# How do you find the domain and range of Y = g(x) = (x-3)/(x+1)?

Apr 2, 2018

The domain is $x \in \mathbb{R} - \left\{- 1\right\}$. The range is $y \in \mathbb{R} - \left\{1\right\}$

#### Explanation:

The denominator must $\ne 0$

Therefore,

$x + 1 \ne 0$

$x \ne - 1$

The domain of $g \left(x\right)$ is $x \in \mathbb{R} - \left\{- 1\right\}$

To find the range, proceed as follows

$y = \frac{x - 3}{x + 1}$

$y x + y = x - 3$

$y x - x = - 3 - y$

$x \left(y - 1\right) = - \left(3 + y\right)$

$x = - \frac{3 + y}{y - 1}$

The denominator is $\ne 0$

$y - 1 \ne 0$

$y \ne 1$

The range of $g \left(x\right)$ is $y \in \mathbb{R} - \left\{1\right\}$

graph{(x-3)/(x+1) [-22.8, 22.83, -11.4, 11.4]}