Find the values of x for which $y = \left[x\left(x-2\right)\right]^2$ is an increasing function.
$y = \left[x\left(x-2\right)\right]^2$
$\therefore$ $\dfrac{dy}{dx}= 2 x \left(x-2\right) \left(x+x-2\right)$
i.e. $\dfrac{dy}{dx}= 4x \left(x-2\right) \left(x-1\right)$
$\dfrac{dy}{dx}=0$ $\implies$ $x=0$, $x=2$, $x=1$
These points divide the real line into four disjoint intervals namely $\left(-\infty,0\right)$, $\left(0,1\right)$, $\left(1,2\right)$ and $\left(2,\infty\right)$
| Interval | Sign of $\dfrac{dy}{dx}$ | Nature of function y |
|---|---|---|
| $\left(-\infty,0\right)$ | -ve | Decreasing |
| $\left(0,1\right)$ | +ve | Increasing |
| $\left(1,2\right)$ | -ve | Decreasing |
| $\left(2,\infty\right)$ | +ve | Increasing |
$\therefore$ y is increasing in the intervals $\left(0,1\right)$ and $\left(2,\infty\right)$.