# How do you find the Domain and Range of f(x)=(5x-3)/(2x+1)?

May 24, 2018

The domain is $x \in \left(- \infty , - \frac{1}{2}\right) \cup \left(- \frac{1}{2} , + \infty\right)$. The range is $y \in \left(- \infty , \frac{5}{2}\right) \cup \left(\frac{5}{2} , + \infty\right)$

#### Explanation:

The denominator must be $\ne 0$

$2 x + 1 \ne 0$

$x \ne - \frac{1}{2}$

The domain is $x \in \left(- \infty , - \frac{1}{2}\right) \cup \left(- \frac{1}{2} , + \infty\right)$

To find the range, proceed as follows :

Let $y = \frac{5 x - 3}{2 x + 1}$

$y \left(2 x + 1\right) = 5 x - 3$

$2 y x + y = 5 x - 3$

$2 y x - 5 x = - y - 3$

$x \left(2 y - 5\right) = - \left(y + 3\right)$

$x = - \frac{y + 3}{2 y - 5}$

Here,

$2 y - 5 \ne 0$

$y \ne \frac{5}{2}$

The range is $y \in \left(- \infty , \frac{5}{2}\right) \cup \left(\frac{5}{2} , + \infty\right)$

graph{(5x-3)/(2x1) [-16.02, 16.02, -8.01, 8.01]}