Question

The function $f\left(x\right)=\frac{x}{2}+\frac{2}{x}$ has a local minimum at

• A. $x=-2$
• B. $x=0$
• C. $x=1$
• D. $x=2$

Answer 1

The correct option is D

$x=2$

Explanation for the correct option

Step 1: Solve for the critical points

Given, $f\left(x\right)=\frac{x}{2}+\frac{2}{x}$

A function has local minima at $x$ if $f\text{'}\left(x\right)=0$ and $f\text{'}\text{'}\left(x\right)>0$

We have,
$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{d}{dx}\left(\frac{x}{2}+\frac{2}{x}\right)\\ & =& \frac{1}{2}-\frac{2}{x^2}\end{array}$

So,
$\begin{array}{cc}& f\text{'}\left(x\right)=0\\ \Rightarrow & \frac{1}{2}-\frac{2}{x^2}=0\\ \Rightarrow & \frac{1}{2}=\frac{2}{x^2}\\ \Rightarrow & x^2=4\\ \Rightarrow & x=\pm 2\end{array}$

Thus, the critical points are $x=\pm 2$

Step 2: Solve for the local minima
$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& \frac{d}{dx}f\text{'}\left(x\right)\\ & =& \frac{d}{dx}\left(\frac{1}{2}-\frac{2}{x^2}\right)\\ & =& 0-\left[\frac{2\left(-2\right)}{x^3}\right]\\ & =& \frac{4}{x^3}\end{array}$

So, at $x=-2$,
$\begin{array}{rcl}f\text{'}\text{'}\left(-2\right)& =& \frac{4}{{\left(-2\right)}^3}\\ & =& -\frac{4}{8}<0\end{array}$

Thus, there is a local maxima at $f\left(-2\right)$

At $x=2$,
$\begin{array}{rcl}f\text{'}\text{'}\left(2\right)& =& \frac{4}{2^3}\\ & =& \frac{4}{8}>0\end{array}$

Thus, there is a local minimum at $x=2$

Hence, option(D) is correct.