How do you find the inverse of $f (x) = \frac{x}{x + 8}$ $f(x) = x / (x + 8)$ ?

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Answer 1 · 2015-11-22T00:43:31

Let $y = f (x)$ $y = f(x)$ and solve for $x$ $x$ to find:

$f^{- 1} (y) = \frac{8 y}{1 - y}$ $f^(-1)(y) = (8y)/(1-y)$

$y = f (x) = \frac{x}{x + 8} = \frac{x + 8 - 8}{x + 8} = 1 - \frac{8}{x + 8}$ $y = f(x) = x/(x+8) = (x+8-8)/(x+8) = 1-8/(x+8)$

Hence (adding $\frac{8}{x + 8} - y$ $8/(x+8)-y$ to both ends):

$\frac{8}{x + 8} = 1 - y$ $8/(x+8) = 1 - y$

Hence (multiplying both sides by $\frac{x + 8}{1 - y}$ $(x+8)/(1-y)$ ):

$x + 8 = \frac{8}{1 - y}$ $x+8 = 8/(1-y)$

So:

$x = \frac{8}{1 - y} - 8 = \frac{8}{1 - y} - \frac{8 - 8 y}{1 - y} = \frac{8 y}{1 - y}$ $x = 8/(1-y) - 8 = 8/(1-y) - (8-8y)/(1-y) = (8y)/(1-y)$

So:

$f^{- 1} (y) = \frac{8 y}{1 - y}$ $f^(-1)(y) = (8y)/(1-y)$

How do you find the inverse of f(x) = x / (x + 8)f(x)=xx+8f(x) = x / (x + 8)?