I am asked to prove whether two functions are inverses of each other using composition.
The functions given are $$f(x)=\dfrac 1{(x-3)}$$ $$ g(x)=\dfrac {(3x+1)}x$$
I've solved the first function (f(x)), and gotten x as the answer (meaning, it is the inverse). However, I'm struggling on solving for the second inverse function. I've shown my work below, of which I'm fairly sure is error-free, but what I'm stuck on is attempting to solve to have 'x' as the answer to g(f(x))=4/(x²+3x). I know that it is the inverse (or, should be), since I checked last night and I believe I found it was. However, I couldn't find the steps to do so anywhere.
Work:
g(x)=(3x+1)/x
g(f(x))=(3(1/(x-3)+1)/x
g(f(x))=(3·(1/(x-3))+1)/x -- a(bc)=a·bc
g(f(x))=(((3·1)/(x-3))+1)/x -- a·(b/c)=(a·b)/c
g(f(x))=(((3·1)/(x-3))+(1(x-3))/(x-3))/x -- a=(a(b))/b
g(f(x))=((3+1)/(x-3))/x
g(f(x))=(((3+1)/(x-3))/x)/1 -- a/b/c=a/b/c/1
g(f(x))=((3+1)/(x-3))*(1/x) -- a/b/c/d=(a/b)*(d/c)
g(f(x))=((3(1/x)+(1/x))/-3(1/x)
g(f(x))=((4(1/x))/-3(1/x)
Aaaand I'm stuck there.