For the function $f(x)=[x+1]-[x]+\frac{1}{x^2-1}$, $f(x)$ has:

As derived previously, $f^{\prime }(x)=\frac{-2x}{(x^2-1{)}^2}$.Critical points occur where $f^{\prime }(x)=0⟹x=0$.$f^{\prime }(x)$ changes sign from positive to negative at $x=0$:For $x<0$, $f^{\prime }(x)>0$.For $x>0$, $f^{\prime }(x)<0$.By the first derivative test, $x=0$ is a point of local maximum.There are no other critical points where $f^{\prime }(x)=0$, and $f(x)$ is not continuous at $\pm 1$.Thus, it has a local maxima but no local minima.