The maximum value of f(x) = (7-x)^4 (2+x)^5 i | vedclass

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(C) To find the maximum value,we first find the derivative $f'(x)$ and set it to $0$.$f(x) = (7-x)^4 (2+x)^5$Using the product rule and chain rule:$f'

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$4^4 5^5$

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(C) To find the maximum value,we first find the derivative $f'(x)$ and set it to $0$.$f(x) = (7-x)^4 (2+x)^5$Using the product rule and chain rule:$f'(x) = 4(7-x)^3(-1)(2+x)^5 + 5(7-x)^4(2+x)^4$$f'(x) = (7-x)^3(2+x)^4 [-4(2+x) + 5(7-x)]$$f'(x) = (7-x)^3(2+x)^4 [-8 - 4x + 35 - 5x]$$f'(x) = (7-x)^3(2+x)^4 [27 - 9x]$Setting $f'(x) = 0$,we get critical points $x = 7, x = -2, x = 3$.For $x \in (-2, 7)$,the function is positive. At $x = 3$,the derivative changes from positive to negative,indicating a local maximum.The maximum value is $f(3) = (7-3)^4 (2+3)^5 = 4^4 	imes 5^5$.

The maximum value of $f(x) = (7-x)^4 (2+x)^5$ is

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