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 $g$.
