Let $g(x)=1+x-[x]$ and $f(x)=\left\{\begin{array}{cc}-1 & x<0 \\ 0 & x=0 \\ 1 & x>1\end{array}\right.$. Then for all x, f {g(x)} is equal to :

x 1 f(x) g(x)

1

$g(x) =1+x-[x]$   ($[x]=x-\{x\}$ (fractional part function)) $⇒g(x)=1+[x]$ so, $0≤\{x\}≤1$ $⇒g(x)≥1$ so $f(g(x))=1$  as $g(x)≥1$ for input ≥ 1, $f = 1$