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?

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    $\begingroup$ The function $r(x)$ is clearly surjective. $r(1)=1$ and $r(-1)=-1$. Since $r$ is a quotient of continuous functions and the denominator of the quotient is non-zero, $r$ is continuous. Thus, the Intermediate Value Theorem guarantees that $r$ takes on every value in $[-1, 1]$. $\endgroup$ – Robert Shore Aug 13 '20 at 2:20
  • $\begingroup$ If the function is bijective, it is inheritable. That doesn't mean that the inverse has a nice representation. $\endgroup$ – Doug M Aug 13 '20 at 2:26
  • $\begingroup$ What means that you solved for the inversen? What did you actually solve? $\endgroup$ – miracle173 Aug 13 '20 at 2:30
  • $\begingroup$ Hört do you know that the function is injective? What is the domain of the function? $\endgroup$ – miracle173 Aug 13 '20 at 8:03
  • $\begingroup$ " Sorry, it should be not injective in the range [−1,1], right?" does not help a reader to understand the problem, so you should remove it . You can post such statement as comments. But you still have not mentioned what the domain is and why you think it is injective. You should edit your post to clarify the question. $\endgroup$ – miracle173 Aug 14 '20 at 10:29

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.

Plot of the 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 $x\in[1,\infty)$.

  • $\begingroup$ I feel like deleting this under peer pressure because of the downvote, and because it disagrees with the other 2 answers: But I do not see my mistake? Could someone please explain to me why this function has an inverse if that inverse is not one to one? $\endgroup$ – user400188 Aug 13 '20 at 3:26
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    $\begingroup$ Personally I am quite confused right now as well. I noticed that the function should not be injective and thus not bijective. $\endgroup$ – Jess Aug 13 '20 at 3:36
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    $\begingroup$ I think writing to use a Plot to prove something is a reason for a downvote. $\endgroup$ – miracle173 Aug 13 '20 at 4:40
  • $\begingroup$ I have edited it now. Thank you for the advice @miracle173. $\endgroup$ – user400188 Aug 13 '20 at 5:58
  • $\begingroup$ I think some users assume that the domain of the function is [-1,2] and not R. $\endgroup$ – miracle173 Aug 13 '20 at 7:32

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

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

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

Using the quadratic formula

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


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

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

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    $\begingroup$ Why not $x=\frac{1+\sqrt{1-y^2}}{y}$? $\endgroup$ – Jess Aug 13 '20 at 2:54
  • $\begingroup$ A good question. Hint: Try it with $y=\frac{1}{2}$. $\endgroup$ – aschepler Aug 13 '20 at 3:20
  • $\begingroup$ It doesn't take values back to the correct domain. It would be the inverse of the function over the domain $(-\infty, -1]\cup [1,\infty)$ $\endgroup$ – Doug M Aug 13 '20 at 17:21

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) $r^{-1}(x)= \frac{1 - \sqrt{1-y^2}}{y}.$


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