Precalculus Examples

$\frac{x + 1}{x^{2} - 8}$

Step 1

Set the denominator in $\frac{x + 1}{x^{2} - 8}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 8 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Add $8$ to both sides of the equation.

$x^{2} = 8$

Step 2.2

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

$x = \pm \sqrt{8}$

Step 2.3

Simplify $\pm \sqrt{8}$ .

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Step 2.3.1

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 2.3.1.1

Factor $4$ out of $8$ .

$x = \pm \sqrt{4 (2)}$

Step 2.3.1.2

Rewrite $4$ as $2^{2}$ .

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

Step 2.3.2

Pull terms out from under the radical.

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

Step 2.4

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

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Step 2.4.1

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

$x = 2 \sqrt{2}$

Step 2.4.2

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

$x = - 2 \sqrt{2}$

Step 2.4.3

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

$x = 2 \sqrt{2}, - 2 \sqrt{2}$

Step 3

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

Interval Notation:

$(- \infty, - 2 \sqrt{2}) \cup (- 2 \sqrt{2}, 2 \sqrt{2}) \cup (2 \sqrt{2}, \infty)$

Set-Builder Notation:

${x | x \neq 2 \sqrt{2}, - 2 \sqrt{2}}$

Step 4