The function f(x) = (x(x - 2))^2 is an increa | vedclass

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

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

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(C) Given the function $f(x) = (x(x - 2))^2 = (x^2 - 2x)^2$.To find the intervals where the function is increasing,we calculate the derivative $f'(x)$:$f'(x) = 2(x^2 - 2x)(2x - 2) = 2x(x - 2) \cdot 2(x - 1) = 4x(x - 1)(x - 2)$.For the function to be increasing,we require $f'(x) > 0$:$4x(x - 1)(x - 2) > 0$.We analyze the sign of $f'(x)$ using the intervals determined by the critical points $x = 0, 1, 2$:$1$. For $x \in (0, 1)$: $f'(x) = 4(+)(-)(-) > 0$.$2$. For $x \in (1, 2)$: $f'(x) = 4(+)(+)(-) $3$. For $x \in (2, \infty)$: $f'(x) = 4(+)(+)(+) > 0$.Thus,$f(x)$ is increasing on $(0, 1) \cup (2, \infty)$.

The function $f(x) = (x(x - 2))^2$ is an increasing function on the set.....

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