2.5 The Absolute Value Function

Introduction

The absolute value function is used to model problems involving distance. Recall that the absolute value of a number gives the distance from the origin to that number on the number line.

For example, the equation $| x - 2 | = 6$ means "the distance between $x$ and $2$ is $6$ units." The number $x$ could be to the left or the right of $2$ on the number line. Thus, the equation has two solutions, $8$ and $minus 4$ , as shown below.

To find the absolute value of a number, we can

_____

To find the absolute value of a number, we can find its distance from 0.

To find the absolute value of a number, we can

change its sign.
subtract it from 0.
find its distance from 0.
square it.

Example 1

Write each statement using absolute value notation. Illustrate the solutions on a number line.

$x$ is three units from the origin.
$p$ is two units from $5$ .
$a$ is within four units of $minus 2$ .

First, restate each statement in terms of distance.

The distance between $x$ and the origin is three units, or $| x | equals 3$ . Thus, $x$ can be $3$ or $minus 3$ .
The distance between $p$ and $5$ is two units, or $| p - 5 | = 2$ . If we count two units on either side of $5$ , we see that $p$ can be $3$ or $7$ .
The distance between $a$ and $minus 2$ is less than four units, or $| a - (- 2) | < 4$ , or $| a plus 2 | less than 4$ . Count four units on either side of $minus 2$ , to find $minus 6$ and $2$ . Then $a$ is between $minus 6$ and $2$ , or $minus 6 less than a less than 2$ .

The notation $| x - 3 | = 5$ means

_____

The notation $| x - 3 | = 5$ means that the distance between $x$ and 3 is 5 units.

The notation $| x - 3 | = 5$ means

$x$ is 3 units bigger than 5.
the distance between $x$ and 5 is 3 units.
5 and 3 are $x$ units apart.
the distance between $x$ and 3 is 5 units.

Write each statement using absolute value notation; then illustrate the solutions on a number line.

$x$ is five units away from $minus 3$ .
_____ $=$ _____
$x$ is at least six units away from $4$ .
_____ $\geq$ _____

Use "abs(x)" to get $| x |$ , etc.

$| x + 3 | = 5$
A number line is shown below.
$| x minus 4 | greater than or equal to 6$
A number line is shown below.

Number lines:

Write each statement using absolute value notation; then illustrate the solutions on a number line.

$x$ is five units away from $minus 3$ .
$x$ is at least six units away from $4$ .

$| x + 3 | = 5$
$| x minus 4 | greater than or equal to 6$

Absolute Value Equations

We can use distances on a number line to solve simple equations such as

$| 3 x - 6 | = 9$

First, we factor out the coefficient of $x$ , to get $| 3 (x - 2) | = 9$ . Because of the multiplicative property of the absolute value, namely that $| a b | equals | a | | b |$ , we can write the left side as

$\begin{aligned} | 3 | | x - 2 | & = 9 \\ 3 | x - 2 | & = 9 & Divide both sides by 3. \\ | x - 2 | & = 3 \end{aligned}$

which tells us that the distance between $x$ and $2$ is $3$ units, so the solutions are $x = - 1$ and $x equals 5$ .

Alternatively, we can use graphs when working with absolute values. For example, we know that the simple equation $| x | equals 5$ has two solutions, $x equals 5$ and $x = - 5$ .

In fact, we can see from the graph at right that the equation $| x | equals k$ has two solutions if $k greater than 0$ , one solution if $k equals 0$ , and no solution if $k less than 0$ .

Which statement is true?

_____

Depending on $x$ , $| 2 x minus 8 |$ can equal $2 x minus 8$ or $8 minus 2 x$ .

Which statement is true?

The graph of $y = | 2 x - 8 |$ has no negative inputs.
The equation $| 2 x - 8 | = - 4$ has two solutions.
Depending on $x, | 2 x minus 8 |$ can equal $2 x minus 8$ or $8 minus 2 x$ .
The graph of $y = | 2 x - 8 |$ is a straight line.

Example 2

Use a graph of $y = | 3 x - 6 |$ to solve the equation $| 3 x - 6 | = 9$ .
Use a graph of $y = | 3 x - 6 |$ to solve the equation $| 3 x - 6 | = - 2$ .

The graph shows the graphs of $y = | 3 x - 6 |$ and $y equals 9$ . We see that there are two points on the graph of $y = | 3 x - 6 |$ that have $y equals 9$ , and those points have $x$ -coordinates $x = - 1$ and $x equals 5$ . We can verify algebraically that the solutions are $minus 1$ and $5$ .
$\begin{aligned} x & = - 1 : | 3 (- 1) - 6 | = | - 9 | = 9 \\ x & = 5 : | 3 (5) - 6 | = | 9 | = 9 \end{aligned}$
There are no points on the graph of $y = | 3 x - 6 |$ with $y = - 2$ , so the equation $| 3 x - 6 | = - 2$ has no solutions.

The graph of $y = | 2 x + 7 |$ is
_____
Use your graph to solve the equation $| 2 x + 7 | = 11$ .
$x =$ _____

Separate different values with a comma.

A graph is also shown below for part (b).
$x = - 9$ or $x equals 2$

A graph for part (b):

Graph $y = | 2 x + 7 |$ for $minus 12 less than or equal to x less than or equal to 8$ .
Use your graph to solve the equation $| 2 x + 7 | = 11$ .

$x = - 9$ or $x equals 2$

To solve an absolute value equation algebraically, we use the definition of absolute value.

Example 3

Solve the equation $| 3 x - 6 | = 9$ algebraically.

We write the piecewise definition of $| 3 x minus 6 |$ .

$| 3 x - 6 | = {\begin{cases} 3 x - 6 & if 3 x - 6 \geq 0, or x \geq 2 \\ - (3 x - 6) & if 3 x - 6 < 0, or x < 2 \end{cases}$

Thus, the absolute value equation $| 3 x - 6 | = 9$ is equivalent to two regular equations:

$3 x - 6 = 9 or - (3 x - 6) = 9$

or, by simplifying the second equation,

$3 x - 6 = 9 or 3 x - 6 = - 9$

Solving these two equations gives us the same solutions we found in Example, namely $x equals 5$ and $minus 1$ .

In general, we have the following strategy for solving absolute value equations.

Solve $| 2 x + 7 | = 11$ algebraically.

$x =$ _____

Separate different values with a comma.

$x = - 9$ or $x equals 2$

Solve $| 2 x + 7 | = 11$ algebraically.

The absolute value equation is equivalent to two regular equations:

$2 x + 7 = 11 or 2 x + 7 = - 11$

whose solutions are $x = - 9$ or $x equals 2$

Explain how the equation $| a x + b | = c$ can have two solutions, one solution, or no solutions.

_____

Explain how the equation $| a x + b | = c$ can have two solutions, one solution, or no solutions.

Absolute Value Inequalities

We can also use graphs to solve absolute value inequalities. Look again at the graph of $y = | 3 x - 6 |$ in figure (a) below.

Because of the V-shape of the graph, all points with $y$ -values less than $9$ lie between the two solutions of $| 3 x - 6 | = 9$ , that is, between $minus 1$ and $5$ . Thus, the solutions of the inequality $| 3 x minus 6 | less than 9$ are $minus 1 less than x less than 5$ . (In the Homework Problems, you will be asked to show this algebraically.)

On the other hand, to solve the inequality $| 3 x minus 6 | greater than 9$ , we look for points on the graph with $y$ -values greater than $9$ . In figure (b), we see that these points have $x$ -values outside the interval between $minus 1$ and $5$ . In other words, the solutions of the inequality $| 3 x minus 6 | greater than 9$ are $x less than minus 1$ or $x greater than 5$ .

Thus, we can solve an absolute value inequality by first solving the related equation.

If $| x minus 3 | greater than 5$ , then

_____

If $| x minus 3 | greater than 5$ , then either $x minus 3 greater than 5$ or $x - 3 < - 5$ .

If $| x minus 3 | greater than 5$ , then

either $x minus 3 greater than 5$ or $x - 3 > - 5$ .
$5 > x - 3 > - 5$ .
either $x minus 3 greater than 5$ or $x - 3 < - 5$ .
$- 5 > x - 3 < 5$ .

Example 4

Solve $| 4 x minus 15 | less than 0.01$

First, we solve the equation $| 4 x - 15 | = 0.01$ . There are two cases:

$\begin{aligned} 4 x - 15 & = 0.01 & or & 4 x - 15 & = - 0.01 \\ 4 x & = 15.01 & 4 x & = 14.99 \\ x & = 3.7525 & x & = 3.7475 \end{aligned}$

Because the inequality symbol is $<$ , the solutions of the inequality are between these two values: $3.7475 less than x less than 3.7525$ . In interval notation, the solutions are $open parenthesis 3.7475, 3.7525 close parenthesis$ .

Solve the inequality $| 2 x plus 7 | less than 11$
_____
Solve the inequality $| 2 x plus 7 | greater than 11$
_____

You may use inequality symbols or enter your answers using interval notation. When using interval notation, use "inf" for $∞$ and use "U" (an upper case letter u) for the union symbol $\cup$ .

$open parenthesis minus 9, 2 close parenthesis$ , or $minus 9 less than x less than 2$
$(- \infty, - 9) \cup (2, \infty)$ ; as inequalities, $x less than minus 9$ or $x greater than 2$

Solve the inequality $| 2 x plus 7 | less than 11$
Solve the inequality $| 2 x plus 7 | greater than 11$

$open parenthesis minus 9, 2 close parenthesis$ , or $minus 9 less than x less than 2$
$(- \infty, - 9) \cup (2, \infty)$ ; as inequalities, $x less than minus 9$ or $x greater than 2$

Using the Absolute Value in Modeling

In the next Example, we use the absolute value function to model a problem about distances.

Example 5

Marlene is driving to a new outlet mall on Highway 17. There is a gas station at Marlene's on-ramp, where she buys gas and resets her odometer to zero before getting on the highway. The mall is only 15 miles from Marlene’s on-ramp, but she mistakenly drives past the mall and continues down the highway. Marlene's distance from the mall is a function of how far she has driven on Highway 17.

Make a table of values showing how far Marlene has driven on Highway 17 and how far she is from the mall.
Make a graph of Marlene’s distance from the mall versus the number of miles she has driven on the highway. Which of the basic graphs from Section 2.2 does your graph most resemble?
Find a piecewise defined formula that describes Marlene’s distance from the mall as a function of the distance she has driven on the highway.

Marlene gets closer to the mall for each mile that she has driven on the highway until she has driven 15 miles, and after that she gets farther from the mall.
Miles on highway $0$ $5$ $10$ $15$ $20$ $25$ $30$
Miles from mall $15$ $10$ $5$ $0$ $5$ $10$ $15$
We plot the points in the table to obtain the graph shown below. This graph looks like the absolute value function, except that the vertex is the point $open parenthesis 15, 0 close parenthesis$ instead of the origin.
Let $x$ represent the number of miles on the highway and $f (x)$ the number of miles from the mall. For $x$ -values less than $15$ , the graph is a straight line with slope $minus 1$ and $y$ -intercept at $open parenthesis 0, 15 close parenthesis$ , so its equation is $y = - x + 15$ . Thus,
$f (x) = - x + 15 when 0 \leq x < 15$
On the other hand, when $x greater than or equal to 15$ , the graph of $f$ is a straight line with slope $1$ that passes through the point $open parenthesis 15, 0 close parenthesis$ . The point-slope form of this line is
$y = 0 + 1 (x - 15)$
so $y = x - 15$ . Thus,
$f (x) = x - 15 when x \geq 15$
Combining the two pieces, we obtain
$f (x) = {\begin{cases} - x + 15 & when 0 \leq x < 15 \\ x - 15 & when x \geq 15 \end{cases}$
The graph of $f (x)$ is a part of the graph of $y = | x - 15 |$ . If we think of the highway as a portion of the real number line, with Marlene’s on-ramp located at the origin, then the outlet mall is located at $15$ . Marlene's coordinate as she drives along the highway is $x$ , and the distance from Marlene to the mall is given by $f (x) = | x - 15 |$ .

Use the graph above from the previous example to determine how far Marlene has driven when she is within 5 miles of the mall. Write and solve an absolute value inequality to verify your answer.
Inequality: _____ $<$ _____ Note: Use "abs(x)" to get $| x |$ .
Solution: _____
Write and solve an absolute value inequality to determine how far Marlene has driven when she is at least $10$ miles from the mall.
Inequality: _____ $\geq$ _____ Note: Use "abs(x)" to get $| x |$ .
Solution: _____

You may enter your solutions using inequality symbols or using interval notation. When using inequalities, type in "<=" for $\leq$ , and type in ">=" for $\geq$ . When using interval notation, type "inf" for $∞$ and type "U" (an upper case letter u) for the union symbol $\cup$ .

$| x minus 15 | less than 5$ ; $10 less than x less than 20$
A graph is below.
$| x minus 15 | greater than or equal to 10$ ; $x less than or equal to 5$ or $x greater than or equal to 25$
A graph is below.

Graphs are below.

Use the graph above from the previous example to determine how far Marlene has driven when she is within 5 miles of the mall. Write and solve an absolute value inequality to verify your answer.
Write and solve an absolute value inequality to determine how far Marlene has driven when she is at least $10$ miles from the mall.

$| x minus 15 | less than 5$ ; $10 less than x less than 20$
$| x minus 15 | greater than or equal to 10$ ; $x less than or equal to 5$ or $x greater than or equal to 25$

Measurement Error

If you weigh a sample in chemistry lab, the scale’s digital readout might show $6.0$ grams. But it is unlikely that the sample weighs exactly $6$ grams; there is always some error in measured values.

Because the scale shows the weight as $6.0$ grams, we know that the true weight of the sample must be between $5.95$ grams and $6.05$ grams: If the weight were less than $5.95$ grams, the scale would round down to $5.9$ grams, and if the weight were more than $6.05$ grams, the scale would round up to $6.1$ grams. We should report the mass of the sample as $6 \pm 0.05$ grams, which tells the reader that the error in the measurement is no more than $0.05$ grams.

We can also describe this measurement error, or error tolerance, using an absolute value inequality. Because the measured mass $m$ can be no more than $0.05$ from $6$ , we write

$| m minus 6 | less than or equal to 0.05$

Note that the solution of this inequality is $5.95 less than or equal to m less than or equal to 6.05$ .

Example 6

The specifications for a computer chip state that its thickness in millimeters must satisfy $| t minus 0.023 | less than 0.001$ . What are the acceptable values for the thickness of the chip?
The safe dosage of a new drug is between $250$ and $450$ milligrams, inclusive. Write the safe dosage as an error tolerance involving absolute values.

The error tolerance can also be stated as $t = 0.023 \pm 0.001$ millimeters, so the acceptable values are between $0.022$ and $0.024$ millimeters.
The safe dosage $d$ satisfies $250 less than or equal to d less than or equal to 450$ , as shown below.

The center of this interval is $350$ , and the endpoints are each $100$ units from the center. Thus, the safe values are within $100$ units of $350$ , or
$| d minus 350 | less than or equal to 100$

Which statement is false?

_____

"The solutions of $| 3 x plus 12 | greater than 6$ form a closed interval" is a false statement.

Which statement is false?

The equation $| 3 x + 12 | = 0$ has one solution.
The statement $| T minus 0.1 | less than 0.05$ describes an error tolerance of 0.05 units.
$| m x plus b | less than k$ is equivalent to $- k < m x + b < k$ .
The solutions of $| 3 x plus 12 | greater than 6$ form a closed interval.

The temperature, $T$ , in a laboratory must remain between $9 degrees C$ and $12 degrees C$ .

Write the error tolerance as an absolute value inequality.
Inequality: _____ $<$ _____ Note: Use "abs(x)" to get $| x |$ .
For a special experiment, the temperature in degrees celsius must satisfy $| T minus 6.7 | less than or equal to 0.03$ . Give the interval of possible temperatures.
Interval: _____

$| T minus 10.5 | less than 1.5$
$6.67 less than or equal to T less than or equal to 6.73$

The temperature, $T$ , in a laboratory must remain between $9 degrees C$ and $12 degrees C$ .

Write the error tolerance as an absolute value inequality.
For a special experiment, the temperature in degrees celsius must satisfy $| T minus 6.7 | less than or equal to 0.03$ . Give the interval of possible temperatures.

$| T minus 10.5 | less than 1.5$
$6.67 less than or equal to T less than or equal to 6.73$

What is error tolerance used for, and why do we need it?

_____

What is error tolerance used for, and why do we need it?

Section Summary

Vocabulary

Look up the definitions of new terms in the Glossary.

Absolute value equation
Absolute value inequality
Error tolerance

CONCEPTS

The absolute value is used to model distance: The distance between two points $x$ and $a$ is given by $| x minus a |$ .
Note
Absolute Value Equations
The equation
$| a x + b | = c (c > 0)$
is equivalent to
$a x + b = c or a x + b = - c$
Note
Absolute Value Inequalities
Suppose the solutions of the equation $| a x + b | = c$ are $r$ and $s$ , with $r < s$ . Then
1. The solutions of $| a x plus b | less than c$ are
  $r < x < s$
2. The solutions of $| a x plus b | greater than c$ are
  $x < r or x > s$
The error tolerance $e$ in a measurement $M$ can be expressed as $| x minus M | less than e$ , or as $x = M \pm e$ . Both indicate that $M - e < x < M + e$ .

STUDY QUESTIONS

Write a function that models the distance between $x$ and a fixed point $k$ on the number line.
For what values of $c$ does the equation $| a x + b | = c$ have one solution? No solution?
If you know that the solutions of $| a x plus b | less than c$ are $minus 3 less than x less than 6$ , what are the solutions of $| a x plus b | greater than c$ ?
What is the center of the interval $[ 220, 238 ]$ ?
What is the center of the interval $[ a, b ]$ ?

SKILLS

Practice each skill in the Homework problems listed.

Use absolute value notation to write statements about distance: #1–8
Use graphs to solve absolute value equations and inequalities: #9–12
Solve absolute value equations: #13–24
Solve absolute value inequalities: #25–40
Express error tolerances using absolute value notation: #41–48
Analyze absolute value functions: #49–56
Model problems about distance using the absolute value function: #57–60

Homework 2.5

In Problems 1–8,

Use absolute value notation to write each expression as an equation or an inequality. (It may be helpful to restate each sentence using the word distance.)
Illustrate the solutions on a number line.

$x$ is six units from the origin.

$| x | equals 6$

$a$ is seven units from the origin.

The distance from $p$ to $minus 3$ is five units.

$| p + 3 | = 5$

The distance from $q$ to $minus 7$ is two units.

$t$ is within three units of $6$ .

$| t minus 6 | less than 3$

$w$ is no more than one unit from $minus 5$ .

$b$ is at least $0.5$ unit from $minus 1$ .

$| b plus 1 | greater than or equal to 0.5$

$m$ is more than $0.1$ unit from $8$ .

Graph $y = | x + 3 |$ . Use your graph to solve the following equations and inequalities.

$| x + 3 | = 2$
$| x plus 3 | less than or equal to 4$
$| x plus 3 | greater than 5$

$x = - 5$ or $x = - 1$
$minus 7 less than or equal to x less than or equal to 1$
$x less than minus 8$ or $x greater than 2$

Graph $y = | x - 2 |$ . Use your graph to solve the following equations and inequalities.

$| x - 2 | = 5$
$| x minus 2 | less than 8$
$| x minus 2 | greater than or equal to 4$

Graph $y = | 2 x - 8 |$ . Use your graph to solve the following equations and inequalities.

$| 2 x - 8 | = 0$
$| 2 x - 8 | = - 2$
$| 2 x - 8 | < - 6$

$x equals 4$
No solution
No solution

Graph $y = | 4 x + 8 |$ . Use your graph to solve the following equations and inequalities.

$| 4 x + 8 | = 0$
$| 4 x plus 8 | less than 0$
$| 4 x + 8 | > - 3$

For Problems 13–24, solve.

$| 2 x - 1 | = 4$

$x = \frac{- 3}{2}$ or $x = \frac{5}{2}$

$| 3 x - 1 | = 5$

$0 = | 7 + 3 q |$

$q = \frac{- 7}{3}$

$| - 11 - 5 t | = 0$

$4 = \frac{| b + 2 |}{3}$

$b = - 14$ or $b equals 10$

$6 | n + 2 | = 9$

$| 2 (w - 7) | = 1$

$w = \frac{13}{2}$ or $w = \frac{15}{2}$

$2 = | \frac{a - 4}{5} |$

$| c - 2 | + 3 = 1$

No solution

$5 = 4 - | h + 3 |$

$- 7 = | 2 m + 3 |$

No solution

$| 5 r - 3 | = - 2$

For Problems 25–36, solve.

$| 2 x plus 6 | less than 3$

$\frac{- 9}{2} < x < \frac{- 3}{2}$

$| 5 minus 3 x | less than or equal to 1$

$7 less than or equal to | 3 minus 2 d |$

$d less than or equal to minus 2$ or $d greater than or equal to 5$

$10 less than | 3 r plus 2 |$

$| 6 s + 15 | > - 3$

All real numbers

$| 8 b - 12 | < - 4$

$| t minus 1.5 | less than 0.1$

$1.4 less than t less than 1.6$

$| z minus 2.6 | less than or equal to 0.1$

$| T minus 3.25 | greater than or equal to 0.05$

$T less than or equal to 3.2$ or $T greater than or equal to 3.3$

$| P minus 0.6 | greater than 0.01$

$- 1 \geq | \frac{n - 3}{2} |$

No solution

$- 0.1 \leq | 9 (p + 2) |$

In Problems 37–40, give an interval of possible values for the measurement.

The length, $l$ , of a rod is given by $| l minus 4.3 | less than 0.001$ , in centimeters.

$4.299 less than l less than 4.301$

The mass, $m$ , of the device shall be $| m minus 450 | less than 4$ , in grams.

The candle will burn for $t$ minutes, where $| t minus 300 | less than or equal to 50$ .

$250 less than or equal to t less than or equal to 350$

The ramp will have angle of inclination $α$ , and $| α minus 10 degrees | less than or equal to 0.5 degrees$ .

In Problems 41–44, write the error tolerance using absolute values.

The chemical compound must be maintained at a temperature, $T$ , between $4.7 degrees$ and $5.3 degrees$ C.

$| T minus 5 | less than 0.3$

The diameter, $d$ , of the hole shall be in the range of $24.98$ to $25.02$ centimeters.

The subject will receive a dosage $D$ from $95$ to $105$ milligrams of the drug.

$| D minus 100 | less than or equal to 5$

The pendulum swings out and back in a time period $t$ between $0.9995$ and $1.0005$ seconds.

An electrical component of a high-tech sensor requires $0.25$ ounce of gold. Assume that the actual amount of gold used, $g$ , is not in error by more than $0.001$ ounce. Write an absolute value inequality for the possible error and show the possible values of $g$ on a number line.

$| g minus 0.25 | less than or equal to 0.001$

In a pasteurization process, milk is to be irradiated for $10$ seconds. The actual period $t$ of irradiation cannot be off by more than $0.8$ second. Write an absolute value inequality for the possible error and show the possible values of $t$ on a number line.

In a lab assignment, a student reports that a chemical reaction required $200$ minutes to complete. Let $t$ represent the actual time of the reaction.

Write an absolute value inequality for $t$ , assuming that the student rounded his answer to the nearest $100$ minutes. Give the smallest and largest possible value for $t$ .
Write an absolute value inequality for $t$ , assuming that the student rounded his answer to the nearest minute. Give the smallest and largest possible value for $t$ .
Write an absolute value inequality for $t$ , assuming that the student rounded his answer to the nearest $0.1$ minute. Give the smallest and largest possible value for $t$ .

What is the shortest time that would round to $200$ minutes? The greatest time?

$| t minus 200 | less than 50$ , $150 less than or equal to t less than 250$
$| t minus 200 | less than 0.5$ , $199.5 less than or equal to t less than 200.5$
$| t minus 200 | less than 0.05$ , $199.95 less than or equal to t less than 200.05$

An espresso machine has a square metal plate. The side of the plate is $2 \pm 0.01$ cm.

Write an absolute value inequality for the length of the side, $s$ . Give the smallest and largest possible value for $s$ .
Compute the smallest and largest possible area of the plate, including units.
Write an absolute value inequality for the area, $A$ .

Write the piecewise definition for $| 3 x minus 6 |$ .
Use your answer to part (a) to write two inequalities that together are equivalent to $| 3 x minus 6 | less than 9$ .
Solve the inequalities in part (b) and check that the solutions agree with the solutions of $| 3 x minus 6 | less than 9$ .
Show that $| 3 x minus 6 | less than 9$ is equivalent to the compound inequality $- 9 < 3 x - 6 < 9$ .

$| 3 x - 6 | = {\begin{cases} - (3 x - 6) & if x < 2 \\ 3 x - 6 & if x \geq 2 \end{cases}$
$- (3 x - 6) \leq 9$ , $3 x minus 6 less than 9$
$minus 1 less than x less than 5$
The solutions are the same.

Write the piecewise definition for $| 3 x minus 6 |$ .
Use your answer to part (a) to write two inequalities that together are equivalent to $| 3 x minus 6 | greater than 9$ .
Solve the inequalities in part (b) and check that the solutions agree with the solutions of $| 3 x minus 6 | greater than 9$ .
Show that $| 3 x minus 6 | greater than 9$ is equivalent to the compound inequality $3 x - 6 < - 9$ or $3 x minus 6 greater than 9$ .

Write the piecewise definition for $| 2 x plus 5 |$ .
Use your answer to part (a) to write two inequalities that together are equivalent to $| 2 x plus 5 | greater than 7$ .
Solve the inequalities in part (b) and check that the solutions agree with the solutions of $| 2 x plus 5 | greater than 7$ .
Show that $| 2 x plus 5 | greater than 7$ is equivalent to the compound inequality $2 x + 5 < - 7$ or $2 x plus 5 greater than 7$ .

$| 2 x + 5 | = {\begin{cases} - (2 x + 5) & if x < \frac{- 5}{2} \\ 2 x + 5 & if x \geq \frac{- 5}{2} \end{cases}$
$- (2 x + 5) > 7$ , $2 x plus 5 greater than 7$
$x less than minus 6$ or $x greater than 1$
The solutions are the same.

Write the piecewise definition for $| 2 x plus 5 |$ .
Use your answer to part (a) to write two inequalities that together are equivalent to $| 2 x plus 5 | less than 7$ .
Solve the inequalities in part (b) and check that the solutions agree with the solutions of $| 2 x plus 5 | less than 7$ .
Show that $| 2 x plus 5 | less than 7$ is equivalent to the compound inequality $- 7 < 2 x + 5 < 7$ .

For Problems 53–56, graph the function and answer the questions.

$f (x) = | x + 4 | + | x - 4 |$

Using your graph, write a piecewise formula for $f (x)$ .
Experiment by graphing $g (x) = | x + p | + | x - q |$ for different positive values of $p$ and $q$ . Make a conjecture about how the graph depends on $p$ and $q$ .
Write a piecewise formula for $g (x) = | x + p | + | x - q |$ .

$f (x) = {\begin{cases} - 2 x, & x < - 4 \\ 8, & - 4 \leq x \leq 4 \\ 2 x, & x > 4 \end{cases}$
The graphs looks like like a trough. The middle horizontal section is $y = p + q$ for $- p \leq x \leq q$ , the left side, $x < - p$ , has slope $minus 2$ and the right side, $x > q$ , has slope $2$ .
$g (x) = {\begin{cases} - 2 x + q - p, & x < - p \\ p + q, & - p \leq x \leq q \\ 2 x + p - q, & x > q \end{cases}$

$f (x) = | x + 4 | - | x - 4 |$

Using your graph, write a piecewise formula for $f (x)$ .
Experiment by graphing $g (x) = | x + p | - | x - q |$ for different positive values of $p$ and $q$ . Make a conjecture about how the graph depends on $p$ and $q$ .

$f (x) = | x + 4 | + | x | + | x - 4 |$

Using your graph, write a piecewise formula for $f (x)$ .
What is the minimum value of $f (x)$ ?
If $p, q greater than or equal to 0$ , what is the minimum value of $g (x) = | x + p | + | x | + | x - q |$ ?

$f (x) = {\begin{cases} - 3 x, & x < - 4 \\ - x + 8, & - 4 \leq x \leq 0 \\ x + 8, & 0 < x < 4 \\ 3 x, & x \geq 4 \end{cases}$
$8$
$p + q$

$f (x) = | x + 4 | - | x | + | x - 4 |$

Using your graph, write a piecewise formula for $f (x)$ .
What is the minimum value of $f (x)$ ?
If $p, q greater than or equal to 0$ , what is the minimum value of $g (x) = | x + p | - | x | + | x - q |$ ?

Problems 57–60 use the absolute value function to model distance. Use the strategy outlined in Problems 57 and 58 to solve Problems 59 and 60.

A small pottery is setting up a workshop to produce mugs. Three machines are located on a long table, as shown in the figure. The potter must use each machine once in the course of producing a mug. Let $x$ represent the coordinate of the potter's station.

Write expressions for the distance from the potter's station to each of the machines.
Write a function that gives the sum of the distances from the potter's station to the three machines.
Graph your function for $minus 20 less than or equal to x less than or equal to 30$ . Where should the potter stand in order to minimize the distance she must walk to the machines?

$| x plus 12 |$ , $| x plus 4 |$ , $| x minus 24 |$
$f (x) = | x + 12 | + | x + 4 | + | x - 24 |$
At $x$ -coordinate $minus 4$

Suppose the pottery in Problem 57 adds a fourth machine to the procedure for producing a mug, located at $x equals 16$ in the figure.

Write and graph a new function for the sum of the potter's distances to the four machines.
Where should the potter stand now to minimize the distance she has to walk while producing a mug?

Richard and Marian are moving to Parkville to take jobs after they graduate. The main road through Parkville runs east and west, crossing a river in the center of town. Richard's job is located $10$ miles east of the river on the main road, and Marian's job is $6$ miles west of the river. There is a health club they both like located $2$ miles east of the river. If they plan to visit the health club every workday, where should Richard and Marian look for an apartment to minimize their total daily driving distance?

$2$ miles east of the river

Romina's Bakery has just signed contracts to provide baked goods for three new restaurants located on Route 28 outside of town. The Coffee Stop is $2$ miles north of town center, Sneaky Pete's is $8$ miles north, and the Sea Shell is $12$ miles south. Romina wants to open a branch bakery on Route 28 to handle the new business, with separate days for each restaurant's weekly delivery. Where should she locate the bakery in order to minimize the distance she must drive for deliveries?

Modeling, Functions, and Graphs by Katherine Yoshiwara (yoshiwarabooks.org), GNU Free Documentation License 1.2 or later. Adapted for the XYZ HTML edition with the authors' permission (recorded 2026-07-04). License: GFDL-1.2-or-later.