<s>[INST] 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. 
 [/INST] 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.
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