Thomas Calculus 13th [Solutions]

238 Chapter 4 Applications of Derivatives
65. Let f ( t ) cos t and consider the interval [0, x ] where x is a real number. f is continuous on [0, x ] and f is
differentiable on (0, x ) since f ( t) sin t f satisfies the conditions of the Mean Value Theorem
f ( x) f (0)
f ( c ) for some c in [0, x] cos x 1 sin c . Since 1 sin c 1 1 sin c 1
x (0)
x
1 cos x 1 1. If x 0, 1 cos x 1 1 x cos x 1 x |cos x 1| x | x | . If x 0, 1 cos x 1 1
x
x
x
x cos x 1 x x cos x 1 x ( x) cos x 1 x |cos x 1| x | x| . Thus, in both cases,
x x If x 0, then |cos0 1| |1 1| |0| |0|, thus |cos x 1| | x | is true for all x.
we have |cos 1| | | .
66. Let f ( x) sin x for a x b . From the Mean Value Theorem there exists a c between a and b such that
sin b sin a sin b sin a sin b sin a
b a b a b a
cosc 1 1 1 |sin b sin a| | b a| .
67. Yes. By Corollary 2 we have f ( x) g( x)
c since f ( x) g ( x ). If the graphs start at the same point x a,
then f ( a) g( a) c 0 f ( x) g( x).
68. Assume f is differentiable and | f ( w) f ( x)| | w x | for all values of w and x. Since f is differentiable,
( ) ( )
f ( x ) exists and f ( x ) lim f w f x using the alternative formula for the derivative. Let g( x) x ,
w x
w x
f ( w) f ( x) f ( w) f ( x)
which is continuous for all x. By Theorem 10 from Chapter 2, | f ( x)| lim lim
w x
w x
w x
w x
| f ( w) f ( x)|
| f ( w) f ( x)|
lim . Since f ( w) f ( x)
w x for all w and x 1 as long as w x . By Theorem 5
w x
| w x|
| w x|
| ( ) ( )|
from Chapter 2, f ( x f w f x
) lim lim 1 1 f
w x
| w x|
( x ) 1 1 f ( x ) 1.
w x
f ( b) f ( a)
69. By the Mean Value Theorem we have
f ( c ) for some point c between a and b. Since b a 0 and
b a
f ( b) f ( a ), we have f ( b) f ( a) 0 f ( c) 0.
70. The condition is that f should be continuous over [ a, b ]. The Mean Value Theorem then guarantees the
f ( b) f ( a)
existence of a point c in ( a, b ) such that
f ( c ). If f is continuous, then it has a minimum and
b a
maximum value on [ a, b ], and min f f ( c ) max f , as required.
71.
72.
4 1 4 2 3 4
f ( x) (1 x cos x) f ( x) (1 x cos x) (4x cos x x sin x)
3 4 2
x (1 x cos x) (4 cos x x sin x) 0 for 0 x 0.1 f ( x ) is decreasing when 0 x 0.1
min f 0.9999 and max f 1. Now we have
1.09999 f (0.1) 1.1.
4 1 4 2 3 4x
(1 x )
f
(0.1) 1
0.1
0.9999 1 0.09999 f (0.1) 1 0.1
f ( x ) (1 x ) f ( x ) (1 x ) ( 4 x ) 0 for 0 x 0.1 f ( x ) is increasing when
0 x 0.1 min f 1 and max f 1.0001. Now we have 1 1.0001
0.1 f (0.1) 2 0.10001 2.1 f (0.1) 2.10001.
3
4 3
f
(0.1) 2
0.1
f ( x) f (1)
x 1
f ( x) f (1)
Mean Value Theorem 0 f ( x) f (1). Therefore ( ) 1
x 1
lim f ( x) f (1) ( ) (1)
1 0 and lim f x f
x
x 1
0.
x 1 x 1
are equal and have the value (1) (1) 0 and (1) 0 (1) 0.
73. (a) Suppose x 1, then by the Mean Value Theorem
(b) Yes. From part (a),
f f f f
0 f ( x) f (1). Suppose x 1, then by the
f x for all x since f (1) 1.
Since f (1) exists, these two one-sided limits
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