The local maximum value of the function f(x)= | vedclass

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(B) Given function is $f(x) = -(x-2)^3(x+2)^2$. To find the local maximum, we first find the derivative $f'(x)$: $f'(x) = -[3(x-2)^2(x+2)^2 + (x-2)^3

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$\frac{12^3 \cdot 8^2}{5^5}$

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(B) Given function is $f(x) = -(x-2)^3(x+2)^2$. To find the local maximum, we first find the derivative $f'(x)$: $f'(x) = -[3(x-2)^2(x+2)^2 + (x-2)^3 \cdot 2(x+2)]$ $f'(x) = -(x-2)^2(x+2) [3(x+2) + 2(x-2)]$ $f'(x) = -(x-2)^2(x+2) [3x + 6 + 2x - 4]$ $f'(x) = -(x-2)^2(x+2)(5x + 2)$. Setting $f'(x) = 0$, we get critical points $x = 2, x = -2, x = -\frac{2}{5}$. Using the first derivative test: For $x For $-2  0$. For $-\frac{2}{5} For $x > 2$, $f'(x) Since $f'(x)$ changes from positive to negative at $x = -\frac{2}{5}$, it is a point of local maximum. The local maximum value is $f(-\frac{2}{5}) = -(-\frac{2}{5}-2)^3(-\frac{2}{5}+2)^2$. $f(-\frac{2}{5}) = -(-\frac{12}{5})^3(\frac{8}{5})^2 = -(-\frac{1728}{125}) \cdot \frac{64}{25} = \frac{1728 \cdot 64}{3125} = \frac{12^3 \cdot 8^2}{5^5}$.

The local maximum value of the function $f(x)=-(x-2)^3(x+2)^2$ is

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