4.4 absolutevalueinequalitiesmsenux2.redwoods.edu/intalgtext/chapter4/section4.pdf · section 4.4...

Section 4.4 Absolute Value Inequalities 391

Version: Fall 2007

4.4 Absolute Value InequalitiesIn the last section, we solved absolute value equations. In this section, we turn ourattention to inequalities involving absolute value.

Solving |x| < aThe solutions of

|x| < a

again depend upon the value and sign of the number a. To solve |x| < a graphically,we must determine where the graph of the left-hand side lies below the graph of theright-hand side of the inequality |x| < a. There are three cases to consider.

• Case I: a < 0

In this case, the graph of y = a lies strictly below the x-axis. As you can see inFigure 1(a), the graph of y = |x| never lies below the graph of y = a. Hence, theinequality |x| < a has no solutions.

• Case II: a = 0

In this case, the graph of y = 0 coincides with the x-axis. As you can see inFigure 1(b), the graph of y = |x| never lies strictly below the x-axis. Hence, theinequality |x| < 0 has no solutions.

• Case III: a > 0

In this case, the graph of y = a lies strictly above the x-axis. In Figure 1(c), thegraph of y = |x| and y = a intersect at x = −a and x = a. In Figure 1(c), wealso see that the graph of y = |x| lies strictly below the graph of y = a when x isin-between −a and a; that is, when −a < x < a.

In Figure 1(c), we’ve dropped dashed vertical lines from the points of intersec-tion of the two graphs to the x-axis. On the x-axis, we’ve shaded the solution of|x| < a, that is, −a < x < a.

x

y y=|x|

y=a

x

y y=|x|

y=a

0x

y y=|x|

y=a

−a a

(a) (b) (c)Figure 1. The solution of |x| < a has three cases.

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392 Chapter 4 Absolute Value Functions

Version: Fall 2007

This discussion leads to the following key property.

Property 1. The solution of |x| < a depends upon the value and sign of a.

• Case I: a < 0

The inequality |x| < a has no solution.

• Case II: a = 0

The inequality |x| < 0 has no solution.

• Case III: a > 0

The inequality |x| < a has solution set {x : −a < x < a}.

Let’s look at some examples.

I Example 2. Solve the inequality |x| < −5 for x.

The graph of the left-hand side of |x| < −5 is the “V” of Figure 1(a). The graphof the right-hand side of |x| < −5 is a horizontal line located 5 units below the x-axis.This is the situation shown in Figure 1(a). The graph of y = |x| is therefore neverbelow the graph of y = −5. Thus, the inequality |x| < −5 has no solution.

An alternate approach is to consider the fact that the absolute value of x is alwaysnonnegative and can never be less than −5. Thus, the inequality |x| < −5 has nosolution.

I Example 3. Solve the inequality |x| < 0 for x.

This is the case shown in Figure 1(b). The graph of y = |x| is never strictly belowthe x-axis. Thus, the inequality |x| < 0 has no solution.

I Example 4. Solve the inequality |x| < 8 for x.

The graph of the left-hand side of |x| < 8 is the “V” of Figure 1(c). The graph ofthe right-hand side of |x| < 8 is a horizontal line located 8 units above the x-axis. Thisis the situation depicted in Figure 1(c). The graphs intersect at (−8, 8) and (8, 8) andthe graph of y = |x| lies strictly below the graph of y = 8 for values of x in-between−8 and 8. Thus, the solution of |x| < 8 is −8 < x < 8.

It helps the intuition if you check the results of the last example. Note that numbersbetween −8 and 8, such as −7.75, −3 and 6.8 satisfy the inequality,

| − 7.75| < 8 and | − 3| < 8 and |6.8| < 8,


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while values that do not lie between −8 and 8 do not satisfy the inequality. For example,none of the numbers −9.3, 8.2, and 11.7 lie between −8 and 8, and each of the followingis a false statement.

| − 9.3| < 8 and |8.2| < 8 and |11.7| < 8 (all are false)

If you reflect upon these results, they will help cement the notion that the solution of|x| < 8 is all values of x satisfying −8 < x < 8.

I Example 5. Solve the inequality |5− 2x| < −3 for x.

If the inequality were |x| < −3, we would not hesitate. This is the situation depictedin Figure 1(a) and the inequality |x| < −3 has no solutions. The reasoning appliedto |x| < −3 works equally well for the inequality |5− 2x| < −3. The left-hand side ofthis inequality must be nonnegative, so its graph must lie on or above the x-axis. Theright-hand side of |5 − 2x| < −3 is a horizontal line located 3 units below the x-axis.Therefore, the graph of y = |5− 2x| can never lie below the graph of y = −3 and theinequality |5− 2x| < −3 has no solution.

We can verify this result with the graphing calculator. Load the left- and right-handsides of |5− 2x| < −3 into Y1 and Y2, respectively, as shown in Figure 2(a). From theZOOM menu, select 6:ZStandard to produce the image shown in Figure 2(b).

As predicted, the graph of y = |5− 2x| never lies below the graph of y = −3, so theinequality |5− 2x| < −3 has no solution.

(a) (b)Figure 2. Using the graphing calculator to solve the in-equality |5− 2x| < −3.

I Example 6. Solve the inequality |5− 2x| < 0 for x.

We know that the left-hand side of the inequality |5 − 2x| < 0 has the “V” shapeindicated in Figure 1(b). The graph “touches” the x-axis when |5− 2x| = 0, or when

5− 2x = 0−2x = −5

x = 52.

However, the graph of y = |5 − 2x| never falls below the x-axis, so the inequality|5− 2x| < 0 has no solution.


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Intuitively, it should be clear that the inequality |5−2x| < 0 has no solution. Indeed,the left-hand side of this inequality is always nonnegative, and can never be strictlyless than zero.

I Example 7. Solve the inequality |5− 2x| < 3 for x.

In this example, the graph of the right-hand side of the inequality |5− 2x| < 3 is ahorizontal line located 3 units above the x-axis. The graph of the left-hand side of theinequality has the “V” shape shown in Figure 3(b) and (c). You can use the intersectutility on the graphing calculator to find the points of intersection of the graphs ofy = |5 − 2x| and y = 3, as we have done in Figures 3(b) and (c). Note that thecalculator indicates two points of intersection, one at x = 1 and a second at x = 4.

(a) (b) (c)Figure 3. Using the graphing calculator to solve the inequality |5− 2x| < 3.

The graph of y = |5− 2x| falls below the graph of y = 3 for all values of x between 1and 4. Hence, the solution of the inequality |5 − 2x| < 3 is the set of all x satisfying1 < x < 4; i.e. {x : 1 < x < 4}.

Expectations. We need a way of summarizing this graphing calculator approachon our homework paper. First, draw a reasonable facsimile of your calculator’sviewing window on your homework paper. Use a ruler to draw all lines. Completethe following checklist.

• Label each axis, in this case with x and y.• Scale each axis. To do this, press the WINDOW button on your calculator, then

report the values of xmin, xmax, ymin, and ymax on the appropriate axis.• Label each graph with its equation.• Drop dashed vertical lines from the points of intersection to the x-axis. Shade

and label the solution set of the inequality on the x-axis.

Following the guidelines in the above checklist, we obtain the image in Figure 4.


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x10

y10

−10

−10

y = 3

y = |5 − 2x|

1 4

Figure 4. Reporting a graphical solution of|5− 2x| < 3.

Algebraic Approach. Let’s now explore an algebraic solution of the inequality|5− 2x| < 3. Much as |x| < 3 implies that −3 < x < 3, the inequality

|5− 2x| < 3

requires that

−3 < 5− 2x < 3.

We can subtract 5 from all three members of this last inequality, then simplify.

−3− 5 < 5− 2x− 5 < 3− 5−8 < −2x < −2

Divide all three members of this last inequality by −2, reversing the inequality symbolsas you go.

4 > x > 1

We prefer that our inequalities read from “small-to-large,” so we write

1 < x < 4.

This form matches the order of the shaded solution on the number line in Figure 4,which we found using the graphing calculator.

The algebraic technique of this last example leads us to the following property.

Property 8. If a > 0, then the inequality |x| < a is equivalent to the inequality−a < x < a.

This property provides a simple method for solving inequalities of the form |x| < a.Let’s apply this algebraic technique in the next example.


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I Example 9. Solve the inequality |4x+ 5| < 7 for x.

The first step is to use Property 8 to write that

|4x+ 5| < 7

is equivalent to the inequality

−7 < 4x+ 5 < 7.

From here, we can solve for x by first subtracting 5 from all three members, thendividing through by 4.

−12 < 4x < 2

−3 < x < 12

We can sketch the solution on a number line.

−3 1/2

And we can describe the solution in both interval and set-builder notation as follows.(−3, 1

2

)={x : −3 < x < 1

2

}

Assuming that a > 0, the inequality |x| ≤ a requires that we find where the absolutevalue of x is either “less than” a or “equal to” a. We know that |x| < a when −a < x < aand we know that |x| = a when x = −a or x = a. Thus, the solution of |x| ≤ a is the“union” of these two solutions.

This argument leads to the following property.

Property 10. If a > 0, then the inequality |x| ≤ a is equivalent to the inequality−a ≤ x ≤ a.

I Example 11. Solve the inequality 5− 3|x− 4| ≥ −4 for x.

At first glance, the inequality

5− 3|x− 4| ≥ −4

has a form quite dissimilar from what we’ve done thus far. However, let’s subtract 5from both sides of the inequality.

−3|x− 4| ≥ −9


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Now, let’s divide both sides of this last inequality by −3, reversing the inequality sign.

|x− 4| ≤ 3

Aha! Familiar ground. Using Property 10, this last inequality is equivalent to

−3 ≤ x− 4 ≤ 3,

and when we add 4 to all three members, we have the solution.

1 ≤ x ≤ 7

We can sketch the solution on a number line.

1 7

And we can describe the solution with interval and set-builder notation.

[1, 7] = {x : 1 ≤ x ≤ 7}

Solving |x| > aThe solutions of |x| > a again depend upon the value and sign of a. To solve |x| > agraphically, we must determine where the graph of y = |x| lies above the graph ofy = a. Again, we consider three cases.

• Case I: a < 0

In this case, the graph of y = a lies strictly below the x-axis. Therefore, the graphof y = |x| in Figure 5(a) always lies above the graph of y = a. Hence, all realnumbers are solutions of the inequality |x| > a.

• Case II: a = 0

In this case, the graph of y = 0 coincides with the x-axis. As shown in Figure 5(b),the graph of y = |x| will lie strictly above the graph of y = 0 for all values of xwith one exception, namely, x cannot equal zero. Hence, every real number exceptx = 0 is a solution of |x| > 0. In Figure 5(b), we’ve shaded the solution of |x| > 0,namely the set of all real numbers except x = 0.

• Case III: a > 0

In this case, the graph of y = a lies strictly above the x-axis. In Figure 5(c), thegraph of y = |x| intersects the graph of y = a at x = −a and x = a. In Figure 5(c),we see that the graph of y = |x| lies strictly above the graph of y = a if x is lessthan −a or greater than a.

In Figure 5(c), we’ve dropped dashed vertical lines from the points of intersec-tion to the x-axis. On the x-axis, we’ve shaded the solution of |x| > a, namely theset of all real numbers x such that x < −a or x > a.


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x

y y=|x|

y=a

x

y y=|x|

y=a

0x

y y=|x|

y=a

−a a

(a) (b) (c)Figure 5. The solution of |x| > a has three cases.

This discussion leads to the following property.

Property 12. The solution of |x| > a depends upon the value and sign of a.

• Case I: a < 0

All real numbers are solutions of the inequality |x| > a.

• Case II: a = 0

All real numbers, with the exception of x = 0, are solutions of |x| > 0.

• Case III: a > 0

The inequality |x| > a has solution set {x : x < −a or x > a}.

I Example 13. State the solution of each of the following inequalities.

a. |x| > −5 b. |x| > 0 c. |x| > 4

Solution:

a. The solution of |x| > −5 is all real numbers.b. The solution of |x| > 0 is all real numbers except zero.c. The solution of |x| > 4 is the set of all real numbers less than −4 or greater than 4.

I Example 14. Solve the inequality |4− x| > −5 for x.

The left-hand side of the inequality |4 − x| > −5 is nonnegative, so the graph ofy = |4 − x| must lie above or on the x-axis. The graph of the right-hand side of|4− x| > −5 is a horizontal line located 5 units below the x-axis. Therefore, the graph


Version: Fall 2007

of y = |4 − x| always lies above the graph of y = −5. Thus, all real numbers aresolutions of the inequality |4− x| > −5.

We can verify our thinking with the graphing calculator. Load the left- and right-hand sides of the inequality |4 − x| > −5 into Y1 and Y2, respectively, as shown inFigure 6(a). From the ZOOM menu, select 6:ZStandard to produce the image shownin Figure 6(b).

As predicted, the graph of y = |4 − x| lies above the graph of y = −5 for all realnumbers.

(a) (b)Figure 6. Using the graphingcalculator to solve |4 − x| > −5.

Intuitively, the absolute value of any number is always nonnegative, so |4−x| > −5 forall real values of x.

I Example 15. Solve the inequality |4− x| > 0 for x.

As we saw in Figure 6(b), the graph of y = |4− x| lies on or above the x-axis forall real numbers. It “touches” the x-axis at the “vertex” of the “V,” where

|4− x| = 0.

This can occur only if4− x = 0−x = −4x = 4.

Thus, the graph of y = |4 − x| is strictly above the x-axis for all real numbers exceptx = 4. That is, the solution of |4− x| > 0 is {x : x 6= 4}.

I Example 16. Solve the inequality |4− x| > 5 for x.

In this example, the graph of the right-hand side of |4 − x| > 5 is a horizontalline located 5 units above the x-axis. The graph of y = |4 − x| has the “V” shapeshown in Figure 6(c). You can use the intersect utility on the graphing calculator toapproximate the points of intersection of the graphs of y = |4 − x| and y = 5, as wehave done in Figure 7(c) and (d). The calculator indicates two points of intersection,one at x = −1 and a second at x = 9.


Version: Fall 2007

(a) (b) (c) (d)Figure 7. Using the graphing calculator to solve the inequality |4 − x| > 5.

The graph of y = |4 − x| lies above the graph of y = 5 for all values of x that lieeither to the left of −1 or to the right of 9. Hence, the solution of |4− x| > 5 is the set{x : x < −1 or x > 9}.

Following the guidelines established in Example 7, we create the image shown inFigure 8 on our homework paper. Note that we’ve labeled each axis, scaled each axiswith xmin, xmax, ymin, and ymax, labeled each graph with its equation, and shadedand labeled the solution on x-axis.

x15

y10

−5

−10

y = 5

y = |4 − x|

−1 9

Figure 8. Reporting a graphical solu-tion of |4− x| > 5.

Algebraic Approach. Let’s explore an algebraic solution of |4− x| > 5. In muchthe same manner that |x| > 5 leads to the conditions x < −5 or x > 5, the inequality

|4− x| > 5

requires that

4− x < −5 or 4− x > 5.

We can solve each of these independently by first subtracting 4 from each side ofthe inequality, then multiplying both sides of each inequality by −1, reversing eachinequality as we do so.


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4− x < −5 or 4− x > 5

− x < −9 − x > 1

x > 9 x < −1

We prefer to write this solution in the order

x < −1 or x > 9,

as it then matches the order of the graphical solution shaded in Figure 8. That is, thesolution set is {x : x < −1 or x > 9}.

The algebraic technique of this last example leads to the following property.

Property 17. If a > 0, then the inequality |x| > a is equivalent to the compoundinequality x < −a or x > a.

This property provides a simple algebraic technique for solving inequalities of theform |x| > a, when a > 0. Let’s concentrate on this technique in the examples thatfollow.

I Example 18. Solve the inequality |4x− 3| > 1 for x.

The first step is to use Property 17 to write that

|4x− 3| > 1

is equivalent to

4x− 3 < −1 or 4x− 3 > 1.

We can now solve each inequality independently. We begin by adding 3 to both sidesof each inequality, then we divide both sides of the resulting inequalities by 4.

4x− 3 < −1 or 4x− 3 > 1

4x < 2 4x > 4

x <12

x > 1

We can sketch the solutions on a number line.

1/2 1


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And we can describe the solution using interval and set-builder notation.

(−∞, 1/2) ∪ (1,∞) = {x : x < 1/2 or x > 1}

Again, let a > 0. As we did with |x| ≤ a, we can take the union of the solutions of|x| = a and |x| > a to find the solution of |x| ≥ a. This leads to the following property.

Property 19. If a > 0, then the inequality |x| ≥ a is equivalent to the inequalityx ≤ −a or x ≥ a.

I Example 20. Solve the inequality 3|1− x| − 4 ≥ |1− x| for x.

Again, at first glance, the inequality

3|1− x| − 4 ≥ |1− x|

looks unlike any inequality we’ve attempted to this point. However, if we subtract|1− x| from both sides of the inequality, then add 4 to both sides of the inequality, weget

3|1− x| − |1− x| ≥ 4.

On the left, we have like terms. Note that 3|1−x|−|1−x| = 3|1−x|−1|1−x| = 2|1−x|.Thus,

2|1− x| ≥ 4.

Divide both sides of the last inequality by 2.

|1− x| ≥ 2

We can now use Property 19 to write

1− x ≤ −2 or 1− x ≥ 2.

We can solve each of these inequalities independently. First, subtract 1 from both sidesof each inequality, then multiply both sides of each resulting inequality by −1, reversingeach inequality as you go.

1− x ≤ −2 or 1− x ≥ 2

− x ≤ −3 − x ≥ 1

x ≥ 3 x ≤ −1

We prefer to write this in the order

x ≤ −1 or x ≥ 3.


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We can sketch the solutions on a number line.

−1 3

And we can describe the solutions using interval and set-builder notation.

(−∞,−1] ∪ [3,∞) = {x : x ≤ −1 or x ≥ 3}

Revisiting DistanceIf a and b are any numbers on the real line, then the distance between a and b is foundby taking the absolute value of their difference. That is, the distance d between a andb is calculated with d = |a − b|. More importantly, we’ve learned to pronounce thesymbolism |a − b| as “the distance between a and b.” This pronunciation is far moreuseful than saying “the absolute value of a minus b.”

I Example 21. Solve the inequality |x− 3| < 8 for x.

This inequality is pronounced “the distance between x and 3 is less than 8.” Drawa number line, locate 3 on the line, then note two points that are 8 units away from 3.

3−5 11

8 8

Now, we need to shade the points that are less than 8 units from 3.

3−5 11

Hence, the solution of the inequality |x− 3| < 8 is

(−5, 11) = {x : −5 < x < 11}.

I Example 22. Solve the inequality |x+ 5| > 2 for x.

First, write the inequality as a difference.

|x− (−5)| > 2

This last inequality is pronounced “the distance between x and −5 is greater than 2.”Draw a number line, locate −5 on the number line, then note two points that are 2units from −5.


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−5−7 −3

2 2

Now, we need to shade the points that are greater than 2 units from −5.

−5−7 −3

Hence, the solution of the inequality |x+ 5| > 2 is

(−∞,−7) ∪ (−3,∞) = {x : x < −7 or x > −3}.

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