Answer
a) $\dfrac{amx+b}{cmx+d}$
b) $\dfrac{amx+bm}{cx+d}$
c) $D_{f\circ g}=\left(-\infty,-\dfrac{d}{cm}\right)\cup\left(-\dfrac{d}{cm},\infty\right)$
$D_{g\circ f}=\left(-\infty,-\dfrac{d}{c}\right)\cup\left(-\dfrac{d}{c},\infty\right)$
d) $m=1$
Work Step by Step
We are given the functions:
$f(x)=\dfrac{ax+b}{cx+d}$
$g(x)=mx$
a) Determine $f\circ g$:
$(f\circ g)(x)=f(g(x))=f(mx)=\dfrac{amx+b}{cmx+d}$
b) Determine $g\circ f$:
$(g\circ f)(x)=g(f(x))=g\left(\dfrac{ax+b}{cx+d}\right)=m\cdot \dfrac{ax+b}{cx+d}=\dfrac{amx+bm}{cx+d}$
c) We have:
$D_f=\left(-\infty,-\dfrac{d}{c}\right)\cup\left(-\dfrac{d}{c},\infty\right)$
$D_g=(-\infty,\infty)$
The domain of $f\circ g$ contains all the real values of $x$ except those for which $g(x)=-\dfrac{d}{c}$.
$mx=-\dfrac{d}{c}$
$x=-\dfrac{d}{cm}$
$D_{f\circ g}=\left(-\infty,-\dfrac{d}{cm}\right)\cup\left(-\dfrac{d}{cm},\infty\right)$
The domain of $g\circ f$ contains all the real values of $x$ except $-\dfrac{d}{c}$.
$D_{g\circ f}=\left(-\infty,-\dfrac{d}{c}\right)\cup\left(-\dfrac{d}{c},\infty\right)$
d) The domains of the two functions must be the same:
$-\dfrac{d}{cm}=-\dfrac{d}{c}\Rightarrow m=1$
In order for $f\circ g=g\circ f$, we must have:
$(f\circ g)(x)=(g\circ f)(x)$ for any real $x$ in the domain:
$\dfrac{amx+b}{cmx+d}=\dfrac{amx+bm}{cx+d}$
But as $m=1$, the equality is true for all $x$ in the domain.