The absolute maximum value of the function f( | vedclass

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Question

(B) Given the function $f(x) = 2x^3 - 3x^2 - 36x + 9$ on the interval $[-3, 3]$.First, find the derivative: $f'(x) = 6x^2 - 6x - 36$.Set $f'(x) = 0$ t

Vedclass · Accepted Answer

$53$

Vedclass · Answer

(B) Given the function $f(x) = 2x^3 - 3x^2 - 36x + 9$ on the interval $[-3, 3]$.First, find the derivative: $f'(x) = 6x^2 - 6x - 36$.Set $f'(x) = 0$ to find critical points:$6(x^2 - x - 6) = 0 \Rightarrow 6(x - 3)(x + 2) = 0$.The critical points are $x = 3$ and $x = -2$, both of which lie within the interval $[-3, 3]$.Now, evaluate $f(x)$ at the critical points and the endpoints:$f(-3) = 2(-3)^3 - 3(-3)^2 - 36(-3) + 9 = -54 - 27 + 108 + 9 = 36$.$f(-2) = 2(-2)^3 - 3(-2)^2 - 36(-2) + 9 = -16 - 12 + 72 + 9 = 53$.$f(3) = 2(3)^3 - 3(3)^2 - 36(3) + 9 = 54 - 27 - 108 + 9 = -72$.The absolute maximum value is the largest of these values, which is $53$.

The absolute maximum value of the function $f(x) = 2x^3 - 3x^2 - 36x + 9$ defined on the interval $[-3, 3]$ is:

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