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

is equal to –1

$f(x)=\frac{x^2-1}{x^2+1}, x \in R$

$\Rightarrow f'(x)=\frac{\left(x^2+1\right) 2 x-\left(x^2-1\right) 2 x}{\left(x^2+1\right)^2}=\frac{4 x}{\left(x^2+1\right)^2}$

∴  $f'(x)=0 \Rightarrow x=0$

$f''(x)=\frac{\left(x^2+1\right)^2 . 4-4 x . 2\left(x^2+1\right) . 2 x}{\left(x^2+1\right)^4}$

∴  $f''(0)=\frac{4-0}{(0+1)^4}=4>0$

∴ x = 0 is a point of minima and minimum value is = $\frac{0-1}{0+1}$ = -1