Answer
See explanations.
$x\ne1/2$, $x\ne-3/2$
Work Step by Step
Step 1. Given $f(x)=\frac{x-5}{2x+3}$ and $g(x)=\frac{3x+5}{1-2x}$, we have $f(g(x))=\frac{(\frac{3x+5}{1-2x})-5}{2(\frac{3x+5}{1-2x})+3}=\frac{3x+5-5+10x}{6x+10+3-6x}=x$ with $x\ne1/2$
Step 2. We have $g(f(x))=\frac{3(\frac{x-5}{2x+3})+5}{1-2(\frac{x-5}{2x+3})}=\frac{3x-15+10x+15}{2x+3-2x+10}=x$ with $x\ne-3/2$
Step 3. Thus $f$ and $g$ are inverse functions.
Step 4. For $f(x)$, we have $x\ne-3/2$ and for $g(x)$, we have $x\ne1/2$ which are also part of the composite requirements.
