Answer
See the verification below.
Work Step by Step
The functions $f(x)$ and $g(x)$ are known as inverse functions. Thus, we simplify
$(f\circ g) (x)=(g\circ f) (x)$ or, $f[g(x)]=g[f(x)]$
as follows:
$(f\circ g) (x) =f[g(x)]=f [\dfrac{(x-b)}{a}]=a [\dfrac{(x-b)}{a}]+b =x-b+b=x$
and
$(g\circ f) (x) = g[f(x)]=g(ax+b)=\dfrac{1}{a}[(ax+b)-b]=\dfrac{1}{a}(ax)=x$
Hence, verified.
