Let `f:R→R and g:R→R defined by f(x)=x 2 and g(x)=(x+1)`. Show that `gof  =fog.`

Explanation

Given, `f(x)=x^{2}` and `g(x)=(x+1)`



`(g circ f)(x)=g{f(x)}=gleft(x^{2}right)=left(x^{2}+1right)`




`(f circ g)(x)=f{g(x)}=f(x+1)=(x+1)^{2}`



Hence, gof `equiv` fog.