Find the values of $x$ for which $y = {\left[ {x\left( {x - 2} \right)} \right]^2}$ is an increasing function.

We have, $y = {\left( {x\left( {x - 2} \right)} \right)^2}$, $x \in R$
$\Rightarrow$ $y = {\left( {{x^2} - 2x} \right)^2}$ …(i)

Differentiating (i) w.r.t. $x$, we get

$\cfrac{{dy}}{{dx}} = 2\left( {{x^2} - 2x} \right)\cfrac{d}{{dx}}\left( {{x^2} - 2x} \right) = 2\left( {{x^2} - 2x} \right)\left( {2x - 2} \right)$

$= 4x\left( {x - 2} \right)\left( {x - 1} \right) = 4x\left( {x - 1} \right)\left( {x - 2} \right)$

For critical points, put $\cfrac{{dy}}{{dx}} = 0$, we get
$\Rightarrow x\left( {x - 1} \right)\left( {x - 2} \right) = 0 \Rightarrow x = 0,1,2$

The above points divide the real line into following disjoint intervals
$\left( { - \infty ,0} \right),\left( {0,1} \right),\left( {1,2} \right)\,\,{\rm{and}}\,\left( {2,\infty } \right)$

Interval
Sign of
$\cfrac{{dy}}{{dx}} = 4x\left( {x - 1} \right)\left( {x - 2} \right)$

Nature of function
$f$
$\left( { - \infty ,0} \right)$
$\left( { - ve} \right)\left( { - ve} \right)\left( { - ve} \right) = - ve$

Decreasing
$\left( {0,1} \right)$
$\left( { + ve} \right)\left( { - ve} \right)\left( { - ve} \right) = + ve$

Increasing
$\left( {1,2} \right)$
$\left( { + ve} \right)\left( { + ve} \right)\left( { - ve} \right) = - ve$

Decreasing
$\left( {2,\infty } \right)$
$\left( { + ve} \right)\left( { + ve} \right)\left( { + ve} \right) = + ve$

Increasing

Therefore we can say that $y$ is an increasing function in $\left( {0,1} \right) \cup \left( {2,\infty } \right)$