Precalculus Examples

$f (x) = \sqrt{\frac{x + 5}{(x - 7) (x - 2)}}$

Step 1

Set the radicand in $\sqrt{\frac{x + 5}{(x - 7) (x - 2)}}$ greater than or equal to $0$ to find where the expression is defined.

$\frac{x + 5}{(x - 7) (x - 2)} \geq 0$

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.

$x + 5 = 0$

$x - 7 = 0$

$x - 2 = 0$

Step 2.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 2.3

Add $7$ to both sides of the equation.

$x = 7$

Step 2.4

Add $2$ to both sides of the equation.

$x = 2$

Step 2.5

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

$x = - 5$

$x = 7$

$x = 2$

Step 2.6

Consolidate the solutions.

$x = - 5, 7, 2$

Step 2.7

Find the domain of $\frac{x + 5}{(x - 7) (x - 2)}$ .

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Step 2.7.1

Set the denominator in $\frac{x + 5}{(x - 7) (x - 2)}$ equal to $0$ to find where the expression is undefined.

$(x - 7) (x - 2) = 0$

Step 2.7.2

Solve for $x$ .

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Step 2.7.2.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 7 = 0$

$x - 2 = 0$

Step 2.7.2.2

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 2.7.2.2.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 2.7.2.2.2

Add $7$ to both sides of the equation.

$x = 7$

Step 2.7.2.3

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 2.7.2.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 2.7.2.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 2.7.2.4

The final solution is all the values that make $(x - 7) (x - 2) = 0$ true.

$x = 7, 2$

Step 2.7.3

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

$(- \infty, 2) \cup (2, 7) \cup (7, \infty)$

Step 2.8

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 2$

$2 < x < 7$

$x > 7$

Step 2.9

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 2.9.1

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

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Step 2.9.1.1

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Step 2.9.1.2

Replace $x$ with $- 8$ in the original inequality.

$\frac{(- 8) + 5}{((- 8) - 7) ((- 8) - 2)} \geq 0$

Step 2.9.1.3

The left side $- 0.02$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.9.2

Test a value on the interval $- 5 < x < 2$ to see if it makes the inequality true.

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Step 2.9.2.1

Choose a value on the interval $- 5 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

Step 2.9.2.2

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

$\frac{(0) + 5}{((0) - 7) ((0) - 2)} \geq 0$

Step 2.9.2.3

The left side $0.3\overline{571428}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.9.3

Test a value on the interval $2 < x < 7$ to see if it makes the inequality true.

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Step 2.9.3.1

Choose a value on the interval $2 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

Step 2.9.3.2

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

$\frac{(4) + 5}{((4) - 7) ((4) - 2)} \geq 0$

Step 2.9.3.3

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

False

Step 2.9.4

Test a value on the interval $x > 7$ to see if it makes the inequality true.

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Step 2.9.4.1

Choose a value on the interval $x > 7$ and see if this value makes the original inequality true.

$x = 10$

Step 2.9.4.2

Replace $x$ with $10$ in the original inequality.

$\frac{(10) + 5}{((10) - 7) ((10) - 2)} \geq 0$

Step 2.9.4.3

The left side $0.625$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.9.5

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

$x < - 5$ False

$- 5 < x < 2$ True

$2 < x < 7$ False

$x > 7$ True

$x < - 5$ False

$- 5 < x < 2$ True

$2 < x < 7$ False

$x > 7$ True

Step 2.10

The solution consists of all of the true intervals.

$- 5 \leq x < 2$ or $x > 7$

Step 3

Set the denominator in $\frac{x + 5}{(x - 7) (x - 2)}$ equal to $0$ to find where the expression is undefined.

$(x - 7) (x - 2) = 0$

Step 4

Solve for $x$ .

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Step 4.1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 7 = 0$

$x - 2 = 0$

Step 4.2

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 4.2.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 4.2.2

Add $7$ to both sides of the equation.

$x = 7$

Step 4.3

Set $x - 2$ equal to $0$ and solve for $x$ .

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Step 4.3.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Step 4.3.2

Add $2$ to both sides of the equation.

$x = 2$

Step 4.4

The final solution is all the values that make $(x - 7) (x - 2) = 0$ true.

$x = 7, 2$

Step 5

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

Interval Notation:

$[- 5, 2) \cup (7, \infty)$

Set-Builder Notation:

${x | - 5 \leq x < 2, x > 7}$

Step 6