so first lets say they are bijective so they can have inverse (you have to prove this in some exercices but here they ask you about inverse you can assume it s bijective )
1/x-3=y
1=(x-3)y
1/y=x-3
x=1-y/y
x=1/y-1 no because f (x) inverse=1-x/x it s not g(x)
let s see with g(x)
y=3x+1/x
xy=3x+1
3x-xy=-1
x(3-y)=-1
x=-1/3-y x=1/y-3 g(x) inverse=1/x-3 so f(x) is the inverse of g (x)
the domain for g:R\{0}->R because the denumitor can t be 0 x doesn t have to be 0
the domain of f(x) which is the same with g(x) inverse because we just prove that they are the same is f:R|{3}->R because x-3 is 0 for x=3.
