Let $f\left(x\right)=\left\{\begin{array}{cc}\left|x\right|,& 0<\left|x\right|\le 2\\ 1,& x=0\end{array}\right.$. Then, at $x=0, f$ has

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

If $h\left(x\right)=f\left(x\right)+f\left(-x\right)$, then $h\left(x\right)$ has got an extreme value at a point, where $f^{\text{'}}\left(x\right)$ is