Section 3.2
Rolle’s Theorem and the Mean Value Theorem
215
Section 3.2 Rolle's Theorem and the Mean Value Theorem
1. f x
9. f x
1
x
f 1
1. But, f is not continuous on >1, 1@.
f 1
2. Rolle's Theorem does not apply to f x
>S , 3S @ because f is not continuous at
x
cot x 2 over
2S .
3. Rolle's Theorem does not apply to
f x 1 x 1 over >0, 2@ because f is not
differentiable at x
2 x 2 3 4. f x
f 1
f c x
1.
f 1
2 x2 3 6. f x
x
x2 x 2
x
7. f x
x
f c x
8. f x
0 at x
f c x
f 0
§S ·
f¨ ¸
©2¹
f c x
2 cos 2 x
S
4
0.
0 at x
0 at x
4
>0, 3@
0
2 x 3
3
2
x 2 5 x 4, >1, 4@
f 1
3
.
2
0 Ÿ x
3
2
12. f x
1 2
§3
·
¨ x 4 ¸ x 4
2
©
¹
S
0
x 2 3 x,
f 3
1
x
0
§S ·
and f c¨ ¸
©4¹
c-value:
f 4
0
f is continuous on >1, 4@. f is differentiable on
x 4
1, 4. Rolle's Theorem applies.
0 at x
f c x
2x 5
2x 5
0 Ÿ x
13. f x
8
3
f 1
5
2
5
2
c-value:
x 1 x
f 3
0
2 x 3, >1, 3@
f is continuous on >1, 3@. f is differentiable on
x 1
1, 3. Rolle's Theorem applies.
1
1 2
12
3 x x 1
3 x 1
2
1 2 § x
·
3 x 1 ¨ x 1 ¸
2
©
¹
3 x 1
sin 2 x
2 x 3
1.
2
x-intercepts: 1, 0, 0, 0
f c x
f c 1
10. f x
f c x
1
1 2
12
x 4
x x 4
2
x
x 41 2 §¨ x 4 ·¸
©2
¹
3 x
0 at
0, 3. Rolle's Theorem applies.
x-intercepts: 4, 0, 0, 0
f c x
and
0
f is continuous on >0, 3@ and differentiable on
2 x 1
x x 3
2x 3
1
f 0
0.
x-intercepts: 0, 0, 3, 0
f c x
2x 2
11. f x
13
2x 1
f c x
c
1
x-intercepts: 1, 0, 2, 0
f c x
f 1
3 x 1
3
f is not differentiable at x
5. f x
f 3
c
x
x2 2 x 3
1 2 § 3
·
¨ x 1¸
©2
¹
0 at x
f x
x3 6 x 2 11x 6
f c x
3 x 2 12 x 11
3x 2 12 x 11
2
3
c-values:
0 Ÿ x
6r 3
3
6 3 6 3
,
3
3
© 2010 Brooks/Cole, Cengage Learning
216
14.
Chapter 3
f x
Applications of Differentiation
x 3 x
f 3
0
f 1
15. f x
1 , >1, 3@
2
f 8
x 2 3 1, >8, 8@
f 8
3
f is continuous on >1, 3@. f is differentiable on
f is continuous on >8, 8@. f is not differentiable on
1, 3. Rolle's Theorem applies.
8, 8 because
f c x
x
x
x
c-values:
3 2 x 1 x 1
f c0 does not exist. Rolle's Theorem
does not apply.
2
16. f x
1>2 x 6 x 1@
f 0
13 x 5
3 x 3 , >0, 6@
f 6
0
f is continuous on >0, 6@. f is not differentiable on
5
3
0, 6 because
f c3 does not exist. Rolle's Theorem
does not apply.
x2 2x 3
, >1, 3@
x 2
f 3
0
17. f x
f 1
f is continuous on >1, 3@. (Note: the discontinuity, x
2, is not in the interval.) f is differentiable on 1, 3. Rolle's
Theorem applies.
2 2 x 2 x 2 2 x 31
x
f c x
x
2
2
x2 4x 1
x
2
2
x
c-value: 2 18.
0
0
4 r 2 5
2
f 1
f is not continuous on >1, 1@ because f 0 does not exist.
Rolle's Theorem does not apply.
19. f x
sin x, >0, 2S @
f 0
f 2S 0
f is continuous on >0, 2S @. f is differentiable on
0, 2S . Rolle's Theorem applies.
f c x cos x
c-values:
20. f x
f 0
S 3S
2
,
2
cos x, >0, 2S @
f 2S 1
f is continuous on >0, 2S @. f is differentiable on
0, 2S . Rolle's Theorem applies.
f c x
5
5
x2 1
, >1, 1@
x
f 1
0
f x
2 r
21. f x
f 0
6x
S
ª Sº
4 sin 2 x, «0, »
¬ 6¼
§S ·
f¨ ¸
©6¹
0
f is continuous on >0, S 6@. f is differentiable on
0, S 6. Rolle's Theorem applies.
f c x
6
S
6
S
3
2S
1
§ 3 ·
arcsin ¨ ¸
2
© 2S ¹
6
S
8 sin x cos x
0
8 sin x cos x
4 sin 2 x
sin 2 x
x
x | 0.2489
c-value: 0.2489
sin x
c-value: S
© 2010 Brooks/Cole, Cengage Learning
Section 3.2
22. f x
cos 2 x, >S , S @
f S f S Rolle’s Theorem and the Mean Value Theorem
x x 1 3 , >0, 1@
26. f x
f 0
1
f 1
0
f is continuous on >S , S @ and differentiable on
f is continuous on >0, 1@. f is differentiable on
S , S . Rolle's Theorem applies.
0, 1. (Note: f is not differentiable at
f c x
f c x
0
sin 2 x
S
S
S
f 0
S
3
x2
0
f is not continuous on >0, S @ because f S 2 does not
24. f x
sec x, >S , 2S @
f is not continuous on
>S , 2S @ because f 3S 2
27. f x
f 14 f x
25.
f 1
x 1, >1, 1@
f 1
0
1
−1
x tan S x, ª¬ 14 , 14 º¼
f 14
Rolle's Theorem does not apply.
3
9
3
| 0.1925
9
c-value:
sec3S 2 does not exist.
1
1
27
x
exist. Rolle's Theorem does not apply.
0
3 3 x2
3 3 x2
1
3
1
27
x2
tan x, >0, S @
f S 0. ) Rolle's
1
1
S , , 0, , S
2
2
c-values: , 0,
2
2
23. f x
1
1
0
x
x
Theorem applies.
2 sin 2 x
2 sin 2 x
217
14 1
1
4
1
3
4
43
Rolle's Theorem does not apply
0.75
0
f is continuous on >1, 1@. f is not differentiable on
− 0.25
0.25
1, 1 because f c0 does not exist. Rolle's Theorem
does not apply.
− 0.75
1
−1
1
−1
x
Sx
sin
, >1, 0@
2
6
f 0
0
f x
28.
f 1
f is continuous on >1, 0@. f is differentiable on
1, 0. Rolle's Theorem applies.
f c x
cos
Sx
6
x
1 S
Sx
cos
2
6
6
3
0
S
6
S
arccos
3
ªValue needed in 1, 0.º¼
S ¬
| 0.5756 radian
c-value: –0.5756
0.02
−1
0
−0.01
© 2010 Brooks/Cole, Cengage Learning
218
Chapter 3
29. f t Applications of Differentiation
(a) f 1
f 2
f is not differentiable at x
38
f ct must be 0 at some time in 1, 2.
(b) v
f ct 32t 48
3
2
t
x 3 , >0, 6@
36. f x
16t 2 48t 6
37. f x
0
x2 5
1 4
2 1
(a) Slope
sec
1
x ·
§1
10¨ ¸
x 3¹
©x
C 6
(b)
C c x
x y 3
§ 1
·
3
¸
10¨ 2 2
¨ x
x 3 ¸¹
©
1
x2
3
x2 6x 9
2 x2 6x 9
0
(b) f c x
2 x
1
(c) f c
§1·
f¨ ¸
© 2¹
6r
4
(d)
Tangent
−6
6
−1
x 2 x 12
6 0
2 4
(a) Slope
(c2, f (c2))
a
Tangent line
(b) f c x
2x 1
(c) f c
f 1
y
32.
x 4
x y 4
(b, f(b))
b
1
y 0
Secant line:
f
x
Tangent line:
1
0
Ÿ
(d)
a
33. f is not continuous on the interval >0, 6@. ( f is not
continuous at x
y 12
x 1
0
15
Tangent
−15
2.)
34. f is not differentiable at x
smooth at x
2.
35. f x
1
5
Secant
b
c
f
−15
x
x
12
x y 13
f
0
f
Tangent line
(c1, f(c1))
1·
§
¨ x ¸
2¹
©
4 x 2
19
4
Secant
38. f x
t line
y 19
4
7
3r3 3
2
y
Secan
1
5
4
1
2
c
4 x 4 y 21
108
In the interval
33 3
| 4.098 | 410 components
3, 6: c
2
(a, f (a))
x
4 y 19
6r 6 3
4
31.
0
Ÿ
Tangent line:
0
x
x 3
y
25
3
(a) C 3
x 1
y 4
Secant line:
30. C x
3.
2. The graph of f is not
1
, >0, 6@
x 3
f has a discontinuity at x
3.
© 2010 Brooks/Cole, Cengage Learning
Section 3.2
x 2 is continuous on >2, 1@ and differentiable on
39. f x
2, 1.
1 4
3
1 2
f c x
2x
1
x
1
2
x
0, 1.
10
1
10
f c x
3x 2
1
x
r
1
3
3
f 1 f 1
3x 2
3
2
1
x
r
2
2 7
4 x3 8
3
2 2 x
3
2 x
3
2
9
4
2 x
1
3
x
c
00
2
03
9
1
2 2 x
3
1
3
1
3
1
4
1
4
sin x is continuous on >0, S @ and differentiable
47. f x
on 0, S .
differentiable on 0, 2.
c
1 2. The
2 x is continuous on >7, 2@ and
f c x
x 4 8 x is continuous on >0, 2@ and
f c x
2 x 1 is not differentiable at x
f 2 f 7
3
3
2 0
0. The Mean
differentiable on 7, 2.
3
3
2
f 2 f 0
8
27
x 1
is not continuous at x
x
Value Theorem does not apply.
46. f x 3 3
1 1
42. f x 3
8
27
45. f x
differentiable on 1, 1.
3x
§ 2·
¨ ¸
© 3¹
44. f x
x3 2 x is continuous on >1, 1@ and
f c x
1
Mean Value Theorem does not apply.
In the interval 0, 1 : c
41. f x
2 1 3
x
3
f c x
c
f 1 f 0
1
10
x3 is continuous on >0, 1@ and differentiable on
r
x 2 3 is continuous on >0, 1@ and differentiable on
43. f x f 1 f 0
1
1
2
40. f x
c
219
0, 1.
f 1 f 2
c
Rolle’s Theorem and the Mean Value Theorem
0
4 x3 2
0
x3
2
x
3
f S f 0
0 0
S 0
S
f c x
cos x
x
2
c
2
48. f x
x
0
0
S 2
S
2
cos x tan x is not continuous at
S 2. The Mean Value Theorem does not apply.
© 2010 Brooks/Cole, Cengage Learning
220
Chapter 3
Applications of Differentiation
50. f x
x ª 1 º
, ,2
x 1 «¬ 2 »¼
49. f x
(a) – (c)
x 2 sin x, >S , S @
(a) – (c)
2p
1
f
Tangent
tangent
− 2p
− 0.5
secant 2p
f
tangent
2
Secant
− 2p
−1
(b) Secant line:
f S f S slope
S S (b) Secant line:
slope
f 2 f 1 2
2 3 1
2 1 2
2
3
52
(c) f c x
1
x
1
x
1
y
2
2
3
2
x
1 r
1 ª1 ¬
3
2
1 r
6
2
1 6 2º 1
¼
Tangent line: y 1 y 1
6 2
2 6 2
1 2 cos x
1
cos x
0
x
c
§S ·
f¨ ¸
©2¹
In the interval >1 2, 2@: c
f c
x
(c) f c x
2
3
6
6
2
6
2
x 3
3
3
1
2x 5 2 6
3
S
2
2
1
6
6
3
r
S
2
2
2
Tangent lines:
·
2§
6
1¸¸
¨¨ x 3©
2
¹
y
S
§ S·
f ¨ ¸
© 2¹
2
6
1
1 x S y S
2
2
y x 2
3
3
2
y
x 1
3
S S 2S
2
§S
·
y ¨ 2¸
2
©
¹
y
S·
§
1¨ x ¸
2¹
©
x 2
§ S
·
y ¨ 2¸
© 2
¹
y
S·
§
1¨ x ¸
2¹
©
x 2
x , >1, 9@
51. f x (a) – (c)
3
Tangent
Secant
f
1
9
1
(b) Secant line:
f 9 f 1
slope
y 1
y
(c) f c x
f 4
1
4
1
x 1
4
1
3
x 4
4
1
2
x
x
1
4
c
4
2
Tangent line: y 2
y
31
8
9 1
1
x 4
4
1
x 1
4
© 2010 Brooks/Cole, Cengage Learning
Section 3.2
56. f a
x 4 2 x3 x 2 , >0, 6@
52. f x (a) – (c)
Rolle’s Theorem and the Mean Value Theorem
Tangent
6
−100
f 6 f 0
900 0
6
6 0
150 x 0
y 0
y
g b g c x
f c x Ÿ g cc
4 x3 6 x 2 2 x
Tangent line: y 123.6721
150
f x k
g a k g b k g c x
f c x k g cc k g x
f kx
§a·
g¨ ¸
©k¹
g c x
§b·
g¨ ¸
©k¹
kf c kx
§c·
g c¨ ¸
©k¹
kf cc
(c)
4.9t 300
s3 s0
30
255.9 300
3
14.7 m/sec
(b) st is continuous on >0, 3@ and differentiable on
0, 3. Therefore, the Mean Value Theorem applies.
sct 9.8t
14.7
9.8
S 12 S 0
12 0
1.5 sec
S ct (b)
1
2
t
2
2t
t
f a
0
Critical number of g :
57. f x
200 ª¬5 9 14º¼ 200 ª¬5 9 2º¼
12
450
7
§
·
9
¸
200¨
2
¨ 2 t ¸
©
¹
1
28
0
ªa b º
Interval: « , »
¬k k ¼
14.7 m/sec
9 ·
§
200¨ 5 ¸
2 t¹
©
54. S t f cc
450
7
2 7
f a
Critical number of g : c k
2
t
0
Interval: >a k , b k @
150 x 3.8721
150 x 457.143
y
v t f a k
g x
(b)
Using a graphing utility, there is one solution in
0, 6, x c | 3.8721 and f c | 123.6721
(a)
g a
Critical number of g: c
150
150 x
(c) f c x
f x k
Interval: >a, b@
(b) Secant line:
slope
g x
(a)
f
0
(a) vavg
0 where c is in the interval
a, b.
1000
Secant
53. st f b and f cc
221
c
k
x
0
­0,
®
¯1 x, 0 x d 1
No, this does not contradict Rolle's Theorem. f is not
continuous on >0, 1@.
58. No. If such a function existed, then the Mean Value
Theorem would say that there exists c  2, 2 such
that
f cc f 2 f 2
2 2 6 2
4
2.
But, f c x 1 for all x.
2 7 2 | 3.2915 months
S ct is equal to the average value in April.
55. No. Let f x
x 2 on >1, 2@.
f c x
2x
f c0
0 and zero is in the interval 1, 2 but
f 1 z f 2.
© 2010 Brooks/Cole, Cengage Learning
222
Chapter 3
Applications of Differentiation
64. (a) f is continuous on >10, 4@ and changes sign,
59. Let S t be the position function of the plane. If
0 corresponds to 2 P.M., S 0
t
0, S 5.5
f 8
2500
and the Mean Value Theorem says that there exists a
time t0 , 0 t0 5.5, such that
2500 0
| 454.54.
5.5 0
Applying the Intermediate Value Theorem to the
velocity function on the intervals >0, t0 @ and >t0 , 5.5@,
S ct0 ! 0, f 3 0. By the Intermediate Value
Theorem, there exists at least one value of x in
>10, 4@ satisfying f x 0.
v t 0 (b) There exist real numbers a and b such that
10 a b 4 and f a
f b 2.
Therefore, by Rolle's Theorem there exists at least
one number c in 10, 4 such that f cc
0. This
you see that there are at least two times during the
flight when the speed was 400 miles per hour.
0 400 454.54
is called a critical number.
(c)
y
8
60. Let T t be the temperature of the object. Then
T 0
1500q and T 5
4
390q. The average
x
temperature over the interval >0, 5@ is
−8
−4
390 1500
222q F/h.
50
By the Mean Value Theorem, there exist a time t0 ,
0 t0 5, such that T ct0 4
−4
−8
y
(d)
222qF/h.
8
4
61. Let S t be the difference in the positions of the 2
bicyclists, S t S 0
−8
0, there must exist a time
t0  0, 2.25 such that S ct0 At this time, v1 t0 −8
v t0 0.
(e) No, f c did not have to be continuous on >10, 4@.
v2 t0 .
65. f is continuous on >5, 5@ and does not satisfy the
0 correspond to 9:13 A.M. By the Mean Value
Theorem, there exists t0 in 0,
vct0 85 35
1 30
a t0 1
30
such that
conditions of the Mean Value Theorem. Ÿ f is not
differentiable on 5, 5. Example: f x
1500 mi/h 2 .
§ S x ·§
§ S x · ·§ S ·
6 cos¨ ¸¨ sin ¨ ¸ ¸¨ ¸
© 2 ¹©
© 2 ¹ ¹© 2 ¹
§S x · §S x ·
3S cos¨ ¸ sin ¨ ¸
© 2 ¹ © 2 ¹
(a)
x
y
8
§S x ·
3 cos 2 ¨ ¸, f c x
© 2 ¹
63. f x
4
−4
−4
S 2.25
62. Let t
x
S1 t S2 t . Because
f(x) = x
6
(− 5, 5)
(5, 5)
4
2
−4
x
−2
2
4
−2
7
66. f is not continuous on >5, 5@.
− 2p
2p
­1 x, x z 0
®
x
0
¯0,
Example: f x
−7
y
(b) f and f c are both continuous on the entire real line.
(c) Because f 1
f 1
applies on >1, 1@. Because f 1
0 and f 2
Rolle's Theorem does not apply on >1, 2@.
(d) lim f c x
0
lim f c x 0
x o 3
x o 3
f (x) = 1x
4
0, Rolle's Theorem
2
(5, 15)
3,
x
2
(− 5, − 15)
4
−5
© 2010 Brooks/Cole, Cengage Learning
Section 3.2
67. f x Rolle’s Theorem and the Mean Value Theorem
x5 x3 x 1
71. f continuous at x
f is differentiable for all x.
f continuous at x
f 1
2 and f 0
1, so the Intermediate Value
Theorem implies that f has at least one zero c in
>1, 0@, f c 0.
Suppose f had 2 zeros, f c1 f c2 f x
0. Then
f c2 f c1 But, f c x
0.
5 x 4 3 x 2 1 ! 0 for all x. So, f has
2 x5 7 x 1
f is differentiable for all x.
f 0
1 and f 1
8, so the Intermediate Value
f c2 0. Then
Rolle's Theorem would guarantee the existence of a
number a such that
f c a f c2 f c1 0 has
exactly one real solution.
69. f x
3 x 1 sin x
f S 3S 1 0 and f 0
1 ! 0, so the
Intermediate Value Theorem implies that f has at least
one zero c in >S , 0@, f c
0.
Suppose f had 2 zeros, f c1 f c2 0. Then
f c2 f c1 But f c x
f 0
3 cos x ! 0 for all x. So, f x
0 has
1: b 2
f differentiable at x
0: 0
f differentiable at x
1: 2b
2 and d
73. f c x
0
f x
c
f 2
5
2S 2 1
2.
74. f c x
2S 1 ! 0. By the
d 4 Ÿ b
d 2
0
d
4.
4
f x
4x c
f 0
1Ÿ c
75. f c x
c
5.
1
4 x 1.
2x
f x
x2 c
f 1
0 Ÿ 0
1 c Ÿ c
1
x 2 1.
2x 3
f x
x 2 3x c
f 1
0 Ÿ 0
So, f x
2 x 2 cos x
3, f S f continuous at x
76. f c x
0.
exactly one real solution.
70. f x
0: 2
So, f x
Rolle's Theorem would guarantee the existence of a
number a such that
f c a 6. So, c
2
f continuous at x
So, f x
f is differentiable for all x.
1: a
72. f continuous at x
So, f x
0.
10 x 4 7 ! 0 for all x. So, f x
But f c x
2 4
1: a
x
0
­1,
°
0 x d1
®6 x 1,
°x 2 4 x 2, 1 x d 3
¯
So, b
Theorem implies that f has at least one zero c in
>0, 1@, f c 0.
Suppose f had 2 zeros, f c1 5 c
0 d x d1
­6 x 1,
® 2
¯x 4 x 2, 1 x d 3
exactly one real solution.
68. f x b
1: a 1
f differentiable at x
Rolle's Theorem would guarantee the existence of a
number a such that
f c a 0: 1
223
13 c Ÿ c
4
x 2 3x 4.
77. False. f x
1 x has a discontinuity at x
0.
Intermediate Value Theorem, f has at least one zero.
Suppose f had 2 zeros, f c1 f c2 0. Then
Rolle's Theorem would guarantee the existence of a
number a such that
f c a f c2 f c1 But, f c x
0.
2 sin x t 1 for all x. So, f has exactly
one real solution.
78. False. f must also be continuous and differentiable on
each interval. Let
f x
x3 4 x
.
x2 1
79. True. A polynomial is continuous and differentiable
everywhere.
© 2010 Brooks/Cole, Cengage Learning
224
Chapter 3
Applications of Differentiation
80. True
81. Suppose that p x
p x1 p x2 x 2 n 1 ax b has two real roots x1 and x2 . Then by Rolle's Theorem, because
0, there exists c in x1 , x2 such that pcc
0. But pc x
2n
1 x 2 n a z 0, because
n ! 0, a ! 0. Therefore, p x cannot have two real roots.
82. Suppose f x is not constant on a, b. Then there exists x1 and x2 in a, b such that f x1 z f x2 . Then by the Mean
Value Theorem, there exists c in a, b such that
f x2 f x1 f cc
x2 x1
z 0.
This contradicts the fact that f c x
83. If p x
0 for all x in a, b.
Ax 2 Bx C , then
pc x
Ab2
f b f a 2 Ax B
b a
Bb C Aa 2 Ba C Ab a
2
b a
2
B b a b a
b
a ª¬ Ab a Bº¼
b a
Ab a B.
b
Ab a and x
So, 2 Ax
84. (a) f x
x 2 , g x
f 1
g 1
Let h x
hcc
x3 x 2 3x 2
1, f 2
g 2
h 2
0. So, by Rolle's Theorem these exists c  1, 2 such that
0.
c, the tangent line to f is parallel to the tangent line to g.
So, at x
x 3 x 2, hc x
3
(b) Let h x
hcc
4
f x g x. Then, h 1
f cc g cc
h x a 2 which is the midpoint of >a, b@.
3x 2 3
f x g x. Then h a f cc g cc
0 Ÿ x
hb 0 by Rolle's Theorem, there exists c in a, b such that
0.
85. Suppose f x has two fixed points c1 and c2 . Then, by
the Mean Value Theorem, there exists c such that
f c2 f c1 c2 c1
c2 c1
c2 c1
1.
This contradicts the fact that f c x 1 for all x.
86. f x
f c x
1
2
1
c, the tangent line to f is parallel to the tangent line to g.
So, at x
f cc
c
cos x differentiable on f, f.
12 sin x
12 d f c x d
1
2
Ÿ f c x 1
for all real numbers.
87. Let f x
cos x. f is continuous and differentiable for
all real numbers. By the Mean Value Theorem, for any
interval >a, b@, there exists c in a, b such that
f b f a f cc
b a
cos b cos a
b a
cos b cos a
sin cb
cos b cos a
sin c b a
sin c
a
cos b cos a d b a since sin c d 1.
So, from Exercise 62, f has, at most, one fixed point.
x | 0.4502
© 2010 Brooks/Cole, Cengage Learning
Section 3.3
88. Let f x
Increasing and Decreasing Functions and the First Derivative Test
89. Let 0 a b. f x
sin x. f is continuous and differentiable for
in a, b such that
f cc
b a
sinb sin a b
sin b sin a b a cosc
x satisfies the hypotheses of
the Mean Value Theorem on >a, b@. Hence, there exists c
all real numbers. By the Mean Value Theorem, for any
interval >a, b@, there exists c in a, b such that
f b f a 225
f cc
a cosc
So
f b f a 1
2 c
b b a
a
b
a
sin a sin b d a b .
b a
.
b a
1
b a
.
2 c
2 a
Section 3.3 Increasing and Decreasing Functions and the First Derivative Test
1. (a) Increasing: 0, 6 and 8, 9. Largest: 0, 6
4. y
(b) Decreasing: 6, 8 and 9, 10. Largest: 6, 8
2
From the graph, f is increasing on f, 1 and
decreasing on 1, f.
2. (a) Increasing: 4, 5, 6, 7. Largest: 4, 5, 6, 7
Analytically, yc
(b) Decreasing: 3, 1, 1, 4, 5, 6. Largest: 3, 1
3. f x
x 1
x2 6x 8
2
x
1
3
No critical numbers. Discontinuity: x
1
From the graph, f is decreasing on f, 3 and
Test Intervals:
f x 1
1 x f
increasing on 3, f.
Sign of yc:
yc ! 0
yc 0
Conclusion:
Increasing
Decreasing
Analytically, f c x
2 x 6.
Critical number: x
3
Test Intervals:
f x 3
3 x f
Sign of f c x:
fc 0
fc ! 0
Conclusion:
5. y
Decreasing
Increasing
x3
3x
4
From the graph, y is increasing on f, 2 and 2, f, and decreasing on 2, 2.
3x 2
3
4
Analytically, yc
Critical numbers: x
3 2
x 4
4
3
x 2 x 2
4
r2
Test Intervals:
f x 2
2 x 2
2 x f
Sign of yc:
yc ! 0
yc 0
yc ! 0
Conclusion:
Increasing
Decreasing
Increasing
© 2010 Brooks/Cole, Cengage Learning