Find the values of

We have,

$y=[x(x-2)]^{2}=\left[x^{2}-2 x\right]^{2}$

$\therefore \frac{d y}{d x}=y^{\prime}=2\left(x^{2}-2 x\right)(2 x-2)=4 x(x-2)(x-1)$

$\therefore \frac{d y}{d x}=0 \Rightarrow x=0, x=2, x=1$

The points $x=0, x=1$, and $x=2$ divide the real line into four disjoint intervals i.e., $(-\infty, 0),(0,1)(1,2)$, and $(2, \infty)$.

In intervals $(-\infty, 0)$ and $(1,2), \frac{d y}{d x}<0$.

$\therefore y$ is strictly decreasing in intervals $(-\infty, 0)$ and $(1,2)$.

However, in intervals $(0,1)$ and $(2, \infty), \frac{d y}{d x}>0$.

$\therefore y$ is strictly increasing in intervals $(0,1)$ and $(2, \infty)$.

$\therefore y$ is strictly increasing for $02$.