In the interval [-2, 4],the absolute maximum | vedclass

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(A) Given the function $f(x) = 2x^3 - 3x^2 - 12x + 5$. To find the critical points,we calculate the derivative $f'(x) = 6x^2 - 6x - 12$. Setting $f'(x

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$4$

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(A) Given the function $f(x) = 2x^3 - 3x^2 - 12x + 5$. To find the critical points,we calculate the derivative $f'(x) = 6x^2 - 6x - 12$. Setting $f'(x) = 0$,we get $6(x^2 - x - 2) = 0$,which factors as $6(x + 1)(x - 2) = 0$. Thus,the critical points are $x = -1$ and $x = 2$,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 absolute maximum is $37$,which occurs at $x = 4$. Therefore,the correct option is $A$.

In the interval $[-2, 4]$,the absolute maximum of $f(x) = 2x^3 - 3x^2 - 12x + 5$ occurs at $x =$

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