# How do you find the domain and range of  f(x)=2/(3x-1)?

May 26, 2018

The domain is $x \in \left(- \infty , \frac{1}{3}\right) \cup \left(\frac{1}{3} , + \infty\right)$. The range is $y \in \left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

The denominator must be $\ne 0$

Therefore,

$3 x - 1 \ne 0$

$\implies$, $x \ne \frac{1}{3}$

The domain is $x \in \left(- \infty , \frac{1}{3}\right) \cup \left(\frac{1}{3} , + \infty\right)$

To find the range, proceed as follows :

Let $y = \frac{2}{3 x - 1}$

$\iff$, $y \left(3 x - 1\right) = 2$

$\iff$, $3 y x - y = 2$

$\iff$, $3 y x = 2 + y$

$\iff$, $x = \frac{2 + y}{3 y}$

The denominator must be $\ne 0$

$3 y \ne 0$

$\implies$, $y \ne 0$

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

graph{2/(3x-1) [-12.66, 12.65, -6.33, 6.33]}