Answer
$$
f(x)=(x^{2}-4)^{\frac{1}{3}}; \quad\quad[-2,3]
$$
Critical numbers in the given interval is: -2, 0, and 2.
The absolute maximum, $1.70997 $ occurs when $x=3 $ and the absolute minimum, $-1.58740 $ occurs when $x=0.$
Work Step by Step
$$
f(x)=(x^{2}-4)^{\frac{1}{3}}; \quad\quad[-2,3]
$$
First look for critical numbers in the interval $(-2,3)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=(\frac{1}{3})(2x)(x^{2}-4)^{\frac{-2}{3}}\\
&=(\frac{2x}{3})(x^{2}-4)^{\frac{-2}{3}}\\
&=\frac{2x}{3(x^{2}-4)^{\frac{-2}{3}}}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{2x}{3(x^{2}-4)^{\frac{2}{3}}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 2x=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0
\end{aligned}
$$
The derivative $f^{\prime}(x) $ is undefined when
$$
\begin{aligned}
3(x^{2}-4)^{\frac{2}{3}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x^{2}-4=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (x-2)(x+2)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=2 \quad \text {or} \quad x=-2 \\
\end{aligned}
$$
So, critical numbers in the given interval is: -2, 0, and 2.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=-2: &\\
f(-2) &=((-2)^{2}-4)^{\frac{1}{3}}\\
&=(0)^{\frac{1}{3}}\\
&=0
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=0: &\\
f(0)&=((0)^{2}-4)^{\frac{1}{3}}\\
&=(-4)^{\frac{1}{3}}\\
&\approx -1.58740,
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=2: &\\
f(2)&=((2)^{2}-4)^{\frac{1}{3}}\\
&=(0)^{\frac{1}{3}}\\
&=0
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=3: &\\
f(3)&=((3)^{2}-4)^{\frac{1}{3}}\\
&=(9-4)^{\frac{1}{3}}\\
&=(5)^{\frac{1}{3}}\\
&\approx 1.70997
\end{aligned}
$$
We conclude that, the absolute maximum, $1.70997 $ occurs when $x=3 $ and the absolute minimum, $-1.58740 $ occurs when $x=0.$
