Section 2.4 Solvin Equations and Inequalities b Graphin 139 2.4 Solvin Equations and Inequalities b Graphin Our emphasis in the chapter has been on unctions and the interpretation o their raphs. In this section, we continue in that vein and turn our eploration to the solution o equations and inequalities b raphin. The equations will have the orm () = (), and the inequalities will have orm () < () and/or () > (). You miht wonder wh we have ailed to mention inequalities havin the orm () () and () (). The reason or this omission is the act that the solution o the inequalit () () is simpl the union o the solutions o () = () and () < (). Ater all, is pronounced less than or equal. Similar comments are in order or the inequalit () (). We will bein b comparin the unction values o two unctions and at various values o in their domains. Comparin Functions Suppose that we evaluate two unctions and at a particular value o. One o three outcomes is possible. Either () = (), or () > (), or () < (). It s prett straihtorward to compare two unction values at a particular value i rules are iven or each unction. Eample 1. Given () = 2 and () = 2+3, compare the unctions at = 2, 0, and 3. Simple calculations reveal the relations. At = 2, so clearl, ( 2) > ( 2). At = 0, so clearl, (0) < (0). Finall, at = 3, so clearl, (3) = (3). ( 2) = ( 2) 2 = 4 and ( 2) = 2( 2) + 3 = 1, (0) = (0) 2 = 0 and (0) = 2(0) + 3 = 3, (3) = (3) 2 = 9 and (3) = 2(3) + 3 = 9, We can also compare unction values at a particular value o b eaminin the raphs o the unctions. For eample, consider the raphs o two unctions and in Fiure 1. 1 Coprihted material. See: http://msenu.redwoods.edu/intaltet/
140 Chapter 2 Functions Fiure 1. Each side o the equation () = () has its own raph. Net, suppose that we draw a dashed vertical line throuh the point o intersection o the raphs o and, then select a value o that lies to the let o the dashed vertical line, as shown in Fiure 2(a). Because the raph o lies above the raph o or all values o that lie to the let o the dashed vertical line, it will be the case that () > () or all such (see Fiure 2(a)). 2 On the other hand, the raph o lies below the raph o or all values o that lie to the riht o the dashed vertical line. Hence, or all such, it will be the case that () < () (see Fiure 2(b)). 3 (,()) () () (,()) (,()) () () (,()) (a) To the let o the vertical dashed line, the raph o lies above the raph o. Fiure 2. Comparin and. (b) To the riht o the vertical dashed line, the raph o lies below the raph o. 2 When thinkin in terms o the vertical direction, reater than is equivalent to sain above. 3 When thinkin in terms o the vertical direction, less than is equivalent to sain below.
Section 2.4 Solvin Equations and Inequalities b Graphin 141 Finall, i we select the -value o the point o intersection o the raphs o and, then or this value o, it is the case that () and () are equal; that is, () = () (see Fiure 3). (), () Let s summarize our indins. Fiure 3. The unction values () and () are equal where the raphs o and intersect. Summar 2. The solution o the equation () = () is the set o all or which the raphs o and intersect. The solution o the inequalit () < () is the set o all or which the raph o lies below the raph o. The solution o the inequalit () > () is the set o all or which the raph o lies above the raph o. Let s look at an eample. Eample 3. Given the raphs o and in Fiure 4(a), use both set-builder and interval notation to describe the solution o the inequalit () < (). Then ind the solutions o the inequalit () > () and the equation () = () in a similar ashion. To ind the solution o () < (), we must locate where the raph o lies below the raph o. We draw a dashed vertical line throuh the point o intersection o the raphs o and (see Fiure 4(b)), then note that the raph o lies below the raph o to the let o this dashed line. Consequentl, the solution o the inequalit () < () is the collection o all that lie to the let o the dashed line. This set is shaded in red (or in a thicker line stle i viewin in black and white) on the -ais in Fiure 4(b).
142 Chapter 2 Functions 2 (a) The raphs o and. Fiure 4. Comparin and. (b) The solution o () < (). Note that the shaded points on the -ais have -values less than 2. Hence, the solution o () < () is (, 2) = { : < 2}. In like manner, the solution o () > () is ound b notin where the raph o lies above the raph o and shadin the correspondin -values on the -ais (see Fiure (a)). The solution o () > () is (2, ), or alternativel, { : > 2}. To ind the solution o () = (), note where the raph o intersects the raph o, then shade the -value o this point o intersection on the -ais (see Fiure (b)). Thereore, the solution o () = () is { : = 2}. This is not an interval, so it is not appropriate to describe this solution with interval notation. 2 2 (a) The solution o () > (). (b) The solution o () = (). Fiure. Further comparisons. Let s look at another eample.
Section 2.4 Solvin Equations and Inequalities b Graphin 143 Eample 4. Given the raphs o and in Fiure 6(a), use both set-builder and interval notation to describe the solution o the inequalit () > (). Then ind the solutions o the inequalit () < () and the equation () = () in a similar ashion. 2 3 (a) The raphs o and Fiure 6. Comparin and. (b) The solution o () > (). To determine the solution o () > (), we must locate where the raph o lies above the raph o. Draw dashed vertical lines throuh the points o intersection o the raphs o and (see Fiure 6(b)), then note that the raph o lies above the raph o between the dashed vertical lines just drawn. Consequentl, the solution o the inequalit () > () is the collection o all that lie between the dashed vertical lines. We have shaded this collection on the -ais in red (or with a thicker line stle or those viewin in black and white) in Fiure 6(b). Note that the points shaded on the -ais in Fiure 6(b) have -values between 2 and 3. Consequentl, the solution o () > () is ( 2, 3) = { : 2 < < 3}. In like manner, the solution o () < () is ound b notin where the raph o lies below the raph o and shadin the correspondin -values on the -ais (see Fiure 7(a)). Thus, the solution o () < () is (, 2) (3, ) = { : < 2 or > 3}. To ind the solution o () = (), note where the raph o intersects the raph o, and shade the -value o each point o intersection on the -ais (see Fiure 7(b)). Thereore, the solution o () = () is { : = 2 or = 3}. Because this solution set is not an interval, it would be inappropriate to describe it with interval notation.
144 Chapter 2 Functions 2 3 2 3 (a) The solution o () < (). Fiure 7. (b) The solution o () = (). Further comparisons. Solvin Equations and Inequalities with the Graphin Calculator We now know that the solution o () = () is the set o all or which the raphs o and intersect. Thereore, the raphin calculator becomes an indispensable tool when solvin equations. Eample. Use a raphin calculator to solve the equation 1.23 4.6 =.28 2.3. (6) Note that equation (6) has the orm () = (), where () = 1.23 4.6 and () =.28 2.3. Thus, our approach will be to draw the raphs o and, then ind the -value o the point o intersection. First, load () = 1.23 4.6 into Y1 and () =.28 2.3 into Y2 in the Y= menu o our raphin calculator (see Fiure 8(a)). Select 6:ZStandard in the ZOOM menu to produce the raphs in Fiure 8(b). (a) (b) Fiure 8. Sketchin the raphs o () = 1.23 4.6 and () =.28 2.3.
Section 2.4 Solvin Equations and Inequalities b Graphin 14 The solution o equation (6) is the -value o the point o intersection o the raphs o and in Fiure 8(b). We will use the intersect utilit in the CALC menu on the raphin calculator to determine the coordinates o the point o intersection. We proceed as ollows: Select 2nd CALC (push the 2nd button, ollowed b the TRACE button), which opens the menu shown in Fiure 9(a). Select :intersect. The calculator responds b placin the cursor on one o the raphs, then asks i ou want to use the selected curve. You respond in the airmative b pressin the ENTER ke on the calculator. The calculator responds b placin the cursor on the second raph, then asks i ou want to use the selected curve. Respond in the airmative b pressin the ENTER ke. The calculator responds b askin ou to make a uess. In this case, there are onl two raphs on the calculator, so an uess is appropriate. 4 Simpl press the ENTER ke to use the current position o the cursor as our uess. (a) (b) (c) (d) Fiure 9. Usin the intersect utilit. The result o this sequence o steps is shown in Fiure 10. The coordinates o the point o intersection are approimatel (2.7486034, 1.179218). The -value o this point o intersection is the solution o equation (6). That is, the solution o 1.23 4.6 =.28 2.3 is approimatel 2.7486034. Fiure 10. The coordinates o the point o intersection. 4 We will see in the case where there are two points o intersection, that the uess becomes more important. It is important to remember that ever time ou pick up our calculator, ou are onl approimatin a solution. 6 Please use a ruler to draw all lines.
146 Chapter 2 Functions Summar 7. Guidelines. You ll need to discuss epectations with our teacher, but we epect our students to summarize their results as ollows. 1. Set up a coordinate sstem. 6 Label and scale each ais with min, ma, min, and ma. 2. Cop the imae in our viewin window onto our coordinate sstem. Label each raph with its equation. 3. Draw a dashed vertical line throuh the point o intersection. 4. Shade and label the solution o the equation on the -ais. The result o ollowin this standard is shown in Fiure 11. 10 10 2.7486034 10 10 Fiure 11. Summarizin the solution o equation (6). =1.23 4.6 =.28 2.3 Let s look at another eample. Eample 8. inequalit Use set-builder and interval notation to describe the solution o the 0.8 2 3 1.23 + 1.2. (9) Note that the inequalit (9) has the orm () (), where () = 0.8 2 3 and () = 1.23 + 1.2. Load () = 0.8 2 3 and () = 1.23 + 1.2 into Y1 and Y2 in the Y= menu, respectivel, as shown in Fiure 12(a). Select 6:ZStandard rom the ZOOM menu to produce the raphs shown in Fiure 12(b). To ind the points o intersection o the raphs o and, we ollow the same sequence o steps as we did in Eample up to the point where the calculator asks ou to make a uess (i.e., 2nd CALC, :intersect, First curve ENTER, Second curve ENTER). Because there are two points o intersection, when the calculator asks ou to
Section 2.4 Solvin Equations and Inequalities b Graphin 147 (a) (b) Fiure 12. The raphs o () = 0.8 2 3 and () = 1.23 + 1.2. make a uess, ou must move our cursor (with the arrow kes) so that it is closer to the point o intersection ou wish to ind than it is to the other point o intersection. Usin this technique produces the two points o intersection ound in Fiures 13(a) and (b). (a) (b) Fiure 13. The points o intersection o the raphs o and. The approimate coordinates o the irst point o intersection are ( 1.626682, 0.708192). The second point o intersection has approimate coordinates (3.0737411,.030701). It is important to remember that ever time ou pick up our calculator, ou are onl ettin an approimation. It is possible that ou will et a slihtl dierent result or the points o intersection. For eample, ou miht et ( 1.62668, 0.708187) or our point o intersection. Based on the position o the cursor when ou marked the curves and made our uess, ou can et slihtl dierent approimations. Note that this second solution is ver nearl the same as the one we ound, dierin onl in the last ew decimal places, and is perectl acceptable as an answer. We now summarize our results b creatin a coordinate sstem, labelin the aes, and scalin the aes with the values o the window parameters min, ma, min, and ma. We cop the imae in our viewin window onto this coordinate sstem, labelin each raph with its equation. We then draw dashed vertical lines throuh each point o intersection, as shown in Fiure 14. We are solvin the inequalit 0.8 2 3 1.23 + 1.2. The solution will be the union o the solutions o 0.8 2 3 > 1.23 + 1.2 and 0.8 2 3 = 1.23 + 1.2. To solve 0.8 2 3 > 1.23 + 1.2, we note where the raph o = 0.8 2 3 lies above the raph o = 1.23 + 1.2 and shade the correspondin -values
148 Chapter 2 Functions =0.8 2 3 10 10 3.0737411 1.626682 10 =1.23+1.2 10 Fiure 14. Summarizin the solution o 0.8 2 3 1.23 + 1.2. on the -ais. In this case, the raph o = 0.8 2 3 lies above the raph o = 1.23 + 1.2 or values o that lie outside o our dashed vertical lines. To solve 0.8 2 3 = 1.23 + 1.2, we note where the raph o = 0.8 2 3 intersects the raph o = 1.23 + 1.2 and shade the correspondin -values on the -ais. This is wh the points at 1.626682 and 3.0737411 are illed. Thus, all values o that are either less than or equal to 1.626682 or reater than or equal to 3.0737411 are solutions. That is, the solution o inequalit 0.8 2 3 > 1.23 + 1.2 is approimatel (, 1.626682] [3.0737411, ) = { : 1.626682 or 3.0737411}. Comparin Functions with Zero When we evaluate a unction at a particular value o, onl one o three outcomes is possible. Either () = 0, or () > 0, or () < 0. That is, either () equals zero, or () is positive, or () is neative. There are no other possibilities. We could start resh, takin a completel new approach, or we can build on what we alread know. We choose the latter approach. Suppose that we are asked to compare () with zero? Is it equal to zero, is it reater than zero, or is it smaller than zero? We set () = 0. Now, i we want to compare the unction with zero, we need onl compare with, which we alread know how to do. To ind where () = (), we note where the raphs o and intersect, to ind where () > (), we note where the raph o lies above the raph o, and inall, to ind where () < (), we simpl note where the raph o lies below the raph o.
Section 2.4 Solvin Equations and Inequalities b Graphin 149 However, the raph o () = 0 is a horizontal line coincident with the -ais. Indeed, () = 0 is the equation o the -ais. This arument leads to the ollowin ke results. Summar 10. The solution o () = 0 is the set o all or which the raph o intersects the -ais. The solution o () > 0 is the set o all or which the raph o lies strictl above the -ais. The solution o () < 0 is the set o all or which the raph o lies strictl below the -ais. For eample: To ind the solution o () = 0 in Fiure 1(a), we simpl note where the raph o crosses the -ais in Fiure 1(a). Thus, the solution o () = 0 is = 1. To ind the solution o () > 0 in Fiure 1(b), we simpl note where the raph o lies above the -ais in Fiure 1(b), which is to the riht o the vertical dashed line throuh = 1. Thus, the solution o () > 0 is (1, ) = { : > 1}. To ind the solution o () < 0 in Fiure 1(c), we simpl note where the raph o lies below the -ais in Fiure 1(c), which is to the let o the vertical dashed line at = 1. Thus, the solution o () < 0 is (, 1) = { : < 1}. 1 1 1 (a) The solution o () = 0 (b) The solution o () > 0 (c) The solution o () < 0 Fiure 1. We net deine some important terminolo. Comparin the unction with zero. Deinition 11. I (a) = 0, then a is called a zero o the unction. The raph o will intercept the -ais at (a, 0), a point called the -intercept o the raph o. Your calculator has a utilit that will help ou to ind the zeros o a unction.
10 Chapter 2 Functions Eample 12. Use a raphin calculator to solve the inequalit 0.2 2 1.24 3.84 0. Note that this inequalit has the orm () 0, where () = 0.2 2 1.24 3.84. Our strate will be to draw the raph o, then determine where the raph o lies below or on the -ais. We proceed as ollows: First, load the unction () = 0.2 2 1.24 3.84 into the Y1 in the Y= menu o our calculator. Select 6:ZStandard rom the ZOOM menu to produce the imae in Fiure 16(a). Press 2nd CALC to open the menu shown in Fiure 16(b), then select 2:zero to start the utilit that will ind a zero o the unction (an -intercept o the raph). The calculator asks or a Let Bound, so use our arrow kes to move the cursor slihtl to the let o the letmost -intercept o the raph, as shown in Fiure 16(c). Press ENTER to record this Let Bound. The calculator then asks or a Riht Bound, so use our arrow kes to move the cursor slihtl to the riht o the -intercept, as shown in Fiure 16(d). Press ENTER to record this Riht Bound. (a) (b) (c) (d) Fiure 16. Findin a zero or -intercept with the calculator. The calculator responds b markin the let and riht bounds on the screen, as shown in Fiure 17(a), then asks ou to make a reasonable startin uess or the zero or -intercept. You ma use the arrow kes to move our cursor to an point, so lon as the cursor remains between the let- and riht-bound marks on the viewin window. We usuall just leave the cursor where it is and press the ENTER to record this uess. We suest ou do that as well. The calculator responds b indin the coordinates o the -intercept, as shown in Fiure 17(b). Note that the -coordinate o the -intercept is approimatel 2.17931. Repeat the procedure to ind the coordinates o the rihtmost -intercept. The result is shown in Fiure 17(c). Note that the -coordinate o the intercept is approimatel 7.1179306. The inal step is the interpretation o results and recordin o our solution on our homework paper. Reerrin to the Summar 7 Guidelines, we come up with the raph shown in Fiure 18.
Section 2.4 Solvin Equations and Inequalities b Graphin 11 (a) (b) (c) Fiure 17. Findin a zero or -intercept with the calculator. 10 ()= 0.2 2 1.24 3.84 10 2.17931 7.1179306 10 10 Fiure 18. The solution o 0.2 2 1.24 3, 84 0. Several comments are in order. Notin that () = 0.2 2 1.24 3.84, we note: 1. The solutions o () = 0 are the points where the raph crosses the -ais. That s wh the points ( 2.17931, 0) and (7.1179306, 0) are shaded and illed in Fiure 18. 2. The solutions o () < 0 are those values o or which the raph o alls strictl below the -ais. This occurs or all values o between 2.17931 and 7.1179306. These points are also shaded on the -ais in Fiure 18. 3. Finall, the solution o () 0 is the union o these two shadins, which we describe in interval and set-builder notation as ollows: [ 2.17931, 7.1179306] = { : 2.17931 7.1179306}