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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.

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Regarding your last sentence, why not apply Newton's method to $f(x) + g(x) - y = 0$ instead of to that complicated expression? – Rahul Aug 26 '11 at 10:24
up vote 4 down vote accepted

No, there is no easy way to go from $f^{-1}$ and $g^{-1}$ to $(f + g)^{-1}$.

For example, take $f(x) = x$ and $g(x) = \log x$, both of which have elementary inverses. However, $h(x) = f(x) + g(x) = x + \log x$ has the inverse $h^{-1}(x) = W(e^x)$, where $W$ is the Lambert W function which cannot be expressed in terms of elementary functions.

Another example: $f(x) = x$ and $g(x) = x^5$. The inverse of $h(x) = x + x^5$ can only be expressed in terms of the Bring radical.

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