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: 60

Answer

(a). $f'(x) = \frac{1}{3} · x^{−2/3} · (x+4)+x^{1/3} = \frac{x+4} {3x^2/3} + \frac{3x}{3x^{2/3}} =\frac{4x+4}{3x^{2/3}}$ . This expression is zero when $x = −1$, and is undefined when $x = 0$ (although $0$ is in the domain of $f$.) So $(0, 0)$ and $(−1,−3)$ are the critical points. (b). $f(−27) = 69$, $f(−1) = −3$, $f(0) = 0$, and $f(27) = 93$. So the absolute maximum is $93$ and the absolute minimum is $−3$.

Work Step by Step

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