What is the inverse function of $f (x) = \frac{2 x - 1}{x - 1}$ $f(x) = (2x-1)/(x-1)$ ?

Question

1202 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-05T18:12:03

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

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

Subtract $2$ $2$ from both ends to get:

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

Multiply both sides by $(x - 1)$ $(x-1)$ to get:

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

Divide both sides by $(y - 2)$ $(y-2)$ to get:

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

Add $1$ $1$ to both sides to get:

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

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

What is the inverse function of f(x) = (2x-1)/(x-1)f(x)=2x−1x−1f(x) = (2x-1)/(x-1)?