What is the inverse of #f(x) = (4x-1)/(2x+3#?

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Answer 1 · 2015-10-04T21:25:58

#f^(-1)(y) = (1+3y)/(4-2y)#

Let #y = f(x) = (4x - 1)/(2x+3) = (4x+6-7)/(2x+3) = 2 - 7/(2x+3)#

Add #7/(2x+3)# to both ends and subtract #y# from both ends to get:

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

Multiply both sides by #(2x+3)# and divide both sides by #(2-y)# to get:

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

Subtract #3# from both sides to get:

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

Divide both sides by #2# to get:

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

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

#= (1+3y)/(4-2y)#

So:

#f^(-1)(y) = (1+3y)/(4-2y)#