# How to find the inverse function of $f(x)=\frac x{x-1}$

How do I find the inverse of $$f(x)=\frac x{x-1}$$? I have attached my work below. I am getting stuck at $$xy-x=y$$. I am not sure what to do from there.

• Put everything with $y$ on one side and factor it out of the expression.
– Sil
Sep 18 '20 at 2:19
• Note: this function is not defined when $x=1$ Sep 18 '20 at 2:22
• In your title and first line $x/x-1=0$ because you divide before you subtract. If $x=0$ it is undefined. Sep 18 '20 at 2:22

$$xy-x=y$$

$$xy-y=x$$

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

$$y=\dfrac x{x-1}$$

\begin{align} y&=\frac{x}{x-1}\\ y&=\frac{x-1+1}{x-1}\\ y&=1+\frac1{x-1}\\ y-1&=\frac1{x-1}\\ x-1&=\frac1{y-1}\\ x&=1+\frac1{y-1}\\ x&=\frac{y-1+1}{y-1}\\ x&=\frac{y}{y-1} \end{align}
$$xy - x = y$$ $$xy - y = x$$ $$y(x-1) = x$$ $$y = \frac{x}{x-1}$$
Normally when finding an inverse we would specify a domain, because such an inverse may not exist everywhere. In our case we should at least assume we are working on the domain $$\mathbb{R} \setminus \{1\}$$. With that in mind let's try to rearrange things. You correctly started with $$x = \frac{y}{y-1}.$$ You got to $$xy - x= y$$ so let's go from there. Then $$-x = y - xy.$$ Factoring gives, $$-x = (1-x)y.$$ So then dividing through by $$(1-x)$$ (this is possible since we assumed $$x\neq 1$$) gives us $$y = - \frac{x}{1-x}.$$ So let's finally just check this. Let's call this function we just found $$g(x)$$ and work out $$f(g(x))$$. $$f(g(x)) = \frac{g(x)}{g(x) - 1} = \frac{\frac{x}{x-1}}{\frac{x}{x-1} -1},$$ and I will leave you the task of checking this just gives $$x$$ back. So it turns out it is its own inverse.