Answer
(a). $f'(x) = \frac{1}{3}
· x^{−2/3} · (x+4)+x^{1/3} = \frac{x+4}
{3x^2/3} + \frac{3x}{3x^{2/3}} =\frac{4x+4}{3x^{2/3}}$ . This expression is zero when $x = −1$, and is undefined when $x = 0$ (although $0$ is in the domain of $f$.) So $(0, 0)$ and $(−1,−3)$ are the critical points.
(b). $f(−27) = 69$, $f(−1) = −3$, $f(0) = 0$, and $f(27) = 93$. So the absolute maximum is $93$ and the absolute minimum is $−3$.
Work Step by Step
See the graph for explanation.
