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

##### 2 Answers
Oct 25, 2017

$- \infty < x < \infty$

$- 1 \le y \le 1$

#### Explanation:

The domain is the set of real values that $x$ can take to give a real value.

The range is the set of real values you can get out of the equation.

With fractions you often have to make sure that the denominator is not $0$, because you can't divide by $0$. However, here the denominator cannot equal $0$, because if

${x}^{2} + 9 = 0$

${x}^{2} = - 9$

$x = \sqrt{- 9}$, which doesn't exist as a real number.

Therefore, we know we can put pretty much anything into the equation.

The domain is $- \infty < x < \infty$.

The range is found by recognising that $\left\mid {x}^{2} + 9 \right\mid \ge \left\mid x + 3 \right\mid$ for any real value of $x$, which means that $\left\mid \frac{x + 3}{{x}^{2} + 9} \right\mid \le 1$

This means that the range is

$- 1 \le y \le 1$

Oct 25, 2017

The domain is $x \in \mathbb{R}$ and the range is $y \in \left[- 0.069 , 0.402\right]$

#### Explanation:

The domain is $x \in \mathbb{R}$ as the denominator is

$\left({x}^{2} + 9\right) > 0 , \forall x \in \mathbb{R}$

For the range, proceed as follows,

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

Then,

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

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

This is a quadratic equation in $x$

In order for this equation to have solutions, the discriminant $\Delta \ge 0$

Therefore,

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

$1 - 36 {y}^{2} + 12 y \ge 0$

$- 36 {y}^{2} + 12 y + 1 \ge 0$

$y = \frac{- 12 \pm \sqrt{{12}^{2} - 4 \left(- 36\right) \left(1\right)}}{2 \cdot - 36}$

$y = \frac{- 12 \pm \sqrt{288}}{- 72} = - \left(\frac{- 1 \pm \sqrt{2}}{6}\right)$

${y}_{1} = \frac{1 + \sqrt{2}}{6} = 0.402$

${y}_{2} = \frac{1 - \sqrt{2}}{6} = - 0.069$

Therefore,

The range is $y \in \left[- 0.069 , 0.402\right]$

You can cofirm this with a sign chart and a graph

graph{(x+3)/(x^2+9) [-7.9, 7.9, -3.95, 3.95]}