Question

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

Answer 1

$y=\left[x\right(x-2){]}^2=[x^2-2x{]}^2$
$\therefore \frac{dy}{dx}=y^{\prime }=2(x^2-2x)(2x-2)=4x(x-2)(x-1)$
$\therefore \frac{dy}{dx}=0\Rightarrow x=0,x=2,x=1$
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 $(1,2)$ , $\frac{dy}{dx}<0$
$\therefore$ $y$ is strictly decreasing in intervals $(-\infty ,0)$ and $(1,2)$
However, in intervals $(0,1)$ and $(2,\infty ),\frac{dy}{dx}>0$.
$\therefore$ $y$ is strictly increasing in intervals $(0,1)$ and $(2,\infty )$.
$\therefore$ $y$ is strictly increasing for $0<x<1$ and $x>2.$