# Inverse of a sum of functions

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 $$(f(g(x)))^{-1} = g^{-1}(f^{-1}(x))$$

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

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 $$(f+g)^{-1}(x)=\frac{x-(b+d)}{a+c}.$$

• Is there a case where there is a formula? Suppose that both f and g are orientation preserving homeomorphisms? – IUissopretty Mar 22 at 18:16
• 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. – st.math Mar 22 at 18:24

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.