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

Answer

(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 $0$.

Work Step by Step

See the graph for explanation.
Small 1561533991
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.