# How do you find the domain and range of f(x) = (2x+1) / (2x^2 + 5x + 2)?

Sep 8, 2017

The domain is $x \in \mathbb{R} / \left\{- 2\right\}$
The range is $y \in \mathbb{R} / \left\{0\right\}$

#### Explanation:

The denominator must de $\ne 0$

So,

$2 {x}^{2} + 5 x + 2 = \left(2 x + 1\right) \left(x + 2\right) \ne 0$

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

Therefore,

The domain is $x \in \mathbb{R} / \left\{- 2\right\}$

To find the range, we proceed as follows

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

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

$y x + 2 y = 1$

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

The denominator must be $\ne 0$

Therefore,

The range is $y \in \mathbb{R} / \left\{0\right\}$

graph{(2x+1)/(2x^2+5x+2) [-18.02, 18.03, -9.01, 9.01]}