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: