Let f(x)=(x+3)^2(x-2)^3 for x in [-4,4]. If M | vedclass

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(C) Given $f(x)=(x+3)^2(x-2)^3$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f^{\prime}(x) = \frac{d}{dx}[(x+3)^2(x-2)^3]$

Vedclass · Accepted Answer

$608$

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(C) Given $f(x)=(x+3)^2(x-2)^3$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f^{\prime}(x) = \frac{d}{dx}[(x+3)^2(x-2)^3]$$f^{\prime}(x) = 2(x+3)(x-2)^3 + 3(x-2)^2(x+3)^2$$f^{\prime}(x) = (x+3)(x-2)^2 [2(x-2) + 3(x+3)]$$f^{\prime}(x) = (x+3)(x-2)^2 [2x - 4 + 3x + 9]$$f^{\prime}(x) = (x+3)(x-2)^2 (5x + 5)$$f^{\prime}(x) = 5(x+3)(x-2)^2 (x+1)$Setting $f^{\prime}(x) = 0$,we get critical points $x = -3, -1, 2$.Now,we evaluate $f(x)$ at the critical points and the endpoints of the interval $[-4, 4]$:$f(-4) = (-4+3)^2(-4-2)^3 = (-1)^2(-6)^3 = 1 	imes (-216) = -216$$f(-3) = (-3+3)^2(-3-2)^3 = 0$$f(-1) = (-1+3)^2(-1-2)^3 = (2)^2(-3)^3 = 4 	imes (-27) = -108$$f(2) = (2+3)^2(2-2)^3 = 0$$f(4) = (4+3)^2(4-2)^3 = (7)^2(2)^3 = 49 	imes 8 = 392$Comparing these values,the maximum value $M = 392$ and the minimum value $m = -216$.Therefore,$M - m = 392 - (-216) = 392 + 216 = 608$.

Let $f(x)=(x+3)^2(x-2)^3$ for $x \in [-4,4]$. If $M$ and $m$ are the maximum and minimum values of $f$ respectively in $[-4,4]$,then the value of $M-m$ is:

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