Answer
See proof
Work Step by Step
We are given the functions:
$f(x)=ax+b$
$g(x)=\dfrac{1}{a}(x-b)$
Determine $f\circ g$:
$(f\circ g)(x)=f(g(x))=f\left(\dfrac{1}{a}(x-b)\right)=a\cdot \dfrac{1}{a}(x-b)+b=x-b+b=x$
Determine $g\circ f$:
$(g\circ f)(x)=g(f(x))=g\left(ax+b\right)=\dfrac{1}{a}(ax+b-b))=\dfrac{1}{a}\cdot ax=x$
We got:
$(f\circ g)(x)=(g\circ f)(x)=x$