Precalculus Examples

$\frac{x^{2} - 1}{x} < 0$

Step 1

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

$x = 0$

$x^{2} - 1 = 0$

Step 2

Add $1$ to both sides of the equation.

$x^{2} = 1$

Step 3

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

$x = \pm \sqrt{1}$

Step 4

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

Step 5.2

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

$x = - 1$

Step 5.3

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

$x = 1, - 1$

Step 6

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

$x = 0$

$x = 1, - 1$

Step 7

Consolidate the solutions.

$x = 0, 1, - 1$

Step 8

Find the domain of $\frac{x^{2} - 1}{x}$ .

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Step 8.1

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

$x = 0$

Step 8.2

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

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

Step 9

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 0$

$0 < x < 1$

$x > 1$

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 $x < - 1$ to see if it makes the inequality true.

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Step 10.1.1

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Step 10.1.2

Replace $x$ with $- 4$ in the original inequality.

$\frac{{(- 4)}^{2} - 1}{- 4} < 0$

Step 10.1.3

The left side $- 3.75$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.2

Test a value on the interval $- 1 < x < 0$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $- 1 < x < 0$ and see if this value makes the original inequality true.

$x = - 0.5$

Step 10.2.2

Replace $x$ with $- 0.5$ in the original inequality.

$\frac{{(- 0.5)}^{2} - 1}{- 0.5} < 0$

Step 10.2.3

The left side $1.5$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 10.3

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

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Step 10.3.1

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Step 10.3.2

Replace $x$ with $0.5$ in the original inequality.

$\frac{{(0.5)}^{2} - 1}{0.5} < 0$

Step 10.3.3

The left side $- 1.5$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 10.4

Test a value on the interval $x > 1$ to see if it makes the inequality true.

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Step 10.4.1

Choose a value on the interval $x > 1$ and see if this value makes the original inequality true.

$x = 4$

Step 10.4.2

Replace $x$ with $4$ in the original inequality.

$\frac{{(4)}^{2} - 1}{4} < 0$

Step 10.4.3

The left side $3.75$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 10.5

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

$x < - 1$ True

$- 1 < x < 0$ False

$0 < x < 1$ True

$x > 1$ False

$x < - 1$ True

$- 1 < x < 0$ False

$0 < x < 1$ True

$x > 1$ False

Step 11

The solution consists of all of the true intervals.

$x < - 1$ or $0 < x < 1$

Step 12

Convert the inequality to interval notation.

$(- \infty, - 1) \cup (0, 1)$

Step 13