How do you find the inverse of #f(x)= (2x+1)/(x-3)#?

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Answer 1 · 2015-07-17T17:18:07

Let #y = f(x) = (2x+1)/(x-3) = (2x-6+7)/(x-3) = 2+7/(x-3)#

Then #x = 7/(y-2)+3#

So #f^(-1)(y) = 7/(y-2)+3#

Let:

#y = f(x) = (2x+1)/(x-3)#

#=(2x-6+7)/(x-3)#

#=(2(x-3) + 7)/(x-3)#

#=2+7/(x-3)#

Note that the domain of #f(x)# is #(-oo,3) uu (3, oo)# and the range is #(-oo,2) uu (2, oo)#

Subtract #2# from both ends to get:

#y - 2 = 7/(x-3)#

Multiply both sides by #(x-3)# to get:

#(y-2)(x-3) = 7#

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

#x - 3 = 7/(y-2)#

Add #3# to both sides to get:

#x = 7/(y-2)+3#

So

#f^(-1)(y) = 7/(y-2)+3#

The domain of #f^(-1)(y)# is #(-oo,2) uu (2,oo)# and its range is #(-oo,3) uu (3,oo)#