Find the maximum value

Solution:

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

$\Rightarrow f^{\prime}(x)=6 x^{2}-24$

For a local maximum or a local minimum, we must have

$f^{\prime}(x)=0$

$\Rightarrow 6 x^{2}-24=0$

$\Rightarrow 6 x^{2}=24$

$\Rightarrow x^{2}=4$

$\Rightarrow x=\pm 2$

Thus, the critical points of $f$ in the interval $[1,3]$ are 1,2 and 3 .

Now,

$f(1)=2(1)^{3}-24(1)+107=85$

$f(2)=2(2)^{3}-24(2)+107=75$

$f(3)=2(3)^{3}-24(3)+107=89$

Hence, the absolute maximum value when $x=3$ in the interval $[1,3]$ is 89 .

Again, the critical points of $f$ in the interval $[-3,-1]$ are $-1,-2$ and $-3$.

So,

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

$f(-2)=2(-2)^{3}-24(-2)+107=139$

$f(-1)=2(-1)^{3}-24(-1)+107=129$

Hence, the absolute maximum value when $x=-2$ is 139 .