Question

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

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

Answer 1

The correct option is C $x=0$

Let $f\left(x\right)=x^2+\frac{1}{1+x^2}$
On differentiating w.r.t. $x$, we get $f^{\prime }\left(x\right)=2x-\frac{1}{(1+x^2{)}^2}\cdot 2x$
For a minimum, put $f^{\prime }\left(x\right)=0$
$\Rightarrow 2x\left(\frac{(1+x^2{)}^2-1}{(1+x^2{)}^2}\right)=0$
$\Rightarrow x=0$

So, the function has the minimum value at $x=0$