Functions

Let $g(x)=1+x-[x]$ and ${ }^{\prime}$

$$ f(x)= \begin{cases}-1, & x<0 \\ 0, & x=0,[x] \text { denotes the greatest integer less } \\ 1, & x>0\end{cases} $$

than or equal to $x$. Then for all $x, f(g(x))=$