Calculus: Early Transcendentals (2nd Edition)

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: 56


(a). $f'(x) = \frac{1}{2\sqrt{x−2}}$ , which is never zero. Also, $f(x)$ exists on $(2,∞)$, so there are no critical points. (b). $f(2) = 0$ and $f(6) = 2$, so the absolute maximum of this function on the given interval is $2$ and the absolute minimum is $0$.

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