Section11.4Absolute Value Functions Chapter Review

Subsection11.4.1Introduction to Absolute Value Functions

In Section 11.1 we covered the definition of absolute value, what the graphs of absolute value functions look like, the fact that $\sqrt{x^2}=\abs{x}\text{,}$ and applications of absolute values.

Example11.4.1Evaluating Absolute Value Functions

Given that $h(x)=\abs{9-4x}\text{,}$ evaluate the following expressions.

1. $h(-1)\text{.}$

2. $h(4)\text{.}$

Explanation
1. \begin{aligned}[t] h(\substitute{-1})\amp=\abs{9-4(\substitute{-1})}\\ \amp=\abs{9+4}\\ \amp=\abs{13}\\ \amp=13 \end{aligned}

2. \begin{aligned}[t] h(\substitute{4})\amp=\abs{9-4(\substitute{4})}\\ \amp=\abs{9-16}\\ \amp=\abs{-7}\\ \amp=7 \end{aligned}

Example11.4.2Graphing Absolute Value Functions

Absolute value functions always make “V” shaped graphs. We usually use technology to make graphs to help speed up the process. Use technology to make a graph of $y=\abs{2x-6}-4\text{.}$

Explanation

To make a graph of a function, we often use technology to generate a table of values for that function. Then we use the graph that the technology creates to thoughtfully connect the points.

 $x$ $y=\abs{2x-6}-4$ $-1$ $4$ $0$ $2$ $1$ $0$ $2$ $-2$ $2$ $-4$ $2$ $-2$ $2$ $0$
Example11.4.5The Alternate Definition of Absolute Value: $\abs{x}=\sqrt{x^2}$

Simplify the following expressions using the fact that $\abs{x}=\sqrt{x^2}\text{.}$

1. $\sqrt{x^{14}}$

2. $\sqrt{x^2-12x+36}$

Explanation
1. \begin{aligned}[t] \sqrt{x^{14}}\amp=\sqrt{\left(\highlight{x^7}\right)^2}\\ \amp=\abs{\highlight{x^7}} \end{aligned}

We know from exponent rules that $\left(x^7\right)^2=x^{14}\text{.}$ Note that $\highlight{x^7}$ will be negative whenever $x$ is a negative number, so the absolute value is meaningful.

2. \begin{aligned}[t] \sqrt{x^2-12x+36}\amp=\sqrt{(\highlight{x-6})^2}\\ \amp=\abs{\highlight{x-6}} \end{aligned}

Note that $\highlight{x-6}$ can be negative for certain values of $x\text{,}$ so the absolute value is meaningful.

Example11.4.6An Application of Absolute Value

Mariam arrived at school one day only to realize that she had left her favorite pencil on her porch at home. She hopped on her bicycle and headed back to get it. Her distance from her home, $d(t)$ in yards, can be modeled as a function of the time, $t$ in seconds, since she left school:

\begin{equation*} d(t)=\abs{5t-300} \end{equation*}

Use this function to answer the following questions.

1. Find and interpret the meaning of $d(0)\text{.}$

2. Using technology, make a graph of $y=d(t)\text{.}$

3. Using your graph, find out how long it took Mariam to get to her home to get her pencil and get back to school.

Explanation
1. \begin{aligned}[t] d(\substitute{0})\amp=\abs{5(\substitute{0})-300}\\ \amp=\abs{-300}\\ \amp=300 \end{aligned}

This means that just as Mariam was leaving her school, she was $300$ yards from her home.

2. Mariam was back at a $y$-value of $300$ at $t=120\text{.}$ We should assume that she is back at her school again here. So it took her $120$ seconds, which is $2$ minutes.

Subsection11.4.2Compound Inequalities

In Section 11.2 we defined the union of intervals, what compound inequalities are, and how to solve both “or” inequalities and triple inequalities.

Example11.4.7Unions of Intervals

Draw a representation of the union of the sets $(-\infty,-1]$ and $(2,\infty)\text{.}$

Explanation

First we make a number line with both intervals drawn to understand what both sets mean.

The two intervals should be viewed as a single object when stating the union, so here is the picture of the union. It looks the same, but now it is a graph of a single set.

Example11.4.10“Or” Compound Inequalities

Solve the compound inequality.

\begin{equation*} 5z+12\le 7\text{ or } 3-9z\lt -2 \end{equation*}
Explanation

First we will solve each inequality for $z\text{.}$

\begin{align*} 5z+12\amp\le 7\amp\text{ or } \amp\amp3-9z\amp\lt -2\\ 5z\amp\le -5\amp\text{ or } \amp\amp-9z\amp\lt -5\\ z\amp\le -1\amp\text{ or } \amp\amp z\amp\gt \frac{5}{9} \end{align*}

The solution set to the compound inequality is:

\begin{equation*} \left(-\infty,-1\right]\cup\left(\frac{5}{9},\infty\right) \end{equation*}
Example11.4.11Three-Part Inequalities

Solve the three-part inequality $-4\le 20-6x\lt 32\text{.}$

Explanation

This is a three-part inequality. The goal is to isolate $x$ in the middle and whatever you do to one “side,” you have to do to the other two “sides.”

\begin{align*} -4\amp\le 20-6x\lt 32\\ -4\subtractright{20}\amp\le 20-6x\subtractright{20}\lt 32\subtractright{20}\\ -24\amp\le -6x\lt 12\\ \divideunder{-24}{-6}\amp\mathbin{\highlight{\ge}} \divideunder{-6x}{-6}\mathbin{\highlight{\gt}} \divideunder{12}{-6}\\ 4\amp\ge x\gt -2 \end{align*}

The solutions to the three-part inequality $4\ge x\gt -2$ are those numbers that are trapped between $-2$ and $4\text{,}$ including $4$ but not $-2\text{.}$ The solution set in interval notation is $\left(-2,4\right]\text{.}$

Example11.4.12Solving Compound Inequalities Graphically

Figure 11.4.13 shows a graph of $y=G(x)\text{.}$ Use the graph do the following.

1. Solve $G(x)\lt-2\text{.}$

2. Solve $G(x)\ge1\text{.}$

3. Solve $-1\le G(x)\lt1\text{.}$

Explanation
1. To solve $G(x)\lt-2\text{,}$ we first draw a dotted line (since it's a less-than, not a less-than-or-equal) at $y=-2\text{.}$ Then we examine the graph to find out where the graph of $y=G(x)$ is underneath the line $y=-2\text{.}$ Our graph is below the line $y=-2$ for $x$-values less than $-5\text{.}$ So the solution set is $(-\infty,-5)\text{.}$

2. To solve $G(x)\ge1\text{,}$ we first draw a solid line (since it's a greater-than-or-equal) at $y=1\text{.}$ Then we examine the graph to find out what parts of the graph of $y=G(x)$ are above the line $y=1\text{.}$ Our graph is above (or on) the line $y=1$ for $x$-values between $-2$ and $0$ as well as $x$-values bigger than $4\text{.}$ So the solution set is $[-2,0]\cup[4,\infty)\text{.}$

3. To solve $-1\lt G(x)\le1\text{,}$ we first draw a solid line at $y=1$ and dotted line at $y=-1\text{.}$ Then we examine the graph to find out what parts of the graph of $y=G(x)$ are trapped between the two lines we just drew. Our graph is between those values for $x$-values between $-4$ and $-2$ as well as $x$-values between $0$ and $2$ as well as as well as $x$-values between $2$ and $4\text{.}$ We use the solid and hollow dots on the graph to decide whether or not to include those values. So the solution set is $(-4,-2]\cup[0,2)\cup(2,4]\text{.}$

Example11.4.17Application of Compound Inequalities

Mishel wanted to buy some mulch for their spring garden. Each cubic yard of mulch cost $\27$ and delivery for any size load was $\40\text{.}$ If they wanted to spend between $\200$ and $\300\text{,}$ set up and solve a compound inequality to solve for the number of cubic yards, $x\text{,}$ that they could buy.

Explanation

Since the mulch costs $\27$ per cubic yard and delivery is $\40\text{,}$ the formula for the cost of $x$ yards of mulch is $27x+40\text{.}$ Since Mishel wants to spend between $\200$ and $\300\text{,}$ we just trap their cost between these two values.

\begin{align*} 200\amp\lt27x+40\lt300\\ 200\subtractright{40}\amp\lt27x+40\subtractright{40}\lt300\subtractright{40}\\ 160\amp\lt27x\lt260\\ \divideunder{160}{27}\amp\lt\divideunder{27x}{27}\lt\divideunder{260}{27}\\ 5.93\amp\lt x\lt9.63\\ \amp\text{Note: these values are approximate} \end{align*}

Most companies will only sell whole number cubic yards of mulch, so we have to round appropriately. Since Mishel wants to spend more than $\200\text{,}$ we have to round our lower value from $5.93$ up to $6$ cubic yards.

If we round the $9.63$ up to $10\text{,}$ then the total cost will be $27\cdot10+40=310$ (which represents $\310$), which is more than Mishel wanted to spend. So we actually have to round down to $9$cubic yards to stay below the $\300$ maximum.

In conclusion, Mishel could buy $6\text{,}$ $7\text{,}$ $8\text{,}$ or $9$ cubic yards of mulch to stay between $\200$ and $\300\text{.}$

Subsection11.4.3Absolute Value Equations and Inequalities

In Section 11.3 we covered how to solve equations when an absolute value is equal to a number and when an absolute value is equal to an absolute value. We also covered how to solve inequalities when an absolute value is less than a number and when an absolute value is greater than a number.

Example11.4.18Solving an Equation with an Absolute Value

Solve the absolute value equation $\abs{9-4x}=17$ using Fact 11.3.6.

Explanation

The equation $\abs{9-4x}=17$ breaks into two pieces, each of which needs to be solved independently.

\begin{align*} 9-4x\amp=17\amp\amp\text{or}\amp 9-4x\amp=-17\\ -4x\amp=8\amp\amp\text{or}\amp -4x\amp=-26\\ \divideunder{-4x}{-4}\amp=\divideunder{8}{-4}\amp\amp\text{or}\amp \divideunder{-4x}{-4}\amp=\divideunder{-26}{-4}\\ x\amp=-2\amp\amp\text{or}\amp x\amp=\frac{13}{2} \end{align*}

The solution set is $\left\{-2,\frac{13}{2}\right\}\text{.}$

Example11.4.19Solving an Equation with Two Absolute Values

Solve the absolute value equation $\abs{7-3x}=\abs{6x-5}$ using Fact 11.3.11.

Explanation

The equation $\abs{7-3x}=\abs{6x-5}$ breaks into two pieces, each of which needs to be solved independently.

\begin{align*} 7-3x\amp=6x-5\amp\amp\text{or}\amp 7-3x\amp=-(6x-5)\\ 7-3x\amp=6x-5\amp\amp\text{or}\amp 7-3x\amp=-6x+5\\ 12-3x\amp=6x\amp\amp\text{or}\amp 2-3x\amp=-6x\\ 12\amp=9x\amp\amp\text{or}\amp 2\amp=-3x\\ \divideunder{12}{9}\amp=\divideunder{9x}{9}\amp\amp\text{or}\amp \divideunder{2}{-3}\amp=\divideunder{-3x}{-3}\\ \frac{4}{3}\amp=x\amp\amp\text{or}\amp -\frac{2}{3}\amp=x \end{align*}

The solution set is $\left\{\frac{4}{3},-\frac{2}{3}\right\}\text{.}$

Example11.4.20Solving an Absolute Value Less-Than Inequality

Solve the absolute value inequality $4\cdot\abs{7-2x}+1\lt 25$ using Fact 11.3.19.

Explanation

The inequality $4\cdot\abs{7-2x}+1\lt 25$ must be simplified into the form that matches Fact 11.3.19, so we will first isolate the absolute value expression on the left side of the equation:

\begin{align*} 4\cdot\abs{7-2x}+1\amp\lt 25\\ 4\cdot\abs{7-2x}\amp\lt 24\\ \abs{7-2x}\amp\lt 6 \end{align*}

Now that we have the absolute value isolated, we can us Fact 11.3.19 to split it into a triple inequality that we can finish solving:

\begin{align*} -6 \amp \lt 7-2x \lt 6\\ -6\subtractright{7} \amp \lt 7-2x\subtractright{7} \lt 6\subtractright{7}\\ -13 \amp \lt -2x \lt -1\\ \divideunder{-13}{-2} \amp \mathbin{\highlight{\gt}} \divideunder{-2x}{-2} \mathbin{\highlight{\gt}} \divideunder{-1}{-2}\\ \frac{13}{2} \amp \gt x \gt \frac{1}{2} \end{align*}

So, the solution set to the inequality is $\left(\frac{1}{2},\frac{13}{2}\right)\text{.}$

Example11.4.21Solving an Absolute Value Greater-Than Inequality

To solve the absolute value inequality $\abs{13-\frac{3}{2}x} \ge 15$ using Fact 11.3.27.

Explanation

Using Fact 11.3.27, the inequality $\abs{13-\frac{3}{2}x} \ge 15$ breaks down into a compound inequality:

\begin{align*} 13-\frac{3}{2}x \amp \le -15 \amp\amp\text{or}\amp 13-\frac{3}{2}x \amp \ge 15\\ -\frac{3}{2}x \amp \le -28 \amp\amp\text{or}\amp -\frac{3}{2}x \amp \ge 2\\ \multiplyleft{-\frac{2}{3}}\left(-\frac{3}{2}x\right) \amp \mathbin{\highlight{\ge}} \multiplyleft{-\frac{2}{3}}(-28) \amp\amp\text{or}\amp \multiplyleft{-\frac{2}{3}}\left(-\frac{3}{2}x\right) \amp \mathbin{\highlight{\le}} \multiplyleft{-\frac{2}{3}}(2)\\ x \amp \ge \frac{56}{3} \amp\amp\text{or}\amp x \amp \le -\frac{4}{3} \end{align*}

We will write the solution set as $\left(-\infty,-\frac{4}{3}\right]\cup\left[\frac{56}{3},\infty\right)\text{.}$

Subsection11.4.4Exercises

1

Evaluate the following.

$\displaystyle{ 3-5\left\lvert 3-7 \right\rvert + 1 = }$

2

Evaluate the following.

$\displaystyle{ 4-9\left\lvert 1-3 \right\rvert + 1 = }$

3

Given $f(x) = {20-\left|-x+5\right|}\text{,}$ find and simplify $f(18)\text{.}$

$f(18)={}$

4

Given $f(r) = {17-\left|3r-16\right|}\text{,}$ find and simplify $f(19)\text{.}$

$f(19)={}$

5

Find the domain of $K$ where $\displaystyle{K(x)=\lvert {8x-5} \rvert}\text{.}$

6

Find the domain of $f$ where $\displaystyle{f(x)=\lvert {x+4} \rvert}\text{.}$

7

Make a table of values for the function $g$ defined by $g(x)={\left|2x-3\right|}\text{.}$

 $x$ $g(x)$
8

Make a table of values for the function $h$ defined by $h(x)={\left|-2x+1\right|}\text{.}$

 $x$ $h(x)$
9

Graph $y=f(x)\text{,}$ where $f(x)=\frac{1}{2}\left\lvert 4x-5\right\rvert-3\text{.}$

10

Graph $y=f(x)\text{,}$ where $f(x)=\frac{3}{4}\left\lvert 6+x\right\rvert+2\text{.}$

11

Simplify the expression. Do not assume the variables take only positive values.

$\displaystyle{{\sqrt{36r^{2}}}}$

12

Simplify the expression. Do not assume the variables take only positive values.

$\displaystyle{{\sqrt{9m^{2}}}}$

13

Simplify the expression.

$\displaystyle{\sqrt{a^{2}+14a+49}}$

14

Simplify the expression.

$\displaystyle{\sqrt{r^{2}+16r+64}}$

15

The height inside a camping tent when you are $d$ feet from the edge of the tent is given by

\begin{equation*} h={-0.7\!\left|d-6.6\right|+7} \end{equation*}

where $h$ stands for height in feet.

Determine the height when you are:

1. ${7.6\ {\rm ft}}$ from the edge.

The height inside a camping tent when you ${7.6\ {\rm ft}}$ from the edge of the tent is

2. ${3.1\ {\rm ft}}$ from the edge.

The height inside a camping tent when you ${3.1\ {\rm ft}}$ from the edge of the tent is

16

The height inside a camping tent when you are $d$ feet from the edge of the tent is given by

\begin{equation*} h={-0.5\!\left|d-6\right|+4.5} \end{equation*}

where $h$ stands for height in feet.

Determine the height when you are:

1. ${8.3\ {\rm ft}}$ from the edge.

The height inside a camping tent when you ${8.3\ {\rm ft}}$ from the edge of the tent is

2. ${1.3\ {\rm ft}}$ from the edge.

The height inside a camping tent when you ${1.3\ {\rm ft}}$ from the edge of the tent is

17

Solve the compound inequality algebraically.

$\displaystyle{-4x-3\geq5 \quad \text{and} \quad -14x-7\geq-5}$

18

Solve the compound inequality algebraically.

$\displaystyle{-18x+14\geq-6 \quad \text{and} \quad 6x+4>9}$

19

Solve the compound inequality algebraically.

$\displaystyle{9x-10\geq-17 \quad \text{and} \quad -15x+7\geq-15}$

20

Solve the compound inequality algebraically.

$\displaystyle{-6x+7\geq12 \quad \text{and} \quad 6x+11\leq2}$

21

Solve the compound inequality algebraically.

$\displaystyle{7x+15>11 \quad \text{or} \quad x+15\lt -20}$

22

Solve the compound inequality algebraically.

$\displaystyle{12x-2\geq10 \quad \text{or} \quad 6x+6\lt 6}$

23

Solve the compound inequality algebraically.

$\displaystyle{13x+3\lt -20 \quad \text{or} \quad 10x+1\geq8}$

24

Solve the compound inequality algebraically.

$\displaystyle{19x-6\geq8 \quad \text{or} \quad 5x-16\leq-14}$

25

A graph of $f$ is given. Use the graph alone to solve the compound inequalities.

1. $f(x)\gt -4$

2. $f(x)\leq -4$

26

A graph of $f$ is given. Use the graph alone to solve the compound inequalities.

1. $f(x)\gt 0$

2. $f(x)\leq 0$

27

A graph of $f$ is given. Use the graph alone to solve the compound inequalities.

1. $f(x)\gt 2$

2. $f(x)\leq 2$

28

A graph of $f$ is given. Use the graph alone to solve the compound inequalities.

1. $f(x)\gt -2$

2. $f(x)\leq -2$

29

Solve the following equation.

$\displaystyle{ \left\lvert 5 x - 6 \right\rvert = 5 }$

30

Solve the following equation.

$\displaystyle{ \left\lvert 6 x+ 8 \right\rvert = 9 }$

31

Solve the equation $\left\lvert 4 x - 2\right\rvert =18\text{.}$

32

Solve the equation $\left\lvert 4 x - 3\right\rvert =17\text{.}$

33

Solve: $\displaystyle \left\lvert\frac{2 y - 7}{3}\right\rvert = 1$

34

Solve: $\displaystyle \left\lvert\frac{2 y - 3}{7}\right\rvert = 3$

35

Solve: $\left\lvert\frac{1}{2}a + 7\right\rvert = 3$

36

Solve: $\left\lvert\frac{1}{4}a + 5\right\rvert = 3$

37

Solve: $\left\lvert b + 5\right\rvert - 8 = 2$

38

Solve: $\left\lvert t + 1\right\rvert - 2 = 6$

39

Solve: $\left\lvert5 t - 20\right\rvert + 5 = 5$

40

Solve: $\left\lvert3 x - 3\right\rvert + 3 = 3$

41

The equation $\lvert x\rvert =\lvert y\rvert$ is satisfied if $x=y$ or $x=-y\text{.}$ Use this fact to solve the following equation.

$\displaystyle{\left\lvert x + 6 \right\rvert = \left\lvert x - 5 \right\rvert}$

42

The equation $\lvert x\rvert =\lvert y\rvert$ is satisfied if $x=y$ or $x=-y\text{.}$ Use this fact to solve the following equation.

$\displaystyle{\left\lvert x + 6 \right\rvert = \left\lvert x - 1 \right\rvert}$

43

Solve the equation: $\displaystyle{ \left\lvert 8 x - 4 \right\rvert = \left\lvert 5 x + 5\right\rvert }$

44

Solve the equation: $\displaystyle{ \left\lvert 2 x - 2 \right\rvert = \left\lvert 7 x + 3\right\rvert }$

45

Solve the following equation.

$\displaystyle{\left\lvert {x+4} \right\rvert = \left\lvert6 x +8\right\rvert }$

46

Solve the following equation.

$\displaystyle{\left\lvert {2x-4} \right\rvert = \left\lvert8 x +2\right\rvert }$

47

Solve the inequality.

$\displaystyle{ {\left|10-4x\right|} \geq 15 }$

48

Solve the inequality.

$\displaystyle{ {\left|7-5x\right|} \geq 7 }$

49

Solve the inequality.

$\displaystyle{ {\left|5x-4\right|} \lt 12 }$

50

Solve the inequality.

$\displaystyle{ {\left|6x-10\right|} \lt 5 }$