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Let g(x)=1+x-[x] and f(x)=={(-1, x lt0),...

Let g(x)=1+x-[x] and `f(x)=={(-1, x lt0),(0, x=0),(1, x gt 0):}` Then for all x find f(g(x))

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g(x)=1+x-[x]
` implies g(x)=1+{x} [as x-[x]={x}]
i.e., g(x) is greater than 1. So, f(g(x))=1. Since, f(x)=1 for all `x gt 0`. Thus, f(g(x))=1, for all `x in R`
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