Calculus Examples

$x^{3} - 49 x \geq 0$

Step 1

Convert the inequality to an equation.

$x^{3} - 49 x = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $x$ out of $x^{3} - 49 x$ .

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Step 2.1.1

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

$x \cdot x^{2} - 49 x = 0$

Step 2.1.2

Factor $x$ out of $- 49 x$ .

$x \cdot x^{2} + x \cdot - 49 = 0$

Step 2.1.3

Factor $x$ out of $x \cdot x^{2} + x \cdot - 49$ .

$x (x^{2} - 49) = 0$

Step 2.2

Rewrite $49$ as $7^{2}$ .

$x (x^{2} - 7^{2}) = 0$

Step 2.3

Factor.

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Step 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 $b = 7$ .

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

Step 2.3.2

Remove unnecessary parentheses.

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

Step 3

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

$x = 0$

$x + 7 = 0$

$x - 7 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

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

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Step 5.1

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

$x + 7 = 0$

Step 5.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 6

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

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Step 6.1

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

$x - 7 = 0$

Step 6.2

Add $7$ to both sides of the equation.

$x = 7$

Step 7

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

$x = 0, - 7, 7$

Step 8

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 0$

$0 < x < 7$

$x > 7$

Step 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 9.1

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

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Step 9.1.1

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

$x = - 10$

Step 9.1.2

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

${(- 10)}^{3} - 49 \cdot - 10 \geq 0$

Step 9.1.3

The left side $- 510$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.2

Test a value on the interval $- 7 < x < 0$ to see if it makes the inequality true.

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Step 9.2.1

Choose a value on the interval $- 7 < x < 0$ and see if this value makes the original inequality true.

$x = - 4$

Step 9.2.2

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

${(- 4)}^{3} - 49 \cdot - 4 \geq 0$

Step 9.2.3

The left side $132$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.3

Test a value on the interval $0 < x < 7$ to see if it makes the inequality true.

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Step 9.3.1

Choose a value on the interval $0 < x < 7$ and see if this value makes the original inequality true.

$x = 4$

Step 9.3.2

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

${(4)}^{3} - 49 \cdot 4 \geq 0$

Step 9.3.3

The left side $- 132$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.4

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

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Step 9.4.1

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

$x = 10$

Step 9.4.2

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

${(10)}^{3} - 49 \cdot 10 \geq 0$

Step 9.4.3

The left side $510$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.5

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

$x < - 7$ False

$- 7 < x < 0$ True

$0 < x < 7$ False

$x > 7$ True

$x < - 7$ False

$- 7 < x < 0$ True

$0 < x < 7$ False

$x > 7$ True

Step 10

The solution consists of all of the true intervals.

$- 7 \leq x \leq 0$ or $x \geq 7$

Step 11

Convert the inequality to interval notation.

$[- 7, 0] \cup [7, \infty)$

Step 12