University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 4 - Section 4.1 - Extreme Values of Functions - Exercises - Page 215: 27

Answer

$h(x)$ has the absolute maximum value of $2$ at $x=8$ and the absolute minimum value of $-1$ at $x=-1$ in the defined domain.
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Work Step by Step

$$h(x)=\sqrt[3]x=x^{1/3}\hspace{1cm}-1\le x\le8$$ 1) Find the derivative $h'(x)$: $$h'(x)=\frac{1}{3}x^{-2/3}$$ Here we have $h'(x)=0$ only when $x=0$. Therefore, $x=0$ is the only critical point of $h(x)$. 2) Evaluate $h'(x)$ at the critical points and endpoints: - For $x=-1$: $$h(-1)=\sqrt[3]{-1}=-1$$ - For $x=0$: $$h(0)=\sqrt[3]{0}=0$$ - For $x=8$: $$h(8)=\sqrt[3]{8}=2$$ This means $h(x)$ has the absolute maximum value of $2$ at $x=8$ and the absolute minimum value of $-1$ at $x=-1$ in the defined domain.
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