<s>[INST] Question about inverse function $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ This was from iranian university entrance exam .Suppose $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ find $f^{-1}(x)+f^{-1}(\frac{1}{x}),x \neq 0$.
It is easy to find $f^{-1}$ and solve this like below ...
$$y=\frac{1}{2}(x+\sqrt{x^2+4}) \\(2y-x)^2=(\sqrt{x^2+4})^2\\4y^2+x^2-4xy=x^2+4\\4y^2-4=4xy\\x=\frac{y^2-1}{y}=y-\frac{1}{y} \\ \to f^{-1}(x)=x-\frac{1}{x}\\
f^{-1}(x)+f^{-1}(\frac{1}{x})=x-\frac{1}{x}+(\frac{1}{x}-\frac{1}{\frac{1}{x}})=0\\$$ And now my question is ...
  is there an other method to solve this question ?
I was thinking about $f(x)f(-x)=1$ but I can't go anymore ...
Any hint ,or other Idea ?
  Thanks in advanced. 
 [/INST] The short version:
$$f(x)f(-x)=1\implies f^{-1}(f(x))+f^{-1}\left(\frac1{f(x)}\right)=f^{-1}(f(x))+f^{-1}\left(f(-x)\right)=0.$$
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