Let $f(x)=2 x^{3}-24 x+107$
$
\therefore f^{\prime}(x)=6 x^{2}-24=6\left(x^{2}-4\right)
$
Now, for local maxima and local minima we have $f^{\prime}(x)=0$
$
\begin{array}{l}
\Rightarrow 6\left(x^{2}-4\right)=0 \\
\Rightarrow x^{2}=4 \\
\Rightarrow x=\pm 2
\end{array}
$
We first consider the interval $[1,3]$.
Then, we evaluate the value of $f$ at the critical point $x=2 \in[1,3]$ and at the end points of the interval $[1,3]$.
$
\begin{array}{l}
f(2)=2\left(2^{3}\right)-24(2)+107=75 \\
f(1)=2(1)^{3}-24(1)+107=85 \\
f(3)=2(3)^{3}-24(3)+107=89
\end{array}
$
Hence, the absolute maximum value of $\mathrm{f}(\mathrm{x})$ in the interval $[1,3]$ is 89 occurring at $\mathrm{x}=3$,
Next, we consider the interval $[-3,-1]$.
Evaluate the value of $f$ at the critical point $x=-2 \in[-3,-1]$
$
\begin{array}{l}
f(-3)=2(-3)^{3}-24(-3)+107=125 \\
f(-2)=2(-2)^{3}-24(-3)+107=139 \\
f(-1)=2(-1)^{3}-24(-2)+107=129
\end{array}
$
Hence, the absolute maximum value of $\mathrm{f}$ is 139 and occurs when $\mathrm{x}=-2$.