2.6 Solving Inequalities Algebraically and Graphically
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1 7_006.qp //07 8:0 AM Page 9 Section.6 Solving Inequalities Algebraically and Graphically 9.6 Solving Inequalities Algebraically and Graphically Properties of Inequalities Simple inequalities were reviewed in Section P.. There, the inequality symbols <,, >, and were used to compare two numbers and to denote subsets of real numbers. For instance, the simple inequality denotes all real numbers that are greater than or equal to. In this section, you will study inequalities that contain more involved statements such as 5 7 > 9 and As with an equation, you solve an inequality in the variable by finding all values of for which the inequality is true. These values are solutions of the inequality and are said to satisfy the inequality. For instance, the number 9 is a solution of the first inequality listed above because 59 7 > > 6. On the other hand, the number 7 is not a solution because > > 0. The set of all real numbers that are solutions of an inequality is the solution set of the inequality. The set of all points on the real number line that represent the solution set is the graph of the inequality. Graphs of many types of inequalities consist of intervals on the real number line. The procedures for solving linear inequalities in one variable are much like those for solving linear equations. To isolate the variable, you can make use of the properties of inequalities. These properties are similar to the properties of equality, but there are two important eceptions. When each side of an inequality is multiplied or divided by a negative number, the direction of the inequality symbol must be reversed in order to maintain a true statement. Here is an eample. < 5 > 5 6 <. Original inequality Multiply each side by and reverse the inequality. 6 > 5 Simplify. Two inequalities that have the same solution set are equivalent inequalities. For instance, the inequalities < 5 and < are equivalent. To obtain the second inequality from the first, you can subtract from each side of the inequality. The properties listed at the top of the net page describe operations that can be used to create equivalent inequalities. What you should learn Use properties of inequalities to solve linear inequalities. Solve inequalities involving absolute values. Solve polynomial inequalities. Solve rational inequalities. Use inequalities to model and solve real-life problems. Why you should learn it An inequality can be used to determine when a real-life quantity eceeds a given level. For instance, Eercises on page 0 show how to use linear inequalities to determine when the number of hours per person spent playing video games eceeded the number of hours per person spent reading newspapers. Image Source/Superstock Prerequisite Skills To review techniques for solving linear inequalities, see Appendi D.
2 7_006.qp /7/06 : AM Page 0 0 Chapter Solving Equations and Inequalities Properties of Inequalities Let a, b, c, and d be real numbers.. Transitive Property a < b and b < c a < c. Addition of Inequalities a < b and c < d a c < b d. Addition of a Constant a < b a c < b c 4. Multiplying by a Constant For c > 0, a < b ac < bc For c < 0, a < b ac > bc Eploration Use a graphing utility to graph f 5 7 and g 9 in the same viewing window. (Use 5 and 5 y 50. ) For which values of does the graph of f lie above the graph of g? Eplain how the answer to this question can be used to solve the inequality in Eample. Each of the properties above is true if the symbol < is replaced by and > is replaced by. For instance, another form of Property is as follows. a b a c b c Solving a Linear Inequality The simplest type of inequality to solve is a linear inequality in one variable, such as > 4. (See Appendi D for help with solving one-step linear inequalities.) Eample Solve 5 7 > 9. Solution Solving a Linear Inequality 5 7 > 9 Write original inequality. 7 > 9 Subtract from each side. > 6 Add 7 to each side. > 8 Divide each side by. So, the solution set is all real numbers that are greater than 8. The interval notation for this solution set is 8,. The number line graph of this solution set is shown in Figure.45. Note that a parenthesis at 8 on the number line indicates that 8 is not part of the solution set. Now try Eercise Figure.45 STUDY TIP Checking the solution set of an inequality is not as simple as checking the solution(s) of an equation because there are simply too many -values to substitute into the original inequality. However, you can get an indication of the validity of the solution set by substituting a few convenient values of. For instance, in Eample, try substituting 5 and 0 into the original inequality. Solution Interval: 8, Note that the four inequalities forming the solution steps of Eample are all equivalent in the sense that each has the same solution set.
3 7_006.qp /7/06 : AM Page Eample Solve 4. Solving an Inequality Section.6 Solving Inequalities Algebraically and Graphically Algebraic Solution Graphical Solution 4 Write original inequality. Use a graphing utility to graph y and Multiply each side by the LCD. Subtract from each side. y 4 in the same viewing window. In Figure.47, you can see that the graphs appear to intersect at the point,. Use the intersect feature of the graphing utility to confirm this. The graph of y 5 0 lies above the graph of y to Subtract from each side. the left of their point of intersection, which implies that Divide each side by 5 and y y for all. reverse the inequality. The solution set is all real numbers that are less than or equal to. The interval notation for this solution set is y = 4,. The number line graph of this solution set is 5 7 shown in Figure.46. Note that a bracket at on the number line indicates that is part of the solution set. y = 0 4 Figure.46 Solution Interval:, ] Now try Eercise 5. 6 Figure.47 Sometimes it is possible to write two inequalities as a double inequality, as demonstrated in Eample. Eample Solving a Double Inequality Solve 6 and 6 <. Algebraic Solution 6 < Write as a double inequality. 6 < 4 Add to each part. < Divide by 6 and simplify. The solution set is all real numbers that are greater than or equal to and less than. The interval notation for this solution set is,. The number line graph of this solution set is shown in Figure Figure.48 Solution Interval: [, Graphical Solution Use a graphing utility to graph y 6, y, and y in the same viewing window. In Figure.49, you can see that the graphs appear to intersect at the points and,,. Use the intersect feature of the graphing utility to confirm this. The graph of y lies above the graph of to the right of, y and the graph of y lies below the graph of y to the left of This implies that y when <,. y < y. y = y 5 = Now try Eercise 7. Figure.49 5 y =
4 7_006.qp /7/06 : AM Page Chapter Solving Equations and Inequalities Inequalities Involving Absolute Values Solving an Absolute Value Inequality Let be a variable or an algebraic epression and let a be a real number such that a 0.. The solutions of are all values of that lie between a and a. < a if and only if a < < a. Double inequality. The solutions of are all values of that are less than a or greater than a. > a < a > a if and only if < a or > a. Compound inequality These rules are also valid if < is replaced by and > is replaced by. Eample 4 Solve each inequality. Solving Absolute Value Inequalities a. 5 < b. 5 > Algebraic Solution 5 < a. Write original inequality. < 5 < Write double inequality. < < 7 Add 5 to each part. The solution set is all real numbers that are greater than and less than 7. The interval notation for this solution set is, 7. The number line graph of this solution set is shown in Figure.50. b. The absolute value inequality 5 > is equivalent to the following compound inequality: 5 < or 5 >. Solve first inequality: 5 < < Write first inequality. Add 5 to each side. Solve second inequality: 5 > > 7 Write second inequality. Add 5 to each side. The solution set is all real numbers that are less than or greater than 7. The interval notation for this solution set is, 7,. The symbol is called a union symbol and is used to denote the combining of two sets. The number line graph of this solution set is shown in Figure.5. units units Figure.50 Figure.5 Now try Eercise. units units You may need to remind your students that corresponds to a or a. a Graphical Solution a. Use a graphing utility to graph and y in the same viewing window. In Figure.5, you can see that the graphs appear to intersect at the points, and 7,. Use the intersect feature of the graphing utility to confirm this. The graph of y lies below the graph of y when < < 7. So, you can approimate the solution set to be all real numbers greater than and less than 7. y = 5 Figure.5 y = 5 y 5 b. In Figure.5, you can see that the graph of y lies above the graph of y when < or when > 7. So, you can approimate the solution set to be all real numbers that are less than or greater than 7. 0
5 7_006.qp /7/06 : AM Page Section.6 Solving Inequalities Algebraically and Graphically Polynomial Inequalities To solve a polynomial inequality such as < 0, use the fact that a polynomial can change signs only at its zeros (the -values that make the polynomial equal to zero). Between two consecutive zeros, a polynomial must be entirely positive or entirely negative. This means that when the real zeros of a polynomial are put in order, they divide the real number line into intervals in which the polynomial has no sign changes. These zeros are the critical numbers of the inequality, and the resulting open intervals are the test intervals for the inequality. For instance, the polynomial above factors as and has two zeros, and, which divide the real number line into three test intervals:,,,, and,. To solve the inequality < 0, you need to test only one value in each test interval. TECHNOLOGY TIP Some graphing utilities will produce graphs of inequalities. For instance, you can graph 5 > by setting the graphing utility to dot mode and entering y 5 >. Using the settings 0 0 and 4 y 4, your graph should look like the graph shown below. Solve the problem algebraically to verify that the solution is, 4,. Finding Test Intervals for a Polynomial To determine the intervals on which the values of a polynomial are entirely negative or entirely positive, use the following steps.. Find all real zeros of the polynomial, and arrange the zeros in increasing order. The zeros of a polynomial are its critical numbers.. Use the critical numbers to determine the test intervals.. Choose one representative -value in each test interval and evaluate the polynomial at that value. If the value of the polynomial is negative, the polynomial will have negative values for every -value in the interval. If the value of the polynomial is positive, the polynomial will have positive values for every -value in the interval. 0 y = + 5 > Eample 5 Investigating Polynomial Behavior To determine the intervals on which is entirely negative and those on which it is entirely positive, factor the quadratic as. The critical numbers occur at and. So, the test intervals for the quadratic are,,,, and,. In each test interval, choose a representative -value and evaluate the polynomial, as shown in the table. Interval -Value Value of Polynomial Sign of Polynomial, 6 Positive, 0 0 Negative, 5 5 Positive 4 5 The polynomial has negative values for every in the interval, and positive values for every in the intervals, and,. This result is shown graphically in Figure.5. Now try Eercise 49. Figure.5 4 y =
6 7_006.qp /7/06 : AM Page 4 4 Chapter Solving Equations and Inequalities To determine the test intervals for a polynomial inequality, the inequality must first be written in general form with the polynomial on one side. Eample 6 Solve 5 >. Solving a Polynomial Inequality Algebraic Solution 5 > 0 4 > 0 Critical Numbers: Test Intervals: 4,, 4, 4,,, Write inequality in general form. Factor. Test: Is 4 > 0? After testing these intervals, you can see that the polynomial 5 is positive on the open intervals, 4 and,. Therefore, the solution set of the inequality is, 4,. Now try Eercise 55. Graphical Solution First write the polynomial inequality 5 > as 5 > 0. Then use a graphing utility to graph y 5. In Figure.54, you can see that the graph is above the -ais when is less than 4 or when is greater than. So, you can graphically approimate the solution set to be, 4,. 7 Figure.54 4 ( 4, 0) 6 (, 0 ( 5 y = + 5 Eample 7 Solving a Polynomial Inequality Solve > 48. Solution 48 > 0 6 > 0 6 > > 0 Write inequality in general form. Factor by grouping. Distributive Property Factor difference of two squares. 4; The critical numbers are 4,, and and the test intervals are, 4, 4,,, 4, and 4,. Interval -Value Polynomial Value Conclusion, Negative 4, Positive, 4 48 Negative 4, Positive From this you can conclude that the polynomial is positive on the open intervals 4, and 4,. So, the solution set is 4, 4,. Now try Eercise 6. STUDY TIP When solving a quadratic inequality, be sure you have accounted for the particular type of inequality symbol given in the inequality. For instance, in Eample 7, note that the original inequality contained a greater than symbol and the solution consisted of two open intervals. If the original inequality had been 48 the solution would have consisted of the closed interval 4, and the interval 4,.
7 7_006.qp /7/06 : AM Page 5 Section.6 Solving Inequalities Algebraically and Graphically 5 Eample 8 a. The solution set of Unusual Solution Sets 4 > 0 consists of the entire set of real numbers,,. In other words, the value of the quadratic 4 is positive for every real value of, as indicated in Figure.55(a). (Note that this quadratic inequality has no critical numbers. In such a case, there is only one test interval the entire real number line.) b. The solution set of 0 consists of the single real number, because the quadratic has one critical number,, and it is the only value that satisfies the inequality, as indicated in Figure.55(b). c. The solution set of 5 < 0 is empty. In other words, the quadratic 5 is not less than zero for any value of, as indicated in Figure.55(c). d. The solution set of 4 4 > 0 consists of all real numbers ecept the number. In interval notation, this solution set can be written as,,. The graph of 4 4 lies above the -ais ecept at, where it touches it, as indicated in Figure.55(d). y = y = TECHNOLOGY TIP One of the advantages of technology is that you can solve complicated polynomial inequalities that might be difficult, or even impossible, to factor. For instance, you could use a graphing utility to approimate the solution to the inequality < 0. Remind students that they can check the answers to inequality problems in two ways: Algebraically: Substitute -values into the original inequality. Graphically: Sketch the graph of the polynomial written in standard form and note where the graph lies relative to the -ais (, 0) 4 Students can also use a graphing utility to check their answers, as indicated in the Technology Tip on page. (a) (b) y = y = (c) Figure.55 5 (d) (, 0) 6 Now try Eercise 59.
8 7_006.qp /7/06 : AM Page 6 6 Chapter Solving Equations and Inequalities Rational Inequalities The concepts of critical numbers and test intervals can be etended to inequalities involving rational epressions. To do this, use the fact that the value of a rational epression can change sign only at its zeros (the -values for which its numerator is zero) and its undefined values (the -values for which its denominator is zero). These two types of numbers make up the critical numbers of a rational inequality. Eample 9 Solve 7 5. Solving a Rational Inequality Algebraic Solution Write original inequality. Write in general form. Write as single fraction. Simplify. Now, in standard form you can see that the critical numbers are 5 and 8, and you can proceed as follows. Critical Numbers: 5, 8 Test Intervals:, 5, 5, 8, 8, 8 Test: Is 5 0? Interval -Value Polynomial Value Conclusion, 5 5, Negative Positive 9 8 8, 9 Negative By testing these intervals, you can determine that the rational epression 8 5 is negative in the open intervals, 5 and 8,. Moreover, because when 8, you can conclude that the solution set of the inequality is, 5 8,. Now try Eercise Graphical Solution Use a graphing utility to graph 7 y and y 5 in the same viewing window. In Figure.56, you can see that the graphs appear to intersect at the point 8,. Use the intersect feature of the graphing utility to confirm this. The graph of y lies below the graph of y in the intervals, 5 and 8,. So, you can graphically approimate the solution set to be all real numbers less than 5 or greater than or equal to Figure.56 y = 7 5 y = Note in Eample 9 that 5 is not included in the solution set because the inequality is undefined when 5.
9 7_006.qp /7/06 : AM Page 7 Application Section.6 Solving Inequalities Algebraically and Graphically 7 In Section. you studied the implied domain of a function, the set of all -values for which the function is defined. A common type of implied domain is used to avoid even roots of negative numbers, as shown in Eample 0. Eample 0 Finding the Domain of an Epression Find the domain of Solution Because 64 4 is defined only if 64 4 is nonnegative, the domain is given by Write in general form. Divide each side by 4. Factor. The inequality has two critical numbers: 4 and 4. A test shows that in the closed interval 4, 4. The graph of y 64 4, shown in Figure.57, confirms that the domain is 4, 4. Now try Eercise 77. y = ( 4, 0) (4, 0) Figure.57 Eample Height of a Projectile A projectile is fired straight upward from ground level with an initial velocity of 84 feet per second. During what time period will its height eceed 000 feet? Solution In Section.4 you saw that the position of an object moving vertically can be modeled by the position equation s 6t v 0 t s 0 where s is the height in feet and t is the time in seconds. In this case, s 0 0 and v So, you need to solve the inequality 6t t > 000. Using a graphing utility, graph y 6t 84t and y 000, as shown in Figure.58. From the graph, you can determine that 6t 84t > 000 for t between approimately 7.6 and 6.4. You can verify this result algebraically. 6t 84t > 000 Write original inequality. t 4t < 5 Divide by 6 and reverse inequality. t 4t 5 < 0 Write in general form. By the Quadratic Formula the critical numbers are t 9 and t 9, or approimately 7.64 and 6.6. A test will verify that the height of the projectile will eceed 000 feet when 7.64 < t < 6.6; that is, during the time interval 7.64, 6.6 seconds. Now try Eercise y = Figure.58
10 7_006.qp /7/06 : AM Page 8 8 Chapter Solving Equations and Inequalities.6 Eercises See for worked-out solutions to odd-numbered eercises. Vocabulary Check Fill in the blanks.. To solve a linear inequality in one variable, you can use the properties of inequalities, which are identical to those used to solve an equation, with the eception of multiplying or dividing each side by a constant.. It is sometimes possible to write two inequalities as one inequality, called a inequality.. The solutions to a are those values of such that. 4. The solutions to are those values of such that or. a 5. The critical numbers of a rational epression are its and its. In Eercises 6, match the inequality with its graph. [The graphs are labeled (a), (b), (c), (d), (e), and (f).] (a) (b) (c) (d) (e) (f) <. 5. < < < 5 In Eercises 7 0, determine whether each value of is a solution of the inequality. Inequality Values 7. 5 > 0 (a) (b) (c) 5 (d) 8. 5 < (a) (b) 5 (c) 4 (d) 0 9. < (a) 0 (b) 5 (c) (d) (a) (b) (c) 4 (d) In Eercises 0, solve the inequality and sketch the solution on the real number line. Use a graphing utility to verify your solution graphically.. 0 < > 5. 4 < 4. 7 < > 7. 8 < 8. 0 < < 0. 0 < 5 < 4 Graphical Analysis In Eercises 4, use a graphing utility to approimate the solution < 6. < In Eercises 5 8, use a graphing utility to graph the equation and graphically approimate the values of that satisfy the specified inequalities. Then solve each inequality algebraically. Equation Inequalities 5. y (a) y (b) y 0 6. y 8 (a) y (b) y 0 7. y (a) 0 y (b) y 0 8. y (a) y 5 (b) y 0
11 7_006.qp /7/06 : AM Page 9 Section.6 Solving Inequalities Algebraically and Graphically 9 In Eercises 9 6, solve the inequality and sketch the solution on the real number line. Use a graphing utility to verify your solutions graphically In Eercises 7 and 8, use a graphing utility to graph the equation and graphically approimate the values of that satisfy the specified inequalities. Then solve each inequality algebraically. Equation Inequalities 7. (a) y (b) y 4 In Eercises 9 46, use absolute value notation to define the interval (or pair of intervals) on the real number line > 0 7 < > < y 8. (a) y 4 (b) y y All real numbers within 0 units of All real numbers no more than 8 units from All real numbers at least 5 units from 46. All real numbers more than units from In Eercises 47 5, determine the intervals on which the polynomial is entirely negative and those on which it is entirely positive In Eercises 5 6, solve the inequality and graph the solution on the real number line. Use a graphing utility to verify your solution graphically. 5. < < > < 6 In Eercises 6 and 64, use the graph of the function to solve the equation or inequality. (a) f g (b) f ~ g (c) f > g 6. y 64. y (, 5) y = f() 8 y = g() 6 (, ) 4 y = g() (, ) y = f() In Eercises 65 and 66, use a graphing utility to graph the equation and graphically approimate the values of that satisfy the specified inequalities. Then solve each inequality algebraically. Equation Inequalities 65. y (a) y 0 (b) y 66. y 6 6 (a) y 0 (b) y 6 In Eercises 67 70, solve the inequality and graph the solution on the real number line. Use a graphing utility to verify your solution graphically > < 0 4 < 0 0 In Eercises 7 and 7, use a graphing utility to graph the equation and graphically approimate the values of that satisfy the specified inequalities. Then solve each inequality algebraically. Equation Inequalities 7. y (a) y 0 (b) y 6 7. y 5 (a) y (b) y 0 4 In Eercises 7 78, find the domain of in the epression
12 7_006.qp /7/06 : AM Page 0 0 Chapter Solving Equations and Inequalities 79. Population The graph models the population P (in thousands) of Las Vegas, Nevada from 990 to 004, where t is the year, with t 0 corresponding to 990. Also shown is the line y 000. Use the graphs of the model and the horizontal line to write an equation or an inequality that could be solved to answer the question. Then answer the question. (Source: U.S. Census Bureau) Population (in thousands) (a) In what year does the population of Las Vegas reach one million? (b) Over what time period is the population of Las Vegas less than one million? greater than one million? 80. Population The graph models the population P (in thousands) of Pittsburgh, Pennsylvania from 99 to 004, where t is the year, with t corresponding to 99. Also shown is the line y 450. Use the graphs of the model and the horizontal line to write an equation or an inequality that could be solved to answer the question. Then answer the question. (Source: U.S. Census Bureau) Population (in thousands) P P y = P(t) y = 450 y = P(t) y = Year (0 990) Year ( 99) (a) In what year did the population of Pittsburgh equal.45 million? (b) Over what time period is the population of Pittsburgh less than.45 million? greater than.45 million? 8. Height of a Projectile A projectile is fired straight upward from ground level with an initial velocity of 60 feet per second. (a) At what instant will it be back at ground level? (b) When will the height eceed 84 feet? 8. Height of a Projectile A projectile is fired straight upward from ground level with an initial velocity of 8 feet per second. (a) At what instant will it be back at ground level? (b) When will the height be less than 8 feet? 4 t t 8. Education The numbers D of doctorate degrees (in thousands) awarded to female students from 990 to 00 in the United States can be approimated by the model D 0.065t 0.755t 4.06, 0 t, where t is the year, with t 0 corresponding to 990. (Source: U.S. National Center for Education Statistics) (a) Use a graphing utility to graph the model. (b) Use the zoom and trace features to find when the number of degrees was between 5 and 0 thousand. (c) Algebraically verify your results from part (b). (d) According to the model, will the number of degrees eceed 0 thousand? If so, when? If not, eplain. 84. Data Analysis You want to determine whether there is a relationship between an athlete s weight (in pounds) and the athlete s maimum bench-press weight y (in pounds). Sample data from athletes is shown below. 65, 70, 84, 85, 50, 00, 0, 55, 96, 05, 40, 95, 0, 90, 70, 75, 85, 95, 90, 85, 0, 50, 60, 50 (a) Use a graphing utility to plot the data. (b) A model for this data is y. 6. Use a graphing utility to graph the equation in the same viewing window used in part (a). (c) Use the graph to estimate the value of that predict a maimum bench-press weight of at least 00 pounds. (d) Use the graph to write a statement about the accuracy of the model. If you think the graph indicates that an athlete s weight is not a good indicator of the athlete s maimum bench-press weight, list other factors that might influence an individual s maimum bench-press weight. Leisure Time In Eercises 85 88, use the models below which approimate the annual numbers of hours per person spent reading daily newspapers N and playing video games V for the years 000 to 005, where t is the year, with t 0 corresponding to 000. (Source: Veronis Suhler Stevenson) Daily Newspapers: N.5t 79.6, 0 } t } 5 Video Games: V.7t 57.9, 0 } t } Solve the inequality Vt 65. Eplain what the solution of the inequality represents. 86. Solve the inequality Nt 75. Eplain what the solution of the inequality represents. 87. Solve the equation Vt Nt. Eplain what the solution of the equation represents. 88. Solve the inequality Vt > Nt. Eplain what the solution of the inequality represents.
13 7_006.qp /7/06 : AM Page Section.6 Solving Inequalities Algebraically and Graphically Music In Eercises 89 9, use the following information. Michael Kasha of Florida State University used physics and mathematics to design a new classical guitar. He used the model for the frequency of the vibrations on a circular plate v.6t d E where v is the frequency (in vibrations per second), t is the plate thickness (in millimeters), d is the diameter of the plate, E is the elasticity of the plate material, and is the density of the plate material. For fied values of d, E, and, the graph of the equation is a line, as shown in the figure. Frequency (vibrationsper second) v 00 t 4 Plate thickness (millimeters) 89. Estimate the frequency when the plate thickness is millimeters. 90. Estimate the plate thickness when the frequency is 600 vibrations per second. 9. Approimate the interval for the plate thickness when the frequency is between 00 and 400 vibrations per second. 9. Approimate the interval for the frequency when the plate thickness is less than millimeters. In Eercises 9 and 94, (a) write equations that represent each option, (b) use a graphing utility to graph the options in the same viewing window, (c) determine when each option is the better choice, and (d) eplain which option you would choose. 9. Cellular Phones You are trying to decide between two different cellular telephone contracts, option A and option B. Option A has a monthly fee of $ plus $0.5 per minute. Option B has no monthly fee but charges $0.0 per minute. All other monthly charges are identical. 94. Moving You are moving from your home to your dorm room, and the moving company has offered you two options. The charges for gasoline, insurance, and all other incidental fees are equal. Option A: $00 plus $8 per hour to move all of your belongings from your home to your dorm room. Option B: $4 per hour to move all of your belongings from your home to your dorm room. Synthesis True or False? In Eercises 95 and 96, determine whether the statement is true or false. Justify your answer. 95. If 0 8, then 0 and The solution set of the inequality 6 0 is the entire set of real numbers. In Eercises 97 and 98, consider the polynomial a b and the real number line (see figure). 97. Identify the points on the line where the polynomial is zero. 98. In each of the three subintervals of the line, write the sign of each factor and the sign of the product. For which -values does the polynomial possibly change signs? 99. Proof The arithmetic mean of a and b is given by a b. Order the statements of the proof to show that if a < b, then a < a b < b. i. a < a b < b ii. a < b iii. a < a b < b iv. a < b 00. Proof The geometric mean of a and b is given by ab. Order the statements of the proof to show that if 0 < a < b, then a < ab < b. i. a < ab < b ii. 0 < a < b iii. a < ab < b Skills Review In Eercises 0 04, sketch a graph of the function. 0. f 6 0. f f 5 6 a In Eercises 05 08, find the inverse function. f y 06. y y y Make a Decision To work an etended application analyzing the number of heart disease deaths per 00,000 people in the United States, visit this tetbook s Online Study Center. (Data Source: U.S. National Center for Health Statistics) b
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