Find the maximum value of $2x^{3}-24x+107$ in the interval $[1,3]$. Also,find the maximum value of the same function in the interval $[-3,-1]$.
(N/A) Let $f(x) = 2x^{3}-24x+107$.
$\therefore f'(x) = 6x^{2}-24 = 6(x^{2}-4)$.
Now,$f'(x) = 0 \Rightarrow 6(x^{2}-4) = 0 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$.
Case $1$: Interval $[1,3]$.
We evaluate $f(x)$ at the critical point $x = 2 \in [1,3]$ and at the endpoints $x = 1$ and $x = 3$.
$f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75$.
$f(1) = 2(1) - 24(1) + 107 = 2 - 24 + 107 = 85$.
$f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89$.
Thus,the absolute maximum value of $f(x)$ in $[1,3]$ is $89$ at $x = 3$.
Case $2$: Interval $[-3,-1]$.
We evaluate $f(x)$ at the critical point $x = -2 \in [-3,-1]$ and at the endpoints $x = -3$ and $x = -1$.
$f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 125$.
$f(-1) = 2(-1) - 24(-1) + 107 = -2 + 24 + 107 = 129$.
$f(-2) = 2(-8) - 24(-2) + 107 = -16 + 48 + 107 = 139$.
Thus,the absolute maximum value of $f(x)$ in $[-3,-1]$ is $139$ at $x = -2$.
$\therefore f'(x) = 6x^{2}-24 = 6(x^{2}-4)$.
Now,$f'(x) = 0 \Rightarrow 6(x^{2}-4) = 0 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$.
Case $1$: Interval $[1,3]$.
We evaluate $f(x)$ at the critical point $x = 2 \in [1,3]$ and at the endpoints $x = 1$ and $x = 3$.
$f(2) = 2(8) - 24(2) + 107 = 16 - 48 + 107 = 75$.
$f(1) = 2(1) - 24(1) + 107 = 2 - 24 + 107 = 85$.
$f(3) = 2(27) - 24(3) + 107 = 54 - 72 + 107 = 89$.
Thus,the absolute maximum value of $f(x)$ in $[1,3]$ is $89$ at $x = 3$.
Case $2$: Interval $[-3,-1]$.
We evaluate $f(x)$ at the critical point $x = -2 \in [-3,-1]$ and at the endpoints $x = -3$ and $x = -1$.
$f(-3) = 2(-27) - 24(-3) + 107 = -54 + 72 + 107 = 125$.
$f(-1) = 2(-1) - 24(-1) + 107 = -2 + 24 + 107 = 129$.
$f(-2) = 2(-8) - 24(-2) + 107 = -16 + 48 + 107 = 139$.
Thus,the absolute maximum value of $f(x)$ in $[-3,-1]$ is $139$ at $x = -2$.
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