Why is the domain different on the inverse of this function?

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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?

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2
On BEST ANSWER

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 $y=0$.

3
On

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 $y$.