Answer
$$
f(x)=(x^{2}-16)^{\frac{2}{3}}; \quad\quad[-5,8]
$$
Critical numbers in the given interval is: -4, 0, and 4.
The absolute maximum, $13.20770 $ occurs when $x=8 $ and the absolute minimum, $0$ occurs when $x=-4$ and $x=4$
Work Step by Step
$$
f(x)=(x^{2}-16)^{\frac{2}{3}}; \quad\quad[-5,8]
$$
First look for critical numbers in the interval $(-5,8)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=(\frac{2}{3})(2x)(x^{2}-16)^{\frac{-1}{3}}\\
&=(\frac{4x}{3})(x^{2}-16)^{\frac{-1}{3}}\\
&=\frac{4x}{3(x^{2}-16)^{\frac{1}{3}}}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{4x}{3(x^{2}-16)^{\frac{1}{3}}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 4x=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0
\end{aligned}
$$
The derivative $f^{\prime}(x) $ is undefined when
$$
\begin{aligned}
3(x^{2}-16)^{\frac{1}{3}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x^{2}-16=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (x-4)(x+4)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=4 \quad \text {or} \quad x=-4 \\
\end{aligned}
$$
So, critical numbers in the given interval is: -4, 0, and 4.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=-5: &\\
f(-5) &=((-5)^{2}-16)^{\frac{2}{3}}\\
&=(25-16)^{\frac{2}{3}}\\
&=(9)^{\frac{2}{3}}\\
&\approx \:4.32674
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=-4: &\\
f(-4) &=((-4)^{2}-16)^{\frac{2}{3}}\\
&=(0)^{\frac{1}{3}}\\
&=0
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=0: &\\
f(0) &=((0)^{2}-16)^{\frac{2}{3}}\\
&=(-16)^{\frac{2}{3}}\\
&\approx 6.34960
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=4: &\\
f(4) &=((4)^{2}-16)^{\frac{2}{3}}\\
&=(0)^{\frac{1}{3}}\\
&=0
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=8: &\\
f(8)&=((8)^{2}-16)^{\frac{2}{3}}\\
&=(64-16)^{\frac{2}{3}}\\
&=(48)^{\frac{2}{3}}\\
&\approx 13.20770
\end{aligned}
$$
We conclude that, the absolute maximum, $13.20770 $ occurs when $x=8 $ and the absolute minimum, $0$ occurs when $x=-4$ and $x=4$
