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. 
 A: The short version:
$$f(x)f(-x)=1\implies f^{-1}(f(x))+f^{-1}\left(\frac1{f(x)}\right)=f^{-1}(f(x))+f^{-1}\left(f(-x)\right)=0.$$
A: Disclaimer: Just offering a geometrical perspective. OP's solution and Anurag's solutions are much more elegant.
If we are given the quadratic equation $$g(z)=az^2+bz+c,$$
we know that the bigger root is $$\frac{-b+\sqrt{b^2-4ac}}{2a}$$
View $x$ as a parameter, when we are given $x$,  we
let $b = -x$, $a = 1$, $c=-1$.
The function $f(x)$ returns the bigger root of  $$g(z)=z^2-xz-1$$
This quadratic equation has vertical intercept $-1$ and it is symmetric about $z=\frac{x}{2}$.
Suppose the bigger root is $y_b$ and the smaller root is $y_s$.
We know that  $y_by_s=-1$, the smaller root $y_s=\frac{-1}{y_b}$
Hence this quadratic equation passes through $(0,-1), (y_b,0)$ and $(-\frac{1}{y_b},0)$
Now let's try to understand what does $f^{-1}(y)$ do geometricallly. $f^{-1}(y)$ construct the quadratic equation that passes through $(0,-1), (y,0)$ and $(-\frac1y, 0)$, (let me call this quadratic curve $Q_1$)look for the symmetrical line for the quadratic curve $z=\frac{x}{2}$ and returns $x$.
$$x=f^{-1}(y)$$ 
(actually we know that $x=y-\frac{1}{y}$)
Now let's try to understand what does $f^{-1}(\frac1y)$ do geometrically.$f^{-1}(\frac1y)$ construct the quadratic equation that passes through $(0,-1),(\frac1y,0)$ and $(-y,0)$, (let me call this quadratic curve $Q_2$), look for the symmetrical line for the quadratic curve $z = \frac{\hat{x}}{2}$ and returns $\hat{x}$ 
$$\hat{x}=f^{-1}(\frac1y)$$
(again, actually we know that $\hat{x}=\frac1y-y$)
$Q_2$ is actually just the reflection of $Q_1$ with reflection axis $z=0$.
Hence $$\hat{x}=-x$$
$$x+\hat{x}=0$$
$$f^{-1}(y)+f^{-1}(\frac1y)=0$$
A: $f(2\sinh\theta) = \frac{1}{2}\left(2\sinh\theta+2\cosh\theta\right) = e^\theta $ implies that the range of $f$ is $\mathbb{R}^+$ and
$$ f(x) = \exp\text{arcsinh }\frac{x}{2} \to \\
\qquad f^{-1}(x)=2\sinh\ln x=x-\frac{1}{x}\tag{1}$$
so $ f^{-1}(x)+f^{-1}\left(\frac{1}{x}\right) = \color{red}{0}$ for any $x\in\mathbb{R}^+$.
A: Perhaps your observation can be made to work (note $f(x) \neq 0$):
\begin{align*}
f(x)f(-x) &  =1\\
f(-x) & = \frac{1}{f(x)}\\
f^{-1}(f(-x)) & = f^{-1}\left(\frac{1}{f(x)}\right)\\
-x & = f^{-1}\left(\frac{1}{f(x)}\right)\\
f^{-1}\left(\frac{1}{f(x)}\right)+x & = 0\\
f^{-1}\left(\frac{1}{f(x)}\right)+f^{-1}(f(x)) & = 0\\
f^{-1}\left(\frac{1}{y}\right)+f^{-1}(y) & = 0
\end{align*}
