What is the inverse function of #f(x) = (2x-1)/(x-1)#?

Question

1091 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-05T18:12:03

#f^(-1)(y) = (y-1)/(y-2)#

Let #y = f(x) = (2x - 1)/(x-1) = (2x-2+1)/(x-1) = 2+1/(x-1)#

Subtract #2# from both ends to get:

#y - 2 = 1/(x-1)#

Multiply both sides by #(x-1)# to get:

#(y-2)(x-1) = 1#

Divide both sides by #(y-2)# to get:

#x-1 = 1/(y-2)#

Add #1# to both sides to get:

#x = 1+1/(y-2) = (y-2)/(y-2)+1/(y-2) = (y-1)/(y-2)#

So #f^(-1)(y) = (y-1)/(y-2)#