Thomas Calculus 13th [Solutions]

224 Chapter 4 Applications of Derivatives
39. The first derivative f ( x ) 1 1
2
x x
has a zero at
x = 1.
Critical point value: f(1) = 1 + ln 1 = 1
Endpoint values: f(0.5) = 2 + ln 0.5 1.307;
f (4) 1 ln 4 1.636;
4
Absolute maximum value is 1 ln 4 at x = 4;
4
Absolute Minimum value is 1 at x = 1;
Local maximum at 1
2 , 2 ln 2
40.
2 2
x
x
g( x) e g ( x) 2xe a critical point at
x = 0;
4
g( 2) e , g(0) = 1, and
g(1)
the absolute maximum is 1 at x = 0 and the
4
absolute minimum is e at x = 2
e
1
41.
42.
43.
44.
45.
4/3 4 1/3
f ( x) x f ( x)
x a critical point at x 0; f ( 1) 1, f (0) 0, f (8) 16 the absolute
3
maximum is 16 at x 8 and the absolute minimum is 0 at x 0
5/3 5 2/3
f ( x) x f ( x)
x a critical point at x 0; f ( 1) 1, f (0) 0, f (8) 32 the absolute
3
maximum is 32 at x 8 and the absolute minimum is 1 at x 1
3/5 3 2/5
g( ) g ( )
a critical point at 0; g( 32) 8, g(0) 0, g (1) 1 the absolute
5
maximum is 1 at 1 and the absolute minimum is 8 at 32
2/3 1/3
h( ) 3 h ( ) 2 a critical point at 0; h( 27) 27, h(0) 0, h(8)
12 the absolute
maximum is 27 at 27 and the absolute minimum is 0 at 0
2
y x 6x 7 y 2x 6 2x 6 0 x 3. The critical point is x 3.
46.
2 3 2
f ( x) 6 x x f ( x) 12x 3x
x 0 and x 4.
2
12x 3x 0 3 x(4 x) 0 x 0 or x 4. The critical points are
47.
48.
49.
3
2
3 2
2
f ( x) x(4 x) f ( x)
x[3(4 x) ( 1)] (4 x)
(4 x) [ 3 x (4 x)]
(4 x) (4 4 x)
2
2
4(4 x) (1 x)
4(4 x) (1 x) 0 x 1 or x 4. The critical points are x 1 and x 4.
g( x) ( x
2
1) ( x
2
3) g ( x)
( x
2
1) 2( x 3)(1) 2( x 1)(1) ( x
2
3) 2( x 3) ( x 1)[( x 1) ( x 3)]
4( x 3)( x 1) ( x 2) 4( x 3)( x 1)( x 2) 0 x 3 or x 1 or x 2. The critical points are x 1, x 2,
and x 3.
3
3 3
2
y x 2 y 2x
2 2x
2 2x
2 0 2 3 2 0 1; 2x
2
2
x x undefined x 0 x 0.
x
2 2
2 2
x x x
x
The domain of the function is ( , 0) (0, ), thus x 0 is not the domain, so the only critical point is x 1.
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