What is the inverse function for $f (x) = \frac{x}{x - 2}$ $f(x) = x/(x-2)$ ?

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Answer 1 · 2015-09-22T23:09:53

$f^{- 1} (x) = \frac{2 x}{x - 1}$ $f^(-1)(x) = (2 x) / (x - 1)$

Setting $y = f (x)$ $y = f(x)$

$y = \frac{x}{x - 2}$ $y = x/(x-2)$

this may be rearranged (intermediate steps shown) as follows

$y (x - 2) = x$ $y(x-2) = x$

$x \cdot y - 2 y = x$ $x*y - 2 y = x$

$x \cdot y - x = 2 y$ $x*y - x = 2 y$

$x (y - 1) = 2 y$ $x(y - 1) = 2 y$

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

This expression shows $x$ $x$ in terms of $y$ $y$ .

That is, it is the inverse of function $f (x)$ $f(x)$ .

That is,

$f^{- 1} (x) = \frac{2 x}{x - 1}$ $f^(-1)(x) = (2 x) / (x - 1)$

as required.

What is the inverse function for f(x) = x/(x-2)f(x)=xx−2f(x) = x/(x-2)?