Question

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

Answer 1

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

$\Rightarrow y=[x^2-2x{]}^2$

$\Rightarrow y=x^4+(2x{)}^2-2(x^2\left)\right(2x)$

$\Rightarrow y=x^4+4x^2-4x^3$

Differentiating w.r.t $x,$

$\frac{dy}{dx}=4x^3+8x-12x^2$

$\Rightarrow \frac{dy}{dx}=4x(x^2-3x+2)$

$\Rightarrow \frac{dy}{dx}=4x(x-2)(x-1)$

$\frac{dy}{dx}=0$

$\Rightarrow 4x(x-1)(x-2)=0$

$\Rightarrow x=0,1,2$

Plotting points on real number line,

Intervals	Sign of $\frac{dy}{dx}=4x(x-1)(x-2)$	Natural of function $y$
$(-\infty ,0)$	$(-)(-)(-)<0$	Strictly decreasing
$(0,1)$	$(+)(-)(-)>0$	Strictly increasing
$(1,2)$	$(+)(+)(-)<0$	Strictly decreasing
$(2,\infty )$	$(+)(+)(+)>0$	Strictly increasing

Thus, the function is strictly increasing for $0<x<1\text{and}x>2$.