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If $h(x) = f(x) + g(x)$, what is $h^{-1}(x)$ in terms of $f^{-1}(x)$ and $g^{-1}(x)$ ?

Also, what are other useful inverse identities that you can give me? I know the basics like $(f(g(x)))^{-1} = g^{-1}(f^{-1}(x))$

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Assuming $f$, $g$ and $h$ all have inverses, $h^{-1}(y) = f^{-1}(t)$ where $f^{-1}(t) = g^{-1}(y-t)$.

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There is no general answer to this. Let $f,g:\mathbb{R}\rightarrow\mathbb{R}$ with $f(x)=x$, $g(x)=-x$. Those functions have inverse functions, but $h:\mathbb{R}\rightarrow\mathbb{R}$ with $h(x)=f(x)+g(x)=0$, is not bijective.

When looking at special cases like linear functions $f,g:\mathbb{R}\rightarrow\mathbb{R}$, with $f+g\not\equiv 0$, we can find a formula: Let $f(x)=ax+b$ and $g(x)=cx+d$ with $a,c\neq 0$. Then, $(f+g)(x)=(a+c)x+(b+d)$ which has an inverse function $$(f+g)^{-1}(x)=\frac{x-(b+d)}{a+c}.$$

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  • $\begingroup$ Is there a case where there is a formula? Suppose that both f and g are orientation preserving homeomorphisms? $\endgroup$ – IUissopretty Mar 22 at 18:16
  • $\begingroup$ No. In the field $\mathbb{F}_2$, consider the function $f$ with $f(0)=0$, $f(1)=1$. Then, this function is a homeomorphism, but $g(x):=f(x)+f(x)=0$ for all $x\in\mathbb{F}_2$, so this is also not bijective. $\endgroup$ – st.math Mar 22 at 18:24
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There is no such relation.

Consider $f(x)=x,g(x)=e^x$. If you try to invert $h(x)$, the equation

$$y=x+e^x$$ has no closed-form solution, so nothing in terms of $f^{-1}(y)=y$ and $g^{-1}(y)=\ln y$.

This is not an isolated case, it is the rule.

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