Given $f (x) = \frac{2 x - 1}{x - 1}$ $f(x) = (2x-1)/(x-1)$ how do you find f^-1(x)?

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Answer 1 · 2017-10-18T18:54:51

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

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

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

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

$y x + y = 2 x - 1$ $yx+y=2x-1$

subtract $2 x$ $2x$ :
$(y x - 2 x) + y = - 1$ $(yx-2x)+y=-1$
$x (y - 2) + y = - 1$ $x(y-2)+y=-1$

subtract $y$ $y$ :

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

divide by $(y - 2)$ $(y-2)$ :

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

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

Given f(x)=2x−1x−1f(x) = (2x-1)/(x-1) how do you find f^-1(x)?