How do you find the inverse of $g (x) = \frac{x + 2}{x - 3}$ $g(x)= (x + 2) / (x - 3)$ ?

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Answer 1 · 2015-06-29T21:02:08

Find the inverse of $g (x)$ $g(x)$ by simplifying it to have one occurrence of $x$ $x$ then rearranging to express $x$ $x$ in terms of $g (x)$ $g(x)$

$g^{- 1} (y) = 3 + \frac{5}{y - 1}$ $g^-1(y) = 3+5/(y-1)$

Let $y = g (x)$ $y = g(x)$

Then

$y = \frac{x + 2}{x - 3} = \frac{x - 3 + 5}{x - 3} = 1 + \frac{5}{x - 3}$ $y = (x+2)/(x-3) = (x-3+5)/(x-3) = 1+5/(x-3)$

Subtract $1$ $1$ from both ends to get:

$y - 1 = \frac{5}{x - 3}$ $y - 1 = 5/(x-3)$

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

$(y - 1) (x - 3) = 5$ $(y-1)(x-3) = 5$

Divide both sides by $(y - 1)$ $(y-1)$ to get:

$x - 3 = \frac{5}{y - 1}$ $x - 3 = 5/(y-1)$

Add $3$ $3$ to both sides to get:

$x = 3 + \frac{5}{y - 1}$ $x = 3 + 5/(y-1)$

So $g^{- 1} (y) = 3 + \frac{5}{y - 1}$ $g^-1(y) = 3+5/(y-1)$

How do you find the inverse of g(x)=x+2x−3g(x)= (x + 2) / (x - 3)?