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# How do you find the inverse of  f(x) = (2x-1)/(x-1) and is it a function?

May 14, 2016

• ${f}^{-} 1 \left(x\right) = \frac{- x + 1}{2 - x}$
• yes, it is a function

#### Explanation:

Determining the Inverse Function
Given,

$f \left(x\right) = \frac{2 x - 1}{x - 1}$

Substitute $y$ for $f \left(x\right)$.

$y = \frac{2 x - 1}{x - 1}$

Swap the $x$ and $y$.

$x = \frac{2 y - 1}{y - 1}$

Solve for $y$.

$x \left(y - 1\right) = 2 y - 1$

$x y - x = 2 y - 1$

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

Factor out $y$ from the left side.

$y \left(2 - x\right) = - x + 1$

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

Rewrite $y$ as ${f}^{-} 1 \left(x\right)$.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{f}^{-} 1 \left(x\right) = \frac{- x + 1}{2 - x}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Determining Whether the Inverse Function Is a Function
Graphically, ${f}^{-} 1 \left(x\right) = \frac{- x + 1}{2 - x}$ would look like:

graph{(-x+1)/(2-x) [-10, 10, -5, 5]}

In the graph above, you can see that the $x$ and $y$ values approach the vertical and horizontal asymptotes. Since it resembles that of an exponential graph, there is only one $y$ value for an $x$ value.

$\therefore$, the inverse function is a function.