Thomas Calculus 13th [Solutions]

252 Chapter 4 Applications of Derivatives
x
x
x
77. f ( x) e 2 x f ( x) e 2; f ( x) 0 e 2 x ln 2; f(0) = 1, the absolute maximum;
f(ln 2) = 2 2 ln 2 0.613706, the absolute minimum; f(1) = e 2 0.71828, a relative or local maximum
since f is increasing on the interval (ln 2, 1).
sin( x/2)
78. The function f ( x) 2e has a maximum whenever sin x 1 and a minimum whenever sin x 1.
2
2
Therefore the maximums occur at x = + 2k(2 ) and the minimums occur at x = 3 + 2k(2 ), where k is any
integer. The maximum is 2e 5.43656 and the minimum is 2 0.73576.
e
2 2 2
79. f ( x) x ln 1 f ( x) 2x ln 1 x 1 ( x ) 2x ln 1 x x(2ln x 1); f ( x) 0 x = 0 or
x x 1
x
x
ln x 1 . Since x = 0 is not in the domain of f,
2
f ( x ) 0 for x 1 . Therefore,
e
assumed at x 1 .
e
1/2
x e 1 . Also, ( ) 0
e
x
x
80. (a) Let f ( x) e x 1 f ( x) e 1 a critical point at x = 0
x
on (0, ); since f(0) = 0 it follows that ( ) 1 0
f x for 0 x 1 and
e
f 1 1 1 1/2
ln ln 1 ln 1
e e e e e
2e e 2e
is the absolute maximum value of f
f x e x for x 0 1
f | , so f is increasing
0
x
e x for x 0.
x 2
(b) Let ( ) 1
x
f x e x x 1 f ( x) e x 1 0 for x 0 by part (a), so f is increasing on (0, ); since
2
x 2
f(0) = 0 it follows that ( ) 1
x 2
f x e x x 1 0 for x 0 e 1 x x 1 for x 0.
2
2
81. Let x1 x 2 be two numbers in the domain of an increasing function f. Then, either x1 x 2 or x1 x 2 which
implies f ( x1 ) f ( x 2)
or f ( x1 ) f ( x 2),
since f(x) is increasing. In either case, f ( x1 ) f ( x 2 ) and f is oneto-one.
Similar arguments hold if f is decreasing.
82. f(x) is increasing since x 1 5 1 5
2 x 1 3 x 2 6 3 x 1 6 ;
1
df 1 df 1
dx 3 dx
1
3
3
3 3
83. f(x) is increasing since x2 x1 27x2 27 x1
;
1
df 2 df 1 1 1 2/3
81x
x
dx dx 81x 9x
9
2
1 1/3 2/3
x
3
3 1 1/3 1 1 1/3
y 27 x x y f ( x) x ;
3 3
84. f(x) is decreasing since
1
3 3
x2 x1 1 8x2 1 8 x1
;
df 2 df 1 1 1 2/3
24 x
(1 x)
dx dx 24x 6(1 x)
6
85. f(x) is decreasing since
1
2
1 1/3
2/3
(1 x)
2
3 3
x2 x1 (1 x2 ) (1 x1
) ;
df
2 df
2/3
3(1 x)
1 1 1 x
dx dx 2 2/3
3(1 x) 1/3 3 3
1 x
x
3 1 1/3 1 1 1/3
y 1 8 x x (1 y) f ( x) (1 x) ;
2 2
3 1/3 1 1/3
y (1 x) x 1 y f ( x) 1 x ;
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