Precalculus Examples

$f (x) = \sqrt{x (x - 17)}$

Step 1

Set the radicand in $\sqrt{x (x - 17)}$ greater than or equal to $0$ to find where the expression is defined.

$x (x - 17) \geq 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$x = 0$

$x - 17 = 0$

Step 2.2

Set $x$ equal to $0$ .

$x = 0$

Step 2.3

Set $x - 17$ equal to $0$ and solve for $x$ .

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Step 2.3.1

Set $x - 17$ equal to $0$ .

$x - 17 = 0$

Step 2.3.2

Add $17$ to both sides of the equation.

$x = 17$

Step 2.4

The final solution is all the values that make $x (x - 17) \geq 0$ true.

$x = 0, 17$

Step 2.5

Use each root to create test intervals.

$x < 0$

$0 < x < 17$

$x > 17$

Step 2.6

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 2.6.1

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

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Step 2.6.1.1

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

$x = - 2$

Step 2.6.1.2

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

$(- 2) ((- 2) - 17) \geq 0$

Step 2.6.1.3

The left side $38$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.2

Test a value on the interval $0 < x < 17$ to see if it makes the inequality true.

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Step 2.6.2.1

Choose a value on the interval $0 < x < 17$ and see if this value makes the original inequality true.

$x = 2$

Step 2.6.2.2

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

$(2) ((2) - 17) \geq 0$

Step 2.6.2.3

The left side $- 30$ is less than the right side $0$ , which means that the given statement is false.

False

Step 2.6.3

Test a value on the interval $x > 17$ to see if it makes the inequality true.

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Step 2.6.3.1

Choose a value on the interval $x > 17$ and see if this value makes the original inequality true.

$x = 20$

Step 2.6.3.2

Replace $x$ with $20$ in the original inequality.

$(20) ((20) - 17) \geq 0$

Step 2.6.3.3

The left side $60$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 2.6.4

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

$x < 0$ True

$0 < x < 17$ False

$x > 17$ True

$x < 0$ True

$0 < x < 17$ False

$x > 17$ True

Step 2.7

The solution consists of all of the true intervals.

$x \leq 0$ or $x \geq 17$

Step 3

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

Interval Notation:

$(- \infty, 0] \cup [17, \infty)$

Set-Builder Notation:

${x | x \leq 0, x \geq 17}$

Step 4