Answer
(a). $f'(x) = \frac{e^x−e^{−x}}
{2}$ , which is zero when $e^x = e^{−x}$ or
$x =-x$, so only for $x = 0$.
(b). There is a local (and absolute) minimum at $x = 0$.
Work Step by Step
See the graph for explanation of the above.
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 4 - Applications of the Derivative - 4.1 Maxima and Minima - 4.1 Exercises - Page 243: 31
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