If f(x) = 2x^3 - 3x^2 - 12x + 5 and x in [-2, | vedclass

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(D) To find the maximum value of $f(x) = 2x^3 - 3x^2 - 12x + 5$ on the interval $[-2, 4]$,we first find the critical points by setting the derivative

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$4$

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(D) To find the maximum value of $f(x) = 2x^3 - 3x^2 - 12x + 5$ on the interval $[-2, 4]$,we first find the critical points by setting the derivative to zero.$f'(x) = 6x^2 - 6x - 12$Setting $f'(x) = 0$ gives $6(x^2 - x - 2) = 0$,which factors as $6(x - 2)(x + 1) = 0$.Thus,the critical points are $x = 2$ and $x = -1$,both of which lie within the interval $[-2, 4]$.Now,we evaluate the function at the critical points and the endpoints of the interval:$f(-2) = 2(-8) - 3(4) - 12(-2) + 5 = -16 - 12 + 24 + 5 = 1$$f(-1) = 2(-1) - 3(1) - 12(-1) + 5 = -2 - 3 + 12 + 5 = 12$$f(2) = 2(8) - 3(4) - 12(2) + 5 = 16 - 12 - 24 + 5 = -15$$f(4) = 2(64) - 3(16) - 12(4) + 5 = 128 - 48 - 48 + 5 = 37$Comparing these values,the maximum value is $37$,which occurs at $x = 4$.

If $f(x) = 2x^3 - 3x^2 - 12x + 5$ and $x \in [-2, 4]$,then the maximum value of the function occurs at which value of $x$?

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