What is the inverse function of $f(x)=x/(1-x^2)$

Question

Can you give me a hint for how the inverse function of $f\colon (-1,1)\to \mathbb{R}\colon f(x)=\frac{x}{1-x^2}$ looks?

I need to show a homeomorphism!

Answer 1

$\begin{align} f(x)&=\frac{x}{1-x^2} \\ f(x)(1-x^2)&=x \\ f(x)-f(x)x^2&=x \\ 0&=f(x)x^2+x-f(x) \\ x&=\frac{-1\pm\sqrt{1+4f(x)^2}}{2f(x)} \end{align}$

Note, as mentioned in the comments below, that this does not give an inverse function, as a function cannot map two values to one point. Rather, you need to choose the correct inverse and demonstrate that it is indeed an inverse over the interval you are considering.

Answer 2

You need to show that $f$ is continuous, injective, surjective, and has a continuous inverse. The first two are rather simple and I think you can show that. Showing that $f$ is surjective amounts to using the intermediate value theorem (hint, $f$ is monotonic increasing on $(-1,1)$ and unbounded). To show that $f$ has a continuous inverse, you just need to show that $f$ is an open map - that is $f$ maps open sets to open sets. You should first show that $f$ maps open intervals to open intervals - use the fact, again, that $f$ is monotonic increasing here.