Let f : R → R and g : R → R be defined by


Let $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined by $f(x)=x+1$ and $g(x)=x-1 .$ Show that fog $=g \circ f=I_{R}$.


Given, $f: R \rightarrow R$ and $g: R \rightarrow R$

$\Rightarrow$ fog $: R \rightarrow R$ and gof $: R \rightarrow R$ (Also, we know that $I_{R}: R \rightarrow R$ )

So, the domains of all fog, gof and IR are the same.

$(f o g)(x)=f(g(x))=f(x-1)=x-1+1=x=I_{R}(x) \quad \ldots(1)$

$(g o f)(x)=g(f(x))=g(x+1)=x+1-1=x=I_{R}(x) \quad \ldots(2)$

From $(1)$ and $(2)$,

$(f o g)(x)=(g o f)(x)=I_{R}(x), \forall x \in R$

Hence, $f o g=g o f=I_{R}$

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