Does the inverse of this function exist? $r(x)=\frac{2x}{1+x^2}$
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?
 A: $y = \frac {2x}{1+x^2}$
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 $[-1,1] \to [-1,1]$
A: 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 $x\in[1,\infty)$.
A: 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) $r^{-1}(x)= \frac{1 - \sqrt{1-y^2}}{y}.$
