I have recently asked a question related to an inverse function which was not so obvious to calculate:
Inverse function of $y=W(e^{ax+b})-W(e^{cx+d})+zx$
Now I would like to learn;
Given $$f(x)=f_1(x)+f_2(x)$$
What are the conditions imposed on $f_1$ and $f_2$ such that
$${f}^{-1}(x)={f_1}^{-1}(x)+{f_2}^{-1}(x)$$
and when is such an approximation is not good at all?
Thanks alot in advance-
Answer
Well that would mean $x = f_1^{-1}\left(f_1(x)+f_2(x)\right) + f_2^{-1}\left(f_1(x)+f_2(x)\right)$ so it'd really be strange.
Assuming there are solution, the set of all solutions $S$:
$$S = \left\{ ( f_1, f_2 ) \in \left( \mathbb{R}^{\mathbb{R}} \right)^2, f_1 : \mathbb{R} \leftrightarrow \mathbb{R}, f_2 : \mathbb{R} \leftrightarrow \mathbb{R}, {f}^{-1}(x)={f_1}^{-1}(x)+{f_2}^{-1}(x) \right\}$$
The neutral element for $+$ would be $( x \mapsto 0, x \mapsto 0 )$ but $x \mapsto 0$ isn't bijective so it can't be in a pair of your set.
The neutral element for $\times$ would be $( x \mapsto 1, x \mapsto 1 )$ but $x \mapsto 1$ isn't bijective so it can't be in a pair of your set.
The neutral element for $\circ$ would be $( x \mapsto x, x \mapsto x )$ but $\frac{1}{2}x \not= x + x$ so it's not in the set either.
So the set would be unlikely to have any interesting properties and you probably wouldn't be able to determine if a function is in the set before calculating its inverse, making it completely useless...
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