Show that the function $f(x)=x/(x^2-1)$ is bijective. 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?
 A: 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.
A: By using the quadratic formula one has
$$f^{-1}(x)=\frac{1-\sqrt{4x^2+1}}{2x}$$
for the given range of $x$.
A: Suppose $\;a\;$ is in the function's image. Let us find out conditions on this element: there exists $\;x\in\Bbb R\setminus\{-1,1\}\;$ s.t.:
$$\frac x{x^2-1}=a\implies ax^2-x-a=0\implies \text{this quadratic's equation discriminant is non-negative}:$$
$$\Delta=1+4a^2\implies \Delta\ge1>0\;\;\forall\,a\in\Bbb R$$
and you now proved your function is surjective...and you didn't have to find out what the inverse is.
Last task: can you prove now that the $\;x\;$ that solves the above is in $\;(-1,1) \;$ , as it must be?
A: Let $y$ be a real number not equal to $0$. Let $f(x)=x/y+1-x^{2}$. Then $f(-1)=-1/y$ and $f(1)=1/y$. Hence there exists $x$ between $-1$ and $1$ with $f(x)=0$.  This gives $x/y+1=x^{2}$ or $y=f(x)$. Clearly, $0=f(0)$. Hence $f$ is surjective. 
A: The function $f(x)=\frac{x}{x^2-1}$ is continuous and strictly (monotonically) decreasing in $(-1,1)$:
$$f'(x)=\frac{-x^2-1}{(x^2-1)^2}<0$$
and:
$$\lim_{x\to -1^+} f(x)=+\infty; \lim_{x\to 1^-} f(x)=-\infty.$$
Therefore, $f(x)$ is surjective for $x\in(-1,1)$. 
Here is the graph:
$\hspace{2cm}$
A: Let $y\in \mathbb{R}$. If $y=0$ then $f(0)=y$ with $0\in(-1,1)$. Suppose  $y\neq 0$. Let $x_1,x_2$ be defined by
$$x_1=\frac{1+\sqrt{1+4y^2}}{2y}\hspace{1cm}x_2=\frac{1-\sqrt{1+4y^2}}{2y}$$
Then $x_1 x_2=-1$, so one of them must lie in the interval $[-1,1]$. Since $x_1, x_2$ are roots of $yx^2-x-y$, we see that none of them can be equal to $1$. Hence none of them can be equal to $-1$ either. Therefore one of them must lie in $(-1,1)$. Let $z$ be this one. Then $yz^2-z-y=0$ and $z\in(-1,1)$. This implies that $f(z)=\frac{z}{z^2-1}=y$.
