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. 
 A: 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 $f(f^{-1}(\mathbb{R}))=(-\infty,-2]\cup [2,\infty)$.
A: 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.
A: Write it as $x^2-xy+1=0$ and use quadratic formula for $x$ you have relation for x in terms of $y$
A: Don't need inverse and looking for inverse will lead you astray.
As for every $x \ne 0; f(x)$ exists and $f(x) = c \in f(\mathbb R) \subset \mathbb R$ for some $c \in f(\mathbb R)$. So $x \in f^{-1}(\mathbb R)$.  If $x = 0$, $f(x)$ is undefined so $x \not \in f^{-1}(\mathbb R)$.
So $f^{-1}(\mathbb R) = \mathbb R - \{0\}$.
Basically for any $f$, $f^{-1}(X)$ is an, albeit confusing, way to state the preimage part of the domain that gets mapped to X.  As $X = \mathbb R$, the pre-image is the entire domain.
So all this question is asking is to find:
$f(\mathbb R -\{0\})$.  In other words to find the codomain/image of the function.  A bit of fussing and you can figure it is $(-\infty, 2] \cup [2, \infty)$.
====
Bit of fussing: For $x > 0$. If $x = 1$, $x + \frac 1x = 2$.  If $x \ne 1$ then either $x > 1; \frac 1x < 1$ or $x< 1; \frac 1x > 1$.  Wolog assume $x = 1 + \delta > 1$ then $1 - \delta^2 < 1 \implies (1 - \delta)(1 + \delta) < 1 \implies 1- \delta < \frac 1{1+\delta} \implies x + \frac 1x = 1 + \delta + \frac 1{1+\delta} > 1 + \delta + 1 - \delta = 2$.  
So $f(\mathbb R^+) \subset [2,\infty)$. 
As $f$ is continuous on $\mathbb R^+$ and for all $y > 2; f(y) > y$ it follows $\exists c;2 < c < y; f(c) = y$ and $f(\mathbb R^+) = [2,\infty)$.
As $f(-x) = -x - 1/x = -(x+1/x) = -f(x)$, $f(\mathbb R - \{0\}) = (-\infty, 2] \cup[2,\infty)$
