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: 38

Answer

(a). $f'(x) = \frac{4}{3} {(x + 1)}^{1/3}$, which is zero for $x = −1$. So $(−1, 0)$ is the only critical point. (b). We have that $f(−9) = 16$, $f(−1) = 0$, and $f(7) ≈ 16$, so the maximum value of $f$ on this interval is $16$ and the minimum is $0$.

Work Step by Step

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