# 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?

• Make quadratic and find discriminant.. I think that's the usual way – Scáthach Feb 28 '19 at 8:02
• Show that $a(x^2-1)-x=0$ for some $a\in(-1,1)$. – Wuestenfux Feb 28 '19 at 8:03

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.

• +1 for demonstrating that the solution is in $(-1,+1)$. And also, for separating $x=0$ as a special case. – stressed out Feb 28 '19 at 8:29

By using the quadratic formula one has $$f^{-1}(x)=\frac{1-\sqrt{4x^2+1}}{2x}$$ for the given range of $$x$$.

• How do you overrule the solution with $+$ sign? Namely, $$f^{-1}(x)=\frac{1+\sqrt{4x^2+1}}{2x}$$ – stressed out Feb 28 '19 at 8:09
• $f(-1)=+\infty \Rightarrow f^{-1}(+\infty)=-1$ – Peter Foreman Feb 28 '19 at 8:11
• Shouldn't we also show that $-1< \frac{1-\sqrt{4x^2+1}}{2x} < +1$? – stressed out Feb 28 '19 at 8:17
• @PeterForeman I still don't get how you overruled the other possibility. – Salman Qureshi Feb 28 '19 at 8:39

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.

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?

• Shouldn't it be $-1<x<+1$ instead of $x \in \mathbb{R} \setminus \{-1,+1\}$? – stressed out Feb 28 '19 at 8:19
• @stressedout You're completely right...but the proof of this is almost the same. Editing now, thanks. – DonAntonio Feb 28 '19 at 8:21
• Why this proves that $f$ is surjective? Is that because we have proved that the equation $x/(x^2-1)=a$ always have two real solutions for all $a$ and so $f$ "covers" all $\mathbb{R}$? Does the "for all $a\in\mathbb{R}$" come from the fact that you've proved that the determinant is positive independently from $a$ so the equality holds for all $a$? Thanks. – ZaWarudo Jun 13 at 8:10

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:

$$\hspace{2cm}$$ 