1 + -,/~].

f’(t) : 1 - -t2 t2 t2

+ = + + +_ 4__-~-~_(’x,/~+ 1~= 4"~-2+4- "/~ =2+ 3"~

x/~ - 1[,.~" +1) 1

Fort > 0, f is decreasing and f(t) > f(-1 + ~/~) = 2 +

Fort< 0, f is increasing on (-w/~ - 1,-x/-~), then decreasing on (-~/~, 0). So f(t)< f(-xi~) =1 - 2w/~.

Finally, f(t)l >_ 2xl~ - 1.

(You can verify this easily with a graphing utility.)

Section 3.4 Concavity and the Second Derivative Test

1. The graph off is increasing and concave upwards:f’>0, f"> 0

3. The graph off is decreasing and concave downward:f’ < 0, f" < 0

2. The graph off is increasing and concave downwards:f’ > 0, f" < 0

4. The graph off is decreasing and concave upward:f’ < 0, f"> 0

© 2010 Brooks/Cole, Cengage Learning

Section 3.4 Concavity and the Second Derivative Test 251

y=x2-x-2

y’ = 2x - 1

y" = 2

Concave upward: (-0%

y = -x3 + 3x2 - 2

y’ = -3x2 + 6x

y" = -6x + 6

Concave upward: (-% 1)

Concave downward: (1, oo)

g(x) = 3x2 - x3g’(x) = 6x- 3x2

g"(x) = 6 - 6x

Concave upward: (-% 1)

Concave downward: (1, m)

h(x) = x5 - 5x + 2hi(x) = 5X4 -- 5

h"(x) = 2Ox3

Concave upward: (0, m)

Concave downward: (-% O)

9. f(x) = -x3 + 6x2 - 9x-1

f’(x) = -3x2 + 12x- 9

f"(x) = -6x + 12 = -6(x- 2)

Concave upward: (-m, 2)

Concave downward: (2, m)

10, y(x) = X5 q- 5X4- 40x2

f’(x) = 5X4 q- 20X3 -- 80X

f"(x) = 20x3 + 60x2 - 80

= 20(X3 + 3X2- 4)

= 20(X- 1)(X + 2)2

Test Interval: -oo < x < -2 -2<x<1 l<x<oo

Sign of f"" f" < 0 f" < 0 f" > 0

Conclusion: Concave downward Concave downward Concave upward

Concave upward: (1,

Concave downward: (-m, 1)

24 x~-11. f(x) - x2 +12 12. f(x) = x2 +1

-48xf’-(x2 + 12)2

Concave upward: (-oo,-2), (2,

Concave downward: (-2, 2)

f"(x) = -2(3x2 -1)


252 Chapter 3 Applications of Differentiation

x2 +113. f(x) - x2 - 1

-4x

f"- 4(3x2 + 1)


Concave downward: (-1,1)

14. = 1 5 135X)y "5-~(--3X + 40X3 +

y’ = -~70(--15X4 + 120X2 + 135)

y"=--}x(x- 2)(x + 2)

Concave upward: (-m,-2), (0, 2)

Concave downward: (-2, 0), (2,

15.x2 +4g(x) = 7"- x2

16x

(4 - x2)2

(4- x2)’ (2- x)3(2 + x)3

Concave upward: @2, 2)

Concave downward: (-0%-2), (2, m)

16.x2 - 1

h(x) - -2x -1

h’(x) = 2(x2- x + 1)

(2x -1)2

-6h"(x) -

(2x -1)3

Concave upward: (-m, ~)

Concave downward: (-~, o@

17.

18.

19.

y = 2x-tanx, - ,

y’ = 2-sec2x

y" = -2 sec2x tan x

Concave downward: /O, ~)

y = x+ 2cscx, (-~,

y’ = 1-2cscxcotx

y"= -2 csc x(-csc2x)- 2 cot x(-csc x cot x)

= 2(CSC3X + CSC X cot2 X)

Concave upward: (0, ~r)

Concave downward: (-~r, O)

1 4f(x) =-~x + 2x3

f’(x) = 2x3 + 6x2

f"(x) = 6x2 + 12x = 6x(x + 2)

f"(x) = 0 when x = O, -2


Concave downward: (-2, O)

Points of inflection: (-2,-8) and(O, O)

20. f(x) = --X4 q- 24X2

f’(x) = --4X3 + 48X

f"(X) =--12X2 + 48 =12(4- X2) = 12(2 + X)(2- X)

f"(x)=O for X=--2,2

Concave upward: (-2, 2)

Concave downward: (-oo,-2), (2,

Points of inflection: (-2, 80), (2, 80)

21. f(x) = x3 - 6x2 + 12x

f’(x) = 3x2 - 12x + 12

f"(x) = 6(x- 2) = 0 when x = 2.

Concave upward: (2,

Concave downward: (-m, 2)

Point of inflection: (2, 8)

© 2010 Brooks/Cole, Cengage Lea’rning


22. f(x) = 2x3 - 3x2 - 12x + 5

f’(x) = 6x2 - 6x - 12

f"(x) = 12x - 6

f"(x) = 12x - 6 = 0 when x = _12’

Point of inflection: (½, -~-)

Test interval: -~<x<±2 ~<x<~2

Sign of f"(x)" S"(x) < 0 S"(x) > 0Conclusion: Concave downward Concave upward

23.1

f(x) = -~x4 - 2x2

f’(x) = x3 - 4x

f"(x) = 3x2 - 4

+2f"(x) = 3x2 - 4 = 0 when x = _ x/-5"

Points of inflection: --- xi~’ "

Test interval:2 2 2-oo < x < 2

/-5

Sign of f"(x): s"(x) > o S"(x) < o S"(x) > oConclusion: Concave upward Concave downward Concave upward

24. f(x) = 2X4 - 8x + 3

f’(x) = 8x3 -8

f"(x) = 24x2 = 0 when x = O.

However, (0, 3)is not a point of inflection because f"(x) > 0 for allx.

Concave upward: (-0% oo)

25. f(x) = x(x - 4)3

f’(x) = xI3(x - 4)21+ (x - 4)3 (x-4)2(4x-4)

f"(x) = 4(x - 1)E2(x - 4)] + 4(x - 4)2 = 4(x - 4)E2(x - 1) + (x - 4)] = 4(x - 4)(3x - 6) = 12(x - 4)(x - 2)

f"(x) = 12(x - 4)(x - 2) = 0when x = 2, 4.

Test interval: -~<x<2 2<x<4 4<x<~

Sign of f"(x): f"(x) > o s"(.) < o S"(x) > oConclusion: Concave upward Concave downward Concave upward

Points of inflection: (2,-16), (4, 0)

26. f(x) = (x- 2)3(x,- 1)

f’(x) = (x- 2)2(4x- 5)

f"(x) = 6(x- 2)(2x- 3)3f"(x) = 0 when x = -, 22

Concave upward: (-oo, ~) and (2, oo)

Concave downward: (~, 2)

Points of inflection: (~,-~61, (2, O)

27. f(x) = x-,,/Tx + 3, Domain:[-3, m)

f’(x) = x (x + 3)-’/2 +~+3 - 2~x+3

6~x + 3 - 3(x + 2)(x + 3)-1/2S"(x) =

4(x + 3)

_ 3(x + 4)

4(x + 3)f"(x) > 0 on the entire domain off(except for

x = -3, for which f"(x) is undefined). There are no

points of inflection.

Concave upward: (-3,



28. f(x) = xx/-~- x Domain:x < 9

3(6 - x):’(~) - ~-94~- 7~

3(x -12):"(x) 4(9 - x)3/2

Concave downward: (-0% 9)

No point of inflection

29.

-8x:’(x) = (: ÷ 0~s"(x) : 8(3: -0

(: ÷ 0~

:"(x) = 0 for x : -5-

x+l30. f(x)- -..~x" Domain: x > 0

x-1f’(x)- ~/~

3-xf"(x) - -42/~

31.

Test intervals: 0<x<3 3<x<m

Sign of f"(x): f" > 0 f" < 0

Conclusion: Concave upward Concave downward

f(x) = sin~,0 < x < 4rr

:’(x) = -i cos

f"(x) = -~ sin

f"(x) = 0 when x = 0, 2~r, 4n’.

Point of inflection: (2x, 0)

Test interval: 0<x<2n" 2x < x < 4x

Sign of f"(x): f" < 0 f" > 0

Conclusion: Concave downward Concave upward


32.

33.

34.

35.


rCx~ = 2 csc- 0 < x < 2x2’

S’(x) -- -3 csc 3x cot 3x2 2

= ~ + csc ~ cot~ ~ 0 for ~y x in the domain off.2 2

Concave upward: /0,

Concave downward’. (-~, ~~r-)


sec x- ,0 < x < 4x

sec3(x-~)+ sec(x- ~)tan2(x- @) ~: 0 for anyx inthe domain off.

Concave upward: (0, x), (2¢r, 3n’)

Concave downward: (n-, 2re), (3n’, 4n’)


f(x) = sinx+cosx, 0_< x < 2n"

f’(x) = cos x - sin x

f"(x) = sin x - cos x

3~ 7xf"(x) = 0 when x - ,

4 4

3~ 3~ 77/"Test interval: 0<x<-- ~<x<~ --<x<2n"

4 4 4 4

Sign of f"(x): f"(x) < 0 f"(x) > 0 S"(-) 0Conclusion: Concave downward Concave upward Concave downward

Points of inflection: 0), (-~, o)

f(x) = 2sinx+sin2x, 0 < x < 2n"

/’(x) = 2 cos x + 2 cos 2x

f"(x) = -2 sin x - 4 sin 2x = -2 sin x(1 + 4 cos x)

f"(x) = 0 when x = 0, 1.823, n’, 4.460.

Test interval: 0 < x < 1.823 1.823 < x < ~r 4.460 4.460 < x < 2n"

Sign of/"(x): f" < 0 f" > 0 f" < 0 f" > 0

Conclusion: Concave downward Concave upward Concave downward Concave upward

Points of inflection: (1.823,1.452), (n’, 0), (4.46,-1.452)



36. f(x) = x + 2 cos x, [0, 2n-]

f’(x) = 1-2sinx

f"(x) = -2 cos x ,a- 3;rt"

f"X = 0whenx = 2’ 2

n" 3n- 3~Test intervals: 0<x<-- --<x<-- <x<2~

2 2 2 2

Sign of f"(x)" f" < 0

Conclusion: Concave downward Concave upward Concave downward

Points of inflection: (@, ~), (~f, 3@)

37. f(x) = (x- 5)2

f’(x) = 2(X- 5)

s"(x) = 2Critical number: x = 5

f"(5) > 0

Therefore, (5, 0) is a relative minimum.

38. y(x) = -(x- 5)2f’(x) = -2(x- 5)

f"(x) = -2Critical number: x = 5

f"(5) < 0

Therefore, (5, 0) is a relative maximum.

39. f(x) = 6x- x2

f’(x) = 6-2x

f"(x) = -2

Critical number: x = 3

f"(3) < 0


40.

41.

42.

f(x) = x3 -3x2 +3

f’(x) = 3x2- 6x = 3x(x- 2)

f"(x) = 6x - 6 = 6(x - 1)

Critical numbers: x = 0, x = 2

f"(0) = -6 < 0


f"(2) = 6 > 0

Therefore, (2, -1) is a relative minimum.

f(x) = x3 -5x2 +7x

f’(x) = 3x2 - 10x + 7 = (3x - 7)(x - 1)

f"(x) = 6x- 10

7Critical numbers: x = 3’ 1

Therefore, (~, 4~-~) is arelative minimum.

f"(1) = -4 < 0

Therefore; (1, 3) is a relative maximum.

f(x) = x2 +3x-8

f’(x) = 2x+3

f"(x) = 2

Critical number: x = __32

Therefore, (--},-~) is a relative minimum.

43. f(x) = x4- 4x3 + 2

f’(x) = 4x3 - lZx2 = 4xZ(x- 3)

f"(x) = 12x2 - 24x = lZx(x- 2)

Critical numbers: x = 0, x = 3

However, f"(0) = 0, so you must use the First

Derivative Test. f’(x) < 0 on the intervals (-0% 0)and

(0, 3); so, (0, 2)is not an extremum, f"(3) > 0 so

(3, -25) is a relative minimum.


Section 3.4

44. f(x) = -x4 + 4x3 + 8x2

f’(x) = -4x3 + 12x: + 16x : -4x(x i 4)(x + 1)

f"(x) = -12x2 + 24x + 16 = -4(3x2- 6x- 4)

Critical numbers: x = -1, 0, 4

f"(-1) = -20

Therefore (-1, 3) is a relative maximum.

f"(0) = 16


f"(4) = -80


45. g(x)= x:(6- x)3

g’(x) = x(x - 6)2(12 - 5x)

g"(x) = 4(6- x)(5x~ - 24x + 18)

Critical numbers: x = 0, !~, 6

g"(0) = 432 > 0


g"(~) =-155.52 <0

Therefore, (~, 268.7)is a relative maximum.

g"(6) = 0

Test fails. By the First Derivative Test, (6, 0) is not an

extremum.

46. 1 2):(x 4)2g(x) =-~(~ + -

g’(x) = -(x- 4)(x- 1)(x + 2)2

g"(x) = 3 + 3x-2

Critical numbers: x = -2,1, 4

g"(-2) = -9 < 0

(-2, 0) is a relative maximum.

9->0g"(1) = 2

(1, -10.125) is a relative minimum.

g"(4) = -9 < 0

(4, O) is a relative maximum.

Concavity and the Second Derivative Test 257 ¯

47,

48.

49.

50.

51,

f(x) = X2/3 -- 3

f’(x) -2

3xlD

f"(x) - 29X4/3

Critical number: x = 0

However, f"(0) is undefined, so you must use the First

Derivative Test. Because f’(x) < 0 on (-0% 0) and

f’(x) > 0 on (0, m), (0,-3) is a relative minimum.

f(x) = x/-x~ +1

xf’(x) - ~x2 +1

f"(x) - 1

Critical number: x -- 0

f"(0) = 1 > 0


4= x +-X

4 x: - 4S’(x) = -x2 X2

8s"(x) = 7Critical numbers: x = +2

f"(-2) < 0

Therefore, (-2, -4) is a relative maximum.

f"(2) > 0


f(x) -x

x-1

f’(x) --1

(x- i):There are no critical numbers and x = 1 is not in thedomain. There are no relative extrema.

f(x) = Cosx-x, 0 < x < 4x

f’(x) =-sinx-1 < 0Therefore, f is non-increasing and there are no relativeextrema.



52. f(x) = 2sinx +cos2x, O < x < 2z

f’(x) = 2 cos x - 2 sin 2x = 2 cos x - 4 sin x cos x

= 2 cos x(1 - 2 sin x) = 0 when x -

f"(x) = -2 sin x - 4 cos 2x

Relative maxima: (~, -~), (-~, ~/

Relative minima: /~, 1), (~-f, - 3)

53. s(-) -- 0.2x2(x- 3)3, {-1, 4](a) S’(x) : 0.2x(5x- 6)(x- 3)2

f"(x) = (x - 3)(4x2 - 9.6x + 3.6)

: 0.4(x- 3)(10x2- 24x + 9)

(b) f"(0) < 0 ~ (0, 0)is a relative maximum.

f"(~) > 0 ~ (1.2,-1.6796) is a relative minimum.

Points of inflection:

(3, 0), (0.4652,-0.7048), (1.9348, -0.9049)

(c)

f~

f is increasing when f’ > 0 and decreasing whenf’ < 0. f is concave upward when f" > 0 and

concave downward when f" < 0.

y’(x)-f’(x) = 0 when x = O, x = +2.

+

f"(x) = 0 when x =9

2

. (b) f"(O) > 0 ~ (0, O)is a relative minimum.

"+ (+2, 4x/~) aref (_2) < 0 ~ relative maxima.

Points of inflection: (+1.2758, 3.4035)

(c) yf

-3 3

f,

The graph of f is increasing when f’ > 0 anddecreasing when f’ < 0. f is concave upward

when f" > 0 and concave downward whenf" < 0.

1 155. f(x] = sin x - - sin 3x + - sin 5x, [0, n’]

3 5 .I

(a) f’(x) = cosx - cos3x + cos5x

() - =7f’x = 0whenx = 6,x ,x = ~.6

f"(x) = -sin x + 3 sin 3x - 5 sin 5X

f"(x) = 0 when x = -~,x

x = 1.1731, x = 1.9685

(b)

(c)

f < 0 ~ ,1.53333 is a relative

maximum.

Points of inflection: (@, 0.2667), (1.1731, 0.9638),

(1.9685, 0.9637),/~-~, 0.2667)

Note: (0, O)and (n-, O)are not points of inflection

because they are endpoints.

The graph of f is increasing when f’ > 0 anddecreasing when f’ < 0. f is concave upward when

f" > 0 and concave downward when f" < 0.



~6. f(x) = ~x sin x, [0, 2~r]

sin x(a) = cos x +Critical numbers: x ,~ 1.84, 4.82

COS X COS Xf"(x) = -~i-~x sin x + ~ + ff~x

2 cos x (4x~ + 1) sin x

4x cos x - (4x~ + 1) sin x

2x~-~x

sin x

2x,~x

(b) Relative maximum: (1.84,1.85)

Relative minimum: (4.82, -3.09)

Points of inflection: (0.75, 0.83), (3.42,-0.72)

-4-

fis increasing when f’ > 0 and decreasing when f’ < 0. f is concave upward

when f" > 0 and concave downward when f" < 0.

27. (a)

(b)

Y

4

3

2

1

1 2 3 4

f’ < 0 meansfdecreasing

f’ increasing means concave upward

4-

3"

2-

l"

__1 2 3 4

f’ > 0meansfincreasing

f’ increasing means concave upward

(b)

l"

f’ < 0meansfdecreasing

f’ decreasing means concave downward

3-

2

I

1 2 3 4

f’ > 0 meansfincreasing

f’ decreasing means concave downward



59. Answers will vary. Sample answer:

Let f(x) = x4.

f"(x) = 12x2

f"(O) = O, but (0, O)is not a point of inflection.),

60. (a) The rate of change of sales is increasing.

S" > 0

(b) The rate of change of sales is decreasing.

S’ > 0, S" < 0

(c) The rate of change of sales is constant.

S’ = C,S" = 0

(d) Sales are steady.

S = C,S’ = 0, S" = 0

(e) Sales are declining, but at a lower rate.

S’ < 0, S" > 0

(f) Sales have bottomed out and have started to rise.

S’>0

61.

2-

62.

63.

64.

65.

66.

67.

68.

-3-

(o, o)

~ x



’ /~,,~. ’~

f" is linear.

f’ is quadratic.

fis cubic.

fconcave upward on (-oo, 3), downward on (3, oo).

70. (a)

/lI~-t

1o

(b) Because the depth d is always increasing, there areno relative extrema, f’(x) > 0

(c) The rate of change of d is decreasing until you reachthe widest point of the jug, then the rate increases untilyou reach the narrowest part of the jug’s neck, then therate decreases until you reach the top of the jug.

71.(a) n =1" n = 2: n = 3: n = 4:

f(x) = x- 2 f(x) = (x- 2)2 f(x) = (x- 2)3 f(x) = (x- 2)4

f’(x) = 1 f’(x) = 2(x- 2) f’(x) = 3(x- 2)2 f’(x) = 4(x- 2)3

f"(x) = 0 f"(x) = 2 f"(x) = 6(x- 2) f"(x) = 12(x- 2)2

No point of inflection No point of inflection Point of inflection: (2, 0) No point of inflection

Relative minimum: (2, 0) Relative minimum: (2, 0)

-6 -6 -6 -6

Conclusion: If n _> 3 and n is odd, then (2, 0) is point of inflection. Ifn >_ 2 and n is even, then (2, 0) is a relative minimum.

(b) Let f(x) = (x- 2)", f’(x) = n(x- 2)"-l, f"(x) = n(n -1)(x- 2)’’-2.

For n >_ 3 and odd, n - 2 is also odd and the concavity changes at x = 2.

For n > 4 and even, n - 2 is also even and the concavity does not change at x = 2.

So, x = 2 is point of inflection if and only if n > 3 is odd.

72. (a) f(x) : 3~x

s’(.) ==


(b) f"(x) does not exist at x = 0.

Y

2-

l- (~0,0)¯ ’ ~ ’ ": ’ I I >-x

-3-

© 2010 Brooks/Cole, Cengage Learning....


73. f(x) : ax3 +bx2 +cx+ d

Relative maximum: (3, 3)

Relative minimum: (5, 1)


f’(x) = 3ax2 + 2bx + c, f"(x) = 6ax + 2b

f(3) = 27a+9b +3e+d = 3 198a+16b+2c =-2 ~ 49a+8b+c =-1

f(5) 125a + 25b + 5c + d = 1

f’(3) = 27a + 6b + c = 0, f"(4) = 24a + 2b = 0

49a+8b +e =-1 24a+ 2b = 0

27a +6b +c = 0 22a+2b =-1

22a+2b =-1 2a = 1

a = ½, b =-6, c = .~.,d =-24

f(x) : ½x3 - 6x~ + ~x - 24

74. f(x) : ax3 + bx2 + cx + d

Relative maximum: (2, 4)

Relative minimum: (4, 2)


f’(x) = 3ax2 + 2bx + c, f"(x) = 6ax + 2b

f(2) = 8a+4b+2c+d = 4 !,56a+12b+2c =-2 ~ 28a+6b+c =-1

f(4) = 64a+16b+4c+d = 2

f’(2) = 12a+4b+c = 0, f’(4) = 48a+Sb+c = 0, f"(3) = 18a+2b = 0

28a+6b+c = -1 18a+2b = 0

12a+4b +c = 0 16a+2b = -1

16a+2b =-1 2a = 1

a = ½, b =--92, c = 12, d =-613 92f(x) = -~x --~x +12x-6

75. f(x) = ax3 + bx2 + cx + ~tMaximum: (-4,1)

Minimum: (0, 0)

(a) f’(x) = 3ax~ + 2~,x + c, f"(x) = 6ax + 2~,f(o) = o ~ d = of(-4) = 1 ~ -64a + 16b- 4c = 1

f’(-4) = 0 ~ 48a - 8b+ c = 0

f’(o) = o ~ c = oSolving this system yields a = ~ and b = 6a = ~16"

(b) The plane would be descending at the greatest rate at the point of inflection.

() ~ ’=o. x=-2.f" x = 6ax + 2b = x + g

Two miles ~om touchdown.



76. (a) line OA: y = -0.06x slope:-0.06

line CB: y = 0.04x + 50 slope: 0.04

f(x) : ax3 + bx2 + cx + d

f’(x) = 3ax2 + 2bx + c

(-1000, 60): 60 : (-1000)3a + (1000)2b -1000c + d

-0.06 = (1000)2 3a - 2000b + c

(1000, 90): 90 = (lO00)3a + (lO00)2b + lO00c + d

0.04 = (1000)2 3a + 2000b + c

(-lO0O, 60)A

y

15o-

(o, 5o)

(1000,90)B

The solution to this system of four equations is a = -1.25 × 10-8, b = 0.000025, c = 0.0275, and d = 50.

(b) y = -1.25 x 10-8x3 + 0.000025x2 + 0.0275x + 50lOO

-1100 1100

-10

(C) o.1

-0,1

11oo

(d) The steepest part of the road is 6% at the point A.

77. D = 2x4 - 5Lx3 + 3L2x~-

D’=8x3-15Lx2 + 6L2x = x(8x2 -15Lx + 6L2) =0

x = Oorx = - - . L16

By the Second Derivative Test, the deflection ismaximum when

x = L ~ 0.578L.

5.755T3 8.521T2 0.654T78. S - + + 0.99987,108 106 ~

0<T<25

(a) The maximum occurs when T ~ 4°andS = 0.999999.

(b)1.001-

1.000-

0.999-

0.998-

0.997-

0.996 -

I I I I I I~’-T5 10 15 20 25 30

(c) S(20°) ~ 0.9982

79. C = 0.5x2 + 15x + 5000

-- C 5000C - - 0.5x+15+--x x

__

C = average cost per unit--

dC 5000-- = 0.5 - 0whenx = 100dx x2

By the First Derivative Test, C is minimized whenx = 100 units.

300,00080. C = 2x +~X

C’ = 2 300,000 = 0 whenx = 100-,J~ ~ 387x2

By the First Derivative Test, is C minimized whenx = 387 units.



5000t281. S - 8+t2,0 < t < 3

(a)t 0.5 1 1.5 2 2.5 3

S 151.5 555.6 1097.6 1666.7 2193.0 2647.1

Increasing at greatest rate when 1.5 < t < 2

(b) 3ooo

oo

Increasing at greatest rate when t ~ 1.5.

5000t~(c) S = 8 + t~

1.633 yrs.

82. S-

(a)

100t2

65 + t:’ t > 0lOO

o

(b) S’(t)- 13,000t

(65 + t:)2

S"(t) 13,000(65- 3t:)= --0~t(65 + tz)3

= 4.65

S is concave upwards on (0, 4.65), concave downwards on (4.65, 30).

(c) S’(t) > 0for t > 0.

As t increases, the speed increases, but at a slower rate.



83,

84,

S’(-)

~,’(.)

= 2(sin x + cos x), f(~)= 2~/~

= 2(cos x- sin x),

: x- os i-

=0

P2"(x) = -2~/~ -4

The values of f, 4, P2, and their first derivatives are equal at x = z/4. The values of the second derivatives off and P2 are

equal at x = z/4. The approximations worsen as you move away from x = z/4.

s(x)S’(x)s"(x)

e2(x)4’(x)4"(x)

= 2(sin x + cos x), f(O) = 2

= 2(cos x- sin x), f’(O) = 2

= 2(-sin x - cos x), f"(O) = -2

= 2+2(x-0)= 2(1+x)

=2

= 2 + 2(x-O) + ½(-2)(x- O)2 = 2 + 2x- x2

= 2-2x

=-2

4

The values of f, P~, P2, and their first derivatives are equal at x = 0. The values of the second derivatives of f and P2 are

equal at x = 0. The approximations worsen as you move away from x = 0.

8~. f(x) = ~- x, f(O) = 111

f’(O) = -7f’(x) i

2.,,/-~- x’1

f"(O) 1f"(x) = 4(1- x)a/2’= -~-

pl.(X) : 1 _t_ (ii~)(x _ O) :1 x2

14(x) = --

2

P2(x) = 1+

14’(x) -21

~"(x) -4

1 1 x x2

2 85

The values of f, P~, P2, and their first derivatives are equal at x = 0. The values of the second derivatives off and P2 are

equal at x -- 0. The approximations worsen as you move away from x = 0.

Cen


86. f(x)- "/71, f(2) : ~X--

87.

88.

-(x + 1) 3 3,/5f’(x) = 2~x(x- 1)2’ f’(2) = --~ -

4

3x2 + 6x - 1 23 23x/~f"(x) = 4x312(x _ 1)3,f"(2) = ~

16

5,/5(x-2) - -x+--

4 2

v (x) = 45 +345

P2’ (x) = 4

p2"(x) - 23-4t~16

+ 23"/~(x- 2)16

The values of f, P~, P2 and their first derivatives are equal at x = 2. The values of the second derivatives off and P2 are equal

at x = 2. The approximations worsen as you move away from x = 2.

3

f’(x) = -7 cos + sin = -7 cos + sin

f"(x) =- 7gsin +Tgcos -Tgcos =-Tsin = 0

1X ~ ~

Point of inflection: (zl--, 0)

When x > l/n-, f" < O, so the graph is concave downward.1

f(x) = x(x-6)2 = x3 -12x2 +36x

f’(x) = 3x2- 24x + 36 = 3(x- 2)(x- 6)= 0

f"(x) = 6x- 24 = 6(x-4) = 0

Relative extrema: (2, 32) and (6, 0)

Point of inflection (4,16) is midway between the relative extrema off



89. Assume the zero of fare all real. Then express the function as f(x) = a(x - rl)(X - r2)(x - r3)where ~i, r2, and r3 are the

distinct zeros off From the Product Rule for a function involving three factors, we have

S’(~) : a[(~ - ,~)(~ - ~). (~ - ~,)(~ - ~). (~ - ~)(~_ ~)]

f"(x) = aE(x - rl) + (x - r2) + (x - rl) + (x- r3)+ (x - r2)+ (x - r3)~ = aE6x- 2(rl + r2 + r3)~.

Consequently, f"(x) = 0 if

2(~ + r2 + r3) rl + r2 + r3 (Average ofrl, r2, and r3).X ~ -- --

6 3

90. ptx)" " = ~x~ + t)x~ + cx +d

p’(x) = 3ax~ + 2bx + c

p"(x) = 6ax + 2b

6ax + 2b = 0

bX --

3a

The sign of p"(x)changes at x = -b/3a. Therefore, (-b/3a, p(-b/3a))is a point of inflection.

~9a2) 27a2 3a

When p(x) = x3 -3x2 +2, a = 1, b =-3, c = 0, and d = 2.

-(-3)x0- 3(1)

2(-3)3 (-3)(0)Y0 - 27(1)2 3(1) + 2 =-2-0+2 = 0

+d

The point of inflection of p(x)= x3- 3x~ + 2is (x0, Y0)= (1, 0).

91. True. Let y = ax3 ÷ bx2 + cx + d, a ~ O. Then

y" = 6ax + 2b = 0 when x = -(b/3a), and the

concavity changes at this point.

92. False. f(x) = 1Ix has a discontinuity at x = 0.

95. fand g are concave upward on (a, b) implies that f’ and

g’ are increasing on (a, b), and f" > 0 and g" > 0.

So, (f + g)" > 0 ~ f + g is concave upward on

(a, b) by Theorem 3.7.

93. False. Concavity is determined by f". For example, let

f(x) = x and c = 2. f’(c) = f’(2) > 0, butfis not

concave upward at c = 2.

94. False. For example, let f(x) = (x - 2)4.

96. f, g are positive, increasing, and concave upward on(a,b) ~ f(x) > O,f’(x) > 0and f"(x) > 0, and

g(x) > O, g’(x) > 0 and g"(x) >. 0 on (a, b).For

x e (a,b),

(fg)’(x) = f’(x)g(x) + f(x)g’(x)

(fg)"(x) = f"(x)g(x)+ 2f’(x)g’(x)+ f(x)g"(x) > 0

So, fg is concave upward on (a, b).


g(x) = 3x2 - x3

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