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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?

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

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    $\begingroup$ +1 for demonstrating that the solution is in $(-1,+1)$. And also, for separating $x=0$ as a special case. $\endgroup$ – stressed out Feb 28 '19 at 8:29
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By using the quadratic formula one has $$f^{-1}(x)=\frac{1-\sqrt{4x^2+1}}{2x}$$ for the given range of $x$.

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

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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?

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  • $\begingroup$ Shouldn't it be $-1<x<+1$ instead of $x \in \mathbb{R} \setminus \{-1,+1\}$? $\endgroup$ – stressed out Feb 28 '19 at 8:19
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    $\begingroup$ @stressedout You're completely right...but the proof of this is almost the same. Editing now, thanks. $\endgroup$ – DonAntonio Feb 28 '19 at 8:21
  • $\begingroup$ 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. $\endgroup$ – ZaWarudo Jun 13 at 8:10
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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:

$\hspace{2cm}$enter image description here

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