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I have the function

$$f(x) = \frac{x}{x^2 - 1}$$

The domain of this function is $(-1,1)$ and the range is $\mathbb{R}$.

When I find the inverse of this, this becomes

$$f^{-1}(x) = \frac{1 + \sqrt{4x^2+1}}{2x}$$

The domain seems to be different, now that I cannot have zero in the denominator. This should match up with the range of the initial function. Why does this happen?

Parcly Taxel
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Trumpetplayer0098
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  • $f^{-1}(0)=0$ exists and is well defined. Review your derivation of $f^{-1}$ and recheck the conditions at each step. – dxiv Aug 25 '16 at 01:38
  • In general, the domain of the inverse function will be exactly the range of the original function. Note that the range is all values the function will actually output, as opposed to the codomain, which is simply the space into which the function outputs. (You may not have heard the term codomain yet, in which case it's nothing to worry about for now). I do not believe you have the correct expression for $f^{-1}$ (and unfortunately I suspect I cannot give you a correct expression since I doubt that there is a nice closed-form representation for it). – Justin Benfield Aug 25 '16 at 01:43
  • @dxiv I did, I confirmed that even on wolframalpha the inverse I have is correct – Trumpetplayer0098 Aug 25 '16 at 01:54
  • @Trumpetplayer0098 Your formula doesn't work for $x=0$ so it's not correct. See the David's answer why. – dxiv Aug 25 '16 at 01:56
  • You have the correct inverse for the domain$(-\infty,1)\cup (1, \infty)$ but it is not correct for $(-1,1)$ What do you think you should change inside that interval? Now at $x=0$, you are still undefined. But the limit exists. And since you know $f(0) = 0$, you can define it for that point. – Doug M Aug 25 '16 at 01:57

2 Answers2

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Best way to think of it: write your function as $y$ in terms of $x$, your aim is to find $x$ in terms of $y$. We have $$y=\frac{x}{x^2-1}\quad\Leftrightarrow\quad yx^2-x-y=0\ .$$ If $y=0$ obviously $x=0$, if $y\ne0$ we can use the quadratic formula to get $$x=\frac{1\pm\sqrt{1+4y^2}}{2y}\ .$$ There must be only one value of $x$ so we need to decide whether to take the $+$ sign or the $-$ sign. One way to do this is to note that $x$ will be a continuous function of $y$ (draw the graph of $y$ in terms of $x$ to see this), and $$\frac{1+\sqrt{1+4y^2}}{2y}\to\infty\quad\hbox{as}\quad y\to0\ ,$$ so we must reject the $+$ sign (note that you got this wrong).

So the inverse function is $$f^{-1}(y)=\cases{0&if $y=0$\cr \frac{1-\sqrt{1+4y^2}}{2y}&if $y\ne0$,\cr}$$ or if you prefer $$f^{-1}(x)=\cases{0&if $x=0$\cr \frac{1-\sqrt{1+4x^2}}{2x}&if $x\ne0$.\cr}$$

Alternatively we can rationalise the numerator to get $$f^{-1}(y)=-\frac{2y}{1+\sqrt{1+4y^2}}$$ which is valid both for $y\ne0$ and for $y=0$.
David
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  • That makes sense, I was moreorless expecting it to have single function that would have the exact range. I guess you kind of have to be aware that switching the Xs and Ys won't provide the entire answer, tricky question – Trumpetplayer0098 Aug 25 '16 at 01:56
  • You can get a single formula, see the bit I just added to my answer. – David Aug 25 '16 at 01:58
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I can't seem to express $f^{-1}$ in a nice form, using the standard trick of replacing $f(x)$ by $y$ and then switch the $y$'s and $x$'s, I obtain:

$x=y/(y^2-1)$

Now my goal is isolate $y$ so that I can write $y=$'stuff' where $y$ is now $f^{-1}(x)$.

In an effort to isolate, $y$, I multiply, both sides by $y^2-1$ to obtain:

$x(y^2-1)=y$

Next I distribute the $x$ and then subtract $y$ from both sides, yielding:

$xy^2-x-y=0$ Now I reorder terms, and factor out a common $y$

$y(xy-1)-x=0$

And now I believe the problem is becoming evident: There is no easy way to isolate the $y$'s from the $x$'s so that I can get the $x$'s to the other side and complete my attempt to solve for $y$.

Justin Benfield
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    I believe @Trumpetplayer0098 used the quadratic formula on $$xy^{2} - y - x = 0$$ with $a = x, b = 1, c = -x$ – Shai Aug 25 '16 at 01:51
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    @Shai +1 except that for $x=0$ that's not a quadratic. – dxiv Aug 25 '16 at 01:54
  • Ah, and that would explain some of the issues with the resulting expression, as you actually get two (and a third at $x=0$) formulas, each one will be valid for part of the original function's graph. – Justin Benfield Aug 25 '16 at 01:55