If $f\left(x\right)=\frac{x^2-1}{x^2+1},$ for every real number $x,$ then the minimum value of $f\left(x\right)$

We have,

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

Clearly, $f\left(x\right)$will be minimum when $\frac{2}{x^2+1}$ is maximum i.e.

when $x^2+1$ is minimum. Obviously, $x^2+1$ is minimum at $x=0.$

$\therefore$Minimum value of $f\left(x\right)$ is $f\left(0\right)=-1\text{. }$