Find the inverse function of $f(x) = \dfrac{x}{1-x^2}$

Question

The function $F : (-1, 1) \to \mathbb {R}$ is defined by $F(x) = \dfrac{x}{1-x^2}$. Example 5 page 106 at Munkers' Topology says that its inverse is $$G(y) = \dfrac{2y}{1+(1+4y^2)^{\frac12}}.$$ I checked this and the result is that $G(F(x))=x$ and $F(G(y))=y$.

However by the method of finding the inverse of a function (exchanging variables, solving for the other variable, re-exchanging variables), I found out that the the inverse function of $f(x) = \dfrac{x}{1-x^2}$ is $$G(y) = \dfrac{-1 \pm (1+4y^2)^{\frac12}}{2y}$$ which is consistent with http://www.wolframalpha.com/ but does not result in $G(F(x))=x$ and $F(G(y))=y$.

Why is so?! Thank you.

Answer 1

The value $\frac{-1-\sqrt{1+4y^2}}{2y}$ is not in $(-1,1)$. That's because $\sqrt{1+4y^2}>2|y|$. So you get two solutions, but only one of them is in $(-1,1)$.

Now you just need to realize that:

$$\frac{2y}{1+\sqrt{1+4y^2}}=\frac{-1+\sqrt{1+4y^2}}{2y}$$

You can multiply that out to see that these are equal.

Your form is the "rationalized" form of the answer.

Answer 2

Note that $$\frac{-1+\sqrt{1+4y^2}}{2y}=\frac{-1-(1+4y^2)}{2y(-1-\sqrt{1+4y^2})}=\frac{-2y}{-1-\sqrt{1+4y^2}}=\frac{2y}{1+\sqrt{1+4y^2}}$$ where we multiply numerator and denominator by $(-1-\sqrt{1+4y^2})$.

Answer 3

Exactly what does the problem in Munkres say? The function, as given, does not have an "inverse" since there are values of y that are given by more than one x value. I suspect that what this problem is asking for the inverse of the function defined by that formula with a limited range for x.

It looks to me like the given solution is for the branch of the function with x between -1 and 1.