250
Chapter 3
Applications of Differentiation
108. First observe that
sin x
cos x
1
1
cos x
sin x
cos x sin x
tan x cot x sec x csc x
sin 2 x cos 2 x sin x cos x
sin x cos x
1 sin x cos x § sin x cos x 1 ·
¨
¸
sin x cos x © sin x cos x 1 ¹
sin x
cos x 1
sin x cos xsin x cos x 1
2
2 sin x cos x
sin x cos xsin x cos x 1
2
sin x cos x 1
Let t
f t sin x cos x 1. The expression inside the absolute value sign is
sin x cos x sin x
2
sin x cos x 1
cos x 1 1 t 1
2
sin x cos x 1
2
t
S·
§
Because sin ¨ x ¸
4¹
©
sin x cos
S
4
cos x sin
S
4
2
sin x cos x,
2
sin x cos x  ª¬
2, 2 º¼ and
sin x cos x 1  ª¬1 t
f ct f 1 2
1
1 t t2 2
t2
2
t2
2 1
4 2§
¨
2 1 ¨©
2 º¼.
2, 1 2 t t
2
1 2 1·
¸
2 1 ¸¹
2
2 1
4 2 2 4
1
2 2
For t ! 0, f is decreasing and f t ! f 1 For t 0, f is increasing on 2
2
2 1, 2
2
2 3 2
23 2
2 , then decreasing on 2, 0 . So f t f 2
1 2 2.
Finally, f t t 2 2 1.
(You can verify this easily with a graphing utility.)
Section 3.4 Concavity and the Second Derivative Test
1. The graph of f is increasing and concave upwards:
f c ! 0, f cc ! 0
3. The graph of f is decreasing and concave downward:
f c 0, f cc 0
2. The graph of f is increasing and concave downwards:
f c ! 0, f cc 0
4. The graph of f is decreasing and concave upward:
f c 0, f cc ! 0
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
5.
Concavity and the Second Derivative Test
g x
3x 2 x3
2x 1
g c x
6 x 3x 2
2
g cc x
6 6x
y
x2 x 2
yc
ycc
7.
Concave upward: f, f
6.
y
x3 3x 2 2
yc
3 x 2 6 x
ycc
6 x 6
251
Concave upward: f, 1
Concave downward: 1, f
8.
Concave upward: f, 1
Concave downward: 1, f
h x x5 5 x 2
hc x
5x4 5
hcc x
20 x3
Concave upward: 0, f
Concave downward: f, 0
9.
f x
x3 6 x 2 9 x 1
f c x
3x 2 12 x 9
f cc x
6 x 12
6 x 2
Concave upward: f, 2
Concave downward: 2, f
10.
f x
x5 5 x 4 40 x 2
f c x
5 x 4 20 x3 80 x
f cc x
20 x3 60 x 2 80
20 x3 3 x 2 4
20 x 1 x 2
2
Test Interval:
f x 2
2 x 1
1 x f
Sign of f cc :
f cc 0
f cc 0
f cc ! 0
Conclusion:
Concave downward
Concave downward
Concave upward
Concave upward: 1, f
Concave downward: f, 1
11. f x
fc
f cc
24
x 12
48 x
12.
2
x2
12
f c x
2
144 4 x 2 x
2
12
f x
3
Concave upward: f, 2, 2, f
Concave downward: 2, 2
x2
x 1
2x
2
x 2 1
23 x 2 1
f cc x
3
x 2 1
2
21 3x 1 x
2
1
3x
3
§
3
3·
Concave upward: ¨¨ ,
¸
3 ¸¹
© 3
§
3· § 3 ·
Concave downward: ¨¨ f, , f ¸¸
¸, ¨
3 ¸¹ ¨© 3
©
¹
© 2010 Brooks/Cole, Cengage Learning
252
Chapter 3
13. f x
fc
Applications of Differentiation
x2 1
x2 1
4 x
17.
x 1
43 x 2 1
3
x 2 1
2
2
f cc
18.
3x 40 x 135 x
15x 4 120 x 2 135
y
1
270
yc
1
270
ycc
92 x x 2 x 2
2 sec2 x
ycc
2 sec 2 x tan x
5
3
S , S y
x 2 csc x,
yc
1 2 csc x cot x
ycc
2 csc x csc 2 x 2 cot xcsc x cot x
2csc3 x csc x cot 2 x
Concave upward: f, 2, 0, 2
Concave upward: 0, S Concave downward: 2, 0, 2, f
Concave downward: S , 0
g x
g c x
x2 4
4 x2
16 x
19.
4 x 2 163x 2 4
g cc x
3
4 x 2 16.
yc
§ S·
Concave downward: ¨ 0, ¸
© 2¹
Concave downward: 1, 1
15.
§ S S·
2 x tan x, ¨ , ¸
© 2 2¹
§ S ·
Concave upward: ¨ , 0 ¸
© 2 ¹
Concave upward: f, 1, 1, f
14.
y
2
163x 2 4
2
x 2 x
3
f x
1 4
x
2
f c x
2 x3 6 x 2
f cc x
6 x 2 12 x
f cc x
0 when x
2 x3
6 x x 2
0, 2
Concave upward: f, 2, 0, f
3
Concave upward: 2, 2
Concave downward: 2, 0
Concave downward: f, 2, 2, f
Points of inflection: 2, 8 and 0, 0
h x hc x
hcc x
20.
x2 1
2x 1
2 x 2 x 1
2 x
1
2
6
2 x
1
x 4 24 x 2
f c x
4 x3 48 x
f cc x
12 x 2 48
f cc x
0
for x
12 4 x 2 12 2 x 2 x
2, 2
Concave upward: 2, 2
3
Concave downward: f, 2, 2, f
1·
§
Concave upward: ¨ f, ¸
2¹
©
§1 ·
Concave downward: ¨ , f ¸
©2 ¹
f x
Points of inflection: 2, 80, 2, 80
21.
f x
x3 6 x 2 12 x
f c x
3 x 2 12 x 12
f cc x
6 x 2
0 when x
2.
Concave upward: 2, f
Concave downward: f, 2
Point of inflection: 2, 8
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
22.
f x
2 x3 3 x 2 12 x 5
f c x
6 x 2 6 x 12
f cc x
12 x 6
f cc x
12 x 6
f c x
1 4
x 2x2
4
x3 4 x
f cc x
3x 2 4
f cc x
3x 2 4
f x
f x Sign of f cc x :
f cc x 0
1
2
1
2
253
x f
f cc x ! 0
Concave downward
Concave upward
12 , 132 Point of inflection:
23.
Test interval:
Conclusion:
1.
2
0 when x
Concavity and the Second Derivative Test
r
0 when x
Test interval:
f x Sign of f cc x:
f cc x ! 0
Conclusion:
2
.
3
2
3
2
x 3
2
3
f cc x 0
Concave upward
2
x f
3
f cc x ! 0
Concave downward
Concave upward
20 ·
§ 2
, ¸
Points of inflection: ¨ r
9¹
3
©
24.
f x
2x4 8x 3
f c x
8 x3 8
f cc x
24 x 2
0 when x
0.
However, 0, 3 is not a point of inflection because f cc x t 0 for all x.
Concave upward: f, f
25.
f x
f c x
x x 4
3
2
3
x ª3 x 4 º x 4
¬
¼
x
4 4 x 4
2
f cc x
4 x 1 ª¬2 x 4º¼ 4 x 4
f cc x
12 x 4 x 2
0 when x
2
4 x 4ª¬2 x 1 x 4º¼
f x 2
2 x 4
4 x f
Sign of f cc x:
f cc x ! 0
f cc x 0
f cc x ! 0
Concave upward
12 x 4 x 2
2, 4.
Test interval:
Conclusion:
4 x 43 x 6
Concave downward
Concave upward
Points of inflection: 2, 16, 4, 0
26.
2 x 1
f x
f x
x
f c x
f cc x
x 2 4 x 5
6 x 2 2 x 3
f c x
f cc x
0 when x
3
,2
2
f cc x
3
27.
2
·
2¸
¹
1·
§3
Points of inflection: ¨ , ¸, 2, 0
16 ¹
©2
1 2
§1·
x¨ ¸ x 3
© 2¹
6
x3
x 3 3 x 2 x 3
4 x 3
3 x 2
2
x 3
1 2
3 x 4
3·
§
Concave upward: ¨ f, ¸ and 2, f
2¹
©
§3
Concave downward: ¨ ,
©2
x 3, Domain: >3, f
x
4 x 3
32
f cc x ! 0 on the entire domain of f (except for
x
3, for which f cc x is undefined). There are no
points of inflection.
Concave upward: 3, f
© 2010 Brooks/Cole, Cengage Learning
254
28.
Chapter 3
f x
f c x
f cc x
Applications of Differentiation
x 9 x
Domain: x d 9
29.
36 x
f c x
2 9 x
4
x 1
8 x
2
x 2 1
83 x 2 1
f cc x
3
x 2 1
3 x 12
49 x
f x
32
Concave downward: f, 9
No point of inflection
f cc x
2
0 for x
r
3
3
§
3· § 3 ·
, f ¸¸
Concave upward: ¨¨ f, ¸, ¨
3 ¸¹ ¨© 3
©
¹
§
3
3·
Concave downward: ¨¨ ,
¸
3 ¸¹
© 3
§
3
,
Points of inflection: ¨¨ 3
©
30.
f x
f c x
f cc x
0 x 3
3 x f
Sign of f cc x :
f cc ! 0
f cc 0
Conclusion:
f c x
·
3¸¸
¹
§ 4 3·
¨¨ 3,
¸
3 ¸¹
©
Test intervals:
f x
§ 3
,
¨¨
© 3
x 1
, Domain: x ! 0
x
x 1
2 x3 2
3 x
4 x5 2
4 ·
§
Point of inflection: ¨ 3,
¸
3¹
©
31.
·
3¸¸ and
¹
Concave upward
Concave downward
x
sin , 0 d x d 4S
2
1
§ x·
cos¨ ¸
2
© 2¹
1
§ x·
sin ¨ ¸
4
© 2¹
f cc x
f cc x
0 when x
0, 2S , 4S .
Point of inflection: 2S , 0
Test interval:
0 x 2S
2S x 4S
Sign of f cc x:
f cc 0
f cc ! 0
Conclusion:
Concave downward
Concave upward
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
32.
f x
f c x
f cc x
Concavity and the Second Derivative Test
255
3x
, 0 x 2S
2
3x
3x
3 csc
cot
2
2
9 § 3 3x
3x
3x ·
csc
cot 2 ¸ z 0 for any x in the domain of f .
¨ csc
2©
2
2
2¹
2 csc
§ 2S · § 4S
·
Concave upward: ¨ 0,
¸, ¨ , 2S ¸
© 3 ¹ © 3
¹
§ 2S 4S ·
Concave downward: ¨ ,
¸
© 3 3 ¹
No point of inflection
33.
f x
S·
§
sec¨ x ¸, 0 x 4S
2¹
©
f c x
S· §
S·
§
sec¨ x ¸ tan ¨ x ¸
2¹
2¹
©
©
f cc x
S·
S·
S·
§
§
§
sec3 ¨ x ¸ sec¨ x ¸ tan 2 ¨ x ¸ z 0 for any x in the domain of f .
2¹
2¹
2¹
©
©
©
Concave upward: 0, S , 2S , 3S Concave downward: S , 2S , 3S , 4S No point of inflection
34.
f x
sin x cos x, 0 d x d 2S
f c x
cos x sin x
f cc x
sin x cos x
f cc x
0 when x
3S 7S
,
.
4 4
3S
4
Test interval:
0 x Sign of f cc x:
f cc x 0
Conclusion:
Concave downward
3S
7S
x 4
4
7S
x 2S
4
f cc x ! 0
f cc x 0
Concave upward
Concave downward
§ 3S · § 7S ·
Points of inflection: ¨ , 0 ¸, ¨ , 0 ¸
© 4 ¹ © 4 ¹
35.
f x
2 sin x sin 2 x, 0 d x d 2S
f c x
2 cos x 2 cos 2 x
f cc x
2 sin x 4 sin 2 x
f cc x
0 when x
2 sin x1 4 cos x
0, 1.823, S , 4.460.
Test interval:
0 x 1.823
1.823 x S
S x 4.460
4.460 x 2S
Sign of f cc x:
f cc 0
f cc ! 0
f cc 0
f cc ! 0
Conclusion:
Concave downward
Concave upward
Concave downward
Concave upward
Points of inflection: 1.823, 1.452, S , 0, 4.46, 1.452
© 2010 Brooks/Cole, Cengage Learning
256
36.
Chapter 3
Applications of Differentiation
f x
x 2 cos x, >0, 2S @
f c x
1 2 sin x
f cc x
2 cos x
f cc x
0 when x
S 3S
2
,
2
.
Test intervals:
0 x Sign of f cc x:
f cc 0
Conclusion:
S
S
2
2
Concave downward
x 3S
2
f cc ! 0
Concave upward
3S
x 2S
2
f cc 0
Concave downward
§ S S · § 3S 3S ·
Points of inflection: ¨ , ¸, ¨ ,
¸
©2 2¹ © 2 2 ¹
37.
f x
f c x
x 5
2 x 5
f cc x
2
2
41.
Critical number: x
5
f x
x 5
3x 2 6 x
f cc x
6x 6
f c x
2 x 5
f cc x
2
f cc 2
5
42.
Therefore, 5, 0 is a relative maximum.
6 x x2
f c x
6 2x
f cc x
2
6 x 1
0, x
2
6 0
6 ! 0
Therefore, 2, 1 is a relative minimum.
f cc5 0
f x
3x x 2
Therefore, 0, 3 is a relative maximum.
2
Critical number: x
f x
x3 5 x 2 7 x
f c x
3x 2 10 x 7
f cc x
6 x 10
§7·
f cc¨ ¸
© 3¹
f cc3 0
Therefore, 3, 9 is a relative maximum.
x 2 3x 8
f c x
f cc x
7 x 1
4 ! 0
§ 7 49 ·
Therefore, ¨ ,
¸ is a relative minimum.
© 3 27 ¹
f cc1
4 0
3
f x
3 x
7
,1
3
Critical numbers: x
Critical number: x
40.
f c x
f cc0
Therefore, 5, 0 is a relative minimum.
39.
x3 3 x 2 3
Critical numbers: x
f cc5 ! 0
38.
f x
Therefore, 1, 3 is a relative maximum.
f x
x 4 4 x3 2
2x 3
f c x
4 x3 12 x 2
4 x 2 x 3
2
f cc x
12 x 2 24 x
12 x x 2
32
Critical number: x
Critical numbers: x
However, f cc0
f cc 32 ! 0
43.
Therefore, 32 , 41
is a relative minimum.
4
0, x
3
0, so you must use the First
Derivative Test. f c x 0 on the intervals f, 0 and
0, 3; so, 0, 2 is not an extremum.
3, 25 is a relative minimum.
f cc3 ! 0 so
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
44.
f x
x 4 4 x3 8 x 2
f c x
4 x3 12 x 2 16 x
f cc x
12 x 24 x 16
47.
4 x x 4 x 1
43x 6 x 4
2
20
80
f c x ! 0 on 0, f, 0, 3 is a relative minimum.
48.
g x
x 2 6 x
g c x
x x 6 12 5 x
g cc x
46 x5 x 2 24 x 18
f cc x
0,
g cc6
49.
125 , 268.7 is a relative maximum.
g c x
g cc x
f x
x 4
x
f c x
1
4
x2
x2 4
x2
r2
2
3 2
3 3x x
2
2, 1, 4
f cc 2 ! 0
Therefore, 2, 4 is a relative minimum.
50. f x
f c x
9 0
2, 0 is a relative maximum.
9
! 0
2
1, 10.125 is a relative minimum.
g cc 4
1 ! 0
Therefore, 2, 4 is a relative maximum.
1
x 22 x 42
8
x 4 x 1 x 2
Critical numbers: x
g cc1
0
f cc 2 0
g cc 2
32
8
x3
Critical numbers: x
extremum.
g x
1
f cc x
0
Test fails. By the First Derivative Test, 6, 0 is not an
46.
x2
Therefore, 0, 1 is a relative minimum.
155.52 0
Therefore,
x 1
1
f cc0
6
Therefore, 0, 0 is a relative minimum.
x
2
Critical number: x
432 ! 0
12
5
x2 1
f c x
2
Critical numbers: x
g cc
f x
3
12 ,
5
0
Derivative Test. Because f c x 0 on f, 0 and
Therefore, 4, 128 is a relative maximum.
g cc0
2
3 x1 3
2
43
9x
257
However, f cc0 is undefined, so you must use the First
Therefore, 0, 0 is a relative minimum.
45.
f c x
Critical number: x
16
f cc 4
x2 3 3
f cc x
Therefore 1, 3 is a relative maximum.
f cc0
f x
2
1, 0, 4
Critical numbers: x
f cc 1
Concavity and the Second Derivative Test
9 0
x
x 1
1
x
1
2
There are no critical numbers and x 1 is not in the
domain. There are no relative extrema.
51.
f x
f c x
cos x x, 0 d x d 4S
sin x 1 d 0
Therefore, f is non-increasing and there are no relative
extrema.
4, 0 is a relative maximum.
© 2010 Brooks/Cole, Cengage Learning
258
Chapter 3
52. f x
Applications of Differentiation
(b) f cc0 ! 0 Ÿ 0, 0 is a relative minimum.
2 sin x cos 2 x, 0 d x d 2S
f c x
2 cos x 2 sin 2 x
2 cos x 4 sin x cos x
2 cos x1 2 sin x
f cc x
, , , .
6 2 6 2
2 sin x 4 cos 2 x
§S ·
f cc¨ ¸
©6¹
§S ·
f cc¨ ¸
©2¹
§ 5S ·
f cc¨ ¸
© 6 ¹
§ 3S ·
f cc¨ ¸
© 2 ¹
Points of inflection: r1.2758, 3.4035
f
6
0
! 0
x
−3
3
0
f'
f ''
! 0
55. f x
0.2 x 2 x 3 , >1, 4@
3
f c x
0.2 x5 x 6 x 3
f cc x
(b) f cc0 (a)
2
x 34 x 2 9.6 x 3.6
0.4 x 310 x 2 24 x 9
0 Ÿ 0, 0 is a relative maximum.
56 ! 0 Ÿ 1.2, –1.6796 is a relative minimum.
Points of inflection:
3, 0, 0.4652, 0.7048, 1.9348, 0.9049
y
sin x 1
1
sin 3x sin 5 x,
3
5
f c x
cos x cos 3x cos 5 x
f c x
0 when x
f cc x
S
5S
,x
,x
.
6
2
6
sin x 3 sin 3 x 5 sin 5 x
f cc x
0 when x
2
1.9685, 0.9637, §¨
x
4
f
f is increasing when f c ! 0 and decreasing when
f c 0. f is concave upward when f cc ! 0 and
concave downward when f cc 0.
(a)
f c x
f c x
f cc x
f cc x
5S
,
6
5S
·
, 0.2667 ¸
6
©
¹
1
x2
,x
§S
·
Points of inflection: ¨ , 0.2667 ¸, 1.1731, 0.9638,
©6
¹
f″
54. f x
S
§S ·
§S
·
(b) f cc¨ ¸ 0 Ÿ ¨ , 1.53333¸ is a relative
2
2
© ¹
©
¹
maximum.
f′
−2 −1
>0, S @
S
6
x | 1.1731, x | 1.9685
f cc
(c)
−6
The graph of f is increasing when f c ! 0 and
decreasing when f c 0. f is concave upward
when f cc ! 0 and concave downward when
f cc 0.
§ S · § 3S
·
Relative minima: ¨ , 1¸, ¨ , 3¸
©2 ¹ © 2
¹
(a)
y
(c)
§ S 3 · § 5S 3 ·
Relative maxima: ¨ , ¸, ¨ , ¸
© 6 2¹ © 6 2¹
53. f x
f cc r2 0 Ÿ r2, 4 2 are relative maxima.
S S 5S 3S
0 when x
6 x 2 , ª¬
6,
f
x
−2
π
4
π
2
π
f′
−4
−6
f″
−8
r 2.
0, x
The graph of f is increasing when f c ! 0 and
decreasing when f c 0. f is concave upward when
f cc ! 0 and concave downward when f cc 0.
6 x 4 9 x 2 12
0 when x
y
(c)
2
6 º¼
6 x2
6 x 2 because they are endpoints.
4
3 x 4 x 2 0 when x
Note: 0, 0 and S , 0 are not points of inflection
32
r
9
33
2
.
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
Concavity and the Second Derivative Test
259
2 x sin x, >0, 2S @
56. f x
(a) f c x
sin x
2x
2 x cos x Critical numbers: x | 1.84, 4.82
f cc x
cos x
cos x
sin x
2x
2x
2x 2x
2 x sin x 4 x2 1 sin x
2 cos x
2x
2x 2x
4 x cos x 4 x 2 1 sin x
2x 2x
(b) Relative maximum: 1.84, 1.85
Relative minimum: 4.82, 3.09
Points of inflection: 0.75, 0.83, 3.42, 0.72
(c)
y
4
f′
f
2
x
π
2
f ''
−2
−4
f is increasing when f c ! 0 and decreasing when f c 0. f is concave upward
when f cc ! 0 and concave downward when f cc 0.
y
57. (a)
58. (a)
y
4
4
3
3
2
2
1
1
x
1
(b)
2
3
x
4
1
2
3
4
f c 0 means f decreasing
f c 0 means f decreasing
f c increasing means concave upward
f c decreasing means concave downward
(b)
y
4
y
4
3
3
2
2
1
1
x
1
2
3
4
x
1
2
3
4
f c ! 0 means f increasing
f c ! 0 means f increasing
f c increasing means concave upward
f c decreasing means concave downward
© 2010 Brooks/Cole, Cengage Learning
260
Chapter 3
Applications of Differentiation
59. Answers will vary. Sample answer:
3
f x
Let
x4.
f cc x
f cc0
y
64.
f ''
2
12 x
2
f
1
x
0, but 0, 0 is not a point of inflection.
−1
1
3
−1
y
−2
6
f'
−3
5
4
y
65.
3
2
4
1
x
−3
−2
−1
1
2
2
3
(2, 0) (4, 0)
x
2
60. (a) The rate of change of sales is increasing.
4
6
S cc ! 0
(b) The rate of change of sales is decreasing.
S c ! 0, S cc 0
y
66.
2
(c) The rate of change of sales is constant.
Sc
C , S cc
0
1
(0, 0)
(d) Sales are steady.
x
C, Sc
S
(2, 0)
−1
0, S cc
1
3
0
(e) Sales are declining, but at a lower rate.
S c 0, S cc ! 0
(f) Sales have bottomed out and have started to rise.
Sc ! 0
3
y
61.
y
67.
2
f
1
(2, 0)
2
(4, 0)
x
1
f′
f″
2
3
4
5
x
−2
1
−1
y
68.
3
y
62.
f
2
f'
3
f ''
(0, 0)
(2, 0)
x
−1
1
3
x
−2
−1
−1
3
−1
y
63.
f″
f′
f
4
x
−2
2
−2
−4
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
69.
Concavity and the Second Derivative Test
70. (a)
y
261
d
12
f
x
−4
8
12
f″
−8
t
f cc is linear.
10
f c is quadratic.
(b) Because the depth d is always increasing, there are
no relative extrema. f c x ! 0
f is cubic.
f concave upward on f, 3, downward on 3, f.
71. (a) n
1:
n
2:
n
3:
n
f x
x
f c x
x 2
2 x 2
f cc x
f c x
3 x 2
2
f cc x
6 x 2
f x
x 2
f x
f c x
1
f cc x
0
No point of inflection
(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.
2
2
3
2
Point of inflection: 2, 0
No point of inflection
Relative minimum: 2, 0
f x
x
f c x
4 x 2
f cc x
12 x 2
2
4
3
2
No point of inflection
Relative minimum: 2, 0
6
6
6
4:
6
f(x) = (x − 2)3
−9
−9
9
9
−9
9
f(x) = (x − 2)2
Point of
inflection
f(x) = x − 2
−6
−6
−6
−9
9
f(x) = (x − 2)4
−6
Conclusion: If n t 3 and n is odd, then 2, 0 is point of inflection. If n t 2 and n is even, then 2, 0 is a relative minimum.
(b) Let f x
x
2 , f c x
n x 2
n
n 1
, f cc x
n n 1 x 2
For n t 3 and odd, n 2 is also odd and the concavity changes at x
n2
.
2.
For n t 4 and even, n 2 is also even and the concavity does not change at x
2 is point of inflection if and only if n t 3 is odd.
So, x
72. (a)
2.
f x
3
x
f c x
1 x 2 3
3
f cc x
92 x 5 3
Point of inflection: 0, 0
(b) f cc x does not exist at x
0.
y
3
2
1
(0, 0)
x
−6
−4
−2
2
4
6
−2
−3
© 2010 Brooks/Cole, Cengage Learning
262
Chapter 3
73. f x
Applications of Differentiation
ax3 bx 2 cx d
Relative maximum: 3, 3
Relative minimum: 5, 1
Point of inflection: 4, 2
f c x
3ax 2 2bx c, f cc x
f 3
6ax 2b
3 ½°
¾ 98a 16b 2c
125a 25b 5c d
1°¿
27 a 6b c
0, f cc 4
24a 2b
0
27 a 9b 3c d
f 5
f c3
49a 8b c
1
24a 2b
0
0
22a 2b
1
27 a 6b c
22a 2b
1,
2
a
f x
2a
6, c
b
1 x3
2
74. f x
1
45
,
2
6x2 2 Ÿ 49a 8b c
1
1
24
d
24
45
x
2
ax3 bx 2 cx d
Relative maximum: 2, 4
Relative minimum: 4, 2
Point of inflection: 3, 3
f c x
3ax 2 2bx c, f cc x
f 2
8a 4b 2c d
6ax 2b
4 °½
¾ 56a 12b 2c
64a 16b 4c d
2°¿
f 4
f c 2
12a 4b c
0, f c 4
48a 8b c
28a 6b c
1
18a 2b
0
12a 4b c
16a 2b
0
1
16a 2b
2a
1
1
1,
2
a
f x
75. f x
b
1 x3
2
92 , c
0, f cc3
18a 2b
1
0
6
12, d
9 2
x
2
2 Ÿ 28a 6b c
12 x 6
ax3 bx 2 cx d
Maximum: 4, 1
Minimum: 0, 0
(a) f c x
f 0
f 4
f c 4
f c0
f cc x
3ax 2 2bx c,
0 Ÿ d
1Ÿ
0 Ÿ
0 Ÿ
0
64a 16b 4c
48a 8b c
c
1
32
Solving this system yields a
f x
1 x3
32
6ax 2b
1
0
0
and b
6a
3
.
16
3 2
x
16
(b) The plane would be descending at the greatest rate at the point of inflection.
f cc x
6ax 2b
3
x
16
3
8
0 Ÿ x
2.
Two miles from touchdown.
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
76. (a) line OA : y
line CB : y
0.06 x
slope: 0.06
0.04 x 50
slope: 0.04
ax3 bx 2 cx d
f c x
3ax 2 2bx c
(1000, 90)
B
100
1000
0.06
1000, 90:
150
10003 a
60
263
y
f x
1000, 60:
Concavity and the Second Derivative Test
90
1000
0.04
1000
3
2
2
(−1000, 60)
A
1000 b 1000c d
2
C
(0, 50)
3a 2000b c
x
−1000
O
1000
a 1000 b 1000c d
2
3a 2000b c
1.25 u 108 , b
The solution to this system of four equations is a
8 3
0.000025, c
0.0275, and d
50.
1.25 u 10 x 0.000025 x 0.0275 x 50
(b) y
2
100
−1100
1100
−10
(c)
0.1
−1100
1100
− 0.1
(d) The steepest part of the road is 6% at the point A.
77. D
2 x 4 5 Lx 3 3L2 x 2
Dc
8 x 15Lx 6 L x
x
3
2
0 or x
(c) S 20q | 0.9982
x8 x 15Lx 6 L
2
2
§ 15 r 33 ·
¨¨
¸¸ L
16
©
¹
15 L r 33L
16
By the Second Derivative Test, the deflection is
maximum when
x
§ 15 33 ·
¨¨
¸¸ L | 0.578 L.
16
©
¹
5.755T 3
8.521T 2
0.654T
78. S
0.99987,
8
10
106
104
0 T 25
(a) The maximum occurs when T | 4q and
S | 0.999999.
(b)
S
2
0
79.
C
0.5 x 2 15 x 5000
C
C
x
C
average cost per unit
dC
dx
0.5 0.5 x 15 5000
x2
5000
x
0 when x
100
By the First Derivative Test, C is minimized when
x 100 units.
80. C
Cc
300,000
x
300,000
2 x2
2x 0 when x
100 15 | 387
By the First Derivative Test, is C minimized when
x | 387 units.
1.001
1.000
0.999
0.998
0.997
0.996
T
5
10
15
20
25
30
© 2010 Brooks/Cole, Cengage Learning
264
Chapter 3
Applications of Differentiation
5000t 2
,0 d t d 3
8 t2
81. S
(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)
3000
0
3
0
Increasing at greatest rate when t | 1.5.
(c)
S
S ct 5000t 2
8 t2
80,000t
8 t 2 80,0008 3t 2 S cct 3
8 t 2 S cct 8
. So, t
3
r
0 for t
2 6
| 1.633 yrs.
3
100t 2
,t ! 0
65 t 2
82. S
(a)
2
100
0
35
0
(b) S ct 13,000t
65 t 2 13,00065 3t 2 S cct 3
65 t 2 2
0 Ÿ t
4.65
S is concave upwards on 0, 4.65, concave downwards on 4.65, 30.
(c) S ct ! 0 for t ! 0.
As t increases, the speed increases, but at a slower rate.
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
83.
f x
2sin x cos x,
§S ·
f¨ ¸
©4¹
2 2
f c x
2cos x sin x,
§S ·
f c¨ ¸
©4¹
0
f cc x
§S ·
2 sin x cos x, f cc¨ ¸
©4¹
P1 x
S·
§
2 2 0¨ x ¸
4¹
©
P1c x
0
P2 x
S· 1
S·
§
§
2 2 0¨ x ¸ 2 2 ¨ x ¸
4
2
4¹
©
¹
©
P2c x
S·
§
2 2 ¨ x ¸
4¹
©
P2cc x
2 2
Concavity and the Second Derivative Test
265
2 2
2 2
2
2 2 S·
§
2¨ x ¸
4¹
©
2
4
P1
− 2p
2p
f
P2
−4
S 4. The values of the second derivatives of f and P2 are
S 4. The approximations worsen as you move away from x S 4.
The values of f , P1 , P2 , and their first derivatives are equal at x
equal at x
84.
f x
2sin x cos x,
f 0
2
f c x
2cos x sin x,
f c0
2
f cc x
2 sin x cos x,
f cc0
2
P1 x
2 2 x 0
P1c x
2
P2 x
2 2 x 0 P2c x
2 2x
P2cc x
2
21 x
4
1
2
2 x
0
2
2 2 x x2
−6
85.
−4
1 x,
f 0
1
1
,
2 1 x
1
f c0
1
2
f cc0
1
4
f c x
f cc x
P1 x
x
§ 1·
1 ¨ ¸ x 0 1 2
© 2¹
1
2
1§ 1 ·
2
§ 1·
1 ¨ ¸ x 0 ¨ ¸ x 0
2© 4 ¹
© 2¹
P1c x
P2 x
P2c x
P2cc x
41 x
32
,
1
0.
x
x2
2
8
5
P1
x
1
2
4
1
4
f
The values of f , P1 , P2 , and their first derivatives are equal at x
equal at x
0. The values of the second derivatives of f and P2 are
0. The approximations worsen as you move away from x
f x
6
P1
The values of f , P1 , P2 , and their first derivatives are equal at x
equal at x
P2
f
−8
4
P2
−3
0. The values of the second derivatives of f and P2 are
0. The approximations worsen as you move away from x
0.
© 2010 Brooks/Cole, Cengage Learning
266
Chapter 3
x
,
x 1
f x
86.
Applications of Differentiation
f 2
x 1
f c x
f c 2
3x 2 6 x 1
,
3
4 x3 2 x 1
f cc 2
23
8 2
2
§ 3 2·
2 ¨¨ ¸ x 2
4 ¸¹
©
P1 x
P1c x
3
2 2
,
x x 1
2
f cc x
2
3 2
4
23 2
16
3 2
5 2
x 4
2
3 2
4
§ 3 2·
1 § 23 2 ·
2
2 ¨¨ ¸¸ x 2 ¨¨
¸¸ x 2
4
2
16
©
¹
©
¹
P2 x
2 3 2
23 2
x 2
4
16
P2c x
P2cc x
23 2
16
The values of f , P1 , P2 and their first derivatives are equal at x
at x
3 2
23 2
x 2 x 22
4
32
2. The approximations worsen as you move away from x
2. The values of the second derivatives of f and P2 are equal
2.
3
P1
P2
f
−1
5
−1
f x
§1·
x sin ¨ ¸
© x¹
f c x
ª 1
§ 1 ·º
§1·
x « 2 cos¨ ¸» sin ¨ ¸
x
x
©
¹
© x¹
¬
¼
f cc x
1ª 1
1
1
§ 1 ·º
§1·
§1·
« 2 sin ¨ ¸» 2 cos¨ ¸ 2 cos¨ ¸
x¬x
x
x
x
x
© ¹¼
© ¹
© x¹
1
87.
x
1
§1·
§1·
cos¨ ¸ sin ¨ ¸
x
© x¹
© x¹
1
§1·
sin ¨ ¸
x3
© x¹
0
S
§1 ·
Point of inflection: ¨ , 0 ¸
©S ¹
When x ! 1 S , f cc 0, so the graph is concave downward.
1
−1
1
( π1 , 0(
−1
88.
f x
x x 6
f c x
3 x 2 24 x 36
f cc x
6 x 24
2
x3 12 x 2 36 x
3 x 2 x 6
6 x 4
0
0
Relative extrema: 2, 32 and 6, 0
Point of inflection 4, 16 is midway between the relative extrema of f.
© 2010 Brooks/Cole, Cengage Learning
Section 3.4
Concavity and the Second Derivative Test
89. Assume the zero of f are all real. Then express the function as f x
267
a x r1 x r2 x r3 where r1 , r2 , and r3 are the
distinct zeros of f. From the Product Rule for a function involving three factors, we have
f c x
a ª¬ x r1 x r2 x r1 x r3 x r2 x r3 º¼
f cc x
a ª¬ x r1 x r2 x r1 x r3 x r2 x r3 º¼
Consequently, f cc x
x
90.
2 r1 r2 r3 6
0 if
r1 r2 r3
3
p x
ax3 bx 2 cx d
pc x
3ax 2 2bx c
pcc x
6ax 2b
6ax 2b
0
x
b 3a. Therefore, b 3a, p b 3a is a point of inflection.
§ b3 ·
§ b2 ·
§ b·
a¨ b¨ 2 ¸ c ¨ ¸ d
3¸
27
a
9
a
© 3a ¹
©
¹
©
¹
When p x
x0
y0
x3 3 x 2 2, a
3
2 3
3
2
30
31
2
2 0 2
ycc
0, and d
ax3 bx 2 cx d , a z 0. Then
6ax 2b
0 when x
b 3a, and the
1 x has a discontinuity at x
0.
93. False. Concavity is determined by f cc. For example, let
f x
x and c
2. f cc
f c 2 ! 0, but f is not
concave upward at c
2.
94. False. For example, let f x
1, 0.
95. f and g are concave upward on a, b implies that f c and
g c are increasing on a, b, and f cc ! 0 and g cc ! 0 .
So, f g cc ! 0 Ÿ f g is concave upward on
concavity changes at this point.
92. False. f x
2.
0
x3 3 x 2 2 is x0 , y0 The point of inflection of p x
91. True. Let y
3, c
1, b
2b3
bc
d
27 a 2
3a
1
31
271
Average of r1, r2 , and r3 .
b
3a
The sign of pcc x changes at x
§ b·
p¨ ¸
© 3a ¹
a ª¬6 x 2 r1 r2 r3 º¼.
x
2 .
4
a, b by Theorem 3.7.
96. f, g are positive, increasing, and concave upward on
a, b Ÿ f x ! 0, f c x t 0 and f cc x ! 0, and
g x ! 0, g c x t 0 and g cc x ! 0 on a, b. For
x  a, b,
fg c x
f c x g x f x g c x
fg cc x
f cc x g x 2 f c x g c x f x g cc x ! 0
So, fg is concave upward on a, b.
© 2010 Brooks/Cole, Cengage Learning
268
Chapter 3
Applications of Differentiation
Section 3.5 Limits at Infinity
2x2
x2 2
1. f x
4. f x
No vertical asymptotes
Horizontal asymptote: y
Horizontal asymptote: y
2
Matches (a)
2x
5. f x
x2 2
No vertical asymptotes
Horizontal asymptotes: y
4 sin x
x2 1
No vertical asymptotes
r2
Horizontal asymptote: y
Matches (c)
0
f 1 ! 1
x
x 2
3. f x
x2
x4 1
No vertical asymptotes
2
Matches (f )
2. f x
2 Matches (b)
2
6. f x
No vertical asymptotes
Horizontal asymptote: y
0
f 1 1
2 x 2 3x 5
x2 1
No vertical asymptotes
Horizontal asymptote: y
Matches (d)
2
Matches (e)
7. f x
4x 3
2x 1
x
100
101
102
103
104
105
106
f(x)
7
2.26
2.025
2.0025
2.0003
2
2
lim f x
2
xof
10
− 10
10
− 10
8. f x
2x2
x 1
x
100
101
102
103
104
105
106
f(x)
1
18.18
198.02
1998.02
19,998
199,998
1,999,998
lim f x
xof
f
Limit does not exist 20
0
10
−2
© 2010 Brooks/Cole, Cengage Learning