Trouble finding the inverse of $f(x) = x + \frac{1}{x}$ . Let $ f: \Bbb R - \{0\} \rightarrow \Bbb R \;\text{ given by } f(x) = x + \frac{1}{x} . \text{Find} $

$f(f^{-1}(\Bbb R))$ , $\Bbb R = \text{real numbers}$. 

For this problem I know one needs to find the inverse in order to solve.
$ y = x+ \frac{1}{x} \Rightarrow y^2 = \left( x+\frac{1}{x} \right)^{2}$
$y^2 = x^2+2 + \frac{1}{x^2}  $
one should arrive at the conclusion. $$f(x)^{-1} = \frac{x \pm \sqrt{x^2-4} }{2}$$
Yet I am having trouble at arriving at that conclusion. Could anyone show me how to do it arithmetically. 

You might not need to compute an inverse in order to solve this, but it may help to think about it. I would say
$$\begin{align*}
f(x) &= x + \frac{1}{x}\\
&= \frac{x^2}{x} + \frac{1}{x}\\
&= \frac{x^2 + 1}{x}\\
\\
y &\equiv \frac{x^2 + 1}{x}\\
xy &= x^2 + 1\\
0 &= x^2 - xy + 1\\
\end{align*}$$
Then solve for $x$ as a function of $y$ using, for example, the quadratic formula.
