Find the maximum value of

Let $f(x)=2 x^{3}-24 x+107$

$\therefore f^{\prime}(x)=6 x^{2}-24=6\left(x^{2}-4\right)$

Now,

$f^{\prime}(x)=0 \Rightarrow 6\left(x^{2}-4\right)=0 \Rightarrow x^{2}=4 \Rightarrow x=\pm 2$

We first consider the interval [1, 3].

Then, we evaluate the value of f at the critical point x = 2 ∈ [1, 3] and at the end points of the interval [1, 3].

$f(2)=2(8)-24(2)+107=16-48+107=75$

$f(1)=2(1)-24(1)+107=2-24+107=85$

$f(3)=2(27)-24(3)+107=54-72+107=89$

Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3.

Next, we consider the interval [−3, −1].

Evaluate the value of $f$ at the critical point $x=-2 \in[-3,-1]$ and at the end points of the interval $[1,3]$.

$f(-3)=2(-27)-24(-3)+107=-54+72+107=125$

$f(-1)=2(-1)-24(-1)+107=-2+24+107=129$

$f(-2)=2(-8)-24(-2)+107=-16+48+107=139$

Hence, the absolute maximum value of $f(x)$ in the interval $[-3,-1]$ is 139 occurring at $x=-2$.