2.2 Absolute Value Functions

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, $vertical bar 5 vertical bar equals 5$ and $| - 5 | = 5$ , since each is $5$ units from $0$ on the number line.

Coordinate-plane figure. — **Figure 2.25**

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

The absolute value of a real number $x$ , denoted $| x |$ , is given by

$| x | = {\begin{matrix} - x, & if & x < 0 \\ x, & if & x \geq 0 \end{matrix}$

In Definition 2.4, we define $| x |$ using a piecewise-defined function. (See page in Section 1.4.) To check that this definition agrees with what we previously understood as absolute value, note that since $5 greater than or equal to 0$ , to find $vertical bar 5 vertical bar$ we use the rule $| x | = x$ , so $vertical bar 5 vertical bar equals 5$ . Similarly, since $minus 5 less than 0$ , 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.

A dashed V with its corner at the origin sits behind a solid V whose corner is a marked point positioned by two sliders, while a third slider makes the solid V's arms steeper or shallower and flips the V to open downward when it goes negative. Sliding the corner around the grid moves the solid V rigidly, changing the steepness spreads or narrows the two arms symmetrically about the vertical line through the corner, and every setting keeps the two arms mirror images of each other. Adjustable parameters: Steepness a (a) = 1, Corner x-coordinate h (h) = 0, Corner y-coordinate k (k) = 0. Viewing window: x from -11.18 to 11.18, y from -5.91 to 7.91. — The dashed gray V is y = |x|, the parent of every absolute value function in this section, with its corner at the origin. The solid V is a|x − h| + k: the marked corner sits at (h, k), so h and k relocate it, a changes the steepness of both arms, and a negative a turns the V upside down. Because |x − h| is a distance, the graph is symmetric about the vertical line x = h no matter how the other dials are set, which is the picture behind solving |x − h| = c by walking c units to either side of h.

Theorem

Properties of Absolute Value: Let $a$ , $b$ and $x$ be real numbers and let $n$ be an integer.¹ Then

Product Rule: $| a b | = | a | | b |$
Power Rule: $| a^{n} | = {| a |}^{n}$ whenever $a^{n}$ is defined
Quotient Rule: $| \frac{a}{b} | = \frac{| a |}{| b |}$ , provided $b not equal to 0$

Equality Properties:

$vertical bar x vertical bar equals 0$ if and only if $x equals 0$ .
For $c greater than 0$ , $| x | = c$ if and only if $x = c$ or $- x = c$ .
For $c less than 0$ , $| x | = c$ has no solution.

The proofs of the Product and Quotient Rules in Theorem 2.1 boil down to checking four cases: when both $a$ and $b$ are positive; when they are both negative; when one is positive and the other is negative; and when one or both are zero.

For example, suppose we wish to show that $| a b | = | a | | b |$ . We need to show that this equation is true for all real numbers $a$ and $b$ . If $a$ and $b$ are both positive, then so is $a b$ . Hence, $| a | = a$ , $| b | = b$ and $| a b | = a b$ . Hence, the equation $| a b | = | a | | b |$ is the same as $a b = a b$ which is true. If both $a$ and $b$ are negative, then $a b$ is positive. Hence, $| a | = - a$ , $| b | = - b$ and $| a b | = a b$ . The equation $| a b | = | a | | b |$ becomes $a b = (- a) (- b)$ , which is true. Suppose $a$ is positive and $b$ is negative. Then $a b$ is negative, and we have $| a b | = - a b$ , $| a | = a$ and $| b | = - b$ . The equation $| a b | = | a | | b |$ reduces to $- a b = a (- b)$ which is true. A symmetric argument shows the equation $| a b | = | a | | b |$ holds when $a$ is negative and $b$ is positive. Finally, if either $a$ or $b$ (or both) are zero, then both sides of $| a b | = | a | | b |$ are zero, so the equation holds in this case, too. All of this rhetoric has shown that the equation $| a b | = | a | | b |$ holds true in all cases.

The proof of the Quotient Rule is very similar, with the exception that $b not equal to 0$ . The Power Rule can be shown by repeated application of the Product Rule. The `Equality Properties' can be proved using Definition 2.4 and by looking at the cases when $x greater than or equal to 0$ , in which case $| x | = x$ , or when $x less than 0$ , in which case $| x | = - x$ . For example, if $c greater than 0$ , and $| x | = c$ , then if $x greater than or equal to 0$ , we have $x = | x | = c$ . If, on the other hand, $x less than 0$ , then $- x = | x | = c$ , so $x = - c$ . The remaining properties are proved similarly and are left for the Exercises. Our first example reviews how to solve basic equations involving absolute value using the properties listed in Theorem 2.1.

Example 1

Solve each of the following equations.

$| 3 x - 1 | = 6$
$3 - | x + 5 | = 1$
$3 | 2 x + 1 | - 5 = 0$
$4 - | 5 x + 3 | = 5$
$| x | = x^{2} - 6$
$| x - 2 | + 1 = x$

The equation $| 3 x - 1 | = 6$ is of the form $| x | = c$ for $c greater than 0$ , so by the Equality Properties, $| 3 x - 1 | = 6$ is equivalent to $3 x - 1 = 6$ or $3 x - 1 = - 6$ . 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.
$\begin{matrix} 3 - | x + 5 | & = & 1 \\ - | x + 5 | & = & - 2 & subtract 3 \\ | x + 5 | & = & 2 & divide by - 1 \end{matrix}$
From the Equality Properties, we have $x + 5 = 2$ or $x + 5 = - 2$ , and get our solutions to be $x = - 3$ or $x = - 7$ . We leave it to the reader to check both answers in the original equation.
As in the previous example, we first isolate the absolute value in the equation $3 | 2 x + 1 | - 5 = 0$ and get $| 2 x + 1 | = \frac{5}{3}$ . Using the Equality Properties, we have $2 x + 1 = \frac{5}{3}$ or $2 x + 1 = - \frac{5}{3}$ . 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 $4 - | 5 x + 3 | = 5$ , we get $| 5 x + 3 | = - 1$ . 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 $x squared minus 6$ 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 Definition 2.4. For $x less than 0$ , $| x | = - x$ , so for $x less than 0$ , 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 $x equals 2$ . Since only $x = - 3$ satisfies $x less than 0$ , this is the answer we keep. For $x greater than or equal to 0$ , $| x | = x$ , 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 $x equals 3$ or $x = - 2$ , and since only $x equals 3$ satisfies $x greater than or equal to 0$ , this is the one we keep. Hence, our two solutions to $| x | = x^{2} - 6$ are $x = - 3$ and $x equals 3$ .
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 $vertical bar x minus 2 vertical bar$ , so we replace ` $x$ ' with the quantity ` $open parenthesis x minus 2 close parenthesis$ ' in Definition 2.4 to get
$| x - 2 | = {\begin{matrix} - (x - 2), & if & (x - 2) < 0 \\ (x - 2), & if & (x - 2) \geq 0 \end{matrix}$
Simplifying yields
$| x - 2 | = {\begin{matrix} - x + 2, & if & x < 2 \\ x - 2, & if & x \geq 2 \end{matrix}$
So, for $x less than 2$ , $| x - 2 | = - x + 2$ and our equation $| x - 2 | = x - 1$ becomes $- x + 2 = x - 1$ , which gives $x = \frac{3}{2}$ . Since this solution satisfies $x less than 2$ , we keep it. Next, for $x greater than or equal to 2$ , $| x - 2 | = x - 2$ , 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 $x = \frac{3}{2}$ .

Next, we turn our attention to graphing absolute value functions. Our strategy in the next example is to make liberal use of Definition 2.4 along with what we know about graphing linear functions (from Section 2.1) and piecewise-defined functions (from Section 1.4).

Example 2

Graph each of the following functions.

$f (x) = | x |$
$g (x) = | x - 3 |$
$h (x) = | x | - 3$
$i (x) = 4 - 2 | 3 x + 1 |$

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.

To find the zeros of $f$ , we set $f ⁢ open parenthesis x close parenthesis equals 0$ . We get $vertical bar x vertical bar equals 0$ , which, by Theorem 2.1 gives us $x equals 0$ . Since the zeros of $f$ are the $x$ -coordinates of the $x$ -intercepts of the graph of $y = f (x)$ , we get $open parenthesis 0, 0 close parenthesis$ as our only $x$ -intercept. To find the $y$ -intercept, we set $x equals 0$ , and find $y = f (0) = 0$ , so that $open parenthesis 0, 0 close parenthesis$ is our $y$ -intercept as well.² Using Definition 2.4, we get
$f (x) = | x | = {\begin{matrix} - x, & if & x < 0 \\ x, & if & x \geq 0 \end{matrix}$
Hence, for $x less than 0$ , we are graphing the line $y = - x$ ; for $x greater than or equal to 0$ , we have the line $y = x$ . Proceeding as we did in Section 1.6, we get
Figure 2.26 $f (x) = | x |$ , $x less than 0$
Figure 2.27 $f (x) = | x |$ , $x greater than or equal to 0$
Notice that we have an `open circle' at $open parenthesis 0, 0 close parenthesis$ in the graph when $x less than 0$ . As we have seen before, this is due to the fact that the points on $y = - x$ approach $open parenthesis 0, 0 close parenthesis$ 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 $x greater than or equal to 0$ , we get to fill in the point at $open parenthesis 0, 0 close parenthesis$ , which effectively `plugs' the hole indicated by the open circle. Thus we get,
Figure 2.28 $f (x) = | x |$
By projecting the graph to the $x$ -axis, we see that the domain is $open parenthesis minus ∞, ∞ close parenthesis$ . Projecting to the $y$ -axis gives us the range $open bracket 0, ∞ close parenthesis$ . The function is increasing on $open bracket 0, ∞ close parenthesis$ and decreasing on $open parenthesis minus ∞, 0 close bracket$ . The relative minimum value of $f$ is the same as the absolute minimum, namely $0$ which occurs at $open parenthesis 0, 0 close parenthesis$ . 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.
To find the zeros of $g$ , we set $g (x) = | x - 3 | = 0$ . By Theorem 2.1, we get $x - 3 = 0$ so that $x equals 3$ . Hence, the $x$ -intercept is $open parenthesis 3, 0 close parenthesis$ . To find our $y$ -intercept, we set $x equals 0$ so that $y = g (0) = | 0 - 3 | = 3$ , which yields $open parenthesis 0, 3 close parenthesis$ as our $y$ -intercept. To graph $g (x) = | x - 3 |$ , we use Definition 2.4 to rewrite $g$ as
$g (x) = | x - 3 | = {\begin{matrix} - (x - 3), & if & (x - 3) < 0 \\ (x - 3), & if & (x - 3) \geq 0 \end{matrix}$
Simplifying, we get
$g (x) = {\begin{matrix} - x + 3, & if & x < 3 \\ x - 3, & if & x \geq 3 \end{matrix}$
As before, the open circle we introduce at $open parenthesis 3, 0 close parenthesis$ from the graph of $y = - x + 3$ is filled by the point $open parenthesis 3, 0 close parenthesis$ from the line $y = x - 3$ . We determine the domain as $open parenthesis minus ∞, ∞ close parenthesis$ and the range as $open bracket 0, ∞ close parenthesis$ . The function $g$ is increasing on $open bracket 3, ∞ close parenthesis$ and decreasing on $open parenthesis minus ∞, 3 close bracket$ . The relative and absolute minimum value of $g$ is $0$ which occurs at $open parenthesis 3, 0 close parenthesis$ . As before, there is no relative or absolute maximum value of $g$ .
Setting $h ⁢ open parenthesis x close parenthesis equals 0$ to look for zeros gives $| x | - 3 = 0$ . As in Example 2.2.1, we isolate the absolute value to get $vertical bar x vertical bar equals 3$ so that $x equals 3$ or $x = - 3$ . As a result, we have a pair of $x$ -intercepts: $open parenthesis minus 3, 0 close parenthesis$ and $open parenthesis 3, 0 close parenthesis$ . Setting $x equals 0$ gives $y = h (0) = | 0 | - 3 = - 3$ , so our $y$ -intercept is $open parenthesis 0, minus 3 close parenthesis$ . As before, we rewrite the absolute value in $h$ to get
$h (x) = {\begin{matrix} - x - 3, & if & x < 0 \\ x - 3, & if & x \geq 0 \end{matrix}$
Once again, the open circle at $open parenthesis 0, minus 3 close parenthesis$ from one piece of the graph of $h$ is filled by the point $open parenthesis 0, minus 3 close parenthesis$ from the other piece of $h$ . From the graph, we determine the domain of $h$ is $open parenthesis minus ∞, ∞ close parenthesis$ and the range is $open bracket minus 3, ∞ close parenthesis$ . On $open bracket 0, ∞ close parenthesis$ , $h$ is increasing; on $open parenthesis minus ∞, 0 close bracket$ it is decreasing. The relative minimum occurs at the point $open parenthesis 0, minus 3 close parenthesis$ on the graph, and we see $minus 3$ is both the relative and absolute minimum value of $h$ . Also, $h$ has no relative or absolute maximum value.
Figure 2.29 $g (x) = | x - 3 |$
Figure 2.30 $h (x) = | x | - 3$
As before, we set $i ⁢ open parenthesis x close parenthesis equals 0$ to find the zeros of $i$ and get $4 - 2 | 3 x + 1 | = 0$ . Isolating the absolute value term gives $| 3 x + 1 | = 2$ , so either $3 x + 1 = 2$ or $3 x + 1 = - 2$ . We get $x = \frac{1}{3}$ or $x = - 1$ , so our $x$ -intercepts are $(\frac{1}{3}, 0)$ and $open parenthesis minus 1, 0 close parenthesis$ . Substituting $x equals 0$ gives $y = i (0) = 4 - 2 | 3 (0) + 1 | = 2$ , for a $y$ -intercept of $open parenthesis 0, 2 close parenthesis$ . Rewriting the formula for $i (x)$ without absolute values gives
$i (x) = {\begin{matrix} 4 - 2 (- (3 x + 1)), & if & (3 x + 1) < 0 \\ 4 - 2 (3 x + 1), & if & (3 x + 1) \geq 0 \end{matrix} = {\begin{matrix} 6 x + 6, & if & x < - \frac{1}{3} \\ - 6 x + 2, & if & x \geq - \frac{1}{3} \end{matrix}$
The usual analysis near the trouble spot $x = - \frac{1}{3}$ gives the `corner' of this graph is $(- \frac{1}{3}, 4)$ , and we get the distinctive ` $∨$ ' shape:
Figure 2.31
$i (x) = 4 - 2 | 3 x + 1 |$
The domain of $i$ is $open parenthesis minus ∞, ∞ close parenthesis$ while the range is $open parenthesis minus ∞, 4 close bracket$ . The function $i$ is increasing on $(- \infty, - \frac{1}{3}]$ and decreasing on $[- \frac{1}{3}, \infty)$ . The relative maximum occurs at the point $(- \frac{1}{3}, 4)$ 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.

Note that all of the functions in the previous example bear the characteristic ` $∨$ ' shape of the graph of $y = | x |$ . We could have graphed the functions $g$ , $h$ and $i$ in Example 2.2.2 starting with the graph of $f (x) = | x |$ and applying transformations as in Section 1.7 as our next example illustrates.

Example 3

Graph the following functions starting with the graph of $f (x) = | x |$ and using transformations.

$g (x) = | x - 3 |$
$h (x) = | x | - 3$
$i (x) = 4 - 2 | 3 x + 1 |$

We begin by graphing $f (x) = | x |$ and labeling three points, $open parenthesis minus 1, 1 close parenthesis$ , $open parenthesis 0, 0 close parenthesis$ and $open parenthesis 1, 1 close parenthesis$ .

Figure 2.32

f (x) = | x |

Since $g (x) = | x - 3 | = f (x - 3)$ , Theorem 1.7 tells us to add $3$ 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 and moves the point $open parenthesis minus 1, 1 close parenthesis$ to $open parenthesis 2, 1 close parenthesis$ , $open parenthesis 0, 0 close parenthesis$ to $open parenthesis 3, 0 close parenthesis$ and $open parenthesis 1, 1 close parenthesis$ to $open parenthesis 4, 1 close parenthesis$ . Connecting these points in the classic ` $∨$ ' fashion produces the graph of $y = g (x)$ .
Figure 2.33 $f (x) = | x |$
$add 3 to each x -coordinate → shift right 3 units$
Figure 2.34 $g (x) = f (x - 3) = | x - 3 |$
For $h (x) = | x | - 3 = f (x) - 3$ , Theorem 1.7 tells us to subtract $3$ 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 and moves $open parenthesis minus 1, 1 close parenthesis$ to $(- 1, - 2)$ , $open parenthesis 0, 0 close parenthesis$ to $open parenthesis 0, minus 3 close parenthesis$ and $open parenthesis 1, 1 close parenthesis$ to $open parenthesis 1, minus 2 close parenthesis$ . Connecting these points with the ` $∨$ ' shape produces our graph of $y = h (x)$ .
Figure 2.35 $f (x) = | x |$
$subtract 3 from each y -coordinate → shift down 3 units$
Figure 2.36 $h (x) = f (x) - 3 = | x | - 3$
We re-write $i (x) = 4 - 2 | 3 x + 1 | = 4 - 2 f (3 x + 1) = - 2 f (3 x + 1) + 4$ and apply Theorem 1.7. First, we take care of the changes on the `inside' of the absolute value. Instead of $| x |$ , we have $vertical bar 3 ⁢ x plus 1 vertical bar$ , so, in accordance with Theorem 1.7, we first subtract $1$ from each of the $x$ -values of points on the graph of $y = f (x)$ , then divide each of those new values by $3$ . This effects a horizontal shift left $1$ unit followed by a horizontal shrink by a factor of $3$ . These transformations move $open parenthesis minus 1, 1 close parenthesis$ to $(- \frac{2}{3}, 1)$ , $open parenthesis 0, 0 close parenthesis$ to $(- \frac{1}{3}, 0)$ and $open parenthesis 1, 1 close parenthesis$ to $open parenthesis 0, 1 close parenthesis$ . Next, we take care of what's happening `outside of' the absolute value. Theorem 1.7 instructs us to first multiply each $y$ -value of these new points by $minus 2$ then add $4$ . Geometrically, this corresponds to a vertical stretch by a factor of $2$ , a reflection across the $x$ -axis and finally, a vertical shift up $4$ units. These transformations move $(- \frac{2}{3}, 1)$ to $(- \frac{2}{3}, 2)$ , $(- \frac{1}{3}, 0)$ to $(- \frac{1}{3}, 4)$ , and $open parenthesis 0, 1 close parenthesis$ to $open parenthesis 0, 2 close parenthesis$ . Connecting these points with the usual ` $∨$ ' shape produces our graph of $y = i (x)$ .
Figure 2.37
$\overset{}{\to}$
Figure 2.38
$f (x) = | x |$
$i (x) = - 2 f (3 x + 1) + 4$
$= - 2 | 3 x + 1 | + 4$

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 ` $∨$ ' shape. As the next example illustrates, often there is no substitute for appealing directly to the definition.

Example 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.

$f (x) = \frac{| x |}{x}$
$g (x) = | x + 2 | - | x - 3 | + 1$

We first note that, due to the fraction in the formula of $f (x)$ , $x not equal to 0$ . Thus the domain is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ . To find the zeros of $f$ , we set $f (x) = \frac{| x |}{x} = 0$ . This last equation implies $vertical bar x vertical bar equals 0$ , which, from Theorem 2.1, implies $x equals 0$ . However, $x equals 0$ is not in the domain of $f$ , which means we have, in fact, no $x$ -intercepts. We have no $y$ -intercepts either, since $f ⁢ open parenthesis 0 close parenthesis$ is undefined. Re-writing the absolute value in the function gives
$f (x) = {\begin{matrix} \frac{- x}{x}, & if & x < 0 \\ \frac{x}{x}, & if & x > 0 \end{matrix} = {\begin{matrix} - 1, & if & x < 0 \\ 1, & if & x > 0 \end{matrix}$
To graph this function, we graph two horizontal lines: $y = - 1$ for $x less than 0$ and $y equals 1$ for $x greater than 0$ . We have open circles at $open parenthesis 0, minus 1 close parenthesis$ and $open parenthesis 0, 1 close parenthesis$ (Can you explain why?) so we get
Figure 2.39 $f (x) = \frac{| x |}{x}$
As we found earlier, the domain is $open parenthesis minus ∞, 0 close parenthesis union open parenthesis 0, ∞ close parenthesis$ . The range consists of just two $y$ -values: $open brace minus 1, 1 close brace$ . The function $f$ is constant on $open parenthesis minus ∞, 0 close parenthesis$ and $open parenthesis 0, ∞ close parenthesis$ . The local minimum value of $f$ is the absolute minimum value of $f$ , namely $minus 1$ ; 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. (Can you remember why in light of Definition 1.11? This was explored in the Exercises in Section 1.6.2.)
To find the zeros of $g$ , we set $g ⁢ open parenthesis x close parenthesis equals 0$ . 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 re-writing the function $g$ with two separate applications of Definition 2.4 to remove each instance of the absolute values, one at a time. In the first round we get
$g (x) = {\begin{matrix} - (x + 2) - | x - 3 | + 1, & if & (x + 2) < 0 \\ (x + 2) - | x - 3 | + 1, & if & (x + 2) \geq 0 \end{matrix} = {\begin{matrix} - x - 1 - | x - 3 |, & if & x < - 2 \\ x + 3 - | x - 3 |, & if & x \geq - 2 \end{matrix}$
Given that
$| x - 3 | = {\begin{matrix} - (x - 3), & if & (x - 3) < 0 \\ x - 3, & if & (x - 3) \geq 0 \end{matrix} = {\begin{matrix} - x + 3, & if & x < 3 \\ x - 3, & if & x \geq 3 \end{matrix},$
we need to break up the domain again at $x equals 3$ . Note that if $x less than minus 2$ , then $x less than 3$ , so we replace $vertical bar x minus 3 vertical bar$ with $- x + 3$ for that part of the domain, too. Our completed revision of the form of $g$ yields
$g (x) = {\begin{matrix} - x - 1 - (- x + 3), & if & x < - 2 \\ x + 3 - (- x + 3), & if & x \geq - 2 and x < 3 \\ x + 3 - (x - 3), & if & x \geq 3 \end{matrix} = {\begin{matrix} - 4, & if & x < - 2 \\ 2 x, & if & - 2 \leq x < 3 \\ 6, & if & x \geq 3 \end{matrix}$

To solve $g ⁢ open parenthesis x close parenthesis equals 0$ , we see that the only piece which contains a variable is $g ⁢ open parenthesis x close parenthesis equals 2 ⁢ x$ for $minus 2 less than or equal to x less than 3$ . Solving $2 ⁢ x equals 0$ gives $x equals 0$ . Since $x equals 0$ is in the interval $open bracket minus 2, 3 close parenthesis$ , we keep this solution and have $open parenthesis 0, 0 close parenthesis$ as our only $x$ -intercept. Accordingly, the $y$ -intercept is also $open parenthesis 0, 0 close parenthesis$ . To graph $g$ , we start with $x less than minus 2$ and graph the horizontal line $y = - 4$ with an open circle at $(- 2, - 4)$ . For $minus 2 less than or equal to x less than 3$ , we graph the line $y equals 2 ⁢ x$ and the point $(- 2, - 4)$ patches the hole left by the previous piece. An open circle at $open parenthesis 3, 6 close parenthesis$ completes the graph of this part. Finally, we graph the horizontal line $y equals 6$ for $x greater than or equal to 3$ , and the point $open parenthesis 3, 6 close parenthesis$ fills in the open circle left by the previous part of the graph. The finished graph is

Figure 2.40

g (x) = | x + 2 | - | x - 3 | + 1

The domain of $g$ is all real numbers, $open parenthesis minus ∞, ∞ close parenthesis$ , and the range of $g$ is all real numbers between $minus 4$ and $6$ inclusive, $open bracket minus 4, 6 close bracket$ . The function is increasing on $open bracket minus 2, 3 close bracket$ and constant on $(- \infty, - 2]$ and $open bracket 3, ∞ close parenthesis$ . The relative minimum value of $f$ is $minus 4$ 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 $x less than minus 2$ and $x greater than 3$ yields both a relative minimum and relative maximum. The point $(- 2, - 4)$ , however, gives only a relative minimum and the point $open parenthesis 3, 6 close parenthesis$ yields only a relative maximum. (Recall the Exercises in Section 1.6.2 which dealt with constant functions.)

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.

Exercises

In Exercises -, solve the equation.

$vertical bar x vertical bar equals 6$
$| 3 x - 1 | = 10$
$| 4 - x | = 7$
$4 - | x | = 3$
$2 | 5 x + 1 | - 3 = 0$
$| 7 x - 1 | + 2 = 0$
$\frac{5 - | x |}{2} = 1$
$\frac{2}{3} | 5 - 2 x | - \frac{1}{2} = 5$
$| x | = x + 3$
$| 2 x - 1 | = x + 1$
$4 - | x | = 2 x + 1$
$| x - 4 | = x - 5$
$| x | = x^{2}$
$| x | = 12 - x^{2}$
$| x^{2} - 1 | = 3$

Prove that if $| f (x) | = | g (x) |$ then either $f (x) = g (x)$ or $f (x) = - g (x)$ . Use that result to solve the equations in Exercises -.

$| 3 x - 2 | = | 2 x + 7 |$
$| 3 x + 1 | = | 4 x |$
$| 1 - 2 x | = | x + 1 |$
$| 4 - x | - | x + 2 | = 0$
$| 2 - 5 x | = 5 | x + 1 |$
$3 | x - 1 | = 2 | x + 1 |$

In Exercises -, graph the function. 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.

$f (x) = | x + 4 |$
$f (x) = | x | + 4$
$f ⁢ open parenthesis x close parenthesis equals vertical bar 4 ⁢ x vertical bar$
$f (x) = - 3 | x |$
$f (x) = 3 | x + 4 | - 4$
$f (x) = \frac{1}{3} | 2 x - 1 |$
$f (x) = \frac{| x + 4 |}{x + 4}$
$f (x) = \frac{| 2 - x |}{2 - x}$
$f (x) = x + | x | - 3$
$f (x) = | x + 2 | - x$
$f (x) = | x + 2 | - | x |$
$f (x) = | x + 4 | + | x - 2 |$
With the help of your classmates, find an absolute value function whose graph is given below.
Figure 2.41
With help from your classmates, prove the second, third and fifth parts of Theorem 2.1.
Prove The Triangle Inequality: For all real numbers $a$ and $b,$ $| a + b | \leq | a | + | b |$ .

Answers

$x = - 6$ or $x equals 6$
$x = - 3$ or $x = \frac{11}{3}$
$x = - 3$ or $x equals 11$
$x = - 1$ or $x equals 1$
$x = - \frac{1}{2}$ or $x = \frac{1}{10}$
no solution
$x = - 3$ or $x equals 3$
$x = - \frac{13}{8}$ or $x = \frac{53}{8}$
$x = - \frac{3}{2}$
$x equals 0$ or $x equals 2$
$x equals 1$
no solution
$x = - 1$ , $x equals 0$ or $x equals 1$
$x = - 3$ or $x equals 3$
$x = - 2$ or $x equals 2$
$x = - 1$ or $x equals 9$
$x = - \frac{1}{7}$ or $x equals 1$
$x equals 0$ or $x equals 2$
$x equals 1$
$x = - \frac{3}{10}$
$x = \frac{1}{5}$ or $x equals 5$
$f (x) = | x + 4 |$ $f (- 4) = 0$ $x$ -intercept $open parenthesis minus 4, 0 close parenthesis$ $y$ -intercept $open parenthesis 0, 4 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket 0, ∞ close parenthesis$ Decreasing on $(- \infty, - 4]$ Increasing on $open bracket minus 4, ∞ close parenthesis$ Relative and absolute min. at $open parenthesis minus 4, 0 close parenthesis$ No relative or absolute maximum
Figure 2.42
$f (x) = | x | + 4$ No zeros No $x$ -intercepts $y$ -intercept $open parenthesis 0, 4 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket 4, ∞ close parenthesis$ Decreasing on $open parenthesis minus ∞, 0 close bracket$ Increasing on $open bracket 0, ∞ close parenthesis$ Relative and absolute minimum at $open parenthesis 0, 4 close parenthesis$ No relative or absolute maximum
Figure 2.43
$f ⁢ open parenthesis x close parenthesis equals vertical bar 4 ⁢ x vertical bar$ $f ⁢ open parenthesis 0 close parenthesis equals 0$ $x$ -intercept $open parenthesis 0, 0 close parenthesis$ $y$ -intercept $open parenthesis 0, 0 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket 0, ∞ close parenthesis$ Decreasing on $open parenthesis minus ∞, 0 close bracket$ Increasing on $open bracket 0, ∞ close parenthesis$ Relative and absolute minimum at $open parenthesis 0, 0 close parenthesis$ No relative or absolute maximum
Figure 2.44
$f (x) = - 3 | x |$ $f ⁢ open parenthesis 0 close parenthesis equals 0$ $x$ -intercept $open parenthesis 0, 0 close parenthesis$ $y$ -intercept $open parenthesis 0, 0 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open parenthesis minus ∞, 0 close bracket$ Increasing on $open parenthesis minus ∞, 0 close bracket$ Decreasing on $open bracket 0, ∞ close parenthesis$ Relative and absolute maximum at $open parenthesis 0, 0 close parenthesis$ No relative or absolute minimum
Figure 2.45
$f (x) = 3 | x + 4 | - 4$ $f (- \frac{16}{3}) = 0$ , $f (- \frac{8}{3}) = 0$ $x$ -intercepts $(- \frac{16}{3}, 0)$ , $(- \frac{8}{3}, 0)$ $y$ -intercept $open parenthesis 0, 8 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket minus 4, ∞ close parenthesis$ Decreasing on $(- \infty, - 4]$ Increasing on $open bracket minus 4, ∞ close parenthesis$ Relative and absolute min. at $(- 4, - 4)$ No relative or absolute maximum
Figure 2.46
$f (x) = \frac{1}{3} | 2 x - 1 |$ $f (\frac{1}{2}) = 0$ $x$ -intercepts $(\frac{1}{2}, 0)$ $y$ -intercept $(0, \frac{1}{3})$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket 0, ∞ close parenthesis$ Decreasing on $(- \infty, \frac{1}{2}]$ Increasing on $[\frac{1}{2}, \infty)$ Relative and absolute min. at $(\frac{1}{2}, 0)$ No relative or absolute maximum
Figure 2.47
$f (x) = \frac{| x + 4 |}{x + 4}$ No zeros No $x$ -intercept $y$ -intercept $open parenthesis 0, 1 close parenthesis$ Domain $(- \infty, - 4) \cup (- 4, \infty)$ Range $open brace minus 1, 1 close brace$ Constant on $(- \infty, - 4)$ Constant on $open parenthesis minus 4, ∞ close parenthesis$ Absolute minimum at every point $open parenthesis x, minus 1 close parenthesis$ where $x less than minus 4$ Absolute maximum at every point $open parenthesis x, 1 close parenthesis$ where $x greater than minus 4$ Relative maximum AND minimum at every point on the graph
Figure 2.48
$f (x) = \frac{| 2 - x |}{2 - x}$ No zeros No $x$ -intercept $y$ -intercept $open parenthesis 0, 1 close parenthesis$ Domain $open parenthesis minus ∞, 2 close parenthesis union open parenthesis 2, ∞ close parenthesis$ Range $open brace minus 1, 1 close brace$ Constant on $open parenthesis minus ∞, 2 close parenthesis$ Constant on $open parenthesis 2, ∞ close parenthesis$ Absolute minimum at every point $open parenthesis x, minus 1 close parenthesis$ where $x greater than 2$ Absolute maximum at every point $open parenthesis x, 1 close parenthesis$ where $x less than 2$ Relative maximum AND minimum at every point on the graph
Figure 2.49
Re-write $f (x) = x + | x | - 3$ as $f (x) = {\begin{matrix} - 3 & if & x < 0 \\ 2 x - 3 & if & x \geq 0 \end{matrix}$ $f (\frac{3}{2}) = 0$ $x$ -intercept $(\frac{3}{2}, 0)$ $y$ -intercept $open parenthesis 0, minus 3 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket minus 3, ∞ close parenthesis$ Increasing on $open bracket 0, ∞ close parenthesis$ Constant on $open parenthesis minus ∞, 0 close bracket$ Absolute minimum at every point $open parenthesis x, minus 3 close parenthesis$ where $x less than or equal to 0$ No absolute maximum Relative minimum at every point $open parenthesis x, minus 3 close parenthesis$ where $x less than or equal to 0$ Relative maximum at every point $open parenthesis x, minus 3 close parenthesis$ where $x less than 0$
Figure 2.50
Re-write $f (x) = | x + 2 | - x$ as $f (x) = {\begin{matrix} - 2 x - 2 & if & x < - 2 \\ 2 & if & x \geq - 2 \end{matrix}$ No zeros No $x$ -intercepts $y$ -intercept $open parenthesis 0, 2 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket 2, ∞ close parenthesis$ Decreasing on $(- \infty, - 2]$ Constant on $open bracket minus 2, ∞ close parenthesis$ Absolute minimum at every point $open parenthesis x, 2 close parenthesis$ where $x greater than or equal to minus 2$ No absolute maximum Relative minimum at every point $open parenthesis x, 2 close parenthesis$ where $x greater than or equal to minus 2$ Relative maximum at every point $open parenthesis x, 2 close parenthesis$ where $x greater than minus 2$
Figure 2.51
Re-write $f (x) = | x + 2 | - | x |$ as $f (x) = {\begin{matrix} - 2 & if & x < - 2 \\ 2 x + 2 & if & - 2 \leq x < 0 \\ 2 & if & x \geq 0 \end{matrix}$ $f (- 1) = 0$ $x$ -intercept $open parenthesis minus 1, 0 close parenthesis$ $y$ -intercept $open parenthesis 0, 2 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket minus 2, 2 close bracket$ Increasing on $open bracket minus 2, 0 close bracket$ Constant on $(- \infty, - 2]$ Constant on $open bracket 0, ∞ close parenthesis$ Absolute minimum at every point $open parenthesis x, minus 2 close parenthesis$ where $x less than or equal to minus 2$ Absolute maximum at every point $open parenthesis x, 2 close parenthesis$ where $x greater than or equal to 0$ Relative minimum at every point $open parenthesis x, minus 2 close parenthesis$ where $x less than or equal to minus 2$ and at every point $open parenthesis x, 2 close parenthesis$ where $x greater than 0$ Relative maximum at every point $open parenthesis x, minus 2 close parenthesis$ where $x less than minus 2$ and at every point $open parenthesis x, 2 close parenthesis$ where $x greater than or equal to 0$
Figure 2.52
Re-write $f (x) = | x + 4 | + | x - 2 |$ as $f (x) = {\begin{matrix} - 2 x - 2 & if & x < - 4 \\ 6 & if & - 4 \leq x < 2 \\ 2 x + 2 & if & x \geq 2 \end{matrix}$ No zeros No $x$ -intercept $y$ -intercept $open parenthesis 0, 6 close parenthesis$ Domain $open parenthesis minus ∞, ∞ close parenthesis$ Range $open bracket 6, ∞ close parenthesis$ Decreasing on $(- \infty, - 4]$ Constant on $open bracket minus 4, 2 close bracket$ Increasing on $open bracket 2, ∞ close parenthesis$ Absolute minimum at every point $open parenthesis x, 6 close parenthesis$ where $minus 4 less than or equal to x less than or equal to 2$ No absolute maximum Relative minimum at every point $open parenthesis x, 6 close parenthesis$ where $minus 4 less than or equal to x less than or equal to 2$ Relative maximum at every point $open parenthesis x, 6 close parenthesis$ where $minus 4 less than x less than 2$
Figure 2.53
$f (x) = | | x | - 4 |$

Adapted from Precalculus, 3rd corrected edition, by Carl Stitz and Jeff Zeager (stitz-zeager.com), licensed under CC BY-NC-SA 3.0. Changes were made: reformatted as an accessible XYZ web edition. License: CC-BY-NC-SA-3.0.