Precalculus Examples

$f (x) = \frac{\sqrt{x^{2} - 9}}{x + 4}$

Step 1

Set the radicand in $\sqrt{x^{2} - 9}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} - 9 \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $9$ to both sides of the inequality.

$x^{2} \geq 9$

Step 2.2

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{x^{2}} \geq \sqrt{9}$

Step 2.3

Simplify the equation.

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Step 2.3.1

Simplify the left side.

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Step 2.3.1.1

Pull terms out from under the radical.

$| x | \geq \sqrt{9}$

Step 2.3.2

Simplify the right side.

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Step 2.3.2.1

Simplify $\sqrt{9}$ .

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Step 2.3.2.1.1

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

$| x | \geq \sqrt{3^{2}}$

Step 2.3.2.1.2

Pull terms out from under the radical.

$| x | \geq | 3 |$

Step 2.3.2.1.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$| x | \geq 3$

Step 2.4

Write $| x | \geq 3$ as a piecewise.

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Step 2.4.1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x \geq 0$

Step 2.4.2

In the piece where $x$ is non-negative, remove the absolute value.

$x \geq 3$

Step 2.4.3

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x < 0$

Step 2.4.4

In the piece where $x$ is negative, remove the absolute value and multiply by $- 1$ .

$- x \geq 3$

Step 2.4.5

Write as a piecewise.

${\begin{cases} x \geq 3 & x \geq 0 \\ - x \geq 3 & x < 0 \end{cases}$

Step 2.5

Find the intersection of $x \geq 3$ and $x \geq 0$ .

$x \geq 3$

Step 2.6

Divide each term in $- x \geq 3$ by $- 1$ and simplify.

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Step 2.6.1

Divide each term in $- x \geq 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{3}{- 1}$

Step 2.6.2

Simplify the left side.

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Step 2.6.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{3}{- 1}$

Step 2.6.2.2

Divide $x$ by $1$ .

$x \leq \frac{3}{- 1}$

Step 2.6.3

Simplify the right side.

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Step 2.6.3.1

Divide $3$ by $- 1$ .

$x \leq - 3$

Step 2.7

Find the union of the solutions.

$x \leq - 3$ or $x \geq 3$

Step 3

Set the denominator in $\frac{\sqrt{x^{2} - 9}}{x + 4}$ equal to $0$ to find where the expression is undefined.

$x + 4 = 0$

Step 4

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 5

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

Interval Notation:

$(- \infty, - 4) \cup (- 4, - 3] \cup [3, \infty)$

Set-Builder Notation:

${x | x \leq - 3, x \geq 3, x \neq - 4}$

Step 6