For what values of x does the function f(x) = | vedclass

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(B) Given the function $f(x) = (x - 1)^2 (x - 2)$.First,find the derivative $f'(x)$ using the product rule:$f'(x) = \frac{d}{dx}[(x - 1)^2] \cdot (x -

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$x \in (1, 5/3)$

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(B) Given the function $f(x) = (x - 1)^2 (x - 2)$.First,find the derivative $f'(x)$ using the product rule:$f'(x) = \frac{d}{dx}[(x - 1)^2] \cdot (x - 2) + (x - 1)^2 \cdot \frac{d}{dx}[(x - 2)]$$f'(x) = 2(x - 1)(x - 2) + (x - 1)^2$$f'(x) = (x - 1) [2(x - 2) + (x - 1)]$$f'(x) = (x - 1) (2x - 4 + x - 1)$$f'(x) = (x - 1) (3x - 5)$For the function to be monotonically decreasing,we must have $f'(x) $(x - 1)(3x - 5) The critical points are $x = 1$ and $x = 5/3$.Testing the intervals $(-\infty, 1)$,$(1, 5/3)$,and $(5/3, \infty)$:For $x \in (1, 5/3)$,the product $(x - 1)(3x - 5)$ is negative.Thus,the function decreases on the interval $x \in (1, 5/3)$.

For what values of $x$ does the function $f(x) = (x - 1)^2 (x - 2)$ monotonically decrease?

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