Find the Domain and Range (x+3)/(x^2-9)

$\frac{x + 3}{x^{2} - 9}$

Set the denominator in $\frac{x + 3}{x^{2} - 9}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 9 = 0$

Solve for $x$ .

Add $9$ to both sides of the equation.

$x^{2} = 9$

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty)$

Set-Builder Notation:

${x | x \neq 3, - 3}$

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$(- \infty, - \frac{1}{6}) \cup (- \frac{1}{6}, 0) \cup (0, \infty)$

Set-Builder Notation:

${y | y \neq 0, - \frac{1}{6}}$

Determine the domain and range.

Domain: $(- \infty, - 3) \cup (- 3, 3) \cup (3, \infty), {x | x \neq 3, - 3}$

Range: $(- \infty, - \frac{1}{6}) \cup (- \frac{1}{6}, 0) \cup (0, \infty), {y | y \neq 0, - \frac{1}{6}}$