The maximum value of 2x^3 - 24x + 107 in the | vedclass

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(D) Let $f(x) = 2x^3 - 24x + 107$.First,we find the critical points by setting the derivative $f'(x) = 0$:$f'(x) = 6x^2 - 24 = 0$$6(x^2 - 4) = 0$$x^2

Vedclass · Accepted Answer

$139$

Vedclass · Answer

(D) Let $f(x) = 2x^3 - 24x + 107$.First,we find the critical points by setting the derivative $f'(x) = 0$:$f'(x) = 6x^2 - 24 = 0$$6(x^2 - 4) = 0$$x^2 = 4 \Rightarrow x = 2, -2$.Both critical points $x = 2$ and $x = -2$ lie within the interval $[-3, 3]$.Now,we evaluate $f(x)$ at the critical points and the endpoints of the interval:$f(-3) = 2(-3)^3 - 24(-3) + 107 = 2(-27) + 72 + 107 = -54 + 72 + 107 = 125$.$f(3) = 2(3)^3 - 24(3) + 107 = 2(27) - 72 + 107 = 54 - 72 + 107 = 89$.$f(2) = 2(2)^3 - 24(2) + 107 = 2(8) - 48 + 107 = 16 - 48 + 107 = 75$.$f(-2) = 2(-2)^3 - 24(-2) + 107 = 2(-8) + 48 + 107 = -16 + 48 + 107 = 139$.Comparing these values,the maximum value is $139$.

The maximum value of $2x^3 - 24x + 107$ in the interval $[-3, 3]$ is

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