How do you find the inverse of $y = x/(x+1)$ and is it a function?

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Answer 1 · 2016-05-23T22:17:49

$x = y/(1-y)$ , which is a function.

Isolate $x$ as follows:

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

So, adding $(1/(x+1) - y)$ to both ends we find:

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

Taking reciprocals of both sides we get:

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

Subtracting $1$ from both sides we get:

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

This uniquely determines $x$ for any chosen value of $y$ (except $y=1$ where it's not defined at all), and so is a function.

So if $f(x) = x/(x+1)$ then $f^(-1)(y) = y/(1-y)$

So:

How do you find the inverse of y = x/(x+1)y = x/(x+1) and is it a function?