# What is the domain and range of y= 1/(x+1)?

Jul 5, 2018

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$. The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

The function is

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

As the denominator must be $\ne 0$

Therefore,

$x + 1 \ne 0$

$\implies$, $x \ne - 1$

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$

To calculate the range, proceed as follows :

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

Cross multiply

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

$y x + y = 1$

$y x = 1 - y$

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

As the denominator must be $\ne 0$

$y \ne 0$

The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

graph{1/(x+1) [-16.02, 16.02, -8.01, 8.01]}

Jul 5, 2018

$x \in \left(- \infty , - 1\right) \cup \left(- 1 , \infty\right)$
$y \in \left(- \infty , 0\right) \cup \left(0 , \infty\right)$

#### Explanation:

The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be.

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

$\text{domain is } x \in \left(- \infty , - 1\right) \cup \left(- 1 , \infty\right)$

$\text{to find range, rearrange making x the subject}$

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

$x y + y = 1$

$x y = 1 - y$

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

$y = 0 \leftarrow \textcolor{red}{\text{excluded value}}$

$\text{range is } y \in \left(- \infty , 0\right) \cup \left(0 , \infty\right)$
graph{1/(x+1) [-10, 10, -5, 5]}