I have two functions $f(x)$ and $g(x)$ with $D = [0..1]$ and both $f:D \to D$ and $g:D \to D$. Now I have a third function $h(x)$ with $h : D \to D$ defined as $h(x) = a*f(x) + b*g(x)$.
Is it somehow possible do build $h^{-1}(x)$ in this case given I already know $f^{-1}$ and $g^{-1}$.
I did not manage to go any further than $x = f^{-1}(\frac{y-b*g(x)}{a})$ but of course the $g(x)$ in there is my problem.
I would assume that it is not possible to solve this entirely but I am lacking proof for that assumption.
If I would like to approximate the result and would go for Newton's method this would probably be a nightmare as well shooting for $0 = f^{-1}(\frac{y-b*g(x)}{a}) - x$ so any suggestion is welcome.