Precalculus Examples

$f (x) = \sqrt[4]{x^{2} + 3 x}$

Step 1

Set the radicand in

$\sqrt[4]{x^{2} + 3 x}$ greater than or equal to

$0$ to find where the expression is defined.

$x^{2} + 3 x \geq 0$

Step 2

Solve for

$x$ .

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Step 2.1

Convert the inequality to an equation.

$x^{2} + 3 x = 0$

Step 2.2

Factor

$x$ out of

$x^{2} + 3 x$ .

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Step 2.2.1

Factor

$x$ out of

$x^{2}$ .

$x \cdot x + 3 x = 0$

Step 2.2.2

Factor

$x$ out of

$3 x$ .

$x \cdot x + x \cdot 3 = 0$

Step 2.2.3

Factor

$x$ out of

$x \cdot x + x \cdot 3$ .

$x (x + 3) = 0$

Step 2.3

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$x = 0$

$x + 3 = 0$

Step 2.4

Set

$x$ equal to

$0$ .

$x = 0$

Step 2.5

Set

$x + 3$ equal to

$0$ and solve for

$x$ .

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Step 2.5.1

Set

$x + 3$ equal to

$0$ .

$x + 3 = 0$

Step 2.5.2

Subtract

$3$ from both sides of the equation.

$x = - 3$

Step 2.6

The final solution is all the values that make

$x (x + 3) = 0$ true.

$x = 0, - 3$

Step 2.7

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 0$

$x > 0$

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

$x < - 3$ to see if it makes the inequality true.

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Step 2.8.1.1

Choose a value on the interval

$x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

Step 2.8.1.2

Replace

$x$ with

$- 6$ in the original inequality.

${(- 6)}^{2} + 3 (- 6) \geq 0$

Step 2.8.1.3

The left side

$18$ 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

$- 3 < x < 0$ to see if it makes the inequality true.

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Step 2.8.2.1

Choose a value on the interval

$- 3 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 2.8.2.2

Replace

$x$ with

$- 2$ in the original inequality.

${(- 2)}^{2} + 3 (- 2) \geq 0$

Step 2.8.2.3

The left side

$- 2$ 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

$x > 0$ to see if it makes the inequality true.

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Step 2.8.3.1

Choose a value on the interval

$x > 0$ and see if this value makes the original inequality true.

$x = 2$

Step 2.8.3.2

Replace

$x$ with

$2$ in the original inequality.

${(2)}^{2} + 3 (2) \geq 0$

Step 2.8.3.3

The left side

$10$ 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.

$x < - 3$ True

$- 3 < x < 0$ False

$x > 0$ True

$x < - 3$ True

$- 3 < x < 0$ False

$x > 0$ True

Step 2.9

The solution consists of all of the true intervals.

$x \leq - 3$ or

$x \geq 0$

$x \leq - 3$ or

$x \geq 0$

Step 3

The domain is all values of

$x$ that make the expression defined.

Interval Notation:

$(- \infty, - 3] \cup [0, \infty)$

Set-Builder Notation:

${x | x \leq - 3, x \geq 0}$

Step 4