Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises: 38

Answer

$c=0$ and $c=\pm2$ are critical numbers of $g$.

Work Step by Step

How to find the critical numbers of a function $f$ according to definition 1) Find all numbers of $c$ satisfying either $f'(c)=0$ or $f'(c)$ does not exist. 2) See that all the $c$ we get lie in the domain of $f$ $(D_f)$ or not. - If $c$ lies in $D_f$, $c$ is a critical number of $f$. - If $c$ does not lie in $D_f$, $c$ is not a critical number of $f$. $$g(x)=\sqrt[3]{4-x^2}$$ $$g(x)=(4-x^2)^{1/3}$$ $D_g=R$ 1) Now, we find $g'(x)$ $$g'(x)=\frac{1}{3}(4-x^2)^{-2/3}(4-x^2)'$$ $$g'(x)=\frac{1}{3}(4-x^2)^{-2/3}(-2x)$$ $$g'(x)=-\frac{2}{3}x(4-x^2)^{-2/3}$$ We find $g'(c)=0$, $$g'(c)=0$$ $$-\frac{2}{3}c(4-c^2)^{-2/3}=0$$ $$c\sqrt[3]{\frac{1}{(4-c^2)^2}}=0$$ $$c=0$$ (it might be tempting to think $\sqrt[3]{\frac{1}{(4-c^2)^2}}=0$ also, but a little consideration shows that $\frac{1}{(4-c^2)^2}$ cannot equal $0$). $g'(c)$ would not exist for $$(4-c^2)^2=0$$ $$c=\pm2$$ 2) Examine whether $c$ lies in $D_h$ or not. We see that all of $c=0$ and $c=\pm2\in R$, so they lie in $D_g$. We conclude that $c=0$ and $c=\pm2$ are critical numbers of $g$.
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