Question

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

What is the minimum value of $f\left(x\right)$?

• A. $0$
• B. $1/2$
• C. $-1$
• D. $2$

Answer 1

The correct option is C $-1$

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

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

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

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

$\therefore f\left(0\right)=-1$ is the minimum value of $f$.