I have to show that the function $f:(-1,1)\to \mathbb{R}$
$$f(x)=\frac{x}{x^2-1}$$
is bijective. I have shown that it is injective which is pretty simple. I'll write it out here for future reference.
To show that $f$ is injective, let $x_1,x_2\in(-1,1)$. Assume that $f(x_1)=f(x_2)$. Then
$$\frac{x_1}{x_1^2-1}=\frac{x_2}{x_2^2-1}$$
Multiplying both sides by $(x_1^2-1)(x_2^2-1)$ which we know can't be $0$. On simplifying the equation further, we get
$$(x_1x_2+1)(x_2-x_1)=0$$
This means either $x_2x_1=-1$ or $x_2=x_1$. It isn't possible that $x_2x_1=-1$ because if it were true then $x_2=\cfrac{-1}{x_1}$ and for all the value of $x_1$ in the domain, $x_2$ will go out of the domain. Thus, it must be that $x_2=x_1$. This concludes the injectivity part.
What I am confused about is the surjectivity part. I know that you usually invert the function and then show that for any value $y$ in the codomain, there exists a value $x$ in the domain of the function such that $f(x)=y$. I can't seem to invert this function appropriately. Any hints?
Best Answer
First notice that $$x=0\iff y=0.$$
Then for $|x|<1\land x\ne0$, $$y=\frac x{x^2-1}\iff yx^2-x-y=0.$$
The discriminant is strictly positive so that there are two distinct real roots. And by Vieta, their product is $-1$, so that one lies in $(-1,1)$ and the other not. Hence the function is invertible.