2.5 Absolute Value Equations and Inequalities

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1 660_ch0pp qd 0/6/08 4: PM Page CHAPTER Linear Functions and Equations Writing about Mathematics 0. Suppose the solution to the equation a + b = 0 with a 7 0 is = k. Discuss how the value of k can be used to help solve the linear inequalities a + b 7 0 and a + b 6 0. Illustrate this process graphicall. How would the solution sets change if a 6 0? 0. Describe how to numericall solve the linear inequalit a + b 0. Give an eample. 0. If ou multipl each part of a three-part inequalit b the same negative number, what must ou make sure to do? Eplain b using an eample. 04. Eplain how a linear function, a linear equation, and a linear inequalit are related. Give an eample. EXTENDED AND DISCOVERY EXERCISES. Arithmetic Mean The arithmetic mean of two numbers a + b a and b is given b. Use properties of inequalities to show that if a 6 b, then a 6 a + b 6 b.. Geometric Mean The geometric mean of two numbers a and b is given b ab. Use properties of inequalities to show that if 0 6 a 6 b, then a 6 ab 6 b. CHECKING BASIC CONCEPTS FOR SECTIONS. AND.4. Solve the linear equation 4( - ) = (5 - ) - b (b) using each method. Compare our results. (a) Graphical (b) Numerical (c) Smbolic (c). Solve the inequalit ( - 4) 7 -. Epress the solution set in set-builder notation.. Solve the compound inequalit - -. Use set-builder interval notation. 4. Use the graph to the right to solve each equation and inequalit. Then solve each part smbolicall. Use setbuilder interval notation when possible. (a) -( - ) - - = 0 -( - ) ( - ) = ( ) Absolute Value Equations and Inequalities Evaluate and graph the absolute value function Solve absolute value equations Solve absolute value inequalities Introduction A margin of err can be ver imptant in man aspects of life, including being fired out of a cannon. The most dangerous part of the feat, first done b a human in 875, is to land squarel on a net. F a human cannonball who wants to fl 80 feet in the air and then land in the center of a net with a 60-foot-long safe zone, there is a margin of err of 0 feet. That is, the hizontal distance D traveled b the human cannonball can var between 80-0 = 50 feet and = 0 feet. (Source: Ontario Science Center.) This margin of err can be epressed mathematicall b using the absolute value inequalit ƒ D - 80 ƒ 0. The absolute value is necessar because D can be either less than greater than 80, but b not me than 0 feet.

2 660_ch0pp qd 0/6/08 4: PM Page 47.5 Absolute Value Equations and Inequalities = Figure.56 The Absolute Value Function The Absolute Value Function The graph of = ƒ ƒ is shown in Figure.56. It is V-shaped and cannot be represented b a single linear function. However, it can be represented b the lines = (when Ú 0) and = - (when 6 0). This suggests that the absolute value function can be defined smbolicall using a piecewise-linear function. The absolute value function is decreasing f 0 and increasing f Ú 0. There is another fmula f ƒ ƒ. Consider the following eamples. = 9 = 7 = 49 = 7 - if 6 0 ƒ ƒ = e if Ú 0 and and (-) = 9 =. (-7) = 49 = 7. That is, regardless of whether a real number is positive negative, the epression equals the absolute value of. This statement is summarized b = ƒ ƒ f all real numbers. F eample, = ƒ ƒ, ( - ) = ƒ - ƒ, and () = ƒ ƒ. EXAMPLE Analzing the graph of = ƒ a + b ƒ F each linear function ƒ, graph = ƒ() and = ƒ ƒ()ƒ separatel. Discuss how the absolute value affects the graph of ƒ. (a) ƒ() = + (b) ƒ() = (a) The graphs of = + and = ƒ + ƒ are shown in Figures.57 and.58, respectivel. The graph of is a line with -intercept -. The graph of is V-shaped. The graphs are identical f 7 -. F 6 -, the graph of = ƒ() passes below the -ais, and the graph of = ƒ ƒ()ƒ is the reflection of = ƒ() across the -ais. The graph of = ƒ ƒ()ƒ does not dip below the -ais because an absolute value is never negative. = + = Figure.57 Figure.58 NOTE In general, the graph of = ƒ ƒ() ƒ is a reflection of the graph of = ƒ() across the -ais whenever ƒ() 6 0. Otherwise (whenever ƒ() Ú 0), their graphs are identical.

3 660_ch0pp qd 0/6/08 4: PM Page CHAPTER Linear Functions and Equations Calculat Help To access the absolute value function, see Appendi A (page AP-0). (b) The graphs of = and = ƒ ƒ are shown in Figures.59 and.60. Again, the graph of is V-shaped. The graph of = ƒ ƒ()ƒ is the reflection of ƒ across the -ais whenever the graph of = ƒ() is below the -ais. = = Figure.59 Figure.60 Now Tr Eercises and 7 8 = 5 ( 5, 5) 4 (5, 5) = Figure.6 Eample illustrates the fact that the graph of = ƒ a + b ƒ with a Z 0 is V-shaped and is never located below the -ais. The verte ( point) of the V-shaped graph cresponds to the -intercept, which can be found b solving the linear equation a + b = 0. Absolute Value Equations The equation ƒ ƒ = 5 has two solutions: 5. This fact is shown visuall in Figure.6, where the graph of = ƒ ƒ intersects the hizontal line = 5 at the points ( 5, 5). In general, the solutions to ƒ ƒ = k with k 7 0 are given b = k. Thus if = a + b, then ƒ a + b ƒ = k has two solutions given b a + b = k. These concepts can be illustrated visuall. The graph of = ƒ a + b ƒ with a Z 0 is V-shaped. It intersects the hizontal line = k twice whenever k 7 0, as illustrated in Figure.6. Thus there are two solutions to the equation ƒ a + b ƒ = k. This V-shaped graph intersects the line = 0 once, as shown in Figure.6. As a result, equation ƒ a + b ƒ = 0 has one solution, which cresponds to the -intercept. When k 6 0, the line = k lies below the -ais and there are no points of intersection, as shown in Figure.64. Thus the equation ƒ a + b ƒ = k with k 6 0 has no solutions. = a + b = k, k > 0 = a + b = a + b = 0 = k, k < 0 Figure.6 Two Solutions Figure.6 One Solution Figure.64 No Solutions Absolute Value Equations Let k be a positive number. Then ƒ a + b ƒ = k is equivalent to a + b = k.

4 660_ch0pp qd 0/6/08 4: PM Page 49.5 Absolute Value Equations and Inequalities 49 EXAMPLE Solving absolute value equations Solve each equation. (a) ƒ 4-6 ƒ = 5 (b) ƒ - ƒ = - (c) ƒ - ƒ - 5 = - = = Figure.65 No Solutions (a) The equation ƒ is satisfied when 4-6 = ƒ = = = -5 Equations to solve 4 = 4 = -9 Add 6 to each side. 4 = 8 = - Multipl b. The solutions are - and 8. (b) Because an absolute value is never negative, ƒ - ƒ Ú 0 f all and can never equal -. There are no solutions. This is illustrated graphicall in Figure.65. (c) Because the right side of the equation is a negative number, it might appear at first glance that there were no solutions. However, if we add 5 to each side of the equation, ƒ - ƒ - 5 = - becomes ƒ - ƒ =. This equation is equivalent to - = and has two solutions. - = = 5 - = - = - Equations to solve Add to each side. = 5 = - Divide b. The solutions are - 5 and. Now Tr Eercises, 9, and EXAMPLE Solving an equation with technolog Solve the equation ƒ + 5 ƒ = graphicall, numericall, and smbolicall. Graphical Solution Graph Y = abs(x + 5) and Y =. The V-shaped graph of intersects the hizontal line at the points (-.5, ) and (-.5, ), as shown in Figures.66 and.67. The solutions are -.5 and -.5. solutions to = are -.5 and -.5. Numerical Solution Table Y = abs(x + 5) and Y =, as shown in Figure.68. The Calculat Help To find a point of intersection, see Appendi A (page AP-8). [-9, 9, ] b [-6, 6, ] [-9, 9, ] b [-6, 6, ] Intersection X.5 Figure.66 = + 5 Y = Intersection X.5 Figure.67 = + 5 Y = X Y Y 4.5 = = Y abs(x 5) Figure.68

5 660_ch0pp qd 0/6/08 4: PM Page CHAPTER Linear Functions and Equations Smbolic Solution The equation ƒ + 5 ƒ = is satisfied when + 5 =. + 5 = + 5 = - Equations to solve = - = -7 Subtract 5 from each side. = - = - 7 Divide b. Now Tr Eercise 45 EXAMPLE 4 Describing speed limits with absolute values The lawful speeds S on an interstate highwa satisf ƒ S - 55 ƒ 5. Find the maimum and minimum speed limits b solving the equation ƒ S - 55 ƒ = 5. The equation ƒ S - 55 ƒ = 5 is equivalent to S - 55 = 5. S - 55 = 5 S - 55 = -5 Equations to solve S = 70 S = 40 Add 55 to each side. The maimum speed limit is 70 miles per hour and the minimum is 40 miles per hour. Now Tr Eercise 7 An Equation with Two Absolute Values Sometimes me than one absolute value sign occurs in an equation. F eample, an equation might be in the fm ƒ a b ƒ ƒ c d ƒ. In this case there are two possibilities: either a b c d a b (c d ). This smbolic technique is demonstrated in the net eample. EXAMPLE 5 Solving an equation involving two absolute values Solve the equation ƒ - ƒ = ƒ - ƒ. We must solve both of the following equations. - = - = = - = -( - ) - = = There are two solutions: - and. Now Tr Eercise 5 Absolute Value Inequalities In Figure.69 the solutions to ƒ a + b ƒ = k are labeled s and s. The V-shaped graph of = ƒ a + b ƒ is below the hizontal line = k between s and s, when s 6 6 s. The solution set f the inequalit ƒ a + b ƒ 6 k is green on the -ais. In Figure.70 the V-shaped graph is above the hizontal line = k left of s right of s, that is, when 6 s 7 s. The solution set f the inequalit ƒ a + b ƒ 7 k is green on the -ais.

6 660_ch0pp qd 0/7/08 0: AM Page 5.5 Absolute Value Equations and Inequalities 5 = k = a + b = k = a + b s s a + b < k s s a + b > k Figure.69 Figure.70 (, ) = (, ) = + Figure.7 Note that in both figures equalit (determined b and s ) is the boundar between greater than and less than. F this reason, s and s are called boundar numbers. F eample, the graphs of = ƒ + ƒ and = are shown in Figure.7. These graphs intersect at the points (-, ) and (, ). It follows that the two solutions to ƒ + ƒ = are s = - and s =. The solutions to ƒ + ƒ 6 lie between s = - and s =, which can be written as Furtherme, the solutions to ƒ + ƒ 7 lie outside s = - and s =. This can be written as 6-7. These results are generalized as follows. Absolute Value Inequalities Let the solutions to ƒ a + b ƒ = k be s and s, where s 6 s and k ƒ a + b ƒ 6 k is equivalent to s 6 6 s.. ƒ a + b ƒ 7 k is equivalent to 6 s 7 s. Similar statements can be made f inequalities involving Ú. s NOTE The union smbol ma be used to write 6 s 7 s in interval notation. F eample, 6-7 is written as (- q, -) (, q) in interval notation. This indicates that the solution set includes all real numbers in either (- q, -) (, q). EXAMPLE 6 Solving inequalities involving absolute values smbolicall Solve each inequalit smbolicall. Write the solution set in interval notation. (a) ƒ - 5 ƒ 6 (b) ƒ 5 - ƒ 7 (a) Begin b solving ƒ - 5 ƒ = 6, equivalentl, - 5 = = 6 = - 5 = -6 = - = = - The solutions to ƒ - 5 ƒ = 6 are - and. The solution set f the inequalit ƒ - 5 ƒ 6 includes all real numbers satisfing -. In interval notation this is written as C -, D.

7 660_ch0pp qd 0/6/08 4: PM Page 5 5 CHAPTER Linear Functions and Equations (b) To solve ƒ 5 - ƒ 7, begin b solving ƒ 5 - ƒ =, equivalentl, 5 - =. 5 - = - = - = 5 - = - - = -8 = 8 The solutions to ƒ 5 - ƒ = are and 8. The solution set f ƒ 5 - ƒ 7 includes all real numbers left of right of 8. Thus ƒ 5 - ƒ 7 is equivalent to In interval notation this is written as (- q, ) (8, q). Now Tr Eercises 55 and 6 EXAMPLE 7 Analzing the temperature range in Santa Fe The inequalit ƒ T - 49 ƒ 0 describes the range of monthl average temperatures T in degrees Fahrenheit f Santa Fe, New Meico. (Source: A. Miller and J. Thompson, Elements of Meteolog.) (a) Solve this inequalit graphicall and smbolicall. (b) The high and low monthl average temperatures satisf the absolute value equation ƒ T - 49 ƒ = 0. Use this fact to interpret the results from part (a). (a) Graphical Solution Graph Y = abs(x - 49) and Y = 0, as in Figure.7. The V-shaped graph of intersects the hizontal line at the points (9, 0) and (69, 0). See Figures.7 and.74. The graph of is below the graph of between these two points. Thus the solution set consists of all temperatures T satisfing 9 T 69. [0, 80, 5] b [0, 5, 5] [0, 80, 5] b [0, 5, 5] [0, 80, 5] b [0, 5, 5] = 49 = 49 = 49 = 0 = 0 = 0 Intersection X 9 Y 0 Intersection X 69 Y 0 Figure.7 Figure.7 Figure.74 Smbolic Solution First solve the related equation ƒ T - 49 ƒ = 0. T - 49 = -0 T = 9 T - 49 = 0 T = 69 Thus b our previous discussion ƒ T - 49 ƒ 0 is equivalent to 9 T 69. (b) The solutions to ƒ T - 49 ƒ = 0 are 9 and 69. Therefe the monthl average temperatures in Sante Fe var between a low of 9 F (Januar) and a high of 69 F (Jul). The monthl averages are alwas within 0 degrees of 49 F. Now Tr Eercise 77 An Alternative Method There is a second smbolic method that can be used to solve absolute value inequalities. This method is often used in advanced mathematics courses, such as calculus. It is based on the following two properties.

8 660_ch0pp qd 0/6/08 4: PM Page 5.5 Absolute Value Equations and Inequalities 5 Absolute Value Inequalities (Alternative Method) Let k be a positive number.. ƒ a + b ƒ 6 k is equivalent to -k 6 a + b 6 k.. ƒ a + b ƒ 7 k is equivalent to a + b 6 -k a + b 7 k. Similar statements can be made f inequalities involving Ú. EXAMPLE 8 Using an alternative method Solve each absolute value inequalit. Write our answer in interval notation. (a) ƒ 4-5 ƒ (b) ƒ -4-6 ƒ 7 CLASS DISCUSSION Sketch the graphs of = a + b, = ƒ a + b ƒ, = -k, and = k on one -plane. Now use these graphs to eplain wh the alternative method f solving absolute value inequalities is crect. (a) ƒ 4-5 ƒ is equivalent to the following three-part inequalit »» 5 Equivalent inequalit Subtract 4 from each part. Divide each part b -5; reverse the inequalit. In interval notation the solution is C 5, 7 5 D. (b) ƒ -4-6 ƒ 7 is equivalent to the following compound inequalit Equivalent compound inequalit -4<4-4>8 Add 6 to each side. >- <- Divide each b reverse the inequalit. -4; In interval notation the solution set is (- q, -) (-, q). Now Tr Eercises 57 and 6.5 Putting It All Together The following table summarizes some imptant concepts from this section. Concept Eplanation Eamples Absolute value ƒ() = ƒ ƒ: The output from the absolute value ƒ(-) = ƒ - ƒ = function function is never negative. ƒ() = ƒ ƒ is equivalent to ƒ() =. f () = continued on net page

9 660_ch0pp qd 0/6/08 4: PM Page CHAPTER Linear Functions and Equations continued from previous page Concept Eplanation Eamples Absolute value equations. If k 7 0, then ƒ a + b ƒ = k has two solutions,. Solve ƒ - 5 ƒ = 4. given b a + b = k. - 5 = -4-5 = 4 = = 9 = =. If k = 0, then ƒ a + b ƒ = k has one solution, given b a + b = 0.. Solve ƒ - ƒ = 0. - = 0 implies =.. If k 6 0, then ƒ a + b ƒ = k has no solutions.. ƒ 4-9 ƒ = - has no solutions. Absolute value inequalities Alternative method f solving absolute value inequalities To solve ƒ a + b ƒ 6 k ƒ a + b ƒ 7 k with To solve ƒ - 5 ƒ 6 4 ƒ - 5 ƒ 7 4, first k 7 0, first solve ƒ a + b ƒ = k. Let these solutions be s and s, where s 6 s. solve ƒ - 5 ƒ = 4 to obtain the solutions s = and s = 9.. ƒ a + b ƒ 6 k is equivalent to s 6 6 s.. ƒ - 5 ƒ 6 4 is equivalent to ƒ a + b ƒ 7 k is equivalent to 6 s 7 s.. ƒ - 5 ƒ 7 4 is equivalent to ƒ a + b ƒ 6 k with k 7 0 is equivalent to. ƒ - ƒ 6 5 is solved as follows. -k 6 a + b 6 k ƒ a + b ƒ 7 k with k 7 0 is equivalent to a + b 6 -k a + b 7 k.. ƒ - ƒ 7 5 is solved as follows Eercises Basic Concepts Eercises 8: Let a Z 0.. Solve =.. Solve.. Solve Solve ƒ a + b ƒ = ƒ + ƒ 0... = ƒ - ƒ = ƒ + ƒ = ƒ - ƒ 5. Describe the graph of = ƒ a + b ƒ. 6. Solve ƒ a + b ƒ = Rewrite 6a b using an absolute value. 8. Rewrite (a + b) b using an absolute value. Absolute Value Graphs Eercises 9 : Graph b hand. (a) Find the -intercept. (b) Determine where the graph is increasing and where it is decreasing. Eercises 8: (Refer to Eample.) Do the following. (a) Graph = ƒ(). (b) Use the graph of = ƒ() to sketch a graph of the equation = ƒ ƒ() ƒ. (c) Determine the -intercept f the graph of the equation = ƒ ƒ() ƒ.. = 4. = 5. = - 6. = = 6-8. = - 4