Does the inverse of this function exist?

Question

$r(x)=\frac{2x}{1+x^2}$

So I know that the range is $[-1,1]$, and the function is injective. It is surjective as well in the range $[-1,1]$.

I'm trying to show whether this function has an inverse. Up till now I should be able to show that the inverse exists since $r(x)$ is bijective.

However, after solving for the inverse I got $r^{-1}(x)=1\pm\sqrt{1-y^2}$, which is a circle, I got a bit confused whether this inverse of $r(x)$ exists or not. Surely I did something wrong midway? It'd be nice if someone can let me know. Thanks!

Edit: I think I just figured it out. The function is not surjective at all in the range $[-1,1]$. Correct me if I'm wrong, thanks!

Edit 2.0: Sorry, it should be not injective in the range $[-1,1]$, right?

Answer 1

The easiest way to check is to plot the function. If it has a turning point within the allowed domain then it will not have an inverse, since the inverse would need to be many to one. i.e. not a function.

From the plot, it is clear that there are two values for $y=\frac{1}{2}$. Solving for $x$, we see that these are $x=2-\sqrt{3}$ and $x=2+\sqrt{3}$

\begin{align} \frac{2x}{1+x^2}&=\frac{1}{2}\\ \frac{x}{1+x^2}&=\frac{1}{4}\\ x&=\frac{1}{4}+\frac{1}{4}x^2\\ 0&=x^2-4x+1\\ \therefore~x&=\frac{4\pm\sqrt{16-4}}{2}\\ &=\frac{4\pm2\sqrt{3}}{2}\\ &=2\pm\sqrt{3} \end{align}

Since the function has two $y$ values for at least one $x$, the function is not bijective and does not have an inverse.

Note: the turning points of the function are at $x=\pm1$, if the domain of the function is restricted to this interval, then it will have an inverse. The same is true if $x\in(-\infty,-1]$ and $x\in[1,\infty)$.

Answer 2

You have miscalculated the inverse. Using the quadratic formula to solve for $x$ in the equation $y(1+x^2)=2x$ yields:

$$x=\frac{1 \pm \sqrt{1-y^2}}{y}.$$

One of those roots is extraneous. The correct answer is (the continuous extension at $x=0$ of) $r^{-1}(x)= \frac{1 - \sqrt{1-y^2}}{y}.$

Answer 3

$y = \frac {2x}{1+x^2}$

If you can isolate $x$ you have your inverse.

$y(1+x^2) = 2x\\ yx^2 - 2x + y = 0$

Using the quadratic formula

$x = \frac {1 \pm \sqrt {1 - y^2}}{y}$

and

$x = f^{-1}(y) = \begin{cases} \frac {1 - \sqrt {1 - y^2}}{y}&y\ne0\\0&y=0\end{cases}$

maps from $[-1,1] \to [-1,1]$