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Section 11.1 Introduction to Absolute Value Functions

This section will introduce the basic concepts behind absolute value functions and their graphs. This information will also be useful at the end of this chapter when we solve absolute value equations and inequalities.

Figure 11.1.1 Alternative Video Lesson

Subsection 11.1.1 Definition of Absolute Value

Recall that in Section 1.3, we defined the absolute value of a number to be the distance between that number and \(0\) on a number line. Also recall that this causes the output of the absolute value function to never be a negative number since we are under the presumption that “distance” is always positive (or zero).

Example 11.1.2
  1. Since the number \(5\) is \(5\) units from \(0\text{,}\) then \(\abs{5}=5\text{.}\)

  2. Since the number \(-3\) is \(3\) units from \(0\text{,}\) then \(\abs{-3}=3\text{.}\)

Example 11.1.3

Yonas takes a \(5\)-block walk north from his home to a food cart. After enjoying dinner, he then walks \(9\) blocks south of the food cart to his favorite movie theater.

  1. How many blocks has Yonas walked in total when he reaches the theater?

  2. How many blocks is Yonas from home when he reaches the theater?

Explanation
  1. Since we only care about total distance, we can ignore the “signs” on the distances walked (either north or south) and simply add the two values together. Mathematically, if we think of north as positive values and south as having negative values, this situation is the same as

    \begin{align*} \abs{5}+\abs{-9}\amp=5+9\\ \amp=14 \end{align*}

    Yonas has walked a total of \(14\) blocks when he reaches the theater.

  2. When he reaches the theater, Yonas's actual position could be thought of as \(5+(-9)\text{.}\) But the actual distance from the theater to his home is better thought of as:

    \begin{align*} \abs{5+(-9)}\amp=\abs{-4}\\ \amp=4 \end{align*}

    Yonas was \(4\) blocks from home when he reached the theater.

Subsection 11.1.2 Evaluating Absolute Value Functions

The formula \(f(x)=\abs{x}\) does satisfy the requirements for \(f\) to be a function because no matter what number you put in for \(x\text{,}\) there is only one measured distance from \(0\) to that value \(x\text{.}\)

Example 11.1.4

Let \(f(x)=\abs{x}\) and \(g(x)=\abs{2x-5}\text{.}\) Evaluate the following expressions.

  1. \(f(34)\)
  2. \(f(-63)\)
  3. \(f(0)\)
  4. \(g(13)\)
  5. \(g(1)\)
Explanation
  1. \(\begin{aligned}[t] f(\highlight{34})\amp=\abs{\highlight{34}}\\ \amp=34 \end{aligned}\)

  2. \(\begin{aligned}[t] f(\highlight{-63})\amp=\abs{\highlight{-63}}\\ \amp=63 \end{aligned}\)

  3. \(\begin{aligned}[t] f(\highlight{0})\amp=\abs{\highlight{0}}\\ \amp=0 \end{aligned}\)

  4. \(\begin{aligned}[t] g(\highlight{13})\amp=\abs{2\cdot\highlight{13}-5}\\ \amp=\abs{21}\\ \amp=21 \end{aligned}\)

  5. \(\begin{aligned}[t] g(\highlight{1})\amp=\abs{2\cdot\highlight{1}-5}\\ \amp=\abs{-3}\\ \amp=3 \end{aligned}\)

Checkpoint 11.1.5

Subsection 11.1.3 Graphs of Absolute Value Functions

Absolute value functions have generally the same shape. They are usually described as “V”-shaped graphs and the tip of the “V” is called the vertex. A few graphs of various absolute value functions are shown in Figure 11.1.6. In general, the domain of an absolute value function (where there is a polynomial inside the absolute value) is \((-\infty,\infty)\text{.}\)

(a)\(y=\abs{x}\)
(b)\(y=-\abs{x+2}\)
(c)\(y=\abs{x-4}-5\)
Figure 11.1.6
Example 11.1.7

Let \(h(x)=-2\abs{x-3}+5\text{.}\) Using technology, create table of values with \(x\)-values from \(-3\) to \(3\text{,}\) using an increment of \(1\text{.}\) Then sketch a graph of \(y=h(x)\text{.}\) State the domain and range of \(h\text{.}\)

Explanation
\(x\) \(y\)
\(-3\) \(-7\)
\(-2\) \(-5\)
\(-1\) \(-3\)
\(0\) \(-1\)
\(1\) \(1\)
\(2\) \(3\)
\(3\) \(5\)
Table 11.1.8 Table for \(y=h(x)\text{.}\)
Figure 11.1.9 Graph of \(y=h(x)\)

The graph indicates that the domain is \((\infty,\infty)\) as it goes to the right and left indefinitely. The range is \((-\infty,5]\text{.}\)

Example 11.1.10

Let \(j(x)=\big\lvert\mathopen{}\abs{x+1}\mathclose{}-2\big\rvert-1\text{.}\) Using technology, create table of values with \(x\)-values from \(-5\) to \(5\text{,}\) using an increment of \(1\) and sketch a graph of \(y=j(x)\text{.}\) State the domain and range of \(j\text{.}\)

Explanation

This is a strange one because it has an absolute value within an absolute value.

\(x\) \(y\)
\(-5\) \(1\)
\(-4\) \(0\)
\(-3\) \(-1\)
\(-2\) \(0\)
\(-1\) \(1\)
\(0\) \(0\)
\(1\) \(-1\)
\(2\) \(0\)
\(3\) \(1\)
\(4\) \(2\)
\(5\) \(3\)
Table 11.1.11 A table of values for \(y=j(x)\text{.}\)
Figure 11.1.12 \(y=\big\lvert\mathopen{}\abs{x+1}\mathclose{}-2\big\rvert-1\)

The graph indicates that the domain is \((\infty,\infty)\) as it goes to the right and left indefinitely. The range is \([-1,\infty)\text{.}\)

Subsection 11.1.4 Another Definition for Absolute Value

How many definitions do we really need? Bear with us because this one is important.

Example 11.1.13

Consider the function \(f\) defined by \(f(x)=\sqrt{x^2}\text{.}\) First, we will evaluate this function at a few arbitrary values: \(3\text{,}\) \(0\text{,}\) and \(-5\text{.}\)

\begin{align*} f(\highlight{3})\amp=\sqrt{\highlight{3}^2}\amp f(\highlight{0})\amp=\sqrt{\highlight{0}^2}\amp f(\highlight{-5})\amp=\sqrt{(\highlight{-5})^2}\\ \amp=\sqrt{9}\amp\amp=\sqrt{0}\amp\amp=\sqrt{25}\\ \amp=3\amp\amp=0\amp\amp=5 \end{align*}

These results should seem familiar: \(f(3)=3\text{,}\) \(f(0)=0\text{,}\) and \(f(-5)=5\text{.}\) The outputs are the same as the inputs, except for a missing negative sign on \(5\text{.}\) It seems like we've seen a function that does that exact same thing already …

Make a quick graph using technology to see what the graph of \(y=f(x)\) looks like.

Explanation

Figure 11.1.14 shows a graph of \(f\) where \(f(x)=\sqrt{x^2}\text{.}\) It looks just like that of \(y=\abs{x}\text{.}\)

Since the graphs of \(y=\sqrt{x^2}\) and \(y=\abs{x}\) match up exactly, that must mean that

\begin{equation*} \abs{x}=\sqrt{x^2} \end{equation*}

This fact will be used later in this chapter and it will continue to pop up in subsequent math courses here and there as important.

Figure 11.1.14 \(y=\sqrt{x^2}\)
Example 11.1.15

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

  1. \(\sqrt{(x-2)^2}\)

  2. \(\sqrt{x^6}\)

  3. \(\sqrt{x^2+10x+25}\)

  4. \(\sqrt{x^4}\)

Explanation
  1. \(\sqrt{(\highlight{x-2})^2}=\abs{\highlight{x-2}}\text{.}\) Note that \(\highlight{x-2}\) might be a negative number depending on the value of \(x\text{,}\) so the absolute value will change those negative numbers to be positive values.

  2. \(\begin{aligned}[t] \sqrt{x^6}\amp=\sqrt{\left(\highlight{x^3}\right)^2}\\ \amp=\abs{\highlight{x^3}} \end{aligned}\)

    We know from exponent rules that \(x^6=\left(x^3\right)^2\text{.}\) Note that \(\highlight{x^3}\) will be negative whenever \(x\) is a negative number, so the absolute value bars must remain.

  3. \(\begin{aligned}[t] \sqrt{x^2+10x+25}\amp=\sqrt{(\highlight{x+5})^2}\\ \amp=\abs{\highlight{x+5}} \end{aligned}\)

    Note again that \(\highlight{x+5}\) can be negative for certain values of \(x\text{,}\) so the absolute value bars must remain.

  4. \(\begin{aligned}[t] \sqrt{x^4}\amp=\sqrt{\left(\highlight{x^2}\right)^2}\\ \amp=\abs{\highlight{x^2}}\\ \amp=x^2 \end{aligned}\)

    Note here that \(\highlight{x^2}\) is never negative. No matter what number you substitute in for \(x\) in \(x^2\text{,}\) you always either get a positive result or zero. So the absolute value around \(x^2\) doesn't have any effect. Absolute values change negative numbers to positive values but leave positive values alone. Thus, it is OK in this case to drop the absolute value bars.

Subsection 11.1.5 Applications Involving Absolute Values

Absolute values are quite useful as models in a variety of real world applications. One example is the path of a billiards (pool) ball: when the ball bounces off one of the side rails, its path is mirrored and creates a “V” shape. The game gets more complicated when more than the rail is hit, but the fundamental mathematics doesn't change: absolute values model the bounces each time.

Here are some more examples. The first one we'll explore involves light reflecting off of a mirror.

Example 11.1.16

When light reflects off of a mirror, the path it takes is in the shape of an absolute value graph. Khenbish was playing with a laser pointer in his bedroom mirror. He set up the laser pointer on his windowsill and the light hit the center of the mirror and reflected onto the corner of his room. He declared that the laser pointer is sitting at the origin, and \(x\) should stand for the horizontal distance from the left wall to the light beam. Shown is a birds-eye view of the situation.

Figure 11.1.17 Birds-Eye View of Khenbish's Room with Laser

After a little bit of work, Khenbish was able to come up with a formula for the light's path:

\begin{equation*} p(x)=5-\frac{5}{4}\abs{x-4} \end{equation*}

where \(p(x)\) stands for the position, in ft, above (for positive values) or below (for negatives) the center line through his room that represents the \(x\)-axis, where \(x\) is also measured in ft. Use technology and a graph of this formula to answer the following questions.

  1. Khenbish's room is 10 ft wide according to Figure 11.1.17 (in the vertical direction in the figure). What is the room's length (in the horizontal direction in the figure)?

  2. How far along the wall is the mirror centered?

  3. If you stood 9 ft from the left wall, how far above or below the room's center line (\(x\)-axis) should you stand to have the laser pointer hit you?

Explanation
Figure 11.1.18 Detailed Birds-Eye View of Khenbish's Room
  1. To find the room's length, first note that since the laser hits the corner of the room, the \(x\) coordinate of the lasers position would tell us the room's width. According to the detailed graph, the \(x\)-coordinate when \(y=-5\) is \(12\text{.}\) So the room must be \(12\) feet wide.

  2. The mirror is centered exactly where the laser hits the wall. This is the vertex of the absolute value graph which, according to the graph, is at the point \((4,5)\text{.}\) This tells us that the mirror is centered \(4\) feet from the left wall.

  3. If you are standing 9 feet from the left wall, the laser's position will be a bit more than one foot behind the rooms center line, by the diagram. While technology can tell us the exact answer, here is how to do this problem algebraically.

    \begin{align*} d(\highlight{9})\amp=5-\frac{5}{4}\abs{\highlight{9}-4}\\ \amp=5-\frac{5}{4}\abs{5}\\ \amp=5-\frac{5}{4}\cdot 5\\ \amp=\frac{20}{4}-\frac{25}{4}\\ \amp=-\frac{5}{4}\\ \amp=-1.25 \end{align*}

    So, it looks like if you stand \(9\) feet from the left wall, you need to stand \(1.25\) feet behind the center line (which would be \(6.25\) feet from the wall with the mirror on it) to be hit by the laser.

Absolute value functions are also used when a value must be within a certain distance or tolerance. For example, a person's body temperature is considered “normal” if it is within \(0.5\) degrees of 98.6 °F, so their temperature could be up to \(0.5\) degrees less than or greater than that temperature. To be within normal range, the difference between the two values must be less than or equal to \(0.5\text{,}\) and it does not matter whether it is positive or negative. We will introduce a function for measuring this in the next example.

Example 11.1.19

The function \(D\) defined by \(D(T)=\abs{T-98.6}\) represents the difference between a person's temperature, T, in Fahrenheit, and 98.6 °F. A person's temperature is considered “normal” if \(D(T)\) is less than or equal to \(0.5\text{.}\) Use \(D(T)\) to determine whether each person's temperature is within the normal range.

  1. LaShonda has a temperature of 98.3 °F.

  2. Castel has a temperature of 99.3 °F.

  3. Daniel has a temperature of 97.3 °F.

Explanation
  1. LaShonda has a temperature of 98.3 °F, so we have:

    \begin{align*} D(\highlight{98.3})\amp=\abs{\highlight{98.3}-98.6}\\ \amp=\abs{-0.3}\\ \amp=0.3 \end{align*}

    Since the value of \(D(98.3)\) is a number smaller than \(0.5\text{,}\) her temperature of 98.3 °F is within the normal range.

  2. If Castel has a temperature of 99.3 °F, then we have:

    \begin{align*} D(\highlight{99.3})\amp=\abs{\highlight{99.3}-98.6}\\ \amp=\abs{0.7}\\ \amp=0.7 \end{align*}

    Since the value of \(D(99.3)\) is a number bigger than \(0.5\text{,}\) their temperature of 99.3 °F is not within the normal range.

  3. Daniel's temperature is 97.3 °F, so we have:

    \begin{align*} D(\highlight{97.3})\amp=\abs{\highlight{97.3}-98.6}\\ \amp=\abs{-1.3}\\ \amp=1.3 \end{align*}

    Since the value of \(D(97.3)\) is a number bigger than \(0.5\text{,}\) his temperature of 97.3 °F is not within the normal range.

Example 11.1.20

The entryway to the Louvre Museum in Paris is through I. M. Pei's metal and glass Louvre Pyramid. This pyramid has a square base and is \(71\) feet high and \(112\) feet wide. The formula \(h(x)=71-\frac{71}{56}\abs{x-56}\) gives the height above ground level of the pyramid at a distance of \(x\) from the left side of the pyramid base.

  1. If you are \(20\) feet from the left edge, how high will the pyramid rise in front of you? Round your result to the nearest tenth of an inch.

  2. How far from the left edge is the center of the pyramid?

  3. Using your previous answer, check that the formula gives you the correct height at the center.

Figure 11.1.21 A Diagram of the Front of the Louvre Pyramid
Explanation
  1. If you are \(20\) feet from the left edge, then \(x\) is \(20\text{.}\) Substituting \(20\) for \(x\) we have

    \begin{align*} h(\substitute{20})\amp=71-\frac{71}{56}\abs{\substitute{20}-56}\\ \amp=71-\frac{71}{56}\abs{-36}\\ \amp=71-\frac{71}{56}\cdot 36\\ \amp\approx 25.4 \end{align*}

    The pyramid is about \(25.4\) feet high at the position \(20\) feet from the left edge.

  2. The center of the pyramid is \(56\) feet from the either edge since it's half of \(112\) feet.

  3. Putting \(x=56\) into the formula for \(h\) gives us

    \begin{align*} h(\substitute{56})\amp=71-\frac{71}{56}\abs{\substitute{56}-56}\\ \amp=71-\frac{71}{56}\abs{0}\\ \amp=71-\frac{71}{56}\cdot 0\\ \amp=71 \end{align*}

    And so the formula does give us the correct maximum height of \(71\) feet at the center of the pyramid.

Subsection 11.1.6 Exercises

Review and Warmup
1

Evaluate the following.

  1. \(\displaystyle{ \left\lvert 3 \right\rvert = }\)

  2. \(\displaystyle{ \left\lvert -4 \right\rvert = }\)

  3. \(\displaystyle{ \left\lvert 0 \right\rvert = }\)

  4. \(\displaystyle{ \left\lvert {18+\left(-8\right)} \right\rvert = }\)

  5. \(\displaystyle{ \left\lvert {-9-\left(-4\right)} \right\rvert = }\)

2

Evaluate the following.

  1. \(\displaystyle{ \left\lvert 4 \right\rvert = }\)

  2. \(\displaystyle{ \left\lvert -8 \right\rvert = }\)

  3. \(\displaystyle{ \left\lvert 0 \right\rvert = }\)

  4. \(\displaystyle{ \left\lvert {12+\left(-1\right)} \right\rvert = }\)

  5. \(\displaystyle{ \left\lvert {-9-\left(-2\right)} \right\rvert = }\)

3

Evaluate the following.

  1. \(\displaystyle{ - \lvert 5-9 \rvert = }\)

  2. \(\displaystyle{ \lvert -5-9 \rvert = }\)

  3. \(\displaystyle{ -3 \lvert 9-5 \rvert = }\)

4

Evaluate the following.

  1. \(\displaystyle{ - \lvert 3-6 \rvert = }\)

  2. \(\displaystyle{ \lvert -3-6 \rvert = }\)

  3. \(\displaystyle{ -3 \lvert 6-3 \rvert = }\)

5

Evaluate the following.

  1. \(\displaystyle{ -3-3\left\lvert 7-1\right\rvert = }\)

  2. \(\displaystyle{ -3-3\left\lvert 1-7\right\rvert = }\)

6

Evaluate the following.

  1. \(\displaystyle{ -2-8\left\lvert 9-4\right\rvert = }\)

  2. \(\displaystyle{ -2-8\left\lvert 4-9\right\rvert = }\)

7

Evaluate the following.

  1. \(\displaystyle{ -1-5\left\lvert 6-3\right\rvert = }\)

  2. \(\displaystyle{ -1-5\left\lvert 3-6\right\rvert = }\)

8

Evaluate the following.

  1. \(\displaystyle{ -10-3\left\lvert 8-2\right\rvert = }\)

  2. \(\displaystyle{ -10-3\left\lvert 2-8\right\rvert = }\)

9

Evaluate the following.

\(\displaystyle{ 1-8\left\lvert 1-2 \right\rvert + 2 = }\)

10

Evaluate the following.

\(\displaystyle{ 2-6\left\lvert 3-8 \right\rvert + 2 = }\)

11

Evaluate the following.

\(\displaystyle{ 4-2\left\lvert -1+(3-5)^{3}\right\rvert = }\)

12

Evaluate the following.

\(\displaystyle{ 5-8\left\lvert -7+(2-5)^{3}\right\rvert = }\)

Function Notation with Absolute Value
13

Given \(H(t) = \left|t - 296\right|\text{,}\) find and simplify \(H(159)\text{.}\)

\(H(159)={}\)

14

Given \(g(r) = \left|r - 261\right|\text{,}\) find and simplify \(g(170)\text{.}\)

\(g(170)={}\)

15

Given \(h(x) = {\left|x+16\right|}\text{,}\) find and simplify \(h(18)\text{.}\)

\(h(18)={}\)

16

Given \(f(x) = {\left|-4x-5\right|}\text{,}\) find and simplify \(f(20)\text{.}\)

\(f(20)={}\)

17

Given \(f(x) = {10-\left|3x-26\right|}\text{,}\) find and simplify \(f(10)\text{.}\)

\(f(10)={}\)

18

Given \(f(r) = {15-\left|4r+14\right|}\text{,}\) find and simplify \(f(11)\text{.}\)

\(f(11)={}\)

19

Given \(g(t) = {t+\left|-2t-7\right|}\text{,}\) find and simplify \(g(12)\text{.}\)

\(g(12)={}\)

20

Given \(K(t) = {t+\left|2t-29\right|}\text{,}\) find and simplify \(K(14)\text{.}\)

\(K(14)={}\)

21

Given \(G(t) = {\left|t^{2}-16\right|}\text{,}\) find and simplify \(G(-5)\text{.}\)

\(G(-5)={}\)

22

Given \(h(t) = {\left|t^{2}-25\right|}\text{,}\) find and simplify \(h(1)\text{.}\)

\(h(1)={}\)

23

Given \(f(t) = {\left|t^{2}-2t-15\right|}\text{,}\) find and simplify \(f(7)\text{.}\)

\(f(7)={}\)

24

Given \(H(t) = {\left|t^{2}-2t-24\right|}\text{,}\) find and simplify \(H(-2)\text{.}\)

\(H(-2)={}\)

Domain
25

Find the domain of \(K\) where \(\displaystyle{K(x)=\lvert {-4x+6} \rvert}\text{.}\)

26

Find the domain of \(f\) where \(\displaystyle{f(x)=\lvert {9x-6} \rvert}\text{.}\)

27

Find the domain of \(f\) where \(\displaystyle{f(x) = 2x - \lvert {2x+3} \rvert}\text{.}\)

28

Find the domain of \(g\) where \(\displaystyle{g(x) = 8x - \lvert {-5x-9} \rvert}\text{.}\)

Tables
29

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

\(x\) \(h(x)\)
30

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

\(x\) \(F(x)\)
31

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

\(x\) \(G(x)\)
32

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

\(x\) \(G(x)\)
33

Make a table of values for the function \(H\) defined by \(H(x)={-\left|2x-3\right|+1}\text{.}\)

\(x\) \(H(x)\)
34

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

\(x\) \(K(x)\)
35

Make a table of values for the function \(f\) defined by \(f(x)=\Big\lvert{-2\!\left|2x-3\right|-1}\Big\rvert\text{.}\)

\(x\) \(f(x)\)
36

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

\(x\) \(f(x)\)
Graphs
37

Graph \(y=f(x)\text{,}\) where \(f(x)=\left\lvert2x-1\right\rvert\text{.}\)

38

Graph \(y=f(x)\text{,}\) where \(f(x)=\left\lvert x-2\right\rvert\text{.}\)

39

Graph \(y=f(x)\text{,}\) where \(f(x)=\left\lvert x^2-2x-1\right\rvert\text{.}\)

40

Graph \(y=f(x)\text{,}\) where \(f(x)=\left\lvert x^2+3x-2\right\rvert\text{.}\)

41

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

42

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

43

Graph \(y=f(x)\text{,}\) where \(f(x)=\big\lvert2\left\lvert3-x\right\rvert-2\big\rvert\text{.}\)

44

Graph \(y=f(x)\text{,}\) where \(f(x)=\big\lvert3 - 2\left\lvert2x - 3\right\rvert\big\rvert\text{.}\)

Absolute Value and Square Roots
45

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

\(\displaystyle{{\sqrt{9z^{2}}}}\)

46

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

\(\displaystyle{{\sqrt{64t^{2}}}}\)

47

Simplify the expression.

\(\displaystyle{{\sqrt{\left(r-43\right)^{2}}}}\)

48

Simplify the expression.

\(\displaystyle{{\sqrt{\left(m-8\right)^{2}}}}\)

49

Simplify the expression.

\(\displaystyle{\sqrt{\left(-13698\right)^2}}\)

50

Simplify the expression.

\(\displaystyle{\sqrt{\left(-15696\right)^2}}\)

51

Simplify the expression.

\(\displaystyle{\sqrt{x^{2}+20x+100}}\)

52

Simplify the expression.

\(\displaystyle{\sqrt{n^{2}+2n+1}}\)

Applications
53

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-4.4\right|+5.5} \end{equation*}

where \(h\) stands for height in feet.

Determine the height when you are:

  1. \({5.9\ {\rm ft}}\) from the edge.

    The height inside a camping tent when you \({5.9\ {\rm ft}}\) from the edge of the tent is

  2. \({3.5\ {\rm ft}}\) from the edge.

    The height inside a camping tent when you \({3.5\ {\rm ft}}\) from the edge of the tent is

54

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-6.6\right|+6.5} \end{equation*}

where \(h\) stands for height in feet.

Determine the height when you are:

  1. \({11.5\ {\rm ft}}\) from the edge.

    The height inside a camping tent when you \({11.5\ {\rm ft}}\) from the edge of the tent is

  2. \({1.2\ {\rm ft}}\) from the edge.

    The height inside a camping tent when you \({1.2\ {\rm ft}}\) from the edge of the tent is

Challenge
55

Write two numbers so that

  • The first number is less than the second number, and

  • The absolute value of the first number is greater than the absolute value of the second number

and