Question

Consider the function $f\left(x\right)=\frac{x^2-1}{x^2+1}$, where $xϵR$.

At what value of $x$ does $f\left(x\right)$ attain minimum value?

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

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

For finding the value of $x$ for which $f\left(x\right)$ attains minimum value

$\frac{df\left(x\right)}{dx}=0$

$\Rightarrow \frac{(x^2+1)\left(2x\right)-(x^2-1)\left(2x\right)}{{(x^2+1)}^2}=0\Rightarrow x=0$