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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. enter image description here

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

$xy-y=x$

$(x-1)y=x$

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

Your function is an involution.

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    $\begingroup$ Awesome thank you for the link. I learned this concept in lecture but I didn't know it was called an involution. That is very interesting and cool! $\endgroup$ Sep 18 '20 at 3:08
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    $\begingroup$ Yes, it is cool! $\endgroup$ Sep 18 '20 at 3:22
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\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}

As it turns out, its inverse is itself!

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$$xy - x = y$$ $$xy - y = x$$ $$y(x-1) = x$$ $$y = \frac{x}{x-1}$$

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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.

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