Answer
$$
f(x)=x+3x^{\frac{2}{3}}; \quad\quad[-10,1]
$$
The critical numbers in the given interval are: -8 and 0.
The absolute maximum, $4 $ occurs when $x=-8, 1 $ and the absolute minimum, $0 $ occurs when $x=0.$
Work Step by Step
$$
f(x)=x+3x^{\frac{2}{3}}; \quad\quad[-10,1]
$$
First look for critical numbers in the interval $(-10,1)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=1+3(\frac{2}{3})x^{\frac{-1}{3}}\\
&=1+(\frac{6}{3})x^{\frac{-1}{3}}\\
&=1+2x^{\frac{-1}{3}}\\
&=\frac{x^{\frac{1}{3}}+2}{x^{\frac{1}{3}}}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{x^{\frac{1}{3}}+2}{x^{\frac{1}{3}}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x^{\frac{1}{3}}+2=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x^{\frac{1}{3}}=-2\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=-8
\end{aligned}
$$
The derivative $f^{\prime}(x) $ is undefined when
$$
\begin{aligned}
x^{\frac{1}{3}}&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0
\end{aligned}
$$
Since $f(0)$ is defined, then $x=0$ is also critical point
Therefore, the critical numbers in the given interval are: -8 and 0.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=-10: &\\
f(-10) &=(-10)+3(-10)^{\frac{2}{3}}\\
&=-10+3\cdot \:10^{\frac{2}{3}}\\
&\approx 3.92476
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=-8: &\\
f(-8) &=(-8)+3(-8)^{\frac{2}{3}}\\
&=4
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=0: &\\
f(0) &=(0)+3(0)^{\frac{2}{3}}\\
&=0
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=1: &\\
f(1) &=(1)+3(1)^{\frac{2}{3}}\\
&=4
\end{aligned}
$$
We conclude that, the absolute maximum, $4 $ occurs when $x=-8, 1 $ and the absolute minimum, $0 $ occurs when $x=0.$