What is the inverse of $f(x) = \dfrac{x}{x^2 - 1}$ I need to find a continuous inverse of $$f(x) = \dfrac{x}{x^2 - 1}$$
Let $y = f(x) = \dfrac{x}{x^2 - 1} \Rightarrow y(x^2-1) = x \Rightarrow yx^2-x = y$
How should I proceed from here?
 A: A plot of the function looks like this 
To be invertible, a function of x and y must pass the horizontal line test. That is, any horizontal line cannot pass through more than one point of the graph. 
This function fails the horizontal line test. But if you break the graph up into the blue and red branches shown (where we consider the two red branches to really be just one branch) each branch does pass the horizontal line test. 
We haven't done any math yet, but we should expect that, when we solve for x in terms of y, we will get two answers. So let's do the math
\begin{align}
   y &= \dfrac{x}{x^2 - 1}\\
  y x^2 - y &= x\\
  y x^2 - x - y &= 0\\
  x &= \dfrac{1 \pm \sqrt{1 + 4y^2}}{2y}
\end{align}
A plot of $ \color{red}{x = \dfrac{1 + \sqrt{1 + 4y^2}}{2y}}$ and
$ \color{blue}{x = \dfrac{1 - \sqrt{1 + 4y^2}}{2y}}$ is shown below.

and the correspondences with the brances of the original function should be obvious.
There are a few things that still bother me. 
First note that $ \color{blue}{x = \dfrac{1 - \sqrt{1 + 4y^2}}{2y}}$ passes through the origin even though it is undefined when $y = 0$. We also see that x is bound between plus and minus one but it is hard to see that by looking at the equation. For these reasons, I think we should write this particular equation in a different, more useful, form.
\begin{align}
  x 
  &= \dfrac{1 - \sqrt{1 + 4y^2}}{2y}\\
  &= \dfrac{1 - \sqrt{1 + 4y^2}}{2y} 
     \cdot
     \dfrac{1 + \sqrt{1 + 4y^2}}{1 + \sqrt{1 + 4y^2}}\\
  &= \dfrac{1 - (1 + 4y^2)}{2y(1 + \sqrt{1 + 4y^2})}\\
  &= \dfrac{-4y^2}{2y(1 + \sqrt{1 + 4y^2})}\\
  x &= \dfrac{-2y}{1 + \sqrt{1 + 4y^2}}
\end{align}
Now it's fairly clear that $x = 0$ when $y = 0$ and, as $y$ approaches minus and plus infinity, $x$ approaches plus and minus one.
A: Let $f(x) = y = \frac{x}{x^{2} - 1}.$ To find the inverse of $f(x),$ we switch $x$ and $y$ in the equation and solve for $y.$ This yields $x = \frac{y}{y^{2} - 1},$ which we can rearrange to get $xy^{2} - y - x = 0.$ This is a quadratic with respect to $y.$ Using the quadratic equation, we get that $y = \boxed{f(x) = \frac{1 \pm \sqrt{1 + 4x^{2}}}{2x}}.$
