Answer
$$
f(x)=5x^{\frac{2}{3}}+2x^{\frac{5}{3}}; \quad\quad[-2,1]
$$
The critical numbers in the given interval are: -1 and 0.
The absolute maximum, $7 $ occurs when $x=1 $ and the absolute minimum, $0 $ occurs when $x=0.$
Work Step by Step
$$
f(x)=5x^{\frac{2}{3}}+2x^{\frac{5}{3}}; \quad\quad[-2,1]
$$
First look for critical numbers in the interval $(-2,1)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=5(\frac{2}{3})x^{\frac{-1}{3}}+2(\frac{5}{3})x^{\frac{2}{3}}\\
&=(\frac{10}{3})x^{\frac{-1}{3}}+(\frac{10}{3})x^{\frac{2}{3}}\\
&=(\frac{10}{3x^{\frac{1}{3}}})+(\frac{10x^{\frac{2}{3}}}{3})\\
&=\frac{10+10x}{3x^{\frac{1}{3}}}\\
&=\frac{10(1+x)}{3x^{\frac{1}{3}}}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{10(1+x)}{3x^{\frac{1}{3}}}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 10(1+x)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=-1
\end{aligned}
$$
The derivative $f^{\prime}(x) $ is undefined when
$$
\begin{aligned}
3x^{\frac{1}{3}}&=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=0
\end{aligned}
$$
Since $f(0)$ is defined, then $x=0$ is also critical point
Therefore, the critical numbers in the given interval are: -1 and 0.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=-2: &\\
f(-2) &=5(-2)^{\frac{2}{3}}+2(-2)^{\frac{5}{3}}\\
&=2^{\frac{2}{3}}\\
&\approx 1.5874
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=-1: &\\
f(-1) &=5(-1)^{\frac{2}{3}}+2(-1)^{\frac{5}{3}}\\
&=3
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=0: &\\
f(0) &=5(0)^{\frac{2}{3}}+2(0)^{\frac{5}{3}}\\
&=0\\
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=1: &\\
f(1) &=5(1)^{\frac{2}{3}}+2(1)^{\frac{5}{3}}\\
&=7
\end{aligned}
$$
We conclude that, the absolute maximum, $7 $ occurs when $x=1 $ and the absolute minimum, $0 $ occurs when $x=0.$
