What is the inverse function of #f(x)= (3x-2)/(x+7)?

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Answer 1 · 2015-09-10T16:13:47

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

Let #y = f(x) = (3x-2)/(x+7) = (3x+21 - 23)/(x+7) = (3(x+7)-23)/(x+7)#

#= 3-23/(x+7)#

Adding #23/(x+7) - y# to both ends we get:

#3-y = 23/(x+7)#

Multiplying both sides by #(x+7)/(3-y)# we get:

#x+7 = 23/(3-y)#

Subtracting #7# from both sides we get:

#x = 23/(3-y)-7 = (23 - 7(3-y))/(3-y) = (7y+2)/(3-y)#

So:

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