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

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

$f\left(x\right)\text{ will be minimum if }2/\left(x^2+1\right)\text{ is maximum i.e. if }$$x^2+1\text{ is least i.e. when }\mathrm{x}=0\text{. Thus minimum value of }$$f\left(x\right)isf\left(0\right)=-1$