Local maximum and local minimum values respec | vedclass

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Question

(B) Given function: $f(x) = (x-1)(x+2)^2$.Find the derivative $f'(x)$ using the product rule:$f'(x) = (1)(x+2)^2 + (x-1)(2)(x+2)$$f'(x) = (x+2)[(x+2)

Vedclass · Accepted Answer

$0, -4$

Vedclass · Answer

(B) Given function: $f(x) = (x-1)(x+2)^2$.Find the derivative $f'(x)$ using the product rule:$f'(x) = (1)(x+2)^2 + (x-1)(2)(x+2)$$f'(x) = (x+2)[(x+2) + 2(x-1)]$$f'(x) = (x+2)(x+2+2x-2) = 3x(x+2)$.Set $f'(x) = 0$ to find critical points:$3x(x+2) = 0 \implies x = 0, x = -2$.Use the first derivative test:For $x  0$ (increasing).For $-2 For $x > 0$,$f'(x) > 0$ (increasing).At $x = -2$,the function changes from increasing to decreasing,so $f(-2)$ is a local maximum:$f(-2) = (-2-1)(-2+2)^2 = (-3)(0) = 0$.At $x = 0$,the function changes from decreasing to increasing,so $f(0)$ is a local minimum:$f(0) = (0-1)(0+2)^2 = (-1)(4) = -4$.Thus,the local maximum and local minimum values are $0$ and $-4$ respectively.

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

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