# How do you find the domain and range of (x-1) / (x-2)?

Mar 15, 2016

The domain is $\left(- \infty , 2\right) \cup \left(2 , \infty\right)$, i.e. $\mathbb{R} \text{\} \left\{2\right\}$

The range is $\left(- \infty , 1\right) \cup \left(1 , \infty\right)$, i.e. $\mathbb{R} \text{\} \left\{1\right\}$

#### Explanation:

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

When $x = 2$, the denominator of $f \left(x\right)$ is $0$, but the numerator is non-zero. So $f \left(x\right)$ is undefined and has a vertical asymptote at $x = 2$.

If we let $y = f \left(x\right) = 1 + \frac{1}{x - 2}$, then:

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

So:

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

So:

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

So:

${f}^{- 1} \left(y\right) = 2 + \frac{1}{y - 1}$

This is well defined for all $y \in \mathbb{R}$ except $y = 1$, where the inverse function ${f}^{- 1} \left(y\right)$ has a vertical asymptote - so the original function has a horizontal asymptote $y = 1$.

Since $y = 1$ is not in the domain of the inverse function, it is not in the range of the original function.

To summarise:

The domain of $\frac{x - 1}{x - 2}$ is $\mathbb{R} \text{\} \left\{2\right\} = \left(- \infty , 2\right) \cup \left(2 , \infty\right)$

The range of $\frac{x - 1}{x - 2}$ is $\mathbb{R} \text{\} \left\{1\right\} = \left(- \infty , 1\right) \cup \left(1 , \infty\right)$

graph{(x-1)/(x-2) [-8.13, 11.87, -3.88, 6.12]}