Find the maximum value of $2{x^3} - 24x + 107$ in the interval $\left[ {1,{\rm{ }}3} \right]$. Find the maximum value of the same function in $[ - 3,\; - 1]$.

Let $f(x) = 2{x^3} - 24x + 107,1 \le x \le 3$
$\Rightarrow f'(x) = 6{x^2} - 24$

For critical points, $f'(x) = 0$
$\Rightarrow 6{x^2} - 24 = 0 \Rightarrow 6({x^2} - 4) = 0 \Rightarrow {x^2} = 4 \Rightarrow x = \pm 2$
$\Rightarrow x = 2 \in \left[ {1,3} \right]$

To find the maximum or minimum value, we have to evaluate $f(1)$,$f(2)$ and $f(3)$
Now, $f(1) = 2 \times {1^3} - 24 \times 1 + 107 = 85$,
$f(2) = 2 \times {2^3} - 24 \times 2 + 107 = 75,$
$f(3) = 2 \times {3^3} - 24 \times 3 + 107 = 89$
Therefore maximum value of $f(x)$ is $89$ at $x = 3$

If we consider the function in the interval $[ - 3,\; - 1]$, then we take $x = - 2 \in [ - 3,\; - 1]$.

So, we evaluate$f( - 1),f( - 2)$ , and$f( - 3)$ .
$f( - 1) = 2{( - 1)^3} - 24( - 1) + 107 = 129$
$f( - 2) = 2{( - 2)^3} - 24( - 2) + 107 = 139$

$f( - 3) = 2{( - 3)^3} - 24( - 3) + 107 = 125$

Therefore maximum value of $f(x)$ is $139$ at $x = - 2$.