ORCCAOpen Resources for Community College Algebra

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.

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

\(x\)	\(y\)
\(-3\)	\(-7\)
\(-2\)	\(-5\)
\(-1\)	\(-3\)
\(0\)	\(-1\)
\(1\)	\(1\)
\(2\)	\(3\)
\(3\)	\(5\)

epspdfpngsvgtex

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

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

\(x\)	\(y\)
\(-5\)	\(1\)
\(-4\)	\(0\)
\(-3\)	\(-1\)
\(-2\)	\(0\)
\(-1\)	\(1\)
\(0\)	\(0\)
\(1\)	\(-1\)
\(2\)	\(0\)
\(3\)	\(1\)
\(4\)	\(2\)
\(5\)	\(3\)

epspdfpngsvgtex

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

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

epspdfpngsvgtex

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.

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.

epspdfpngsvgtex

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.

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.

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.