## Calculus 8th Edition

Critical number of $g(x)$ is $x=0$.
Given, $g(x) = \sqrt[3]{4-x^2}=(4-x^2)^{\frac{1}{3}}$. Differentiating $g$ with respect to $x$ using power and chain rule, we get, $$\frac{dg}{dx}=\frac{1}{3}(4-x^2)^{-\frac{2}{3}}(-2x)=-\frac{2}{3}x(4-x^2)^{-\frac{2}{3}}$$ At critical points, the derivative of the function is equal to $0$. Thus, $\frac{dg}{dx}=0$. $$-\frac{2}{3}x(4-x^2)^{-\frac{2}{3}}=0$$ $$\frac{-\frac{2}{3}x}{(4-x^2)^{-\frac{2}{3}}}=0\implies -\frac{2}{3}x=0$$ $$x=0.$$