Question

The minimum value of $x^2+\frac{1}{1+x^2}$ is at

• A. $x=0$
• B. $x=1$
• C. $x=4$
• D. $x=3$

Answer 1

The correct option is A

$x=0$

Find the minimum value of the given function

Given : $f\left(x\right)=x^2+\frac{1}{1+x^2}$

Differentiate the function with respect to $x$,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& 2x-\frac{1}{{\left(1+x^2\right)}^2}\times 2x\left[\frac{d}{dx}{x}^n=n.x^{n-1}\right]\\ & =& \frac{2x{(1+x^2)}^2-2x}{{\left(1+x^2\right)}^2}\end{array}$

For minima or maxima put $f\text{'}\left(x\right)=0$,

$$\begin{gathered} \frac{2x{(1+x^2)}^2-2x}{{\left(1+x^2\right)}^2}=0 \\ \Rightarrow 2x\left[{\left(1+x^2\right)}^2-1\right]=0 \end{gathered}$$

Therefore, $x=0$ and ${\left(1+x^2\right)}^2-1=0$

${\left(1+x^2\right)}^2=1$

$\Rightarrow$$1+x^2=1$ or $1+x^2=-1$

$\Rightarrow$ $x^2=0$ or $x^2=-2$ [rejected]

Since the square of a number can not be a negative value, thus $x^2=-2$ is rejected.

$\Rightarrow$ $x=0$

Therefore we have only one value of $x$, then the function have minimum value at $x=0$.

Hence option $\left(A\right)$, $x=0$ is the correct answer.