Write the Absolute Value as Piecewise |x^3-10x^2|

$| x^{3} - 10 x^{2} |$

Step 1

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$x^{3} - 10 x^{2} \geq 0$

Step 2

Solve the inequality.

Convert the inequality to an equation.

$x^{3} - 10 x^{2} = 0$

Factor $x^{2}$ out of $x^{3} - 10 x^{2}$ .

Factor $x^{2}$ out of $x^{3}$ .

$x^{2} x - 10 x^{2} = 0$

Factor $x^{2}$ out of $- 10 x^{2}$ .

$x^{2} x + x^{2} \cdot - 10 = 0$

Factor $x^{2}$ out of $x^{2} x + x^{2} \cdot - 10$ .

$x^{2} (x - 10) = 0$

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

$x^{2} = 0$

$x - 10 = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

The final solution is all the values that make $x^{2} (x - 10) = 0$ true.

$x = 0, 10$

Use each root to create test intervals.

$x < 0$

$0 < x < 10$

$x > 10$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 2$

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

${(- 2)}^{3} - 10 {(- 2)}^{2} \geq 0$

The left side $- 48$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 10$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 10$ and see if this value makes the original inequality true.

$x = 5$

Replace $x$ with $5$ in the original inequality.

${(5)}^{3} - 10 {(5)}^{2} \geq 0$

The left side $- 125$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 10$ to see if it makes the inequality true.

Choose a value on the interval $x > 10$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

${(12)}^{3} - 10 {(12)}^{2} \geq 0$

The left side $288$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < 0$ False

$0 < x < 10$ False

$x > 10$ True

$x < 0$ False

$0 < x < 10$ False

$x > 10$ True

The solution consists of all of the true intervals.

$x \geq 10$ or $x = 0$

Combine the intervals.

$x = 0 or x \geq 10$

Step 3

In the piece where $x^{3} - 10 x^{2}$ is non-negative, remove the absolute value.

$x^{3} - 10 x^{2}$

Step 4

To find the interval for the second piece, find where the inside of the absolute value is negative.

$x^{3} - 10 x^{2} < 0$

Step 5

Solve the inequality.

Convert the inequality to an equation.

$x^{3} - 10 x^{2} = 0$

Factor $x^{2}$ out of $x^{3} - 10 x^{2}$ .

Factor $x^{2}$ out of $x^{3}$ .

$x^{2} x - 10 x^{2} = 0$

Factor $x^{2}$ out of $- 10 x^{2}$ .

$x^{2} x + x^{2} \cdot - 10 = 0$

Factor $x^{2}$ out of $x^{2} x + x^{2} \cdot - 10$ .

$x^{2} (x - 10) = 0$

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

$x^{2} = 0$

$x - 10 = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $x - 10$ equal to $0$ and solve for $x$ .

Set $x - 10$ equal to $0$ .

$x - 10 = 0$

Add $10$ to both sides of the equation.

$x = 10$

The final solution is all the values that make $x^{2} (x - 10) = 0$ true.

$x = 0, 10$

Use each root to create test intervals.

$x < 0$

$0 < x < 10$

$x > 10$

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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

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

$x = - 2$

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

${(- 2)}^{3} - 10 {(- 2)}^{2} < 0$

The left side $- 48$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 10$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 10$ and see if this value makes the original inequality true.

$x = 5$

Replace $x$ with $5$ in the original inequality.

${(5)}^{3} - 10 {(5)}^{2} < 0$

The left side $- 125$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 10$ to see if it makes the inequality true.

Choose a value on the interval $x > 10$ and see if this value makes the original inequality true.

$x = 12$

Replace $x$ with $12$ in the original inequality.

${(12)}^{3} - 10 {(12)}^{2} < 0$

The left side $288$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$x < 0$ True

$0 < x < 10$ True

$x > 10$ False

$x < 0$ True

$0 < x < 10$ True

$x > 10$ False

The solution consists of all of the true intervals.

$x < 0$ or $0 < x < 10$

Step 6

In the piece where $x^{3} - 10 x^{2}$ is negative, remove the absolute value and multiply by $- 1$ .

$- (x^{3} - 10 x^{2})$

Step 7

Write as a piecewise.

${\begin{cases} x^{3} - 10 x^{2} & x = 0 or x \geq 10 \\ - (x^{3} - 10 x^{2}) & x < 0 or 0 < x < 10 \end{cases}$

Step 8

Simplify $- (x^{3} - 10 x^{2})$ .

Apply the distributive property.

${\begin{cases} x^{3} - 10 x^{2} & x = 0 or x \geq 10 \\ - x^{3} - (- 10 x^{2}) & x < 0 or 0 < x < 10 \end{cases}$

Multiply $- 10$ by $- 1$ .

${\begin{cases} x^{3} - 10 x^{2} & x = 0 or x \geq 10 \\ - x^{3} + 10 x^{2} & x < 0 or 0 < x < 10 \end{cases}$

Write the Absolute Value as Piecewise |x^3-10x^2|

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