Question

The function $f\left(x\right)={\left[x\left(x-2\right)\right]}^2$ is increasing in the set

• A. $\left(-\infty ,0\right)\cup \left(2,\infty \right)$
• B. $\left(-\infty ,1\right)$
• C. $\left(0,1\right)\cup \left(2,\infty \right)$
• D. $\left(1,2\right)$
• E. $\left(0,2\right)$

Answer 1

The correct option is C

$\left(0,1\right)\cup \left(2,\infty \right)$

Explanation for the correct option:

Step 1: Solve for the critical points

Given function is $f\left(x\right)={\left[x\left(x-2\right)\right]}^2$

We know that a function is increasing at $x$ if $f\text{'}\left(x\right)>0$

We have,
$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}{\left[x\left(x-2\right)\right]}^2\\ & =& 2\left[x\left(x-2\right)\right]\left[1·\left(x-2\right)+x\left(1-0\right)\right]\left[\because \frac{d}{\mathrm{dx}}\left[f\left(g\left(x\right)\right)\right]=f\text{'}\left(g\left(x\right)\right)·g\text{'}\left(x\right)\right]\\ & =& 2x\left(x-2\right)\left(2x-2\right)\\ & =& 4x\left(x-2\right)\left(x-1\right)\end{array}$

The critical points of $f\left(x\right)$ are obtained by setting $f\text{'}\left(x\right)=0$
$$\begin{gathered} \Rightarrow 4x\left(x-2\right)\left(x-1\right)=0 \\ \begin{array}{cccc}& & x-2=0& x-1=0\\ \Rightarrow & x=0& x=2& x=1\end{array} \end{gathered}$$

So $f\left(x\right)$ has critical points at $x=0,1,2$

Step 2: Solve for the required interval

Pick a value of $x$ such that $-\infty <x<0$. Let $x=-1$
$\begin{array}{rcl}f\text{'}\left(-1\right)& =& 4\left(-1\right)\left(-1-2\right)\left(-1-1\right)\\ & =& 4\left(-1\right)\left(-3\right)\left(-2\right)\\ & =& -24<0\end{array}$
Thus, $f\left(x\right)$ is decreasing in the interval $\left(-\infty ,0\right)$

Pick a value of $x$ such that $0<x<1$,. Let $x=\frac{1}{2}$
$\begin{array}{rcl}f\text{'}\left(\frac{1}{2}\right)& =& 4\left(\frac{1}{2}\right)\left(\frac{1}{2}-2\right)\left(\frac{1}{2}-1\right)\\ & =& 2\left(-\frac{3}{2}\right)\left(-\frac{1}{2}\right)\\ & =& \frac{3}{2}>0\end{array}$
Thus, $f\left(x\right)$ is increasing in the interval $\left(0,1\right)$

Pick a value of $x$ such that $1<x<2$. Let $x=\frac{3}{2}$
$\begin{array}{rcl}f\text{'}\left(\frac{3}{2}\right)& =& 4\left(\frac{3}{2}\right)\left(\frac{3}{2}-2\right)\left(\frac{3}{2}-1\right)\\ & =& 2\left(3\right)\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right)\\ & =& -\frac{3}{2}<0\end{array}$

Thus, $f\left(x\right)$ is decreasing in the interval $\left(1,2\right)$

$f\left(x\right)$ is increasing in the interval $\left(2,\infty \right)$ because we observe that it is decreasing in the previous interval and there is a critical point at $x=2$.

Therefore, $f\left(x\right)$ is increasing in the interval $\left(0,1\right)\cup \left(2,\infty \right)$

Hence, option(C) is correct.