Thomas Calculus 13th [Solutions]

220 Chapter 4 Applications of Derivatives
16. f has an absolute max at x 0 but does not have an
absolute min. Since the interval on which f is defined,
1 x 1, is an open interval, we do not meet the
conditions of Theorem 1.
17. f has an absolute max at x 2 but does not have an
absolute min. Since the function is not continuous at
x 1, we do not meet the conditions of Theorem 1.
18. f has an absolute max at x 4 but does not have an
absolute min. Since the function is not continuous at
x 0, we do not meet the conditions of Theorem 1.
19. f has an absolute max at x and an absolute min at
2
x
3
2 . Since the interval on which f is defined,
0 x 2 , is an open interval we do not meet the
conditions of Theorem 1.
20. f has an absolute max at x 0 and an absolute min
at x
2 and x 1 but does not have an absolute
maximum. Since f is defined on a union of halfopen
intervals, we do not meet the conditions of
Theorem 1.
y
( 0, 1)
y
f ( x)
1 0
2
x
21. f ( x) 2 x 5 f ( x ) 2 no critical points;
3 3
f ( 2) 19 , f (3) 3 the absolute maximum
3
is 3 at x 3 and the absolute minimum is 19
3
at x 2
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