$f (x) = x^{2} (x - 2)^{2}$ is increasing in

(- \infty, 0)

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(2, \infty)

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(1,2)

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Solution

The correct option is D $(2, \infty)$
$f^{'} (x) = 2 x (x - 2)^{2} + 2 x^{2} (x - 2)$
$= 2 x (x - 2) (x - 2 + x)$
$= 4 x (x - 2) (x - 1)$
$f^{'} (x) > 0$ for $\forall x ϵ (0, 1) \cup (2, \infty)$

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