The function f(x) = [x(x-2)]^2 is increasing | vedclass

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(D) Given the function $f(x) = [x(x-2)]^2 = (x^2 - 2x)^2$. To find the intervals where the function is increasing,we find the derivative $f'(x)$. $f'(

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$(0, 1) \cup (2, \infty)$

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(D) Given the function $f(x) = [x(x-2)]^2 = (x^2 - 2x)^2$. To find the intervals where the function is increasing,we find the derivative $f'(x)$. $f'(x) = 2(x^2 - 2x)(2x - 2) = 4x(x-2)(x-1)$. For the function to be increasing,we set $f'(x) > 0$. $4x(x-1)(x-2) > 0$. Using the wavy curve method (sign scheme) for the critical points $x = 0, 1, 2$: - For $x \in (0, 1)$,$f'(x) - For $x \in (1, 2)$,$f'(x) > 0$. - For $x \in (2, \infty)$,$f'(x) > 0$. Wait,let's re-evaluate the sign scheme: If $x > 2$,all factors $(x), (x-1), (x-2)$ are positive,so $f'(x) > 0$. If $1 If $0  0$. If $x Thus,$f'(x) > 0$ in $(0, 1) \cup (2, \infty)$. Therefore,the function is increasing in $(0, 1) \cup (2, \infty)$.

The function $f(x) = [x(x-2)]^2$ is increasing in the set

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