Question

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

Answer 1

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

Explanation:

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)`

• The domain of `f(x)` is `(-oo, -1) uu (-1, oo)`

• The domain of `f^(-1)(y)` is `(-oo, 1) uu (1, oo)`

So:

• The range of `f(x)` is `(-oo, 1) uu (1, oo)`

• The range of `f^(-1)(y)` is `(-oo, -1) uu (-1, oo)`