# Let g(x)=1+x−[x] and f(x)

Question:

Let $g(x)=1+x-[x]$ and $f(x)=\left\{\begin{array}{ll}-1, & x<0_{\square} \\ 0, & x=0, \\ 1, & x>0\end{array}\right.$, where $[x]$ denotes the greatest integer less than or equal to $x .$ Then for all $\left.x, f(g)\right)$ is equal to

(a) $x$

(b) 1

(c) $f(x)$

(d) $g(x)$

Solution:

(b) 1

When, $-1 Then,$g(x)=1+x-[x]\quad=1+x-(-1)=2+x\therefore f(g(x))=1$When,$x=0$Then,$g(x)=1+x-[x]=1+x-0=1+x\therefore f(g(x))=1$When,$x>1$Then,$g(x)=1+x-[x]=1+x-1=x\therefore f(g(x))=1+x-1=x\$

Therefore, for each interval f(g(x))=1