Answer
$$
f(x)=x^{2}-8 \ln x ; \quad\quad[1,4]
$$
The critical number in the given interval is: $2$,
The absolute maximum, $4.9096 $ occurs when $x=4 $ and the absolute minimum, $-1.54517 $ occurs when $x=2.$
Work Step by Step
$$
f(x)=x^{2}-8 \ln x ; \quad\quad[1,4]
$$
First look for critical numbers in the interval $(1,4)$.
find the points where the derivative is $0$
$$
\begin{aligned}
f^{\prime}(x) &=(2)x-8(\frac{1}{x})\\
&=2x-\frac{8}{x}\\
&=\frac{2x^{2}-8}{x}\\
\end{aligned}
$$
Solve the equation $f^{\prime}(x)=0 $ to get
$$
\begin{aligned}
f^{\prime}(x)& =\frac{2x^{2}-8}{x}=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 2x^{2}-8=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& 2(x^{2}-4)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (x-2)(x+2)=0\\
\Rightarrow\quad\quad\quad\quad\quad\\
& (x-2)=0 \quad \text {or} \quad(x+2)=0 \\
\Rightarrow\quad\quad\quad\quad\quad\\
& x=2 \quad \text {or} \quad x=-2 \\
\end{aligned}
$$
We know that $x=2 \in [1,4]$ but $-2 \notin [1,4]$
So, critical number in the given interval is: $2$,
also $ f^{\prime}(x) $ is not exist when $x=0 $ but $0 \notin [1,4] $, so $x=0$ is not critical numbers in the given interval.
Now, evaluate the function at the critical numbers and the endpoints:
$$
\begin{aligned}
\text{when } x=1: \\
f(1) &=(1)^{2}-8 \ln (1)\\
&=(1)^{2}-8 (0) \\
&=1
\end{aligned}
$$
$$
\begin{aligned}
\text{when } x=2: \\
f(2) &=(2)^{2}-8 \ln (2)\\
&=4-8\ln \left(2\right)\\
&\approx -1.54517
\end{aligned}
$$
and
$$
\begin{aligned}
\text{when } x=4: \\
f(4) &=(4)^{2}-8 \ln (4)\\
&=16-16\ln \left(2\right)\\
&\approx 4.9096
\end{aligned}
$$
We conclude that, the absolute maximum, $4.9096 $ occurs when $x=4 $ and the absolute minimum, $-1.54517 $ occurs when $x=2.$
