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Math 370 Learning Objectives
Most of the topics in this section are provided for you to review on your own, if needed; however, the following two should be covered in lecture.
- True Definition of the Absolute Value: Understand and use the piecewise definition of the absolute value.
- Graphing Absolute Value Functions: Graph transformations of absolute value functions.
True Definition of the Absolute Value
There are a few ways to describe what is meant by the absolute value \(|x|\) of a real number \(x\). You may have been taught that \(|x|\) is the distance from the real number \(x\) to \(0\) on the number line. So, for example, \(|5| = 5\) and \(|-5| = 5\), since each is \(5\) units from \(0\) on the number line.
Another way to define absolute value is by the equation \(|x| = \sqrt{x^2}\). Using this definition, we have \(|5| = \sqrt{(5)^2} = \sqrt{25} = 5\) and \(|-5| = \sqrt{(-5)^2} = \sqrt{25} = 5\). The long and short of both of these procedures is that \(|x|\) takes negative real numbers and assigns them to their positive counterparts while it leaves positive numbers alone. This last description is the one we shall adopt, and is summarized in the following definition.
Definition: Absolute Value
The absolute value of a real number \(x\), denoted \(|x|\), is given by
\[ |x| = \left\{ \begin{array}{rcl}
x, & \text{ if } & x \geq 0 \\
-x, & \text{ if } & x \lt 0 \\
\end{array} \right. \label{AbsValDefn} \]
Note that we define \(|x|\) using a piecewise-defined function(see Section 1.4). To check that this definition agrees with what we previously understood as absolute value, note that since \(5 \geq 0\), to find \(|5|\) we use the rule \(|x| = x\), so \(|5|=5\). Similarly, since \(-5 < 0\), we use the rule \(|x| = -x\), so that \(|-5| = -(-5) = 5\). This is one of the times when it’s best to interpret the expression"\(-x\)"as "the opposite of \(x\)" as opposed to "negative \(x\)." Before we begin studying absolute value functions, we remind ourselves of the properties of absolute value.
Properties of the Absolute Value
Theorem \(\PageIndex{1}\): Properties of the Absolute Value
Let \(a\), \(b\), and \(x\) be real numbers and let \(n\) be an integer.1Then
Product Rule for Absolute Values2: \(|ab|= |a||b|\)
Power Rule for Absolute Values: \(\left| a^{n} \right| = |a|^{n}\) whenever \(a^{n}\) is defined
Quotient Rule for Absolute Values: \(\left| \dfrac{a}{b} \right| = \dfrac{|a|}{|b|}\), provided \(b \neq 0\)
Equality Properties for Absolute Values:
\(|x| = 0\) if and only if \(x = 0\).
For \(c > 0\), \(|x| = c\) if and only if \(x = c\) or \(-x = c\).
For \(c < 0\), \(|x| = c\) has no solution.
- Proof of the Product Rule
- To prove the Product Rule, we need to show that theequation \( |ab| = |a| |b| \) is true for all real numbers \(a\) and \(b\).
If \(a\) and \(b\) are both positive, then so is \(ab\). Hence, \(|a| = a\), \(|b| = b\) and \(|ab| = ab\). Hence, the equation \(|ab| = |a||b|\) is the same as \(ab = ab\) which is true.
See Also1.9.2: Absolute ValueIf both \(a\) and \(b\) are negative, then \(ab\) is positive. Hence, \(|a| = -a\), \(|b| = -b\) and \(|ab| = ab\). The equation \(|ab| = |a||b|\) becomes \(ab = (-a)(-b)\), which is true.
Suppose \(a\) is positive and \(b\) is negative. Then \(ab\) is negative, and we have \(|ab| = -ab\), \(|a| = a\) and \(|b| = -b\). The equation \(|ab| = |a||b|\) reduces to \(-ab = a(-b)\) which is true. A symmetric argument shows the equation \(|ab| = |a||b|\) holds when \(a\) is negative and \(b\) is positive.
Finally, if either \(a\) or \(b\) (or both) are zero, then both sides of \(|ab| = |a||b|\) are zero, so the equation holds in this case, too.
Thus, the equation \(|ab| = |a||b|\) holds true in all cases.
The proof of the Quotient Rule is very similar to the proof of the Product Rule, with the exception that \(b \neq 0\). The Power Rule can be shown by repeated application of the Product Rule. The Equality Propertiescan be proved using Equation \( \ref{AbsValDefn}\) and by looking at the cases when \(x \geq 0\), in which case \(|x| = x\), or when \(x < 0\), in which case \(|x| = -x\). For example, if \(c > 0\), and \(|x| = c\), then if \(x \geq 0\), we have \(x = |x| = c\). If, on the other hand, \(x < 0\), then \(-x = |x| = c\), so \(x = -c\). The remaining properties are proved similarly and are left for the Exercises.
Subsection Footnotes
1See Section 1.1 if you don’t remember what an integer is.
2Some texts will call this the Product Rule without saying "for Absolute Values"; however, it turns out that we havemanydifferent "Product Rules" in Mathematics. I will try my best to name them accordingly to make them distinct.
Solving Absolute Value Equations
Our first example reviews how to solve basic equations involving absolute value using the properties listed in Theorem \( \PageIndex{1} \).
Example \( \PageIndex{1} \)
- \(|3x-1| = 6\)
- \(3 - |x+5| = 1\)
- \(3|2x+1| - 5 = 0\)
- \(4 - |5x+3| = 5\)
- \(|x| = x^2-6\)
- \(|x-2| + 1 = x\)
Solution
- The equation \(|3x-1| = 6\) is of the form \(|x| = c\) for \(c > 0\), so by the Equality Properties, \(|3x - 1| = 6\) is equivalent to \(3x - 1=6\) or \(3x - 1= -6\). Solving the former, we arrive at \(x = \frac{7}{3}\), and solving the latter, we get \(x = -\frac{5}{3}\). We may check both of these solutions by substituting them into the original equation and showing that the arithmetic works out.
- To use the Equality Properties to solve \(3 - |x+5| = 1\), we first isolate the absolute value.\[ \begin{array}{rcl}
3 - |x+5| & = & 1 \\
-|x+5| & = & -2 \\
|x+5| & = & 2 \\
\end{array} \nonumber \]Hence, \( x + 5 = 2 \) or \( x + 5 = -2 \). In the former case, we get \( x = -3 \), and in the latter case we get \( x = -7 \). - As in the previous example, we first isolate the absolute value in the equation \(3|2x+1| - 5 = 0\) and get \(|2x+1| = \frac{5}{3}\). Using the Equality Properties, we have \(2x+1 = \frac{5}{3}\) or \(2x+1 = -\frac{5}{3}\). Solving the former gives \(x = \frac{1}{3}\) and solving the latter gives \(x = -\frac{4}{3}\). As usual, we may substitute both answers in the original equation to check.
- Upon isolating the absolute value in the equation \(4 - |5x+3| = 5\), we get \(|5x+3| = -1\). At this point, we know there cannot be any real solution, since, by definition, the absolute value of anything is never negative. We are done.
- The equation \(|x| = x^2-6\) presents us with some difficulty, since \(x\) appears both inside and outside of the absolute value. Moreover, there are values of \(x\) for which \(x^2-6\) is positive, negative and zero, so we cannot use the Equality Properties without the risk of introducing extraneous solutions, or worse, losing solutions. For this reason, we break equations like this into cases by rewriting the term in absolute values, \(|x|\), using Equation \( \ref{AbsValDefn} \).
For \(x < 0\), \(|x| = -x\), so for \(x < 0\), the equation \(|x| = x^2-6\) is equivalent to \(-x = x^2-6\). Rearranging this gives us \(x^2+x-6 = 0\), or \((x+3)(x-2) = 0\). We get \(x = -3\) or \(x=2\). Since only \(x=-3\) satisfies \(x<0\), this is the answer we keep.
For \(x \geq 0\), \(|x| = x\), so the equation \(|x| = x^2-6\) becomes \(x = x^2-6\). From this, we get \(x^2-x-6 =0\) or \((x-3)(x+2) = 0\). Our solutions are \(x=3\) or \(x = -2\), and since only \(x=3\) satisfies \(x \geq 0\), this is the one we keep. Hence, our two solutions to \(|x| = x^2-6\) are \(x=-3\) and \(x=3\).
- To solve \(|x-2| + 1 = x\), we first isolate the absolute value and get \(|x-2| = x-1\). Since we see \(x\) both inside and outside of the absolute value, we break the equation into cases. The term with absolute values here is \(|x-2|\), so we replace "\(x\)" with the quantity "\((x-2)\)" in Equation \( \ref{AbsValDefn} \) to get \[ |x-2| = \left\{ \begin{array}{rclcl}
x-2, & \text{ if } & (x-2) \geq 0 & \implies & x \geq 2 \\
-(x-2), & \text{ if } & (x-2) \lt 0 & \implies & x \lt 2 \\
\end{array} \right. \nonumber \]So, for \(x < 2\), \(|x - 2| = -x + 2\) and our equation \(|x - 2| = x - 1\) becomes \(-x + 2 = x - 1\), which gives \(x = \frac{3}{2}\). Since this solution satisfies \(x < 2\), we keep it.Next, for \(x \geq 2\), \(|x - 2| = x - 2\), so the equation \(|x - 2| = x - 1\) becomes \(x - 2 = x - 1\). Here, the equation reduces to \(-2 = -1\), which signifies we have no solutions here. Hence, our only solution is \(x = \frac{3}{2}\).
Graphing Absolute Value Functions
Next, we turn our attention to graphing absolute value functions. Our strategy in the next example is to make liberal use of Equation \( \ref{AbsValDefn} \) along with what we know about graphing linear functions (from Section 2.1) and piecewise-defined functions (from Section 1.4).
Example \( \PageIndex{2} \)
Graph each of the following functions.Find the zeros of each function and the \(x\)- and \(y\)-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute extrema, if they exist.
- \(f(x) = |x|\)
- \(g(x) = |x-3|\)
- \(h(x) = |x| -3\)
- \(i(x) = 4 - 2|3x+1|\)
Solution
- To find the zeros of \(f\), we set \(f(x)= 0\). We get \(|x|=0\), which, by Theorem \( \PageIndex{1} \) gives us \(x=0\). Since the zeros of \(f\) are the \(x\)-coordinates of the \(x\)-intercepts of the graph of \(y=f(x)\), we get \((0,0)\) as our only \(x\)-intercept.
To find the \(y\)-intercept, we set \(x=0\), and find \(y = f(0) = 0\), so that \((0,0)\) is our \(y\)-intercept as well.3Using Equation \( \ref{AbsValDefn}\), we get \[f(x) = |x| = \left\{ \begin{array}{rcl}
x, & \text{ if } & x \geq 0 \\
x, & \text{ if } & x \lt0 \\
\end{array} \right. \nonumber\]Hence, for \(x < 0\), we are graphing the line \(y = -x\); for \(x \geq 0\), we have the line \(y = x\). Proceeding as we did in Section 1.6, we get
Notice that we have an open circleat \((0,0)\) in the graph when \(x < 0\). As we have seen before, this is due to the fact that the points on \(y = -x\) approach \((0,0)\) as the \(x\)-values approach \(0\). Since \(x\) is required to be strictly less than zero on this stretch, the open circle is drawn at the origin. However, notice that when \(x \geq 0\), we get to fill in the point at \((0,0)\), which effectively "plugs"the hole indicated by the open circle. Thus we get,
Figures \( \PageIndex{2}A \) (left) and \( \PageIndex{2}B \) (right)
By projecting the graph to the \(x\)-axis, we see that the domain is \((-\infty, \infty)\). Projecting to the \(y\)-axis gives us the range \([0,\infty)\). The function is increasing on \([0,\infty)\) and decreasing on \((-\infty,0]\). The relative minimum value of \(f\) is the same as the absolute minimum, namely \(0\) which occurs at \((0,0)\). There is no relative maximum value of \(f\). There is also no absolute maximum value of \(f\), since the \(y\)-values on the graph extend infinitely upwards.
Figure \( \PageIndex{3} \) - To find the zeros of \(g\), we set \(g(x) = |x-3|=0\). By Theorem \( \PageIndex{1} \), we get \(x-3=0\) so that \(x=3\). Hence, the \(x\)-intercept is \((3,0)\).
To find our \(y\)-intercept, we set \(x=0\) so that \(y = g(0) = |0-3| = 3\), which yields \((0,3)\) as our \(y\)-intercept.
To graph \(g(x) = |x-3|\), we use Equation \( \ref{AbsValDefn} \) to rewrite \(g\) as \[ g(x) = |x-3| = \left\{ \begin{array}{rclcl}
x-3, & \text{ if } & x-3\geq 0 & \implies & x \geq 3\\
-(x-3), & \text{ if } & x -3\lt0 & \implies & x \lt 3 \\
\end{array} \right. \nonumber \]For the graph, see Figure \( \PageIndex{4}A \). As before, the open circle we introduce at \((3,0)\) from the graph of \(y = -x+3\) is filled by the point \((3,0)\) from the line \(y = x-3\). We determine the domain as \((-\infty, \infty)\) and the range as \([0,\infty)\). The function \(g\) is increasing on \([3,\infty)\) and decreasing on \((-\infty,3]\). The relative and absolute minimum value of \(g\) is \(0\) which occurs at \((3,0)\). As before, there is no relative or absolute maximum value of \(g\). - Setting \(h(x) = 0\) to look for zeros gives \(|x|-3=0\). As in Example \( \PageIndex{1} \), we isolate the absolute value to get \(|x| = 3\) so that \(x =3\) or \(x=-3\). As a result, we have a pair of \(x\)-intercepts: \((-3,0)\) and \((3,0)\).
Setting \(x=0\) gives \(y = h(0) = |0|-3 = -3\), so our \(y\)-intercept is \((0,-3)\).
As before, we rewrite the absolute value in \(h\) to get \[h(x) =\left\{ \begin{array}{rclcl}
x-3, & \text{ if } & x \geq 0 \\
-x-3, & \text{ if } & x \lt 0 \\
\end{array} \right. \nonumber \]See Figure \( \PageIndex{4}B \) for the graph. Once again, the open circle at \((0,-3)\) from one piece of the graph of \(h\) is filled by the point \((0,-3)\) from the other piece of \(h\). From the graph, we determine the domain of \(h\) is \((-\infty, \infty)\) and the range is \([-3,\infty)\). On \([0,\infty)\), \(h\) is increasing; on \((-\infty,0]\) it is decreasing. The relative minimum occurs at the point \((0,-3)\) on the graph, and we see \(-3\) is both the relative and absolute minimum value of \(h\). Also, \(h\) has no relative or absolute maximum value.
Figures \( \PageIndex{4}A \) (left) and \( \PageIndex{4}B \) (right) - As before, we set \(i(x)=0\) to find the zeros of \(i\) and get \(4 - 2|3x+1|=0\). Isolating the absolute value term gives \(|3x+1|=2\), so either \(3x+1 = 2\) or \(3x+1=-2\). We get \(x=\frac{1}{3}\) or \(x=-1\), so our \(x\)-intercepts are \(\left(\frac{1}{3},0\right)\) and \((-1,0)\).
Substituting \(x=0\) gives \(y = i(0) = 4-2|3(0)+1| = 2\), for a \(y\)-intercept of \((0,2)\).
Rewriting the formula for \(i(x)\) without absolute values give \[\begin{array}{rcl}
i(x) & = & \left\{ \begin{array}{rclcl}
4-2(3x+1), & \text{ if } & 3x+1\geq 0 & \implies & x \geq -\dfrac{1}{3} \\
4-2(-(3x+1)), & \text{ if } & 3x+1\lt 0 & \implies & x \lt -\dfrac{1}{3} \\
\end{array} \right. \\
& = & \left\{ \begin{array}{rcl}
-6x+2, & \text{ if } & x \geq -\dfrac{1}{3} \\
6x+6, & \text{ if } & x \lt - \dfrac{1}{3} \\
\end{array} \right. \\
\end{array} \nonumber \]The usual analysis near the trouble spot \(x=-\frac{1}{3}\) gives the "corner" of this graph is \(\left( -\frac{1}{3}, 4\right)\), and we get the distinctive "\(\vee\)" shape:
The domain of \(i\) is \((-\infty, \infty)\) while the range is \((-\infty, 4]\). The function \(i\) is increasing on \(\left(-\infty, -\frac{1}{3}\right]\) and decreasing on \(\left[ -\frac{1}{3}, \infty\right)\). The relative maximum occurs at the point \(\left(-\frac{1}{3}, 4\right)\) and the relative and absolute maximum value of \(i\) is \(4\). Since the graph of \(i\) extends downwards forever more, there is no absolute minimum value. As we can see from the graph, there is no relative minimum, either.
Figure \( \PageIndex{5} \)
Note that all of the functions in the previous example bear the characteristic "\(\vee\)" shape of the graph of \(y=|x|\). We could have graphed the functions \(g\), \(h\) and \(i\) in Example \( \PageIndex{2} \) starting with the graph of \(f(x)=|x|\) and applying transformations as in Section 1.7 as our next example illustrates.
Example \( \PageIndex{3} \)
Graph the following functions starting with the graph of \(f(x) = |x|\) and using transformations.
- \(g(x) = |x-3|\)
- \(h(x) = |x| - 3\)
- \(i(x) = 4-2|3x+1|\)
Solution
We begin by graphing \(f(x) = |x|\) and labeling three points, \((-1,1)\), \((0,0)\) and \((1,1)\).
Figure \(\PageIndex{6} \)
- Since \(g(x) = |x-3| = f(x-3)\), weadd \(3\) to each of the \(x\)-values of the points on the graph of \(y=f(x)\) to obtain the graph of \(y=g(x)\). This shifts the graph of \(y=f(x)\) to the right \(3\) units(see Figure \( \PageIndex{7} \)).
Figure \( \PageIndex{7} \) - For \(h(x) = |x| - 3 = f(x) -3\),wesubtract \(3\) from each of the \(y\)-values of the points on the graph of \(y=f(x)\) to obtain the graph of \(y = h(x)\). This shifts the graph of \(y=f(x)\) down \(3\) units (see Figure \( \PageIndex{8} \)).
Figure \( \PageIndex{8} \) - We begin by rewriting\[i(x) = 4-2|3x+1| = 4-2f(3x+1) = -2f(3x+1) + 4 = -2 f\left( 3 \left( x + \dfrac{1}{3}\right) \right) + 4.\nonumber\]We now go through, step-by-step, following the Order of Transformations.\[ i(x) = -2 f\left( \boxed{3} \left( x + \dfrac{1}{3}\right) \right) + 4 \nonumber \]The coefficient of \( 3 \) on \( x \) compressesthe graph horizontally by a factor of \( \frac{1}{3} \).\[ i(x) = -2 f\left( \boxed{3\left( x + \dfrac{1}{3}\right)} \right) + 4 \nonumber \]We then shift the graph left\( \frac{1}{3} \) units.\[ i(x) = \boxed{-2 f\left( 3\left( x + \dfrac{1}{3}\right)\right)} + 4 \nonumber \]We then reflect the resulting graph about the \( x \)-axis and stretch it vertically (away from the \( x \)-axis) by a factor of \( 2\).\[ i(x) = \boxed{-2 f\left( 3\left( x + \dfrac{1}{3}\right)\right) + 4} \nonumber \]Finally, we shift this resultup\( 4 \) units (see Figure \( \PageIndex{9} \)).
Figure \( \PageIndex{9} \)
While the methods in Section 1.7 can be used to graph an entire family of absolute value functions, not all functions involving absolute values posses the characteristic "\(\vee\)" shape. As the next example illustrates, often there is no substitute for appealing directly to the definition.
Example \( \PageIndex{4} \)
Graph each of the following functions. Find the zeros of each function and the \(x\)- and \(y\)-intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and find the relative and absolute extrema, if they exist.
- \(f(x) = \frac{|x|}{x}\)
- \(g(x) = |x+2| - |x-3|+1\)
Solution
- We first note that, due to the fraction in the formula of \(f(x)\), \(x \neq 0\). Thus, the domain is \((-\infty, 0) \cup (0, \infty)\).
To find the zeros of \(f\), we set \(f(x) = \frac{|x|}{x} = 0\). This last equation implies \(|x|=0\), which, from Theorem \( \PageIndex{1} \), implies \(x=0\). However, \(x=0\) is not in the domain of \(f\), which means we have, in fact, no \(x\)-intercepts. We have no \(y\)-intercepts either, since \(f(0)\) is undefined.
Rewriting the absolute value in the function gives \[\begin{array}{rcl}
f(x) & = & \left\{ \begin{array}{rcl}
\dfrac{x}{x}, & \text{ if } & x \geq 0 \\
\dfrac{-x}{x}, & \text{ if } & x \lt 0 \\
\end{array} \right. \\
& =& \left\{ \begin{array}{rcl}
1, & \text{ if } & x \geq 0 \\
-1, & \text{ if } & x \lt 0 \\
\end{array} \right. \\
\end{array} \nonumber \]To graph this function, we graph two horizontal lines: \(y = -1\) for \(x < 0\) and \(y = 1\) for \(x > 0\). We have open circles at \((0,-1)\) and \((0,1)\) so we get the graph in Figure \( \PageIndex{10} \).
As we found earlier, the domain is \((-\infty, 0)\cup(0,\infty)\). The range consists of just two \(y\)-values: \(\{-1,1\}\). The function \(f\) is constant on \((-\infty,0)\) and \((0,\infty)\). The local minimum value of \(f\) is the absolute minimum value of \(f\), namely \(-1\); the local maximum and absolute maximum values for \(f\) also coincide - they both are \(1\). Every point on the graph of \(f\) is simultaneously a relative maximum and a relative minimum.
Figure \( \PageIndex{10} \) - To find the zeros of \(g\), we set \(g(x) = 0\). The result is \(|x+2|-|x-3| +1 = 0\). Attempting to isolate the absolute value term is complicated by the fact that there are two terms with absolute values. In this case, it easier to proceed using cases by rewriting the function \(g\) with two separate applications of \( \ref{AbsValDefn}\) and to remove each instance of the absolute values, one at a time. In the first round we get \[\begin{array}{rcl}
g(x) & = & \left\{ \begin{array}{rclcl}
x+2- |x-3|+1, & \text{ if } & x+2\geq 0 & \implies & x \geq-2 \\
-(x+2) - |x-3|+1, & \text{ if } & x+2\lt 0 & \implies & x \lt-2 \\
\end{array} \right. \\
& = & \left\{ \begin{array}{rcl}
x + 3- |x-3|, & \text{ if } & x \geq -2 \\
-x - 1 -|x-3|, & \text{ if } & x \lt -2 \\
\end{array} \right. \\
\end{array} \nonumber \]Given that \[ |x-3| =\left\{ \begin{array}{rclcl}
x-3, & \text{ if } & x-3\geq 0 & \implies & x \geq 3 \\
-(x-3), & \text{ if } & x-3\lt 0 & \implies & x \lt 3\\
\end{array} \right., \nonumber \]we need to break up the domain again at \(x=3\). Note that if \(x < -2\), then \(x < 3\), so we replace \(|x-3|\) with \(-x+3\) for that part of the domain, too. Our completed revision of the form of \(g\) yields\[g(x) = \left\{ \begin{array}{rcl}
x+3 - (x-3), & \text{ if } & x \geq 3 \\
x+3-(-x+3), & \text{ if } & x \geq -2 \, \mbox{ and } \, x < 3 \\
-x-1 - (-x+3), & \text{ if } & x<-2 \\
\end{array} \right. = \left\{ \begin{array}{rcl}
6, & \text{ if } & x \geq 3 \\
2x, & \text{ if } & -2 \leq x < 3 \\
-4, & \text{ if } & x<-2 \\
\end{array} \right. \nonumber \]To solve \(g(x)=0\), we see that the only piece which contains a variable is \(g(x) = 2x\) for \(-2 \leq x < 3\). Solving \(2x=0\) gives \(x=0\). Since \(x=0\) is in the interval \([-2,3)\), we keep this solution and have \((0,0)\) as our only \(x\)-intercept. Accordingly, the \(y\)-intercept is also \((0,0)\).To graph \(g\), we start with \(x < -2\) and graph the horizontal line \(y=-4\) with an open circle at \((-2,-4)\). For \(-2 \leq x < 3\), we graph the line \(y=2x\) and the point \((-2,-4)\) patches the hole left by the previous piece. An open circle at \((3,6)\) completes the graph of this part. Finally, we graph the horizontal line \(y=6\) for \(x \geq 3\), and the point \((3,6)\) fills in the open circle left by the previous part of the graph. The finished graph is
The domain of \(g\) is all real numbers, \((-\infty, \infty)\), and the range of \(g\) is all real numbers between \(-4\) and \(6\) inclusive, \([-4,6]\). The function is increasing on \([-2,3]\) and constant on \((-\infty, -2]\) and \([3,\infty)\). The relative minimum value of \(f\) is \(-4\) which matches the absolute minimum. The relative and absolute maximum values also coincide at \(6\). Every point on the graph of \(y=g(x)\) for \(x<-2\) and \(x> 3\) yields both a relative minimum and relative maximum. The point \((-2,-4)\), however, gives only a relative minimum and the point \((3,6)\) yields only a relative maximum.
Figure \( \PageIndex{11} \)
Many of the applications that the authors are aware of involving absolute values also involve absolute value inequalities. For that reason, we save our discussion of applications for Section 2.4.
Subsection Footnotes
3Actually, since functions can have at most one \(y\)-intercept (Do you know why?), as soon as we found \((0, 0)\) as the \(x\)-intercept, we knew this was also the \(y\)-intercept.
