Answer
Absolute maximum $f(8)=2$ and the absolute minimum $f(-1)=-1$.
See graph.
Work Step by Step
Step 1. Given the function $h(x)=\sqrt[3] x, -1\leq x\leq 8$, we have $f(-1)=-1, f(8)=2$,
Step 2. $f'(x)=\frac{1}{3}x^{-2/3}\ne0$; no extrema here.
Step 3. We can identify the absolute maximum as $f(8)=2$ and the absolute minimum as $f(-1)=-1$ on the given interval.
Step 4. See graph.