Precalculus Examples

$\frac{x}{x^{2} - 16} > 0$

Step 1

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x = 0$

$x^{2} - 16 = 0$

Step 2

Add $16$ to both sides of the equation.

$x^{2} = 16$

Step 3

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

$x = \pm \sqrt{16}$

Step 4

Simplify $\pm \sqrt{16}$ .

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Step 4.1

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Step 4.2

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

$x = \pm 4$

Step 5

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

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Step 5.1

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

$x = 4$

Step 5.2

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

$x = - 4$

Step 5.3

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

$x = 4, - 4$

Step 6

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 0$

$x = 4, - 4$

Step 7

Consolidate the solutions.

$x = 0, 4, - 4$

Step 8

Find the domain of $\frac{x}{x^{2} - 16}$ .

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Step 8.1

Set the denominator in $\frac{x}{x^{2} - 16}$ equal to $0$ to find where the expression is undefined.

$x^{2} - 16 = 0$

Step 8.2

Solve for $x$ .

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Step 8.2.1

Add $16$ to both sides of the equation.

$x^{2} = 16$

Step 8.2.2

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

$x = \pm \sqrt{16}$

Step 8.2.3

Simplify $\pm \sqrt{16}$ .

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Step 8.2.3.1

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Step 8.2.3.2

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

$x = \pm 4$

Step 8.2.4

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

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Step 8.2.4.1

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

$x = 4$

Step 8.2.4.2

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

$x = - 4$

Step 8.2.4.3

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

$x = 4, - 4$

Step 8.3

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

$(- \infty, - 4) \cup (- 4, 4) \cup (4, \infty)$

Step 9

Use each root to create test intervals.

$x < - 4$

$- 4 < x < 0$

$0 < x < 4$

$x > 4$

Step 10

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 10.1

Test a value on the interval $x < - 4$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < - 4$ and see if this value makes the original inequality true.

$x = - 6$

Step 10.1.2

Replace $x$ with $- 6$ in the original inequality.

$\frac{- 6}{{(- 6)}^{2} - 16} > 0$

Step 10.1.3

The left side $- 0.3$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $- 4 < x < 0$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 4 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 10.2.2

Replace $x$ with $- 2$ in the original inequality.

$\frac{- 2}{{(- 2)}^{2} - 16} > 0$

Step 10.2.3

The left side $0.1\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.3

Test a value on the interval $0 < x < 4$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $0 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Step 10.3.2

Replace $x$ with $2$ in the original inequality.

$\frac{2}{{(2)}^{2} - 16} > 0$

Step 10.3.3

The left side $- 0.1\overline{6}$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 10.4

Test a value on the interval $x > 4$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

Step 10.4.2

Replace $x$ with $6$ in the original inequality.

$\frac{6}{{(6)}^{2} - 16} > 0$

Step 10.4.3

The left side $0.3$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 10.5

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 4$ False

$- 4 < x < 0$ True

$0 < x < 4$ False

$x > 4$ True

$x < - 4$ False

$- 4 < x < 0$ True

$0 < x < 4$ False

$x > 4$ True

Step 11

The solution consists of all of the true intervals.

$- 4 < x < 0$ or $x > 4$

Step 12

Convert the inequality to interval notation.

$(- 4, 0) \cup (4, \infty)$

Step 13