Precalculus Examples

$y = \log (x^{3} - 36 x)$

Step 1

Set the argument in $\log (x^{3} - 36 x)$ greater than $0$ to find where the expression is defined.

$x^{3} - 36 x > 0$

Step 2

Solve for $x$ .

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Step 2.1

Convert the inequality to an equation.

$x^{3} - 36 x = 0$

Step 2.2

Factor the left side of the equation.

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Step 2.2.1

Factor $x$ out of $x^{3} - 36 x$ .

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Step 2.2.1.1

Factor $x$ out of $x^{3}$ .

$x \cdot x^{2} - 36 x = 0$

Step 2.2.1.2

Factor $x$ out of $- 36 x$ .

$x \cdot x^{2} + x \cdot - 36 = 0$

Step 2.2.1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 36$ .

$x (x^{2} - 36) = 0$

Step 2.2.2

Rewrite $36$ as $6^{2}$ .

$x (x^{2} - 6^{2}) = 0$

Step 2.2.3

Factor.

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Step 2.2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 6$ .

$x ((x + 6) (x - 6)) = 0$

Step 2.2.3.2

Remove unnecessary parentheses.

$x (x + 6) (x - 6) = 0$

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 $0$ .

$x = 0$

$x + 6 = 0$

$x - 6 = 0$

Step 2.4

Set $x$ equal to $0$ .

$x = 0$

Step 2.5

Set $x + 6$ equal to $0$ and solve for $x$ .

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Step 2.5.1

Set $x + 6$ equal to $0$ .

$x + 6 = 0$

Step 2.5.2

Subtract $6$ from both sides of the equation.

$x = - 6$

Step 2.6

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 2.6.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 2.7

The final solution is all the values that make $x (x + 6) (x - 6) = 0$ true.

$x = 0, - 6, 6$

Step 2.8

Use each root to create test intervals.

$x < - 6$

$- 6 < x < 0$

$0 < x < 6$

$x > 6$

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 $x < - 6$ to see if it makes the inequality true.

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Step 2.9.1.1

Choose a value on the interval $x < - 6$ 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.

${(- 8)}^{3} - 36 \cdot - 8 > 0$

Step 2.9.1.3

The left side $- 224$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 2.9.2

Test a value on the interval $- 6 < x < 0$ to see if it makes the inequality true.

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Step 2.9.2.1

Choose a value on the interval $- 6 < x < 0$ and see if this value makes the original inequality true.

$x = - 3$

Step 2.9.2.2

Replace $x$ with $- 3$ in the original inequality.

${(- 3)}^{3} - 36 \cdot - 3 > 0$

Step 2.9.2.3

The left side $81$ 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 $0 < x < 6$ to see if it makes the inequality true.

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Step 2.9.3.1

Choose a value on the interval $0 < x < 6$ and see if this value makes the original inequality true.

$x = 3$

Step 2.9.3.2

Replace $x$ with $3$ in the original inequality.

${(3)}^{3} - 36 \cdot 3 > 0$

Step 2.9.3.3

The left side $- 81$ is not greater than the right side $0$ , which means that the given statement is false.

False

Step 2.9.4

Test a value on the interval $x > 6$ to see if it makes the inequality true.

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Step 2.9.4.1

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Step 2.9.4.2

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

${(8)}^{3} - 36 \cdot 8 > 0$

Step 2.9.4.3

The left side $224$ 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.

$x < - 6$ False

$- 6 < x < 0$ True

$0 < x < 6$ False

$x > 6$ True

$x < - 6$ False

$- 6 < x < 0$ True

$0 < x < 6$ False

$x > 6$ True

Step 2.10

The solution consists of all of the true intervals.

$- 6 < x < 0$ or $x > 6$

Step 3

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

Interval Notation:

$(- 6, 0) \cup (6, \infty)$

Set-Builder Notation:

${x | - 6 < x < 0, x > 6}$

Step 4