Solving x < a. Section 4.4 Absolute Value Inequalities 391

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1 Section 4.4 Absolute Value Inequalities Absolute Value Inequalities In the last section, we solved absolute value equations. In this section, we turn our attention to inequalities involving absolute value. Solving < a The solutions of < a again depend upon the value and sign of the number a. To solve < a graphicall, we must determine where the graph of the left-hand side lies below the graph of the right-hand side of the inequalit < a. There are three cases to consider. Case I: a < 0 In this case, the graph of = a lies strictl below the -ais. As ou can see in Figure 1(a), the graph of = never lies below the graph of = a. Hence, the inequalit < a has no solutions. Case II: a = 0 In this case, the graph of = 0 coincides with the -ais. As ou can see in Figure 1(b), the graph of = never lies strictl below the -ais. Hence, the inequalit < 0 has no solutions. Case III: a > 0 In this case, the graph of = a lies strictl above the -ais. In Figure 1(c), the graph of = and = a intersect at = a and = a. In Figure 1(c), we also see that the graph of = lies strictl below the graph of = a when is in-between a and a; that is, when a < < a. In Figure 1(c), we ve dropped dashed vertical lines from the points of intersection of the two graphs to the -ais. On the -ais, we ve shaded the solution of < a, that is, a < < a. = = = =a =a 0 a a =a 1 (a) (b) (c) Figure 1. The solution of < a has three cases. Coprighted material. See:

2 392 Chapter 4 Absolute Value Functions This discussion leads to the following ke propert. Propert 1. The solution of < a depends upon the value and sign of a. Case I: a < 0 The inequalit < a has no solution. Case II: a = 0 The inequalit < 0 has no solution. Case III: a > 0 The inequalit < a has solution set { : a < < a}. Let s look at some eamples. Eample 2. Solve the inequalit < 5 for. The graph of the left-hand side of < 5 is the V of Figure 1(a). The graph of the right-hand side of < 5 is a horizontal line located 5 units below the -ais. This is the situation shown in Figure 1(a). The graph of = is therefore never below the graph of = 5. Thus, the inequalit < 5 has no solution. An alternate approach is to consider the fact that the absolute value of is alwas nonnegative and can never be less than 5. Thus, the inequalit < 5 has no solution. Eample 3. Solve the inequalit < 0 for. This is the case shown in Figure 1(b). The graph of = is never strictl below the -ais. Thus, the inequalit < 0 has no solution. Eample 4. Solve the inequalit < 8 for. The graph of the left-hand side of < 8 is the V of Figure 1(c). The graph of the right-hand side of < 8 is a horizontal line located 8 units above the -ais. This is the situation depicted in Figure 1(c). The graphs intersect at ( 8, 8) and (8, 8) and the graph of = lies strictl below the graph of = 8 for values of in-between 8 and 8. Thus, the solution of < 8 is 8 < < 8. It helps the intuition if ou check the results of the last eample. Note that numbers between 8 and 8, such as 7.75, 3 and 6.8 satisf the inequalit, 7.75 < 8 and 3 < 8 and 6.8 < 8,

3 Section 4.4 Absolute Value Inequalities 393 while values that do not lie between 8 and 8 do not satisf the inequalit. For eample, none of the numbers 9.3, 8.2, and 11.7 lie between 8 and 8, and each of the following is a false statement. 9.3 < 8 and 8.2 < 8 and 11.7 < 8 (all are false) If ou reflect upon these results, the will help cement the notion that the solution of < 8 is all values of satisfing 8 < < 8. Eample 5. Solve the inequalit 5 2 < 3 for. If the inequalit were < 3, we would not hesitate. This is the situation depicted in Figure 1(a) and the inequalit < 3 has no solutions. The reasoning applied to < 3 works equall well for the inequalit 5 2 < 3. The left-hand side of this inequalit must be nonnegative, so its graph must lie on or above the -ais. The right-hand side of 5 2 < 3 is a horizontal line located 3 units below the -ais. Therefore, the graph of = 5 2 can never lie below the graph of = 3 and the inequalit 5 2 < 3 has no solution. We can verif this result with the graphing calculator. Load the left- and right-hand sides of 5 2 < 3 into Y1 and Y2, respectivel, as shown in Figure 2(a). From the ZOOM menu, select 6:ZStandard to produce the image shown in Figure 2(b). As predicted, the graph of = 5 2 never lies below the graph of = 3, so the inequalit 5 2 < 3 has no solution. (a) (b) Figure 2. Using the graphing calculator to solve the inequalit 5 2 < 3. Eample 6. Solve the inequalit 5 2 < 0 for. We know that the left-hand side of the inequalit 5 2 < 0 has the V shape indicated in Figure 1(b). The graph touches the -ais when 5 2 = 0, or when 5 2 = 0 2 = 5 = 5 2. However, the graph of = 5 2 never falls below the -ais, so the inequalit 5 2 < 0 has no solution.

4 394 Chapter 4 Absolute Value Functions Intuitivel, it should be clear that the inequalit 5 2 < 0 has no solution. Indeed, the left-hand side of this inequalit is alwas nonnegative, and can never be strictl less than zero. Eample 7. Solve the inequalit 5 2 < 3 for. In this eample, the graph of the right-hand side of the inequalit 5 2 < 3 is a horizontal line located 3 units above the -ais. The graph of the left-hand side of the inequalit has the V shape shown in Figure 3(b) and (c). You can use the intersect utilit on the graphing calculator to find the points of intersection of the graphs of = 5 2 and = 3, as we have done in Figures 3(b) and (c). Note that the calculator indicates two points of intersection, one at = 1 and a second at = 4. (a) (b) (c) Figure 3. Using the graphing calculator to solve the inequalit 5 2 < 3. The graph of = 5 2 falls below the graph of = 3 for all values of between 1 and 4. Hence, the solution of the inequalit 5 2 < 3 is the set of all satisfing 1 < < 4; i.e. { : 1 < < 4}. Epectations. We need a wa of summarizing this graphing calculator approach on our homework paper. First, draw a reasonable facsimile of our calculator s viewing window on our homework paper. Use a ruler to draw all lines. Complete the following checklist. Label each ais, in this case with and. Scale each ais. To do this, press the WINDOW button on our calculator, then report the values of min, ma, min, and ma on the appropriate ais. Label each graph with its equation. Drop dashed vertical lines from the points of intersection to the -ais. Shade and label the solution set of the inequalit on the -ais. Following the guidelines in the above checklist, we obtain the image in Figure 4.

5 Section 4.4 Absolute Value Inequalities = 5 2 = Figure 4. Reporting a graphical solution of 5 2 < 3. Algebraic Approach. Let s now eplore an algebraic solution of the inequalit 5 2 < 3. Much as < 3 implies that 3 < < 3, the inequalit 5 2 < 3 requires that 3 < 5 2 < 3. We can subtract 5 from all three members of this last inequalit, then simplif. 3 5 < < < 2 < 2 Divide all three members of this last inequalit b 2, reversing the inequalit smbols as ou go. 4 > > 1 We prefer that our inequalities read from small-to-large, so we write 1 < < 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 eample leads us to the following propert. Propert 8. a < < a. If a > 0, then the inequalit < a is equivalent to the inequalit This propert provides a simple method for solving inequalities of the form < a. Let s appl this algebraic technique in the net eample.

6 396 Chapter 4 Absolute Value Functions Eample 9. Solve the inequalit < 7 for. The first step is to use Propert 8 to write that is equivalent to the inequalit < 7 7 < < 7. From here, we can solve for b first subtracting 5 from all three members, then dividing through b < 4 < 2 3 < < 1 2 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 ) { = : 3 < < 1 } 2 2 Assuming that a > 0, the inequalit a requires that we find where the absolute value of is either less than a or equal to a. We know that < a when a < < a and we know that = a when = a or = a. Thus, the solution of a is the union of these two solutions. This argument leads to the following propert. Propert 10. a a. If a > 0, then the inequalit a is equivalent to the inequalit Eample 11. Solve the inequalit for. At first glance, the inequalit has a form quite dissimilar from what we ve done thus far. However, let s subtract 5 from both sides of the inequalit

7 Section 4.4 Absolute Value Inequalities 397 Now, let s divide both sides of this last inequalit b 3, reversing the inequalit sign. 4 3 Aha! Familiar ground. Using Propert 10, this last inequalit is equivalent to 3 4 3, and when we add 4 to all three members, we have the solution. 1 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] = { : 1 7} Solving > a The solutions of > a again depend upon the value and sign of a. To solve > a graphicall, we must determine where the graph of = lies above the graph of = a. Again, we consider three cases. Case I: a < 0 In this case, the graph of = a lies strictl below the -ais. Therefore, the graph of = in Figure 5(a) alwas lies above the graph of = a. Hence, all real numbers are solutions of the inequalit > a. Case II: a = 0 In this case, the graph of = 0 coincides with the -ais. As shown in Figure 5(b), the graph of = will lie strictl above the graph of = 0 for all values of with one eception, namel, cannot equal zero. Hence, ever real number ecept = 0 is a solution of > 0. In Figure 5(b), we ve shaded the solution of > 0, namel the set of all real numbers ecept = 0. Case III: a > 0 In this case, the graph of = a lies strictl above the -ais. In Figure 5(c), the graph of = intersects the graph of = a at = a and = a. In Figure 5(c), we see that the graph of = lies strictl above the graph of = a if is less than a or greater than a. In Figure 5(c), we ve dropped dashed vertical lines from the points of intersection to the -ais. On the -ais, we ve shaded the solution of > a, namel the set of all real numbers such that < a or > a.

8 398 Chapter 4 Absolute Value Functions = = = =a =a 0 a a =a (a) (b) (c) Figure 5. The solution of > a has three cases. This discussion leads to the following propert. Propert 12. The solution of > a depends upon the value and sign of a. Case I: a < 0 All real numbers are solutions of the inequalit > a. Case II: a = 0 All real numbers, with the eception of = 0, are solutions of > 0. Case III: a > 0 The inequalit > a has solution set { : < a or > a}. Eample 13. State the solution of each of the following inequalities. a. > 5 b. > 0 c. > 4 Solution: a. The solution of > 5 is all real numbers. b. The solution of > 0 is all real numbers ecept zero. c. The solution of > 4 is the set of all real numbers less than 4 or greater than 4. Eample 14. Solve the inequalit 4 > 5 for. The left-hand side of the inequalit 4 > 5 is nonnegative, so the graph of = 4 must lie above or on the -ais. The graph of the right-hand side of 4 > 5 is a horizontal line located 5 units below the -ais. Therefore, the graph

9 Section 4.4 Absolute Value Inequalities 399 of = 4 alwas lies above the graph of = 5. Thus, all real numbers are solutions of the inequalit 4 > 5. We can verif our thinking with the graphing calculator. Load the left- and righthand sides of the inequalit 4 > 5 into Y1 and Y2, respectivel, as shown in Figure 6(a). From the ZOOM menu, select 6:ZStandard to produce the image shown in Figure 6(b). As predicted, the graph of = 4 lies above the graph of = 5 for all real numbers. (a) (b) Figure 6. Using the graphing calculator to solve 4 > 5. Intuitivel, the absolute value of an number is alwas nonnegative, so 4 > 5 for all real values of. Eample 15. Solve the inequalit 4 > 0 for. As we saw in Figure 6(b), the graph of = 4 lies on or above the -ais for all real numbers. It touches the -ais at the verte of the V, where This can occur onl if 4 = 0. 4 = 0 = 4 = 4. Thus, the graph of = 4 is strictl above the -ais for all real numbers ecept = 4. That is, the solution of 4 > 0 is { : 4}. Eample 16. Solve the inequalit 4 > 5 for. In this eample, the graph of the right-hand side of 4 > 5 is a horizontal line located 5 units above the -ais. The graph of = 4 has the V shape shown in Figure 6(c). You can use the intersect utilit on the graphing calculator to approimate the points of intersection of the graphs of = 4 and = 5, as we have done in Figure 7(c) and (d). The calculator indicates two points of intersection, one at = 1 and a second at = 9.

10 400 Chapter 4 Absolute Value Functions (a) (b) (c) (d) Figure 7. Using the graphing calculator to solve the inequalit 4 > 5. The graph of = 4 lies above the graph of = 5 for all values of that lie either to the left of 1 or to the right of 9. Hence, the solution of 4 > 5 is the set { : < 1 or > 9}. Following the guidelines established in Eample 7, we create the image shown in Figure 8 on our homework paper. Note that we ve labeled each ais, scaled each ais with min, ma, min, and ma, labeled each graph with its equation, and shaded and labeled the solution on -ais. 10 = 4 = Figure 8. Reporting a graphical solution of 4 > 5. Algebraic Approach. Let s eplore an algebraic solution of 4 > 5. In much the same manner that > 5 leads to the conditions < 5 or > 5, the inequalit requires that 4 > 5 4 < 5 or 4 > 5. We can solve each of these independentl b first subtracting 4 from each side of the inequalit, then multipling both sides of each inequalit b 1, reversing each inequalit as we do so.

11 Section 4.4 Absolute Value Inequalities < 5 or 4 > 5 < 9 > 1 > 9 We prefer to write this solution in the order < 1 or > 9, < 1 as it then matches the order of the graphical solution shaded in Figure 8. That is, the solution set is { : < 1 or > 9}. The algebraic technique of this last eample leads to the following propert. Propert 17. If a > 0, then the inequalit > a is equivalent to the compound inequalit < a or > a. This propert provides a simple algebraic technique for solving inequalities of the form > a, when a > 0. Let s concentrate on this technique in the eamples that follow. Eample 18. Solve the inequalit 4 3 > 1 for. The first step is to use Propert 17 to write that is equivalent to 4 3 > < 1 or 4 3 > 1. We can now solve each inequalit independentl. We begin b adding 3 to both sides of each inequalit, then we divide both sides of the resulting inequalities b < 1 or 4 3 > 1 4 < 2 4 > 4 < 1 2 > 1 We can sketch the solutions on a number line. 1/2 1

12 402 Chapter 4 Absolute Value Functions And we can describe the solution using interval and set-builder notation. (, 1/2) (1, ) = { : < 1/2 or > 1} Again, let a > 0. As we did with a, we can take the union of the solutions of = a and > a to find the solution of a. This leads to the following propert. Propert 19. If a > 0, then the inequalit a is equivalent to the inequalit a or a. Eample 20. Solve the inequalit for. Again, at first glance, the inequalit looks unlike an inequalit we ve attempted to this point. However, if we subtract 1 from both sides of the inequalit, then add 4 to both sides of the inequalit, we get On the left, we have like terms. Note that = = 2 1. Thus, Divide both sides of the last inequalit b 2. We can now use Propert 19 to write or 1 2. We can solve each of these inequalities independentl. First, subtract 1 from both sides of each inequalit, then multipl both sides of each resulting inequalit b 1, reversing each inequalit as ou go. 1 2 or We prefer to write this in the order 1 or 3. 1

13 Section 4.4 Absolute Value Inequalities 403 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, ) = { : 1 or 3} Revisiting Distance If a and b are an numbers on the real line, then the distance between a and b is found b taking the absolute value of their difference. That is, the distance d between a and b is calculated with d = a b. More importantl, we ve learned to pronounce the smbolism a b as the distance between a and b. This pronunciation is far more useful than saing the absolute value of a minus b. Eample 21. Solve the inequalit 3 < 8 for. This inequalit is pronounced the distance between and 3 is less than 8. Draw a number line, locate 3 on the line, then note two points that are 8 units awa from Now, we need to shade the points that are less than 8 units from Hence, the solution of the inequalit 3 < 8 is ( 5, 11) = { : 5 < < 11}. Eample 22. Solve the inequalit + 5 > 2 for. First, write the inequalit as a difference. ( 5) > 2 This last inequalit is pronounced the distance between and 5 is greater than 2. Draw a number line, locate 5 on the number line, then note two points that are 2 units from 5.

14 404 Chapter 4 Absolute Value Functions Now, we need to shade the points that are greater than 2 units from Hence, the solution of the inequalit + 5 > 2 is (, 7) ( 3, ) = { : < 7 or > 3}.