Answer
(a). $f'(x) = \frac{4}{3} {(x + 1)}^{1/3}$, which is zero for $x = −1$. So $(−1, 0)$ is the only critical point.
(b). We have that $f(−9) = 16$, $f(−1) = 0$, and
$f(7) ≈ 16$, so the maximum value of $f$ on
this interval is $16$ and the minimum is $0$.
Work Step by Step
See the graph for explanation.
