251)
Chapter 3
Applications of Differentiation
108. First observe that
1
sin x cos x + 1
----+
tanx+cotx+secx+cscxcos x sin x cos x sin x
sin2x + cos2x + sin x + cos x
sin x cos x
sin x cos x ~ x + cos x .
l+sinx+cosx(sinx+cosx-~)
(sin x + cos x)~ - 1
sin x cos x[sin x + cos x - 1)
2 sin x cos x
sin x cos x(sin x + cos x - 1)
2
sin x + cosx - 1
Let t = sin x + cos x - 1. The expression inside the absolute value sign is
2
sin x + cos x - 1
2
\(sin x + cos x - 13 + 1 +sin x + cos x - 1
2
=t+l+t
sin x + cos x +
+ cosxsln-= sin x -cos
4
4
Because sin x +
- ~(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
g(x) = 3x2 - x3
y=x2-x-2
y’ = 2x - 1
y" = 2
g’(x) = 6x- 3x2
g"(x) = 6 - 6x
Concave upward: (-0%
Concave upward: (-% 1)
y = -x3 + 3x2 - 2
y’ = -3x2 + 6x
y" = -6x + 6
Concave downward: (1, m)
Concave upward: (-% 1)
h"(x) = 2Ox3
h(x) = x5 - 5x + 2
hi(x) = 5X4 -- 5
Concave downward: (1, oo)
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
5X4 q- 20X3 -- 80X
f"(x) = 20x3 + 60x2 - 80
f’(x)
=
= 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)
11. f(x) - x224+12
f’-
x~-
12. f(x) = x2 +1
-48x
2
(x + 12)2
f"(x) = -2(3x2 -1)
Concave upward: (-oo,-2), (2,
Concave downward: (-2, 2)
© 2010 Brooks/Cole, Cengage Learning
252
Chapter 3 Applications of Differentiation
x2 +1
13. f(x) - x2 - 1
17. y = 2x-tanx, - ,
-4x
y’ = 2-sec2x
y" = -2 sec2x tan x
f"- 4(3x2 + 1)
Concave upward: (-oo,-1), (1,
Concave downward: (-1,1)
14.
=1
5 40X3 +
y "5-~(--3X
+
135X)
y’ = -~70(--15X4 + 120X2 + 135)
y"=--}x(x- 2)(x + 2)
Concave downward: /O, ~)
18. 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: (-m,-2), (0, 2)
Concave upward: (0, ~r)
Concave downward: (-2, 0), (2,
Concave downward: (-~r, O)
x2 +4
15. g(x) = 7"- x2
16x
(4 - x2)2
14
19. f(x) =-~x
+ 2x3
f’(x) = 2x3 + 6x2
f"(x) = 6x2 + 12x = 6x(x + 2)
f"(x) = 0 when x = O, -2
(4- x2)’ (2- x)3(2 + x)3
Concave upward: (-oo,-2), (0,
Concave upward: @2, 2)
Concave downward: (-2, O)
Concave downward: (-0%-2), (2, m)
Points of inflection: (-2,-8) and(O, O)
x2 - 1
16. h(x) - -2x -1
h’(x) = 2(x2- x + 1)
(2x -1)2
-6
h"(x) (2x -1)3
--X4 q- 24X2
f’(x) = --4X3 + 48X
f"(X) =--12X2 + 48 =12(4- X2) = 12(2 + X)(2- X)
20. f(x)
=
f"(x)=O for X=--2,2
Concave upward: (-2, 2)
Concave downward: (-oo,-2), (2,
Concave upward: (-m, ~)
Concave downward: (-~, o@
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
Section 3.4 Concavity and the Second Derivative Test 253
22. f(x) = 2x3 - 3x2 - 12x + 5
f’(x) = 6x2 - 6x - 12
Test interval:
-~<x<±
f"(x) = 12x - 6
Sign of f"(x)"
S"(x) < 0
S"(x) > 0
f"(x) = 12x - 6 = 0 when x =2’
_1
Conclusion:
Concave downward
Concave upward
~<x<~
2
2
Point of inflection: (½, -~-)
1
23. f(x) = -~x4 - 2x2
Test interval:
f’(x)
= x3 -
2
-oo < x <
2
2
2
/-5
4x
f"(x) = 3x2 - 4
Sign of f"(x):
s"(x) > o
S"(x) < o
S"(x) > o
f"(x) = 3x2 - 4 = 0 when x = +2
_ x/-5"
Conclusion:
Concave upward
Concave downward
Concave upward
Points of inflection: --- xi~’
"
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) > o
Conclusion:
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)
3
f"(x) = 0 when x = -, 2
2
Concave upward: (-oo, ~) and (2, oo)
27. f(x) = x-,,/Tx + 3, Domain:[-3, m)
f’(x) = x (x + 3)-’/2 +~+3 - 2~x+3
S"(x) =
6~x + 3 - 3(x + 2)(x + 3)-1/2
4(x + 3)
_ 3(x + 4)
4(x +
3)
Concave downward: (~, 2)
f"(x) > 0 on the entire domain off(except for
Points of inflection: (~,-~61, (2, O)
x = -3, for which f"(x) is undefined). There are no
points of inflection.
Concave upward: (-3,
© 2010 Brooks/Cole, Cengage Learning
254
Chapter 3 Applications of Differentiation
28. f(x) = xx/-~- x Domain:x < 9
29.
3(6 - x)
:’(~) - ~-94~- 7~
3/2
:"(x) 4(93(x
- x)-12)
:’(x) =
-8x
(: ÷ 0~
s"(x) : 8(3: -0
(: ÷ 0~
Concave downward: (-0% 9)
No point of inflection
:"(x) = 0 for x : -5-
x+l
30. f(x)- -..~x" Domain: x > 0
x-1
f’(x)- ~/~
3-x
f"(x) - -42/~
Test intervals:
0<x<3
3<x<m
Sign of f"(x):
f" > 0
f" < 0
Conclusion:
Concave upward
Concave downward
31. 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
© 2010 Brooks/Cole, Cengage Learning
Section 3.4 Concavity and the Second Derivative Test 255
32.
rCx~ = 2 csc- 0 < x < 2x
2’
S’(x) -- -3 csc 3x2 2cot 3x
=
~ + csc ~ cot~
2 2
~ 0 for ~y x in the domain off.
Concave upward: /0,
Concave downward’. (-~, ~~r-)
No point of inflection
33.
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’)
No point of inflection
34. f(x) = sinx+cosx, 0_< x < 2n"
f’(x) = cos x - sin x
f"(x) = sin x - cos x
3~ 7x
f"(x) = 0 when x - ,
44
3~
~<x<~
4 4
--<x<2n"
4
f"(x) < 0
f"(x) > 0
S"(-) 0
Concave downward
Concave upward
Concave downward
Test interval:
0<x<--
Sign of f"(x):
Conclusion:
3~
4
77/"
Points of inflection: 0), (-~, o)
35. 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)
© 2010 Brooks/Cole, Cengage Learning
256 Chapter 3 Applications of Differentiation
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
Test intervals:
0<x<--
2
n"
3n2 2
--<x<--
Sign of f"(x)"
Conclusion:
3~
2
<x<2~
f" < 0
Concave downward
Concave upward
Concave downward
Points of inflection: (@, ~), (~f, 3@)
37. f(x) = (x- 5)2
f’(x) = 2(X- 5)
41. f(x) = x3 -3x2 +3
f’(x) = 3x2- 6x = 3x(x- 2)
s"(x) = 2
f"(x) = 6x - 6 = 6(x - 1)
Critical number: x = 5
Critical numbers: x = 0, x = 2
f"(5) > 0
f"(0) = -6 < 0
Therefore, (5, 0) is a relative minimum.
38. y(x) = -(x- 5)2
f’(x) = -2(x- 5)
f"(x) = -2
Critical number: x = 5
f"(5) < 0
Therefore, (5, 0) is a relative maximum.
39. f(x) = 6x-
x2
Therefore, (0, 3) is a relative maximum.
f"(2) = 6 > 0
Therefore, (2, -1) is a relative minimum.
42. f(x) = x3 -5x2 +7x
f’(x) = 3x2 - 10x + 7 = (3x - 7)(x - 1)
f"(x) = 6x- 10
Critical numbers: x =73’ 1
f’(x) = 6-2x
f"(x) = -2
Critical number: x = 3
f"(3) < 0
Therefore, (3, 9) is a relative maximum.
40. f(x) = x2 +3x-8
Therefore, (~, 4~-~) is arelative minimum.
f"(1) = -4 < 0
Therefore; (1, 3) is a relative maximum.
43. f(x)
= x4- 4x3
f"(x) = 2
+2
3
f’(x) = 4x - lZx2 = 4xZ(x- 3)
f"(x) = 12x2 - 24x = lZx(x- 2)
Critical number: x = __3
2
Critical numbers: x = 0, x = 3
f’(x) = 2x+3
However, f"(0) = 0, so you must use the First
Derivative Test. f’(x) < 0 on the intervals (-0% 0)and
Therefore, (--},-~) is a relative minimum.
(0, 3); so, (0, 2)is not an extremum, f"(3) > 0 so
(3, -25) is a relative minimum.
© 2010 Brooks/Cole, Cengage Learning
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)
Concavity and the Second Derivative Test 257 ¯
47, f(x) = X2/3 -- 3
2
f’(x) 3xlD
Critical numbers: x = -1, 0, 4
2
f"(x) - 9X4/3
f"(-1) = -20
Critical number: x = 0
Therefore (-1, 3) is a relative maximum.
However, f"(0) is undefined, so you must use the First
f"(0) = 16
Derivative Test. Because f’(x) < 0 on (-0% 0) and
f’(x) > 0 on (0, m), (0,-3) is a relative minimum.
Therefore, (0, 0) is a relative minimum.
f"(4) = -80
Therefore, (4, 128) is a relative maximum.
48. f(x) = x/-x~ +1
f’(x) - ~x2x +1
45. g(x)= x:(6- x)3
f"(x) - 1
g’(x) = x(x - 6)2(12 - 5x)
g"(x) = 4(6- x)(5x~ - 24x + 18)
Critical number: x -- 0
Critical numbers: x = 0, !~, 6
f"(0) = 1 > 0
g"(0) = 432 > 0
Therefore, (0, 1) is a relative minimum.
Therefore, (0, 0) is a relative minimum.
g"(~) =-155.52 <0
49.
= x +-4X
Therefore, (~, 268.7)is a relative maximum.
4 x: - 4
S’(x) =
x2 X2
g"(6) = 0
s"(x) = 87
Test fails. By the First Derivative Test, (6, 0) is not an
extremum.
Critical numbers: x = +2
46. g(x) =-~(~
1 2):(x+4)2 g’(x) = -(x- 4)(x- 1)(x + 2)
2
f"(-2) < 0
Therefore, (-2, -4) is a relative maximum.
f"(2) > 0
Therefore, (2, 4) is a relative minimum.
g"(x) = 3 + 3x-
2
Critical numbers: x = -2,1, 4
g"(-2) = -9 < 0
(-2, 0) is a relative maximum.
9
g"(1) = ->0
2
(1, -10.125) is a relative minimum.
g"(4) = -9 < 0
50. f(x) f’(x) -
x
x-1
-1
(x- i):
There are no critical numbers and x = 1 is not in the
domain. There are no relative extrema.
51, f(x) = Cosx-x, 0 < x < 4x
f’(x) =-sinx-1 < 0
Therefore, f is non-increasing and there are no relative
extrema.
(4, O) is a relative maximum.
© 2010 Brooks/Cole, Cengage Learning
258 Chapter 3 Applications of Differentiation
52. f(x) = 2sinx +cos2x, O < x < 2z
f’(x) = 2 cos x - 2 sin 2x = 2 cos x - 4 sin x cos x
. (b) f"(O) > 0 ~ (0, O)is a relative minimum.
f "+
(_2) < 0 ~(+2, 4x/~) arerelative maxima.
= 2 cos x(1 - 2 sin x) = 0 when x f"(x) = -2 sin x - 4 cos 2x
Points of inflection: (+1.2758, 3.4035)
y
(c)
f
-3
3
f,
The graph of f is increasing when f’ > 0 and
decreasing when f’ < 0. f is concave upward
when f" > 0 and concave downward when
f" < 0.
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)
1
1
55. f(x] = sin x - - sin 3x + - sin 5x, [0,.I n’]
3
5
(a) f’(x) = cosx - cos3x + cos5x
,x = ~.
6
f"(x) = -sin x + 3 sin 3x - 5 sin 5X
f’x = 0whenx = 6,x
()
: 0.4(x- 3)(10x2- 24x + 9)
(b) f"(0) < 0 ~ (0, 0)is a relative maximum.
f"(x) = 0 when x = -~,x
f"(~) > 0 ~ (1.2,-1.6796) is a relative minimum.
Points of inflection:
(3, 0), (0.4652,-0.7048), (1.9348, -0.9049)
- =7
x = 1.1731, x = 1.9685
(b) f
(c)
< 0 ~ ,1.53333 is a relative
maximum.
f~
Points of inflection: (@, 0.2667), (1.1731, 0.9638),
(1.9685, 0.9637),/~-~, 0.2667)
f is increasing when f’ > 0 and decreasing when
f’ < 0. f is concave upward when f" > 0 and
concave downward when f" < 0.
Note: (0, O)and (n-, O)are not points of inflection
because they are endpoints.
(c)
y’(x)f’(x) = 0 when x = O, x = +2.
+
f"(x) = 0 when x =
The graph of f is increasing when f’ > 0 and
decreasing when f’ < 0. f is concave upward when
f" > 0 and concave downward when f" < 0.
9
2
© 2010 Brooks/Cole, Cengage Learning
Section 3.4 Concavity and the Second Derivative Test 259
~6. f(x) = ~x sin x, [0, 2~r]
(a) = cos x + sin x
Critical numbers: x ,~ 1.84, 4.82
COS X
COS X
f"(x) = -~i-~x sin x + ~ + ff~x
sin x
2x,~x
2 cos x (4x~ + 1) sin x
4x cos x - (4x~ + 1) 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)
Y
4
3
2
1
l"
1
2
3
4
f’ < 0 meansfdecreasing
f’ < 0meansfdecreasing
f’ increasing means concave upward
f’ decreasing means concave downward
(b)
(b)
43"
3-
2-
2
l"
I
__
1
2
3
4
1
2
3
4
f’ > 0meansfincreasing
f’ > 0 meansfincreasing
f’ increasing means concave upward
f’ decreasing means concave downward
© 2010 Brooks/Cole, Cengage Learning
260
Chapter 3 Applications of Differentiation
59. Answers will vary. Sample answer:
Let f(x) = x4.
64.
f"(x) = 12x2
f"(O) = O, but (0, O)is not a point of inflection.
),
-3-
65.
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
66.
(c) The rate of change of sales is constant.
S’ = C,S" = 0
(o, o)
(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
~
x
67.
61.
2-
68.
62.
63.
© 2010 Brooks/Cole, Cengage Learning
Section 3.4 Concavity and the Second Derivative Test 261
70. (a)
’
/
/~,,~. ’~
lI~-t
1o
f" is linear.
f’ is quadratic.
(b) Because the depth d is always increasing, there are
no relative extrema, f’(x) > 0
fis cubic.
fconcave upward on (-oo, 3), downward on (3, oo).
71.(a) n =1"
n = 2:
(c) The rate of change of d is decreasing until you reach
the widest point of the jug, then the rate increases until
you reach the narrowest part of the jug’s neck, then the
rate decreases until you reach the top of the jug.
n = 3:
n = 4:
f(x) = x- 2
f(x) = (x- 2)2
f(x) = (x- 2)3
f’(x) = 1
f’(x) = 2(x- 2)
f’(x) = 3(x- 2)2
f"(x) = 0
f"(x) = 2
f"(x) = 6(x- 2)
f’(x) = 4(x- 2)3
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)
-6
-6
f(x) = (x- 2)4
Relative minimum: (2, 0)
-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’(.) =
=
Point of inflection: (0, 0)
(b) f"(x) does not exist at x = 0.
Y
2-
l- (~0,0)
¯ ’
~
’ ":
’
I
I >-x
-3-
....
© 2010 Brooks/Cole, Cengage Learning
262
Chapter 3 Applications of Differentiation
73. f(x) : ax3 +bx2 +cx+ d
Relative maximum: (3, 3)
Relative minimum: (5, 1)
Point of inflection: (4, 2)
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
22a+2b =-1
27a +6b +c = 0
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)
Point of inflection: (3, 3)
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
16a+2b = -1
12a+4b +c = 0
= 1
16a+2b
=-1
2a
a = ½, b =--92, c = 12, d =-6
92
f(x) = 13
-~x
--~x +12x-6
75. f(x) = ax3 + bx2 + cx + ~t
Maximum: (-4,1)
Minimum: (0, 0)
(a) f’(x) = 3ax~ + 2~,x + c, f"(x) = 6ax + 2~,
f(o) = o ~ d = o
f(-4) = 1 ~ -64a + 16b- 4c = 1
f’(-4) = 0 ~ 48a - 8b+ c = 0
c=o
f’(o) = o ~
Solving 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.
© 2010 Brooks/Cole, Cengage Learning
Section 3.4 Concavity and the Second Derivative Test 263
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
y
15o-
f’(x) = 3ax2 + 2bx + c
(1000,90)
B
(-lO0O, 60)
A
(-1000, 60): 60 : (-1000)3a + (1000)2b -1000c + d
(o, 5o)
-0.06 = (1000)2 3a - 2000b + c
(1000, 90): 90 = (lO00)3a + (lO00)2b + lO00c + d
0.04 = (1000)2 3a + 2000b + c
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 + 50
lOO
-1100
1100
-10
o.1
(C)
11oo
-0,1
(d) The steepest part of the road is 6% at the point A.
77. D = 2x4 - 5Lx3 + 3L2x~-
(c) S(20°) ~ 0.9982
D’=8x3-15Lx2 + 6L2x = x(8x2 -15Lx + 6L2) =0
x = Oorx =
16
-
.L
By the Second Derivative Test, the deflection is
maximum when
79.
C = 0.5x2 + 15x + 5000
-CC- 0.5x+15+-- 5000
__
x
x
C = average cost per unit
--
x=
L ~ 0.578L.
3
78. S - 5.755T
108
8.521T2 0.654T
+
+ 0.99987,
106 ~
0<T<25
(a) The maximum occurs when T ~ 4°and
S = 0.999999.
(b)
dC= 0.5
-dx
5000
- 0whenx = 100
x2
By the First Derivative Test, C is minimized when
x = 100 units.
80. C = 2x +~ 300,000
X
C’ = 2 300,000
= 0 whenx = 100-,J~ ~ 387
2
x
By the First Derivative Test, is C minimized when
x = 387 units.
1.0011.0000.9990.9980.9970.996 I I I I I I~’-T
5 10 15 20 25 30
© 2010 Brooks/Cole, Cengage Learning
264
Chapter 3 Applications of Differentiation
5000t2
81. S - 8+t2,0 < t < 3
(a)
t
0.5
1
1.5
2
S
151.5
555.6
1097.6 1666.7
2.5
3
2193.0
2647.1
Increasing at greatest rate when 1.5 < t < 2
(b) 3ooo
o
o
Increasing at greatest rate when t ~ 1.5.
5000t~
(c) S = 8 + t~
1.633 yrs.
82. S-
(a)
100t2
65 + t:’ t > 0
lOO
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.
© 2010 Brooks/Cole, Cengage Learning
Section 3.4 Concavity and the Second Derivative Test 265
83,
= 2(sin x + cos x), f(~)= 2~/~
S’(-) = 2(cos x- sin x),
i-
: x- os
~,’(.) =0
-4
P2"(x) = -2~/~
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.
84,
s(x) = 2(sin x + cos x),
S’(x) = 2(cos x- sin x),
s"(x) = 2(-sin x - cos x),
f(O) = 2
f’(O) = 2
f"(O) = -2
= 2+2(x-0)= 2(1+x)
=2
4
= 2 + 2(x-O) + ½(-2)(xe2(x)
4’(x) = 2-2x
4"(x) =-2
O)2 = 2
+ 2x- x2
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,
1
f’(x) i 2.,,/-~- x’
1
f"(x) = 4(1- x)a/2’
f(O) = 1
1
f’(O) = -7
f"(O) 1
= -~-
pl.(X) : 1 _t_ (ii~)(x _ O) :1 x2
1
4(x) = -2
P2(x) = 1+
11
x x2
28
5
1
2
1
~"(x) 4
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.
4’(x) -
Cen
266
Chapter 3 Applications of Differentiation
86. f(x)- "/71,
X--
f(2) : ~
-(x + 1) 1)2’
f’(x) = 2~x(x-
3
f’(2) = --~ -
3,/5
4
f"(x) = 4x312(
1)-31
,
3x2 x+ _6x
f"(2) = ~
(x-2) -
23
23x/~
16
5,/5
4
-x+--
2
v (x) = 45 +
345
P2’ (x) = 4 + 23"/~(x- 2)
16
p2"(x) - 23-4t~
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
87.
f’(x) = -7 cos + sin = -7 cos + sin
f"(x) =- 7gsin
X~~
+Tgcos
-Tgcos
=-Tsin
=0
1
Point of inflection: (zl--, 0)
When x > l/n-, f" < O, so the graph is concave downward.
1
88.
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
© 2010 Brooks/Cole, Cengage Learning
Section 3.4 Concavity and the Second Derivative Test 267
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
X~
90.
2(~ + r2 + r3) rl + r2 + r3 (Average ofrl, r2, and r3).
--6
3
ptx)" " = ~x~ + t)x~ + cxd+
p’(x) = 3ax~ + 2bx + c
p"(x) = 6ax + 2b
6ax + 2b = 0
b
X -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
+d
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
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.
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.
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.
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).
© 2010 Brooks/Cole, Cengage Learning