Minima of $f(x)=\frac{x^2-1}{x^2+1}$ > If $f(x)=\dfrac{x^2-1}{x^2+1}$ for every real $x$ then find the minimum value of $f$ $$ f'(x)=\frac{4x}{(x^2+1)^2}=0\implies x=0\\\ f'(-0.5)<0\quad\&\quad f'(0.5)>0 $$ Seems to me like $x=0$ is a point of inflexion. But, $$ f''(x)=\frac{4(1-3x^2)}{(x^2+1)^3}\\\ f''(0)>0\implies x=0\text{ is a minima} $$ Am I making some stupid mistake here, since it doesnt make sense to me ?

Your calculations are right. You are just confusing the ideas:

$1)$ $f'(0)=0$ then $x=0$ is a critical point.

$2)$ $f'(-0,5)<0$ and $f'(0.5)>0$ implies that $f$ is decreasing at $-0.5$ and increasing at $0.5$. It reinforce the idea that $x=0$ is a minimum. You confirm that when you see that $f''(0)>0$.

A inflection point occours when the concavity changes. It means the sign of $f''$ changes, not the sign of $f'$.