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. 

We do not need to find a formula for the "inverse function" to answer the original question. Indeed since $f(x)$ is not one to one, the inverse function does not exist.
By symmetry, we can begin by confining attention to positive $x$. Note that $$\left(x+\frac{1}{x}\right)^2=\left(x-\frac{1}{x}\right)^2+4.$$
So (for positive $x$) our function is very large near $0$, then decreases steadily until $x=1$, where it reaches a minimum of $2$, then increases steadily, taking on arbitrarily large values. 
It follows that as $x$ ranges over the positive reals, $f(x)$ ranges over all reals $\ge 2$.
Similarly, as $x$ ranges over the negative reals, $f(x)$ ranges over all reals $\le -2$. 
It follows that $f(f^{-1}(\mathbb{R}))=(-\infty,-2]\cup [2,\infty)$.
