# What is the domain and range of  f(x)=(3x-1)/(x^2+9)?

Oct 13, 2017

The domain is $x \in \mathbb{R}$
The range is $f \left(x\right) \in \left[- 0.559 , 0.448\right]$

#### Explanation:

The function is $f \left(x\right) = \frac{3 x - 1}{{x}^{2} + 9}$

$\forall x \in \mathbb{R}$, the denominator is ${x}^{2} + 9 > 0$

Therefore,

The domain is $x \in \mathbb{R}$

To find the range, proceed as follows

Let $y = \frac{3 x - 1}{{x}^{2} + 9}$

Rearranging,

$y {x}^{2} + 9 y = 3 x - 1$

$y {x}^{2} - 3 x + 9 y + 1 = 0$

This is a quadratic equation in ${x}^{2}$, in order for this equation to have solutions, the discriminant $\Delta \ge 0$

$\Delta = {b}^{2} - 4 a c = {\left(- 3\right)}^{2} - \left(4\right) \cdot \left(y\right) \left(9 y + 1\right) \ge 0$

$9 - 36 {y}^{2} - 4 y \ge 0$

$36 {y}^{2} + 4 y - 9 \le 0$

Solving this inequality,

$y = \frac{- 4 \pm \sqrt{{4}^{2} + 4 \cdot 9 \cdot 36}}{2 \cdot 36} = \frac{- 4 \pm \sqrt{1312}}{72}$

${y}_{1} = \frac{- 4 - 36.22}{72} = - 0.559$

${y}_{2} = \frac{- 4 + 36.22}{72} = 0.448$

We can make a sign chart.

The range is $y \in \left[- 0.559 , 0.448\right]$

graph{(3x-1)/(x^2+9) [-10, 10, -5, 5]}