Inverse functions problem

Question

There are two functions $f\colon\mathbb Q \to \mathbb Q \setminus \{-1\}$ and $g\colon\mathbb Q \to \mathbb Q \setminus \{1\}$.

$$g(x) = \frac{f(x)}{f(x)+1}.$$

Prove that if there is a inverse function for $f$, then $g$ is also inverse. And find $g$ inverse function by $f^{-1}$ (the inverse of $f$). Kind of lost, any ideas on how to solve this question?

Answer 1

Let $h(x)=\dfrac x{x+1}$, so

$$g(x)=h(f(x))$$

$f,h$ have inverse function, so is $g$.

en $g^{-1}(x)=f^{-1}(h^{-1}(x))$

Answer 2

Show that g is one to one whenever f is one to one and that g is onto whenever f is onto.