Write the Absolute Value as Piecewise |12-1/(6x)|

$| 12 - \frac{1}{6 x} |$

Step 1

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

$12 - \frac{1}{6 x} \geq 0$

Step 2

Solve the inequality.

Subtract $12$ from both sides of the inequality.

$- \frac{1}{6 x} \geq - 12$

Multiply both sides by $6 x$ .

$- \frac{1}{6 x} (6 x) = - 12 (6 x)$

Simplify.

Simplify the left side.

Cancel the common factor of $6 x$ .

Move the leading negative in $- \frac{1}{6 x}$ into the numerator.

$\frac{- 1}{6 x} (6 x) = - 12 (6 x)$

Cancel the common factor.

$\frac{- 1}{6 x} (6 x) = - 12 (6 x)$

Rewrite the expression.

$- 1 = - 12 (6 x)$

Simplify the right side.

Multiply $6$ by $- 12$ .

$- 1 = - 72 x$

Solve for $x$ .

Rewrite the equation as $- 72 x = - 1$ .

$- 72 x = - 1$

Divide each term in $- 72 x = - 1$ by $- 72$ and simplify.

Divide each term in $- 72 x = - 1$ by $- 72$ .

$\frac{- 72 x}{- 72} = \frac{- 1}{- 72}$

Simplify the left side.

Cancel the common factor of $- 72$ .

Cancel the common factor.

$\frac{- 72 x}{- 72} = \frac{- 1}{- 72}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 72}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{1}{72}$

Find the domain of $12 - \frac{1}{6 x}$ .

Set the denominator in $\frac{1}{6 x}$ equal to $0$ to find where the expression is undefined.

$6 x = 0$

Divide each term in $6 x = 0$ by $6$ and simplify.

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Simplify the right side.

Divide $0$ by $6$ .

$x = 0$

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

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

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{1}{72}$

$x > \frac{1}{72}$

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.

$12 - \frac{1}{6 (- 2)} \geq 0$

The left side $12.08\overline{3}$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < \frac{1}{72}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < \frac{1}{72}$ and see if this value makes the original inequality true.

$x = 0.01$

Replace $x$ with $0.01$ in the original inequality.

$12 - \frac{1}{6 (0.01)} \geq 0$

The left side $- 4.\overline{6}$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{1}{72}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{1}{72}$ and see if this value makes the original inequality true.

$x = 3$

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

$12 - \frac{1}{6 (3)} \geq 0$

The left side $11.9\overline{4}$ 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.

$x < 0$ True

$0 < x < \frac{1}{72}$ False

$x > \frac{1}{72}$ True

$x < 0$ True

$0 < x < \frac{1}{72}$ False

$x > \frac{1}{72}$ True

The solution consists of all of the true intervals.

$x < 0$ or $x \geq \frac{1}{72}$

Step 3

In the piece where $12 - \frac{1}{6 x}$ is non-negative, remove the absolute value.

$12 - \frac{1}{6 x}$

Step 4

Find the domain of $12 - \frac{1}{6 x}$ and find the intersection with $x < 0 or x \geq \frac{1}{72}$ .

Find the domain of $12 - \frac{1}{6 x}$ .

Set the denominator in $\frac{1}{6 x}$ equal to $0$ to find where the expression is undefined.

$6 x = 0$

Divide each term in $6 x = 0$ by $6$ and simplify.

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Simplify the right side.

Divide $0$ by $6$ .

$x = 0$

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

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

Find the intersection of $x < 0 or x \geq \frac{1}{72}$ and $(- \infty, 0) \cup (0, \infty)$ .

$x < 0 or x \geq \frac{1}{72}$

Step 5

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

$12 - \frac{1}{6 x} < 0$

Step 6

Solve the inequality.

Subtract $12$ from both sides of the inequality.

$- \frac{1}{6 x} < - 12$

Multiply both sides by $6 x$ .

$- \frac{1}{6 x} (6 x) = - 12 (6 x)$

Simplify.

Simplify the left side.

Cancel the common factor of $6 x$ .

Move the leading negative in $- \frac{1}{6 x}$ into the numerator.

$\frac{- 1}{6 x} (6 x) = - 12 (6 x)$

Cancel the common factor.

$\frac{- 1}{6 x} (6 x) = - 12 (6 x)$

Rewrite the expression.

$- 1 = - 12 (6 x)$

Simplify the right side.

Multiply $6$ by $- 12$ .

$- 1 = - 72 x$

Solve for $x$ .

Rewrite the equation as $- 72 x = - 1$ .

$- 72 x = - 1$

Divide each term in $- 72 x = - 1$ by $- 72$ and simplify.

Divide each term in $- 72 x = - 1$ by $- 72$ .

$\frac{- 72 x}{- 72} = \frac{- 1}{- 72}$

Simplify the left side.

Cancel the common factor of $- 72$ .

Cancel the common factor.

$\frac{- 72 x}{- 72} = \frac{- 1}{- 72}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 72}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{1}{72}$

Find the domain of $12 - \frac{1}{6 x}$ .

Set the denominator in $\frac{1}{6 x}$ equal to $0$ to find where the expression is undefined.

$6 x = 0$

Divide each term in $6 x = 0$ by $6$ and simplify.

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Simplify the right side.

Divide $0$ by $6$ .

$x = 0$

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

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

Use each root to create test intervals.

$x < 0$

$0 < x < \frac{1}{72}$

$x > \frac{1}{72}$

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.

$12 - \frac{1}{6 (- 2)} < 0$

The left side $12.08\overline{3}$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < \frac{1}{72}$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < \frac{1}{72}$ and see if this value makes the original inequality true.

$x = 0.01$

Replace $x$ with $0.01$ in the original inequality.

$12 - \frac{1}{6 (0.01)} < 0$

The left side $- 4.\overline{6}$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > \frac{1}{72}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{1}{72}$ and see if this value makes the original inequality true.

$x = 3$

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

$12 - \frac{1}{6 (3)} < 0$

The left side $11.9\overline{4}$ 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.

$x < 0$ False

$0 < x < \frac{1}{72}$ True

$x > \frac{1}{72}$ False

$x < 0$ False

$0 < x < \frac{1}{72}$ True

$x > \frac{1}{72}$ False

The solution consists of all of the true intervals.

$0 < x < \frac{1}{72}$

Step 7

In the piece where $12 - \frac{1}{6 x}$ is negative, remove the absolute value and multiply by $- 1$ .

$- (12 - \frac{1}{6 x})$

Step 8

Find the domain of $- (12 - \frac{1}{6 x})$ and find the intersection with $0 < x < \frac{1}{72}$ .

Find the domain of $- (12 - \frac{1}{6 x})$ .

Set the denominator in $\frac{1}{6 x}$ equal to $0$ to find where the expression is undefined.

$6 x = 0$

Divide each term in $6 x = 0$ by $6$ and simplify.

Divide each term in $6 x = 0$ by $6$ .

$\frac{6 x}{6} = \frac{0}{6}$

Simplify the left side.

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6 x}{6} = \frac{0}{6}$

Divide $x$ by $1$ .

$x = \frac{0}{6}$

Simplify the right side.

Divide $0$ by $6$ .

$x = 0$

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

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

Find the intersection of $0 < x < \frac{1}{72}$ and $(- \infty, 0) \cup (0, \infty)$ .

$0 < x < \frac{1}{72}$

Step 9

Write as a piecewise.

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - (12 - \frac{1}{6 x}) & 0 < x < \frac{1}{72} \end{cases}$

Step 10

Simplify $- (12 - \frac{1}{6 x})$ .

Apply the distributive property.

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - 1 \cdot 12 - - \frac{1}{6 x} & 0 < x < \frac{1}{72} \end{cases}$

Multiply $- 1$ by $12$ .

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - 12 - - \frac{1}{6 x} & 0 < x < \frac{1}{72} \end{cases}$

Multiply $- - \frac{1}{6 x}$ .

Multiply $- 1$ by $- 1$ .

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - 12 + 1 \frac{1}{6 x} & 0 < x < \frac{1}{72} \end{cases}$

Multiply $\frac{1}{6 x}$ by $1$ .

${\begin{cases} 12 - \frac{1}{6 x} & x < 0 or x \geq \frac{1}{72} \\ - 12 + \frac{1}{6 x} & 0 < x < \frac{1}{72} \end{cases}$

Write the Absolute Value as Piecewise |12-1/(6x)|

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