ORCCA Absolute Value Functions Chapter Review

Section 12.4 Absolute Value Functions Chapter Review

Subsection 12.4.1 Introduction to Absolute Value Functions

In Section 12.1 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.

Example 12.4.1. Evaluating Absolute Value Functions.

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

$h(-1)\text{.}$
$h(4)\text{.}$

Explanation

$\displaystyle \begin{aligned}[t] h(\substitute{-1})\amp=\abs{9-4(\substitute{-1})}\\ \amp=\abs{9+4}\\ \amp=\abs{13}\\ \amp=13 \end{aligned}$
$\displaystyle \begin{aligned}[t] h(\substitute{4})\amp=\abs{9-4(\substitute{4})}\\ \amp=\abs{9-16}\\ \amp=\abs{-7}\\ \amp=7 \end{aligned}$

Example 12.4.2. Graphing Absolute Value Functions.

Absolute value functions always make “V” shaped graphs. Use a table of values to make a graph of $y=\abs{2x-6}-4\text{.}$

Explanation

To make a graph of a function, generate a table of values for that function. Then, plot and connect the points.

Table 12.4.3. A table of values for $y=\abs{2x-6}-4$

$x$	$y=\abs{2x-6}-4$
$-1$	$4$
$0$	$2$
$1$	$0$
$2$	$-2$
$2$	$-4$
$2$	$-2$
$2$	$0$

A Cartesian graph with the graph of y=abs{2x-6}-4, which looks like a V with the vertex of the V at the point (3,-4). The right side extends as a line with slope 2 and the left side extends with slope -2. — Figure 12.4.4. A graph of $y=\abs{2x-6}-4$

Example 12.4.5. An 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.

Find and interpret the meaning of $d(0)\text{.}$
Using table of values, make a graph of $y=d(t)\text{.}$
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

$\begin{aligned}[t] d(\substitute{0})\amp=\abs{5(\substitute{0})-300}\\ \amp=\abs{-300}\\ \amp=300 \end{aligned}$ This means that just as Mariam was leaving her school, she was $300$ yards from her home.
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.

Subsection 12.4.2 Compound Inequalities

In Section 12.2 we defined the union of intervals, the intersection of intervals, and what compound inequalities are, as well as how to solve “or” inequalities, “and” inequalities, and double inequalities.

Example 12.4.6. Unions 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.

Figure 12.4.7. A number line sketch of $(-\infty,-1]$ as well as $(2,\infty)$

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.

Figure 12.4.8. A number line sketch of $(-\infty,-1]\cup(2,\infty)$

Example 12.4.9.

Find the intersection of the sets $[-3,5)$ and $[2,8)\text{.}$

Explanation

To find the intersection of the sets $[-3,5)$ and $[2,8)\text{,}$ first we draw a number line with both intervals drawn to visualize the overlap of the sets.

a number line where the numbers -3, 0, 2, 5 and 8 are marked; a thick line is above the x-axis from -3 to 5 with a left bracket at -3 and a parenthesis at 5; another thick line is above the first line from 2 to 8 with a left bracket at 2 and a parenthesis at 8. — Figure 12.4.10. A number line sketch of $[-3,5)$ and $[2,8)$

Recall that the intersection of two sets is the set of the numbers in common to both sets. The lines have numbers in common on a long stretch where they overlap. How would we describe all of the numbers where the two lines overlap? In English, we might say that the lines overlap at every number between $2$ and $5\text{.}$ Note that the number $2$ is in both sets, but since $5$ is not in the set $[-3,5)\text{,}$ we should exclude $5$ from the intersection. The description we have come up with is the same as the interval $[2,5)\text{.}$

a number line where the numbers -3, 0, 2, 5 and 8 are marked; a thick line is above the x-axis from 2 to 5 with a left bracket at 2 and a parenthesis at 5. — Figure 12.4.11. A number line sketch of the intersection of $[-3,5)$ and $[2,8)$

In conclusion,

\begin{equation*} [-3,5)\cap[2,8)=[2,5)\text{.} \end{equation*}

Example 12.4.12. “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*}

Example 12.4.13. Double Inequalities.

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

Explanation

This is a double or 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 double 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 $\left(-2,4\right]\text{.}$

Example 12.4.14. Application 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 $\$300\text{.}$

Subsection 12.4.3 Absolute Value Equations and Inequalities

In Section 12.3 we covered how to solve equations when an absolute value is equal to a number. 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.

Example 12.4.15. Solving an Equation with an Absolute Value.

Solve the absolute value equation $\abs{9-4x}=17$ using Fact 12.3.5.

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 $\left\{-2,\frac{13}{2}\right\}\text{.}$

Example 12.4.16. Solving an Absolute Value Less-Than Inequality.

Solve the absolute value inequality $4\cdot\abs{7-2x}+1\lt 25$ using Fact 12.3.8.

Explanation

The inequality $4\cdot\abs{7-2x}+1\lt 25$ must be simplified into the form that matches Fact 12.3.8, 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 12.3.8 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 $\left(\frac{1}{2},\frac{13}{2}\right)\text{.}$

Example 12.4.17. Solving an Absolute Value Greater-Than Inequality.

To solve the absolute value inequality $\abs{13-\frac{3}{2}x} \ge 15$ using Fact 12.3.12.

Explanation

Using Fact 12.3.12, 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 $\left(-\infty,-\frac{4}{3}\right]\cup\left[\frac{56}{3},\infty\right)\text{.}$

Exercises 12.4.4 Exercises

Introduction to Absolute Value Functions

1.

Evaluate the following.

$\displaystyle{ 1-7\left\lvert 5-7 \right\rvert + 6 = }$

2.

Evaluate the following.

$\displaystyle{ 1-4\left\lvert 2-4 \right\rvert + 6 = }$

3.

Given $G(r) = {19-\left|2r-29\right|}\text{,}$ find and simplify $G(13)\text{.}$

$G(13)={}$

4.

Given $F(x) = {16-\left|-3x+11\right|}\text{,}$ find and simplify $F(14)\text{.}$

$F(14)={}$

5.

Find the domain of $F$ where $\displaystyle{F(x)=\lvert {-4x+1} \rvert}\text{.}$

6.

Find the domain of $G$ where $\displaystyle{G(x)=\lvert {10x+10} \rvert}\text{.}$

7.

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

$x$	$H(x)$

8.

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

$x$	$K(x)$

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.

$\displaystyle{{\sqrt{100c^{2}}}}$

12.

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

$\displaystyle{{\sqrt{49x^{2}}}}$

13.

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

\begin{equation*} h={-1.5\!\left|d-4.6\right|+5} \end{equation*}

where $h$ stands for height in feet.

Determine the height when you are:

${5.8\ {\rm ft}}$ from the edge.

The height inside a camping tent when you are ${5.8\ {\rm ft}}$ from the edge of the tent is .
${4\ {\rm ft}}$ from the edge.

The height inside a camping tent when you are ${4\ {\rm ft}}$ from the edge of the tent is .

14.

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

\begin{equation*} h={-1.1\!\left|d-4.8\right|+5} \end{equation*}

where $h$ stands for height in feet.

Determine the height when you are:

${8.2\ {\rm ft}}$ from the edge.

The height inside a camping tent when you are ${8.2\ {\rm ft}}$ from the edge of the tent is .
${2.8\ {\rm ft}}$ from the edge.

The height inside a camping tent when you are ${2.8\ {\rm ft}}$ from the edge of the tent is .