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:

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