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

Jan 29, 2018

$x \in \mathbb{R} , x \ne - 1$
$y \in \mathbb{R} , y \ne 1$

#### Explanation:

The denominator of f(x) cannot be zero as tis would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be.

$\text{solve "x+1=0rArrx=-1larrcolor(red)"excluded value}$

$\text{domain is } x \in \mathbb{R} , x \ne - 1$

$\left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$

$\text{divide terms on numerator/denominator by x}$

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

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{1 + 0}$

$\Rightarrow y = 1 \leftarrow \textcolor{red}{\text{excluded value}}$

$\Rightarrow \text{range is } y \in \mathbb{R} , y \ne 1$

$\left(- \infty , 1\right) \cup \left(1 , + \infty\right) \leftarrow \textcolor{b l u e}{\text{in interval notation}}$
graph{(x-2)/(x+1) [-10, 10, -5, 5]}