Solve for f 4|f-6|<=12

$4 | f - 6 | \leq 12$

Step 1

Divide each term in $4 | f - 6 | \leq 12$ by $4$ and simplify.

Divide each term in $4 | f - 6 | \leq 12$ by $4$ .

$\frac{4 | f - 6 |}{4} \leq \frac{12}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 | f - 6 |}{4} \leq \frac{12}{4}$

Divide $| f - 6 |$ by $1$ .

$| f - 6 | \leq \frac{12}{4}$

Simplify the right side.

Divide $12$ by $4$ .

$| f - 6 | \leq 3$

Step 2

Write $| f - 6 | \leq 3$ as a piecewise.

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

$f - 6 \geq 0$

Add $6$ to both sides of the inequality.

$f \geq 6$

In the piece where $f - 6$ is non-negative, remove the absolute value.

$f - 6 \leq 3$

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

$f - 6 < 0$

Add $6$ to both sides of the inequality.

$f < 6$

In the piece where $f - 6$ is negative, remove the absolute value and multiply by $- 1$ .

$- (f - 6) \leq 3$

Write as a piecewise.

${\begin{cases} f - 6 \leq 3 & f \geq 6 \\ - (f - 6) \leq 3 & f < 6 \end{cases}$

Simplify $- (f - 6) \leq 3$ .

Apply the distributive property.

${\begin{cases} f - 6 \leq 3 & f \geq 6 \\ - f - - 6 \leq 3 & f < 6 \end{cases}$

Multiply $- 1$ by $- 6$ .

${\begin{cases} f - 6 \leq 3 & f \geq 6 \\ - f + 6 \leq 3 & f < 6 \end{cases}$

Step 3

Solve $f - 6 \leq 3$ when $f \geq 6$ .

Move all terms not containing $f$ to the right side of the inequality.

Add $6$ to both sides of the inequality.

$f \leq 3 + 6$

Add $3$ and $6$ .

$f \leq 9$

Find the intersection of $f \leq 9$ and $f \geq 6$ .

$6 \leq f \leq 9$

Step 4

Solve $- f + 6 \leq 3$ when $f < 6$ .

Solve $- f + 6 \leq 3$ for $f$ .

Move all terms not containing $f$ to the right side of the inequality.

Subtract $6$ from both sides of the inequality.

$- f \leq 3 - 6$

Subtract $6$ from $3$ .

$- f \leq - 3$

Divide each term in $- f \leq - 3$ by $- 1$ and simplify.

Divide each term in $- f \leq - 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- f}{- 1} \geq \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{f}{1} \geq \frac{- 3}{- 1}$

Divide $f$ by $1$ .

$f \geq \frac{- 3}{- 1}$

Simplify the right side.

Divide $- 3$ by $- 1$ .

$f \geq 3$

Find the intersection of $f \geq 3$ and $f < 6$ .

$3 \leq f < 6$

Step 5

Find the union of the solutions.

$3 \leq f \leq 9$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$3 \leq f \leq 9$

Interval Notation:

$[3, 9]$

Step 7

Solve for f 4|f-6|<=12

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