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

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

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Answer 1 · 2015-09-10T16:13:47

$f^{- 1} (y) = \frac{7 y + 2}{3 - y}$ $f^(-1)(y) = (7y+2)/(3-y)$

Explanation:

Let $y = f (x) = \frac{3 x - 2}{x + 7} = \frac{3 x + 21 - 23}{x + 7} = \frac{3 (x + 7) - 23}{x + 7}$ $y = f(x) = (3x-2)/(x+7) = (3x+21 - 23)/(x+7) = (3(x+7)-23)/(x+7)$

$= 3 - \frac{23}{x + 7}$ $= 3-23/(x+7)$

Adding $\frac{23}{x + 7} - y$ $23/(x+7) - y$ to both ends we get:

$3 - y = \frac{23}{x + 7}$ $3-y = 23/(x+7)$

Multiplying both sides by $\frac{x + 7}{3 - y}$ $(x+7)/(3-y)$ we get:

$x + 7 = \frac{23}{3 - y}$ $x+7 = 23/(3-y)$

Subtracting $7$ $7$ from both sides we get:

$x = \frac{23}{3 - y} - 7 = \frac{23 - 7 (3 - y)}{3 - y} = \frac{7 y + 2}{3 - y}$ $x = 23/(3-y)-7 = (23 - 7(3-y))/(3-y) = (7y+2)/(3-y)$

So:

$f^{- 1} (y) = \frac{7 y + 2}{3 - y}$ $f^(-1)(y) = (7y+2)/(3-y)$

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

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