Precalculus Examples

$f (x) = \sqrt{\frac{x - 1}{x + 1}}$

Step 1

Set the radicand in $\sqrt{\frac{x - 1}{x + 1}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{x - 1}{x + 1} \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$x - 1 = 0$

$x + 1 = 0$

Step 2.2

Add $1$ to both sides of the equation.

$x = 1$

Step 2.3

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.4

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

$x = 1$

$x = - 1$

Step 2.5

Consolidate the solutions.

$x = 1, - 1$

Step 2.6

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

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Step 2.6.1

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

$x + 1 = 0$

Step 2.6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 2.6.3

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

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

Step 2.7

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$x > 1$

Step 2.8

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 2.8.1

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

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Step 2.8.1.1

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

$x = - 4$

Step 2.8.1.2

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

$\frac{(- 4) - 1}{(- 4) + 1} \geq 0$

Step 2.8.1.3

The left side $1.\overline{6}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.2

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

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Step 2.8.2.1

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

$x = 0$

Step 2.8.2.2

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

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

Step 2.8.2.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.8.3

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

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Step 2.8.3.1

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

$x = 4$

Step 2.8.3.2

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

$\frac{(4) - 1}{(4) + 1} \geq 0$

Step 2.8.3.3

The left side $0.6$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.8.4

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

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

$x < - 1$ True

$- 1 < x < 1$ False

$x > 1$ True

Step 2.9

The solution consists of all of the true intervals.

$x < - 1$ or $x \geq 1$

Step 3

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

$x + 1 = 0$

Step 4

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

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

Interval Notation:

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

Set-Builder Notation:

${x | x < - 1, x \geq 1}$

Step 6