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Assuming two functions are invertible, is it true that the inverse of the sum of the two functions is the sum of the inverses (assuming all functions are well behaved)?

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Let $f(x) = x, g(x) = -x$, both obviously invertible. Then $(f+g)(x) == 0$, which is not invertible.

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As gnometorule pointed in general the sum of invertible is not necessarily invertible.. Anyhow, if two functions are strictly increasing over $\mathbb R$ then their sum is invertible.. Still there is no connection in general between the inverse of the sum and the inverses.

Consider the following simple example:

$$f(X)=X^{5}, g(x)=X \,.$$

Then, both functions are invertible, and so is $f+g$. Anyhow, using Galois Theory, it can be shown that $(f+g)^{-1}(-1)$ is not a number expressible via radicals, let alone in terms of $(-1)$ and $\sqrt[5]{(-1)}=-1$.

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  • $\begingroup$ Hi ! Though I have not taken galois theory,but still I want to know are you referring to differential galois theory? $\endgroup$ May 15 '19 at 21:16
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    $\begingroup$ @NewBornMATH No. Just standard algebraic Galois theory. The "simplest" equation which cannot be solved via radicals is $X^5+X+1=0$. $\endgroup$
    – N. S.
    May 16 '19 at 1:08
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No, this one is wrong. The best invertible function is $f(x)=x$. Then $f^{-1}=x$ and $f(x)+f(x)=2f(x)$, but $$(2f(x))^{-1}=\frac{x}{2}$$

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