Answer
$h(x)$ has the absolute maximum value of $2$ at $x=8$ and the absolute minimum value of $-1$ at $x=-1$ in the defined domain.
Work Step by Step
$$h(x)=\sqrt[3]x=x^{1/3}\hspace{1cm}-1\le x\le8$$
1) Find the derivative $h'(x)$: $$h'(x)=\frac{1}{3}x^{-2/3}$$
Here we have $h'(x)=0$ only when $x=0$. Therefore, $x=0$ is the only critical point of $h(x)$.
2) Evaluate $h'(x)$ at the critical points and endpoints:
- For $x=-1$: $$h(-1)=\sqrt[3]{-1}=-1$$
- For $x=0$: $$h(0)=\sqrt[3]{0}=0$$
- For $x=8$: $$h(8)=\sqrt[3]{8}=2$$
This means $h(x)$ has the absolute maximum value of $2$ at $x=8$ and the absolute minimum value of $-1$ at $x=-1$ in the defined domain.