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(A) Given function is $f(x) = (x - 1)(x + 2)^2$.First,find the derivative $f'(x)$ using the product rule:$f'(x) = 1 \cdot (x + 2)^2 + (x - 1) \cdot 2(

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$0, -4$

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(A) Given function is $f(x) = (x - 1)(x + 2)^2$.First,find the derivative $f'(x)$ using the product rule:$f'(x) = 1 \cdot (x + 2)^2 + (x - 1) \cdot 2(x + 2)$$f'(x) = (x + 2)(x + 2 + 2x - 2) = (x + 2)(3x) = 3x(x + 2)$.Set $f'(x) = 0$ to find critical points:$3x(x + 2) = 0 \implies x = 0$ or $x = -2$.Now,find the second derivative $f''(x) = \frac{d}{dx}(3x^2 + 6x) = 6x + 6$.At $x = -2$: $f''(-2) = 6(-2) + 6 = -6 The local maximum value is $f(-2) = (-2 - 1)(-2 + 2)^2 = 0$.At $x = 0$: $f''(0) = 6(0) + 6 = 6 > 0$,so $x = 0$ is a point of local minimum.The local minimum value is $f(0) = (0 - 1)(0 + 2)^2 = -4$.Thus,the local maximum and minimum values are $0$ and $-4$ respectively.

Local maximum and local minimum values of the function $f(x) = (x - 1)(x + 2)^2$ are

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