Precalculus Examples

$\frac{(x + 11) (x - 9)}{x - 1} \geq 0$

Step 1

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

$x + 11 = 0$

$x - 9 = 0$

$x - 1 = 0$

Step 2

Subtract $11$ from both sides of the equation.

$x = - 11$

Step 3

Add $9$ to both sides of the equation.

$x = 9$

Step 4

Add $1$ to both sides of the equation.

$x = 1$

Step 5

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

$x = - 11$

$x = 9$

$x = 1$

Step 6

Consolidate the solutions.

$x = - 11, 9, 1$

Step 7

Find the domain of $\frac{(x + 11) (x - 9)}{x - 1}$ .

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Step 7.1

Set the denominator in $\frac{(x + 11) (x - 9)}{x - 1}$ equal to $0$ to find where the expression is undefined.

$x - 1 = 0$

Step 7.2

Add $1$ to both sides of the equation.

$x = 1$

Step 7.3

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

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

Step 8

Use each root to create test intervals.

$x < - 11$

$- 11 < x < 1$

$1 < x < 9$

$x > 9$

Step 9

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 9.1

Test a value on the interval $x < - 11$ to see if it makes the inequality true.

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Step 9.1.1

Choose a value on the interval $x < - 11$ and see if this value makes the original inequality true.

$x = - 14$

Step 9.1.2

Replace $x$ with $- 14$ in the original inequality.

$\frac{((- 14) + 11) ((- 14) - 9)}{(- 14) - 1} \geq 0$

Step 9.1.3

The left side $- 4.6$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $- 11 < x < 1$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 11 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

Replace $x$ with $0$ in the original inequality.

$\frac{((0) + 11) ((0) - 9)}{(0) - 1} \geq 0$

Step 9.2.3

The left side $99$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $1 < x < 9$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $1 < x < 9$ and see if this value makes the original inequality true.

$x = 5$

Step 9.3.2

Replace $x$ with $5$ in the original inequality.

$\frac{((5) + 11) ((5) - 9)}{(5) - 1} \geq 0$

Step 9.3.3

The left side $- 16$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.4

Test a value on the interval $x > 9$ to see if it makes the inequality true.

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Step 9.4.1

Choose a value on the interval $x > 9$ and see if this value makes the original inequality true.

$x = 12$

Step 9.4.2

Replace $x$ with $12$ in the original inequality.

$\frac{((12) + 11) ((12) - 9)}{(12) - 1} \geq 0$

Step 9.4.3

The left side $6.\overline{27}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.5

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

$x < - 11$ False

$- 11 < x < 1$ True

$1 < x < 9$ False

$x > 9$ True

$x < - 11$ False

$- 11 < x < 1$ True

$1 < x < 9$ False

$x > 9$ True

Step 10

The solution consists of all of the true intervals.

$- 11 \leq x < 1$ or $x \geq 9$

Step 11

Convert the inequality to interval notation.

$[- 11, 1) \cup [9, \infty)$

Step 12