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?

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 $y = \boxed{f(x) = \frac{1 \pm \sqrt{1 + 4x^{2}}}{2x}}.$
