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HAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES P′A – P′B follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.1404. Estimated proportion for group A: p′ A = Estimated proportion for group B: p′B = x A n A xB nB = 20 200 = 12 200 = 0.1 = 0.06 Graph: Figure 10.7 P′A – P′B = 0.1 – 0.06 = 0.04. Half the p-value is below –0.04, and half is above 0.04. Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value. Make a decision: Since α < p-value, do not reject H0. Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B. Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404 and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw. 10.8 Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance. Example 10.9 A research study was conducted about gender differences in “sexting.” The researcher believed that the proportion of girls involved in “sexting” is less than the proportion of boys involved. The data collected in the spring of 2010 among a random sample of middle and high school students in a large school district in the southern United States This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 539 is summarized in Table 10.9. Is the proportion of girls sending sexts less than the proportion of boys “sexting?” Test at a 1% level of significance. Males Females Sent “sexts” 183 Total number surveyed 2231 156 2169 Table 10.10 Solution 10.9 This is a test of two population proportions. Let M and F be the subscripts for males and females. Then pM and pF are the desired population proportions. Random variable: p′F − p′M = difference in the proportions of males and females who sent “sexts.” H0: pF = pM H0: pF – pM = 0 Ha: pF < pM Ha: pF – pM < 0 The words "less than" tell you the test is left-tailed. Distribution for the test: Since this is a test of two population proportions, the distribution is normal: pc = = 0.077 = 156 + 183 2169 + 2231 xF + x M nF + n M 1 − pc = 0.923 Therefore, p′F – p′ M ∼ N⎛ ⎝0, p′F – p′M follows an approximate normal distribution. ⎛ (0.077)(0.923) ⎝ 1 2169 + 1 2231 ⎞ ⎞ ⎠ ⎠ Calculate the p-value using the normal distribution: p-value = 0.1045 Estimated proportion for females: 0.0719 Estimated proportion for males: 0.082 Graph: Figure 10.8 Decision: Since α < p-value, Do not reject H0 Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending “sexts” is less than the proportion of boys sending “sexts.” 540 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is P = 0.1045 and the test statistic is z = -1.256. Example 10.10 Researchers conducted a study of smartphone use among adults. A cell phone company claimed that iPhone smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5% have an iPhone. Of the 1,343 white cell phone owners randomly sampled, 10% own an iPhone. Test at the 5% level of significance. Is the proportion of white iPhone owners greater than the proportion of African American iPhone owners? Solution 10.10 This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then pW and pA are the desired population proportions. Random variable: p′W – p′A = difference in the proportions of Android and iPhone users. H0: pW = pA H0: pW – pA = 0 Ha: pW > pA Ha: pW – pA > 0 The words "more popular" indicate that the test is right-tailed. Distribution for the test: The distribution is approximately normal: pc = xW + x A nW + n A = 134 + 12 1343 + 232 = 0.1077 1 − pc = 0.8923 Therefore, p′W – p′ A ∽ N ⎛ ⎝0, ⎛ (0.1077)(0.8923) ⎝ 1 1343 + 1 232 ⎞ ⎞ ⎠ ⎠ p′W – p′ A follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.0092 Estimated proportion for group A: 0.10 Estimated proportion for group B: 0.05 Graph: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 541 Figure 10.9 Decision: Since α > p-value, reject the H0. Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use iPhones than African Americans. TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER. Arrow down to Calculate and press ENTER. The P-value is P = 0.0092 and the test statistic is Z = 2.33. 10.10 A concerned group of citizens wanted to know if the proportion of forcible rapes in Texas was different in 2011 than in 2010. Their research showed that of the 113,231 violent crimes in Texas in 2010, 7,622 of them were forcible rapes. In 2011, 7,439 of the 104,873 violent crimes were in the forcible rape category. Test at a 5% significance level. Answer the following questions: a. Is this a test of two means or two proportions? b. Which distribution do you use to perform the test? c. What is the random variable? d. What are the null and alternative hypothesis? Write the null and alternative hypothesis in symbols. e. Is this test right-, left-, or two-tailed? f. What is the p-value? g. Do you reject or not reject the null hypothesis? h. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ____________. 10.4 \| Matched or Paired Samples When using a hypothesis test for matched or paired samples, the following characteristics should be present: 1. Simple random sampling is used. 2. Sample sizes are often small. 3. Two measurements (samples) are drawn from the same pair of individuals or objects. 542 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 4. Differences are calculated from the matched or paired samples. 5. The differences form the sample that is used for the hypothesis test. 6. Either the matched pairs have differences that come from a population that is normal or the number of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal. In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μd, is then tested using a Student's-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences. The test statistic (t-score) is: x¯ t = d − µ d sd ⎛ ⎞ ⎝ ⎠ n Example 10.11 A study was conducted to investigate the effectiveness of hypnotism in reducing pain. Results for randomly selected subjects are shown in Table 10.10. A lower score indicates less pain. The "before" value is matched to an "after" value and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after hypnotism? Test at a 5% significance level. Subject: A B C D E F G H Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0 Table 10.11 Solution 10.11 Corresponding "before" and "after" values form matched pairs. (Calculate "after" – "before.") After Data Before Data Difference 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2 Table 10.12 6.6 6.5 9 10.3 11.3 8.1 6.3 11.6 0.2 -4.1 -1.6 -1.8 -3.2 -2 -2.9 -9.6 The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6} The sample mean and sample standard deviation of the differences are: xd = –3.13 and sd = 2.91 Verify these values. Let µ d be the population mean for the differences. We use the subscript d to denote "differences." ¯ Random variable: X d = the mean difference of the sensory measurements This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 543 H0: μd ≥ 0 The null hypothesis is zero or positive, meaning that there is the same or more pain felt after hypnotism. That means the subject shows no improvement. μd is the population mean of the differences.) Ha: μd < 0 The alternative hypothesis is negative, meaning there is less pain felt after hypnotism. That means the subject shows improvement. The score should be lower after hypnotism, so the difference ought to be negative to indicate improvement. Distribution for the test: The distribution is a Student's t with df = n – 1 = 8 – 1 = 7. Use t7. (Notice that the test is for a single population mean.) Calculate the p-value using the Student's-t distribution: p-value = 0.0095 Graph: Figure 10.10 ¯ X d is the random variable for the differences. The sample mean and sample standard deviation of the differences are: x¯ s¯ d = –3.13 d = 2.91 Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-va
lue. Make a decision: Since α > p-value, reject H0. This means that μd < 0 and there is improvement. Conclusion: At a 5% level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after hypnotism. Hypnotism appears to be effective in reducing pain. NOTE For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then go to a third list and arrow up to the name. Enter 1st list name - 2nd list name. The calculator will do the subtraction, and you will have the differences in the third list. Use your list of differences as the data. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for µ 0 , the name of the list where you put the data, 544 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES and 1 for Freq:. Arrow down to μ: and arrow over to < µ 0 . Press ENTER. Arrow down to Calculate and press ENTER. The p-value is 0.0094, and the test statistic is -3.04. Do these instructions again except, arrow to Draw (instead of Calculate). Press ENTER. 10.11 A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5% level. Subject A B C D E F G H I Before 209 210 205 198 216 217 238 240 222 After 199 207 189 209 217 202 211 223 201 Table 10.13 Example 10.12 A college football coach was interested in whether the college's strength development class increased his players' maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows: Weight (in pounds) Player 1 Player 2 Player 3 Player 4 Amount of weight lifted prior to the class 205 Amount of weight lifted after the class 295 241 252 338 330 368 360 Table 10.14 The coach wants to know if the strength development class makes his players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. Assume the differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x¯ d = 21.3, sd = 46.7 NOTE The data given here would indicate that the distribution is actually right-skewed. The difference 90 may be an extreme outlier? It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are actually negative. Using the difference data, this becomes a test of a single __________ (fill in the blank). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 545 ¯ Define the random variable: X d mean difference in the maximum lift per player. The distribution for the hypothesis test is t3. H0: μd ≤ 0, Ha: μd > 0 Graph: Figure 10.11 Calculate the p-value: The p-value is 0.2150 Decision: If the level of significance is 5%, the decision is not to reject the null hypothesis, because α < p-value. What is the conclusion? At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped to make the players stronger, on average. 10.12 A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data recorded in Table 10.15. Are the scores, on average, higher after the class? Test at a 5% level. SAT Scores Student 1 Student 2 Student 3 Student 4 Score before class 1840 Score after class 1920 1960 2160 1920 2200 2150 2100 Table 10.15 Example 10.13 Seven eighth graders at Kennedy Middle School measured how far they could push the shot-put with their dominant (writing) hand and their weaker (non-writing) hand. They thought that they could push equal distances with either hand. The data were collected and recorded in Table 10.16. Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Dominant Hand 30 26 34 17 19 26 20 Table 10.16 546 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Weaker Hand 28 14 27 18 17 26 16 Table 10.16 Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant. Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x¯ d = 3.71, sd = 4.5. ¯ Random variable: X d = mean difference in the distances between the hands. Distribution for the hypothesis test: t6 H0: μd = 0 Ha: μd ≠ 0 Graph: Figure 10.12 Calculate the p-value: The p-value is 0.0716 (using the data directly). (test statistic = 2.18. p-value = 0.0719 using ⎛ ⎝ x¯ d = 3.71, sd = 4.5. ⎞ ⎠ Decision: Assume α = 0.05. Since α < p-value, Do not reject H0. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot-put. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 547 10.13 Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in Table 10.17. Conduct a hypothesis test to determine whether the mean difference in distances between the dominant and off-hand is significant. Test at the 5% level. Player 1 Player 2 Player 3 Player 4 Player 5 Dominant Hand 120 Off-hand 105 111 109 135 98 140 111 125 99 Table 10.17 548 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 10.1 Hypothesis Testing for Two Means and Two Proportions Class Time: Names: Student Learning Outcomes • The student will select the appropriate distributions to use in each case. • The student will conduct hypothesis tests and interpret the results. Supplies: • • • the business section from two consecutive days’ newspapers three small packages of M&Ms® five small packages of Reese's Pieces® Increasing Stocks Survey Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks, and 32 NASDAQ stocks and complete the following statements. 1. H0: _________ 2. Ha: _________ 3. In words, define the random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 10.13 b. Calculate the p-value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 549 Decreasing Stocks Survey Randomly pick eight stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 10.14 b. Calculate the p-value: 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Candy Survey Buy three small packages of M&Ms and five small packages of Reese's Pieces (same net weight as the M&Ms). Test whether or not the mean number of candy pieces per package is the same for the two brands. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. What distribution should be used for this test? 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: 550 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES Figure 10.15 b. Calculate the p-value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Shoe Survey Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. The distribution to use for the test is ________________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 10.16 b. Calculate the p-value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. This content is avai
lable for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 551 KEY TERMS Degrees of Freedom (df) the number of objects in a sample that are free to vary. Pooled Proportion estimate of the common value of p1 and p2. Standard Deviation A number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation. Variable (Random Variable) a characteristic of interest in a population being studied. Common notation for variables are upper-case Latin letters X, Y, Z,... Common notation for a specific value from the domain (set of all possible values of a variable) are lower-case Latin letters x, y, z,.... For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3, .... Variables in statistics differ from variables in intermediate algebra in two following ways. • The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color, then the domain is {black, blond, gray, green, orange}. • We can tell what specific value x of the random variable X takes only after performing the experiment. CHAPTER REVIEW 10.1 Two Population Means with Unknown Standard Deviations Two population means from independent samples where the population standard deviations are not known ¯ • Random Variable: X ¯ 1 − X 2 = the difference of the sampling means • Distribution: Student's t-distribution with degrees of freedom (variances not pooled) 10.2 Two Population Means with Known Standard Deviations A hypothesis test of two population means from independent samples where the population standard deviations are known (typically approximated with the sample standard deviations), will have these characteristics: ¯ • Random variable: X ¯ 1 − X 2 = the difference of the means • Distribution: normal distribution 10.3 Comparing Two Independent Population Proportions Test of two population proportions from independent samples. • Random variable: p^ A – p^ B = difference between the two estimated proportions • Distribution: normal distribution 10.4 Matched or Paired Samples A hypothesis test for matched or paired samples (t-test) has these characteristics: • Test the differences by subtracting one measurement from the other measurement • Random Variable: x¯ d = mean of the differences • Distribution: Student’s-t distribution with n – 1 degrees of freedom • If the number of differences is small (less than 30), the differences must follow a normal distribution. • Two samples are drawn from the same set of objects. 552 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES • Samples are dependent. FORMULA REVIEW 10.1 Two Population Means with Unknown Standard Deviations Standard error: SE = (s1)2 n1 (s2)2 n2 + Test statistic (t-score): t = ( x¯ Degrees of freedom: 1 − x¯ 2) − (µ 1 − µ 2) (s2)2 n2 + (s1)2 n1 (s1)2 n1 ⎛ ⎜ ⎝ (s1)2 n1 ⎛ ⎞ ⎜ ⎠ ⎝ 2 (s2)2 n2 ⎞ ⎟ ⎠ + 2 ⎞ ⎟ ⎠ + ⎛ ⎝ 1 n2 − 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s2)2 n2 ⎞ ⎟ ⎠ 2 d f = where: ⎛ ⎝ 1 n1 − 1 s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes. x¯ 1 and x¯ 2 are the sample means. Cohen’s d is the measure of effect size: d = x¯ 1 − x¯ s pooled 2 where s pooled = (n1 − 1)s1 2 + (n2 − 1)s2 2 n1 + n2 − 2 Generally µ1 - µ2 = 0. where: σ1 and σ2 are the known population standard deviations. n1 and x¯ and n2 are the sample sizes. x¯ 2 means. μ1 and μ2 are the population means. are the sample 1 10.3 Comparing Two Independent Population Proportions Pooled Proportion: pc = xF + x M nF + n M Distribution for the differences: ⎡ ⎛ ⎣0, pc(1 − pc) ⎝ p′ A − p′B ∼ N 1 n A + 1 nB ⎤ ⎞ ⎦ ⎠ where the null hypothesis is H0: pA = pB or H0: pA – pB = 0. Test Statistic (z-score): z = (p′ A − p′ B) ⎛ pc(1 − pc) ⎝ n A + 1 1 nB ⎞ ⎠ where the null hypothesis is H0: pA = pB or H0: pA − pB = 0. where p′A and p′B are the sample proportions, pA and pB are the population proportions, Pc is the pooled proportion, and nA and nB are the sample sizes. 10.2 Two Population Means with Known Standard Deviations 10.4 Matched or Paired Samples Normal Distributionµ 1 − µ 2, ⎣ (σ1)2 n1 + (σ2)2 n2 ⎤ ⎥ . ⎦ Generally µ1 – µ2 = 0. Test Statistic (z-score): z = ( x¯ 1 − x¯ 2) − (µ 1 − µ 2) (σ2)2 n2 + (σ1)2 n1 Test Statistic (t-score): t = x¯ d − µ d sd ⎛ ⎞ ⎝ ⎠ n where: x¯ d is the mean of the sample differences. μd is the mean of the population differences. sd is the sample standard deviation of the differences. n is the sample size. PRACTICE 10.1 Two Population Means with Unknown Standard Deviations Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for a. b. independent group means, population standard deviations, and/or variances known independent group means, population standard deviations, and/or variances unknown c. matched or paired samples This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 553 d. single mean e. two proportions f. single proportion 1. It is believed that 70% of males pass their drivers test in the first attempt, while 65% of females pass the test in the first attempt. Of interest is whether the proportions are in fact equal. 2. A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this. 3. A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted. 4. The known standard deviation in salary for all mid-level professionals in the financial industry is $11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is $80,000. The sample mean salary for mid-level professionals in Company B is $96,000. Company A and Company B management want to know if their midlevel professionals are paid differently, on average. 5. The average worker in Germany gets eight weeks of paid vacation. 6. According to a television commercial, 80% of dentists agree that Ultrafresh toothpaste is the best on the market. 7. It is believed that the average grade on an English essay in a particular school system for females is higher than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of three, and a random sample of 25 males had a mean score of 76 with a standard deviation of four. 8. The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different? 9. In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico? 10. A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The means hours slept for each person were recorded before starting the medication and after. 11. It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6. 12. Varsity athletes practice five times a week, on average. 13. A sample of 12 in-state graduate school programs at school A has a mean tuition of $64,000 with a standard deviation of $8,000. At school B, a sample of 16 in-state graduate programs has a mean of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different? 14. A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job? 15. A high school principal claims that 30% of student athletes drive themselves to school, while 4% of non-athletes drive themselves to school. In a sample of 20 student athletes, 45% drive themselves to school. In a sample of 35 non-athlete students, 6% drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes? Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions. 16. Are standard deviations known or unknown? 17. What is the random variable? 18. Is this a one-tailed or two-tailed test? 554 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites.
Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. 19. Is this a test of means or proportions? 20. State the null and alternative hypotheses. a. H0: __________ b. Ha: __________ 21. Is this a right-tailed, left-tailed, or two-tailed test? 22. In symbols, what is the random variable of interest for this test? 23. In words, define the random variable of interest for this test. 24. Which distribution (normal or Student's t) would you use for this hypothesis test? 25. Explain why you chose the distribution you did for Exercise 10.24. 26. Calculate the test statistic and p-value. 27. Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value. 28. Find the p-value. 29. At a pre-conceived α = 0.05, what is your: a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence): 30. Does it appear that the means are the same? Why or why not? 10.2 Two Population Means with Known Standard Deviations Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball. Pitcher Sample Mean Speed of Pitches (mph) Population Standard Deviation Wesley 86 Rodriguez 91 Table 10.18 3 7 31. What is the random variable? 32. State the null and alternative hypotheses. 33. What is the test statistic? 34. What is the p-value? 35. At the 1% significance level, what is your conclusion? Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. Plant Group Sample Mean Height of Plants (inches) Population Standard Deviation Food No food Table 10.19 16 14 2.5 1.5 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 555 36. Is the population standard deviation known or unknown? 37. State the null and alternative hypotheses. 38. What is the p-value? 39. Draw the graph of the p-value. 40. At the 1% significance level, what is your conclusion? Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. Sample Mean Melting Temperatures (°F) Population Standard Deviation Alloy Gamma 800 Alloy Zeta 900 Table 10.20 95 105 41. State the null and alternative hypotheses. 42. Is this a right-, left-, or two-tailed test? 43. What is the p-value? 44. Draw the graph of the p-value. 45. At the 1% significance level, what is your conclusion? 10.3 Comparing Two Independent Population Proportions Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. 46. Is this a test of means or proportions? 47. What is the random variable? 48. State the null and alternative hypotheses. 49. What is the p-value? 50. What can you conclude about the two operating systems? Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. 51. Is this a test of means or proportions? 52. State the null and alternative hypotheses. a. H0: _________ b. Ha: _________ 53. Is this a right-tailed, left-tailed, or two-tailed test? How do you know? 54. What is the random variable of interest for this test? 55. In words, define the random variable for this test. 56. Which distribution (normal or Student's t) would you use for this hypothesis test? 57. Explain why you chose the distribution you did for the Exercise 10.56. 556 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 58. Calculate the test statistic. 59. Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value. Figure 10.17 60. Find the p-value. 61. At a pre-conceived α = 0.05, what is your: a. Decision: b. Reason for the decision: c. Conclusion (write out in a complete sentence): 62. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not? 10.4 Matched or Paired Samples Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level. Installation A B C D E F G H Before After Table 10.21 63. What is the random variable? 64. State the null and alternative hypotheses. 65. What is the p-value? 66. Draw the graph of the p-value. 67. What conclusion can you draw about the software patch? Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. Subject A B C D E F Before After Table 10.22 68. State the null and alternative hypotheses. 69. What is the p-value? 70. What is the sample mean difference? 71. Draw the graph of the p-value. 72. What conclusion can you draw about the juggling class? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 557 Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level. Patient A B C D E F Before 161 162 165 162 166 171 After 158 159 166 160 167 169 Table 10.23 73. State the null and alternative hypotheses. 74. What is the test statistic? 75. What is the p-value? 76. What is the sample mean difference? 77. What is the conclusion? HOMEWORK 10.1 Two Population Means with Unknown Standard Deviations DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t-distribution for a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) 78. The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of three English courses with a standard deviation of 0.8. The females took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? 79. A student at a four-year college claims that mean enrollment at four–year colleges is higher than at two–year colleges in the United States. Two surveys are conducted. Of the 35 two–year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. 80. At Rachel’s 11th birthday party, eight girls were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The gir
ls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis. Relaxed time (seconds) Jumping time (seconds) 26 47 30 22 Table 10.24 21 40 28 21 558 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES Relaxed time (seconds) Jumping time (seconds) 23 45 37 29 Table 10.24 25 43 35 32 81. Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is actually lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry level mechanical engineers and 60 entry level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary. 82. Marketing companies have collected data implying that teenage girls use more ring tones on their cellular phones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with cellular phones, the mean number of ring tones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean. Use the information from Appendix C to answer the next four exercises. 83. Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices. 84. Repeat the test in Exercise 10.83, but use Lap 5 data this time. 85. Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5. 86. In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question. “Does Terri Vogel drive faster in races than she does in practices?” Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals. Western Eastern Los Angeles 9 D.C. United 9 FC Dallas 3 Chicago 8 Chivas USA 4 Columbus 7 Real Salt Lake 3 New England 6 Colorado 4 MetroStars 5 San Jose 4 Kansas City 3 Table 10.25 Conduct a hypothesis test to answer the next two exercises. 87. The exact distribution for the hypothesis test is: a. b. c. d. the normal distribution the Student's t-distribution the uniform distribution the exponential distribution 88. If the level of significance is 0.05, the conclusion is: a. There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams b. There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 559 c. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams score. d. Unable to determine 89. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. A concluding statement is: a. There is sufficient evidence to conclude that statistics night students' mean on Exam 2 is better than the statistics day students' mean on Exam 2. b. There is insufficient evidence to conclude that the statistics day students' mean on Exam 2 is better than the statistics night students' mean on Exam 2. c. There is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. d. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. 90. Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon entering prostitution was 18 with a standard deviation of six. The mean age of the 130 United States prostitutes upon entering prostitution was 20 with a standard deviation of eight. Is the mean age of entering prostitution in Canada lower than the mean age in the United States? Test at a 1% significance level. 91. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds. 92. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The “day” subscript refers to the statistics day students. The “night” subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is: a. μday > μnight b. μday < μnight c. μday = μnight d. μday ≠ μnight 10.2 Two Population Means with Known Standard Deviations DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) 93. A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. 94. Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls. 560 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 95. A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same. 96. Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirtyone non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim. 97. A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test. 98. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). Table 10.26 contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference i
n the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). Wife’s Score Husband’s Score Table 10.26 10.3 Comparing Two Independent Population Proportions DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.) 99. A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them. 100. We are interested in whether the proportions of female suicide victims for ages 15 to 24 are the same for the whites and the blacks races in the United States. We randomly pick one year, 1992, to compare the races. The number of suicides estimated in the United States in 1992 for white females is 4,930. Five hundred eighty were aged 15 to 24. The estimate for black females is 330. Forty were aged 15 to 24. We will let female suicide victims be our population. larger + smaller dimension larger dimension 101. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ⎛ ⎝ to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 ⎞ ⎠ 102. A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 561 of 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different? Use the following information to answer the next two exercises. Neuroinvasive West Nile virus is a severe disease that affects a person’s nervous system . It is spread by the Culex species of mosquito. In the United States in 2010 there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1% level of significance, conduct an appropriate hypothesis test. • “2011” subscript: 2011 group. • “2010” subscript: 2010 group 103. This is: a. a test of two proportions b. a test of two independent means c. a test of a single mean d. a test of matched pairs. 104. An appropriate null hypothesis is: a. p2011 ≤ p2010 b. p2011 ≥ p2010 c. μ2011 ≤ μ2010 d. p2011 > p2010 105. The p-value is 0.0022. At a 1% level of significance, the appropriate conclusion is a. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. b. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. c. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. d. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile disease is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile disease. 106. Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders. 107. Adults aged 18 years old and older were randomly selected for a survey on obesity. Adults are considered obese if their body mass index (BMI) is at least 30. The researchers wanted to determine if the proportion of women who are obese in the south is less than the proportion of southern men who are obese. The results are shown in Table 10.27. Test at the 1% level of significance. Number who are obese Sample size Men 42,769 Women 67,169 Table 10.27 155,525 248,775 108. Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. Table 10.28 details the number of tablet owners for each age group. Test at the 1% level of significance. 16–29 year olds 30 years old and older Own a Tablet 69 231 Table 10.28 562 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 16–29 year olds 30 years old and older Sample Size 628 2,309 Table 10.28 109. A group of friends debated whether more men use smartphones than women. They consulted a research study of smartphone use among adults. The results of the survey indicate that of the 973 men randomly sampled, 379 use smartphones. For women, 404 of the 1,304 who were randomly sampled use smartphones. Test at the 5% level of significance. 110. While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey. 111. We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. 112. Joan Nguyen recently claimed that the proportion of college-age males with at least one pierced ear is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 had at least one pierced ear. Out of 92 females, 47 had at least one pierced ear. Do you believe that the proportion of males has reached the proportion of females? 113. Use the data sets found in Appendix C to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds? 114. "To Breakfast or Not to Breakfast?" by Richard Ayore In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what he or she has achieved for the past year and also focuses ahead for more to come. If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine. Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!” And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in Table 10.29, solve our problem. Work hours with breakfast Work hours without breakfast 8 7 9 5 9 8 10 7 Table 10.29 6 5 5 4 7 7 7 5 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 563 Work hours with breakfast Work hours without breakfast 6 9 Table 10.29 6 5 10.4 Matched or Paired Samples DIRECTIONS: For each of the word problems, use
a solution sheet to do the hypothesis test. The solution sheet is found in m47889 (http://cnx.org/content/m47889/latest/) . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t-distribution for the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) 115. Ten individuals went on a low–fat diet for 12 weeks to lower their cholesterol. The data are recorded in Table 10.30. Do you think that their cholesterol levels were significantly lowered? Starting cholesterol level Ending cholesterol level 140 220 110 240 200 180 190 360 280 260 Table 10.30 140 230 120 220 190 150 200 300 300 240 Use the following information to answer the next two exercises. A new AIDS prevention drug was tried on a group of 224 HIV positive patients. Forty-five patients developed AIDS after four years. In a control group of 224 HIV positive patients, 68 developed AIDS after four years. We want to test whether the method of treatment reduces the proportion of patients that develop AIDS after four years or if the proportions of the treated group and the untreated group stay the same. Let the subscript t = treated patient and ut = untreated patient. 116. The appropriate hypotheses are: a. H0: pt < put and Ha: pt ≥ put b. H0: pt ≤ put and Ha: pt > put c. H0: pt = put and Ha: pt ≠ put d. H0: pt = put and Ha: pt < put 117. If the p-value is 0.0062 what is the conclusion (use α = 0.05)? a. The method has no effect. b. There is sufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years. 564 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES c. There is sufficient evidence to conclude that the method increases the proportion of HIV positive patients who develop AIDS after four years. d. There is insufficient evidence to conclude that the method reduces the proportion of HIV positive patients who develop AIDS after four years. Use the following information to answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a “biofeedback exercise program.” Six subjects were randomly selected and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after - before) producing the following results: x¯ d = −10.2 sd = 8.4. Using the data, test the hypothesis that the blood pressure has decreased after the training. 118. The distribution for the test is: t5 t6 a. b. c. N(−10.2, 8.4) d. N(−10.2, 8.4 6 ) 119. If α = 0.05, the p-value and the conclusion are a. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training. b. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training. c. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training. d. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training. 120. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 Mean score after class 80 78 80 93 86 87 86 Table 10.31 The correct decision is: a. Reject H0. b. Do not reject the H0. 121. A local cancer support group believes that the estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. The group compared the estimates of new female breast cancer cases by southern state in 2012 and in 2013. The results are in Table 10.32. Southern States 2012 2013 Alabama Arkansas Florida Georgia Kentucky Louisiana 3,450 3,720 2,150 2,280 15,540 15,710 6,970 7,310 3,160 3,300 3,320 3,630 Mississippi 1,990 2,080 North Carolina 7,090 7,430 Oklahoma 2,630 2,690 Table 10.32 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 565 Southern States 2012 2013 South Carolina 3,570 3,580 Tennessee 4,680 5,070 Texas Virginia Table 10.32 15,050 14,980 6,190 6,280 122. A traveler wanted to know if the prices of hotels are different in the ten cities that he visits the most often. The list of the cities with the corresponding hotel prices for his two favorite hotel chains is in Table 10.33. Test at the 1% level of significance. Hyatt Regency prices in dollars Hilton prices in dollars Cities Atlanta Boston Chicago Dallas Denver Indianapolis Los Angeles New York City Philadelphia 107 358 209 209 167 179 179 625 179 Washington, DC 245 Table 10.33 169 289 299 198 169 214 169 459 159 239 123. A politician asked his staff to determine whether the underemployment rate in the northeast decreased from 2011 to 2012. The results are in Table 10.34. Northeastern States 2011 2012 Connecticut Delaware Maine Maryland Massachusetts New Hampshire New Jersey New York Ohio Pennsylvania Rhode Island Vermont West Virginia Table 10.34 17.3 17.4 19.3 16.0 17.6 15.4 19.2 18.5 18.2 16.5 20.7 14.7 15.5 16.4 13.7 16.1 15.5 18.2 13.5 18.7 18.7 18.8 16.9 22.4 12.3 17.3 566 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES BRINGING IT TOGETHER: HOMEWORK Use the following information to answer the next ten exercises. indicate which of the following choices best identifies the hypothesis test. a. b. independent group means, population standard deviations and/or variances known independent group means, population standard deviations and/or variances unknown c. matched or paired samples d. single mean e. two proportions f. single proportion 124. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. 125. A new chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it. 126. The mean number of English courses taken in a two–year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from nine males and 16 females. 127. A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased. 128. A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and one year, respectively. 129. According to a YWCA Rape Crisis Center newsletter, 75% of rape victims know their attackers. A study is done to verify this. 130. According to a recent study, U.S. companies have a mean maternity-leave of six weeks. 131. A recent drug survey showed an increase in use of drugs and alcohol among local high school students as compared to the national percent. Suppose that a survey of 100 local youths and 100 national youths is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. 132. A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected: Pre-course score Post-course score 1 960 1010 840 1100 1250 860 1330 790 990 1110 Table 10.35 300 920 1100 880 1070 1320 860 1370 770 1040 1200 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 567 Pre-course score Post-course score 740 850 Table 10.35 133. University of Michigan researchers reported in the Journal of the National Cancer Institute that quitting smoking is especially beneficial for those under age 49. In this American Cancer Society study, the risk (probability) of dying of lung cancer was about the same as for those who had never smoked. 134. Lesley E. Tan investigated the relationship between left-handedness vs. right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the children based on this experiment. The experiment produced the means and standard deviations shown Table 10.36. Determine the appropriate test and best distribution to use for that test. Left-handed Right-handed Sample size Sample mean 41 97.5 Sample standard deviation 17.5 41 98.1 19.2 Table 10.36 a. Two independent means, normal distribution b. Two independent means, Student’s-t distribution c. Matched or paired samples, Student’s-t distribution d. Two population proportions, normal distribution 135. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four (4) new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as Table 10.37. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 Mean score after class 80 78 80 93 86 87 86 Table 10.37 This is: a. a tes
t of two independent means. b. a test of two proportions. c. a test of a single mean. d. a test of a single proportion. REFERENCES 10.1 Two Population Means with Unknown Standard Deviations Data from Graduating Engineer + Computer Careers. Available online at http://www.graduatingengineer.com Data from Microsoft Bookshelf. Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013). “List of current United States Senators by Age.” Wikipedia. Available online at http://en.wikipedia.org/wiki/ List_of_current_United_States_Senators_by_age (accessed June 17, 2013). 568 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES “Sectoring sectors.aspx?page=sectors&base=industry (accessed June 17, 2013). Industry Groups.” Nasdaq. Available by online at http://www.nasdaq.com/markets/barchart- “Strip Clubs: Where Prostitution and Trafficking Happen.” Prostitution Research and Education, 2013. Available online at www.prostitutionresearch.com/ProsViolPosttrauStress.html (accessed June 17, 2013). “World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013). 10.2 Two Population Means with Known Standard Deviations Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbulling Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (accessed June 17, 2013). “Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013). Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed June 17, 2013). at “State-Specific Prevalence of Obesity AmongAduls—Unites States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013). “Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/ crime/txcrime.htm (accessed June 17, 2013). 10.3 Comparing Two Independent Population Proportions Data from Educational Resources, December catalog. Data from Hilton Hotels. Available online at http://www.hilton.com (accessed June 17, 2013). Data from Hyatt Hotels. Available online at http://hyatt.com (accessed June 17, 2013). Data from Statistics, United States Department of Health and Human Services. Data from Whitney Exhibit on loan to San Jose Museum of Art. Data from the American Cancer Society. Available online at http://www.cancer.org/index (accessed June 17, 2013). Data from the Chancellor’s Office, California Community Colleges, November 1994. “State States.aspx?ref=interactive (accessed June 17, 2013). States.” Gallup, the of 2013. Available online at http://www.gallup.com/poll/125066/State- “West Nile Virus.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/ncidod/dvbid/ westnile/index.htm (accessed June 17, 2013). SOLUTIONS 1 two proportions 3 matched or paired samples 5 single mean 7 independent group means, population standard deviations and/or variances unknown 9 two proportions 11 independent group means, population standard deviations and/or variances unknown 13 independent group means, population standard deviations and/or variances unknown 15 two proportions 17 The random variable is the difference between the mean amounts of sugar in the two soft drinks. 19 means This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 569 21 two-tailed 23 the difference between the mean life spans of whites and nonwhites 25 This is a comparison of two population means with unknown population standard deviations. 27 Check student’s solution. 29 a. Reject the null hypothesis b. p-value < 0.05 c. There is not enough evidence at the 5% level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites. 31 The difference in mean speeds of the fastball pitches of the two pitchers 33 –2.46 35 At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s. 37 Subscripts: 1 = Food, 2 = No Food H0: μ1 ≤ μ2 Ha: μ1 > μ2 39 Figure 10.18 41 Subscripts: 1 = Gamma, 2 = Zeta H0: μ1 = μ2 Ha: μ1 ≠ μ2 43 0.0062 45 There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma. 47 P′OS1 – P′OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS1 and OS2. 49 0.1018 51 proportions 53 right-tailed 55 The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota. 570 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 57 Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test. 59 Check student’s solution. 61 a. Reject the null hypothesis. b. p-value < alpha c. At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota. 63 the mean difference of the system failures 65 0.0067 67 With a p-value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures. 69 0.0021 71 Figure 10.19 73 H0: μd ≥ 0 Ha: μd < 0 75 0.0699 77 We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective. 79 Subscripts: 1: two-year colleges; 2: four-year colleges a. H0: μ1 ≥ μ2 b. Ha: μ1 < μ2 ¯ c. X ¯ 1 – X 2 d. Student’s-t is the difference between the mean enrollments of the two-year colleges and the four-year colleges. e. test statistic: -0.2480 f. p-value: 0.4019 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges. 81 Subscripts: 1: mechanical engineering; 2: electrical engineering This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 571 a. H0: µ1 ≥ µ2 b. Ha: µ1 < µ2 ¯ c. X ¯ 1 − X 2 is the difference between the mean entry level salaries of mechanical engineers and electrical engineers. d. e. t108 test statistic: t = –0.82 f. p-value: 0.2061 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers. 83 a. H0: µ1 = µ2 b. Ha: µ1 ≠ µ2 ¯ c. X ¯ 1 − X 2 is the difference between the mean times for completing a lap in races and in practices. d. e. t20.32 test statistic: –4.70 f. p-value: 0.0001 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. 85 a. H0: µ1 = µ2 b. Ha: µ1 ≠ µ2 c. d. e. is the difference between the mean times for completing a lap in races and in practices. t40.94 test statistic: –5.08 f. p-value: zero g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. 88 c 572 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES ¯ 90 Test: two independent sample means, population standard deviations unknown. Random variable: X 2 Distribution: H0: μ1 = μ2 Ha: μ1 < μ2 The mean age of entering prostitution in Canada is lower than the mean age in the United States. ¯ 1 − X Figure 10.20 Graph: left-tailed p-value : 0.0151 Decision: Do not reject H0. Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering prostitution in Canada is lower than the mean age in the United States. 92 d 94 Subscripts: 1 = boys, 2 = girls a. H0: µ1 ≤ µ2 b. Ha: µ1 > µ2 c. The random variable is the difference in the mean auto insurance costs for boys and girls. d. normal e. test statistic: z = 2.50 f. p-value: 0.0062 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls. 96 Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans a. H0: µ1 ≥ µ2 b. Ha: µ1 < µ2 c. The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans. d. normal e. test statistic: 6.36 f. p-value: 0 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid s
edans is less than that of hybrid sedans. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 573 98 a. H0: µd = 0 b. Ha: µd < 0 c. The random variable Xd is the average difference between husband’s and wife’s satisfaction level. d. e. t9 test statistic: t = –1.86 f. p-value: 0.0479 g. Check student’s solution h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis, but run another test. iii. Reason for Decision: p-value < alpha iv. Conclusion: This is a weak test because alpha and the p-value are close. However, there is insufficient evidence to conclude that the mean difference is negative. 100 a. H0: PW = PB b. Ha: PW ≠ PB c. The random variable is the difference in the proportions of white and black suicide victims, aged 15 to 24. d. normal for two proportions e. test statistic: –0.1944 f. p-value: 0.8458 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of white and black female suicide victims, aged 15 to 24, are different. 102 Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College a. H0: p1 = p2 b. Ha: p1 ≠ p2 c. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College. d. normal for two proportions e. test statistic: 4.29 f. p-value: 0.00002 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic students at Cabrillo College and Lake Tahoe College are different. 104 a 106 Test: two independent sample proportions. Random variable: p′1 - p′2 Distribution: H0: p1 = p2 574 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES Ha: p1 ≠ p2 The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users. Graph: two-tailed Figure 10.21 p-value : 0.0033 Decision: Reject the null hypothesis. Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older. 108 Test: two independent sample proportions Random variable: p′1 − p′2 Distribution: H0: p1 = p2 Ha: p1 > p2 A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. Graph: righttailed Figure 10.22 p-value: 0.2354 Decision: Do not reject the H0. Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. 110 Subscripts: 1: men; 2: women a. H0: p1 ≤ p2 b. Ha: p1 > p2 c. P′1 − P′2 is the difference between the proportions of men and women who enjoy shopping for electronic equipment. d. normal for two proportions e. test statistic: 0.22 f. p-value: 0.4133 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women. 112 a. H0: p1 = p2 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES 575 b. Ha: p1 ≠ p2 c. P′1 − P′2 is the difference between the proportions of men and women that have at least one pierced ear. d. normal for two proportions e. test statistic: –4.82 f. p-value: zero g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of males and females with at least one pierced ear is different. 114 a. H0: µd = 0 b. Ha: µd > 0 c. The random variable Xd is the mean difference in work times on days when eating breakfast and on days when not eating breakfast. t9 test statistic: 4.8963 d. e. f. p-value: 0.0004 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased. 115 p-value = 0.1494 At the 5% significance level, there is insufficient evidence to conclude that the medication lowered cholesterol levels after 12 weeks. 117 b 119 c ¯ 121 Test: two matched pairs or paired samples (t-test) Random variable: X d Distribution: t12 H0: μd = 0 Ha: μd > 0 The mean of the differences of new female breast cancer cases in the south between 2013 and 2012 is greater than zero. The estimate for new female breast cancer cases in the south is higher in 2013 than in 2012. Graph: right-tailed p-value: 0.0004 Figure 10.23 576 CHAPTER 10 \| HYPOTHESIS TESTING WITH TWO SAMPLES Decision: Reject H0 Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that there was a higher estimate of new female breast cancer cases in 2013 than in 2012. 123 Test: matched or paired samples (t-test) Difference data: {–0.9, –3.7, –3.2, –0.5, 0.6, –1.9, –0.5, 0.2, 0.6, 0.4, 1.7, –2.4, ¯ 1.8} Random Variable: X d Distribution: H0: μd = 0 Ha: μd < 0 The mean of the differences of the rate of underemployment in the northeastern states between 2012 and 2011 is less than zero. The underemployment rate went down from 2011 to 2012. Graph: left-tailed. Figure 10.24 p-value: 0.1207 Decision: Do not reject H0. Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that there was a decrease in the underemployment rates of the northeastern states from 2011 to 2012. 125 e 127 d 129 f 131 e 133 f 135 a This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 577 11 \| THE CHI-SQUARE DISTRIBUTION Figure 11.1 The chi-square distribution can be used to find relationships between two things, like grocery prices at different stores. (credit: Pete/flickr) Introduction Chapter Objectives By the end of this chapter, the student should be able to: Interpret the chi-square probability distribution as the sample size changes. • • Conduct and interpret chi-square goodness-of-fit hypothesis tests. • Conduct and interpret chi-square test of independence hypothesis tests. • Conduct and interpret chi-square homogeneity hypothesis tests. • Conduct and interpret chi-square single variance hypothesis tests. Have you ever wondered if lottery numbers were evenly distributed or if some numbers occurred with a greater frequency? How about if the types of movies people preferred were different across different age groups? What about if a coffee machine was dispensing approximately the same amount of coffee each time? You could answer these questions by conducting a hypothesis test. 578 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION You will now study a new distribution, one that is used to determine the answers to such questions. This distribution is called the chi-square distribution. In this chapter, you will learn the three major applications of the chi-square distribution: 1. 2. 3. the goodness-of-fit test, which determines if data fit a particular distribution, such as in the lottery example the test of independence, which determines if events are independent, such as in the movie example the test of a single variance, which tests variability, such as in the coffee example NOTE Though the chi-square distribution depends on calculators or computers for most of the calculations, there is a table available (see Appendix G). TI-83+ and TI-84 calculator instructions are included in the text. Look in the sports section of a newspaper or on the Internet for some sports data (baseball averages, basketball scores, golf tournament scores, football odds, swimming times, and the like). Plot a histogram and a boxplot using your data. See if you can determine a probability distribution that your data fits. Have a discussion with the class about your choice. 11.1 \| Facts About the Chi-Square Distribution The notation for the chi-square distribution is: 2 χ ∼ χd f where df = degrees of freedom which depends on how chi-square is being used. (If you want to practice calculating chisquare probabilities then use df = n - 1. The degrees of freedom for the three major uses are each calculated differently.) For the χ2 distribution, the population mean is μ = df and the population standard deviation is σ = 2(d f ) . The random variable is shown as χ2, but may be any upper case letter. The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables. χ2 = (Z1)2 + (Z2)2 + ... + (Zk)2 1. The curve is nonsymmetrical and skewed to the right. 2. There is a different chi-square curve for each df. Figure 11.2 3. The test statistic for any test is always greater than or equal to zero. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 4. When df > 90, the chi-square curve approximates the normal distribution. For X ~ χ1,000 2 the mean, μ = df = 1,000 and the standard deviation, σ = 2(1,000) = 44.7. Therefore, X ~ N(1,000, 44.7), approximately. 5. The mean, μ, is located just to the right of the peak. CHAPT
ER 11 \| THE CHI-SQUARE DISTRIBUTION 579 Figure 11.3 11.2 \| Goodness-of-Fit Test In this type of hypothesis test, you determine whether the data "fit" a particular distribution or not. For example, you may suspect your unknown data fit a binomial distribution. You use a chi-square test (meaning the distribution for the hypothesis test is chi-square) to determine if there is a fit or not. The null and the alternative hypotheses for this test may be written in sentences or may be stated as equations or inequalities. The test statistic for a goodness-of-fit test is: (O − E)2 E Σ k where: • O = observed values (data) • E = expected values (from theory) • k = the number of different data cells or categories The observed values are the data values and the expected values are the values you would expect to get if the null hypothesis were true. There are n terms of the form (O − E)2 . E The number of degrees of freedom is df = (number of categories – 1). The goodness-of-fit test is almost always right-tailed. If the observed values and the corresponding expected values are not close to each other, then the test statistic can get very large and will be way out in the right tail of the chi-square curve. NOTE The expected value for each cell needs to be at least five in order for you to use this test. Example 11.1 Absenteeism of college students from math classes is a major concern to math instructors because missing class appears to increase the drop rate. Suppose that a study was done to determine if the actual student absenteeism rate follows faculty perception. The faculty expected that a group of 100 students would miss class according to Table 11.1. 580 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Number of absences per term Expected number of students 0–2 3–5 6–8 9–11 12+ Table 11.1 50 30 12 6 2 A random survey across all mathematics courses was then done to determine the actual number (observed) of absences in a course. The chart in Table 11.2 displays the results of that survey. Number of absences per term Actual number of students 0–2 3–5 6–8 9–11 12+ Table 11.2 35 40 20 1 4 Determine the null and alternative hypotheses needed to conduct a goodness-of-fit test. H0: Student absenteeism fits faculty perception. The alternative hypothesis is the opposite of the null hypothesis. Ha: Student absenteeism does not fit faculty perception. a. Can you use the information as it appears in the charts to conduct the goodness-of-fit test? Solution 11.1 a. No. Notice that the expected number of absences for the "12+" entry is less than five (it is two). Combine that group with the "9–11" group to create new tables where the number of students for each entry are at least five. The new results are in Table 11.2 and Table 11.3. Number of absences per term Expected number of students 0–2 3–5 6–8 9+ Table 11.3 50 30 12 8 Number of absences per term Actual number of students 0–2 35 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 581 Number of absences per term Actual number of students 3–5 6–8 9+ Table 11.4 40 20 5 b. What is the number of degrees of freedom (df)? Solution 11.1 b. There are four "cells" or categories in each of the new tables. df = number of cells – 1 = 4 – 1 = 3 11.1 A factory manager needs to understand how many products are defective versus how many are produced. The number of expected defects is listed in Table 11.5. Number produced Number defective 0–100 101–200 201–300 301–400 401–500 Table 11.5 5 6 7 8 10 A random sample was taken to determine the actual number of defects. Table 11.6 shows the results of the survey. Number produced Number defective 0–100 101–200 201–300 301–400 401–500 Table 11.6 5 7 8 9 11 State the null and alternative hypotheses needed to conduct a goodness-of-fit test, and state the degrees of freedom. 582 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Example 11.2 Employers want to know which days of the week employees are absent in a five-day work week. Most employers would like to believe that employees are absent equally during the week. Suppose a random sample of 60 managers were asked on which day of the week they had the highest number of employee absences. The results were distributed as in Table 11.6. For the population of employees, do the days for the highest number of absences occur with equal frequencies during a five-day work week? Test at a 5% significance level. Monday Tuesday Wednesday Thursday Friday Number of Absences 15 12 9 9 15 Table 11.7 Day of the Week Employees were Most Absent Solution 11.2 The null and alternative hypotheses are: • H0: The absent days occur with equal frequencies, that is, they fit a uniform distribution. • Ha: The absent days occur with unequal frequencies, that is, they do not fit a uniform distribution. If the absent days occur with equal frequencies, then, out of 60 absent days (the total in the sample: 15 + 12 + 9 + 9 + 15 = 60), there would be 12 absences on Monday, 12 on Tuesday, 12 on Wednesday, 12 on Thursday, and 12 on Friday. These numbers are the expected (E) values. The values in the table are the observed (O) values or data. This time, calculate the χ2 test statistic by hand. Make a chart with the following headings and fill in the columns: • Expected (E) values (12, 12, 12, 12, 12) • Observed (O) values (15, 12, 9, 9, 15) • • • (O – E) (O – E)2 (O – E)2 E Now add (sum) the last column. The sum is three. This is the χ2 test statistic. To find the p-value, calculate P(χ2 > 3). This test is right-tailed. (Use a computer or calculator to find the p-value. You should get p-value = 0.5578.) The dfs are the number of cells – 1 = 5 – 1 = 4 Press 2nd DISTR. Arrow down to χ2cdf. Press ENTER. Enter (3,10^99,4). Rounded to four decimal places, you should see 0.5578, which is the p-value. Next, complete a graph like the following one with the proper labeling and shading. (You should shade the right tail.) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 583 Figure 11.4 The decision is not to reject the null hypothesis. Conclusion: At a 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the absent days do not occur with equal frequencies. TI-83+ and some TI-84 calculators do not have a special program for the test statistic for the goodness-offit test. The next example Example 11.3 has the calculator instructions. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. To Clear Lists in the calculators: Go into STAT EDIT and arrow up to the list name area of the particular list. Press CLEAR and then arrow down. The list will be cleared. Alternatively, you can press STAT and press 4 (for ClrList). Enter the list name and press ENTER. 11.2 Teachers want to know which night each week their students are doing most of their homework. Most teachers think that students do homework equally throughout the week. Suppose a random sample of 49 students were asked on which night of the week they did the most homework. The results were distributed as in Table 11.8. Sunday Monday Tuesday Wednesday Thursday Friday Saturday 11 8 10 7 10 5 5 Number of Students Table 11.8 From the population of students, do the nights for the highest number of students doing the majority of their homework occur with equal frequencies during a week? What type of hypothesis test should you use? 584 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Example 11.3 One study indicates that the number of televisions that American families have is distributed (this is the given distribution for the American population) as in Table 11.9. Number of Televisions Percent 0 1 2 3 4+ Table 11.9 10 16 55 11 8 The table contains expected (E) percents. A random sample of 600 families in the far western United States resulted in the data in Table 11.10. Number of Televisions Frequency 0 1 2 3 4+ Table 11.10 66 119 340 60 15 Total = 600 The table contains observed (O) frequency values. At the 1% significance level, does it appear that the distribution "number of televisions" of far western United States families is different from the distribution for the American population as a whole? Solution 11.3 This problem asks you to test whether the far western United States families distribution fits the distribution of the American families. This test is always right-tailed. The first table contains expected percentages. To get expected (E) frequencies, multiply the percentage by 600. The expected frequencies are shown in Table 11.10. Number of Televisions Percent Expected Frequency 0 1 2 3 over 3 Table 11.11 10 16 55 11 8 (0.10)(600) = 60 (0.16)(600) = 96 (0.55)(600) = 330 (0.11)(600) = 66 (0.08)(600) = 48 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 585 Therefore, the expected frequencies are 60, 96, 330, 66, and 48. In the TI calculators, you can let the calculator do the math. For example, instead of 60, enter 0.10*600. H0: The "number of televisions" distribution of far western United States families is the same as the "number of televisions" distribution of the American population. Ha: The "number of televisions" distribution of far western United States families is different from the "number of televisions" distribution of the American population. Distribution for the test: χ4 2 where df = (the number of cells) – 1 = 5 – 1 = 4. NO
TE df ≠ 600 – 1 Calculate the test statistic: χ2 = 29.65 Graph: Figure 11.5 Probability statement: p-value = P(χ2 > 29.65) = 0.000006 Compare α and the p-value: • α = 0.01 • p-value = 0.000006 So, α > p-value. Make a decision: Since α > p-value, reject Ho. This means you reject the belief that the distribution for the far western states is the same as that of the American population as a whole. Conclusion: At the 1% significance level, from the data, there is sufficient evidence to conclude that the "number of televisions" distribution for the far western United States is different from the "number of televisions" distribution for the American population as a whole. 586 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Press STAT and ENTER. Make sure to clear lists L1, L2, and L3 if they have data in them (see the note at the end of Example 11.2). Into L1, put the observed frequencies 66, 119, 349, 60, 15. Into L2, put the expected frequencies .10600, .16600, .55600, .11600, .08*600. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum" (Enter L3). Rounded to 2 decimal places, you should see 29.65. Press 2nd DISTR. Press 7 or Arrow down to 7:χ2cdf and press ENTER. Enter (29.65,1E99,4). Rounded to four places, you should see 5.77E-6 = .000006 (rounded to six decimal places), which is the p-value. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. 11.3 The expected percentage of the number of pets students have in their homes is distributed (this is the given distribution for the student population of the United States) as in Table 11.12. Number of Pets Percent 0 1 2 3 4+ Table 11.12 18 25 30 18 9 A random sample of 1,000 students from the Eastern United States resulted in the data in Table 11.13. Number of Pets Frequency 0 1 2 3 4+ Table 11.13 210 240 320 140 90 At the 1% significance level, does it appear that the distribution “number of pets” of students in the Eastern United States is different from the distribution for the United States student population as a whole? What is the p-value? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 587 Example 11.4 Suppose you flip two coins 100 times. The results are 20 HH, 27 HT, 30 TH, and 23 TT. Are the coins fair? Test at a 5% significance level. Solution 11.4 This problem can be set up as a goodness-of-fit problem. The sample space for flipping two fair coins is {HH, HT, TH, TT}. Out of 100 flips, you would expect 25 HH, 25 HT, 25 TH, and 25 TT. This is the expected distribution. The question, "Are the coins fair?" is the same as saying, "Does the distribution of the coins (20 HH, 27 HT, 30 TH, 23 TT) fit the expected distribution?" Random Variable: Let X = the number of heads in one flip of the two coins. X takes on the values 0, 1, 2. (There are 0, 1, or 2 heads in the flip of two coins.) Therefore, the number of cells is three. Since X = the number of heads, the observed frequencies are 20 (for two heads), 57 (for one head), and 23 (for zero heads or both tails). The expected frequencies are 25 (for two heads), 50 (for one head), and 25 (for zero heads or both tails). This test is right-tailed. H0: The coins are fair. Ha: The coins are not fair. Distribution for the test: χ2 Calculate the test statistic: χ2 = 2.14 2 where df = 3 – 1 = 2. Graph: Figure 11.6 Probability statement: p-value = P(χ2 > 2.14) = 0.3430 Compare α and the p-value: • α = 0.05 • p-value = 0.3430 α < p-value. Make a decision: Since α < p-value, do not reject H0. Conclusion: There is insufficient evidence to conclude that the coins are not fair. Press STAT and ENTER. Make sure you clear lists L1, L2, and L3 if they have data in them. Into L1, put the observed frequencies 20, 57, 23. Into L2, put the expected frequencies 25, 50, 25. Arrow over to list L3 and up to the name area "L3". Enter (L1-L2)^2/L2 and ENTER. Press 2nd QUIT. Press 2nd LIST and arrow over to MATH. Press 5. You should see "sum".Enter L3. Rounded to two decimal places, you should see 2.14. Press 2nd DISTR. Arrow down to 7:χ2cdf (or press 7). Press ENTER. Enter 2.14,1E99,2). Rounded to four places, you should see .3430, which is the p-value. The newer TI-84 calculators have in STAT TESTS the test Chi2 GOF. To run the test, put the observed values (the data) into a first list and the expected values (the values you expect if the null hypothesis is true) 588 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION into a second list. Press STAT TESTS and Chi2 GOF. Enter the list names for the Observed list and the Expected list. Enter the degrees of freedom and press calculate or draw. Make sure you clear any lists before you start. 11.4 Students in a social studies class hypothesize that the literacy rates across the world for every region are 82%. Table 11.14 shows the actual literacy rates across the world broken down by region. What are the test statistic and the degrees of freedom? MDG Region Adult Literacy Rate (%) Developed Regions 99.0 Commonwealth of Independent States 99.5 Northern Africa Sub-Saharan Africa Latin America and the Caribbean Eastern Asia Southern Asia South-Eastern Asia Western Asia Oceania Table 11.14 67.3 62.5 91.0 93.8 61.9 91.9 84.5 66.4 11.3 \| Test of Independence Tests of independence involve using a contingency table of observed (data) values. The test statistic for a test of independence is similar to that of a goodness-of-fit test: (O – E)2 E Σ (i ⋅ j) where: • O = observed values • E = expected values • i = the number of rows in the table • j = the number of columns in the table There are i ⋅ j terms of the form (O – E)2 E . A test of independence determines whether two factors are independent or not. You first encountered the term independence in Section 3.. As a review, consider the following example. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 NOTE The expected value for each cell needs to be at least five in order for you to use this test. CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 589 Example 11.5 Suppose A = a speeding violation in the last year and B = a cell phone user while driving. If A and B are independent then P(A AND B) = P(A)P(B). A AND B is the event that a driver received a speeding violation last year and also used a cell phone while driving. Suppose, in a study of drivers who received speeding violations in the last year, and who used cell phone while driving, that 755 people were surveyed. Out of the 755, 70 had a speeding violation and 685 did not; 305 used cell phones while driving and 450 did not. Let y = expected number of drivers who used a cell phone while driving and received speeding violations. If A and B are independent, then P(A AND B) = P(A)P(B). By substitution, y 755 = ⎛ ⎝ 70 755 ⎛ ⎞ ⎝ ⎠ 305 755 ⎞ ⎠ Solve for y: y = (70)(305) 755 = 28.3 About 28 people from the sample are expected to use cell phones while driving and to receive speeding violations. In a test of independence, we state the null and alternative hypotheses in words. Since the contingency table consists of two factors, the null hypothesis states that the factors are independent and the alternative hypothesis states that they are not independent (dependent). If we do a test of independence using the example, then the null hypothesis is: H0: Being a cell phone user while driving and receiving a speeding violation are independent events. If the null hypothesis were true, we would expect about 28 people to use cell phones while driving and to receive a speeding violation. The test of independence is always right-tailed because of the calculation of the test statistic. If the expected and observed values are not close together, then the test statistic is very large and way out in the right tail of the chi-square curve, as it is in a goodness-of-fit. The number of degrees of freedom for the test of independence is: df = (number of columns - 1)(number of rows - 1) The following formula calculates the expected number (E): E = (row total)(column total) total number surveyed 11.5 A sample of 300 students is taken. Of the students surveyed, 50 were music students, while 250 were not. Ninetyseven were on the honor roll, while 203 were not. If we assume being a music student and being on the honor roll are independent events, what is the expected number of music students who are also on the honor roll? Example 11.6 In a volunteer group, adults 21 and older volunteer from one to nine hours each week to spend time with a disabled senior citizen. The program recruits among community college students, four-year college students, and nonstudents. In Table 11.15 is a sample of the adult volunteers and the number of hours they volunteer per week. 590 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Type of Volunteer 1–3 Hours 4–6 Hours 7–9 Hours Row Total Community College Students 111 Four-Year College Students Nonstudents Column Total 96 91 298 96 133 150 379 48 61 53 162 255 290 294 839 Table 11.15 Number of Hours Worked Per Week by Volunteer Type (Observed) The table contains observed (O) values (data). Is the number of hours volunteered independent of the type of volunteer? Solution 11.6 The observed table and the question at the end of the problem, "Is the number of hours volunteered independent of the type of volunteer?" tell you this is a test of independence. The two factors are number of hours volunteered and type of volunteer. This test is al
ways right-tailed. H0: The number of hours volunteered is independent of the type of volunteer. Ha: The number of hours volunteered is dependent on the type of volunteer. The expected result are in Table 11.15. Type of Volunteer 1-3 Hours 4-6 Hours 7-9 Hours Community College Students 90.57 Four-Year College Students 103.00 Nonstudents 104.42 115.19 131.00 132.81 49.24 56.00 56.77 Table 11.16 Number of Hours Worked Per Week by Volunteer Type (Expected) The table contains expected (E) values (data). For example, the calculation for the expected frequency for the top left cell is E = (row total)(column total) total number surveyed = (255)(298) 839 = 90.57 Calculate the test statistic: χ2 = 12.99 (calculator or computer) 2 Distribution for the test: χ4 df = (3 columns – 1)(3 rows – 1) = (2)(2) = 4 Graph: Figure 11.7 Probability statement: p-value=P(χ2 > 12.99) = 0.0113 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 591 Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0113. α > p-value. Make a decision: Since α > p-value, reject H0. This means that the factors are not independent. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the number of hours volunteered and the type of volunteer are dependent on one another. For the example in Table 11.15, if there had been another type of volunteer, teenagers, what would the degrees of freedom be? Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 3 ENTER 3 ENTER. Enter the table values by row from Table 11.15. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 12.9909 and the p-value = 0.0113. Do the procedure a second time, but arrow down to Draw instead of calculate. 11.6 The Bureau of Labor Statistics gathers data about employment in the United States. A sample is taken to calculate the number of U.S. citizens working in one of several industry sectors over time. Table 11.17 shows the results: Industry Sector Nonagriculture wage and salary Goods-producing, excluding agriculture Services-providing Agriculture, forestry, fishing, and hunting Nonagriculture self-employed and unpaid family worker 2000 2010 2020 Total 13,243 13,044 15,018 41,305 2,457 1,771 1,950 6,178 10,786 11,273 13,068 35,127 240 931 214 894 201 972 655 2,797 Secondary wage and salary jobs in agriculture and private household industries 14 11 11 36 Secondary jobs as a self-employed or unpaid family worker 196 144 152 492 Total Table 11.17 27,867 27,351 31,372 86,590 We want to know if the change in the number of jobs is independent of the change in years. State the null and alternative hypotheses and the degrees of freedom. Example 11.7 De Anza College is interested in the relationship between anxiety level and the need to succeed in school. A random sample of 400 students took a test that measured anxiety level and need to succeed in school. Table 11.18 shows the results. De Anza College wants to know if anxiety level and need to succeed in school are independent events. 592 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Need to Succeed in School High Anxiety Medhigh Anxiety Medium Anxiety Medlow Anxiety Low Anxiety Row Total High Need Medium Need Low Need Column Total 35 18 4 57 42 48 5 95 53 63 11 127 15 33 15 63 10 31 17 58 155 193 52 400 Table 11.18 Need to Succeed in School vs. Anxiety Level a. How many high anxiety level students are expected to have a high need to succeed in school? Solution 11.7 a. The column total for a high anxiety level is 57. The row total for high need to succeed in school is 155. The sample size or total surveyed is 400. E = (row total)(column total) total surveyed = 155 ⋅ 57 400 = 22.09 The expected number of students who have a high anxiety level and a high need to succeed in school is about 22. b. If the two variables are independent, how many students do you expect to have a low need to succeed in school and a med-low level of anxiety? Solution 11.7 b. The column total for a med-low anxiety level is 63. The row total for a low need to succeed in school is 52. The sample size or total surveyed is 400. c. E = (row total)(column total) total surveyed = ________ Solution 11.7 c. E = (row total)(column total) total surveyed = 8.19 d. The expected number of students who have a med-low anxiety level and a low need to succeed in school is about ________. Solution 11.7 d. 8 11.7 Refer back to the information in Try It. How many service providing jobs are there expected to be in 2020? How many nonagriculture wage and salary jobs are there expected to be in 2020? 11.4 \| Test for Homogeneity The goodness–of–fit test can be used to decide whether a population fits a given distribution, but it will not suffice to decide whether two populations follow the same unknown distribution. A different test, called the test for homogeneity, can be used to draw a conclusion about whether two populations have the same distribution. To calculate the test statistic for a test for homogeneity, follow the same procedure as with the test of independence. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 NOTE The expected value for each cell needs to be at least five in order for you to use this test. CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 593 Hypotheses H0: The distributions of the two populations are the same. Ha: The distributions of the two populations are not the same. Test Statistic Use a χ 2 test statistic. It is computed in the same way as the test for independence. Degrees of Freedom (df) df = number of columns - 1 Requirements All values in the table must be greater than or equal to five. Common Uses Comparing two populations. For example: men vs. women, before vs. after, east vs. west. The variable is categorical with more than two possible response values. Example 11.8 Do male and female college students have the same distribution of living arrangements? Use a level of significance of 0.05. Suppose that 250 randomly selected male college students and 300 randomly selected female college students were asked about their living arrangements: dormitory, apartment, with parents, other. The results are shown in Table 11.18. Do male and female college students have the same distribution of living arrangements? Dormitory Apartment With Parents Other Males 72 Females 91 84 86 49 88 45 35 Table 11.19 Distribution of Living Arragements for College Males and College Females Solution 11.8 H0: The distribution of living arrangements for male college students is the same as the distribution of living arrangements for female college students. Ha: The distribution of living arrangements for male college students is not the same as the distribution of living arrangements for female college students. Degrees of Freedom (df): df = number of columns – 1 = 4 – 1 = 3 2 Distribution for the test: χ3 Calculate the test statistic: χ2 = 10.1287 (calculator or computer) Probability statement: p-value = P(χ2 >10.1287) = 0.0175 594 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 2 ENTER 4 ENTER. Enter the table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 10.1287 and the p-value = 0.0175. Do the procedure a second time but arrow down to Draw instead of calculate. Compare α and the p-value: Since no α is given, assume α = 0.05. p-value = 0.0175. α > p-value. Make a decision: Since α > p-value, reject H0. This means that the distributions are not the same. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that the distributions of living arrangements for male and female college students are not the same. Notice that the conclusion is only that the distributions are not the same. We cannot use the test for homogeneity to draw any conclusions about how they differ. 11.8 Do families and singles have the same distribution of cars? Use a level of significance of 0.05. Suppose that 100 randomly selected families and 200 randomly selected singles were asked what type of car they drove: sport, sedan, hatchback, truck, van/SUV. The results are shown in Table 11.20. Do families and singles have the same distribution of cars? Test at a level of significance of 0.05. Sport Sedan Hatchback Truck Van/SUV Family 5 Single 45 Table 11.20 15 65 35 37 17 46 28 7 Example 11.9 Both before and after a recent earthquake, surveys were conducted asking voters which of the three candidates they planned on voting for in the upcoming city council election. Has there been a change since the earthquake? Use a level of significance of 0.05. Table 11.20 shows the results of the survey. Has there been a change in the distribution of voter preferences since the earthquake? Perez Chung Stevens Before 167 After 214 128 197 135 225 Table 11.21 Solution 11.9 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 595 H0: The distribution of voter preferences was the same before and after the earthquake. Ha: The distribution of voter preferences was not the same before and after the earthquake. Degrees of Freedom (df): df = number of columns – 1 = 3 – 1 = 2 2 Distribution for the test: χ2 Calculate the test statistic: χ2 = 3.2603 (calculator or computer) Probability statement: p-value=P(χ2 > 3.2603) = 0.1959 Press the MATRX key and arrow over to EDIT. Press 1:[A]. Press 2 ENTER 3 ENTER. Enter the
table values by row. Press ENTER after each. Press 2nd QUIT. Press STAT and arrow over to TESTS. Arrow down to C:χ2-TEST. Press ENTER. You should see Observed:[A] and Expected:[B]. Arrow down to Calculate. Press ENTER. The test statistic is 3.2603 and the p-value = 0.1959. Do the procedure a second time but arrow down to Draw instead of calculate. Compare α and the p-value: α = 0.05 and the p-value = 0.1959. α < p-value. Make a decision: Since α < p-value, do not reject Ho. Conclusion: At a 5% level of significance, from the data, there is insufficient evidence to conclude that the distribution of voter preferences was not the same before and after the earthquake. 11.9 Ivy League schools receive many applications, but only some can be accepted. At the schools listed in Table 11.22, two types of applications are accepted: regular and early decision. Application Type Accepted Brown Columbia Cornell Dartmouth Penn Yale Regular Early Decision Table 11.22 2,115 1,792 577 627 5,306 1,228 1,734 444 2,685 1,245 1,195 761 We want to know if the number of regular applications accepted follows the same distribution as the number of early applications accepted. State the null and alternative hypotheses, the degrees of freedom and the test statistic, sketch the graph of the p-value, and draw a conclusion about the test of homogeneity. 11.5 \| Comparison of the Chi-Square Tests You have seen the χ2 test statistic used in three different circumstances. The following bulleted list is a summary that will help you decide which χ2 test is the appropriate one to use. • Goodness-of-Fit: Use the goodness-of-fit test to decide whether a population with an unknown distribution "fits" a known distribution. In this case there will be a single qualitative survey question or a single outcome of an experiment 596 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION from a single population. Goodness-of-Fit is typically used to see if the population is uniform (all outcomes occur with equal frequency), the population is normal, or the population is the same as another population with a known distribution. The null and alternative hypotheses are: H0: The population fits the given distribution. Ha: The population does not fit the given distribution. • Independence: Use the test for independence to decide whether two variables (factors) are independent or dependent. In this case there will be two qualitative survey questions or experiments and a contingency table will be constructed. The goal is to see if the two variables are unrelated (independent) or related (dependent). The null and alternative hypotheses are: H0: The two variables (factors) are independent. Ha: The two variables (factors) are dependent. • Homogeneity: Use the test for homogeneity to decide if two populations with unknown distributions have the same distribution as each other. In this case there will be a single qualitative survey question or experiment given to two different populations. The null and alternative hypotheses are: H0: The two populations follow the same distribution. Ha: The two populations have different distributions. 11.6 \| Test of a Single Variance A test of a single variance assumes that the underlying distribution is normal. The null and alternative hypotheses are stated in terms of the population variance (or population standard deviation). The test statistic is: ⎞ ⎠s2 ⎛ ⎝n - 1 σ 2 where: • n = the total number of data • s2 = sample variance • σ2 = population variance You may think of s as the random variable in this test. The number of degrees of freedom is df = n - 1. A test of a single variance may be right-tailed, left-tailed, or two-tailed. Example 11.10 will show you how to set up the null and alternative hypotheses. The null and alternative hypotheses contain statements about the population variance. Example 11.10 Math instructors are not only interested in how their students do on exams, on average, but how the exam scores vary. To many instructors, the variance (or standard deviation) may be more important than the average. Suppose a math instructor believes that the standard deviation for his final exam is five points. One of his best students thinks otherwise. The student claims that the standard deviation is more than five points. If the student were to conduct a hypothesis test, what would the null and alternative hypotheses be? Solution 11.10 Even though we are given the population standard deviation, we can set up the test using the population variance as follows. • H0: σ2 = 52 • Ha: σ2 > 52 11.10 A SCUBA instructor wants to record the collective depths each of his students dives during their checkout. He is interested in how the depths vary, even though everyone should have been at the same depth. He believes the This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 standard deviation is three feet. His assistant thinks the standard deviation is less than three feet. If the instructor were to conduct a test, what would the null and alternative hypotheses be? CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 597 Example 11.11 With individual lines at its various windows, a post office finds that the standard deviation for normally distributed waiting times for customers on Friday afternoon is 7.2 minutes. The post office experiments with a single, main waiting line and finds that for a random sample of 25 customers, the waiting times for customers have a standard deviation of 3.5 minutes. With a significance level of 5%, test the claim that a single line causes lower variation among waiting times (shorter waiting times) for customers. Solution 11.11 Since the claim is that a single line causes less variation, this is a test of a single variance. The parameter is the population variance, σ2, or the population standard deviation, σ. Random Variable: The sample standard deviation, s, is the random variable. Let s = standard deviation for the waiting times. • H0: σ2 = 7.22 • Ha: σ2 < 7.22 The word "less" tells you this is a left-tailed test. Distribution for the test: χ24 2 , where: • n = the number of customers sampled • df = n – 1 = 25 – 1 = 24 Calculate the test statistic: χ 2 = (n − 1)s2 σ 2 = (25 − 1)(3.5)2 7.22 = 5.67 where n = 25, s = 3.5, and σ = 7.2. Graph: Figure 11.8 Probability statement: p-value = P ( χ2 < 5.67) = 0.000042 Compare α and the p-value: α = 0.05; p-value = 0.000042; α > p-value Make a decision: Since α > p-value, reject H0. This means that you reject σ2 = 7.22. In other words, you do not think the variation in waiting times is 7.2 minutes; you think the variation in waiting times is less. Conclusion: At a 5% level of significance, from the data, there is sufficient evidence to conclude that a single line causes a lower variation among the waiting times or with a single line, the customer waiting times vary less than 7.2 minutes. 598 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION In 2nd DISTR, use 7:χ2cdf. The syntax is (lower, upper, df) for the parameter list. For Example 11.11, χ2cdf(-1E99,5.67,24). The p-value = 0.000042. 11.11 The FCC conducts broadband speed tests to measure how much data per second passes between a consumer’s computer and the internet. As of August of 2012, the standard deviation of Internet speeds across Internet Service Providers (ISPs) was 12.2 percent. Suppose a sample of 15 ISPs is taken, and the standard deviation is 13.2. An analyst claims that the standard deviation of speeds is more than what was reported. State the null and alternative hypotheses, compute the degrees of freedom, the test statistic, sketch the graph of the p-value, and draw a conclusion. Test at the 1% significance level. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 599 11.1 Lab 1: Chi-Square Goodness-of-Fit Class Time: Names: Student Learning Outcome • The student will evaluate data collected to determine if they fit either the uniform or exponential distributions. Collect the Data Go to your local supermarket. Ask 30 people as they leave for the total amount on their grocery receipts. (Or, ask three cashiers for the last ten amounts. Be sure to include the express lane, if it is open.) NOTE You may need to combine two categories so that each cell has an expected value of at least five. 1. Record the values. __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ __________ Table 11.23 2. Construct a histogram of the data. Make five to six intervals. Sketch the graph using a ruler and pencil. Scale the axes. Figure 11.9 3. Calculate the following: a. b. x¯ = ________ s = ________ 600 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION c. s2 = ________ Uniform Distribution Test to see if grocery receipts follow the uniform distribution. 1. Using your lowest and highest values, X ~ U (_______, _______) 2. Divide the distribution into fifths. 3. Calculate the following: a. lowest value = _________ b. 20th percentile = _________ c. 40th percentile = _________ d. 60th percentile = _________ e. 80th percentile = _________ f. highest value = _________ 4. For each fifth, count the observed number of receipts and record it. Then determine the expected number of receipts and record that. Fifth Observed Expected 1st 2nd 3rd 4th 5th Table 11.24 5. H0: ________ 6. Ha: ________ 7. What distribution should you use for a hypothesis test? 8. Why did you choose this distribution? 9. Calculate the test statistic. 10. Find the p-value. 11. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value. This content is available for free at http://textbookequity.org/introd
uctory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 601 Figure 11.10 12. State your decision. 13. State your conclusion in a complete sentence. Exponential Distribution Test to see if grocery receipts follow the exponential distribution with decay parameter 1 x¯ . 1. Using 1 x¯ as the decay parameter, X ~ Exp(_________). 2. Calculate the following: a. b. lowest value = ________ first quartile = ________ c. 37th percentile = ________ d. median = ________ e. 63rd percentile = ________ f. 3rd quartile = ________ g. highest value = ________ 3. For each cell, count the observed number of receipts and record it. Then determine the expected number of receipts and record that. Cell Observed Expected 1st 2nd 3rd 4th 5th 6th Table 11.25 4. H0: ________ 5. Ha: ________ 602 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 6. What distribution should you use for a hypothesis test? 7. Why did you choose this distribution? 8. Calculate the test statistic. 9. Find the p-value. 10. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value. Figure 11.11 11. State your decision. 12. State your conclusion in a complete sentence. Discussion Questions 1. Did your data fit either distribution? If so, which? 2. In general, do you think it’s likely that data could fit more than one distribution? In complete sentences, explain why or why not. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 603 11.2 Lab 2: Chi-Square Test of Independence Class Time: Names: Student Learning Outcome • The student will evaluate if there is a significant relationship between favorite type of snack and gender. Collect the Data 1. Using your class as a sample, complete the following chart. Ask each other what your favorite snack is, then total the results. NOTE You may need to combine two food categories so that each cell has an expected value of at least five. sweets (candy & baked goods) ice cream chips & pretzels fruits & vegetables Total male female Total Table 11.26 Favorite type of snack 2. Looking at Table 11.26, does it appear to you that there is a dependence between gender and favorite type of snack food? Why or why not? Hypothesis Test Conduct a hypothesis test to determine if the factors are independent: 1. H0: ________ 2. Ha: ________ 3. What distribution should you use for a hypothesis test? 4. Why did you choose this distribution? 5. Calculate the test statistic. 6. Find the p-value. 7. Sketch a graph of the situation. Label and scale the x-axis. Shade the area corresponding to the p-value. 604 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Figure 11.12 8. State your decision. 9. State your conclusion in a complete sentence. Discussion Questions 1. Is the conclusion of your study the same as or different from your answer to answer to question two under Collect the Data? 2. Why do you think that occurred? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 605 KEY TERMS Contingency Table a table that displays sample values for two different factors that may be dependent or contingent on one another; it facilitates determining conditional probabilities. CHAPTER REVIEW 11.1 Facts About the Chi-Square Distribution The chi-square distribution is a useful tool for assessment in a series of problem categories. These problem categories include primarily (i) whether a data set fits a particular distribution, (ii) whether the distributions of two populations are the same, (iii) whether two events might be independent, and (iv) whether there is a different variability than expected within a population. An important parameter in a chi-square distribution is the degrees of freedom df in a given problem. The random variable in the chi-square distribution is the sum of squares of df standard normal variables, which must be independent. The key characteristics of the chi-square distribution also depend directly on the degrees of freedom. The chi-square distribution curve is skewed to the right, and its shape depends on the degrees of freedom df. For df > 90, the curve approximates the normal distribution. Test statistics based on the chi-square distribution are always greater than or equal to zero. Such application tests are almost always right-tailed tests. 11.2 Goodness-of-Fit Test To assess whether a data set fits a specific distribution, you can apply the goodness-of-fit hypothesis test that uses the chi-square distribution. The null hypothesis for this test states that the data come from the assumed distribution. The test compares observed values against the values you would expect to have if your data followed the assumed distribution. The test is almost always right-tailed. Each observation or cell category must have an expected value of at least five. 11.3 Test of Independence To assess whether two factors are independent or not, you can apply the test of independence that uses the chi-square distribution. The null hypothesis for this test states that the two factors are independent. The test compares observed values to expected values. The test is right-tailed. Each observation or cell category must have an expected value of at least 5. 11.4 Test for Homogeneity To assess whether two data sets are derived from the same distribution—which need not be known, you can apply the test for homogeneity that uses the chi-square distribution. The null hypothesis for this test states that the populations of the two data sets come from the same distribution. The test compares the observed values against the expected values if the two populations followed the same distribution. The test is right-tailed. Each observation or cell category must have an expected value of at least five. 11.5 Comparison of the Chi-Square Tests The goodness-of-fit test is typically used to determine if data fits a particular distribution. The test of independence makes use of a contingency table to determine the independence of two factors. The test for homogeneity determines whether two populations come from the same distribution, even if this distribution is unknown. 11.6 Test of a Single Variance To test variability, use the chi-square test of a single variance. The test may be left-, right-, or two-tailed, and its hypotheses are always expressed in terms of the variance (or standard deviation). FORMULA REVIEW 11.1 Facts About the Chi-Square Distribution χ2 = (Z1)2 + (Z2)2 + … (Zdf)2 chi-square distribution random variable μχ2 = df chi-square distribution population mean ⎝d f ⎞ ⎠ Chi-Square distribution population standard σ χ 2 = 2⎛ deviation 606 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 11.2 Goodness-of-Fit Test (O − E)2 E ∑ k goodness-of-fit test statistic where: O: observed values E: expected values k: number of different data cells or categories df = k − 1 degrees of freedom 11.3 Test of Independence Test of Independence • The number of degrees of freedom is equal to (number of columns - 1)(number of rows - 1). • The test statistic is Σ (i ⋅ j) (O – E)2 E where O = observed values, E = expected values, i = the number of rows in the table, and j = the number of columns in the table. • If the null hypothesis is true, the expected number E = (row total)(column total) total surveyed . 11.4 Test for Homogeneity PRACTICE Homogeneity test statistic where: O = (O − E)2 E ∑ i ⋅ j observed values E = expected values i = number of rows in data contingency table j = number of columns in data contingency table df = (i −1)(j −1) Degrees of freedom 11.6 Test of a Single Variance χ 2 = (n − 1) ⋅ s2 σ 2 Test of a single variance statistic where: n: sample size s: sample standard deviation σ: population standard deviation df = n – 1 Degrees of freedom Test of a Single Variance • Use the test to determine variation. • The degrees of freedom is the number of samples – 1. • The test statistic is (n – 1) ⋅ s2 , where n = the total σ 2 number of data, s2 = sample variance, and σ2 = population variance. • The test may be left-, right-, or two-tailed. 11.1 Facts About the Chi-Square Distribution 1. If the number of degrees of freedom for a chi-square distribution is 25, what is the population mean and standard deviation? 2. If df > 90, the distribution is _____________. If df = 15, the distribution is ________________. 3. When does the chi-square curve approximate a normal distribution? 4. Where is μ located on a chi-square curve? 5. Is it more likely the df is 90, 20, or two in the graph? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 607 Figure 11.13 11.2 Goodness-of-Fit Test Determine the appropriate test to be used in the next three exercises. 6. An archeologist is calculating the distribution of the frequency of the number of artifacts she finds in a dig site. Based on previous digs, the archeologist creates an expected distribution broken down by grid sections in the dig site. Once the site has been fully excavated, she compares the actual number of artifacts found in each grid section to see if her expectation was accurate. 7. An economist is deriving a model to predict outcomes on the stock market. He creates a list of expected points on the stock market index for the next two weeks. At the close of each day’s trading, he records the actual points on the index. He wants to see how well his model matched what actually happened. 8. A personal trainer is putting together a weight-lifting program for her clients. For a 90-day program, she expects each client to lift a specific maximum weight each week. As she goes along, she records the actual maximum weights her clients lifted. She wants to know how well her expectations met with what was observed. Use the
following information to answer the next five exercises: A teacher predicts that the distribution of grades on the final exam will be and they are recorded in Table 11.27. Grade Proportion A B C D 0.25 0.30 0.35 0.10 Table 11.27 The actual distribution for a class of 20 is in Table 11.28. Grade Frequency A B C D 7 7 5 1 Table 11.28 608 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 9. d f = ______ 10. State the null and alternative hypotheses. 11. χ2 test statistic = ______ 12. p-value = ______ 13. At the 5% significance level, what can you conclude? Use the following information to answer the next nine exercises: The following data are real. The cumulative number of AIDS cases reported for Santa Clara County is broken down by ethnicity as in Table 11.29. Ethnicity White Hispanic Number of Cases 2,229 1,157 Black/African-American 457 Asian, Pacific Islander 232 Total = 4,075 Table 11.29 The percentage of each ethnic group in Santa Clara County is as in Table 11.30. Ethnicity White Hispanic Black/AfricanAmerican Asian, Pacific Islander Table 11.30 Percentage of total county population Number expected (round to two decimal places) 1748.18 42.9% 26.7% 2.6% 27.8% Total = 100% 14. If the ethnicities of AIDS victims followed the ethnicities of the total county population, fill in the expected number of cases per ethnic group. Perform a goodness-of-fit test to determine whether the occurrence of AIDS cases follows the ethnicities of the general population of Santa Clara County. 15. H0: _______ 16. Ha: _______ 17. Is this a right-tailed, left-tailed, or two-tailed test? 18. degrees of freedom = _______ 19. χ2 test statistic = _______ 20. p-value = _______ 21. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 609 Figure 11.14 Let α = 0.05 Decision: ________________ Reason for the Decision: ________________ Conclusion (write out in complete sentences): ________________ 22. Does it appear that the pattern of AIDS cases in Santa Clara County corresponds to the distribution of ethnic groups in this county? Why or why not? 11.3 Test of Independence Determine the appropriate test to be used in the next three exercises. 23. A pharmaceutical company is interested in the relationship between age and presentation of symptoms for a common viral infection. A random sample is taken of 500 people with the infection across different age groups. 24. The owner of a baseball team is interested in the relationship between player salaries and team winning percentage. He takes a random sample of 100 players from different organizations. 25. A marathon runner is interested in the relationship between the brand of shoes runners wear and their run times. She takes a random sample of 50 runners and records their run times as well as the brand of shoes they were wearing. Use the following information to answer the next seven exercises: Transit Railroads is interested in the relationship between travel distance and the ticket class purchased. A random sample of 200 passengers is taken. Table 11.31 shows the results. The railroad wants to know if a passenger’s choice in ticket class is independent of the distance they must travel. Traveling Distance Third class Second class First class Total 1–100 miles 101–200 miles 201–300 miles 301–400 miles 401–500 miles Total Table 11.31 21 18 16 12 6 73 14 16 17 14 6 67 6 8 15 21 10 60 41 42 48 47 22 200 26. State the hypotheses. H0: _______ Ha: _______ 27. df = _______ 28. How many passengers are expected to travel between 201 and 300 miles and purchase second-class tickets? 29. How many passengers are expected to travel between 401 and 500 miles and purchase first-class tickets? 610 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 30. What is the test statistic? 31. What is the p-value? 32. What can you conclude at the 5% level of significance? Use the following information to answer the next eight exercises: An article in the New England Journal of Medicine, discussed a study on smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African Americans, 2,745 Native Hawaiians, 12,831 Latinos, 8,378 Japanese Americans and 7,650 whites. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African Americans, 3,062 Native Hawaiians, 4,932 Latinos, 10,680 Japanese Americans, and 9,877 whites. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African Americans, 1,419 Native Hawaiians, 1,406 Latinos, 4,715 Japanese Americans, and 6,062 whites. Of the people smoking at least 31 cigarettes per day, there were 759 African Americans, 788 Native Hawaiians, 800 Latinos, 2,305 Japanese Americans, and 3,970 whites. 33. Complete the table. Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans White TOTALS 1-10 11-20 21-30 31+ TOTALS Table 11.32 Smoking Levels by Ethnicity (Observed) 34. State the hypotheses. H0: _______ Ha: _______ 35. Enter expected values in Table 11.32. Round to two decimal places. Calculate the following values: 36. df = _______ 37. χ 2 test statistic = ______ 38. p-value = ______ 39. Is this a right-tailed, left-tailed, or two-tailed test? Explain why. 40. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade in the region corresponding to the p-value. Figure 11.15 State the decision and conclusion (in a complete sentence) for the following preconceived levels of α. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 611 41. α = 0.05 a. Decision: ___________________ b. Reason for the decision: ___________________ c. Conclusion (write out in a complete sentence): ___________________ 42. α = 0.01 a. Decision: ___________________ b. Reason for the decision: ___________________ c. Conclusion (write out in a complete sentence): ___________________ 11.4 Test for Homogeneity 43. A math teacher wants to see if two of her classes have the same distribution of test scores. What test should she use? 44. What are the null and alternative hypotheses for Exercise 11.43? 45. A market researcher wants to see if two different stores have the same distribution of sales throughout the year. What type of test should he use? 46. A meteorologist wants to know if East and West Australia have the same distribution of storms. What type of test should she use? 47. What condition must be met to use the test for homogeneity? Use the following information to answer the next five exercises: Do private practice doctors and hospital doctors have the same distribution of working hours? Suppose that a sample of 100 private practice doctors and 150 hospital doctors are selected at random and asked about the number of hours a week they work. The results are shown in Table 11.33. 20–30 30–40 40–50 50–60 Private Practice 16 Hospital 8 40 44 38 59 6 39 Table 11.33 48. State the null and alternative hypotheses. 49. df = _______ 50. What is the test statistic? 51. What is the p-value? 52. What can you conclude at the 5% significance level? 11.5 Comparison of the Chi-Square Tests 53. Which test do you use to decide whether an observed distribution is the same as an expected distribution? 54. What is the null hypothesis for the type of test from Exercise 11.53? 55. Which test would you use to decide whether two factors have a relationship? 56. Which test would you use to decide if two populations have the same distribution? 57. How are tests of independence similar to tests for homogeneity? 58. How are tests of independence different from tests for homogeneity? 11.6 Test of a Single Variance Use the following information to answer the next three exercises: An archer’s standard deviation for his hits is six (data is measured in distance from the center of the target). An observer claims the standard deviation is less. 59. What type of test should be used? 60. State the null and alternative hypotheses. 61. Is this a right-tailed, left-tailed, or two-tailed test? Use the following information to answer the next three exercises: The standard deviation of heights for students in a school 612 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION is 0.81. A random sample of 50 students is taken, and the standard deviation of heights of the sample is 0.96. A researcher in charge of the study believes the standard deviation of heights for the school is greater than 0.81. 62. What type of test should be used? 63. State the null and alternative hypotheses. 64. df = ________ Use the following information to answer the next four exercises: The average waiting time in a doctor’s office varies. The standard deviation of waiting times in a doctor’s office is 3.4 minutes. A random sample of 30 patients in the doctor’s office has a standard deviation of waiting times of 4.1 minutes. One doctor believes the variance of waiting times is greater than originally thought. 65. What type of test should be used? 66. What is the test statistic? 67. What is the p-value? 68. What can you conclude at the 5% significance level? HOMEWORK 11.1 Facts About the Chi-Square Distribution Decide whether the following statements are true or false. 69. As the number of degrees of freedom increases, the graph of the chi-square distribution looks more and more symmetrical. 70. The standard deviation of the chi-square distribution is twice the mean. 71. The mean and the median of the chi-square distribution are the same if df = 24. 11.2 Goodness-of-Fit Test For each problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round
expected frequency to two decimal places. 72. A six-sided die is rolled 120 times. Fill in the expected frequency column. Then, conduct a hypothesis test to determine if the die is fair. The data in Table 11.34 are the result of the 120 rolls. Face Value Frequency Expected Frequency 1 2 3 4 5 6 Table 11.34 15 29 16 15 30 15 73. The marital status distribution of the U.S. male population, ages 15 and older, is as shown in Table 11.35. Marital Status Percent Expected Frequency never married married widowed Table 11.35 31.3 56.1 2.5 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 613 Marital Status Percent Expected Frequency divorced/separated 10.1 Table 11.35 Suppose that a random sample of 400 U.S. young adult males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population. Calculate the frequency one would expect when surveying 400 people. Fill in Table 11.36, rounding to two decimal places. Marital Status Frequency never married married widowed 140 238 2 divorced/separated 20 Table 11.36 Use the following information to answer the next two exercises: The columns in Table 11.37 contain the Race/Ethnicity of U.S. Public Schools for a recent year, the percentages for the Advanced Placement Examinee Population for that class, and the Overall Student Population. Suppose the right column contains the result of a survey of 1,000 local students from that year who took an AP Exam. Race/Ethnicity AP Examinee Population Overall Student Population Survey Frequency Asian, Asian American, or Pacific Islander Black or African-American Hispanic or Latino 10.2% 8.2% 15.5% American Indian or Alaska Native 0.6% White Not reported/other Table 11.37 59.4% 6.1% 5.4% 14.5% 15.9% 1.2% 61.6% 1.4% 113 94 136 10 604 43 74. Perform a goodness-of-fit test to determine whether the local results follow the distribution of the U.S. overall student population based on ethnicity. 75. Perform a goodness-of-fit test to determine whether the local results follow the distribution of U.S. AP examinee population, based on ethnicity. 76. The City of South Lake Tahoe, CA, has an Asian population of 1,419 people, out of a total population of 23,609. Suppose that a survey of 1,419 self-reported Asians in the Manhattan, NY, area yielded the data in Table 11.38. Conduct a goodness-of-fit test to determine if the self-reported sub-groups of Asians in the Manhattan area fit that of the Lake Tahoe area. Race Lake Tahoe Frequency Manhattan Frequency Asian Indian 131 Chinese 118 Filipino 1,045 Table 11.38 174 557 518 614 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Race Lake Tahoe Frequency Manhattan Frequency Japanese Korean 80 12 Vietnamese 9 Other 24 Table 11.38 54 29 21 66 Use the following information to answer the next two exercises: UCLA conducted a survey of more than 263,000 college freshmen from 385 colleges in fall 2005. The results of students' expected majors by gender were reported in The Chronicle of Higher Education (2/2/2006). Suppose a survey of 5,000 graduating females and 5,000 graduating males was done as a follow-up last year to determine what their actual majors were. The results are shown in the tables for Exercise 11.77 and Exercise 11.78. The second column in each table does not add to 100% because of rounding. 77. Conduct a goodness-of-fit test to determine if the actual college majors of graduating females fit the distribution of their expected majors. Major Women - Expected Major Women - Actual Major Arts & Humanities 14.0% Biological Sciences 8.4% Business Education Engineering 13.1% 13.0% 2.6% Physical Sciences 2.6% Professional 18.9% Social Sciences 13.0% Technical Other Undecided Table 11.39 0.4% 5.8% 8.0% 670 410 685 650 145 125 975 605 15 300 420 78. Conduct a goodness-of-fit test to determine if the actual college majors of graduating males fit the distribution of their expected majors. Major Men - Expected Major Men - Actual Major Arts & Humanities 11.0% Biological Sciences 6.7% Business Education Engineering 22.7% 5.8% 15.6% Physical Sciences 3.6% Professional Social Sciences Technical Table 11.40 9.3% 7.6% 1.8% 600 330 1130 305 800 175 460 370 90 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 615 Major Other Undecided Table 11.40 Men - Expected Major Men - Actual Major 8.2% 6.6% 400 340 Read the statement and decide whether it is true or false. 79. In a goodness-of-fit test, the expected values are the values we would expect if the null hypothesis were true. 80. In general, if the observed values and expected values of a goodness-of-fit test are not close together, then the test statistic can get very large and on a graph will be way out in the right tail. 81. Use a goodness-of-fit test to determine if high school principals believe that students are absent equally during the week or not. 82. The test to use to determine if a six-sided die is fair is a goodness-of-fit test. 83. In a goodness-of fit test, if the p-value is 0.0113, in general, do not reject the null hypothesis. 84. A sample of 212 commercial businesses was surveyed for recycling one commodity; a commodity here means any one type of recyclable material such as plastic or aluminum. Table 11.41 shows the business categories in the survey, the sample size of each category, and the number of businesses in each category that recycle one commodity. Based on the study, on average half of the businesses were expected to be recycling one commodity. As a result, the last column shows the expected number of businesses in each category that recycle one commodity. At the 5% significance level, perform a hypothesis test to determine if the observed number of businesses that recycle one commodity follows the uniform distribution of the expected values. Business Type Office Retail/ Wholesale Food/ Restaurants Manufacturing/ Medical Hotel/Mixed Table 11.41 Number in class Observed Number that recycle one commodity Expected number that recycle one commodity 35 48 53 52 24 19 27 35 21 9 17.5 24 26.5 26 12 85. Table 11.42 contains information from a survey among 499 participants classified according to their age groups. The second column shows the percentage of obese people per age class among the study participants. The last column comes from a different study at the national level that shows the corresponding percentages of obese people in the same age classes in the USA. Perform a hypothesis test at the 5% significance level to determine whether the survey participants are a representative sample of the USA obese population. Age Class (Years) Obese (Percentage) Expected USA average (Percentage) 20–30 31–40 41–50 51–60 61–70 Table 11.42 75.0 26.5 13.6 21.9 21.0 32.6 32.6 36.6 36.6 39.7 616 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 11.3 Test of Independence For each problem, use a solution sheet to solve the hypothesis test problem. Go to the Appendix (http://cnx.org/ content/m47157/latest/) for the chi-square solution sheet. Round expected frequency to two decimal places. 86. A recent debate about where in the United States skiers believe the skiing is best prompted the following survey. Test to see if the best ski area is independent of the level of the skier. U.S. Ski Area Beginner Intermediate Advanced Tahoe Utah Colorado Table 11.43 20 10 10 30 30 40 40 60 50 87. Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver’s family (that is, whether car size and family size are independent). To test this, suppose that 800 car owners were randomly surveyed with the results in Table 11.44. Conduct a test of independence. Family Size Sub & Compact Mid-size Full-size Van & Truck 1 2 3–4 5+ Table 11.44 20 20 20 20 35 50 50 30 40 70 100 70 35 80 90 70 88. College students may be interested in whether or not their majors have any effect on starting salaries after graduation. Suppose that 300 recent graduates were surveyed as to their majors in college and their starting salaries after graduation. Table 11.45 shows the data. Conduct a test of independence. Major < $50,000 $50,000 – $68,999 $69,000 + English 5 Engineering 10 Nursing Business 10 10 Psychology 20 Table 11.45 20 30 15 20 30 5 60 15 30 20 89. Some travel agents claim that honeymoon hot spots vary according to age of the bride. Suppose that 280 recent brides were interviewed as to where they spent their honeymoons. The information is given in Table 11.46. Conduct a test of independence. Location 20–29 30–39 40–49 50 and over Niagara Falls 15 Poconos Europe 15 10 Table 11.46 25 25 25 25 25 15 20 10 5 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 617 Location 20–29 30–39 40–49 50 and over Virgin Islands 20 25 15 5 Table 11.46 90. A manager of a sports club keeps information concerning the main sport in which members participate and their ages. To test whether there is a relationship between the age of a member and his or her choice of sport, 643 members of the sports club are randomly selected. Conduct a test of independence. Sport 18 - 25 26 - 30 31 - 40 41 and over racquetball 42 tennis 58 swimming 72 Table 11.47 58 76 60 30 38 65 46 65 33 91. A major food manufacturer is concerned that the sales for its skinny french fries have been decreasing. As a part of a feasibility study, the company conducts research into the types of fries sold across the country to determine if the type of fries sold is independent of the area of the country. The results of the study are shown in Table 11.48. Conduct a test of independence. Type of Fries Northeast South Central
West skinny fries curly fries steak fries Table 11.48 70 100 20 50 60 40 20 15 10 25 30 10 92. According to Dan Lenard, an independent insurance agent in the Buffalo, N.Y. area, the following is a breakdown of the amount of life insurance purchased by males in the following age groups. He is interested in whether the age of the male and the amount of life insurance purchased are independent events. Conduct a test for independence. Age of Males None < $200,000 $200,000–$400,000 $401,001–$1,000,000 $1,000,001+ 20–29 30–39 40–49 50+ Table 11.49 40 35 20 40 15 5 0 30 40 20 30 15 0 20 0 15 5 10 30 10 93. Suppose that 600 thirty-year-olds were surveyed to determine whether or not there is a relationship between the level of education an individual has and salary. Conduct a test of independence. Annual Salary Not a high school graduate High school graduate College graduate Masters or doctorate < $30,000 15 $30,000–$40,000 20 Table 11.50 25 40 10 70 5 30 618 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Annual Salary Not a high school graduate High school graduate College graduate Masters or doctorate $40,000–$50,000 10 $50,000–$60,000 5 $60,000+ 0 Table 11.50 20 10 5 40 20 10 55 60 150 Read the statement and decide whether it is true or false. 94. The number of degrees of freedom for a test of independence is equal to the sample size minus one. 95. The test for independence uses tables of observed and expected data values. 96. The test to use when determining if the college or university a student chooses to attend is related to his or her socioeconomic status is a test for independence. 97. In a test of independence, the expected number is equal to the row total multiplied by the column total divided by the total surveyed. 98. An ice cream maker performs a nationwide survey about favorite flavors of ice cream in different geographic areas of the U.S. Based on Table 11.51, do the numbers suggest that geographic location is independent of favorite ice cream flavors? Test at the 5% significance level. Strawberry Chocolate Vanilla Rocky Road Mint Chocolate Chip Pistachio 12 10 8 15 45 21 32 31 28 22 22 27 30 112 101 19 11 8 8 46 15 15 15 15 60 8 6 7 6 27 Row total 97 96 96 102 391 U.S. region/ Flavor West Midwest East South Column Total Table 11.51 99. Table 11.52 provides a recent survey of the youngest online entrepreneurs whose net worth is estimated at one million dollars or more. Their ages range from 17 to 30. Each cell in the table illustrates the number of entrepreneurs who correspond to the specific age group and their net worth. Are the ages and net worth independent? Perform a test of independence at the 5% significance level. Age Group\ Net Worth Value (in millions of US dollars) 1–5 6–24 ≥25 Row Total 17–25 26–30 Column Total Table 11.52 8 6 7 5 14 12 5 9 14 20 20 40 100. A 2013 poll in California surveyed people about taxing sugar-sweetened beverages. The results are presented in Table 11.53, and are classified by ethnic group and response type. Are the poll responses independent of the participants’ ethnic group? Conduct a test of independence at the 5% significance level. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 619 AsianAmerican White/NonHispanic AfricanAmerican 433 234 43 710 41 24 16 71 Opinion/ Ethnicity Against tax In Favor of tax No opinion 48 54 16 Column Total 118 Table 11.53 11.4 Test for Homogeneity Latino 160 147 19 272 Row Total 628 459 84 1171 For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 101. A psychologist is interested in testing whether there is a difference in the distribution of personality types for business majors and social science majors. The results of the study are shown in Table 11.54. Conduct a test of homogeneity. Test at a 5% level of significance. Open Conscientious Extrovert Agreeable Neurotic Business 41 Social Science 72 52 75 46 63 61 80 58 65 Table 11.54 102. Do men and women select different breakfasts? The breakfasts ordered by randomly selected men and women at a popular breakfast place is shown in Table 11.55. Conduct a test for homogeneity at a 5% level of significance. French Toast Pancakes Waffles Omelettes Men 47 Women 65 Table 11.55 35 59 28 55 53 60 103. A fisherman is interested in whether the distribution of fish caught in Green Valley Lake is the same as the distribution of fish caught in Echo Lake. Of the 191 randomly selected fish caught in Green Valley Lake, 105 were rainbow trout, 27 were other trout, 35 were bass, and 24 were catfish. Of the 293 randomly selected fish caught in Echo Lake, 115 were rainbow trout, 58 were other trout, 67 were bass, and 53 were catfish. Perform a test for homogeneity at a 5% level of significance. 104. In 2007, the United States had 1.5 million homeschooled students, according to the U.S. National Center for Education Statistics. In Table 11.56 you can see that parents decide to homeschool their children for different reasons, and some reasons are ranked by parents as more important than others. According to the survey results shown in the table, is the distribution of applicable reasons the same as the distribution of the most important reason? Provide your assessment at the 5% significance level. Did you expect the result you obtained? Reasons for Homeschooling Applicable Reason (in thousands of respondents) Most Important Reason (in thousands of respondents) Concern about the environment of other schools 1,321 309 Row Total 1,630 Table 11.56 620 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION Reasons for Homeschooling Dissatisfaction with academic instruction at other schools To provide religious or moral instruction Child has special needs, other than physical or mental Nontraditional approach to child’s education Other reasons (e.g., finances, travel, family time, etc.) Column Total Table 11.56 Applicable Reason (in thousands of respondents) Most Important Reason (in thousands of respondents) 1,096 1,257 315 984 485 5,458 258 540 55 99 216 1,477 Row Total 1,354 1,797 370 1,083 701 6,935 105. When looking at energy consumption, we are often interested in detecting trends over time and how they correlate among different countries. The information in Table 11.57 shows the average energy use (in units of kg of oil equivalent per capita) in the USA and the joint European Union countries (EU) for the six-year period 2005 to 2010. Do the energy use values in these two areas come from the same distribution? Perform the analysis at the 5% significance level. European Union United States Row Total Year 2010 2009 2008 2007 2006 2005 3,413 3,302 3,505 3,537 3,595 3,613 7,164 7,057 7,488 7,758 7,697 7,847 Column Total 45,011 20,965 Table 11.57 10,557 10,359 10,993 11,295 11,292 11,460 65,976 106. The Insurance Institute for Highway Safety collects safety information about all types of cars every year, and publishes a report of Top Safety Picks among all cars, makes, and models. Table 11.58 presents the number of Top Safety Picks in six car categories for the two years 2009 and 2013. Analyze the table data to conclude whether the distribution of cars that earned the Top Safety Picks safety award has remained the same between 2009 and 2013. Derive your results at the 5% significance level. Year \ Car Type 2009 2013 Column Total Table 11.58 Small 12 31 43 MidSize 22 30 52 Large Small SUV Mid-Size SUV Large SUV 10 19 29 10 11 21 27 29 56 6 4 10 Row Total 87 124 211 11.5 Comparison of the Chi-Square Tests This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 621 For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 107. Is there a difference between the distribution of community college statistics students and the distribution of university statistics students in what technology they use on their homework? Of some randomly selected community college students, 43 used a computer, 102 used a calculator with built in statistics functions, and 65 used a table from the textbook. Of some randomly selected university students, 28 used a computer, 33 used a calculator with built in statistics functions, and 40 used a table from the textbook. Conduct an appropriate hypothesis test using a 0.05 level of significance. Read the statement and decide whether it is true or false. 108. If df = 2, the chi-square distribution has a shape that reminds us of the exponential. 11.6 Test of a Single Variance Use the following information to answer the next twelve exercises: Suppose an airline claims that its flights are consistently on time with an average delay of at most 15 minutes. It claims that the average delay is so consistent that the variance is no more than 150 minutes. Doubting the consistency part of the claim, a disgruntled traveler calculates the delays for his next 25 flights. The average delay for those 25 flights is 22 minutes with a standard deviation of 15 minutes. 109. Is the traveler disputing the claim about the average or about the variance? 110. A sample standard deviation of 15 minutes is the same as a sample variance of __________ minutes. 111. Is this a right-tailed, left-tailed, or two-tailed test? 112. H0: __________ 113. df = ________ 114. chi-square test statistic = ________ 115. p-value = ________ 116. Graph the situation. Label and scale the horizontal axis. Mark the mean and test statistic. Shade the p-value. 117. Let α = 0.05 Decision: ________ Conclusion (write out in a complete sentence.): ________ 118. How did you know to test the variance instead of the mean? 119. If an additional test were done on the claim of the average de
lay, which distribution would you use? 120. If an additional test were done on the claim of the average delay, but 45 flights were surveyed, which distribution would you use? For each word problem, use a solution sheet to solve the hypothesis test problem. Go to Appendix E for the chi-square solution sheet. Round expected frequency to two decimal places. 121. A plant manager is concerned her equipment may need recalibrating. It seems that the actual weight of the 15 oz. cereal boxes it fills has been fluctuating. The standard deviation should be at most 0.5 oz. In order to determine if the machine needs to be recalibrated, 84 randomly selected boxes of cereal from the next day’s production were weighed. The standard deviation of the 84 boxes was 0.54. Does the machine need to be recalibrated? 122. Consumers may be interested in whether the cost of a particular calculator varies from store to store. Based on surveying 43 stores, which yielded a sample mean of $84 and a sample standard deviation of $12, test the claim that the standard deviation is greater than $15. 123. Isabella, an accomplished Bay to Breakers runner, claims that the standard deviation for her time to run the 7.5 mile race is at most three minutes. To test her claim, Rupinder looks up five of her race times. They are 55 minutes, 61 minutes, 58 minutes, 63 minutes, and 57 minutes. 124. Airline companies are interested in the consistency of the number of babies on each flight, so that they have adequate safety equipment. They are also interested in the variation of the number of babies. Suppose that an airline executive believes the average number of babies on flights is six with a variance of nine at most. The airline conducts a survey. The results of the 18 flights surveyed give a sample average of 6.4 with a sample standard deviation of 3.9. Conduct a hypothesis test of the airline executive’s belief. 125. The number of births per woman in China is 1.6 down from 5.91 in 1966. This fertility rate has been attributed to the law passed in 1979 restricting births to one per woman. Suppose that a group of students studied whether or not the standard deviation of births per woman was greater than 0.75. They asked 50 women across China the number of births they had had. The results are shown in Table 11.59. Does the students’ survey indicate that the standard deviation is greater than 0.75? 622 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION # of births Frequency 0 1 2 3 Table 11.59 5 30 10 5 126. According to an avid aquarist, the average number of fish in a 20-gallon tank is 10, with a standard deviation of two. His friend, also an aquarist, does not believe that the standard deviation is two. She counts the number of fish in 15 other 20-gallon tanks. Based on the results that follow, do you think that the standard deviation is different from two? Data: 11; 10; 9; 10; 10; 11; 11; 10; 12; 9; 7; 9; 11; 10; 11 127. The manager of "Frenchies" is concerned that patrons are not consistently receiving the same amount of French fries with each order. The chef claims that the standard deviation for a ten-ounce order of fries is at most 1.5 oz., but the manager thinks that it may be higher. He randomly weighs 49 orders of fries, which yields a mean of 11 oz. and a standard deviation of two oz. 128. You want to buy a specific computer. A sales representative of the manufacturer claims that retail stores sell this computer at an average price of $1,249 with a very narrow standard deviation of $25. You find a website that has a price comparison for the same computer at a series of stores as follows: $1,299; $1,229.99; $1,193.08; $1,279; $1,224.95; $1,229.99; $1,269.95; $1,249. Can you argue that pricing has a larger standard deviation than claimed by the manufacturer? Use the 5% significance level. As a potential buyer, what would be the practical conclusion from your analysis? 129. A company packages apples by weight. One of the weight grades is Class A apples. Class A apples have a mean weight of 150 g, and there is a maximum allowed weight tolerance of 5% above or below the mean for apples in the same consumer package. A batch of apples is selected to be included in a Class A apple package. Given the following apple weights of the batch, does the fruit comply with the Class A grade weight tolerance requirements. Conduct an appropriate hypothesis test. (a) at the 5% significance level (b) at the 1% significance level Weights in selected apple batch (in grams): 158; 167; 149; 169; 164; 139; 154; 150; 157; 171; 152; 161; 141; 166; 172; BRINGING IT TOGETHER: HOMEWORK 130. a. Explain why a goodness-of-fit test and a test of independence are generally right-tailed tests. b. If you did a left-tailed test, what would you be testing? REFERENCES 11.1 Facts About the Chi-Square Distribution Data from Parade Magazine. “HIV/AIDS Epidemiology Santa Clara County.”Santa Clara County Public Health Department, May 2011. 11.2 Goodness-of-Fit Test Data from the U.S. Census Bureau Data from the College Board. Available online at http://www.collegeboard.com. Data from the U.S. Census Bureau, Current Population Reports. Ma, Y., E.R. Bertone, E.J. Stanek III, G.W. Reed, J.R. Hebert, N.L. Cohen, P.A. Merriam, I.S. Ockene, “Association between Eating Patterns and Obesity in a Free-living US Adult Population.” American Journal of Epidemiology volume 158, no. 1, pages 85-92. Ogden, Cynthia L., Margaret D. Carroll, Brian K. Kit, Katherine M. Flegal, “Prevalence of Obesity in the United States, 2009–2010.” NCHS Data Brief no. 82, January 2012. Available online at http://www.cdc.gov/nchs/data/databriefs/db82.pdf (accessed May 24, 2013). This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 623 Stevens, Barbara J., “Multi-family and Commercial Solid Waste and Recycling Survey.” Arlington Count, VA. Available online at http://www.arlingtonva.us/departments/EnvironmentalServices/SW/file84429.pdf (accessed May 24,2013). 11.3 Test of Independence DiCamilo, Mark, Mervin Field, “Most Californians See a Direct Linkage between Obesity and Sugary Sodas. Two in Three Voters Support Taxing Sugar-Sweetened Beverages If Proceeds are Tied to Improving School Nutrition and Physical Activity Programs.” The Field Poll, released Feb. 14, 2013. Available online at http://field.com/fieldpollonline/subscribers/ Rls2436.pdf (accessed May 24, 2013). Harris Interactive, “Favorite Flavor of Ice Cream.” Available online at http://www.statisticbrain.com/favorite-flavor-of-icecream (accessed May 24, 2013) “Youngest Online Entrepreneurs List.” Available online at http://www.statisticbrain.com/youngest-online-entrepreneur-list (accessed May 24, 2013). 11.4 Test for Homogeneity Data from the Insurance Institute for Highway Safety, 2013. Available online at www.iihs.org/iihs/ratings (accessed May 24, 2013). “Energy use (kg of oil equivalent per capita).” The World Bank, 2013. Available online at http://data.worldbank.org/ indicator/EG.USE.PCAP.KG.OE/countries (accessed May 24, 2013). “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubsearch/ pubsinfo.asp?pubid=2009030 (accessed May 24, 2013). “Parent and Family Involvement Survey of 2007 National Household Education Survey Program (NHES),” U.S. Department of Education, National Center for Education Statistics. Available online at http://nces.ed.gov/pubs2009/ 2009030_sup.pdf (accessed May 24, 2013). 11.6 Test of a Single Variance “AppleInsider Price Guides.” Apple Insider, 2013. Available online at http://appleinsider.com/mac_price_guide (accessed May 14, 2013). Data from the World Bank, June 5, 2012. SOLUTIONS 1 mean = 25 and standard deviation = 7.0711 3 when the number of degrees of freedom is greater than 90 5 df = 2 7 a goodness-of-fit test 9 3 11 2.04 13 We decline to reject the null hypothesis. There is not enough evidence to suggest that the observed test scores are significantly different from the expected test scores. 15 H0: the distribution of AIDS cases follows the ethnicities of the general population of Santa Clara County. 17 right-tailed 19 88,621 21 Graph: Check student’s solution. Decision: Reject the null hypothesis. Reason for the Decision: p-value < alpha Conclusion (write out in complete sentences): The make-up of AIDS cases does not fit the ethnicities of the general population of Santa Clara County. 624 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 23 a test of independence 25 a test of independence 27 8 29 6.6 31 0.0435 33 Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans 12,831 8,378 4,932 1,406 800 10,680 4,715 2,305 White Totals 7,650 9,877 6,062 3,970 41,490 35,065 15,273 8,622 19,969 26,078 27,559 10,0450 9,886 6,514 1,671 759 18,830 2,745 3,062 1,419 788 8,014 1-10 11-20 21-30 31+ Totals Table 11.60 35 White 11383.01 9620.27 4190.23 2365.49 Smoking Level Per Day African American Native Hawaiian Latino Japanese Americans 7777.57 6573.16 2863.02 1616.25 3310.11 2797.52 1218.49 687.87 8248.02 10771.29 6970.76 9103.29 3036.20 3965.05 1714.01 2238.37 1-10 11-20 21-30 31+ Table 11.61 37 10,301.8 39 right 41 a. Reject the null hypothesis. b. p-value < alpha c. There is sufficient evidence to conclude that smoking level is dependent on ethnic group. 43 test for homogeneity 45 test for homogeneity 47 All values in the table must be greater than or equal to five. 49 3 51 0.00005 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 625 53 a goodness-of-fit test 55 a test for independence 57 Answers will vary. Sample answer: Tests of independence and tests for homogeneity both calculate the test statistic the same way ∑ (i j) (O - E)2 E . In
addition, all values must be greater than or equal to five. 59 a test of a single variance 61 a left-tailed test 63 H0: σ2 = 0.812; Ha: σ2 > 0.812 65 a test of a single variance 67 0.0542 69 true 71 false 73 Marital Status Percent Expected Frequency never married married widowed 31.3 56.1 2.5 divorced/separated 10.1 Table 11.62 125.2 224.4 10 40.4 a. The data fits the distribution. b. The data does not fit the distribution. c. 3 d. chi-square distribution with df = 3 e. 19.27 f. 0.0002 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Reject null iii. Reason for decision: p-value < alpha iv. Conclusion: Data does not fit the distribution. 75 a. H0: The local results follow the distribution of the U.S. AP examinee population b. Ha: The local results do not follow the distribution of the U.S. AP examinee population c. df = 5 d. chi-square distribution with df = 5 e. chi-square test statistic = 13.4 626 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION f. p-value = 0.0199 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Reject null when a = 0.05 iii. Reason for Decision: p-value < alpha iv. Conclusion: Local data do not fit the AP Examinee Distribution. v. Decision: Do not reject null when a = 0.01 vi. Conclusion: There is insufficient evidence to conclude that local data do not follow the distribution of the U.S. AP examinee distribution. 77 a. H0: The actual college majors of graduating females fit the distribution of their expected majors b. Ha: The actual college majors of graduating females do not fit the distribution of their expected majors c. df = 10 d. chi-square distribution with df = 10 e. test statistic = 11.48 f. p-value = 0.3211 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Do not reject null when a = 0.05 and a = 0.01 iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the distribution of actual college majors of graduating females fits the distribution of their expected majors. 79 true 81 true 83 false 85 a. H0: Surveyed obese fit the distribution of expected obese b. Ha: Surveyed obese do not fit the distribution of expected obese c. df = 4 d. chi-square distribution with df = 4 e. test statistic = 54.01 f. p-value = 0 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5% level of significance, from the data, there is sufficient evidence to conclude that the surveyed obese do not fit the distribution of expected obese. 87 a. H0: Car size is independent of family size. b. Ha: Car size is dependent on family size. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 627 c. df = 9 d. chi-square distribution with df = 9 e. test statistic = 15.8284 f. p-value = 0.0706 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that car size and family size are dependent. 89 a. H0: Honeymoon locations are independent of bride’s age. b. Ha: Honeymoon locations are dependent on bride’s age. c. df = 9 d. chi-square distribution with df = 9 e. test statistic = 15.7027 f. p-value = 0.0734 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that honeymoon location and bride age are dependent. 91 a. H0: The types of fries sold are independent of the location. b. Ha: The types of fries sold are dependent on the location. c. df = 6 d. chi-square distribution with df = 6 e. test statistic =18.8369 f. p-value = 0.0044 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5% significance level, There is sufficient evidence that types of fries and location are dependent. 93 a. H0: Salary is independent of level of education. b. Ha: Salary is dependent on level of education. c. df = 12 d. chi-square distribution with df = 12 e. test statistic = 255.7704 628 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION f. p-value = 0 g. Check student’s solution. h. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p-value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that salary and level of education are dependent. 95 true 97 true 99 a. H0: Age is independent of the youngest online entrepreneurs’ net worth. b. Ha: Age is dependent on the net worth of the youngest online entrepreneurs. c. df = 2 d. chi-square distribution with df = 2 e. test statistic = 1.76 f. p-value 0.4144 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that age and net worth for the youngest online entrepreneurs are dependent. 101 a. H0: The distribution for personality types is the same for both majors b. Ha: The distribution for personality types is not the same for both majors c. df = 4 d. chi-square with df = 4 e. test statistic = 3.01 f. p-value = 0.5568 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the distribution of personality types is different for business and social science majors. 103 a. H0: The distribution for fish caught is the same in Green Valley Lake and in Echo Lake. b. Ha: The distribution for fish caught is not the same in Green Valley Lake and in Echo Lake. c. 3 d. chi-square with df = 3 e. 11.75 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 629 f. p-value = 0.0083 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is evidence to conclude that the distribution of fish caught is different in Green Valley Lake and in Echo Lake 105 a. H0: The distribution of average energy use in the USA is the same as in Europe between 2005 and 2010. b. Ha: The distribution of average energy use in the USA is not the same as in Europe between 2005 and 2010. c. df = 4 d. chi-square with df = 4 e. test statistic = 2.7434 f. p-value = 0.7395 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the average energy use values in the US and EU are not derived from different distributions for the period from 2005 to 2010. 107 a. H0: The distribution for technology use is the same for community college students and university students. b. Ha: The distribution for technology use is not the same for community college students and university students. c. 2 d. chi-square with df = 2 e. 7.05 f. p-value = 0.0294 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the distribution of technology use for statistics homework is not the same for statistics students at community colleges and at universities. 110 225 112 H0: σ2 ≤ 150 114 36 116 Check student’s solution. 118 The claim is that the variance is no more than 150 minutes. 120 a Student's t- or normal distribution 122 630 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION a. H0: σ = 15 b. Ha: σ > 15 c. df = 42 d. chi-square with df = 42 e. test statistic = 26.88 f. p-value = 0.9663 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Do not reject null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is insufficient evidence to conclude that the standard deviation is greater than 15. 124 a. H0: σ ≤ 3 b. Ha: σ > 3 c. df = 17 d. chi-square distribution with df = 17 e. test statistic = 28.73 f. p-value = 0.0371 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the standard deviation is greater than three. 126 a. H0: σ = 2 b. Ha: σ ≠ 2 c. df = 14 d. chi-square distiribution with df = 14 e. chi-square test statistic = 5.2094 f. p-value = 0.0346 g. Check student’s solution. h. i. Alpha = 0.05 ii. Decision: Reject the null hypothesis iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the standard deviation is different than 2. 128 The sample standard deviation is $34.29. H0 : σ2 = 252 Ha : σ2 > 252 df = n – 1 = 7. test statistic: x2 = x7 2 = (n – 1)s2 252 ⎠ = 1 – P⎛ ⎞ 2 > 13.169 ⎝x7 (8 – 1)(34.29)2 252 ⎞ 2 ≤ 13.169 ⎠ = 0.0681 = p-value: P⎛ ⎝x7 = 13.169 ; Alpha: 0.05 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION 631 Decision: Do not reject the null hypothesis. Reason for decision: p-value > alpha Conclusion: At the 5% level, there is insufficient evidence to conclude that the variance is more than 625. 130 a. The test statistic is always positive and if the expected and observed values are not close together, the test statistic is large and the null hypothesis will be rejected. b. Testing to see if the data fits the distribution “too well” or is too perfect. 632 CHAPTER 11 \| THE CHI-SQUARE DISTRIBUTION This content is availa
ble for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 633 12 \| LINEAR REGRESSION AND CORRELATION Figure 12.1 Linear regression and correlation can help you determine if an auto mechanic’s salary is related to his work experience. (credit: Joshua Rothhaas) Introduction Chapter Objectives By the end of this chapter, the student should be able to: • Discuss basic ideas of linear regression and correlation. • Create and interpret a line of best fit. • Calculate and interpret the correlation coefficient. • Calculate and interpret outliers. Professionals often want to know how two or more numeric variables are related. For example, is there a relationship between the grade on the second math exam a student takes and the grade on the final exam? If there is a relationship, what is the relationship and how strong is it? In another example, your income may be determined by your education, your profession, your years of experience, and your ability. The amount you pay a repair person for labor is often determined by an initial amount plus an hourly fee. 634 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION The type of data described in the examples is bivariate data — "bi" for two variables. In reality, statisticians use multivariate data, meaning many variables. In this chapter, you will be studying the simplest form of regression, "linear regression" with one independent variable (x). This involves data that fits a line in two dimensions. You will also study correlation which measures how strong the relationship is. 12.1 \| Linear Equations Linear regression for two variables is based on a linear equation with one independent variable. The equation has the form: y = a + bx where a and b are constant numbers. The variable x is the independent variable, and y is the dependent variable. Typically, you choose a value to substitute for the independent variable and then solve for the dependent variable. Example 12.1 The following examples are linear equations. y = 3 + 2x y = –0.01 + 1.2x 12.1 Is the following an example of a linear equation? y = –0.125 – 3.5x The graph of a linear equation of the form y = a + bx is a straight line. Any line that is not vertical can be described by this equation. Example 12.2 Graph the equation y = –1 + 2x. Figure 12.2 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 635 12.2 Is the following an example of a linear equation? Why or why not? Figure 12.3 Example 12.3 Aaron's Word Processing Service (AWPS) does word processing. The rate for services is $32 per hour plus a $31.50 one-time charge. The total cost to a customer depends on the number of hours it takes to complete the job. Find the equation that expresses the total cost in terms of the number of hours required to complete the job. Solution 12.3 Let x = the number of hours it takes to get the job done. Let y = the total cost to the customer. The $31.50 is a fixed cost. If it takes x hours to complete the job, then (32)(x) is the cost of the word processing only. The total cost is: y = 31.50 + 32x 12.3 Emma’s Extreme Sports hires hang-gliding instructors and pays them a fee of $50 per class as well as $20 per student in the class. The total cost Emma pays depends on the number of students in a class. Find the equation that expresses the total cost in terms of the number of students in a class. Slope and Y-Intercept of a Linear Equation For the linear equation y = a + bx, b = slope and a = y-intercept. From algebra recall that the slope is a number that describes the steepness of a line, and the y-intercept is the y coordinate of the point (0, a) where the line crosses the y-axis. Figure 12.4 Three possible graphs of y = a + bx. (a) If b > 0, the line slopes upward to the right. (b) If b = 0, the line is horizontal. (c) If b < 0, the line slopes downward to the right. 636 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Example 12.4 Svetlana tutors to make extra money for college. For each tutoring session, she charges a one-time fee of $25 plus $15 per hour of tutoring. A linear equation that expresses the total amount of money Svetlana earns for each session she tutors is y = 25 + 15x. What are the independent and dependent variables? What is the y-intercept and what is the slope? Interpret them using complete sentences. Solution 12.4 The independent variable (x) is the number of hours Svetlana tutors each session. The dependent variable (y) is the amount, in dollars, Svetlana earns for each session. The y-intercept is 25 (a = 25). At the start of the tutoring session, Svetlana charges a one-time fee of $25 (this is when x = 0). The slope is 15 (b = 15). For each session, Svetlana earns $15 for each hour she tutors. 12.4 Ethan repairs household appliances like dishwashers and refrigerators. For each visit, he charges $25 plus $20 per hour of work. A linear equation that expresses the total amount of money Ethan earns per visit is y = 25 + 20x. What are the independent and dependent variables? What is the y-intercept and what is the slope? Interpret them using complete sentences. 12.2 \| Scatter Plots Before we take up the discussion of linear regression and correlation, we need to examine a way to display the relation between two variables x and y. The most common and easiest way is a scatter plot. The following example illustrates a scatter plot. Example 12.5 In Europe and Asia, m-commerce is popular. M-commerce users have special mobile phones that work like electronic wallets as well as provide phone and Internet services. Users can do everything from paying for parking to buying a TV set or soda from a machine to banking to checking sports scores on the Internet. For the years 2000 through 2004, was there a relationship between the year and the number of m-commerce users? Construct a scatter plot. Let x = the year and let y = the number of m-commerce users, in millions. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 637 x (year) y (# of users) 0.5 20.0 33.0 47.0 2000 2002 2003 2004 Table 12.1 (a) Table showing the number of mcommerce users (in millions) by year. Figure 12.5 (b) Scatter plot showing the number of m-commerce users (in millions) by year. To create a scatter plot: 1. Enter your X data into list L1 and your Y data into list L2. 2. Press 2nd STATPLOT ENTER to use Plot 1. On the input screen for PLOT 1, highlight On and press ENTER. (Make sure the other plots are OFF.) 3. For TYPE: highlight the very first icon, which is the scatter plot, and press ENTER. 4. For Xlist:, enter L1 ENTER and for Ylist: L2 ENTER. 5. For Mark: it does not matter which symbol you highlight, but the square is the easiest to see. Press ENTER. 6. Make sure there are no other equations that could be plotted. Press Y = and clear any equations out. 7. Press the ZOOM key and then the number 9 (for menu item "ZoomStat") ; the calculator will fit the window to the data. You can press WINDOW to see the scaling of the axes. 638 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 12.5 Amelia plays basketball for her high school. She wants to improve to play at the college level. She notices that the number of points she scores in a game goes up in response to the number of hours she practices her jump shot each week. She records the following data: X (hours practicing jump shot) Y (points scored in a game) 5 7 9 10 11 12 Table 12.2 15 22 28 31 33 36 Construct a scatter plot and state if what Amelia thinks appears to be true. A scatter plot shows the direction of a relationship between the variables. A clear direction happens when there is either: • High values of one variable occurring with high values of the other variable or low values of one variable occurring with low values of the other variable. • High values of one variable occurring with low values of the other variable. You can determine the strength of the relationship by looking at the scatter plot and seeing how close the points are to a line, a power function, an exponential function, or to some other type of function. For a linear relationship there is an exception. Consider a scatter plot where all the points fall on a horizontal line providing a "perfect fit." The horizontal line would in fact show no relationship. When you look at a scatterplot, you want to notice the overall pattern and any deviations from the pattern. The following scatterplot examples illustrate these concepts. Figure 12.6 Figure 12.7 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 639 Figure 12.8 In this chapter, we are interested in scatter plots that show a linear pattern. Linear patterns are quite common. The linear relationship is strong if the points are close to a straight line, except in the case of a horizontal line where there is no relationship. If we think that the points show a linear relationship, we would like to draw a line on the scatter plot. This line can be calculated through a process called linear regression. However, we only calculate a regression line if one of the variables helps to explain or predict the other variable. If x is the independent variable and y the dependent variable, then we can use a regression line to predict y for a given value of x 12.3 \| The Regression Equation Data rarely fit a straight line exactly. Usually, you must be satisfied with rough predictions. Typically, you have a set of data whose scatter plot appears to "fit" a straight line. This is called a Line of Best Fit or Least-Squares Line. If you know a person's pinky (smallest) finger length, do you think you could predict that person's height?
Collect data from your class (pinky finger length, in inches). The independent variable, x, is pinky finger length and the dependent variable, y, is height. For each set of data, plot the points on graph paper. Make your graph big enough and use a ruler. Then "by eye" draw a line that appears to "fit" the data. For your line, pick two convenient points and use them to find the slope of the line. Find the y-intercept of the line by extending your line so it crosses the y-axis. Using the slopes and the y-intercepts, write your equation of "best fit." Do you think everyone will have the same equation? Why or why not? According to your equation, what is the predicted height for a pinky length of 2.5 inches? Example 12.6 A random sample of 11 statistics students produced the following data, where x is the third exam score out of 80, and y is the final exam score out of 200. Can you predict the final exam score of a random student if you know the third exam score? 640 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION x (third exam score) y (final exam score) 65 67 71 71 66 75 67 70 71 69 69 175 133 185 163 126 198 153 163 159 151 159 Table 12.3 (a) Table showing the scores on the final exam based on scores from the third exam. Figure 12.9 (b) Scatter plot showing the scores on the final exam based on scores from the third exam. 12.6 SCUBA divers have maximum dive times they cannot exceed when going to different depths. The data in Table 12.4 show different depths with the maximum dive times in minutes. Use your calculator to find the least squares regression line and predict the maximum dive time for 110 feet. X (depth in feet) Y (maximum dive time) 50 60 70 80 90 100 Table 12.4 80 55 45 35 25 22 The third exam score, x, is the independent variable and the final exam score, y, is the dependent variable. We will plot a regression line that best "fits" the data. If each of you were to fit a line "by eye," you would draw different lines. We can use what is called a least-squares regression line to obtain the best fit line. Consider the following diagram. Each point of data is of the the form (x, y) and each point ofthe line of best fit using leastsquares linear regression has the form (x, ŷ). The ŷ is read "y hat" and is the estimated value of y. It is the value of y obtained using the regression line. It is not generally equal to y from data. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 641 Figure 12.10 The term y0 – ŷ0 = ε0 is called the "error" or residual. It is not an error in the sense of a mistake. The absolute value of a residual measures the vertical distance between the actual value of y and the estimated value of y. In other words, it measures the vertical distance between the actual data point and the predicted point on the line. If the observed data point lies above the line, the residual is positive, and the line underestimates the actual data value for y. If the observed data point lies below the line, the residual is negative, and the line overestimates that actual data value for y. In the diagram in Figure 12.10, y0 – ŷ0 = ε0 is the residual for the point shown. Here the point lies above the line and the residual is positive. ε = the Greek letter epsilon For each data point, you can calculate the residuals or errors, yi - ŷi = εi for i = 1, 2, 3, ..., 11. Each \|ε\| is a vertical distance. For the example about the third exam scores and the final exam scores for the 11 statistics students, there are 11 data points. Therefore, there are 11 ε values. If yousquare each ε and add, you get 11 (ε1)2 + (ε2)2 + ... + (ε11)2 = Σ i = 1 ε2 This is called the Sum of Squared Errors (SSE). Using calculus, you can determine the values of a and b that make the SSE a minimum. When you make the SSE a minimum, you have determined the points that are on the line of best fit. It turns out that the line of best fit has the equation: where a = y¯ − b x¯ and b = Σ(x − x¯ )(y − y¯ ) Σ(x − x¯ ) 2 . y^ = a + bx The sample means of the x values and the y values are x¯ and y¯ , respectively. The best fit line always passes through the point ( x¯ , y¯ ) . The slope b can be written as b = r⎛ ⎝ s y s x ⎞ ⎠ where sy = the standard deviation of the y values and sx = the standard deviation of the x values. r is the correlation coefficient, which is discussed in the next section. Least Squares Criteria for Best Fit The process of fitting the best-fit line is called linear regression. The idea behind finding the best-fit line is based on the assumption that the data are scattered about a straight line. The criteria for the best fit line is that the sum of the squared errors (SSE) is minimized, that is, made as small as possible. Any other line you might choose would have a higher SSE than the best fit line. This best fit line is called the least-squares regression line . 642 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION NOTE Computer spreadsheets, statistical software, and many calculators can quickly calculate the best-fit line and create the graphs. The calculations tend to be tedious if done by hand. Instructions to use the TI-83, TI-83+, and TI-84+ calculators to find the best-fit line and create a scatterplot are shown at the end of this section. THIRD EXAM vs FINAL EXAM EXAMPLE: The graph of the line of best fit for the third-exam/final-exam example is as follows: Figure 12.11 The least squares regression line (best-fit line) for the third-exam/final-exam example has the equation: y^ = − 173.51 + 4.83x REMINDER Remember, it is always important to plot a scatter diagram first. If the scatter plot indicates that there is a linear relationship between the variables, then it is reasonable to use a best fit line to make predictions for y given x within the domain of x-values in the sample data, but not necessarily for x-values outside that domain. You could use the line to predict the final exam score for a student who earned a grade of 73 on the third exam. You should NOT use the line to predict the final exam score for a student who earned a grade of 50 on the third exam, because 50 is not within the domain of the x-values in the sample data, which are between 65 and 75. UNDERSTANDING SLOPE The slope of the line, b, describes how changes in the variables are related. It is important to interpret the slope of the line in the context of the situation represented by the data. You should be able to write a sentence interpreting the slope in plain English. INTERPRETATION OF THE SLOPE: The slope of the best-fit line tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average. THIRD EXAM vs FINAL EXAM EXAMPLE Slope: The slope of the line is b = 4.83. Interpretation: For a one-point increase in the score on the third exam, the final exam score increases by 4.83 points, on average. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 643 Using the Linear Regression T Test: LinRegTTest 1. In the STAT list editor, enter the X data in list L1 and the Y data in list L2, paired so that the corresponding (x,y) values are next to each other in the lists. (If a particular pair of values is repeated, enter it as many times as it appears in the data.) 2. On the STAT TESTS menu, scroll down with the cursor to select the LinRegTTest. (Be careful to select LinRegTTest, as some calculators may also have a different item called LinRegTInt.) 3. On the LinRegTTest input screen enter: Xlist: L1 ; Ylist: L2 ; Freq: 1 4. On the next line, at the prompt β or ρ, highlight "≠ 0" and press ENTER 5. Leave the line for "RegEq:" blank 6. Highlight Calculate and press ENTER. Figure 12.12 The output screen contains a lot of information. For now we will focus on a few items from the output, and will return later to the other items. The second line says y = a + bx. Scroll down to find the values a = –173.513, and b = 4.8273; the equation of the best fit line is ŷ = –173.51 + 4.83x The two items at the bottom are r2 = 0.43969 and r = 0.663. For now, just note where to find these values; we will discuss them in the next two sections. Graphing the Scatterplot and Regression Line 1. We are assuming your X data is already entered in list L1 and your Y data is in list L2 2. Press 2nd STATPLOT ENTER to use Plot 1 3. On the input screen for PLOT 1, highlight On, and press ENTER 4. For TYPE: highlight the very first icon which is the scatterplot and press ENTER 5. Indicate Xlist: L1 and Ylist: L2 6. For Mark: it does not matter which symbol you highlight. 7. Press the ZOOM key and then the number 9 (for menu item "ZoomStat") ; the calculator will fit the window to the data 8. To graph the best-fit line, press the "Y=" key and type the equation –173.5 + 4.83X into equation Y1. (The X key is immediately left of the STAT key). Press ZOOM 9 again to graph it. 9. Optional: If you want to change the viewing window, press the WINDOW key. Enter your desired window using Xmin, Xmax, Ymin, Ymax 644 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION NOTE Another way to graph the line after you create a scatter plot is to use LinRegTTest. 1. Make sure you have done the scatter plot. Check it on your screen. 2. Go to LinRegTTest and enter the lists. 3. At RegEq: press VARS and arrow over to Y-VARS. Press 1 for 1:Function. Press 1 for 1:Y1. Then arrow down to Calculate and do the calculation for the line of best fit. 4. Press Y = (you will see the regression equation). 5. Press GRAPH. The line will be drawn." The Correlation Coefficient r Besides looking at the scatter plot and seeing that a line seems reasonable, how can you tell if the line is a good predictor? Use the correlation coefficient as another indicator (besides the scatterplot) of the strength of the relationship between x and
y. The correlation coefficient, r, developed by Karl Pearson in the early 1900s, is numerical and provides a measure of strength and direction of the linear association between the independent variable x and the dependent variable y. The correlation coefficient is calculated as r = nΣ(xy) − (Σx)(Σy) ⎡ ⎡ ⎣nΣx2 − (Σx)2⎤ ⎣nΣy2 − (Σy)2⎤ ⎦ ⎦ where n = the number of data points. If you suspect a linear relationship between x and y, then r can measure how strong the linear relationship is. What the VALUE of r tells us: • The value of r is always between –1 and +1: –1 ≤ r ≤ 1. • The size of the correlation r indicates the strength of the linear relationship between x and y. Values of r close to –1 or to +1 indicate a stronger linear relationship between x and y. • • If r = 0 there is absolutely no linear relationship between x and y (no linear correlation). If r = 1, there is perfect positive correlation. If r = –1, there is perfect negativecorrelation. In both these cases, all of the original data points lie on a straight line. Of course,in the real world, this will not generally happen. What the SIGN of r tells us • A positive value of r means that when x increases, y tends to increase and when x decreases, y tends to decrease (positive correlation). • A negative value of r means that when x increases, y tends to decrease and when x decreases, y tends to increase (negative correlation). • The sign of r is the same as the sign of the slope, b, of the best-fit line. NOTE Strong correlation does not suggest that x causes y or y causes x. We say "correlation does not imply causation." Figure 12.13 (a) A scatter plot showing data with a positive correlation. 0 < r < 1 (b) A scatter plot showing data with a negative correlation. –1 < r < 0 (c) A scatter plot showing data with zero correlation. r = 0 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 645 The formula for r looks formidable. However, computer spreadsheets, statistical software, and many calculators can quickly calculate r. The correlation coefficient r is the bottom item in the output screens for the LinRegTTest on the TI-83, TI-83+, or TI-84+ calculator (see previous section for instructions). The Coefficient of Determination The variable r2 is called the coefficient of determination and is the square of the correlation coefficient, but is usually stated as a percent, rather than in decimal form. It has an interpretation in the context of the data: • r 2 , when expressed as a percent, represents the percent of variation in the dependent (predicted) variable y that can be explained by variation in the independent (explanatory) variable x using the regression (best-fit) line. • 1 – r 2 , when expressed as a percentage, represents the percent of variation in y that is NOT explained by variation in x using the regression line. This can be seen as the scattering of the observed data points about the regression line. Consider the third exam/final exam example introduced in the previous section • The line of best fit is: ŷ = –173.51 + 4.83x • The correlation coefficient is r = 0.6631 • The coefficient of determination is r2 = 0.66312 = 0.4397 • Interpretation of r2 in the context of this example: • Approximately 44% of the variation (0.4397 is approximately 0.44) in the final-exam grades can be explained by the variation in the grades on the third exam, using the best-fit regression line. • Therefore, approximately 56% of the variation (1 – 0.44 = 0.56) in the final exam grades can NOT be explained by the variation in the grades on the third exam, using the best-fit regression line. (This is seen as the scattering of the points about the line.) 12.4 \| Testing the Significance of the Correlation Coefficient The correlation coefficient, r, tells us about the strength and direction of the linear relationship between x and y. However, the reliability of the linear model also depends on how many observed data points are in the sample. We need to look at both the value of the correlation coefficient r and the sample size n, together. We perform a hypothesis test of the "significance of the correlation coefficient" to decide whether the linear relationship in the sample data is strong enough to use to model the relationship in the population. The sample data are used to compute r, the correlation coefficient for the sample. If we had data for the entire population, we could find the population correlation coefficient. But because we have only have sample data, we cannot calculate the population correlation coefficient. The sample correlation coefficient, r, is our estimate of the unknown population correlation coefficient. The symbol for the population correlation coefficient is ρ, the Greek letter "rho." ρ = population correlation coefficient (unknown) r = sample correlation coefficient (known; calculated from sample data) The hypothesis test lets us decide whether the value of the population correlation coefficient ρ is "close to zero" or "significantly different from zero". We decide this based on the sample correlation coefficient r and the sample size n. If the test concludes that the correlation coefficient is significantly different from zero, we say that the correlation coefficient is "significant." • Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero. • What the conclusion means: There is a significant linear relationship between x and y. We can use the regression line to model the linear relationship between x and y in the population. If the test concludes that the correlation coefficient is not significantly different from zero (it is close to zero), we say that correlation coefficient is "not significant". • Conclusion: "There is insufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero." • What the conclusion means: There is not a significant linear relationship between x and y. Therefore, we CANNOT use the regression line to model a linear relationship between x and y in the population. 646 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION NOTE • • • If r is significant and the scatter plot shows a linear trend, the line can be used to predict the value of y for values of x that are within the domain of observed x values. If r is not significant OR if the scatter plot does not show a linear trend, the line should not be used for prediction. If r is significant and if the scatter plot shows a linear trend, the line may NOT be appropriate or reliable for prediction OUTSIDE the domain of observed x values in the data. PERFORMING THE HYPOTHESIS TEST • Null Hypothesis: H0: ρ = 0 • Alternate Hypothesis: Ha: ρ ≠ 0 WHAT THE HYPOTHESES MEAN IN WORDS: • Null Hypothesis H0: The population correlation coefficient IS NOT significantly different from zero. There IS NOT a significant linear relationship(correlation) between x and y in the population. • Alternate Hypothesis Ha: The population correlation coefficient IS significantly DIFFERENT FROM zero. There IS A SIGNIFICANT LINEAR RELATIONSHIP (correlation) between x and y in the population. DRAWING A CONCLUSION: There are two methods of making the decision. The two methods are equivalent and give the same result. • Method 1: Using the p-value • Method 2: Using a table of critical values In this chapter of this textbook, we will always use a significance level of 5%, α = 0.05 NOTE Using the p-value method, you could choose any appropriate significance level you want; you are not limited to using α = 0.05. But the table of critical values provided in this textbook assumes that we are using a significance level of 5%, α = 0.05. (If we wanted to use a different significance level than 5% with the critical value method, we would need different tables of critical values that are not provided in this textbook.) METHOD 1: Using a p-value to make a decision To calculate the p-value using LinRegTTEST: On the LinRegTTEST input screen, on the line prompt for β or ρ, highlight "≠ 0" The output screen shows the p-value on the line that reads "p =". (Most computer statistical software can calculate the p-value.) If the p-value is less than the significance level (α = 0.05): • Decision: Reject the null hypothesis. • Conclusion: "There is sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is significantly different from zero." If the p-value is NOT less than the significance level (α = 0.05) • Decision: DO NOT REJECT the null hypothesis. • Conclusion: "There is insufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is NOT significantly different from zero." You will use technology to calculate the p-value. The following describes the calculations to compute the test statistics and the p-value: The p-value is calculated using a t-distribution with n - 2 degrees of freedom. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 647 The formula for the test statistic is . The value of the test statistic, t, is shown in the computer or calculator output along with the p-value. The test statistic t has the same sign as the correlation coefficient r. The p-value is the combined area in both tails. An alternative way to calculate the p-value (p) given by LinRegTTest is the command 2*tcdf(abs(t),10^99, n-2) in 2nd DISTR. THIRD-EXAM vs FINAL-EXAM EXAMPLE: p-value method • Consider the third exam/final exam example. • The line of best fit is: ŷ = -173.51 + 4.83x with r = 0.6631 and there are n =
11 data points. • Can the regression line be used for prediction? Given a third exam score (x value), can we use the line to predict the final exam score (predicted y value)? H0: ρ = 0 Ha: ρ ≠ 0 α = 0.05 • The p-value is 0.026 (from LinRegTTest on your calculator or from computer software). • The p-value, 0.026, is less than the significance level of α = 0.05. • Decision: Reject the Null Hypothesis H0 • Conclusion: There is sufficient evidence to conclude that there is a significant linear relationship between the third exam score (x) and the final exam score (y) because the correlation coefficient is significantly different from zero. Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores. METHOD 2: Using a table of Critical Values to make a decision The 95% Critical Values of the Sample Correlation Coefficient Table can be used to give you a good idea of whether the computed value of r is significant or not. Compare r to the appropriate critical value in the table. If r is not between the positive and negative critical values, then the correlation coefficient is significant. If r is significant, then you may want to use the line for prediction. Example 12.7 Suppose you computed r = 0.801 using n = 10 data points.df = n - 2 = 10 - 2 = 8. The critical values associated with df = 8 are -0.632 and + 0.632. If r < negative critical value or r > positive critical value, then r issignificant. Since r = 0.801 and 0.801 > 0.632, r is significant and the line may be usedfor prediction. If you view this example on a number line, it will help you. Figure 12.14 r is not significant between -0.632 and +0.632. r = 0.801 > +0.632. Therefore, r is significant. 12.7 For a given line of best fit, you computed that r = 0.6501 using n = 12 data points and the critical value is 0.576. Can the line be used for prediction? Why or why not? 648 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Example 12.8 Suppose you computed r = –0.624 with 14 data points. df = 14 – 2 = 12. The critical values are –0.532 and 0.532. Since –0.624 < –0.532, r is significant and the line can be used for prediction Figure 12.15 r = –0.624-0.532. Therefore, r is significant. 12.8 For a given line of best fit, you compute that r = 0.5204 using n = 9 data points, and the critical value is 0.666. Can the line be used for prediction? Why or why not? Example 12.9 Suppose you computed r = 0.776 and n = 6. df = 6 – 2 = 4. The critical values are –0.811 and 0.811. Since –0.811 < 0.776 < 0.811, r is not significant, and the line should not be used for prediction. Figure 12.16 -0.811 < r = 0.776 < 0.811. Therefore, r is not significant. 12.9 For a given line of best fit, you compute that r = –0.7204 using n = 8 data points, and the critical value is = 0.707. Can the line be used for prediction? Why or why not? THIRD-EXAM vs FINAL-EXAM EXAMPLE: critical value method Consider the third exam/final exam example. The line of best fit is: ŷ = –173.51+4.83x with r = 0.6631 and there are n = 11 data points. Can the regression line be used for prediction? Given a third-exam score (x value), can we use the line to predict the final exam score (predicted y value)? H0: ρ = 0 Ha: ρ ≠ 0 α = 0.05 • Use the "95% Critical Value" table for r with df = n – 2 = 11 – 2 = 9. • The critical values are –0.602 and +0.602 • Since 0.6631 > 0.602, r is significant. • Decision: Reject the null hypothesis. • Conclusion:There is sufficient evidence to conclude that there is a significant linear relationship between the third exam score (x) and the final exam score (y) because the correlation coefficient is significantly different from zero. Because r is significant and the scatter plot shows a linear trend, the regression line can be used to predict final exam scores. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 649 Example 12.10 Suppose you computed the following correlation coefficients. Using the table at the end of the chapter, determine if r is significant and the line of best fit associated with each r can be used to predict a y value. If it helps, draw a number line. a. b. c. d. r = –0.567 and the sample size, n, is 19. The df = n – 2 = 17. The critical value is –0.456. –0.567 < –0.456 so r is significant. r = 0.708 and the sample size, n, is nine. The df = n – 2 = 7. The critical value is 0.666. 0.708 > 0.666 so r is significant. r = 0.134 and the sample size, n, is 14. The df = 14 – 2 = 12. The critical value is 0.532. 0.134 is between –0.532 and 0.532 so r is not significant. r = 0 and the sample size, n, is five. No matter what the dfs are, r = 0 is between the two critical values so r is not significant. 12.10 For a given line of best fit, you compute that r = 0 using n = 100 data points. Can the line be used for prediction? Why or why not? Assumptions in Testing the Significance of the Correlation Coefficient Testing the significance of the correlation coefficient requires that certain assumptions about the data are satisfied. The premise of this test is that the data are a sample of observed points taken from a larger population. We have not examined the entire population because it is not possible or feasible to do so. We are examining the sample to draw a conclusion about whether the linear relationship that we see between x and y in the sample data provides strong enough evidence so that we can conclude that there is a linear relationship between x and y in the population. The regression line equation that we calculate from the sample data gives the best-fit line for our particular sample. We want to use this best-fit line for the sample as an estimate of the best-fit line for the population. Examining the scatterplot and testing the significance of the correlation coefficient helps us determine if it is appropriate to do this. The assumptions underlying the test of significance are: • There is a linear relationship in the population that models the average value of y for varying values of x. In other words, the expected value of y for each particular value lies on a straight line in the population. (We do not know the equation for the line for the population. Our regression line from the sample is our best estimate of this line in the population.) • The y values for any particular x value are normally distributed about the line. This implies that there are more y values scattered closer to the line than are scattered farther away. Assumption (1) implies that these normal distributions are centered on the line: the means of these normal distributions of y values lie on the line. • The standard deviations of the population y values about the line are equal for each value of x. In other words, each of these normal distributions of y values has the same shape and spread about the line. • The residual errors are mutually independent (no pattern). • The data are produced from a well-designed, random sample or randomized experiment. 650 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Figure 12.17 The y values for each x value are normally distributed about the line with the same standard deviation. For each x value, the mean of the y values lies on the regression line. More y values lie near the line than are scattered further away from the line. 12.5 \| Prediction Recall the third exam/final exam example. We examined the scatterplot and showed that the correlation coefficient is significant. We found the equation of the best-fit line for the final exam grade as a function of the grade on the third-exam. We can now use the least-squares regression line for prediction. Suppose you want to estimate, or predict, the mean final exam score of statistics students who received 73 on the third exam. The exam scores (x-values) range from 65 to 75. Since 73 is between the x-values 65 and 75, substitute x = 73 into the equation. Then: y^ = − 173.51 + 4.83(73) = 179.08 We predict that statistics students who earn a grade of 73 on the third exam will earn a grade of 179.08 on the final exam, on average. Example 12.11 Recall the third exam/final exam example. a. What would you predict the final exam score to be for a student who scored a 66 on the third exam? Solution 12.11 a. 145.27 b. What would you predict the final exam score to be for a student who scored a 90 on the third exam? Solution 12.11 b. The x values in the data are between 65 and 75. Ninety is outside of the domain of the observed x values in the data (independent variable), so you cannot reliably predict the final exam score for this student. (Even though it is possible to enter 90 into the equation for x and calculate a corresponding y value, the y value that you get will not be reliable.) To understand really how unreliable the prediction can be outside of the observed x values observed in the data, make the substitution x = 90 into the equation. y^ = –173.51 + 4.83 ⎛ ⎝90 ⎞ ⎠ = 261.19 The final-exam score is predicted to be 261.19. The largest the final-exam score can be is 200. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 651 NOTE The process of predicting inside of the observed x values observed in the data is called interpolation. The process of predicting outside of the observed x values observed in the data is called extrapolation. 12.11 Data are collected on the relationship between the number of hours per week practicing a musical instrument and scores on a math test. The line of best fit is as follows: ŷ = 72.5 + 2.8x What would you predict the score on a math test would be for a student who practices a musical instrument for five hours a week? 12.6 \| Outliers In some data sets, there are values (observed data points) called outliers. Outliers are observed data points that are far from the l
east squares line. They have large "errors", where the "error" or residual is the vertical distance from the line to the point. Outliers need to be examined closely. Sometimes, for some reason or another, they should not be included in the analysis of the data. It is possible that an outlier is a result of erroneous data. Other times, an outlier may hold valuable information about the population under study and should remain included in the data. The key is to examine carefully what causes a data point to be an outlier. Besides outliers, a sample may contain one or a few points that are called influential points. Influential points are observed data points that are far from the other observed data points in the horizontal direction. These points may have a big effect on the slope of the regression line. To begin to identify an influential point, you can remove it from the data set and see if the slope of the regression line is changed significantly. Computers and many calculators can be used to identify outliers from the data. Computer output for regression analysis will often identify both outliers and influential points so that you can examine them. Identifying Outliers We could guess at outliers by looking at a graph of the scatterplot and best fit-line. However, we would like some guideline as to how far away a point needs to be in order to be considered an outlier. As a rough rule of thumb, we can flag any point that is located further than two standard deviations above or below the best-fit line as an outlier. The standard deviation used is the standard deviation of the residuals or errors. We can do this visually in the scatter plot by drawing an extra pair of lines that are two standard deviations above and below the best-fit line. Any data points that are outside this extra pair of lines are flagged as potential outliers. Or we can do this numerically by calculating each residual and comparing it to twice the standard deviation. On the TI-83, 83+, or 84+, the graphical approach is easier. The graphical procedure is shown first, followed by the numerical calculations. You would generally need to use only one of these methods. Example 12.12 In the third exam/final exam example, you can determine if there is an outlier or not. If there is an outlier, as an exercise, delete it and fit the remaining data to a new line. For this example, the new line ought to fit the remaining data better. This means the SSE should be smaller and the correlation coefficient ought to be closer to 1 or –1. Solution 12.12 652 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Graphical Identification of Outliers With the TI-83, 83+, 84+ graphing calculators, it is easy to identify the outliers graphically and visually. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance were equal to 2s or more, then we would consider the data point to be "too far" from the line of best fit. We need to find and graph the lines that are two standard deviations below and above the regression line. Any points that are outside these two lines are outliers. We will call these lines Y2 and Y3: As we did with the equation of the regression line and the correlation coefficient, we will use technology to calculate this standard deviation for us. Using the LinRegTTest with this data, scroll down through the output screens to find s = 16.412. Line Y2 = –173.5 + 4.83x –2(16.4) and line Y3 = –173.5 + 4.83x + 2(16.4) where ŷ = –173.5 + 4.83x is the line of best fit. Y2 and Y3 have the same slope as the line of best fit. Graph the scatterplot with the best fit line in equation Y1, then enter the two extra lines as Y2 and Y3 in the "Y="equation editor and press ZOOM 9. You will find that the only data point that is not between lines Y2 and Y3 is the point x = 65, y = 175. On the calculator screen it is just barely outside these lines. The outlier is the student who had a grade of 65 on the third exam and 175 on the final exam; this point is further than two standard deviations away from the best-fit line. Sometimes a point is so close to the lines used to flag outliers on the graph that it is difficult to tell if the point is between or outside the lines. On a computer, enlarging the graph may help; on a small calculator screen, zooming in may make the graph clearer. Note that when the graph does not give a clear enough picture, you can use the numerical comparisons to identify outliers. Figure 12.18 12.12 Identify the potential outlier in the scatter plot. The standard deviation of the residuals or errors is approximately 8.6. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 653 Figure 12.19 Numerical Identification of Outliers In Table 12.5, the first two columns are the third-exam and final-exam data. The third column shows the predicted ŷ values calculated from the line of best fit: ŷ = –173.5 + 4.83x. The residuals, or errors, have been calculated in the fourth column of the table: observed y value−predicted y value = y − ŷ. s is the standard deviation of all the y − ŷ = ε values where n = the total number of data points. If each residual is calculated and squared, and the results are added, we get the SSE. The standard deviation of the residuals is calculated from the SSE as: s = SSE n − 2 NOTE We divide by (n – 2) because the regression model involves two estimates. Rather than calculate the value of s ourselves, we can find s using the computer or calculator. For this example, the calculator function LinRegTTest found s = 16.4 as the standard deviation of the residuals 35; –17; 16; –6; –19; 9; 3; –1; –10; –9; –1. x y ŷ y – ŷ 65 175 140 175 – 140 = 35 67 133 150 133 – 150= –17 71 185 169 185 – 169 = 16 71 163 169 163 – 169 = –6 66 126 145 126 – 145 = –19 75 198 189 198 – 189 = 9 67 153 150 153 – 150 = 3 70 163 164 163 – 164 = –1 71 159 169 159 – 169 = –10 Table 12.5 654 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION x y ŷ y – ŷ 69 151 160 151 – 160 = –9 69 159 160 159 – 160 = –1 Table 12.5 We are looking for all data points for which the residual is greater than 2s = 2(16.4) = 32.8 or less than –32.8. Compare these values to the residuals in column four of the table. The only such data point is the student who had a grade of 65 on the third exam and 175 on the final exam; the residual for this student is 35. How does the outlier affect the best fit line? Numerically and graphically, we have identified the point (65, 175) as an outlier. We should re-examine the data for this point to see if there are any problems with the data. If there is an error, we should fix the error if possible, or delete the data. If the data is correct, we would leave it in the data set. For this problem, we will suppose that we examined the data and found that this outlier data was an error. Therefore we will continue on and delete the outlier, so that we can explore how it affects the results, as a learning experience. Compute a new best-fit line and correlation coefficient using the ten remaining points: On the TI-83, TI-83+, TI-84+ calculators, delete the outlier from L1 and L2. Using the LinRegTTest, the new line of best fit and the correlation coefficient are: ŷ = –355.19 + 7.39x and r = 0.9121 The new line with r = 0.9121 is a stronger correlation than the original (r = 0.6631) because r = 0.9121 is closer to one. This means that the new line is a better fit to the ten remaining data values. The line can better predict the final exam score given the third exam score. Numerical Identification of Outliers: Calculating s and Finding Outliers Manually If you do not have the function LinRegTTest, then you can calculate the outlier in the first example by doing the following. First, square each \|y – ŷ\| The squares are 352; 172; 162; 62; 192; 92; 32; 12; 102; 92; 12 Then, add (sum) all the \|y – ŷ\| squared terms using the formula 11 ⎛ Σ i = 1 ⎝\|yi − y^ 2 ⎠ i\|⎞ = 11 Σ i = 1 εi 2 (Recall that yi – ŷi = εi.) = 352 + 172 + 162 + 62 + 192 + 92 + 32 + 12 + 102 + 92 + 12 = 2440 = SSE. The result, SSE is the Sum of Squared Errors. Next, calculate s, the standard deviation of all the y – ŷ = ε values where n = the total number of data points. The calculation is s = SSE n – 2 . For the third exam/final exam problem, s = 2440 11 – 2 = 16.47 . Next, multiply s by 1.9: (1.9)(16.47) = 31.29 31.29 is almost 2 standard deviations away from the mean of the y – ŷ values. If we were to measure the vertical distance from any data point to the corresponding point on the line of best fit and that distance is at least 1.9s, then we would consider the data point to be "too far" from the line of best fit. We call that point a potential outlier. For the example, if any of the \|y – ŷ\| values are at least 31.29, the corresponding (x, y) data point is a potential outlier. For the third exam/final exam problem, all the \|y – ŷ\|'s are less than 31.29 except for the first one which is 35. 35 > 31.29 That is, \|y – ŷ\| ≥ (1.9)(s) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 The point which corresponds to \|y – ŷ\| = 35 is (65, 175). Therefore, the data point (65,175) is a potential outlier. For this example, we will delete it. (Remember, we do not always delete an outlier.) CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 655 NOTE When outliers are deleted, the researcher should either record that data was deleted, and why, or the researcher should provide results both with and without the deleted data. If data is erroneous and the correct values are known (e.g., student one actually scored a 70 instead of a 65), then this correction can be made to the data. The next step is to compute a new best-fit line using the ten remaining points. The new line of best fit and the correlation coefficient are: ŷ = –355.19 + 7.39x and r = 0.912
1 Example 12.13 Using this new line of best fit (based on the remaining ten data points in the third exam/final exam example (http://cnx.org/content/m47117/1.3/##element-22) ), what would a student who receives a 73 on the third exam expect to receive on the final exam? Is this the same as the prediction made using the original line? Solution 12.13 Using the new line of best fit, ŷ = –355.19 + 7.39(73) = 184.28. A student who scored 73 points on the third exam would expect to earn 184 points on the final exam. The original line predicted ŷ = –173.51 + 4.83(73) = 179.08 so the prediction using the new line with the outlier eliminated differs from the original prediction. 12.13 The data points for the graph from the third exam/final exam example (http://cnx.org/content/ m47117/1.3/##element-22) are as follows: (1, 5), (2, 7), (2, 6), (3, 9), (4, 12), (4, 13), (5, 18), (6, 19), (7, 12), and (7, 21). Remove the outlier and recalculate the line of best fit. Find the value of ŷ when x = 10. 656 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Example 12.14 The Consumer Price Index (CPI) measures the average change over time in the prices paid by urban consumers for consumer goods and services. The CPI affects nearly all Americans because of the many ways it is used. One of its biggest uses is as a measure of inflation. By providing information about price changes in the Nation's economy to government, business, and labor, the CPI helps them to make economic decisions. The President, Congress, and the Federal Reserve Board use the CPI's trends to formulate monetary and fiscal policies. In the following table, x is the year and y is the CPI. x y x y 1915 10.1 1969 36.7 1926 17.7 1975 49.3 1935 13.7 1979 72.6 1940 14.7 1980 82.4 1947 24.1 1986 109.6 1952 26.5 1991 130.7 1964 31.0 1999 166.6 Table 12.6 Data a. Draw a scatterplot of the data. b. Calculate the least squares line. Write the equation in the form ŷ = a + bx. c. Draw the line on the scatterplot. d. Find the correlation coefficient. Is it significant? e. What is the average CPI for the year 1990? Solution 12.14 a. See Figure 12.19. b. ŷ = –3204 + 1.662x is the equation of the line of best fit. c. r = 0.8694 d. The number of data points is n = 14. Use the 95% Critical Values of the Sample Correlation Coefficient table at the end of Chapter 12. n – 2 = 12. The corresponding critical value is 0.532. Since 0.8694 > 0.532, r is significant. ŷ = –3204 + 1.662(1990) = 103.4 CPI e. Using the calculator LinRegTTest, we find that s = 25.4 ; graphing the lines Y2 = –3204 + 1.662X – 2(25.4) and Y3 = –3204 + 1.662X + 2(25.4) shows that no data values are outside those lines, identifying no outliers. (Note that the year 1999 was very close to the upper line, but still inside it.) Figure 12.20 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 657 NOTE In the example, notice the pattern of the points compared to the line. Although the correlation coefficient is significant, the pattern in the scatterplot indicates that a curve would be a more appropriate model to use than a line. In this example, a statistician should prefer to use other methods to fit a curve to this data, rather than model the data with the line we found. In addition to doing the calculations, it is always important to look at the scatterplot when deciding whether a linear model is appropriate. If you are interested in seeing more years of data, visit the Bureau of Labor Statistics CPI website ftp://ftp.bls.gov/ pub/special.requests/cpi/cpiai.txt; our data is taken from the column entitled "Annual Avg." (third column from the right). For example you could add more current years of data. Try adding the more recent years: 2004: CPI = 188.9; 2008: CPI = 215.3; 2011: CPI = 224.9. See how it affects the model. (Check: ŷ = –4436 + 2.295x; r = 0.9018. Is r significant? Is the fit better with the addition of the new points?) 12.14 The following table shows economic development measured in per capita income PCINC. Year PCINC Year PCINC 1870 1880 1890 1900 1910 340 499 592 757 927 Table 12.7 1920 1050 1930 1170 1940 1364 1950 1836 1960 2132 a. What are the independent and dependent variables? b. Draw a scatter plot. c. Use regression to find the line of best fit and the correlation coefficient. d. e. Interpret the significance of the correlation coefficient. Is there a linear relationship between the variables? f. Find the coefficient of determination and interpret it. g. What is the slope of the regression equation? What does it mean? h. Use the line of best fit to estimate PCINC for 1900, for 2000. i. Determine if there are any outliers. 95% Critical Values of the Sample Correlation Coefficient Table Degrees of Freedom: n – 2 Critical Values: (+ and –) 1 2 3 Table 12.8 0.997 0.950 0.878 658 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Degrees of Freedom: n – 2 Critical Values: (+ and –) 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 40 50 60 70 80 90 100 Table 12.8 0.811 0.754 0.707 0.666 0.632 0.602 0.576 0.555 0.532 0.514 0.497 0.482 0.468 0.456 0.444 0.433 0.423 0.413 0.404 0.396 0.388 0.381 0.374 0.367 0.361 0.355 0.349 0.304 0.273 0.250 0.232 0.217 0.205 0.195 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 659 12.1 Regression (Distance from School) Class Time: Names: Student Learning Outcomes • The student will calculate and construct the line of best fit between two variables. • The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Use eight members of your class for the sample. Collect bivariate data (distance an individual lives from school, the cost of supplies for the current term). 1. Complete the table. Distance from school Cost of supplies this term Table 12.9 2. Which variable should be the dependent variable and which should be the independent variable? Why? 3. Graph “distance” vs. “cost.” Plot the points on the graph. Label both axes with words. Scale both axes. Figure 12.21 Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to four decimal places. 660 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 1. Calculate the following: a. a = ______ b. b = ______ c. correlation = ______ d. n = ______ e. equation: ŷ = ______ f. Is the correlation significant? Why or why not? (Answer in one to three complete sentences.) 2. Supply an answer for the following senarios: a. For a person who lives eight miles from campus, predict the total cost of supplies this term: b. For a person who lives eighty miles from campus, predict the total cost of supplies this term: 3. Obtain the graph on your calculator or computer. Sketch the regression line. Figure 12.22 Discussion Questions 1. Answer each question in complete sentences. a. Does the line seem to fit the data? Why? b. What does the correlation imply about the relationship between the distance and the cost? 2. Are there any outliers? If so, which point is an outlier? 3. Should the outlier, if it exists, be removed? Why or why not? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 661 12.2 Regression (Textbook Cost) Class Time: Names: Student Learning Outcomes • The student will calculate and construct the line of best fit between two variables. • The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Survey ten textbooks. Collect bivariate data (number of pages in a textbook, the cost of the textbook). 1. Complete the table. Number of pages Cost of textbook Table 12.10 2. Which variable should be the dependent variable and which should be the independent variable? Why? 3. Graph “pages” vs. “cost.” Plot the points on the graph in Analyze the Data. Label both axes with words. Scale both axes. Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to four decimal places. 1. Calculate the following: a. a = ______ b. b = ______ c. correlation = ______ d. n = ______ e. equation: y = ______ f. Is the correlation significant? Why or why not? (Answer in complete sentences.) 2. Supply an answer for the following senarios: a. For a textbook with 400 pages, predict the cost. b. For a textbook with 600 pages, predict the cost. 3. Obtain the graph on your calculator or computer. Sketch the regression line. 662 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Figure 12.23 Discussion Questions 1. Answer each question in complete sentences. a. Does the line seem to fit the data? Why? b. What does the correlation imply about the relationship between the number of pages and the cost? 2. Are there any outliers? If so, which point(s) is an outlier? 3. Should the outlier, if it exists, be removed? Why or why not? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 663 12.3 Regression (Fuel Efficiency) Class Time: Names: Student Learning Outcomes • The student will calculate and construct the line of best fit between two variables. • The student will evaluate the relationship between two variables to determine if that relationship is significant. Collect the Data Use the most recent April issue of Consumer Reports. It will give the total fuel efficiency (in miles per gallon) and weight (in pounds) of new model cars with automatic transmissions. We will use this data to determine the relationship, if any, between the fuel efficiency of a car and its weight. 1. Using your random number generator, randomly select
20 cars from the list and record their weights and fuel efficiency into Table 12.11. Weight Fuel Efficiency Table 12.11 2. Which variable should be the dependent variable and which should be the independent variable? Why? 3. By hand, do a scatterplot of “weight” vs. “fuel efficiency”. Plot the points on graph paper. Label both axes with words. Scale both axes accurately. 664 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Figure 12.24 Analyze the Data Enter your data into your calculator or computer. Write the linear equation, rounding to 4 decimal places. 1. Calculate the following: a. a = ______ b. b = ______ c. correlation = ______ d. n = ______ e. equation: ŷ = ______ 2. Obtain the graph of the regression line on your calculator. Sketch the regression line on the same axes as your scatter plot. Discussion Questions 1. Is the correlation significant? Explain how you determined this in complete sentences. 2. 3. Is the relationship a positive one or a negative one? Explain how you can tell and what this means in terms of weight and fuel efficiency. In one or two complete sentences, what is the practical interpretation of the slope of the least squares line in terms of fuel efficiency and weight? 4. For a car that weighs 4,000 pounds, predict its fuel efficiency. Include units. 5. Can we predict the fuel efficiency of a car that weighs 10,000 pounds using the least squares line? Explain why or why not. 6. Answer each question in complete sentences. a. Does the line seem to fit the data? Why or why not? b. What does the correlation imply about the relationship between fuel efficiency and weight of a car? Is this what you expected? 7. Are there any outliers? If so, which point is an outlier? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 665 KEY TERMS Coefficient of Correlation a measure developed by Karl Pearson (early 1900s) that gives the strength of association between the independent variable and the dependent variable; the formula is: r = n∑ xy − (∑ x)(∑ y) 2 [n∑ x2 − (∑ x) ][n∑ y2 − (∑ y) 2 ] where n is the number of data points. The coefficient cannot be more then 1 and less then –1. The closer the coefficient is to ±1, the stronger the evidence of a significant linear relationship between x and y. Outlier an observation that does not fit the rest of the data CHAPTER REVIEW 12.1 Linear Equations The most basic type of association is a linear association. This type of relationship can be defined algebraically by the equations used, numerically with actual or predicted data values, or graphically from a plotted curve. (Lines are classified as straight curves.) Algebraically, a linear equation typically takes the form y = mx + b, where m and b are constants, x is the independent variable, y is the dependent variable. In a statistical context, a linear equation is written in the form y = a + bx, where a and b are the constants. This form is used to help readers distinguish the statistical context from the algebraic context. In the equation y = a + bx, the constant b that multiplies the x variable (b is called a coefficient) is called as the slope. The slope describes the rate of change between the independent and dependent variables; in other words, the rate of change describes the change that occurs in the dependent variable as the independent variable is changed. In the equation y = a + bx, the constant a is called as the y-intercept. Graphically, the y-intercept is the y coordinate of the point where the graph of the line crosses the y axis. At this point x = 0. The slope of a line is a value that describes the rate of change between the independent and dependent variables. The slope tells us how the dependent variable (y) changes for every one unit increase in the independent (x) variable, on average. The y-intercept is used to describe the dependent variable when the independent variable equals zero. Graphically, the slope is represented by three line types in elementary statistics. 12.2 Scatter Plots Scatter plots are particularly helpful graphs when we want to see if there is a linear relationship among data points. They indicate both the direction of the relationship between the x variables and the y variables, and the strength of the relationship. We calculate the strength of the relationship between an independent variable and a dependent variable using linear regression. 12.3 The Regression Equation A regression line, or a line of best fit, can be drawn on a scatter plot and used to predict outcomes for the x and y variables in a given data set or sample data. There are several ways to find a regression line, but usually the least-squares regression line is used because it creates a uniform line. Residuals, also called “errors,” measure the distance from the actual value of y and the estimated value of y. The Sum of Squared Errors, when set to its minimum, calculates the points on the line of best fit. Regression lines can be used to predict values within the given set of data, but should not be used to make predictions for values outside the set of data. The correlation coefficient r measures the strength of the linear association between x and y. The variable r has to be between –1 and +1. When r is positive, the x and y will tend to increase and decrease together. When r is negative, x will increase and y will decrease, or the opposite, x will decrease and y will increase. The coefficient of determination r2, is equal to the square of the correlation coefficient. When expressed as a percent, r2 represents the percent of variation in the dependent variable y that can be explained by variation in the independent variable x using the regression line. 12.4 Testing the Significance of the Correlation Coefficient Linear regression is a procedure for fitting a straight line of the form ŷ = a + bx to data. The conditions for regression are: • Linear In the population, there is a linear relationship that models the average value of y for different values of x. 666 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION • Independent The residuals are assumed to be independent. • Normal The y values are distributed normally for any value of x. • Equal variance The standard deviation of the y values is equal for each x value. • Random The data are produced from a well-designed random sample or randomized experiment. The slope b and intercept a of the least-squares line estimate the slope β and intercept α of the population (true) regression line. To estimate the population standard deviation of y, σ, use the standard deviation of the residuals, s. s = SEE n − 2 . The variable ρ (rho) is the population correlation coefficient. To test the null hypothesis H0: ρ = hypothesized value, use a linear regression t-test. The most common null hypothesis is H0: ρ = 0 which indicates there is no linear relationship between x and y in the population. The TI-83, 83+, 84, 84+ calculator function LinRegTTest can perform this test (STATS TESTS LinRegTTest). 12.5 Prediction After determining the presence of a strong correlation coefficient and calculating the line of best fit, you can use the least squares regression line to make predictions about your data. 12.6 Outliers To determine if a point is an outlier, do one of the following: 1. Input the following equations into the TI 83, 83+,84, 84+: y1 = a + bx y2 = (2s)a + bx y3 = − (2s)a + bx where s is the standard deviation of the residuals If any point is above y2or below y3 then the point is considered to be an outlier. 2. Use the residuals and compare their absolute values to 1.9s where s is the standard deviation of the residuals. If the absolute value of any residual is greater than or equal to 1.9s, then the corresponding point is an outlier. 3. Note: The calculator function LinRegTTest (STATS TESTS LinRegTTest) calculates s. To determine if a point is an influential point, graph the least-squares line with the point included, then graph the leastsquares line with the point excluded. If the graph changes by a considerable amount, the point is influential. FORMULA REVIEW 12.1 Linear Equations y = a + bx where a is the y-intercept and b is the slope. The variable x is the independent variable and y is the dependent variable. 12.4 Testing the Significance of the Correlation Coefficient Least Squares Line or Line of Best Fit: a = y-intercept b = slope Standard deviation of the residuals: s = SEE n − 2 . where SSE = sum of squared errors n = the number of data points y^ = a + bx where PRACTICE 12.1 Linear Equations This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 667 Use the following information to answer the next three exercises. A vacation resort rents SCUBA equipment to certified divers. The resort charges an up-front fee of $25 and another fee of $12.50 an hour. 1. What are the dependent and independent variables? 2. Find the equation that expresses the total fee in terms of the number of hours the equipment is rented. 3. Graph the equation from Exercise 12.2. Use the following information to answer the next two exercises. A credit card company charges $10 when a payment is late, and $5 a day each day the payment remains unpaid. 4. Find the equation that expresses the total fee in terms of the number of days the payment is late. 5. Graph the equation from Exercise 12.4. 6. Is the equation y = 10 + 5x – 3x2 linear? Why or why not? 7. Which of the following equations are linear? a. y = 6x + 8 b. y + 7 = 3x c. y – x = 8x2 d. 4y = 8 8. Does the graph show a linear equation? Why or why not? Figure 12.25 Table 12.12 contains real data for the first two decades of AIDS reporting. Year # AIDS cases diagnosed # AIDS deaths Pre-1981 91 1981 1982 1983 1984 1985 1986 1987 1988 319 1,170 3,076 6,240 11,776 19,032 28,564 35,447 29 121
453 1,482 3,466 6,878 11,987 16,162 20,868 Table 12.12 Adults and Adolescents only, United States 668 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 42,674 48,634 59,660 78,530 78,834 71,874 68,505 59,347 47,149 38,393 25,174 25,522 25,643 26,464 27,591 31,335 36,560 41,055 44,730 49,095 49,456 38,510 20,736 19,005 18,454 17,347 17,402 16,371 Total 802,118 489,093 Table 12.12 Adults and Adolescents only, United States 9. Use the columns "year" and "# AIDS cases diagnosed. Why is “year” the independent variable and “# AIDS cases diagnosed.” the dependent variable (instead of the reverse)? Use the following information to answer the next two exercises. A specialty cleaning company charges an equipment fee and an hourly labor fee. A linear equation that expresses the total amount of the fee the company charges for each session is y = 50 + 100x. 10. What are the independent and dependent variables? 11. What is the y-intercept and what is the slope? Interpret them using complete sentences. Use the following information to answer the next three questions. Due to erosion, a river shoreline is losing several thousand pounds of soil each year. A linear equation that expresses the total amount of soil lost per year is y = 12,000x. 12. What are the independent and dependent variables? 13. How many pounds of soil does the shoreline lose in a year? 14. What is the y-intercept? Interpret its meaning. Use the following information to answer the next two exercises. The price of a single issue of stock can fluctuate throughout the day. A linear equation that represents the price of stock for Shipment Express is y = 15 – 1.5x where x is the number of hours passed in an eight-hour day of trading. 15. What are the slope and y-intercept? Interpret their meaning. 16. If you owned this stock, would you want a positive or negative slope? Why? 12.2 Scatter Plots 17. Does the scatter plot appear linear? Strong or weak? Positive or negative? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 669 Figure 12.26 18. Does the scatter plot appear linear? Strong or weak? Positive or negative? Figure 12.27 19. Does the scatter plot appear linear? Strong or weak? Positive or negative? 670 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Figure 12.28 12.3 The Regression Equation Use the following information to answer the next five exercises. A random sample of ten professional athletes produced the following data where x is the number of endorsements the player has and y is the amount of money made (in millions of dollars). 12 9 9 3 13 4 10 Table 12.13 20. Draw a scatter plot of the data. 21. Use regression to find the equation for the line of best fit. 22. Draw the line of best fit on the scatter plot. 23. What is the slope of the line of best fit? What does it represent? 24. What is the y-intercept of the line of best fit? What does it represent? 25. What does an r value of zero mean? 26. When n = 2 and r = 1, are the data significant? Explain. 27. When n = 100 and r = -0.89, is there a significant correlation? Explain. 12.4 Testing the Significance of the Correlation Coefficient 28. When testing the significance of the correlation coefficient, what is the null hypothesis? 29. When testing the significance of the correlation coefficient, what is the alternative hypothesis? 30. If the level of significance is 0.05 and the p-value is 0.04, what conclusion can you draw? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 671 12.5 Prediction Use the following information to answer the next two exercises. An electronics retailer used regression to find a simple model to predict sales growth in the first quarter of the new year (January through March). The model is good for 90 days, where x is the day. The model can be written as follows: ŷ = 101.32 + 2.48x where ŷ is in thousands of dollars. 31. What would you predict the sales to be on day 60? 32. What would you predict the sales to be on day 90? Use the following information to answer the next three exercises. A landscaping company is hired to mow the grass for several large properties. The total area of the properties combined is 1,345 acres. The rate at which one person can mow is as follows: ŷ = 1350 – 1.2x where x is the number of hours and ŷ represents the number of acres left to mow. 33. How many acres will be left to mow after 20 hours of work? 34. How many acres will be left to mow after 100 hours of work? 35. How many hours will it take to mow all of the lawns? (When is ŷ = 0?) Table 12.14 contains real data for the first two decades of AIDS reporting. Year # AIDS cases diagnosed # AIDS deaths Pre-1981 91 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 319 1,170 3,076 6,240 11,776 19,032 28,564 35,447 42,674 48,634 59,660 78,530 78,834 71,874 68,505 59,347 47,149 38,393 25,174 25,522 25,643 26,464 29 121 453 1,482 3,466 6,878 11,987 16,162 20,868 27,591 31,335 36,560 41,055 44,730 49,095 49,456 38,510 20,736 19,005 18,454 17,347 17,402 16,371 Table 12.14 Adults and Adolescents only, United States 672 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Total 802,118 489,093 Table 12.14 Adults and Adolescents only, United States 36. Graph “year” versus “# AIDS cases diagnosed” (plot the scatter plot). Do not include pre-1981 data. 37. Perform linear regression. What is the linear equation? Round to the nearest whole number. 38. Write the equations: a. Linear equation: __________ b. a = ________ c. b = ________ d. r = ________ e. n = ________ 39. Solve. a. When x = 1985, ŷ = _____ b. When x = 1990, ŷ =_____ c. When x = 1970, ŷ =______ Why doesn’t this answer make sense? 40. Does the line seem to fit the data? Why or why not? 41. What does the correlation imply about the relationship between time (years) and the number of diagnosed AIDS cases reported in the U.S.? 42. Plot the two given points on the following graph. Then, connect the two points to form the regression line. Figure 12.29 Obtain the graph on your calculator or computer. 43. Write the equation: ŷ= ____________ 44. Hand draw a smooth curve on the graph that shows the flow of the data. 45. Does the line seem to fit the data? Why or why not? 46. Do you think a linear fit is best? Why or why not? 47. What does the correlation imply about the relationship between time (years) and the number of diagnosed AIDS cases reported in the U.S.? 48. Graph “year” vs. “# AIDS cases diagnosed.” Do not include pre-1981. Label both axes with words. Scale both axes. 49. Enter your data into your calculator or computer. The pre-1981 data should not be included. Why is that so? Write the linear equation, rounding to four decimal places: 50. Calculate the following: a. a = _____ b. b = _____ c. correlation = _____ d. n = _____ This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 673 12.6 Outliers Use the following information to answer the next four exercises. The scatter plot shows the relationship between hours spent studying and exam scores. The line shown is the calculated line of best fit. The correlation coefficient is 0.69. Figure 12.30 51. Do there appear to be any outliers? 52. A point is removed, and the line of best fit is recalculated. The new correlation coefficient is 0.98. Does the point appear to have been an outlier? Why? 53. What effect did the potential outlier have on the line of best fit? 54. Are you more or less confident in the predictive ability of the new line of best fit? 55. The Sum of Squared Errors for a data set of 18 numbers is 49. What is the standard deviation? 56. The Standard Deviation for the Sum of Squared Errors for a data set is 9.8. What is the cutoff for the vertical distance that a point can be from the line of best fit to be considered an outlier? HOMEWORK 12.1 Linear Equations 57. For each of the following situations, state the independent variable and the dependent variable. a. A study is done to determine if elderly drivers are involved in more motor vehicle fatalities than other drivers. The number of fatalities per 100,000 drivers is compared to the age of drivers. Insurance companies base life insurance premiums partially on the age of the applicant. b. A study is done to determine if the weekly grocery bill changes based on the number of family members. c. d. Utility bills vary according to power consumption. e. A study is done to determine if a higher education reduces the crime rate in a population. 58. Piece-rate systems are widely debated incentive payment plans. In a recent study of loan officer effectiveness, the following piece-rate system was examined: % of goal reached < 80 80 100 120 Incentive n/ a $4,000 with an additional $125 added per percentage point from 81–99% $6,500 with an additional $125 added per percentage point from 101–119% $9,500 with an additional $125 added per percentage point starting at 121% Table 12.15 674 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION If a loan officer makes 95% of his or her goal, write the linear function that applies based on the incentive plan table. In context, explain the y-intercept and slope. 12.2 Scatter Plots 59. The Gross Domestic Product Purchasing Power Parity is an indication of a country’s currency value compared to another country. Table 12.16 shows the GDP PPP of Cuba as compared to US dollars. Construct a scatter plot of the data. Year Cuba’s PPP Year Cuba’s PPP 1999 1,700 2006 4,000 2000 1,700 2007 11,000 2002 2,300 2008 9,500 2003 2,900 2009 9,700 2004 3,000 2010 9,900 2005 3,500 Table 12.16 60. The following table shows the poverty rates
and cell phone usage in the United States. Construct a scatter plot of the data Year Poverty Rate Cellular Usage per Capita 2003 12.7 2005 12.6 2007 2009 12 12 Table 12.17 54.67 74.19 84.86 90.82 61. Does the higher cost of tuition translate into higher-paying jobs? The table lists the top ten colleges based on mid-career salary and the associated yearly tuition costs. Construct a scatter plot of the data. Mid-Career Salary (in thousands) Yearly Tuition School Princeton Harvey Mudd CalTech 137 135 127 US Naval Academy 122 West Point MIT Lehigh University NYU-Poly Babson College Stanford Table 12.18 120 118 118 117 117 114 28,540 40,133 39,900 0 0 42,050 43,220 39,565 40,400 54,506 62. If the level of significance is 0.05 and the p-value is 0.06, what conclusion can you draw? 63. If there are 15 data points in a set of data, what is the number of degree of freedom? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 675 12.3 The Regression Equation 64. What is the process through which we can calculate a line that goes through a scatter plot with a linear pattern? 65. Explain what it means when a correlation has an r2 of 0.72. 66. Can a coefficient of determination be negative? Why or why not? 12.4 Testing the Significance of the Correlation Coefficient 67. If the level of significance is 0.05 and the p-value is 0.06, what conclusion can you draw? 68. If there are 15 data points in a set of data, what is the number of degree of freedom? 12.5 Prediction 69. Recently, the annual number of driver deaths per 100,000 for the selected age groups was as follows: Age Number of Driver Deaths per 100,000 17.5 22 29.5 44.5 64.5 80 38 36 24 20 18 28 Table 12.19 a. For each age group, pick the midpoint of the interval for the x value. (For the 75+ group, use 80.) b. Using “ages” as the independent variable and “Number of driver deaths per 100,000” as the dependent variable, make a scatter plot of the data. c. Calculate the least squares (best–fit) line. Put the equation in the form of: ŷ = a + bx d. Find the correlation coefficient. Is it significant? e. Predict the number of deaths for ages 40 and 60. f. Based on the given data, is there a linear relationship between age of a driver and driver fatality rate? g. What is the slope of the least squares (best-fit) line? Interpret the slope. 70. Table 12.20 shows the life expectancy for an individual born in the United States in certain years. Year of Birth Life Expectancy 1930 1940 1950 1965 1973 1982 1987 1992 2010 Table 12.20 59.7 62.9 70.2 69.7 71.4 74.5 75 75.7 78.7 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the ordered pairs. 676 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION c. Calculate the least squares line. Put the equation in the form of: ŷ = a + bx d. Find the correlation coefficient. Is it significant? e. Find the estimated life expectancy for an individual born in 1950 and for one born in 1982. f. Why aren’t the answers to part e the same as the values in Table 12.20 that correspond to those years? g. Use the two points in part e to plot the least squares line on your graph from part b. h. Based on the data, is there a linear relationship between the year of birth and life expectancy? i. Are there any outliers in the data? j. Using the least squares line, find the estimated life expectancy for an individual born in 1850. Does the least squares line give an accurate estimate for that year? Explain why or why not. k. What is the slope of the least-squares (best-fit) line? Interpret the slope. 71. The maximum discount value of the Entertainment® card for the “Fine Dining” section, Edition ten, for various pages is given in Table 12.21 Page number Maximum value ($) 4 14 25 32 43 57 72 85 90 Table 12.21 16 19 15 17 19 15 16 15 17 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the ordered pairs. c. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx d. Find the correlation coefficient. Is it significant? e. Find the estimated maximum values for the restaurants on page ten and on page 70. f. Does it appear that the restaurants giving the maximum value are placed in the beginning of the “Fine Dining” section? How did you arrive at your answer? g. Suppose that there were 200 pages of restaurants. What do you estimate to be the maximum value for a restaurant listed on page 200? Is the least squares line valid for page 200? Why or why not? h. i. What is the slope of the least-squares (best-fit) line? Interpret the slope. 72. Table 12.22 gives the gold medal times for every other Summer Olympics for the women’s 100-meter freestyle (swimming). Year Time (seconds) 1912 1924 1932 1952 1960 1968 1976 Table 12.22 82.2 72.4 66.8 66.8 61.2 60.0 55.65 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 677 Year Time (seconds) 1984 1992 2000 2008 Table 12.22 55.92 54.64 53.8 53.1 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least squares line. Put the equation in the form of: ŷ = a + bx. e. Find the correlation coefficient. Is the decrease in times significant? f. Find the estimated gold medal time for 1932. Find the estimated time for 1984. g. Why are the answers from part f different from the chart values? h. Does it appear that a line is the best way to fit the data? Why or why not? i. Use the least-squares line to estimate the gold medal time for the next Summer Olympics. Do you think that your answer is reasonable? Why or why not? # letters in name Year entered the Union Rank for entering the Union Area (square miles) 73. State Alabama Colorado Hawaii Iowa Maryland Missouri 7 8 6 4 8 8 New Jersey 9 Ohio South Carolina Utah Wisconsin Table 12.23 4 13 4 9 1819 1876 1959 1846 1788 1821 1787 1803 1788 1896 1848 22 38 50 29 7 24 3 17 8 45 30 52,423 104,100 10,932 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 We are interested in whether or not the number of letters in a state name depends upon the year the state entered the Union. a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx. e. Find the correlation coefficient. What does it imply about the significance of the relationship? f. Find the estimated number of letters (to the nearest integer) a state would have if it entered the Union in 1900. Find the estimated number of letters a state would have if it entered the Union in 1940. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Use the least-squares line to estimate the number of letters a new state that enters the Union this year would have. Can the least squares line be used to predict it? Why or why not? 12.6 Outliers 678 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 74. The height (sidewalk to roof) of notable tall buildings in America is compared to the number of stories of the building (beginning at street level). Height (in feet) Stories 1,050 428 362 529 790 401 380 1,454 1,127 700 Table 12.24 57 28 26 40 60 22 38 110 100 46 a. Using “stories” as the independent variable and “height” as the dependent variable, make a scatter plot of the data. b. Does it appear from inspection that there is a relationship between the variables? c. Calculate the least squares line. Put the equation in the form of: ŷ = a + bx d. Find the correlation coefficient. Is it significant? e. Find the estimated heights for 32 stories and for 94 stories. f. Based on the data in Table 12.24, is there a linear relationship between the number of stories in tall buildings and the height of the buildings? g. Are there any outliers in the data? If so, which point(s)? h. What is the estimated height of a building with six stories? Does the least squares line give an accurate estimate of height? Explain why or why not. i. Based on the least squares line, adding an extra story is predicted to add about how many feet to a building? j. What is the slope of the least squares (best-fit) line? Interpret the slope. 75. Ornithologists, scientists who study birds, tag sparrow hawks in 13 different colonies to study their population. They gather data for the percent of new sparrow hawks in each colony and the percent of those that have returned from migration. Percent return:74; 66; 81; 52; 73; 62; 52; 45; 62; 46; 60; 46; 38 Percent new:5; 6; 8; 11; 12; 15; 16; 17; 18; 18; 19; 20; 20 a. Enter the data into your calculator and make a scatter plot. b. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. c. Explain in words what the slope and y-intercept of the regression line tell us. d. How well does the regression line fit the data? Explain your response. e. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. f. An ecologist wants to predict how many birds will join another colony of sparrow hawks to which 70% of the adults from the previous year have returned. What is the prediction? 76. The following table shows data on average per capita wine consumption and heart disease rate in a random sample of 10 countries. Yearly wine consumption in liters 2.5 3.9 2.9 2.4 2.9 0.8 9.1 2.7 0.8 0.7
Death from heart diseases 221 167 131 191 220 297 71 172 211 300 Table 12.25 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 679 a. Enter the data into your calculator and make a scatter plot. b. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. c. Explain in words what the slope and y-intercept of the regression line tell us. d. How well does the regression line fit the data? Explain your response. e. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. f. Do the data provide convincing evidence that there is a linear relationship between the amount of alcohol consumed and the heart disease death rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question. 77. The following table consists of one student athlete’s time (in minutes) to swim 2000 yards and the student’s heart rate (beats per minute) after swimming on a random sample of 10 days: Swim Time Heart Rate 34.12 35.72 34.72 34.05 34.13 35.73 36.17 35.57 35.37 35.57 Table 12.26 144 152 124 140 152 146 128 136 144 148 a. Enter the data into your calculator and make a scatter plot. b. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot from part a. c. Explain in words what the slope and y-intercept of the regression line tell us. d. How well does the regression line fit the data? Explain your response. e. Which point has the largest residual? Explain what the residual means in context. Is this point an outlier? An influential point? Explain. 78. A researcher is investigating whether non-white minorities commit a disproportionate number of homicides. He uses demographic data from Detroit, MI to compare homicide rates and the number of the population that are white males. White Males Homicide rate per 100,000 people 558,724 538,584 519,171 500,457 482,418 465,029 448,267 432,109 416,533 Table 12.27 8.6 8.9 8.52 8.89 13.07 14.57 21.36 28.03 31.49 680 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION White Males Homicide rate per 100,000 people 401,518 387,046 373,095 359,647 Table 12.27 37.39 46.26 47.24 52.33 a. Use your calculator to construct a scatter plot of the data. What should the independent variable be? Why? b. Use your calculator’s regression function to find the equation of the least-squares regression line. Add this to your scatter plot. c. Discuss what the following mean in context. i. The slope of the regression equation ii. The y-intercept of the regression equation iii. The correlation r iv. The coefficient of determination r2. d. Do the data provide convincing evidence that there is a linear relationship between the number of white males in the population and the homicide rate? Carry out an appropriate test at a significance level of 0.05 to help answer this question. Mid-Career Salary (in thousands) Yearly Tuition 79. School Princeton Harvey Mudd CalTech 137 135 127 US Naval Academy 122 West Point MIT Lehigh University NYU-Poly Babson College Stanford Table 12.28 120 118 118 117 117 114 28,540 40,133 39,900 0 0 42,050 43,220 39,565 40,400 54,506 Using the data to determine the linear-regression line equation with the outliers removed. Is there a linear correlation for the data set with outliers removed? Justify your answer. REFERENCES 12.1 Linear Equations Data from the Centers for Disease Control and Prevention. Data from the National Center for HIV, STD, and TB Prevention. 12.5 Prediction Data from the Centers for Disease Control and Prevention. Data from the National Center for HIV, STD, and TB Prevention. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Data from the United States Census Bureau. Available online at http://www.census.gov/compendia/statab/cats/ transportation/motor_vehicle_accidents_and_fatalities.html CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 681 Data from the National Center for Health Statistics. 12.6 Outliers Data from the House Ways and Means Committee, the Health and Human Services Department. Data from Microsoft Bookshelf. Data from the United States Department of Labor, the Bureau of Labor Statistics. Data from the Physician’s Handbook, 1990. Data from the United States Department of Labor, the Bureau of Labor Statistics. BRINGING IT TOGETHER: HOMEWORK 80. The average number of people in a family that received welfare for various years is given in Table 12.29. Year Welfare family size 1969 1973 1975 1979 1983 1988 1991 Table 12.29 4.0 3.6 3.2 3.0 3.0 3.0 2.9 a. Using “year” as the independent variable and “welfare family size” as the dependent variable, draw a scatter plot of the data. b. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx c. Find the correlation coefficient. Is it significant? d. Pick two years between 1969 and 1991 and find the estimated welfare family sizes. e. Based on the data in Table 12.29, is there a linear relationship between the year and the average number of people in a welfare family? f. Using the least-squares line, estimate the welfare family sizes for 1960 and 1995. Does the least-squares line give an accurate estimate for those years? Explain why or why not. g. Are there any outliers in the data? h. What is the estimated average welfare family size for 1986? Does the least squares line give an accurate estimate for that year? Explain why or why not. i. What is the slope of the least squares (best-fit) line? Interpret the slope. 81. The percent of female wage and salary workers who are paid hourly rates is given in Table 12.30 for the years 1979 to 1992. Year Percent of workers paid hourly rates 1979 1980 1981 1982 Table 12.30 61.2 60.7 61.3 61.3 682 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Year Percent of workers paid hourly rates 1983 1984 1985 1986 1987 1990 1992 Table 12.30 61.8 61.7 61.8 62.0 62.7 62.8 62.9 a. Using “year” as the independent variable and “percent” as the dependent variable, draw a scatter plot of the data. b. Does it appear from inspection that there is a relationship between the variables? Why or why not? c. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx d. Find the correlation coefficient. Is it significant? e. Find the estimated percents for 1991 and 1988. f. Based on the data, is there a linear relationship between the year and the percent of female wage and salary earners who are paid hourly rates? g. Are there any outliers in the data? h. What is the estimated percent for the year 2050? Does the least-squares line give an accurate estimate for that year? Explain why or why not. i. What is the slope of the least-squares (best-fit) line? Interpret the slope. Use the following information to answer the next two exercises. The cost of a leading liquid laundry detergent in different sizes is given in Table 12.31. Size (ounces) Cost ($) Cost per ounce 16 32 64 200 Table 12.31 3.99 4.99 5.99 10.99 82. 83. a. Using “size” as the independent variable and “cost” as the dependent variable, draw a scatter plot. b. Does it appear from inspection that there is a relationship between the variables? Why or why not? c. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx d. Find the correlation coefficient. Is it significant? e. f. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Are there any outliers in the given data? i. If the laundry detergent were sold in a 40-ounce size, find the estimated cost. If the laundry detergent were sold in a 90-ounce size, find the estimated cost. Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would you cost? Why or why not? j. What is the slope of the least-squares (best-fit) line? Interpret the slope. a. Complete Table 12.31 for the cost per ounce of the different sizes. b. Using “size” as the independent variable and “cost per ounce” as the dependent variable, draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 683 If the laundry detergent were sold in a 40-ounce size, find the estimated cost per ounce. If the laundry detergent were sold in a 90-ounce size, find the estimated cost per ounce. e. Find the correlation coefficient. Is it significant? f. g. h. Does it appear that a line is the best way to fit the data? Why or why not? i. Are there any outliers in the the data? j. Is the least-squares line valid for predicting what a 300-ounce size of the laundry detergent would cost per ounce? Why or why not? k. What is the slope of the least-squares (best-fit) line? Interpret the slope. 84. According to a flyer by a Prudential Insurance Company representative, the costs of approximate probate fees and taxes for selected net taxable estates are as follows: Net Taxable Estate ($) Approximate Probate Fees and Taxes ($) 600,000 750,000 1,000,000 1,500,000 2,000,000 2,500,000 3,000,000 Table 12.32 30,000 92,500 203,000 438,000 688,000 1,037,000 1,350,000 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx. e. Find the correlation coefficient. Is it significant? f. Find the es
timated total cost for a next taxable estate of $1,000,000. Find the cost for $2,500,000. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Are there any outliers in the data? i. Based on these results, what would be the probate fees and taxes for an estate that does not have any assets? j. What is the slope of the least-squares (best-fit) line? Interpret the slope. 85. The following are advertised sale prices of color televisions at Anderson’s. Size (inches) Sale Price ($) 9 20 27 31 35 40 60 Table 12.33 147 197 297 447 1177 2177 2497 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx e. Find the correlation coefficient. Is it significant? 684 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION f. Find the estimated sale price for a 32 inch television. Find the cost for a 50 inch television. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Are there any outliers in the data? i. What is the slope of the least-squares (best-fit) line? Interpret the slope. 86. Table 12.34 shows the average heights for American boy s in 1990. Age (years) Height (cm) birth 2 3 5 7 10 14 Table 12.34 50.8 83.8 91.4 106.6 119.3 137.1 157.5 a. Decide which variable should be the independent variable and which should be the dependent variable. b. Draw a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx e. Find the correlation coefficient. Is it significant? f. Find the estimated average height for a one-year-old. Find the estimated average height for an eleven-year-old. g. Does it appear that a line is the best way to fit the data? Why or why not? h. Are there any outliers in the data? i. Use the least squares line to estimate the average height for a sixty-two-year-old man. Do you think that your answer is reasonable? Why or why not? j. What is the slope of the least-squares (best-fit) line? Interpret the slope. # letters in name Year entered the Union Ranks for entering the Union Area (square miles) 87. State Alabama Colorado Hawaii Iowa Maryland Missouri 7 8 6 4 8 8 New Jersey 9 Ohio South Carolina Utah Wisconsin Table 12.35 4 13 4 9 1819 1876 1959 1846 1788 1821 1787 1803 1788 1896 1848 22 38 50 29 7 24 3 17 8 45 30 52,423 104,100 10,932 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 We are interested in whether there is a relationship between the ranking of a state and the area of the state. a. What are the independent and dependent variables? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 685 b. What do you think the scatter plot will look like? Make a scatter plot of the data. c. Does it appear from inspection that there is a relationship between the variables? Why or why not? d. Calculate the least-squares line. Put the equation in the form of: ŷ = a + bx e. Find the correlation coefficient. What does it imply about the significance of the relationship? f. Find the estimated areas for Alabama and for Colorado. Are they close to the actual areas? g. Use the two points in part f to plot the least-squares line on your graph from part b. h. Does it appear that a line is the best way to fit the data? Why or why not? i. Are there any outliers? j. Use the least squares line to estimate the area of a new state that enters the Union. Can the least-squares line be used to predict it? Why or why not? k. Delete “Hawaii” and substitute “Alaska” for it. Alaska is the forty-ninth, state with an area of 656,424 square miles. l. Calculate the new least-squares line. m. Find the estimated area for Alabama. Is it closer to the actual area with this new least-squares line or with the previous one that included Hawaii? Why do you think that’s the case? n. Do you think that, in general, newer states are larger than the original states? SOLUTIONS 1 dependent variable: fee amount; independent variable: time 3 Figure 12.31 5 Figure 12.32 7 y = 6x + 8, 4y = 8, and y + 7 = 3x are all linear equations. 686 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 9 The number of AIDS cases depends on the year. Therefore, year becomes the independent variable and the number of AIDS cases is the dependent variable. 11 The y-intercept is 50 (a = 50). At the start of the cleaning, the company charges a one-time fee of $50 (this is when x = 0). The slope is 100 (b = 100). For each session, the company charges $100 for each hour they clean. 13 12,000 pounds of soil 15 The slope is –1.5 (b = –1.5). This means the stock is losing value at a rate of $1.50 per hour. The y-intercept is $15 (a = 15). This means the price of stock before the trading day was $15. 17 The data appear to be linear with a strong, positive correlation. 19 The data appear to have no correlation. 21 ŷ = 2.23 + 1.99x 23 The slope is 1.99 (b = 1.99). It means that for every endorsement deal a professional player gets, he gets an average of another $1.99 million in pay each year. 25 It means that there is no correlation between the data sets. 27 Yes, there are enough data points and the value of r is strong enough to show that there is a strong negative correlation between the data sets. 29 Ha: ρ ≠ 0 31 $250,120 33 1,326 acres 35 1,125 hours, or when x = 1,125 37 Check student’s solution. 39 a. When x = 1985, ŷ = 25,52 b. When x = 1990, ŷ = 34,275 c. When x = 1970, ŷ = –725 Why doesn’t this answer make sense? The range of x values was 1981 to 2002; the year 1970 is not in this range. The regression equation does not apply, because predicting for the year 1970 is extrapolation, which requires a different process. Also, a negative number does not make sense in this context, where we are predicting AIDS cases diagnosed. 41 Also, the correlation r = 0.4526. If r is compared to the value in the 95% Critical Values of the Sample Correlation Coefficient Table, because r > 0.423, r is significant, and you would think that the line could be used for prediction. But the scatter plot indicates otherwise. 43 y^ = 3,448,225 + 1750x 45 There was an increase in AIDS cases diagnosed until 1993. From 1993 through 2002, the number of AIDS cases diagnosed declined each year. It is not appropriate to use a linear regression line to fit to the data. 47 Since there is no linear association between year and # of AIDS cases diagnosed, it is not appropriate to calculate a linear correlation coefficient. When there is a linear association and it is appropriate to calculate a correlation, we cannot say that one variable “causes” the other variable. 49 We don’t know if the pre-1981 data was collected from a single year. So we don’t have an accurate x value for this figure. Regression equation: ŷ (#AIDS Cases) = –3,448,225 + 1749.777 (year) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 687 Coefficients Intercept –3,448,225 X Variable 1 1,749.777 Table 12.36 51 Yes, there appears to be an outlier at (6, 58). 53 The potential outlier flattened the slope of the line of best fit because it was below the data set. It made the line of best fit less accurate is a predictor for the data. 55 s = 1.75 57 a. b. c. d. e. independent variable: age; dependent variable: fatalities independent variable: # of family members; dependent variable: grocery bill independent variable: age of applicant; dependent variable: insurance premium independent variable: power consumption; dependent variable: utility independent variable: higher education (years); dependent variable: crime rates 59 Check student’s solution. 61 For graph: check student’s solution. Note that tuition is the independent variable and salary is the dependent variable. 63 13 65 It means that 72% of the variation in the dependent variable (y) can be explained by the variation in the independent variable (x). 67 We do not reject the null hypothesis. There is not sufficient evidence to conclude that there is a significant linear relationship between x and y because the correlation coefficient is not significantly different from zero. 69 a. Age Number of Driver Deaths per 100,000 16–19 20–24 25–34 35–54 55–74 75+ Table 12.37 38 36 24 20 18 28 b. Check student’s solution. c. ŷ = 35.5818045 – 0.19182491x d. r = –0.57874 For four df and alpha = 0.05, the LinRegTTest gives p-value = 0.2288 so we do not reject the null hypothesis; there is not a significant linear relationship between deaths and age. 688 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION Using the table of critical values for the correlation coefficient, with four df, the critical value is 0.811. The correlation coefficient r = –0.57874 is not less than –0.811, so we do not reject the null hypothesis. e. if age = 40, ŷ (deaths) = 35.5818045 – 0.19182491(40) = 27.9 if age = 60, ŷ (deaths) = 35.5818045 – 0.19182491(60) = 24.1 f. For entire dataset, there is a linear relationship for the ages up to age 74. The oldest age group shows an increase in deaths from the prior group, which is not consistent with the younger ages. g. slope = –0.19182491 71 a. We wonder if the better discounts appear earlier in the book so we select page as X and discount as Y. b. Check student’s solution. c. ŷ = 17.21757 – 0.01412x d. r = – 0.2752 For seven df and alpha = 0.05, using LinRegTTest p-value = 0.4736 so we do not reject; there is a not a significant linear relationship between page and discount. Using the table of critical values for the correlation coefficient, with seven df, the critical value is 0.666.
The correlation coefficient xi = –0.2752 is not less than 0.666 so we do not reject. e. page 10: 17.08 page 70: 16.23 f. There is not a significant linear correlation so it appears there is no relationship between the page and the amount of the discount. g. page 200: 14.39 h. No, using the regression equation to predict for page 200 is extrapolation. i. slope = –0.01412 As the page number increases by one page, the discount decreases by $0.01412 73 a. Year is the independent or x variable; the number of letters is the dependent or y variable. b. Check student’s solution. c. no d. ŷ = 47.03 – 0.0216x e. –0.4280 f. 6; 5 g. No, the relationship does not appear to be linear; the correlation is not significant. h. current year: 2013: 3.55 or four letters; this is not an appropriate use of the least squares line. It is extrapolation. 75 a. and b. Check student’s solution. c. The slope of the regression line is -0.3179 with a y-intercept of 32.966. In context, the y-intercept indicates that when there are no returning sparrow hawks, there will be almost 31% new sparrow hawks, which doesn’t make sense since if there are no returning birds, then the new percentage would have to be 100% (this is an example of why we do not extrapolate). The slope tells us that for each percentage increase in returning birds, the percentage of new birds in the colony decreases by 0.3179%. d. If we examine r2, we see that only 50.238% of the variation in the percent of new birds is explained by the model and the correlation coefficient, r = 0.71 only indicates a somewhat strong correlation between returning and new percentages. e. The ordered pair (66, 6) generates the largest residual of 6.0. This means that when the observed return percentage is 66%, our observed new percentage, 6%, is almost 6% less than the predicted new value of 11.98%. If we remove this data pair, we see only an adjusted slope of -0.2723 and an adjusted intercept of 30.606. In other words, even though this data generates the largest residual, it is not an outlier, nor is the data pair an influential point. f. If there are 70% returning birds, we would expect to see y = -0.2723(70) + 30.606 = 0.115 or 11.5% new birds in the colony. 77 a. Check student’s solution. b. Check student’s solution. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 689 c. We have a slope of –1.4946 with a y-intercept of 193.88. The slope, in context, indicates that for each additional minute added to the swim time, the heart rate will decrease by 1.5 beats per minute. If the student is not swimming at all, the y-intercept indicates that his heart rate will be 193.88 beats per minute. While the slope has meaning (the longer it takes to swim 2,000 meters, the less effort the heart puts out), the y-intercept does not make sense. If the athlete is not swimming (resting), then his heart rate should be very low. d. Since only 1.5% of the heart rate variation is explained by this regression equation, we must conclude that this association is not explained with a linear relationship. e. The point (34.72, 124) generates the largest residual of –11.82. This means that our observed heart rate is almost 12 beats less than our predicted rate of 136 beats per minute. When this point is removed, the slope becomes 1.6914 with the y-intercept changing to 83.694. While the linear association is still very weak, we see that the removed data pair can be considered an influential point in the sense that the y-intercept becomes more meaningful. 79 If we remove the two service academies (the tuition is $0.00), we construct a new regression equation of y = –0.0009x + 160 with a correlation coefficient of 0.71397 and a coefficient of determination of 0.50976. This allows us to say there is a fairly strong linear association between tuition costs and salaries if the service academies are removed from the data set. 81 a. Check student's solution. b. yes c. ŷ = −266.8863+0.1656x d. 0.9448; Yes e. 62.8233; 62.3265 f. yes g. yes; (1987, 62.7) h. 72.5937; no i. slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656. 83 a. Size (ounces) Cost ($) cents/oz 16 32 64 200 Table 12.38 3.99 4.99 5.99 10.99 24.94 15.59 9.36 5.50 b. Check student’s solution. c. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size. d. ŷ = 20.2368 – 0.0819x e. r = –0.8086 f. 40-oz: 16.96 cents/oz g. 90-oz: 12.87 cents/oz h. The relationship is not linear; the least squares line is not appropriate. i. no outliers j. No, you would be extrapolating. The 300-oz size is outside the range of x. k. slope = –0.08194; for each additional ounce in size, the cost per ounce decreases by 0.082 cents. 690 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 85 a. Size is x, the independent variable, price is y, the dependent variable. b. Check student’s solution. c. The relationship does not appear to be linear. d. ŷ = –745.252 + 54.75569x e. r = 0.8944, yes it is significant f. 32-inch: $1006.93, 50-inch: $1992.53 g. No, the relationship does not appear to be linear. However, r is significant. h. yes, the 60-inch TV i. For each additional inch, the price increases by $54.76 87 a. Let rank be the independent variable and area be the dependent variable. b. Check student’s solution. c. There appears to be a linear relationship, with one outlier. d. ŷ (area) = 24177.06 + 1010.478x e. r = 0.50047, r is not significant so there is no relationship between the variables. f. Alabama: 46407.576 Colorado: 62575.224 g. Alabama estimate is closer than Colorado estimate. h. If the outlier is removed, there is a linear relationship. i. There is one outlier (Hawaii). j. rank 51: 75711.4; no k. Alabama Colorado Alaska Iowa Maryland Missouri New Jersey Ohio 7 8 6 4 8 8 9 4 1819 1876 1959 1846 1788 1821 1787 1803 South Carolina 13 1788 Utah Wisconsin 4 9 1896 1848 Table 12.39 22 38 51 29 7 24 3 17 8 45 30 52,423 104,100 656,424 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 l. ŷ = –87065.3 + 7828.532x m. Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier. n. yes, with the exception of Hawaii 81 a. Check student's solution. b. yes This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION 691 c. ŷ = −266.8863+0.1656x d. 0.9448; Yes e. 62.8233; 62.3265 f. yes g. yes; (1987, 62.7) h. 72.5937; no i. slope = 0.1656. As the year increases by one, the percent of workers paid hourly rates tends to increase by 0.1656. 83 a. Size (ounces) Cost ($) cents/oz 16 32 64 200 Table 12.40 3.99 4.99 5.99 10.99 24.94 15.59 9.36 5.50 b. Check student’s solution. c. There is a linear relationship for the sizes 16 through 64, but that linear trend does not continue to the 200-oz size. d. ŷ = 20.2368 – 0.0819x e. r = –0.8086 f. 40-oz: 16.96 cents/oz g. 90-oz: 12.87 cents/oz h. The relationship is not linear; the least squares line is not appropriate. i. no outliers j. No, you would be extrapolating. The 300-oz size is outside the range of x. k. slope = –0.08194; for each additional ounce in size, the cost per ounce decreases by 0.082 cents. 85 a. Size is x, the independent variable, price is y, the dependent variable. b. Check student’s solution. c. The relationship does not appear to be linear. d. ŷ = –745.252 + 54.75569x e. r = 0.8944, yes it is significant f. 32-inch: $1006.93, 50-inch: $1992.53 g. No, the relationship does not appear to be linear. However, r is significant. h. yes, the 60-inch TV i. For each additional inch, the price increases by $54.76 87 a. Let rank be the independent variable and area be the dependent variable. b. Check student’s solution. c. There appears to be a linear relationship, with one outlier. 692 CHAPTER 12 \| LINEAR REGRESSION AND CORRELATION d. ŷ (area) = 24177.06 + 1010.478x e. r = 0.50047, r is not significant so there is no relationship between the variables. f. Alabama: 46407.576 Colorado: 62575.224 g. Alabama estimate is closer than Colorado estimate. h. If the outlier is removed, there is a linear relationship. i. There is one outlier (Hawaii). j. rank 51: 75711.4; no k. Alabama Colorado Alaska Iowa Maryland Missouri New Jersey Ohio 7 8 6 4 8 8 9 4 1819 1876 1959 1846 1788 1821 1787 1803 South Carolina 13 1788 Utah Wisconsin 4 9 1896 1848 Table 12.41 22 38 51 29 7 24 3 17 8 45 30 52,423 104,100 656,424 56,276 12,407 69,709 8,722 44,828 32,008 84,904 65,499 l. ŷ = –87065.3 + 7828.532x m. Alabama: 85,162.404; the prior estimate was closer. Alaska is an outlier. n. yes, with the exception of Hawaii This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 693 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Figure 13.1 One-way ANOVA is used to measure information from several groups. Introduction Chapter Objectives By the end of this chapter, the student should be able to: Interpret the F probability distribution as the number of groups and the sample size change. • • Discuss two uses for the F distribution: one-way ANOVA and the test of two variances. • Conduct and interpret one-way ANOVA. • Conduct and interpret hypothesis tests of two variances. Many statistical applications in psychology, social science, business administration, and the natural sciences involve several groups. For example, an environmentalist is interested in knowing if the average amount of pollution varies in several bodies of water. A sociologist is interested in knowing if the amount of income a person earns varies according to his or her upbringing. A consumer looking for a new car might compare the average gas mileage of several models. 694 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA For hyp
othesis tests comparing averages between more than two groups, statisticians have developed a method called "Analysis of Variance" (abbreviated ANOVA). In this chapter, you will study the simplest form of ANOVA called single factor or one-way ANOVA. You will also study the F distribution, used for one-way ANOVA, and the test of two variances. This is just a very brief overview of one-way ANOVA. You will study this topic in much greater detail in future statistics courses. One-Way ANOVA, as it is presented here, relies heavily on a calculator or computer. 13.1 \| One-Way ANOVA The purpose of a one-way ANOVA test is to determine the existence of a statistically significant difference among several group means. The test actually uses variances to help determine if the means are equal or not. In order to perform a oneway ANOVA test, there are five basic assumptions to be fulfilled: 1. Each population from which a sample is taken is assumed to be normal. 2. All samples are randomly selected and independent. 3. The populations are assumed to have equal standard deviations (or variances). 4. The factor is a categorical variable. 5. The response is a numerical variable. The Null and Alternative Hypotheses The null hypothesis is simply that all the group population means are the same. The alternative hypothesis is that at least one pair of means is different. For example, if there are k groups: H0: μ1 = μ2 = μ3 = ... = μk Ha: At least two of the group means μ1, μ2, μ3, ..., μk are not equal. The graphs, a set of box plots representing the distribution of values with the group means indicated by a horizontal line through the box, help in the understanding of the hypothesis test. In the first graph (red box plots), H0: μ1 = μ2 = μ3 and the three populations have the same distribution if the null hypothesis is true. The variance of the combined data is approximately the same as the variance of each of the populations. If the null hypothesis is false, then the variance of the combined data is larger which is caused by the different means as shown in the second graph (green box plots). Figure 13.2 (a) H0 is true. All means are the same; the differences are due to random variation. (b) H0 is not true. All means are not the same; the differences are too large to be due to random variation. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 695 13.2 \| The F Distribution and the F-Ratio The distribution used for the hypothesis test is a new one. It is called the F distribution, named after Sir Ronald Fisher, an English statistician. The F statistic is a ratio (a fraction). There are two sets of degrees of freedom; one for the numerator and one for the denominator. For example, if F follows an F distribution and the number of degrees of freedom for the numerator is four, and the number of degrees of freedom for the denominator is ten, then F ~ F4,10. NOTE The F distribution is derived from the Student's t-distribution. The values of the F distribution are squares of the corresponding values of the t-distribution. One-Way ANOVA expands the t-test for comparing more than two groups. The scope of that derivation is beyond the level of this course. To calculate the F ratio, two estimates of the variance are made. 1. Variance between samples: An estimate of σ2 that is the variance of the sample means multiplied by n (when the sample sizes are the same.). If the samples are different sizes, the variance between samples is weighted to account for the different sample sizes. The variance is also called variation due to treatment or explained variation. 2. Variance within samples: An estimate of σ2 that is the average of the sample variances (also known as a pooled variance). When the sample sizes are different, the variance within samples is weighted. The variance is also called the variation due to error or unexplained variation. • SSbetween = the sum of squares that represents the variation among the different samples • SSwithin = the sum of squares that represents the variation within samples that is due to chance. To find a "sum of squares" means to add together squared quantities that, in some cases, may be weighted. We used sum of squares to calculate the sample variance and the sample standard deviation in Section 2.. MS means " mean square." MSbetween is the variance between groups, and MSwithin is the variance within groups. Calculation of Sum of Squares and Mean Square • k = the number of different groups • nj = the size of the jth group • sj = the sum of the values in the jth group • n = total number of all the values combined (total sample size: ∑nj) • x = one value: ∑x = ∑sj • Sum of squares of all values from every group combined: ∑x2 • Between group variability: SStotal = ∑x2 – ⎝∑ x2⎞ ⎛ ⎠ n • Total sum of squares: ∑x2 – 2 ⎛ ⎝∑ x⎞ n ⎠ • Explained variation: (∑ s j)2 ⎡ ⎢(sj)2 n n j ⎣ ⎤ ⎥ − ⎦ ∑ sum of squares representing variation among the different samples: SSbetween = • Unexplained variation: sum of squares representing variation within samples due to chance: SSwithin = SStotal – SSbetween • df's for different groups (df's for the numerator): df = k – 1 • Equation for errors within samples (df's for the denominator): dfwithin = n – k • Mean square (variance estimate) explained by the different groups: MSbetween = SSbetween d fbetween 696 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA • Mean square (variance estimate) that is due to chance (unexplained): MSwithin = SSwithin d fwithin MSbetween and MSwithin can be written as follows: • MSbetween = SSbetween d fbetween = SSbetween k − 1 • MSwithin = SSwithin d fwithin = SSwithin n − k The one-way ANOVA test depends on the fact that MSbetween can be influenced by population differences among means of the several groups. Since MSwithin compares values of each group to its own group mean, the fact that group means might be different does not affect MSwithin. The null hypothesis says that all groups are samples from populations having the same normal distribution. The alternate hypothesis says that at least two of the sample groups come from populations with different normal distributions. If the null hypothesis is true, MSbetween and MSwithin should both estimate the same value. NOTE The null hypothesis says that all the group population means are equal. The hypothesis of equal means implies that the populations have the same normal distribution, because it is assumed that the populations are normal and that they have equal variances. F-Ratio or F Statistic F = MSbetween MSwithin If MSbetween and MSwithin estimate the same value (following the belief that H0 is true), then the F-ratio should be approximately equal to one. Mostly, just sampling errors would contribute to variations away from one. As it turns out, MSbetween consists of the population variance plus a variance produced from the differences between the samples. MSwithin is an estimate of the population variance. Since variances are always positive, if the null hypothesis is false, MSbetween will generally be larger than MSwithin.Then the F-ratio will be larger than one. However, if the population effect is small, it is not unlikely that MSwithin will be larger in a given sample. The foregoing calculations were done with groups of different sizes. If the groups are the same size, the calculations simplify somewhat and the F-ratio can be written as: F-Ratio Formula when the groups are the same size F = 2 n ⋅ s x¯ s2 pooled where ... • n = the sample size • dfnumerator = k – 1 • dfdenominator = n – k • • s2 pooled = the mean of the sample variances (pooled variance) 2 = the variance of the sample means s x¯ Data are typically put into a table for easy viewing. One-Way ANOVA results are often displayed in this manner by computer software. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 697 SS(Factor) SS(Error) SS(Total MS(Factor) = SS(Factor)/(k – 1) F = MS(Factor)/MS(Error) MS(Error) = SS(Error)/(n – k) Factor (Between) Error (Within) Total Table 13.1 Example 13.1 Three different diet plans are to be tested for mean weight loss. The entries in the table are the weight losses for the different plans. The one-way ANOVA results are shown in Table 13.2. Plan 1: n1 = 4 Plan 2: n2 = 3 Plan 3: n3 = 3 5 4.5 4 3 Table 13.2 3.5 7 4.5 8 4 3.5 s1 = 16.5, s2 =15, s3 = 15.7 Following are the calculations needed to fill in the one-way ANOVA table. The table is used to conduct a hypothesis test. SS(between) = ∑ (s j) ⎞ ⎠ ⎛ ⎝∑ s j n = 2 s1 4 2 s2 3 + + 2 s3 3 (s1 + s2 + s3)2 10 − where n1 = 4, n2 = 3, n3 = 3 and n = n1 + n2 + n3 = 10 = (16.5)2 4 + (15)2 3 + (5.5)2 3 − (16.5 + 15 + 15.5)2 10 SS(between) = 2.2458 ⎝∑ x⎞ n S(total) = ∑ x2 − ⎛ ⎠ 2 = ⎝52 + 4.52 + 42 + 32 + 3.52 + 72 + 4.52 + 82 + 42 + 3.52⎞ ⎛ ⎠ (5 + 4.5 + 4 + 3 + 3.5 + 7 + 4.5 + 8 + 4 + 3.5)2 10 − = 244 − 472 10 = 244 − 220.9 SS(total) = 23.1 SS(within) = SS(total) − SS(between) = 23.1 − 2.2458 SS(within) = 20.8542 698 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA One-Way ANOVA Table: The formulas for SS(Total), SS(Factor) = SS(Between) and SS(Error) = SS(Within) as shown previously. The same information is provided by the TI calculator hypothesis test function ANOVA in STAT TESTS (syntax is ANOVA(L1, L2, L3) where L1, L2, L3 have the data from Plan 1, Plan 2, Plan 3 respectively). Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (Between) SS(Factor) = SS(Between) = 2.2458 k – 1 = 3 groups – 1 = 2 SS(Error) = SS(Within) = 20.8542 SS(Total) = 2.2458 + 20.8542 = 23.1 n – k = 10 total data – 3 groups = 7 n – 1 = 10 total data – 1 = 9 Error (Within) Total Table 13.3 F = MS(Factor)/MS(Error) = 1.1229/2.9792
= 0.3769 MS(Factor) = SS(Factor)/(k – 1) = 2.2458/2 = 1.1229 MS(Error) = SS(Error)/(n – k) = 20.8542/7 = 2.9792 13.1 As part of an experiment to see how different types of soil cover would affect slicing tomato production, Marist College students grew tomato plants under different soil cover conditions. Groups of three plants each had one of the following treatments • bare soil • a commercial ground cover • black plastic • straw • compost All plants grew under the same conditions and were the same variety. Students recorded the weight (in grams) of tomatoes produced by each of the n = 15 plants: Bare: n1 = 3 Ground Cover: n2 = 3 Plastic: n3 = 3 Straw: n4 = 3 Compost: n5 = 3 6,583 8,560 3,830 7,285 6,897 9,230 6,277 7,818 8,677 2,625 2,997 4,915 Table 13.4 5,348 5,682 5,482 Create the one-way ANOVA table. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 The one-way ANOVA hypothesis test is always right-tailed because larger F-values are way out in the right tail of the F-distribution curve and tend to make us reject H0. CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 699 Notation The notation for the F distribution is F ~ Fdf(num),df(denom) where df(num) = dfbetween and df(denom) = dfwithin The mean for the F distribution is µ = d f (num) d f (denom) – 1 13.3 \| Facts About the F Distribution Here are some facts about the F distribution. 1. The curve is not symmetrical but skewed to the right. 2. There is a different curve for each set of dfs. 3. The F statistic is greater than or equal to zero. 4. As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal. 5. Other uses for the F distribution include comparing two variances and two-way Analysis of Variance. Two-Way Analysis is beyond the scope of this chapter. Figure 13.3 Example 13.2 Let’s return to the slicing tomato exercise in Try It. The means of the tomato yields under the five mulching conditions are represented by μ1, μ2, μ3, μ4, μ5. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5%, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest. Solution 13.2 The null and alternative hypotheses are: H0: μ1 = μ2 = μ3 = μ4 = μ5 Ha: μi ≠ μj some i ≠ j The one-way ANOVA results are shown in Table 13.4 700 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F 36,648,561 4 20,446,726 10 = 9,162,140 9,162,140 2,044,672.6 = 4.4810 = 2,044,672.6 Source of Variation Factor (Between) 36,648,561 5 – 1 = 4 Error (Within) 20,446,726 15 – 5 = 10 Total 57,095,287 15 – 1 = 14 Table 13.5 Distribution for the test: F4,10 df(num) = 5 – 1 = 4 df(denom) = 15 – 5 = 10 Test statistic: F = 4.4810 Figure 13.4 Probability Statement: p-value = P(F > 4.481) = 0.0248. Compare α and the p-value: α = 0.05, p-value = 0.0248 Make a decision: Since α > p-value, we reject H0. Conclusion: At the 5% significance level, we have reasonably strong evidence that differences in mean yields for slicing tomato plants grown under different mulching conditions are unlikely to be due to chance alone. We may conclude that at least some of mulches led to different mean yields. To find these results on the calculator: Press STAT. Press 1:EDIT. Put the data into the lists L1, L2, L3, L4, L5. Press STAT, and arrow over to TESTS, and arrow down to ANOVA. Press ENTER, and then enter L1, L2, L3, L4, L5). Press ENTER. You will see that the values in the foregoing ANOVA table are easily produced by the calculator, including the test statistic and the p-value of the test. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 701 The calculator displays: F = 4.4810 p = 0.0248 (p-value) Factor df = 4 SS = 36648560.9 MS = 9162140.23 Error df = 10 SS = 20446726 MS = 2044672.6 13.2 MRSA, or Staphylococcus aureus, can cause a serious bacterial infections in hospital patients. Table 13.6 shows various colony counts from different patients who may or may not have MRSA. Conc = 0.6 Conc = 0.8 Conc = 1.0 Conc = 1.2 Conc = 1.4 9 66 98 Table 13.6 16 93 82 22 147 120 30 199 148 27 168 132 Plot of the data for the different concentrations: Figure 13.5 Test whether the mean number of colonies are the same or are different. Construct the ANOVA table (by hand or by using a TI-83, 83+, or 84+ calculator), find the p-value, and state your conclusion. Use a 5% significance level. 702 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Example 13.3 Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table 13.7. Sorority 1 Sorority 2 Sorority 3 Sorority 4 2.17 1.85 2.83 1.69 3.33 2.63 1.77 3.25 1.86 2.21 2.63 3.78 4.00 2.55 2.45 3.79 3.45 3.08 2.26 3.18 Table 13.7 MEAN GRADES FOR FOUR SORORITIES Using a significance level of 1%, is there a difference in mean grades among the sororities? Solution 13.3 Let μ1, μ2, μ3, μ4 be the population means of the sororities. Remember that the null hypothesis claims that the sorority groups are from the same normal distribution. The alternate hypothesis says that at least two of the sorority groups come from populations with different normal distributions. Notice that the four sample sizes are each five. NOTE This is an example of a balanced design, because each factor (i.e., sorority) has the same number of observations. H0: μ1 = μ2 = μ3 = μ4 Ha: Not all of the means μ1, μ2, μ3, μ4 are equal. Distribution for the test: F3,16 where k = 4 groups and n = 20 samples in total df(num)= k – 1 = 4 – 1 = 3 df(denom) = n – k = 20 – 4 = 16 Calculate the test statistic: F = 2.23 Graph: This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 703 Figure 13.6 Probability statement: p-value = P(F > 2.23) = 0.1241 Compare α and the p-value: α = 0.01 p-value = 0.1241 α < p-value Make a decision: Since α < p-value, you cannot reject H0. Conclusion: There is not sufficient evidence to conclude that there is a difference among the mean grades for the sororities. Put the data into lists L1, L2, L3, and L4. Press STAT and arrow over to TESTS. Arrow down to F:ANOVA. Press ENTER and Enter (L1,L2,L3,L4). The calculator displays the F statistic, the p-value and the values for the one-way ANOVA table: F = 2.2303 p = 0.1241 (p-value) Factor df = 3 SS = 2.88732 MS = 0.96244 Error df = 16 SS = 6.9044 MS = 0.431525 704 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 13.3 Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table 13.8. Basketball Baseball Hockey Lacrosse 3.6 2.9 2.5 3.3 3.8 2.1 2.6 3.9 3.1 3.4 4.0 2.0 2.6 3.2 3.2 2.0 3.6 3.9 2.7 2.5 Table 13.8 GPAs FOR FOUR SPORTS TEAMS Use a significance level of 5%, and determine if there is a difference in GPA among the teams. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 705 Example 13.4 A fourth grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother's garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data (in inches) in Table 13.9. Tommy's Plants Tara's Plants Nick's Plants 24 21 23 30 23 Table 13.9 25 31 23 20 28 23 27 22 30 20 Does it appear that the three media in which the bean plants were grown produce the same mean height? Test at a 3% level of significance. Solution 13.4 This time, we will perform the calculations that lead to the F' statistic. Notice that each group has the same number of plants, so we will use the formula F' = 2 . n ⋅ s x¯ s2 pooled First, calculate the sample mean and sample variance of each group. Tommy's Plants Tara's Plants Nick's Plants 24.2 11.7 25.4 18.3 24.4 16.3 Sample Mean Sample Variance Table 13.10 Next, calculate the variance of the three group means (Calculate the variance of 24.2, 25.4, and 24.4). Variance of the group means = 0.413 = s x¯ 2 Then MSbetween = ns x¯ 2 = (5)(0.413) where n = 5 is the sample size (number of plants each child grew). Calculate the mean of the three sample variances (Calculate the mean of 11.7, 18.3, and 16.3). Mean of the sample variances = 15.433 = s2 pooled Then MSwithin = s2 pooled = 15.433. The F statistic (or F ratio) is F = MSbetween MSwithin = 2 ns x¯ s2 pooled = (5)(0.413) 15.433 = 0.134 The dfs for the numerator = the number of groups – 1 = 3 – 1 = 2. The dfs for the denominator = the total number of samples – the number of groups = 15 – 3 = 12 The distribution for the test is F2,12 and the F statistic is F = 0.134 The p-value is P(F > 0.134) = 0.8759. 706 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Decision: Since α = 0.03 and the p-value = 0.8759, do not reject H0. (Why?) Conclusion: With a 3% level of significance, from the sample data, the evidence is not sufficient to conclude that the mean heights of the bean plants are different. To calculate the p-value: Press 2nd DISTR Arrow down to Fcdf(and press ENTER. Enter 0.134, E99, 2, 12) Press ENTER The p-value is 0.8759. 13.4 Another fourth grader also grew bean plants, but this time in a jelly-like mass. The heights were (
in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 13.4. From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1% level of significance. Use one of the solution sheets in Appendix E. 13.4 \| Test of Two Variances Another of the uses of the F distribution is testing two variances. It is often desirable to compare two variances rather than two averages. For instance, college administrators would like two college professors grading exams to have the same variation in their grading. In order for a lid to fit a container, the variation in the lid and the container should be the same. A supermarket might be interested in the variability of check-out times for two checkers. In order to perform a F test of two variances, it is important that the following are true: 1. The populations from which the two samples are drawn are normally distributed. 2. The two populations are independent of each other. Unlike most other tests in this book, the F test for equality of two variances is very sensitive to deviations from normality. If the two distributions are not normal, the test can give higher p-values than it should, or lower ones, in ways that are unpredictable. Many texts suggest that students not use this test at all, but in the interest of completeness we include it here. Suppose we sample randomly from two independent normal populations. Let σ1 2 and s2 s1 variances, we use the F ratio: 2 be the sample variances. Let the sample sizes be n1 and n2. Since we are interested in comparing the two sample 2 and σ2 2 be the population variances and This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 707 F = ⎡ ⎢ ⎣ ⎡ ⎢ ⎣ (s1)2 (σ1)2 (s2)2 (σ2)2 ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ F has the distribution F ~ F(n1 – 1, n2 – 1) where n1 – 1 are the degrees of freedom for the numerator and n2 – 1 are the degrees of freedom for the denominator. If the null hypothesis is σ1 2 = σ2 2 , then the F Ratio becomes F = ⎡ ⎢ ⎣ ⎡ ⎢ ⎣ (s1)2 (σ1)2 (s2)2 (σ2)2 ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ = (s1)2 (s2)2 . NOTE The F ratio could also be (s2)2 (s1)2 . It depends on Ha and on which sample variance is larger. If the two populations have equal variances, then s1 2 and s2 2 are close in value and F = (s1)2 (s2)2 is close to one. But if the two population variances are very different, s1 variance causes the ratio (s1)2 (s2)2 2 and s2 2 tend to be very different, too. Choosing s1 2 as the larger sample to be greater than one. If s1 2 and s2 2 are far apart, then F = (s1)2 (s2)2 is a large number. Therefore, if F is close to one, the evidence favors the null hypothesis (the two population variances are equal). But if F is much larger than one, then the evidence is against the null hypothesis. A test of two variances may be left, right, or two-tailed. Example 13.5 Two college instructors are interested in whether or not there is any variation in the way they grade math exams. They each grade the same set of 30 exams. The first instructor's grades have a variance of 52.3. The second instructor's grades have a variance of 89.9. Test the claim that the first instructor's variance is smaller. (In most colleges, it is desirable for the variances of exam grades to be nearly the same among instructors.) The level of significance is 10%. Solution 13.5 Let 1 and 2 be the subscripts that indicate the first and second instructor, respectively. n1 = n2 = 30. H0: σ1 2 = σ2 2 and Ha: σ1 2 2 < σ2 Calculate the test statistic: By the null hypothesis (σ1 2 = σ2 2 ) , the F statistic iss1)2 (σ1)2 (s2)2 (σ2)2 ⎤ ⎥ ⎦ ⎤ ⎥ ⎦ = (s1)2 (s2)2 = 52.3 89.9 = 0.5818 Distribution for the test: F29,29 where n1 – 1 = 29 and n2 – 1 = 29. Graph: This test is left tailed. 708 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Draw the graph labeling and shading appropriately. Figure 13.7 Probability statement: p-value = P(F < 0.5818) = 0.0753 Compare α and the p-value: α = 0.10 α > p-value. Make a decision: Since α > p-value, reject H0. Conclusion: With a 10% level of significance, from the data, there is sufficient evidence to conclude that the variance in grades for the first instructor is smaller. Press STAT and arrow over to TESTS. Arrow down to D:2-SampFTest. Press ENTER. Arrow to Stats (89.9) , and 30. Press ENTER after and press ENTER. For Sx1, n1, Sx2, and n2, enter each. Arrow to σ1: and < σ2. Press ENTER. Arrow down to Calculate and press ENTER. F = 0.5818 and p-value = 0.0753. Do the procedure again and try Draw instead of Calculate. (52.3) , 30, This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 709 13.5 The New York Choral Society divides male singers up into four categories from highest voices to lowest: Tenor1, Tenor2, Bass1, Bass2. In the table are heights of the men in the Tenor1 and Bass2 groups. One suspects that taller men will have lower voices, and that the variance of height may go up with the lower voices as well. Do we have good evidence that the variance of the heights of singers in each of these two groups (Tenor1 and Bass2) are different? Tenor1 Bass2 Tenor 1 Bass 2 Tenor 1 Bass 2 69 72 71 66 76 74 71 66 68 72 75 67 75 74 72 72 74 72 Table 13.11 67 70 65 72 70 68 64 73 66 72 74 70 66 68 75 68 70 72 68 67 64 67 70 70 69 72 71 74 75 710 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 13.1 One-Way ANOVA Class Time: Names: Student Learning Outcome • The student will conduct a simple one-way ANOVA test involving three variables. Collect the Data 1. Record the price per pound of eight fruits, eight vegetables, and eight breads in your local supermarket. Fruits Vegetables Breads Table 13.12 2. Explain how you could try to collect the data randomly. Analyze the Data and Conduct a Hypothesis Test 1. Compute the following: a. Fruit: i. ii. x¯ = ______ s x = ______ iii. n = ______ b. Vegetables: i. ii. x¯ = ______ s x = ______ iii. n = ______ c. Bread: i. ii. x¯ = ______ s x = ______ iii. n = ______ 2. Find the following: a. df(num) = ______ b. df(denom) = ______ This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 3. State the approximate distribution for the test. 4. Test statistic: F = ______ 5. Sketch a graph of this situation. CLEARLY, label and scale the horizontal axis and shade the region(s) CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 711 corresponding to the p-value. 6. p-value = ______ 7. Test at α = 0.05. State your decision and conclusion. 8. a. Decision: Why did you make this decision? b. Conclusion (write a complete sentence). c. Based on the results of your study, is there a need to investigate any of the food groups’ prices? Why or why not? 712 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA KEY TERMS Analysis of Variance also referred to as ANOVA, is a method of testing whether or not the means of three or more populations are equal. The method is applicable if: • all populations of interest are normally distributed. • • the populations have equal standard deviations. samples (not necessarily of the same size) are randomly and independently selected from each population. The test statistic for analysis of variance is the F-ratio. One-Way ANOVA a method of testing whether or not the means of three or more populations are equal; the method is applicable if: • all populations of interest are normally distributed. • • the populations have equal standard deviations. samples (not necessarily of the same size) are randomly and independently selected from each population. The test statistic for analysis of variance is the F-ratio. Variance mean of the squared deviations from the mean; the square of the standard deviation. For a set of data, a deviation can be represented as x – x¯ where x is a value of the data and x¯ variance is equal to the sum of the squares of the deviations divided by the difference of the sample size and one. is the sample mean. The sample CHAPTER REVIEW 13.1 One-Way ANOVA Analysis of variance extends the comparison of two groups to several, each a level of a categorical variable (factor). Samples from each group are independent, and must be randomly selected from normal populations with equal variances. We test the null hypothesis of equal means of the response in every group versus the alternative hypothesis of one or more group means being different from the others. A one-way ANOVA hypothesis test determines if several population means are equal. The distribution for the test is the F distribution with two different degrees of freedom. Assumptions: 1. Each population from which a sample is taken is assumed to be normal. 2. All samples are randomly selected and independent. 3. The populations are assumed to have equal standard deviations (or variances). 13.2 The F Distribution and the F-Ratio Analysis of variance compares the means of a response variable for several groups. ANOVA compares the variation within each group to the variation of the mean of each group. The ratio of these two is the F statistic from an F distribution with (number of groups – 1) as the numerator degrees of freedom and (number of observations – number of groups) as the denominator degrees of freedom. These statistics are summarized in the ANOVA table. 13.3 Facts About the F Distribution The graph of the F distribution is always positive and skewed right, though the shape can be mounded or exponential depending on the combination of numerator and denominator degrees of freedom. The F statistic
is the ratio of a measure of the variation in the group means to a similar measure of the variation within the groups. If the null hypothesis is correct, then the numerator should be small compared to the denominator. A small F statistic will result, and the area under the F curve to the right will be large, representing a large p-value. When the null hypothesis of equal group means is incorrect, then the numerator should be large compared to the denominator, giving a large F statistic and a small area (small p-value) to the right of the statistic under the F curve. When the data have unequal group sizes (unbalanced data), then techniques from Section 13.2 need to be used for hand calculations. In the case of balanced data (the groups are the same size) however, simplified calculations based on group This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 713 means and variances may be used. In practice, of course, software is usually employed in the analysis. As in any analysis, graphs of various sorts should be used in conjunction with numerical techniques. Always look of your data! 13.4 Test of Two Variances The F test for the equality of two variances rests heavily on the assumption of normal distributions. The test is unreliable if this assumption is not met. If both distributions are normal, then the ratio of the two sample variances is distributed as an F statistic, with numerator and denominator degrees of freedom that are one less than the samples sizes of the corresponding two groups. A test of two variances hypothesis test determines if two variances are the same. The distribution for the hypothesis test is the F distribution with two different degrees of freedom. Assumptions: 1. The populations from which the two samples are drawn are normally distributed. 2. The two populations are independent of each other. FORMULA REVIEW 13.2 The F Distribution and the F-Ratio SSbetween = ∑ (s j) ⎞ ⎠ ⎛ ⎝∑ s j n SStotal = ∑ x2 − 2 ⎛ ⎝∑ x⎞ ⎠ n SSwithin = SStotal − SSbetween dfbetween = df(num) = k – 1 dfwithin = df(denom) = n – k MSbetween = SSbetween d fbetween MSwithin = SSwithin d fwithin F = MSbetween MSwithin F ratio when the groups are the same size: F = Mean of the F distribution: µ = d f (num) d f (denom) − 1 PRACTICE 13.1 One-Way ANOVA where: • k = the number of groups • nj = the size of the jth group • sj = the sum of the values in the jth group • n = the total number of all values (observations) combined • x = one value (one observation) from the data • • s x¯ s2 2 = the variance of the sample means pooled = the mean of the sample variances (pooled variance) 13.4 Test of Two Variances F has the distribution F ~ F(n1 – 1, n2 – 1 ns x¯ s2 pooled If σ1 = σ2, then F = 2 2 s1 s2 Use the following information to answer the next five exercises. There are five basic assumptions that must be fulfilled in order to perform a one-way ANOVA test. What are they? 1. Write one assumption. 2. Write another assumption. 3. Write a third assumption. 4. Write a fourth assumption. 714 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 5. Write the final assumption. 6. State the null hypothesis for a one-way ANOVA test if there are four groups. 7. State the alternative hypothesis for a one-way ANOVA test if there are three groups. 8. When do you use an ANOVA test? 13.2 The F Distribution and the F-Ratio Use the following information to answer the next eight exercises. Groups of men from three different areas of the country are to be tested for mean weight. The entries in the table are the weights for the different groups. The one-way ANOVA results are shown in Table 13.13. Group 1 Group 2 Group 3 216 198 240 187 176 202 213 284 228 210 Table 13.13 170 165 182 197 201 9. What is the Sum of Squares Factor? 10. What is the Sum of Squares Error? 11. What is the df for the numerator? 12. What is the df for the denominator? 13. What is the Mean Square Factor? 14. What is the Mean Square Error? 15. What is the F statistic? Use the following information to answer the next eight exercises. Girls from four different soccer teams are to be tested for mean goals scored per game. The entries in the table are the goals per game for the different teams. The one-way ANOVA results are shown in Table 13.14. Team 1 Team 2 Team 3 Team Table 13.14 16. What is SSbetween? 17. What is the df for the numerator? 18. What is MSbetween? 19. What is SSwithin? 20. What is the df for the denominator? 21. What is MSwithin? 22. What is the F statistic? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 23. Judging by the F statistic, do you think it is likely or unlikely that you will reject the null hypothesis? CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 715 13.3 Facts About the F Distribution 24. An F statistic can have what values? 25. What happens to the curves as the degrees of freedom for the numerator and the denominator get larger? Use the following information to answer the next seven exercise. Four basketball teams took a random sample of players regarding how high each player can jump (in inches). The results are shown in Table 13.15. Team 1 Team 2 Team 3 Team 4 Team 5 36 42 51 32 35 38 48 50 39 38 44 46 41 39 40 Table 13.15 26. What is the df(num)? 27. What is the df(denom)? 28. What are the Sum of Squares and Mean Squares Factors? 29. What are the Sum of Squares and Mean Squares Errors? 30. What is the F statistic? 31. What is the p-value? 32. At the 5% significance level, is there a difference in the mean jump heights among the teams? Use the following information to answer the next seven exercises. A video game developer is testing a new game on three different groups. Each group represents a different target market for the game. The developer collects scores from a random sample from each group. The results are shown in Table 13.16 Group A Group B Group C 101 108 98 107 111 151 149 160 112 126 101 109 198 186 160 Table 13.16 33. What is the df(num)? 34. What is the df(denom)? 35. What are the SSbetween and MSbetween? 36. What are the SSwithin and MSwithin? 37. What is the F Statistic? 38. What is the p-value? 39. At the 10% significance level, are the scores among the different groups different? Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. 716 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 x¯ = ________ s2 = ________ Table 13.17 ________ ________ ________ ________ ________ ________ ________ ________ Enter the data into your calculator or computer. 40. p-value = ______ State the decisions and conclusions (in complete sentences) for the following preconceived levels of α. 41. α = 0.05 a. Decision: ____________________________ b. Conclusion: ____________________________ 42. α = 0.01 a. Decision: ____________________________ b. Conclusion: ____________________________ 13.4 Test of Two Variances Use the following information to answer the next two exercises. There are two assumptions that must be true in order to perform an F test of two variances. 43. Name one assumption that must be true. 44. What is the other assumption that must be true? Use the following information to answer the next five exercises. Two coworkers commute from the same building. They are interested in whether or not there is any variation in the time it takes them to drive to work. They each record their times for 20 commutes. The first worker’s times have a variance of 12.1. The second worker’s times have a variance of 16.9. The first worker thinks that he is more consistent with his commute times and that his commute time is shorter. Test the claim at the 10% level. 45. State the null and alternative hypotheses. 46. What is s1 in this problem? 47. What is s2 in this problem? 48. What is n? 49. What is the F statistic? 50. What is the p-value? 51. Is the claim accurate? Use the following information to answer the next four exercises. Two students are interested in whether or not there is variation in their test scores for math class. There are 15 total math tests they have taken so far. The first student’s grades have a standard deviation of 38.1. The second student’s grades have a standard deviation of 22.5. The second student thinks his scores are lower. 52. State the null and alternative hypotheses. 53. What is the F Statistic? 54. What is the p-value? 55. At the 5% significance level, do we reject the null hypothesis? This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 717 Use the following information to answer the next three exercises. Two cyclists are comparing the variances of their overall paces going uphill. Each cyclist records his or her speeds going up 35 hills. The first cyclist has a variance of 23.8 and the second cyclist has a variance of 32.1. The cyclists want to see if their variances are the same or different. 56. State the null and alternative hypotheses. 57. What is the F Statistic? 58. At the 5% significance level, what can we say about the cyclists’ variances? HOMEWORK 13.1 One-Way ANOVA 59. Three different traffic routes are tested for mean driving time. The entries in the table are the driving times in minutes on the three different routes. The one-way ANOVA results are shown in Table 13.18. Route 1 Route 2 Route 3 30 32 27 35 27 29 28 36 Table 13
.18 16 41 22 31 State SSbetween, SSwithin, and the F statistic. 60. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 x¯ = ________ s2 = ________ Table 13.19 ________ ________ ________ ________ ________ ________ ________ ________ State the hypotheses. H0: ____________ Ha: ____________ 13.2 The F Distribution and the F-Ratio Use the following information to answer the next three exercises. Suppose a group is interested in determining whether teenagers obtain their drivers licenses at approximately the same average age across the country. Suppose that the following data are randomly collected from five teenagers in each region of the country. The numbers represent the age at which teenagers obtained their drivers licenses. 718 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Northeast South West Central East 16.3 16.1 16.4 16.5 16.9 16.5 16.4 16.2 16.4 16.5 16.6 16.1 16.2 16.6 16.5 16.4 17.1 17.2 16.6 16.8 x¯ = ________ s2 = ________ Table 13.20 ________ ________ ________ ________ ________ ________ ________ ________ H0: µ1 = µ2 = µ3 = µ4 = µ5 Hα: At least any two of the group means µ1, µ2, …, µ5 are not equal. 61. degrees of freedom – numerator: df(num) = _________ 62. degrees of freedom – denominator: df(denom) = ________ 63. F statistic = ________ 13.3 Facts About the F Distribution DIRECTIONS Use a solution sheet to conduct the following hypothesis tests. The solution sheet can be found in Appendix E. 64. Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat's weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again, and the net gain in grams is recorded. Using a significance level of 10%, test the hypothesis that the three formulas produce the same mean weight gain. Linda's rats Tuan's rats Javier's rats 43.5 39.4 41.3 46.0 38.2 47.0 40.5 38.9 46.3 44.2 51.2 40.9 37.9 45.0 48.6 Table 13.21 Weights of Student Lab Rats 65. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are in Table 13.22. Using a 5% significance level, test the hypothesis that the three mean commuting mileages are the same. working-class professional (middle incomes) professional (wealthy) 17.8 26.7 49.4 Table 13.22 16.5 17.4 22.0 8.5 6.3 4.6 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 working-class professional (middle incomes) professional (wealthy) CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 719 9.4 65.4 47.1 19.5 51.2 Table 13.22 7.4 9.4 2.1 6.4 13.9 12.6 11.0 28.6 15.4 9.3 66. Examine the seven practice laps from Table 13.1. Determine whether the mean lap time is statistically the same for the seven practice laps, or if there is at least one lap that has a different mean time from the others. Use the following information to answer the next two exercises. Table 13.23 lists the number of pages in four different types of magazines. home decorating news health computer 172 286 163 205 197 Table 13.23 87 94 123 106 101 82 153 87 103 96 104 136 98 207 146 67. Using a significance level of 5%, test the hypothesis that the four magazine types have the same mean length. 68. Eliminate one magazine type that you now feel has a mean length different from the others. Redo the hypothesis test, testing that the remaining three means are statistically the same. Use a new solution sheet. Based on this test, are the mean lengths for the remaining three magazines statistically the same? 69. A researcher wants to know if the mean times (in minutes) that people watch their favorite news station are the same. Suppose that Table 13.24 shows the results of a study. CNN FOX Local 72 37 56 60 51 45 12 18 38 23 35 15 43 68 50 31 22 Table 13.24 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 70. Are the means for the final exams the same for all statistics class delivery types? Table 13.25 shows the scores on final exams from several randomly selected classes that used the different delivery types. 720 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Online Hybrid Face-to-Face 83 73 84 81 72 84 77 80 81 Table 13.25 80 78 84 81 86 79 82 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 71. Are the mean number of times a month a person eats out the same for whites, blacks, Hispanics and Asians? Suppose that Table 13.26 shows the results of a study. White Black Hispanic Asian 4 1 5 2 6 8 2 4 6 Table 13.26 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 72. Are the mean numbers of daily visitors to a ski resort the same for the three types of snow conditions? Suppose that Table 13.27 shows the results of a study. Powder Machine Made Hard Packed 1,210 1,080 1,537 941 Table 13.27 2,107 1,149 862 1,870 1,528 1,382 2,846 1,638 2,019 1,178 2,233 Assume that all distributions are normal, the four population standard deviations are approximately the same, and the data were collected independently and randomly. Use a level of significance of 0.05. 73. Sanjay made identical paper airplanes out of three different weights of paper, light, medium and heavy. He made four airplanes from each of the weights, and launched them himself across the room. Here are the distances (in meters) that his planes flew. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 721 Paper Type/Trial Trial 1 Trial 2 Trial 3 Trial 4 Heavy Medium Light Table 13.28 5.1 meters 3.1 meters 4.7 meters 5.3 meters 4 meters 3.5 meters 4.5 meters 6.1 meters 3.1 meters 3.3 meters 2.1 meters 1.9 meters Figure 13.8 a. Take a look at the data in the graph. Look at the spread of data for each group (light, medium, heavy). Does it seem reasonable to assume a normal distribution with the same variance for each group? Yes or No. b. Why is this a balanced design? c. Calculate the sample mean and sample standard deviation for each group. d. Does the weight of the paper have an effect on how far the plane will travel? Use a 1% level of significance. Complete the test using the method shown in the bean plant example in Example 13.4. ◦ variance of the group means __________ ◦ MSbetween= ___________ ◦ mean of the three sample variances ___________ ◦ MSwithin = _____________ ◦ F statistic = ____________ ◦ df(num) = __________, df(denom) = ___________ ◦ number of groups _______ ◦ number of observations _______ ◦ p-value = __________ (P(F > _______) = __________) ◦ Graph the p-value. ◦ decision: _______________________ ◦ conclusion: _______________________________________________________________ 74. DDT is a pesticide that has been banned from use in the United States and most other areas of the world. It is quite effective, but persisted in the environment and over time became seen as harmful to higher-level organisms. Famously, egg shells of eagles and other raptors were believed to be thinner and prone to breakage in the nest because of ingestion of DDT in the food chain of the birds. An experiment was conducted on the number of eggs (fecundity) laid by female fruit flies. There are three groups of flies. One group was bred to be resistant to DDT (the RS group). Another was bred to be especially susceptible to DDT (SS). Finally there was a control line of non-selected or typical fruitflies (NS). Here are the data: RS SS NS RS SS NS 12.8 38.4 35.4 22.4 23.1 22.6 Table 13.29 722 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA RS SS NS RS SS NS 21.6 32.9 27.4 27.5 29.4 40.4 14.8 48.5 19.3 20.3 16 34.4 23.1 20.9 41.8 38.7 20.1 30.4 34.6 11.6 20.3 26.4 23.3 14.9 19.7 22.3 37.6 23.7 22.9 51.8 22.6 30.2 36.9 26.1 22.5 33.8 29.6 33.4 37.3 29.5 15.1 37.9 16.4 26.7 28.2 38.6 31 29.5 20.3 39 23.4 44.4 16.9 42.4 29.3 12.8 33.7 23.2 16.1 36.6 14.9 14.6 29.2 23.6 10.8 47.4 27.3 12.2 41.7 Table 13.29 The values are the average number of eggs laid daily for each of 75 flies (25 in each group) over the first 14 days of their lives. Using a 1% level of significance, are the mean rates of egg selection for the three strains of fruitfly different? If so, in what way? Specifically, the researchers were interested in whether or not the selectively bred strains were different from the nonselected line, and whether the two selected lines were different from each other. Here is a chart of the three groups: Figure 13.9 75. The data shown is the recorded body temperatures of 130 subjects as estimated from available histograms. Traditionally we are taught that the normal human body temperature is 98.6 F. This is not quite correct for everyone. Are the mean temperatures among the four groups different? Calculate 95% confidence intervals for the mean body temperature in each group and comment about the confidence intervals. FL FH ML MH FL FH ML MH 9
6.4 96.8 96.3 96.9 98.4 98.6 98.1 98.6 96.7 97.7 96.7 97 98.7 98.6 98.1 98.6 Table 13.30 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 723 FL FH ML MH FL FH ML MH 97.2 97.8 97.1 97.1 98.7 98.6 98.2 98.7 97.2 97.9 97.2 97.1 98.7 98.7 98.2 98.8 97.4 98 97.3 97.4 98.7 98.7 98.2 98.8 97.6 98 97.4 97.5 98.8 98.8 98.2 98.8 97.7 98 97.4 97.6 98.8 98.8 98.3 98.9 97.8 98 97.4 97.7 98.8 98.8 98.4 99 97.8 98.1 97.5 97.8 98.8 98.9 98.4 99 97.9 98.3 97.6 97.9 99.2 99 98.5 99 97.9 98.3 97.6 98 99.3 99 98.5 99.2 98 98.3 97.8 98 99.1 98.6 99.5 98.2 98.4 97.8 98 99.1 98.6 98.2 98.4 97.8 98.3 99.2 98.7 98.2 98.4 97.9 98.4 99.4 99.1 98.2 98.4 98 98.2 98.5 98 98.2 98.6 98 98.4 98.6 98.6 99.9 99.3 100 99.4 100.8 Table 13.30 13.4 Test of Two Variances 76. Three students, Linda, Tuan, and Javier, are given five laboratory rats each for a nutritional experiment. Each rat’s weight is recorded in grams. Linda feeds her rats Formula A, Tuan feeds his rats Formula B, and Javier feeds his rats Formula C. At the end of a specified time period, each rat is weighed again and the net gain in grams is recorded. Linda's rats Tuan's rats Javier's rats 43.5 39.4 41.3 46.0 38.2 Table 13.31 47.0 40.5 38.9 46.3 44.2 51.2 40.9 37.9 45.0 48.6 Determine whether or not the variance in weight gain is statistically the same among Javier’s and Linda’s rats. Test at a significance level of 10%. 77. A grassroots group opposed to a proposed increase in the gas tax claimed that the increase would hurt working-class people the most, since they commute the farthest to work. Suppose that the group randomly surveyed 24 individuals and asked them their daily one-way commuting mileage. The results are as follows. working-class professional (middle incomes) professional (wealthy) 17.8 26.7 Table 13.32 16.5 17.4 8.5 6.3 724 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA working-class professional (middle incomes) professional (wealthy) 49.4 9.4 65.4 47.1 19.5 51.2 Table 13.32 22.0 7.4 9.4 2.1 6.4 13.9 4.6 12.6 11.0 28.6 15.4 9.3 Determine whether or not the variance in mileage driven is statistically the same among the working class and professional (middle income) groups. Use a 5% significance level. 78. Refer to the data from Table 13.1. Examine practice laps 3 and 4. Determine whether or not the variance in lap time is statistically the same for those practice laps. Use the following information to answer the next two exercises. The following table lists the number of pages in four different types of magazines. home decorating news health computer 172 286 163 205 197 Table 13.33 87 94 123 106 101 82 153 87 103 96 104 136 98 207 146 79. Which two magazine types do you think have the same variance in length? 80. Which two magazine types do you think have different variances in length? 81. Is the variance for the amount of money, in dollars, that shoppers spend on Saturdays at the mall the same as the variance for the amount of money that shoppers spend on Sundays at the mall? Suppose that the Table 13.34 shows the results of a study. Saturday Sunday Saturday Sunday 75 18 150 94 62 73 Table 13.34 44 58 61 19 99 60 89 62 0 124 50 31 118 137 82 39 127 141 73 82. Are the variances for incomes on the East Coast and the West Coast the same? Suppose that Table 13.35 shows the results of a study. Income is shown in thousands of dollars. Assume that both distributions are normal. Use a level of significance of 0.05. This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 725 East West 71 126 42 51 44 90 88 38 47 30 82 75 52 115 67 Table 13.35 83. Thirty men in college were taught a method of finger tapping. They were randomly assigned to three groups of ten, with each receiving one of three doses of caffeine: 0 mg, 100 mg, 200 mg. This is approximately the amount in no, one, or two cups of coffee. Two hours after ingesting the caffeine, the men had the rate of finger tapping per minute recorded. The experiment was double blind, so neither the recorders nor the students knew which group they were in. Does caffeine affect the rate of tapping, and if so how? Here are the data: 0 mg 100 mg 200 mg 0 mg 100 mg 200 mg 242 244 247 242 246 248 245 248 247 243 Table 13.36 246 250 248 246 245 245 248 248 244 242 246 247 250 246 244 248 252 250 248 250 84. King Manuel I, Komnenus ruled the Byzantine Empire from Constantinople (Istanbul) during the years 1145 to 1180 A.D. The empire was very powerful during his reign, but declined significantly afterwards. Coins minted during his era were found in Cyprus, an island in the eastern Mediterranean Sea. Nine coins were from his first coinage, seven from the second, four from the third, and seven from a fourth. These spanned most of his reign. We have data on the silver content of the coins: First Coinage Second Coinage Third Coinage Fourth Coinage 6.9 9.0 6.6 8.1 9.3 9.2 8.6 5.9 6.8 6.4 7.0 6.6 7.7 7.2 6.9 6.2 Table 13.37 4.9 5.5 4.6 4.5 5.3 5.6 5.5 5.1 6.2 5.8 5.8 726 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Did the silver content of the coins change over the course of Manuel’s reign? Here are the means and variances of each coinage. The data are unbalanced. First Second Third Fourth Mean 6.7444 8.2429 4.875 5.6143 Variance 0.2953 1.2095 0.2025 0.1314 Table 13.38 85. The American League and the National League of Major League Baseball are each divided into three divisions: East, Central, and West. Many years, fans talk about some divisions being stronger (having better teams) than other divisions. This may have consequences for the postseason. For instance, in 2012 Tampa Bay won 90 games and did not play in the postseason, while Detroit won only 88 and did play in the postseason. This may have been an oddity, but is there good evidence that in the 2012 season, the American League divisions were significantly different in overall records? Use the following data to test whether the mean number of wins per team in the three American League divisions were the same or not. Note that the data are not balanced, as two divisions had five teams, while one had only four. Division Team Wins East East East East East NY Yankees 95 Baltimore 93 Tampa Bay 90 Toronto Boston 73 69 Table 13.39 Division Team Wins Central Detroit 88 Central Chicago Sox 85 Central Kansas City 72 Central Cleveland Central Minnesota 68 66 Table 13.40 Division Team Wins West West West West Oakland Texas 94 93 LA Angels 89 Seattle 75 Table 13.41 REFERENCES 13.2 The F Distribution and the F-Ratio Tomato Data, Marist College School of Science (unpublished student research) This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 727 13.3 Facts About the F Distribution Data from a fourth grade classroom in 1994 in a private K – 12 school in San Jose, CA. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets: Data for Fruitfly Fecundity. London: Chapman & Hall, 1994. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 50. Hand, D.J., F. Daly, A.D. Lunn, K.J. McConway, and E. Ostrowski. A Handbook of Small Datasets. London: Chapman & Hall, 1994, pg. 118. “MLB Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012. Mackowiak, P. A., Wasserman, S. S., and Levine, M. M. (1992), "A Critical Appraisal of 98.6 Degrees F, the Upper Limit of the Normal Body Temperature, and Other Legacies of Carl Reinhold August Wunderlich," Journal of the American Medical Association, 268, 1578-1580. 13.4 Test of Two Variances “MLB Vs. Division Standings – 2012.” Available online at http://espn.go.com/mlb/standings/_/year/2012/type/vs-division/ order/true. SOLUTIONS 1 Each population from which a sample is taken is assumed to be normal. 3 The populations are assumed to have equal standard deviations (or variances). 5 The response is a numerical value. 7 Ha: At least two of the group means μ1, μ2, μ3 are not equal. 9 4,939.2 11 2 13 2,469.6 15 3.7416 17 3 19 13.2 21 0.825 23 Because a one-way ANOVA test is always right-tailed, a high F statistic corresponds to a low p-value, so it is likely that we will reject the null hypothesis. 25 The curves approximate the normal distribution. 27 ten 29 SS = 237.33; MS = 23.73 31 0.1614 33 two 35 SS = 5,700.4; MS = 2,850.2 37 3.6101 39 Yes, there is enough evidence to show that the scores among the groups are statistically significant at the 10% level. 728 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 43 The populations from which the two samples are drawn are normally distributed. 45 H0: σ1 = σ2 Ha: σ1 < σ2 or H0: σ1 2 = σ2 2 Ha: σ1 2 2 < σ2 47 4.11 49 0.7159 51 No, at the 10% level of significance, we do not reject the null hypothesis and state that the data do not show that the variation in drive times for the first worker is less than the variation in drive times for the second worker. 53 2.8674 55 Reject the null hypothesis. There is enough evidence to say that the variance of the grades for the first student is higher than the variance in the grades for the second student. 57 0.7414 59 SSbetween = 26 SSwithin = 441 F = 0.2653 62 df(denom) = 15 64 a. H0: µL = µT = µJ b. at least any two of the means are different c. df(num) = 2; df(denom) = 12 d. F distribution e. 0.67 f. 0.5305 g. Check student’s solution. h. Decision: Do not reject null hypothesis; Conclusion: There is insufficient evidence to conclude that the means are different. 66 a. H0: µ1 = µ2 = µ3 = µ4 = µ5 = µ6 = µ7 b. At least two mean lap times are different. c. df(num) = 6; df(denom) = 98 d. F distribution e. 1.69 f. 0.1319 g. Check student’s solution. h. Decision: Do not reject null hypothesis;
Conclusion: There is insufficient evidence to conclude that the mean lap times are different. 68 a. Ha: µd = µn = µh b. At least any two of the magazines have different mean lengths. c. df(num) = 2, df(denom) = 12 d. F distribtuion e. F = 15.28 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 729 f. p-value = 0.001 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the Null Hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the mean lengths of the magazines are different. 70 a. H0: μo = μh = μf b. At least two of the means are different. c. df(n) = 2, df(d) = 13 d. F2,13 e. 0.64 f. 0.5437 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: The mean scores of different class delivery are not different. 72 a. H0: μp = μm = μh b. At least any two of the means are different. c. df(n) = 2, df(d) = 12 d. F2,12 e. 3.13 f. 0.0807 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is not sufficient evidence to conclude that the mean numbers of daily visitors are different. 74 The data appear normally distributed from the chart and of similar spread. There do not appear to be any serious outliers, so we may proceed with our ANOVA calculations, to see if we have good evidence of a difference between the three groups. H0: μ1 = μ2 = μ3; Ha: μi ≠ μj some i ≠ j. Define μ1, μ2, μ3, as the population mean number of eggs laid by the three groups of fruit flies. F statistic = 8.6657; p-value = 0.0004 730 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA Figure 13.10 Decision: Since the p-value is less than the level of significance of 0.01, we reject the null hypothesis. Conclusion: We have good evidence that the average number of eggs laid during the first 14 days of life for these three strains of fruitflies are different. Interestingly, if you perform a two sample t-test to compare the RS and NS groups they are significantly different (p = 0.0013). Similarly, SS and NS are significantly different (p = 0.0006). However, the two selected groups, RS and SS are not significantly different (p = 0.5176). Thus we appear to have good evidence that selection either for resistance or for susceptibility involves a reduced rate of egg production (for these specific strains) as compared to flies that were not selected for resistance or susceptibility to DDT. Here, genetic selection has apparently involved a loss of fecundity. 76 a. H0 : σ1 2 2 = σ2 b. Ha : σ1 2 2 ≠ σ1 c. df(num) = 4; df(denom) = 4 d. F4, 4 e. 3.00 f. 2(0.1563) = 0.3126. Using the TI-83+/84+ function 2-SampFtest, you get the test statistic as 2.9986 and p-value directly as 0.3127. If you input the lists in a different order, you get a test statistic of 0.3335 but the p-value is the same because this is a two-tailed test. g. Check student't solution. h. Decision: Do not reject the null hypothesis; Conclusion: There is insufficient evidence to conclude that the variances are different. 78 a. H0: σ1 2 2 = σ2 b. Ha: σ1 2 2 ≠ σ1 c. df(n) = 19, df(d) = 19 d. F19,19 e. 1.13 f. 0.786 g. Check student’s solution. h. i. Alpha:0.05 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA 731 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is not sufficient evidence to conclude that the variances are different. 80 The answers may vary. Sample answer: Home decorating magazines and news magazines have different variances. 82 a. H0: = σ1 2 2 = σ2 b. Ha: σ1 2 2 ≠ σ1 c. df(n) = 7, df(d) = 6 d. F7,6 e. 0.8117 f. 0.7825 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: p-value > alpha iv. Conclusion: There is not sufficient evidence to conclude that the variances are different. 84 Here is a strip chart of the silver content of the coins: Figure 13.11 While there are differences in spread, it is not unreasonable to use ANOVA techniques. Here is the completed ANOVA table: Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (Between) Error (Within) Total Table 13.42 37.748 11.015 48.763 4 – 1 = 3 27 – 4 = 23 27 – 1 = 26 12.5825 0.4789 26.272 732 CHAPTER 13 \| F DISTRIBUTION AND ONE-WAY ANOVA P(F > 26.272) = 0; Reject the null hypothesis for any alpha. There is sufficient evidence to conclude that the mean silver content among the four coinages are different. From the strip chart, it appears that the first and second coinages had higher silver contents than the third and fourth. 85 Here is a stripchart of the number of wins for the 14 teams in the AL for the 2012 season. Figure 13.12 While the spread seems similar, there may be some question about the normality of the data, given the wide gaps in the middle near the 0.500 mark of 82 games (teams play 162 games each season in MLB). However, one-way ANOVA is robust. Here is the ANOVA table for the data: Source of Variation Sum of Squares (SS) Degrees of Freedom (df) Mean Square (MS) F Factor (Between) 344.16 Error (Within) Total Table 13.43 1,219.55 1,563.71 3 – 1 = 2 14 – 3 = 11 14 – 1 = 13 172.08 110.87 26.272 1.5521 P(F > 1.5521) = 0.2548 Since the p-value is so large, there is not good evidence against the null hypothesis of equal means. We decline to reject the null hypothesis. Thus, for 2012, there is not any have any good evidence of a significant difference in mean number of wins between the divisions of the American League. Appendices are in Volume 2 846 INDEX INDEX Symbols α, 477 A absolute value of a residual, 641 alternative hypothesis, 470 Analysis of Variance, 712 Area to the left, 343 Area to the right, 343 assumption, 475 average, 11 Average, 46, 394 B balanced design, 702 bar graph, 17 Bernoulli Trial, 235 Bernoulli Trials, 258 binomial distribution, 385, 427, 475 Binomial Distribution, 441, 497 Binomial Experiment, 258 binomial probability distribution, 235 Binomial Probability Distribution, 258 bivariate, 634 Blinding, 37, 46 Box plot, 121 Box plots, 94 box-and-whisker plots, 94 box-whisker plots, 94 C Categorical Variable, 46 Categorical variables, 11 central limit theorem, 369, 371, 378, 481 Central Limit Theorem, 394, 497 central limit theorem for means, 373 central limit theorem for sums, 375 chi-square distribution, 578 Cluster Sampling, 46 Coefficient of Correlation, 665 coefficient of determination, 645 Cohen's d, 532 complement, 165 conditional probability, 165, 298 Conditional Probability, 198, 316 confidence interval, 412, 422 Confidence Interval (CI), 441, 497 confidence intervals, 427 Confidence intervals, 469 confidence level, 413, 427 Confidence Level (CL), 441 contingency table, 180, 198, 588 Contingency Table, 605 continuity correction factor, 385 continuous, 14 Continuous Random Variable, 46 continuous random variable, 301 control group, 37 Control Group, 46 Convenience Sampling, 46 critical value, 345 cumulative distribution function (CDF), 302 Cumulative relative frequency, 30 Cumulative Relative Frequency, 46 D data, 9 Data, 11, 46 decay parameter, 316 degrees of freedom, 423 Degrees of Freedom (df), 441, 551 degrees of freedom (df), 527 Dependent Events, 198 descriptive statistics, 10 discrete, 14 Discrete Random Variable, 46 double-blind experiment, 37 Double-blinding, 46 E Empirical Rule, 340 empirical rule, 412 equal standard deviations, 694 Equally likely, 164 Equally Likely, 198 equally likely, 293 error bound, 427 error bound for a population mean, 413, 423 Error Bound for a Population Mean (EBM), 441 Error Bound for a Population Proportion (EBP), 441 event, 164 Event, 198 expected value, 228 Expected Value, 258 expected values, 579 experiment, 164 Experiment, 198 experimental unit, 37 Experimental Unit, 46 explanatory variable, 37 Explanatory Variable, 46 exponential distribution, 301, 381 Exponential Distribution, 316, 394 extrapolation, 651 F F distribution, 695 F ratio, 695 fair, 164 first quartile, 87 First Quartile, 121 frequency, 30, 76 Frequency, 46, 121 Frequency Polygon, 121 Frequency Table, 121 G geometric distribution, 242 Geometric Distribution, 258 Geometric Experiment, 258 goodness-of-fit test, 579 H histogram, 76 Histogram, 121 hypergeometric experiment, 260 Hypergeometric Experiment, 258 hypergeometric probability, 244 Hypergeometric Probability, 258 hypotheses, 470 Hypothesis, 497 hypothesis test, 475, 478, 498 hypothesis testing, 470 Hypothesis Testing, 497 I independent, 168, 176 Independent Events, 198 Independent groups, 526 inferential statistics, 10, 412 Inferential Statistics, 441 influential points, 651 informed consent, 39 Informed Consent, 46 Institutional Review Board, 46 Institutional Review Boards (IRB), 39 interpolation, 651 interquartile range, 87 Interquartile Range, 121 Interval, 121 interval scale, 29 L law of large numbers, 164, 378 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16 INDEX 847 Least-Squares Line, 639 least-squares regression line, 640 level of measurement, 29 Level of Significance of the Test , 497 Line of Best Fit, 639 linear regression, 639 long-term relative frequency, 164 Lurking Variable, 46 lurking variables, 37 M margin of error, 412 matched pairs, 526 mean, 11, 98, 228, 370, 373, 379 Mean, 121, 258, 394 Mean of a Probability Distribution, 258 mean square, 695 median, 86, 98 Median, 121 memoryless property, 316 Midpoint, 121 mode, 100 Mode, 121 multivariate, 634 mutually exclusive, 170, 176 Mutually Exclusive, 198 N nominal scale, 29 Nonsampling Error, 46 normal app
roximation to the binomial, 385 Normal Distribution, 355, 394, 441, 497 normal distribution, 423, 427, 474 Normal distribution, 534 normally distributed, 370, 375, 475 null hypothesis, 470, 475, 475, 477 Numerical Variable, 46 Numerical variables, 11 O observed values, 579 One-Way ANOVA, 712 ordinal scale, 29 outcome, 164 Outcome, 198 outlier, 69, 87 Outlier, 121, 665 outliers, 651 P p-value, 475, 477, 478, 499, 497 paired data set, 84 Paired Data Set, 121 parameter, 11, 412 Parameter, 46, 441 Pareto chart, 17 Pearson, 10 Percentile, 121 percentile, 377 percentiles, 86 pie chart, 17 placebo, 37 Placebo, 46 point estimate, 412 Point Estimate, 441 Poisson distribution, 316 Poisson probability distribution, 247, 260 Poisson Probability Distribution, 259 Pooled Proportion, 551 population, 11, 27 Population, 46 population variance, 596 potential outlier, 654 Probability, 10, 46, 198 probability, 164 probability density function, 288 probability distribution function, 226 Probability Distribution Function (PDF), 259 proportion, 11 Proportion, 46 Q Qualitative data, 13 Qualitative Data, 47 quantitative continuous data, 14 Quantitative data, 14 Quantitative Data, 47 quantitative discrete data, 14 quartiles, 86 Quartiles, 87, 121 R random assignment, 37 Random Assignment, 47 Random Sampling, 47 random variable, 226 Random variable, 528 Random Variable, 534 Random Variable (RV), 259 ratio scale, 29 relative frequency, 30, 76 Relative Frequency, 47, 121 replacement, 168 Representative Sample, 47 residual, 641 response variable, 37 Response Variable, 47 S sample, 11 Sample, 47 sample mean, 371 sample size, 371, 375 sample space, 164, 175, 186 Sample Space, 198 samples, 27 sampling, 11 Sampling Bias, 47 sampling distribution, 101 Sampling Distribution, 394 Sampling Error, 47 sampling variability of a statistic, 109 Sampling with Replacement, 47, 198 Sampling without Replacement, 47, 198 simple random sample, 475 Simple Random Sampling, 47 single population mean, 474 single population proportion, 474 Skewed, 121 standard deviation, 108, 422, 474, 475, 476, 480 Standard Deviation, 121, 441, 497, 551 Standard Deviation of a Probability Distribution, 259 standard error, 526 Standard Error of the Mean, 394 standard error of the mean., 371 standard normal distribution, 338 Standard Normal Distribution, 355 statistic, 11 Statistic, 47 statistics, 9 Stratified Sampling, 47 Student's t-distribution, 423, 474, 475 Student's t-Distribution, 441, 497 Sum of Squared Errors (SSE), 641 sum of squares, 695 Systematic Sampling, 47 T test for homogeneity, 592 test of a single variance, 596 test of independence, 588 848 INDEX test statistic, 534 The AND Event, 198 The Complement Event, 198 The Conditional Probability of A GIVEN B, 198 The Conditional Probability of One Event Given Another Event, 198 The Law of Large Numbers, 259 The Or Event, 198 The OR of Two Events, 198 The standard deviation, 534 third quartile, 87 treatments, 37 Treatments, 47 tree diagram, 185 Tree Diagram, 198 Type 1 Error, 497 Type 2 Error, 498 Type I error, 472, 477 Type II error, 472 U unfair, 165 Uniform Distribution, 316, 394 uniform distribution, 379 V variable, 11 Variable, 47 Variable (Random Variable), 551 variance, 109 Variance, 121, 712 Variance between samples, 695 Variance within samples, 695 variances, 694 Variation, 27 Venn diagram, 191 Venn Diagram, 199 Z z-score, 355, 423 z-scores, 338 This content is available for free at http://textbookequity.org/introductory-statistics or at http://cnx.org/content/col11562/1.16
onrepeating, and are nonterminating: {h \| h is not a rational number}. _ n ∣ m and n are integers and n ≠ 0 } m Example 5 Differentiating the Sets of Numbers Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q'). a. √ — 36 8 _ b. 3 c. √ — 73 d. −6 e. 3.2121121112 … Solution — a. √ 36 = 6 _ 8 _ #6 = 2. b. 3 c. √ — 73 d. −6 e. 3.2121121112... ' × × Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 6 CHAPTER 1 PREREQUISITES Try It #5 Classify each number as being a natural number (N), whole number (W), integer (I), rational number (Q), and/or irrational number (Q'). a. − 35 _ 7 b. 0 c. √ — 169 d. √ — 24 e. 4.763763763 … Performing Calculations Using the Order of Operations When we multiply a number by itself, we square it or raise it to a power of 2. For example, 4 2 = 4 ∙ 4 = 16. We can raise any number to any power. In general, the exponential notation a n means that the number or variable a is used as a factor n times. n factors In this notation, a n is read as the nth power of a, where a is called the base and n is called the exponent. A term in exponential notation may be part of a mathematical expression, which is a combination of numbers and operations. 2 _ For example, 24 + 6 ∙ − 4 2 is a mathematical expression. 3 To evaluate a mathematical expression, we perform the various operations. However, we do not perform them in any random order. We use the order of operations. This is a sequence of rules for evaluating such expressions. Recall that in mathematics we use parentheses ( ), brackets [ ], and braces { } to group numbers and expressions so that anything appearing within the symbols is treated as a unit. Additionally, fraction bars, radicals, and absolute value bars are treated as grouping symbols. When evaluating a mathematical expression, begin by simplifying expressions within grouping symbols. The next step is to address any exponents or radicals. Afterward, perform multiplication and division from left to right and finally addition and subtraction from left to right. Let’s take a look at the expression provided. 2 _ 24 + 6 ∙ − 4 2 3 There are no grouping symbols, so we move on to exponents or radicals. The number 4 is raised to a power of 2, so simplify 4 2 as 16. 2 _ − 4 2 24 + 6 ∙ 3 2 _ − 16 24 + 6 ∙ 3 Next, perform multiplication or division, left to right. 2 _ 24 + 6 ∙ − 16 3 24 + 4 − 16 Lastly, perform addition or subtraction, left to right. 2 _ Therefore, 24 + 6 ∙ − 4 2 = 12. 3 24 + 4 − 16 28 − 16 12 For some complicated expressions, several passes through the order of operations will be needed. For instance, there may be a radical expression inside parentheses that must be simplified before the parentheses are evaluated. Following the order of operations ensures that anyone simplifying the same mathematical expression will get the same result. order of operations Operations in mathematical expressions must be evaluated in a systematic order, which can be simplified using the acronym PEMDAS: P(arentheses) E(xponents) M(ultiplication) and D(ivision) A(ddition) and S(ubtraction) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.1 REAL NUMBERS: ALGEBRA ESSENTIALS 7 How To… Given a mathematical expression, simplify it using the order of operations. 1. Simplify any expressions within grouping symbols. 2. Simplify any expressions containing exponents or radicals. 3. Perform any multiplication and division in order, from left to right. 4. Perform any addition and subtraction in order, from left to right. Example 6 Using the Order of Operations Use the order of operations to evaluate each of the following expressions. b. 5 2 − 4 ______ 7 − √ — 11 − 2 c(4 − 1) e. 7(5 ∙ 3) − 2[(6 − 3) − 4 2 ] + 1 a. (3 ∙ 2) 2 − 4(6 + 2) d. 14 − Solution a. (3 ∙ 2) 2 − 4(6 + 2) = (6) 2 − 4(8) = 36 − 4(8) = 36 − 32 = 4 Simplify parentheses. Simplify exponent. Simplify multiplication. Simplify subtraction. 5 2 __ b. 7 − √ — 11 − ______ 7 5 2 − 4 ______ 7 25 − 4 ______ 7 21 ___ Simplify grouping symbols (radical). Simplify radical. Simplify exponent. Simplify subtraction in numerator. Simplify division. Simplify subtraction. Note that in the first step, the radical is treated as a grouping symbol, like parentheses. Also, in the third step, the fraction bar is considered a grouping symbol so the numerator is considered to be grouped. c. 6 − \|5 − 8\| + 3(4 − 1) = 6 − \|−3\| + 3(3) = 6 − 3 + 3(3 = 12 Simplify inside grouping symbols. Simplify absolute value. Simplify multiplication. Simplify subtraction. Simplify addition. d. 14 − = Simplify exponent. = 14 − 3 ∙ 2 _ 2 ∙ 5 − 9 14 − 6 _ 10 − 9 8 _ = 1 = 8 Simplify products. Simplify differences. Simplify quotient. In this example, the fraction bar separates the numerator and denominator, which we simplify separately until the last step. e 7(5 ∙ 3) − 2[(6 − 3) − 4 2 ] + 1 = 7(15) − 2[(3) − 4 2 ] + 1 = 7(15) − 2(3 − 16) + 1 = 7(15) − 2(−13) + 1 = 105 + 26 + 1 = 132 Simplify inside parentheses. Simplify exponent. Subtract. Multiply. Add. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 8 CHAPTER 1 PREREQUISITES Try It #6 Use the order of operations to evaluate each of the following expressions5 − 4) 2 a. √ 1 1 __ __ · 9 2 [. 3 2 b __________ 9 − 6 e. [ (3 − 8) 2 − 4] − (3 − 8) c. \|1.8 − 4.3\| + 0.4 √ — 15 + 10 Using Properties of Real Numbers For some activities we perform, the order of certain operations does not matter, but the order of other operations does. For example, it does not make a difference if we put on the right shoe before the left or vice-versa. However, it does matter whether we put on shoes or socks first. The same thing is true for operations in mathematics. Commutative Properties The commutative property of addition states that numbers may be added in any order without affecting the sum. a + b = b + a We can better see this relationship when using real numbers. and (−2) + 7 = 5 7 + (−2) = 5 Similarly, the commutative property of multiplication states that numbers may be multiplied in any order without affecting the product. a ∙ b = b ∙ a Again, consider an example with real numbers. (−11) ∙ (−4) = 44 and (−4) ∙ (−11) = 44 It is important to note that neither subtraction nor division is commutative. For example, 17 − 5 is not the same as 5 − 17. Similarly, 20 ÷ 5 ≠ 5 ÷ 20. Associative Properties The associative property of multiplication tells us that it does not matter how we group numbers when multiplying. We can move the grouping symbols to make the calculation easier, and the product remains the same. Consider this example. a(bc) = (ab)c (3 ∙ 4) ∙ 5 = 60 and 3 ∙ (4 ∙ 5) = 60 The associative property of addition tells us that numbers may be grouped differently without affecting the sum. a + (b + c) = (a + b) + c This property can be especially helpful when dealing with negative integers. Consider this example. [15 + (−9)] + 23 = 29 and 15 + [(−9) + 23] = 29 Are subtraction and division associative? Review these examples. 8 − (3 − 15) $#(8 − 3) − 15 8 − ( − 12) $ 5 − 15 20 ≠ −10 64 ÷ (8 ÷ 4) $ (64 ÷ 8) ÷ 4 64 ÷ 2 $ 8 ÷ 4 32 ≠ 2 As we can see, neither subtraction nor division is associative. Distributive Property The distributive property states that the product of a factor times a sum is the sum of the factor times each term in the sum. a ∙ (b + c) = a ∙ b + a ∙ c Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.1 REAL NUMBERS: ALGEBRA ESSENTIALS 9 This property combines both addition and multiplication (and is the only property to do so). Let us consider an example. 4 ∙ [12 + (−7)] = 4 ∙ 12 + 4 ∙ (−7) = 48 + (−28) = 20 Note that 4 is outside the grouping symbols, so we distribute the 4 by multiplying it by 12, multiplying it by −7, and adding the products. To be more precise when describing this property, we say that multiplication distributes over addition. The reverse is not true, as we can see in this example. 6 + (3 ∙ 5) $ (6 + 3) ∙ (6 + 5) 6 + (15) $ (9) ∙ (11) 21 ≠ 99 Multiplication does not distribute over subtraction, and division distributes over neither addition nor subtraction. A special case of the distributive property occurs when a sum of terms is subtracted. a − b = a + (−b) For example, consider the difference 12 − (5 + 3). We can rewrite the difference of the two terms 12 and (5 + 3) by turning the subtraction expression into addition of the opposite. So instead of subtracting (5 + 3), we add the opposite. 12 + (−1) ∙ (5 + 3) Now, distribute −1 and simplify the result. 12 − (5 + 3) = 12 + (−1) ∙ (5 + 3) = 12 + [(−1) ∙ 5 + (−1) ∙ 3] = 12 + (−8) = 4 This seems like a lot of trouble for a simple sum, but it illustrates a powerful result that will be useful once we introduce algebraic terms. To subtract a sum of terms, change the sign of each term and add the results. With this in mind, we can rewrite the last example. 12 − (5 + 3) = 12 + (−5 − 3) = 12 + (−8) = 4 Identity Properties The identity property of addition states that there is a unique number, called the additive identity (0) that, when added to a number, results in the original number. a + 0 = a The identity property of multiplication states that there is a unique number, called the multiplicative identity (1) that, when multiplied by a number, results in the original number. For example, we have (−6) + 0 = −6 and 23 ∙ 1 = 23. There are no exceptions for these properties; they work for every real number, including 0 and 1. a ∙ 1 = a Inverse Properties The inverse property of addition states that, for every real number a, there is a unique number, called the additive inverse (or opposite), denoted−a, that, when added to the original number, results in the additive identity, 0. For example, if a = −8, the additive inverse is 8, since (−8) + 8 = 0. a + (−a) = 0 The inverse property of multiplication holds for all real numbers except 0 because the reciprocal of 0 is not define
d. The property states that, for every real number a, there is a unique number, called the multiplicative inverse (or 1 _ a , that, when multiplied by the original number, results in the multiplicative identity, 1. reciprocal), denoted 1 _ a = 1 a ∙ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 10 CHAPTER 1 PREREQUISITES 3 2 1 _ _ a , is − _ For example, if a = − because , the reciprocal, denoted properties of real numbers The following properties hold for real numbers a, b, and c. Commutative Property Associative Property Distributive Property Identity Property Inverse Property Addition b + c) = (a + b) + c Multiplication a ∙ b = b ∙ a a(bc) = (ab)c a ∙ (b + c) = a ∙ b + a ∙ c There exists a unique real number called the additive identity, 0, such that, for any real number a a + 0 = a Every real number a has an additive inverse, or opposite, denoted −a, such that a + (−a) = 0 There exists a unique real number called the multiplicative identity, 1, such that, for any real number a a ∙ 1 = a Every nonzero real number a has a multiplicative inverse, or reciprocal, 1 _ a , such that denoted 1 a ∙ ( _ a ) = 1 Example 7 Using Properties of Real Numbers Use the properties of real numbers to rewrite and simplify each expression. State which properties apply. a. 3 ∙ 6 + 3 ∙ 4 b. (5 + 8) + (−8) c. 6 − (15 + 9) Solution a. b. c. d6 + 4) = 3 ∙ 10 = 30 (5 + 8) + (−8) = 5 + [8 + (−8)] = 5 + 0 = 5 6 − (15 + 9) = 6 + [(−15) + (−9)] = 6 + (−24) = −18 2 7 4 7 2 4 ) ∙ ( ) = ∙ ( __ __ __ __ __ __ ∙ ∙ 7 4 7 3 4 3 2 7 4 ) ∙ = ( __ __ __ ∙ 4 3 7 2 __ = 1 ∙ 3 2 __ = 3 7 2 4 ) ∙ ( __ __ __ ∙ d. 4 3 7 e. 100 ∙ [0.75 + (−2.38)] Distributive property Simplify. Simplify. Associative property of addition Inverse property of addition Identity property of addition Distributive property Simplify. Simplify. Commutative property of multiplication Associative property of multiplication Inverse property of multiplication Identity property of multiplication e. 100 ∙ [0.75 + (−2.38)] = 100 ∙ 0.75 + 100 ∙ (−2.38) = 75 + (−238) = −163 Distributive property Simplify. Simplify. Try It #7 Use the properties of real numbers to rewrite and simplify each expression. State which properties apply. a. ( − ) ] b. 5 · (6.2 + 0.4) c. 18 − (7 − 15) d. 23 ) · [ 11 · ( − 17 _ 18 17 _ 18 5 _ 23 ) ] e. 6 ċ (−3) + 6 ċ 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.1 REAL NUMBERS: ALGEBRA ESSENTIALS 11 Evaluating Algebraic Expressions So far, the mathematical expressions we have seen have involved real numbers only. In mathematics, we may see 4 2 m 3 n 2 . In the expression x + 5, 5 is called a constant because it does not vary π r 3 , or √ _ expressions such as x + 5, 3 — and x is called a variable because it does. (In naming the variable, ignore any exponents or radicals containing the variable.) An algebraic expression is a collection of constants and variables joined together by the algebraic operations of addition, subtraction, multiplication, and division. We have already seen some real number examples of exponential notation, a shorthand method of writing products of the same factor. When variables are used, the constants and variables are treated the same way. (−3) 5 = (−3) ∙ (−3) ∙ (−3) ∙ (−3) ∙ (−3) (2 ∙ 7) 3 = (2 ∙ 7) ∙ (2 ∙ 7) ∙ (2 ∙ 7yz) 3 = (yz) ∙ (yz) ∙ (yz) In each case, the exponent tells us how many factors of the base to use, whether the base consists of constants or variables. Any variable in an algebraic expression may take on or be assigned different values. When that happens, the value of the algebraic expression changes. To evaluate an algebraic expression means to determine the value of the expression for a given value of each variable in the expression. Replace each variable in the expression with the given value, then simplify the resulting expression using the order of operations. If the algebraic expression contains more than one variable, replace each variable with its assigned value and simplify the expression as before. Example 8 Describing Algebraic Expressions List the constants and variables for each algebraic expression. a. 3 c. √ — 2 m 3 n 2 Solution Constants Variables a. 3 5 4 _ , π 3 c, n Try It #8 List the constants and variables for each algebraic expression. a. 2πr(r + h) b. 2(L + W) c. 4 y 3 + y Example 9 Evaluating an Algebraic Expression at Different Values Evaluate the expression 2x − 7 for each value for x. 1 _ c. x = 2 b. x = 1 a. x = 0 Solution a. Substitute 0 for x. b. Substitute 1 for x. d. x = −4 2x − 7 = 2(0) − 7 = 0 − 7 = −7 2x − 7 = 2(1) − 7 = 2 − 7 = −5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 12 CHAPTER 1 PREREQUISITES 1 __ for x. c. Substitute 2 d. Substitute −4 for x. 1 ) − 7 2x − 7 = 2 ( __ 2 = 1 − 7 = −6 2x − 7 = 2(−4) − 7 = −8 − 7 = −15 Try It #9 Evaluate the expression 11 − 3y for each value for y. a. y = 2 b. y = 0 2 _ c. y =# 3 d. y = −5 Example 10 Evaluating Algebraic Expressions Evaluate each expression for the given values. a. x + 5 for x = −5 b. t _ 2t−1 for t = 10 4 _ π r 3 for r = 5 c. 3 d. a + ab + b for a = 11, b = −8 e. √ — 2 m 3 n 2 for m = 2, n = 3 Solution a. Substitute −5 for x. b. Substitute 10 for t. c. Substitute 5 for r. x + 5 = (−5) + 5 = 0 t _ 2t − 1 = (10) _ 2(10) − 1 = 10 _ 20 − 1 = 10 _ 19 4 4 _ _ πr 3 = π(5)3 3 3 4 _ = π(125) 3 = 500 _ π 3 d. Substitute 11 for a and −8 for b. a + ab + b = (11) + (11)(−8) + (−8) = 11 − 88 − 8 = −85 e. Substitute 2 for m and 3 for n. √ Try It #10 Evaluate each expression for the given values. — 2m3n2 = √ = √ = √ = 12 — 2(2)3(3)2 — 2(8)(9) — 144 a. y + 3 _ y − 3 for y = 5 d. (p 2q)3 for p = −2, q = 3 b. 7 − 2t for t = −2 c. 1 _ πr 2 for r = 11 3 2 1 _ _ e. 4(m − n) − 5(n − m) for m = , n = 3 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.1 REAL NUMBERS: ALGEBRA ESSENTIALS 13 Formulas An equation is a mathematical statement indicating that two expressions are equal. The expressions can be numerical or algebraic. The equation is not inherently true or false, but only a proposition. The values that make the equation true, the solutions, are found using the properties of real numbers and other results. For example, the equation 2x + 1 = 7 has the unique solution x = 3 because when we substitute 3 for x in the equation, we obtain the true statement 2(3) + 1 = 7. A formula is an equation expressing a relationship between constant and variable quantities. Very often, the equation is a means of finding the value of one quantity (often a single variable) in terms of another or other quantities. One of the most common examples is the formula for finding the area A of a circle in terms of the radius r of the circle: A = πr 2. For any value of r, the area A can be found by evaluating the expression πr 2. Example 11 Using a Formula A right circular cylinder with radius r and height h has the surface area S (in square units) given by the formula S = 2πr(r + h). See Figure 4. Find the surface area of a cylinder with radius 6 in. and height 9 in. Leave the answer in terms of π. r h Solution Evaluate the expression 2πr(r + h) for r = 6 and h = 9. Figure 4 Right circular cylinder S = 2πr(r + h) = 2π(6)[(6) + (9)] = 2π(6)(15) = 180π The surface area is 180π square inches. Try It #11 A photograph with length L and width W is placed in a matte of width 8 centimeters (cm). The area of the matte (in square centimeters, or cm2) is found to be A = (L + 16)(W + 16) − L ∙ W. See Figure 5. Find the area of a matte for a photograph with length 32 cm and width 24 cm. 8 cm 24 cm 8 cm 32 cm Figure 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 14 CHAPTER 1 PREREQUISITES Simplifying Algebraic Expressions Sometimes we can simplify an algebraic expression to make it easier to evaluate or to use in some other way. To do so, we use the properties of real numbers. We can use the same properties in formulas because they contain algebraic expressions. Example 12 Simplifying Algebraic Expressions Simplify each algebraic expression. a. 3x − 2y + x − 3y − 7 b. 2r − 5(3 − r) + 4 5 2 t + 2s ) s ) − ( c. ( 4t − _ _ 4 3 d. 2mn − 5m + 3mn + n Solution a. b. 3x − 2y + x − 3y − 7 = 3x + x − 2y − 3y − 7 = 4x − 5y − 7 2r − 5(3 − r) + 4 = 2r − 15 + 5r + 4 = 2r + 5y − 15 + 4 = 7r − 11 5 5 2 2 t + 2s ) = 4t − s ) − ( c. 4t − − 2s 4 4 3 3 5 2 _ _ s − 2s t − = 4t − 4 3 = t − 10 _ 3 13 _ s 4 Commutative property of addition Simplify. Distributive property Commutative property of addition Simplify. Distributive property Commutative property of addition Simplify. d. mn − 5m + 3mn + n = 2mn + 3mn − 5m + n = 5mn − 5m + n Commutative property of addition Simplify. Try It #12 Simplify each algebraic expression. t t Example 13 Simplifying a Formula c. 4p(q − 1) + q(1 − p) d. 9r − (s + 2r) + (6 − s) A rectangle with length L and width W has a perimeter P given by P = L + W + L + W. Simplify this expression. Solution = 2L + 2W P = 2(L + W) Commutative property of addition Simplify. Distributive property Try It #13 If the amount P is deposited into an account paying simple interest r for time t, the total value of the deposit A is given by A = P + Prt. Simplify the expression. (This formula will be explored in more detail later in the course.) Access these online resources for additional instruction and practice with real numbers. • Simplify an Expression (http://openstaxcollege.org/l/simexpress) • Evaluate an Expression1 (http://openstaxcollege.org/l/ordofoper1) • Evaluate an Expression2 (http://openstaxcollege.org/l/ordofoper2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.1 SECTION EXERCISES 15 1.1 SECTION EXERCISES VERBAL — 1. Is √ 2 an example of a rational terminating, rational repeating, or irrational number? Tell why it fits that category. 2. What is the order of operations? What acronym is used to describe the order of operations, and what does it stand for? 3. What do the As
sociative Properties allow us to do when following the order of operations? Explain your answer. NUMERIC For the following exercises, simplify the given expression. 4. 10 + 2 · (5 − 3) 5. 6 ÷ 2 − (81 ÷ 32) 6. 18 + (6 − 8)3 7. −2 · [16 ÷ (8 − 4)2] 2 8. 4 − 6 + 2 · 7 9. 3(5 − 8) 10. 4 + 6 − 10 ÷ 2 11. 12 ÷ (36 ÷ 9) + 6 12. (4 + 5)2 ÷ 3 13. 3 − 12 · 2 + 19 14. 2 + 8 · 7 ÷ 4 15. 5 + (6 + 4) − 11 16. 9 − 18 ÷ 32 17. 14 · 3 ÷ 7 − 6 18. 9 − (3 + 11) · 2 19. 6 + 2 · 2 − 1 20. 64 ÷ (8 + 4 · 2) 21. 9 + 4(22) 22. (12 ÷ 3 · 3)2 23. 25 ÷ 52 − 7 24. (15 − 7) · (3 − 7) 25. 2 · 4 − 9(−1) 1 _ 26. 42 − 25 · 5 27. 12(3 − 1) ÷ 6 ALGEBRAIC For the following exercises, solve for the variable. 28. 8(x + 3) = 64 29. 4y + 8 = 2y 30. (11a + 3) − 18a = −4 31. 4z − 2z(1 + 4) = 36 32. 4y(7 − 2)2 = −200 33. −(2x)2 + 1 = −3 34. 8(2 + 4) − 15b = b 35. 2(11c − 4) = 36 36. 4(3 − 1)x = 4 1 _ (8w − 42) = 0 37. 4 For the following exercises, simplify the expression. 38. 4x + x(13 − 7) 39. 2y − (4)2 y − 11 a __ 40. 23 (64) − 12a ÷ 6 41. 8b − 4b(3) + 1 42. 5l ÷ 3l · (9 − 6) 43. 7z − 3 + z · 62 44. 4 · 3 + 18x ÷ 9 − 12 45. 9(y + 8) − 27 9 t − 4 ) 2 46. ( _ 6 47. 6 + 12b − 3 · 6b 48. 18y − 2(1 + 7y) 50. 8(3 − m) + 1(−8) 51. 9x + 4x(2 + 3) − 4(2x + 3x) 4 ) 49. ( _ 9 2 · 27x 52. 52 − 4(3x) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 16 CHAPTER 1 PREREQUISITES REAL-WORLD APPLICATIONS For the following exercises, consider this scenario: Fred earns $40 mowing lawns. He spends $10 on mp3s, puts half of what is left in a savings account, and gets another $5 for washing his neighbor’s car. 53. Write the expression that represents the number of dollars Fred keeps (and does not put in his savings account). Remember the order of operations. For the following exercises, solve the given problem. 55. According to the U.S. Mint, the diameter of a quarter is 0.955 inches. The circumference of the quarter would be the diameter multiplied by π. Is the circumference of a quarter a whole number, a rational number, or an irrational number? 54. How much money does Fred keep? 56. Jessica and her roommate, Adriana, have decided to share a change jar for joint expenses. Jessica put her loose change in the jar first, and then Adriana put her change in the jar. We know that it does not matter in which order the change was added to the jar. What property of addition describes this fact? For the following exercises, consider this scenario: There is a mound of g pounds of gravel in a quarry. Throughout the day, 400 pounds of gravel is added to the mound. Two orders of 600 pounds are sold and the gravel is removed from the mound. At the end of the day, the mound has 1,200 pounds of gravel. 57. Write the equation that describes the situation. 58. Solve for g. For the following exercise, solve the given problem. 59. Ramon runs the marketing department at his company. His department gets a budget every year, and every year, he must spend the entire budget without going over. If he spends less than the budget, then his department gets a smaller budget the following year. At the beginning of this year, Ramon got $2.5 million for the annual marketing budget. He must spend the budget such that 2,500,000 − x = 0. What property of addition tells us what the value of x must be? TECHNOLOGY For the following exercises, use a graphing calculator to solve for x. Round the answers to the nearest hundredth. 3 _ 60. 0.5(12.3)2 − 48x = 5 EXTENSIONS 61. (0.25 − 0.75)2x − 7.2 = 9.9 62. If a whole number is not a natural number, what must the number be? 64. Determine whether the statement is true or false: The product of a rational and irrational number is always irrational. 63. Determine whether the statement is true or false: The multiplicative inverse of a rational number is also rational. 65. Determine whether the simplified expression is rational or irrational: √ — −18 − 4(5)(−1) . 66. Determine whether the simplified expression is 67. The division of two whole numbers will always result rational or irrational: √ — −16 + 4(5) + 5 . in what type of number? 68. What property of real numbers would simplify the following expression: 4 + 7(x − 1)? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 17 LEARNING OBJECTIVES In this section students will: • Use the product rule of exponents. • Use the quotient rule of exponents. • Use the power rule of exponents. • Use the zero exponent rule of exponents. • Use the negative rule of exponents. • Find the power of a product and a quotient. • Simplify exponential expressions. • Use scientiﬁc notation. 1. 2 EXPONENTS AND SCIENTIFIC NOTATION Mathematicians, scientists, and economists commonly encounter very large and very small numbers. But it may not be obvious how common such figures are in everyday life. For instance, a pixel is the smallest unit of light that can be perceived and recorded by a digital camera. A particular camera might record an image that is 2,048 pixels by 1,536 pixels, which is a very high resolution picture. It can also perceive a color depth (gradations in colors) of up to 48 bits per frame, and can shoot the equivalent of 24 frames per second. The maximum possible number of bits of information used to film a one-hour (3,600-second) digital film is then an extremely large number. Using a calculator, we enter 2,048 · 1,536 · 48 · 24 · 3,600 and press ENTER. The calculator displays 1.304596316E13. What does this mean? The “E13” portion of the result represents the exponent 13 of ten, so there are a maximum of approximately 1.3 · 1013 bits of data in that one-hour film. In this section, we review rules of exponents first and then apply them to calculations involving very large or small numbers. Using the Product Rule of Exponents Consider the product x 3 ∙ x 4. Both terms have the same base, x, but they are raised to different exponents. Expand each expression, and then rewrite the resulting expression. 3 factors 4 factors factors = The result is that . Notice that the exponent of the product is the sum of the exponents of the terms. In other words, when multiplying exponential expressions with the same base, we write the result with the common base and add the exponents. This is the product rule of exponents. Now consider an example with real numbers. am · an = am + n 23 · 24 = 23 + 4 = 27 We can always check that this is true by simplifying each exponential expression. We find that 23 is 8, 24 is 16, and 27 is 128. The product 8 ∙ 16 equals 128, so the relationship is true. We can use the product rule of exponents to simplify expressions that are a product of two numbers or expressions with the same base but different exponents. the product rule of exponents For any real number a and natural numbers m and n, the product rule of exponents states that am · an = am + n Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 18 CHAPTER 1 PREREQUISITES Example 1 Using the Product Rule Write each of the following products with a single base. Do not simplify further. a. t 5 ∙ t 3 b. (−3)5 ∙ (−3) c. x 2 ∙ x 5 ∙ x 3 Solution Use the product rule to simplify each expression. a. t 5 ∙ t3 = t5 + 3 = t8 b. (−3)5 ∙ (−3) = (−3)5 ∙ (−3)1 = (−3)5 + 1 = (−3)6 c. x 2 ∙ x 5 ∙ x 3 At first, it may appear that we cannot simplify a product of three factors. However, using the associative property of multiplication, begin by simplifying the first twox 2 · x 5) · x 3 = (x 2 + 5)· = x10 Notice we get the same result by adding the three exponents in one step = x10 Try It #1 Write each of the following products with a single base. Do not simplify further. a. ( 2 _ y ) · ( c. t3 · t6 · t5 Using the Quotient Rule of Exponents The quotient rule of exponents allows us to simplify an expression that divides two numbers with the same base but ym ___ yn , where m > n. different exponents. In a similar way to the product rule, we can simplify an expression such as y 9 _ y 5 . Perform the division by canceling common factors. Consider the example y9 ___ y5 = ___ ___ % __________ 1 = y 4 Notice that the exponent of the quotient is the difference between the exponents of the divisor and dividend. = am − n am ___ an y9 __ y5 = y9 − 5 = y4 In other words, when dividing exponential expressions with the same base, we write the result with the common base and subtract the exponents. For the time being, we must be aware of the condition m > n. Otherwise, the difference m − n could be zero or negative. Those possibilities will be explored shortly. Also, instead of qualifying variables as nonzero each time, we will simplify matters and assume from here on that all variables represent nonzero real numbers. the quotient rule of exponents For any real number a and natural numbers m and n, such that m > n, the quotient rule of exponents states that am ___ an = am − n Example 2 Using the Quotient Rule Write each of the following products with a single base. Do not simplify further. a. (−2)14 ______ (−2)9 b. t 23 __ t 15 c. — 5 2 ) ( z √ ________ 2 z √ — Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 19 Solution Use the quotient rule to simplify each expression. a. = (−2)14 − 9 = (−2)5 b. = t 23 − 15 = t 8 (−2)14 ______ (−2)9 t23 __ t15 2 ) ( z √ ________ 2 z √ — — 5 c ) Try It #2 Write each of the following products with a single base. Do not simplify further. a. s75 __ s68 b. (−3)6 _____ −3 c. (ef 2)5 _ (ef 2)3 Using the Power Rule of Exponents Suppose an exponential expression is raised to some power. Can we simplify the result? Yes. To do this, we use the power rule of exponents. Consider the expression (x 2)3. The expression inside the parentheses is multiplied twice because it has an exponent of 2. Then the result is multiplied three times because the entire expression has an exponent of 3. 3 factors (x 2) 3 = (x 2) · (x 2) · (x 2) 3 factors x · x ) · ( 2 factors factors = ( 2
factors = The exponent of the answer is the product of the exponents: (x 2. In other words, when raising an exponential expression to a power, we write the result with the common base and the product of the exponents. (a m)n = am ∙ n Be careful to distinguish between uses of the product rule and the power rule. When using the product rule, different terms with the same bases are raised to exponents. In this case, you add the exponents. When using the power rule, a term in exponential notation is raised to a power. In this case, you multiply the exponents. Product Rule Power Rule 53 ∙ 54 = 53 + 3a)7 ∙ (3a)10 = (3a)7 + 10 = 57 = x 7 = (3a)17 but but but (53)4 = 5 3 ∙ 4 (x 5)2 = x 5 ∙ 2 ((3a)7)10 = (3a)7 ∙ 10 = 512 = x 10 = (3a)70 the power rule of exponents For any real number a and positive integers m and n, the power rule of exponents states that (am)n = am ∙ n Example 3 Using the Power Rule Write each of the following products with a single base. Do not simplify further. a. (x 2) 7 b. ((2t)5) 3 c. ((−3)5) 11 Solution Use the power rule to simplify each expression. a. (x 2) 7 = x 2 ∙ 7 = x 14 b. ((2t)5)3 = (2t)5 ∙ 3 = (2t)15 c. ((−3)5)11 = (−3)5 ∙ 11 = (−3)55 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 20 CHAPTER 1 PREREQUISITES Try It #3 Write each of the following products with a single base. Do not simplify further. c. ((−g)4)4 a. ((3y)8)3 b. (t 5)7 Using the Zero Exponent Rule of Exponents Return to the quotient rule. We made the condition that m > n so that the difference m − n would never be zero or negative. What would happen if m = n? In this case, we would use the zero exponent rule of exponents to simplify the expression to 1. To see how this is done, let us begin with an example. t 8 __ t 8 % t8 __ % t8 = = 1 If we were to simplify the original expression using the quotient rule, we would have t 8 __ t 8 If we equate the two answers, the result is t0 = 1. This is true for any nonzero real number, or any variable representing a real number. a0 = 1 The sole exception is the expression 00. This appears later in more advanced courses, but for now, we will consider the value to be undefined. = t 8 − 8 = t 0 the zero exponent rule of exponents For any nonzero real number a, the zero exponent rule of exponents states that a0 = 1 Example 4 Using the Zero Exponent Rule Simplify each expression using the zero exponent rule of exponents. c 3 __ a. c 3 b. −3x 5 _____ x 5 c. (j 2k)4 _ (j 2k) ∙ (j 2k)3 d. 5(rs2)2 _ (rs2)2 Solution Use the zero exponent and other rules to simplify each expression. c3 __ a. c3 = c3 − 3 = c0 b. = −3 ∙ −3x5 _____ x5 x5 __ x5 = −3 ∙ x5 − 5 = −3 ∙ x0 = −3 ∙ 1 = −3 c. (j2k)4 _ (j2k) ∙ (j2k)3 = = (j2k)4 _ (j2k)1 + 3 (j2k)4 _ (j2k)4 = (j2k)4 − 4 = (j2k)0 = 1 d. 5(rs2)2 ______ (rs2)2 = 5(rs2)2 − 2 = 5(rs2)0 = 5 ∙ 1 = 5 Use the product rule in the denominator. Simplify. Use the quotient rule. Simplify. Use the quotient rule. Simplify. Use the zero exponent rule. Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 21 Try It #4 Simplify each expression using the zero exponent rule of exponents. t 7 __ a. t 7 b. (de2)11 _ 2(de2)11 c Using the Negative Rule of Exponents Another useful result occurs if we relax the condition that m > n in the quotient rule even further. For example, can we h3 _ h5 ? When m < n —that is, where the difference m − n is negative—we can use the negative rule of exponents simplify to simplify the expression to its reciprocal. Divide one exponential expression by another with a larger exponent. Use our example, h3 _ h5 . h3 _ h5 = h · h · h __ __ % h2 If we were to simplify the original expression using the quotient rule, we would have h3 _ h5 = h3 − 5 = h−2 Putting the answers together, we have h−2 = a nonzero real number. 1 _ h2 . This is true for any nonzero real number, or any variable representing A factor with a negative exponent becomes the same factor with a positive exponent if it is moved across the fraction bar—from numerator to denominator or vice versa. a−n = 1 _ an and an = 1 _ a−n We have shown that the exponential expression an is defined when n is a natural number, 0, or the negative of a natural number. That means that an is defined for any integer n. Also, the product and quotient rules and all of the rules we will look at soon hold for any integer n. the negative rule of exponents For any nonzero real number a and natural number n, the negative rule of exponents states that a−n = 1 _ an Example 5 Using the Negative Exponent Rule Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents. a. θ 3 _ θ 10 Solution b. z 2 ċ z _ z 4 c. (−5t 3)4 _ (−5t 3)8 a. b. c. θ 3 1 _ _ θ 10 = θ 3 − 10 = θ−7 = θ7 = z3 − 4 = z−1 = z 4 z 4 (−5t 3)4 1 _ _ (−5t3)8 = (−5t 3)4 − 8 = (−5t 3)−4 = (−5t 3)4 = = Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 22 CHAPTER 1 PREREQUISITES Try It #5 Write each of the following quotients with a single base. Do not simplify further. Write answers with positive exponents. a. (−3t)2 ______ (−3t)8 b. f 47 _____ f 49 ⋅ f c. 2k 4 _ 5k 7 Example 6 Using the Product and Quotient Rules Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents. −7z_ (−7z)5 b. (−x)5 ∙ (−x)−5 a. b 2 ∙ b−8 c. Solution 1 __ a. b 2 ∙ b−8 = b 2 − 8 = b −6 = b 6 b. (−x)5 ∙ (−x)−5 = (−x)5 − 5 = (−x)0 = 1 c. −7z _ (−7z)5 = (−7z)1 _ (−7z)5 = (−7z)1 − 5 = (−7z)−4 = 1 _ (−7z)4 Try It #6 Write each of the following products with a single base. Do not simplify further. Write answers with positive exponents. a. t−11 ⋅ t 6 b. 2512 ____ 2513 Finding the Power of a Product To simplify the power of a product of two exponential expressions, we can use the power of a product rule of exponents, which breaks up the power of a product of factors into the product of the powers of the factors. For instance, consider (pq)3. We begin by using the associative and commutative properties of multiplication to regroup the factors. 3 factors (pq)3 = (pq) · (pq) · (pq factors = factors = p3 · q3 In other words, (pq)3 = p3 · q3. the power of a product rule of exponents For any nonzero real number a and natural number n, the negative rule of exponents states that (ab)n = an bn Example 7 Using the Power of a Product Rule Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents. a. (ab 2)3 c. (−2w3)3 e. (e−2f 2)7 b. (2t)15 d. 1 ______ (−7z)4 Solution Use the product and quotient rules and the new definitions to simplify each expression. a. (ab 2)3 = (a)3 ∙ (b2)3 = a1 ∙ 3 ∙ b2 ∙ 3 = a3b6 b. (2t)15 = (2)15 ∙ (t)15 = 215t15 = 32,768t15 c. (−2w3)3 = (−2)3 ∙ (w3)3 = −8 ∙ w 3 ∙ 3 = −8w9 d. 1 _ (−7z)4 = 1 _ (−7)4 ∙ (z)4 = 1 _ 2,401z 4 e. (e−2f 2)7 = (e−2)7 ∙ (f 2)7 = e−−14f 14 = f 14 _ e14 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 23 Try It #7 Simplify each of the following products as much as possible using the power of a product rule. Write answers with positive exponents. a. (g 2h3)5 b. (5t)3 c. (−3y5)3 d. 1 _______ (a6b7)3 e. (r 3s−2)4 Finding the Power of a Quotient To simplify the power of a quotient of two expressions, we can use the power of a quotient rule, which states that the power of a quotient of factors is the quotient of the powers of the factors. For example, let’s look at the following example. (e −2f 2)7 = f 14 _ e14 Let’s rewrite the original problem differently and look at the result. 7 (e −2f 2)7 = ( f 2 _ e2 ) It appears from the last two steps that we can use the power of a product rule as a power of a quotient rule. = f 14 _ e14 (e −2f 2)7 = ( 7 f 2 _ e2 ) = = = (f 2)7 _ (e 2)7 f 2 · 7 _ e2 · 7 f 14 _ e14 the power of a quotient rule of exponents For any real numbers a and b and any integer n, the power of a quotient rule of exponents states that n = a _ ) ( b an __ bn Example 8 Using the Power of a Quotient Rule Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents. a. ( 3 z11 ) 4 _ 6 b. ( p _ q3 ) c. ( 27 −1 t2 ) _ d. ( j3k−2)4 e. (m−2n−2)3 Solution 27 3 = 6 = 43 z11 ) a. ( 64 64 4 _ _ _ _ z11 ∙ 3 = (z11)3 = z33 p6 p1 ∙ 6 p6 q3 ) b. ( p _ _ _ _ q3 ∙ 6 = (q3)6 = q18 (−1)27 −1 −1 −1 −1 c. ( t2 ) _ _ _ _ _ (t2)27 = t54 = t2 ∙ 27 = t54 d. ( j3k−2)4 = ( j 12 j 3 ∙ 4 (j 3)4 j 3 k2 ) k2)4 = k8 = ( 13 e. (m−2n−2)3 = ( 1 1 (m2n2)3 ) = m2n2 ) _ _ _ (m2)3(n2)3 = 4 = = 3 1 _ m2 ∙ 3 ∙ n2 ∙ 3 = 1 _ m6n6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 24 CHAPTER 1 PREREQUISITES Try It #8 Simplify each of the following quotients as much as possible using the power of a quotient rule. Write answers with positive exponents. a. ( _ c ) b5 e. (c−5d−3)4 d. (p−4q3)8 c. ( 35 3 4 u8 ) b. ( 5 _ −1 w 3 ) _ Simplifying Exponential Expressions Recall that to simplify an expression means to rewrite it by combing terms or exponents; in other words, to write the expression more simply with fewer terms. The rules for exponents may be combined to simplify expressions. Example 9 Simplifying Exponential Expressions Simplify each expression and write the answer with positive exponents only. a. (6m2n−1) 3 4 e. ( x2 √ — 2 ) ( x2 √ Solution b. 175 ∙ 17−4 ∙ 17−3 — −4 2 ) f. (3w2)5 _ (6w−2)2 c. ( 2 u−1v ) ____ v −1 d. (−2a3b−1)(5a−2b2) a. (6m2n−1) 3 = (6)3(m2)3(n−1)3 The power of a product rule = 63m2 ∙ 3n−1 ∙ 3 = 216m6n−3 216m6 _ n3 = b. 175 ∙ 17−4 ∙ 17−3 = 175 − 4 − 3 = 17−2 1 ___ = 172 or 1 ___ 289 c. ( u−1v _ v−1 ) 2 = = (u−1v)2 _ (v−1)2 u−2v2 _ v−2 = u−2v2−(−2) = u−2v 4 v 4 _ = u2 d. (−2a3b−1)(5a−2b2) = −2 ∙ 5 ∙ a3 ∙ a−2 ∙ b−1 ∙ b2 e10 ∙ a3 − 2 ∙ b−1 + 2 = −10ab −. (3w2)5 _ (6w−2)2 = = 35 ∙ (w 2)5 _ 62 ∙ (w−2)2 35w 2 ∙ 5 _ 62 w −2 ∙ 2 243w10 _ 36w −#4 27w 10 − (−4) _ 4 27w14 _ 4
= = = The power rule Simplify. The negative exponent rule The product rule Simplify. The negative exponent rule The power of a quotient rule The power of a product rule The quotient rule Simplify. The negative exponent rule Commutative and associative laws of multiplication The product rule Simplify. The product rule Simplify. The zero exponent rule The power of a product rule The power rule Simplify. The quotient rule and reduce fraction Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 25 Try It #9 Simplify each expression and write the answer with positive exponents only. a. (2uv−2)−3 b. x 8 · x−12 · x 2 c. ( e2f −3 _ f −1 ) d. (9r −5s3)(3r 6s−4) −3 4 tw−2 ) e. ( _ 9 3 4 tw−2 ) ( _ 9 f. (2h2k)4 _ (7h−1k2)2 Using Scientiﬁc Notation Recall at the beginning of the section that we found the number 1.3 × 1013 when describing bits of information in digital images. Other extreme numbers include the width of a human hair, which is about 0.00005 m, and the radius of an electron, which is about 0.00000000000047 m. How can we effectively work read, compare, and calculate with numbers such as these? A shorthand method of writing very small and very large numbers is called scientific notation, in which we express numbers in terms of exponents of 10. To write a number in scientific notation, move the decimal point to the right of the first digit in the number. Write the digits as a decimal number between 1 and 10. Count the number of places n that you moved the decimal point. Multiply the decimal number by 10 raised to a power of n. If you moved the decimal left as in a very large number, n is positive. If you moved the decimal right as in a small large number, n is negative. For example, consider the number 2,780,418. Move the decimal left until it is to the right of the first nonzero digit, which is 2. 2,780418 6 places left 2.780418 We obtain 2.780418 by moving the decimal point 6 places to the left. Therefore, the exponent of 10 is 6, and it is positive because we moved the decimal point to the left. This is what we should expect for a large number. 2.780418 × 106 Working with small numbers is similar. Take, for example, the radius of an electron, 0.00000000000047 m. Perform the same series of steps as above, except move the decimal point to the right. 0.00000000000047 00000000000004.7 13 places right Be careful not to include the leading 0 in your count. We move the decimal point 13 places to the right, so the exponent of 10 is 13. The exponent is negative because we moved the decimal point to the right. This is what we should expect for a small number. 4.7 × 10−13 scientific notation A number is written in scientific notation if it is written in the form a × 10n, where 1 ≤ \|a\| < 10 and n is an integer. Example 10 Converting Standard Notation to Scientiﬁc Notation Write each number in scientific notation. a. Distance to Andromeda Galaxy from Earth: 24,000,000,000,000,000,000,000 m b. Diameter of Andromeda Galaxy: 1,300,000,000,000,000,000,000 m c. Number of stars in Andromeda Galaxy: 1,000,000,000,000 d. Diameter of electron: 0.00000000000094 m e. Probability of being struck by lightning in any single year: 0.00000143 Solution a. 24,000,000,000,000,000,000,000 m ← 22 places 2.4 × 1022 m b. 1,300,000,000,000,000,000,000 m ← 21 places 1.3 × 1021 m Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 26 CHAPTER 1 PREREQUISITES c. 1,000,000,000,000 ← 12 places 1 × 1012 d. 0.00000000000094 m → 13 places 9.4 × 10−13 m e. 0.00000143 → 6 places 1.43 × 10−6 Analysis Observe that, if the given number is greater than 1, as in examples a–c, the exponent of 10 is positive; and if the number is less than 1, as in examples d–e, the exponent is negative. Try It #10 Write each number in scientific notation. a. U.S. national debt per taxpayer (April 2014): $152,000 b. World population (April 2014): 7,158,000,000 c. World gross national income (April 2014): $85,500,000,000,000 d. Time for light to travel 1 m: 0.00000000334 s e. Probability of winning lottery (match 6 of 49 possible numbers): 0.0000000715 Converting from Scientiﬁc to Standard Notation To convert a number in scientific notation to standard notation, simply reverse the process. Move the decimal n places to the right if n is positive or n places to the left if n is negative and add zeros as needed. Remember, if n is positive, the value of the number is greater than 1, and if n is negative, the value of the number is less than one. Example 11 Converting Scientiﬁc Notation to Standard Notation Convert each number in scientific notation to standard notation. a. 3.547 × 1014 b. −2 × 106 c. 7.91 × 10−7 d. −8.05 × 10−12 Solution a. 3.547 × 1014 3.54700000000000 → 14 places 354,700,000,000,000 b. −2 × 106 −2.000000 → 6 places −2,000,000 c. 7.91 × 10−7 0000007.91 ← 7 places 0.000000791 d. −8.05 × 10−12 −000000000008.05 ← 12 places −0.00000000000805 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 EXPONENTS AND SCIENTIFIC NOTATION 27 Try It #11 Convert each number in scientific notation to standard notation. a. 7.03 × 105 c. −3.9 × 10−13 b. −8.16 × 1011 d. 8 × 10−6 Using Scientiﬁc Notation in Applications Scientific notation, used with the rules of exponents, makes calculating with large or small numbers much easier than doing so using standard notation. For example, suppose we are asked to calculate the number of atoms in 1 L of water. Each water molecule contains 3 atoms (2 hydrogen and 1 oxygen). The average drop of water contains around 1.32 × 1021 molecules of water and 1 L of water holds about 1.22 × 104 average drops. Therefore, there are approximately 3 × (1.32 × 1021) × (1.22 × 104) ≈ 4.83 × 1025 atoms in 1 L of water. We simply multiply the decimal terms and add the exponents. Imagine having to perform the calculation without using scientific notation! When performing calculations with scientific notation, be sure to write the answer in proper scientific notation. For example, consider the product (7 × 104) × (5 × 106) = 35 × 1010. The answer is not in proper scientific notation because 35 is greater than 10. Consider 35 as 3.5 × 10. That adds a ten to the exponent of the answer. (35) × 1010 = (3.5 × 10) × 1010 = 3.5 × (10 × 1010) = 3.5 × 1011 Example 12 Using Scientiﬁc Notation Perform the operations and write the answer in scientific notation. a. (8.14 × 10−7) (6.5 × 1010) b. (4 × 105) ÷ (−1.52 × 109) c. (2.7 × 105) (6.04 × 1013) d. (1.2 × 108) ÷ (9.6 × 105) e. (3.33 × 104) (−1.05 × 107) (5.62 × 105) Solution a. (8.14 × 10−7) (6.5 × 1010) = (8.14 × 6.5) (10−7 × 1010) = (52.91) (103) b. (4 × 105) ÷ (−1.52 × 109) = ( = 5.291 × 104 4 ) ( ______ −1.52 105 ) ___ 109 ≈ (−2.63) (10−4) = −2.63 × 10−4 c. (2.7 × 105) (6.04 × 1013) = (2.7 × 6.04) (105 × 1013) = (16.308) (1018) d. (1.2 × 108) ÷ (9.6 × 105) = ( = 1.6308 × 1019 108 ) ) ( ___ 105 = (0.125) (103) 1.2 ___ 9.6 = 1.25 × 102 Commutative and associative properties of multiplication Product rule of exponents Scientific notation Commutative and associative properties of multiplication Quotient rule of exponents Scientific notation Commutative and associative properties of multiplication Product rule of exponents Scientific notation Commutative and associative properties of multiplication Quotient rule of exponents Scientific notation e. (3.33 × 104)(−1.05 × 107) (5.62 × 105) = [3.33 × (−1.05) × 5.62] (104 × 107 × 105) ≈ (−19.65) (1016) = −1.965 × 1017 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 28 CHAPTER 1 PREREQUISITES Try It #12 Perform the operations and write the answer in scientific notation. a. (−7.5 × 108)(1.13 × 10−2) b. (1.24 × 1011) ÷ (1.55 × 1018) c. (3.72 × 109)(8 × 103) d. (9.933 × 1023) ÷ (−2.31 × 1017) e. (−6.04 × 109)(7.3 × 102)(−2.81 × 102) Example 13 Applying Scientiﬁc Notation to Solve Problems In April 2014, the population of the United States was about 308,000,000 people. The national debt was about $17,547,000,000,000. Write each number in scientific notation, rounding figures to two decimal places, and find the amount of the debt per U.S. citizen. Write the answer in both scientific and standard notations. Solution The population was 308,000,000 = 3.08 × 108. The national debt was $17,547,000,000,000 ≈ $1.75 × 1013. To find the amount of debt per citizen, divide the national debt by the number of citizens. (1.75 × 1013) ÷ (3.08 × 108) = ( 1013 ) ____ 108 1.75 ____ 3.08 ) × ( ≈ 0.57 × 105 The debt per citizen at the time was about $5.7 × 104, or $57,000. = 5.7 × 104 Try It #13 An average human body contains around 30,000,000,000,000 red blood cells. Each cell measures approximately 0.000008 m long. Write each number in scientific notation and find the total length if the cells were laid end-to-end. Write the answer in both scientific and standard notations. Access these online resources for additional instruction and practice with exponents and scientific notation. • Exponential Notation (http://openstaxcollege.org/l/exponnot) • Properties of Exponents (http://openstaxcollege.org/l/exponprops) • Zero Exponent (http://openstaxcollege.org/l/zeroexponent) • Simplify Exponent Expressions (http://openstaxcollege.org/l/exponexpres) • Quotient Rule for Exponents (http://openstaxcollege.org/l/quotofexpon) • Scientiﬁc Notation (http://openstaxcollege.org/l/scientiﬁcnota) • Converting to Decimal Notation (http://openstaxcollege.org/l/decimalnota) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.2 SECTION EXERCISES 29 1.2 SECTION EXERCISES VERBAL 1. Is 23 the same as 32? Explain. 3. What is the purpose of scientific notation? 2. When can you add two exponents? 4. Explain what a negative exponent does. NUMERIC For the following exercises, simplify the given expression. Write answers with positive exponents. 8. 44 ÷ 4 7. 32 · 33 6. 15−2 5. 92 −2 9. (22) 2 13. (80) 10. (5 − 8)0 14. 5−2 ÷ 52
11. 113 ÷ 114 12. 65 · 6−7 For the following exercises, write each expression with a single base. Do not simplify further. Write answers with positive exponents. 15. 42 · 43 ÷ 4−4 18. 106 ÷ (1010) 17. (123 · 12) 16. −2 10 612 ___ 69 19. 7−6 · 7−3 20. (33 ÷ 34) 5 For the following exercises, express the decimal in scientific notation. 21. 0.0000314 22. 148,000,000 For the following exercises, convert each number in scientific notation to standard notation. 23. 1.6 × 1010 24. 9.8 × 10−9 ALGEBRAIC For the following exercises, simplify the given expression. Write answers with positive exponents. x−3 _ 2 25. a3a2 _ a 26. mn2 _ m−2 −5 28. ( y 2 ) 27. ( b3c4) 29. ab 2 ÷ d−3 33. p−4q2 _ p 2q−3 37. 52m ÷ 50m 41. (b−3c) 3 −1 30. (w0x5) 34. (l × w)2 — 2 x ) 38. ( 16 √ _ y−1 42. (x 2y13 ÷ y0) 2 31. m4 ___ n0 3 35. (y7) ÷ x 14 39. 23 _____ (3a)−2 −2 43. (9z3) y 32. y−4 (y2) 2 2 a __ ) 36. ( 23 40. (ma6) 2 1 _ m3a2 REAL-WORLD APPLICATIONS 44. To reach escape velocity, a rocket must travel at the rate of 2.2 × 106 ft/min. Rewrite the rate in standard notation. 45. A dime is the thinnest coin in U.S. currency. A dime’s thickness measures 2.2 × 106 m. Rewrite the number in standard notation. 46. The average distance between Earth and the Sun is 92,960,000 mi. Rewrite the distance using scientific notation. 47. A terabyte is made of approximately 1,099,500,000,000 bytes. Rewrite in scientific notation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 CHAPTER 1 PREREQUISITES 49. One picometer is approximately 3.397 × 10−11 in. Rewrite this length using standard notation. 48. The Gross Domestic Product (GDP) for the United States in the first quarter of 2014 was $1.71496 × 1013. Rewrite the GDP in standard notation. 50. The value of the services sector of the U.S. economy in the first quarter of 2012 was $10,633.6 billion. Rewrite this amount in scientific notation. TECHNOLOGY For the following exercises, use a graphing calculator to simplify. Round the answers to the nearest hundredth. 2 51. ( 123 m33 ) ______ 4−3 52. 173 ÷ 152x3 EXTENSIONS For the following exercises, simplify the given expression. Write answers with positive exponents. 32 ) 53. ( __ a3 −2 2 a4 ) ( __ 22 54. (62 − 24) −5 2 x __ ÷ ( ) y 55. m2n3 _____ a2c−3 · a−7n−2 ______ m2c4 56. ( x 6y 3 _____ · x 3y−3 y −7 ___ x −3 ) 10 2 57. ( (ab2c)−3 ) _______ b−3 58. Avogadro’s constant is used to calculate the number of particles in a mole. A mole is a basic unit in chemistry to measure the amount of a substance. The constant is 6.0221413 × 1023. Write Avogadro’s constant in standard notation. 59. Planck’s constant is an important unit of measure in quantum physics. It describes the relationship between energy and frequency. The constant is written as 6.62606957 × 10−34. Write Planck’s constant in standard notation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 RADICALS AND RATIONAL EXPRESSIONS 31 LEARNING OBJECTIVES In this section, you will: • Evaluate square roots. • Use the product rule to simplify square roots. • Use the quotient rule to simplify square roots. • Add and subtract square roots. • Rationalize denominators. • Use rational roots. 1. 3 RADICALS AND RATIONAL EXPRESSIONS A hardware store sells 16-ft ladders and 24-ft ladders. A window is located 12 feet above the ground. A ladder needs to be purchased that will reach the window from a point on the ground 5 feet from the building. To find out the length of ladder needed, we can draw a right triangle as shown in Figure 1, and use the Pythagorean Theorem. 12 feet c 5 feet Figure 1 a2 + b2 = c2 52 + 122 = c2 169 = c2 Now, we need to find out the length that, when squared, is 169, to determine which ladder to choose. In other words, we need to find a square root. In this section, we will investigate methods of finding solutions to problems such as this one. Evaluating Square Roots When the square root of a number is squared, the result is the original number. Since 42 = 16, the square root of 16 is 4. The square root function is the inverse of the squaring function just as subtraction is the inverse of addition. To undo squaring, we take the square root. In general terms, if a is a positive real number, then the square root of a is a number that, when multiplied by itself, gives a. The square root could be positive or negative because multiplying two negative numbers gives a positive number. The principal square root is the nonnegative number that when multiplied by itself equals a. The square root obtained using a calculator is the principal square root. The principal square root of a is written as √ the radicand, and the entire expression is called a radical expression. — a . The symbol is called a radical, the term under the symbol is called Radical Radicand — 25 √ Radical expression principal square root The principal square root of a is the nonnegative number that, when multiplied by itself, equals a. It is written as a radical expression, with a symbol called a radical over the term called the radicand: √ a . — Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 CHAPTER 1 PREREQUISITES — 25 = ±5? Q & A… Does √ No. Although both 52 and (−5)2 are 25, the radical symbol implies only a nonnegative root, the principal square root. The principal square root of 25 is √ 25 = 5. — Example 1 Evaluating Square Roots Evaluate each expression. — 100 b. √ a. √ — — 16 √ Solution c. √ — 25 + 144 d. √ — 49 − √ — 81 — a. √ b. √ c. √ d. √ — — — 100 = 10 because 102 = 100 √ 16 = √ 25 + 144 = √ 49 − √ — — — — 4 = 2 because 42 = 16 and 22 = 4 169 = 13 because 132 = 169 81 = 7 − 9 = −2 because 72 = 49 and 92 = 81 Q & A… For √ No. √ — 25 + 144 , can we find the square roots before adding? 144 = 5 + 12 = 17. This is not equivalent to √ 25 + √ — — — 25 + 144 = 13. The order of operations requires us to add the terms in the radicand before finding the square root. Try It #1 a. √ — 225 b. √ — — 81 √ c. √ — 25 − 9 d. √ — 36 + √ — 121 Using the Product Rule to Simplify Square Roots To simplify a square root, we rewrite it such that there are no perfect squares in the radicand. There are several properties of square roots that allow us to simplify complicated radical expressions. The first rule we will look at is the product rule for simplifying square roots, which allows us to separate the square root of a product of two numbers 5 . We can also use into the product of two separate rational expressions. For instance, we can rewrite √ the product rule to express the product of multiple radical expressions as a single radical expression. 15 as √ 3 ∙ √ — — — the product rule for simplifying square roots If a and b are nonnegative, the square root of the product ab is equal to the product of the square roots of a and b. — ab = √ — a · √ — b √ How To… Given a square root radical expression, use the product rule to simplify it. 1. Factor any perfect squares from the radicand. 2. Write the radical expression as a product of radical expressions. 3. Simplify. Example 2 Using the Product Rule to Simplify Square Roots Simplify the radical expression. a. √ — 300 b. √ — 162a5b4 Solution a. √ — — — 100 ∙ 3 100 ∙ √ 3 3 — √ 10 √ Factor perfect square from radicand. Write radical expression as product of radical expressions. Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 RADICALS AND RATIONAL EXPRESSIONS 33 b. √ — — 81a4b4 ∙ 2a 81a4b4 ∙ √ 2a 2a √ 9a2b2 √ — — Factor perfect square from radicand. Write radical expression as product of radical expressions. Simplify. Try It #2 Simplify √ — 50x2y3z . How To… Given the product of multiple radical expressions, use the product rule to combine them into one radical expression. 1. Express the product of multiple radical expressions as a single radical expression. 2. Simplify. Example 3 Using the Product Rule to Simplify the Product of Multiple Square Roots Simplify the radical expression. — 12 ∙ √ — 3 √ Solution — 12 ∙ 3 — 36 √ √ 6 Express the product as a single radical expression. Simplify. Try It #3 Simplify √ — 50x · √ — 2x assuming x > 0. Using the Quotient Rule to Simplify Square Roots Just as we can rewrite the square root of a product as a product of square roots, so too can we rewrite the square root of a quotient as a quotient of square roots, using the quotient rule for simplifying square roots. It can be helpful to separate the numerator and denominator of a fraction under a radical so that we can take their square roots separately. ___ We can rewrite √ 5 __ 2 as — 5 √ ____ . 2 √ — the quotient rule for simplifying square roots The square root of the quotient a _ is equal to the quotient of the square roots of a and b, where b ≠ 0. b a √ ____ b √ ___ √ a __ b = — — How To… Given a radical expression, use the quotient rule to simplify it. 1. Write the radical expression as the quotient of two radical expression. 2. Simplify the numerator and denominator. Example 4 Using the Quotient Rule to Simplify Square Roots Simplify the radical expression. Solution — — 5 √ _____ 36 √ 5 √ ____ 6 — ____ 5 ___ 36 √ Write as quotient of two radical expressions. Simplify denominator. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 34 CHAPTER 1 PREREQUISITES Try It #4 ____ Simplify √ 2x2 ____ 9y4 . Example 5 Using the Quotient Rule to Simplify an Expression with Two Square Roots Simplify the radical expression. Solution _______ 234x11y √ ______ 26x7y — 9x4 √ 3x 2 — 234x11y √ _________ 26x7y √ — Combine numerator and denominator into one radical expression. Simplify fraction. Simplify square root. Try It #5 Simplify — 9a5 b 14 √ ________ . — 3a4b5 √ Adding and Subtracting Square Roots We can add or subtract radical expressions only when they have the same radicand and when they have the same — 2 . However, it is often possible to radical type such as square roots. For example, the sum of √ — simplify radical expressions, and tha
t may change the radicand. The radical expression √ 18 can be written with a 2 in the radicand, as 3 √ — 2 and 3 √ — 2 is 4 √ — 2 , so √ 2 + 3 √ 2 = 4 √ 18 = √ 2 + √ — 2 . — — — — How To… Given a radical expression requiring addition or subtraction of square roots, solve. 1. Simplify each radical expression. 2. Add or subtract expressions with equal radicands. Example 6 Adding Square Roots Add 5 √ — 12 + 2 √ — 3 . Solution We can rewrite 5 √ the expression becomes 5(2) √ 12 as 5 √ — — — 4 · 3 . According the product rule, this becomes 5 √ — 4 √ — 3 . The square root of √ — 4 is 2, so 3 , which is 10 √ 3 . Now the terms have the same radicand so we can add. — 10 √ — 3 + 2 √ — 3 = 12 √ — 3 Try It #6 Add √ — 5 + 6 √ — 20 . Example 7 Subtracting Square Roots Subtract 20 √ — 72a3b4c − 14 √ — 8a3b4c . Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 RADICALS AND RATIONAL EXPRESSIONS 35 Solution Rewrite each term so they have equal radicands. 20 √ — 72a3b4c = 20 √ — — — 4 √ 9 √ 2 √ = 20(3)(2) ∣ a ∣ b2 √ 2ac = 120 ∣ a ∣ b2 √ — — a √ 2ac — — a2 √ — 2 (b2) √ — c 14 √ — 8a3b4c = 14 √ — a2 √ — 2 (b2) √ — c — — 2 √ 4 √ = 14(2) ∣ a ∣ b2 √ = 28 ∣ a ∣ b2 √ — 2ac — a √ 2ac — Now the terms have the same radicand so we can subtract. — — 120 ∣ a ∣ b2 √ 2ac − 28 ∣ a ∣ b2 √ 2ac = 92 ∣ a ∣ b2 √ — 2ac Try It #7 Subtract 3 √ — 80x − 4 √ — 45x . Rationalizing Denominators When an expression involving square root radicals is written in simplest form, it will not contain a radical in the denominator. We can remove radicals from the denominators of fractions using a process called rationalizing the denominator. We know that multiplying by 1 does not change the value of an expression. We use this property of multiplication to change expressions that contain radicals in the denominator. To remove radicals from the denominators of fractions, multiply by the form of 1 that will eliminate the radical. For a denominator containing a single term, multiply by the radical in the denominator over itself. In other words, if the denominator is b √ — c , multiply by — c √ ____ . — c √ For a denominator containing the sum of a rational and an irrational term, multiply the numerator and denominator by the conjugate of the denominator, which is found by changing the sign of the radical portion of the denominator. If the denominator is a + b √ c , then the conjugate is a − b √ — c . — How To… Given an expression with a single square root radical term in the denominator, rationalize the denominator. 1. Multiply the numerator and denominator by the radical in the denominator. 2. Simplify. Example 8 Rationalizing a Denominator Containing a Single Term Write in simplest form. — 3 2 √ ______ 10 3 √ — Solution The radical in the denominator is √ — 10 . So multiply the fraction by . Then simplify. — 10 √ _____ 10 √ — — 3 2 √ ______ · 10 3 √ — — — 10 √ _____ 10 √ — 30 2 √ ______ 30 30 √ _____ 15 — Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 36 Try It #8 Write — 3 12 √ ______ in simplest form. — 2 √ CHAPTER 1 PREREQUISITES How To… Given an expression with a radical term and a constant in the denominator, rationalize the denominator. 1. Find the conjugate of the denominator. 2. Multiply the numerator and denominator by the conjugate. 3. Use the distributive property. 4. Simplify. Example 9 Rationalizing a Denominator Containing Two Terms Write 4 _______ 1 + √ — in simplest form. 5 Solution Begin by finding the conjugate of the denominator by writing the denominator and changing the sign. So the conjugate — 5 . Then multiply the fraction by — 1 − √ _______ 1 − √ 5 . — 5 4 _______ ∙ 1 + √ — 5 1 − √ _______ ________ −4 — 5 − 1 √ Use the distributive property. Simplify. of 1 + √ — 5 is 1 − √ Try It #9 Write 7 _______ 2 + √ — in simplest form. 3 Using Rational Roots Although square roots are the most common rational roots, we can also find cuberoots, 4th roots, 5th roots, and more. Just as the square root function is the inverse of the squaring function, these roots are the inverse of their respective power functions. These functions can be useful when we need to determine the number that, when raised to a certain power, gives a certain number. Understanding n th Roots Suppose we know that a3 = 8. We want to find what number raised to the 3rd power is equal to 8. Since 23 = 8, we say that 2 is the cube root of 8. The nth root of a is a number that, when raised to the nth power, gives a. For example, −3 is the 5th root of −243 because (−3)5 = −243. If a is a real number with at least one nth root, then the principal nth root of a is the number with the same sign as a that, when raised to the nth power, equals a. The principal nth root of a is written as expression, n is called the index of the radical. — n √ a , where n is a positive integer greater than or equal to 2. In the radical principal nth root √ If a is a real number with at least one nth root, then the principal nth root of a, written as the same sign as a that, when raised to the nth power, equals a. The index of the radical is n. n — a , is the number with Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 RADICALS AND RATIONAL EXPRESSIONS 37 Example 10 Simplifying n th Roots Simplify each of the following: — √ −32 b. a, 024 Solution 5 a. √ — −32 = −2 because (−2)5 = −32 c. − 3 √ — 8x6 ___ 125 d. 8 √ 4 — 4 3 − √ — 48 b. First, express the product as a single radical expression. 4 √ — 4,096 = 8 because 84 = 4,096 c. — 3 8x6 − √ ________ 125 √ — 3 Write as quotient of two radical expressions. −2x 2 _____ 5 Simplify. 4 √ d Simplify to get equal radicands. Add. Try It #10 Simplify. 3 a. √ — −216 — 4 √ b. 3 √ 80 _______ 5 — 4 3 c. 6 √ — 9,000 + 7 √ 3 — 576 Using Rational Exponents Radical expressions can also be written without using the radical symbol. We can use rational (fractional) exponents. The index must be a positive integer. If the index n is even, then a cannot be negative We can also have rational exponents with numerators other than 1. In these cases, the exponent must be a fraction in lowest terms. We raise the base to a power and take an nth root. The numerator tells us the power and the denominator tells us the root √ — am All of the properties of exponents that we learned for integer exponents also hold for rational exponents. rational exponents Rational exponents are another way to express principal nth roots. The general form for converting between a radical expression with a radical symbol and one with a rational exponent is — am How To… Given an expression with a rational exponent, write the expression as a radical. 1. Determine the power by looking at the numerator of the exponent. 2. Determine the root by looking at the denominator of the exponent. 3. Using the base as the radicand, raise the radicand to the power and use the root as the index. Example 11 Writing Rational Exponents as Radicals 2 __ Write 343 as a radical. Simplify. 3 Solution The 2 tells us the power and the 3 tells us the root. 2 = ( _ 343 3 √ 3 — 343 ) 2 = √ 3 — 3432 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 38 CHAPTER 1 PREREQUISITES — 3 √ 343 = 7 because 7 3 = 343. Because the cube root is easy to find, it is easiest to find the cube root before We know that squaring for this problem. In general, it is easier to find the root first and then raise it to a power. 2 _ = ( 3 343 2 = 72 = 49 343 ) √ — 3 Try It #11 5 _ Write 9 as a radical. Simplify. 2 Example 12 Writing Radicals as Rational Exponents 7 Write using a rational exponent. 4 _ — a2 √ Solution 2 . We get 4 ___ _ The power is 2 and the root is 7, so the rational exponent will be 7 4 _ — a2 √ −2 _ . = 4a 7 7 . Using properties of exponents, we get 2 _ a 7 Try It #12 Write x √ — (5y)9 using a rational exponent. Example 13 Simplifying Rational Exponents Simplify: 3 _ 1 _ ) ( 3x a. 5 ( 2x ) 5 4 Solution 1 _ 3 __ 5 4 x a. 30 x 3 _ + 1 _ 5 4 30x 19 _ 20 30x 1 _ 2 9 b. ( ) ___ 16 ____ √ 9 ___ 16 — 9 √ _____ 16 √ 3 _ 4 — b. ( 1 __ − 2 16 ) ___ 9 Multiply the coefficient. Use properties of exponents. Simplify. Use definition of negative exponents. Rewrite as a radical. Use the quotient rule. Simplify. Try It #13 1 _ 6 _ ) . ( 14x 5 3 Simplify 8x Access these online resources for additional instruction and practice with radicals and rational exponents. • Radicals (http://openstaxcollege.org/l/introradical) • Rational Exponents (http://openstaxcollege.org/l/rationexpon) • Simplify Radicals (http://openstaxcollege.org/l/simpradical) • Rationalize Denominator (http://openstaxcollege.org/l/rationdenom) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.3 SECTION EXERCISES 39 1.3 SECTION EXERCISES VERBAL 1. What does it mean when a radical does not have 2. Where would radicals come in the order of an index? Is the expression equal to the radicand? Explain. operations? Explain why. 3. Every number will have two square roots. What is 4. Can a radical with a negative radicand have a real the principal square root? square root? Why or why not? NUMERIC For the following exercises, simplify each expression. 5. √ — 256 9. √ — 196 13. √ ____ 81 ___ 5 18 ______ — 162 √ 17. 21. √ — 150 ____ 4 _ 225 25. √ 29. 8 ________ 1 − √ 17 — — 4 √ 33. 15 √ 125 _________ 5 — 4 6. √ — — √ 256 7. √ — 4(9 + 16) 10. √ — 1 14. √ — 800 18. √ — 192 11. √ — 98 15. √ — 169 + √ — 144 19. 14 √ — 6 − 6 √ — 24 — 121 — 8. √ 289 − √ ___ 27 12. √ _ 64 ___ 8 16. √ _ 50 20. 15 √ — 5 + 7 √ — 45 ____ 96 _ 100 ____ 405 _ 324 22. √ 26. √ 23. ( √ — 42 ) ( √ — 30 ) 24. 12 √ — 3 − 4 √ — 75 ____ 360 _ 361 27. √ 28. 5 _______ 1 + √ 3 — 4 30. √ — 16 3 31. √ — 128 + 3 √ 3 — 2 — −32 ____ 243 32. 5 √ 3 34. 3 √ — −432 + √ 3 — 16 ALGEBRAIC For the following exercises, simplify each expression. 35. √ — 400x4 36. √ — 4y2 37. √ — 49p 1 _ 38. (144p2q6) 2 289 — 5 _ 2 39. m √ ______ 225x3 _ 49x ____ 32 _ 14d 43. √ 47. √ 40. 9 √ — 3m2 + √ — 27 41. 3 √ —
ab2 − b √ — a 44. 3 √ — 44z + √ — 99z 45. √ — 50y8 3 _ 48. q √ 2 — 63p 49. — √ 8 ________ 1 − √ 3x — 3 _ 2 51. w √ — 3 _ 32 − w 2 √ — 50 52. √ — 108x4 + √ — 27x4 53. — 12x √ ________ 3 2 + 2 √ — 55. √ — 125n10 56. √ ____ 42q ____ 36q3 57. √ ______ 81m _ 361m2 42. — 2n 4 √ _______ 16n4 √ — 46. √ — 490bc2 50. √ ______ 20 _ 121d4 54. √ — 147k3 58. √ — 72c − 2 √ — 2c Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 40 CHAPTER 1 PREREQUISITES 59. √ ______ 144 _ 324d2 3 60. √ — 24x6 + √ 3 — 81x6 61. 4 √ — 162x6 _____ 16x4 3 62. √ — 64y 3 63. √ — 128z3 − √ 3 — −16z3 5 64. √ — 1,024c10 REAL-WORLD APPLICATIONS 65. A guy wire for a suspension bridge runs from the ground diagonally to the top of the closest pylon to make a triangle. We can use the Pythagorean Theorem to find the length of guy wire needed. The square of the distance between the wire on the ground and the pylon on the ground is 90,000 feet. The square of the height of the pylon is 160,000 feet. So the length of the guy wire can be found by 90,000 + 160,000 . What is the length evaluating √ of the guy wire? — EXTENSIONS For the following exercises, simplify each expression. 67. — — 8 − √ 16 √ ___________ − 16 68. 4 _________ 1 _ 3 8 66. A car accelerates at a rate of 6 − — 4 √ ____ m/s2 where t t √ — is the time in seconds after the car moves from rest. Simplify the expression. — 69. mn3 √ ________ · c−3 a2 √ — a−7n−2 _______ m2c4 √ — 70. a _______ a − √ c — 71. — — 64y + 4 √ x √ __ 128y √ — y — — 72. ( 250x2 √ ________ 100b3 √ )( — b 7 √ _______ 125x √ — ) 3 — ______________ — 256 64 + √ √ __ 256 64 + √ √ — — 4 73. √ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.4 POLYNOMIALS 41 LEARNING OBJECTIVES Identify the degree and leading coefﬁcient of polynomials. In this section, you will: • • Add and subtract polynomials. • Multiply polynomials. • Use FOIL to multiply binomials. • Perform operations with polynomials of several variables. 1. 4 POLYNOMIALS Earl is building a doghouse, whose front is in the shape of a square topped with a triangle. There will be a rectangular door through which the dog can enter and exit the house. Earl wants to find the area of the front of the doghouse so that he can purchase the correct amount of paint. Using the measurements of the front of the house, shown in Figure 1, we can create an expression that combines several variable terms, allowing us to solve this problem and others like it. 1 foot x First find the area of the square in square feet. Then find the area of the triangle in square feet. 3 2 feet 2x Figure 1 A = s2 = (2x)2 = 4x2 1 _ A = bh 2 3 1 ) (2x Next find the area of the rectangular door in square feet. A = lw = x · 1 = x The area of the front of the doghouse can be found by adding the areas of the square and the triangle, and then 3 1 _ _ x − x ft 2, or 4x 2 + subtracting the area of the rectangle. When we do this, we get 4x 2 + x ft 2. 2 2 In this section, we will examine expressions such as this one, which combine several variable terms. Identifying the Degree and Leading Coefﬁcient of Polynomials The formula just found is an example of a polynomial, which is a sum of or difference of terms, each consisting of a variable raised to a nonnegative integer power. A number multiplied by a variable raised to an exponent, such as 384π, is known as a coefficient. Coefficients can be positive, negative, or zero, and can be whole numbers, decimals, or fractions. Each product ai x i, such as 384πw, is a term of a polynomial. If a term does not contain a variable, it is called a constant. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 42 CHAPTER 1 PREREQUISITES A polynomial containing only one term, such as 5x 4, is called a monomial. A polynomial containing two terms, such as 2x − 9, is called a binomial. A polynomial containing three terms, such as −3x 2 + 8x − 7, is called a trinomial. We can find the degree of a polynomial by identifying the highest power of the variable that occurs in the polynomial. The term with the highest degree is called the leading term because it is usually written first. The coefficient of the leading term is called the leading coefficient. When a polynomial is written so that the powers are descending, we say that it is in standard form. Leading coefficient Degree an x n + ... + a2 x 2 + a1 x + a0 Leading term polynomials A polynomial is an expression that can be written in the form an x n + ... + a2 x 2 + a1 x + a0 Each real number ai is called a coefficient. The number a0 that is not multiplied by a variable is called a constant. Each product ai x i is a term of a polynomial. The highest power of the variable that occurs in the polynomial is called the degree of a polynomial. The leading term is the term with the highest power, and its coefficient is called the leading coefficient. How To… Given a polynomial expression, identify the degree and leading coefficient. 1. Find the highest power of x to determine the degree. 2. Identify the term containing the highest power of x to find the leading term. 3. Identify the coefficient of the leading term. Example 1 Identifying the Degree and Leading Coefﬁcient of a Polynomial For the following polynomials, identify the degree, the leading term, and the leading coefficient. a. 3 + 2x2 − 4x3 b. 5t5 − 2t3 + 7t c. 6p − p3 − 2 Solution a. The highest power of x is 3, so the degree is 3. The leading term is the term containing that degree, −4x 3. The leading coefficient is the coefficient of that term, −4. b. The highest power of t is 5, so the degree is 5. The leading term is the term containing that degree, 5t5. The leading coefficient is the coefficient of that term, 5. c. The highest power of p is 3, so the degree is 3. The leading term is the term containing that degree, −p3, The leading coefficient is the coefficient of that term, −1. Try It #1 Identify the degree, leading term, and leading coefficient of the polynomial 4x 2 − x 6 + 2x − 6. Adding and Subtracting Polynomials We can add and subtract polynomials by combining like terms, which are terms that contain the same variables raised to the same exponents. For example, 5x 2 and −2x 2 are like terms, and can be added to get 3x 2, but 3x and 3x 2 are not like terms, and therefore cannot be added. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.4 POLYNOMIALS 43 How To… Given multiple polynomials, add or subtract them to simplify the expressions. 1. Combine like terms. 2. Simplify and write in standard form. Example 2 Adding Polynomials Find the sum. (12x 2 + 9x − 21) + (4x 3 + 8x 2 − 5x + 20) Solution 4x 3 + (12x 2 + 8x 2) + (9x − 5x) + (−21 + 20) Combine like terms. 4x 3 + 20x 2 + 4x − 1 Simplify. Analysis We can check our answers to these types of problems using a graphing calculator. To check, graph the problem as given along with the simplified answer. The two graphs should be equivalent. Be sure to use the same window to compare the graphs. Using different windows can make the expressions seem equivalent when they are not. Try It #2 Find the sum. (2x 3 + 5x 2 − x + 1) + (2x 2 − 3x − 4) Example 3 Subtracting Polynomials Find the difference. (7x 4 − x 2 + 6x + 1) − (5x 3 − 2x 2 + 3x + 2) Solution 7x 4 − 5x 3 + (−x 2 + 2x 2) + (6x − 3x) + (1 − 2) Combine like terms. 7x 4 − 5x 3 + x 2 + 3x − 1 Simplify. Analysis Note that finding the difference between two polynomials is the same as adding the opposite of the second polynomial to the first. Try It #3 Find the difference. (−7x 3 − 7x 2 + 6x − 2) − (4x 3 − 6x 2 − x + 7) Multiplying Polynomials Multiplying polynomials is a bit more challenging than adding and subtracting polynomials. We must use the distributive property to multiply each term in the first polynomial by each term in the second polynomial. We then combine like terms. We can also use a shortcut called the FOIL method when multiplying binomials. Certain special products follow patterns that we can memorize and use instead of multiplying the polynomials by hand each time. We will look at a variety of ways to multiply polynomials. Multiplying Polynomials Using the Distributive Property To multiply a number by a polynomial, we use the distributive property. The number must be distributed to each term of the polynomial. We can distribute the 2 in 2(x + 7) to obtain the equivalent expression 2x + 14. When multiplying polynomials, the distributive property allows us to multiply each term of the first polynomial by each term of the second. We then add the products together and combine like terms to simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 44 CHAPTER 1 PREREQUISITES How To… Given the multiplication of two polynomials, use the distributive property to simplify the expression. 1. Multiply each term of the first polynomial by each term of the second. 2. Combine like terms. 3. Simplify. Example 4 Multiplying Polynomials Using the Distributive Property Find the product. (2x + 1) (3x 2 − x + 4) Solution 2x (3x 2 − x + 4) + 1(3x 2 − x + 4) (6x 3 − 2x 2 + 8x ) + (3x 2 − x + 4) Use the distributive property. Multiply. 6x 3 + (−2x 2 + 3x 2) + (8x − x) + 4 Combine like terms. 6x 3 + x 2 + 7x + 4 Simplify. Analysis We can use a table to keep track of our work, as shown in Table 1. Write one polynomial across the top and the other down the side. For each box in the table, multiply the term for that row by the term for that column. Then add all of the terms together, combine like terms, and simplify. 2x +1 −x 3x 2 6x 3 −2x 2 −x 3x 2 Table 1 +4 8x 4 Try It #4 Find the product. (3x + 2) (x3 − 4x2 + 7) Using FOIL to Multiply Binomials A shortcut called FOIL is sometimes used to find the product of two binomials. It is called FOIL because we multiply the first terms, the outer terms, the inner terms, and then the last terms of each binomial. First terms Last terms Inner terms Outer terms The FOIL method arises out of the
distributive property. We are simply multiplying each term of the first binomial by each term of the second binomial, and then combining like terms. How To… Given two binomials, use FOIL to simplify the expression. 1. Multiply the first terms of each binomial. 2. Multiply the outer terms of the binomials. 3. Multiply the inner terms of the binomials. 4. Multiply the last terms of each binomial. 5. Add the products. 6. Combine like terms and simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.4 POLYNOMIALS 45 Example 5 Using FOIL to Multiply Binomials Use FOIL to find the product. (2x − 10)(3x + 3) Solution Find the product of the first terms. 2x − 18 3x + 3 2x · 3x = 6x2 Find the product of the outer terms. 2x − 18 3x + 3 2x · 3 = 6x Find the product of the inner terms. 2x − 18 3x + 3 −18 · 3x = −54x Find the product of the last terms. 2x − 18 3x + 3 −18 · 3 = −54 6x2 + 6x − 54x − 54 Add the products. 6x2 + (6x − 54x) − 54 Combine like terms. 6x2 − 48x − 54 Simplify. Try It #5 Use FOIL to find the product. (x + 7)(3x − 5) Perfect Square Trinomials Certain binomial products have special forms. When a binomial is squared, the result is called a perfect square trinomial. We can find the square by multiplying the binomial by itself. However, there is a special form that each of these perfect square trinomials takes, and memorizing the form makes squaring binomials much easier and faster. Let’s look at a few perfect square trinomials to familiarize ourselves with the form. (x + 5)2 = x2 + 10x + 25 (x − 3)2 = x2 − 6x + 9 (4x − 1)2 = 4x2 − 8x + 1 Notice that the first term of each trinomial is the square of the first term of the binomial and, similarly, the last term of each trinomial is the square of the last term of the binomial. The middle term is double the product of the two terms. Lastly, we see that the first sign of the trinomial is the same as the sign of the binomial. perfect square trinomials When a binomial is squared, the result is the first term squared added to double the product of both terms and the last term squared. (x + a)2 = (x + a)(x + a) = x2 + 2ax + a2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 46 CHAPTER 1 PREREQUISITES How To… Given a binomial, square it using the formula for perfect square trinomials. 1. Square the first term of the binomial. 2. Square the last term of the binomial. 3. For the middle term of the trinomial, double the product of the two terms. 4. Add and simplify. Example 6 Expanding Perfect Squares Expand (3x − 8)2. Solution Begin by squaring the first term and the last term. For the middle term of the trinomial, double the product of the two terms. (3x)2 − 2(3x)(8) + (−8)2 9x2 − 48x + 64 Simplify. Try It #6 Expand (4x − 1)2. Difference of Squares Another special product is called the difference of squares, which occurs when we multiply a binomial by another binomial with the same terms but the opposite sign. Let’s see what happens when we multiply (x + 1)(x − 1) using the FOIL method. (x + 1)(x − 1) = x2 − x + x − 1 = x2 − 1 The middle term drops out, resulting in a difference of squares. Just as we did with the perfect squares, let’s look at a few examples. (x + 5)(x − 5) = x2 − 25 (x + 11)(x − 11) = x2 − 121 (2x + 3)(2x − 3) = 4x2 − 9 Because the sign changes in the second binomial, the outer and inner terms cancel each other out, and we are left only with the square of the first term minus the square of the last term. Q & A… Is there a special form for the sum of squares? No. The difference of squares occurs because the opposite signs of the binomials cause the middle terms to disappear. There are no two binomials that multiply to equal a sum of squares. difference of squares When a binomial is multiplied by a binomial with the same terms separated by the opposite sign, the result is the square of the first term minus the square of the last term. (a + b)(a − b) = a2 − b2 How To… Given a binomial multiplied by a binomial with the same terms but the opposite sign, find the difference of squares. 1. Square the first term of the binomials. 2. Square the last term of the binomials. 3. Subtract the square of the last term from the square of the first term. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.4 POLYNOMIALS 47 Example 7 Multiplying Binomials Resulting in a Difference of Squares Multiply (9x + 4)(9x − 4). Solution Square the first term to get (9x)2 = 81x2. Square the last term to get 42 = 16. Subtract the square of the last term from the square of the first term to find the product of 81x2 − 16. Try It #7 Multiply (2x + 7)(2x − 7). Performing Operations with Polynomials of Several Variables We have looked at polynomials containing only one variable. However, a polynomial can contain several variables. All of the same rules apply when working with polynomials containing several variables. Consider an example: (a + 2b)(4a − b − c) a(4a − b − c) + 2b(4a − b − c) Use the distributive property. 4a2 − ab − ac + 8ab − 2b2 − 2bc Multiply. 4a2 + (− ab + 8ab) − ac − 2b2 − 2bc Combine like terms. 4a2 + 7ab − ac − 2bc − 2b2 Simplify. Example 8 Multiplying Polynomials Containing Several Variables Multiply (x + 4)(3x − 2y + 5). Solution Follow the same steps that we used to multiply polynomials containing only one variable. x(3x − 2y + 5) + 4(3x − 2y + 5) Use the distributive property. 3x2 − 2xy + 5x + 12x − 8y + 20 Multiply. 3x2 − 2xy + (5x + 12x) − 8y + 20 Combine like terms. 3x2 − 2xy + 17x − 8y + 20 Simplify. Try It #8 Multiply (3x − 1)(2x + 7y − 9). Access these online resources for additional instruction and practice with polynomials. • Adding and Subtracting Polynomials (http://openstaxcollege.org/l/addsubpoly) • Multiplying Polynomials (http://openstaxcollege.org/l/multiplpoly) • Special Products of Polynomials (http://openstaxcollege.org/l/specialpolyprod) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 48 CHAPTER 1 PREREQUISITES 1.4 SECTION EXERCISES VERBAL 1. Evaluate the following statement: The degree of a 2. Many times, multiplying two binomials with two polynomial in standard form is the exponent of the leading term. Explain why the statement is true or false. variables results in a trinomial. This is not the case when there is a difference of two squares. Explain why the product in this case is also a binomial. 3. You can multiply polynomials with any number of terms and any number of variables using four basic steps over and over until you reach the expanded polynomial. What are the four steps? 4. State whether the following statement is true and explain why or why not: A trinomial is always a higher degree than a monomial. ALGEBRAIC For the following exercises, identify the degree of the polynomial. 5. 7x − 2x2 + 13 9. x2 + 4x + 4 6. 14m3 + m2 − 16m + 8 10. 6y4 − y5 + 3y − 4 For the following exercises, find the sum or difference. 11. (12x2 + 3x) −(8x2 −19) 13. (6w2 + 24w + 24) − (3w − 6w + 3) 15. (11b4 − 6b3 + 18b2 − 4b + 8) − (3b3 + 6b2 + 3b) For the following exercises, find the product. 17. (4x + 2) (6x − 4) 21. (9v − 11) (11v − 9) 18. (14c2 + 4c) (2c2 − 3c) 22. (4t2 + 7t) (−3t2 + 4) For the following exercises, expand the binomial. 24. (4x + 5)2 28. (2m − 3)2 25. (3y − 7)2 29. (3y − 6)2 For the following exercises, multiply the binomials. 32. (9a − 4)(9a + 4) 31. (4c + 1)(4c − 1) 36. (14p + 7)(14p − 7) 35. (4 + 4m)(4 − 4m) 7. −625a8 + 16b4 8. 200p − 30p2m + 40m3 12. (4z3 + 8z2 − z) + (−2z2 + z + 6) 14. (7a3 + 6a2 − 4a − 13) + (−3a3 − 4a2 + 6a + 17) 16. (49p2 − 25) + (16p4 − 32p2 + 16) 19. (6b2 − 6) (4b2 − 4) 23. (8n − 4) (n2 + 9) 20. (3d − 5)(2d + 9) 26. (12 − 4x)2 30. (9b + 1)2 27. (4p + 9)2 33. (15n − 6)(15n + 6) 37. (11q − 10)(11q + 10) 34. (25b + 2)(25b − 2) For the following exercises, find the sum or difference. 38. (2x2 + 2x + 1) (4x − 1) 41. (y − 2) (y2 − 4y − 9) 44. (4m − 13) (2m2 − 7m + 9) 47. (4t − 5u)2 50. (b2 − 1) (a2 + 2ab + b2) 39. (4t2 + t − 7) (4t2 − 1) 42. (6k − 5) (6k2 + 5k − 1) 45. (a + b) (a − b) 48. (9m + 4n − 1) (2m + 8) 51. (4r − d)(6r + 7d) 40. (x − 1) (x2 − 2x + 1) 43. (3p2 + 2p − 10) (p − 1) 46. (4x − 6y) (6x − 4y) 49. (4t − x)(t − x + 1) 52. (x + y) (x2 − xy + y2) REAL-WORLD APPLICATIONS 53. A developer wants to purchase a plot of land to build a house. The area of the plot can be described by the following expression: (4x + 1)(8x − 3) where x is measured in meters. Multiply the binomials to find the area of the plot in standard form. EXTENSIONS For the following exercises, perform the given operations. 55. (4t − 7)2 (2t + 1) − (4t2 + 2t + 11) 57. (a2 + 4ac + 4c2) (a2 − 4c2) 54. A prospective buyer wants to know how much grain a specific silo can hold. The area of the floor of the silo is (2x + 9)2. The height of the silo is 10x + 10, where x is measured in feet. Expand the square and multiply by the height to find the expression that shows how much grain the silo can hold. 56. (3b + 6)(3b − 6) (9b2 − 36) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 FACTORING POLYNOMIALS 49 LEARNING OBJECTIVES In this section, you will: • Factor the greatest common factor of a polynomial. • Factor a trinomial. • Factor by grouping. • Factor a perfect square trinomial. • Factor a difference of squares. • Factor the sum and difference of cubes. • Factor expressions using fractional or negative exponents. 1. 5 FACTORING POLYNOMIALS Imagine that we are trying to find the area of a lawn so that we can determine how much grass seed to purchase. The lawn is the green portion in Figure 1. 4 4 4 4 10x Figure 1 4 6x The area of the entire region can be found using the formula for the area of a rectangle. A = lw = 10x · 6x = 60x2 units 2 The areas of the portions that do not require grass seed need to be subtracted from the area of the entire region. The two square regions each have an area of A = s2 = 42 = 16 units 2. The other rectangular region has one sid
e of length 10x − 8 and one side of length 4, giving an area of A = lw = 4(10x − 8) = 40x − 32 units 2. So the region that must be subtracted has an area of 2(16) + 40x − 32 = 40x units 2. The area of the region that requires grass seed is found by subtracting 60x2 − 40x units 2. This area can also be expressed in factored form as 20x(3x − 2) units 2. We can confirm that this is an equivalent expression by multiplying. Many polynomial expressions can be written in simpler forms by factoring. In this section, we will look at a variety of methods that can be used to factor polynomial expressions. Factoring the Greatest Common Factor of a Polynomial When we study fractions, we learn that the greatest common factor (GCF) of two numbers is the largest number that divides evenly into both numbers. For instance, 4 is the GCF of 16 and 20 because it is the largest number that divides evenly into both 16 and 20 The GCF of polynomials works the same way: 4x is the GCF of 16x and 20x 2 because it is the largest polynomial that divides evenly into both 16x and 20x 2. When factoring a polynomial expression, our first step should be to check for a GCF. Look for the GCF of the coefficients, and then look for the GCF of the variables. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 50 CHAPTER 1 PREREQUISITES greatest common factor The greatest common factor (GCF) of polynomials is the largest polynomial that divides evenly into the polynomials. How To… Given a polynomial expression, factor out the greatest common factor. 1. Identify the GCF of the coefficients. 2. Identify the GCF of the variables. 3. Combine to find the GCF of the expression. 4. Determine what the GCF needs to be multiplied by to obtain each term in the expression. 5. Write the factored expression as the product of the GCF and the sum of the terms we need to multiply by. Example 1 Factoring the Greatest Common Factor Factor 6x 3y 3 + 45x 2y 2 + 21xy. Solution First, find the GCF of the expression. The GCF of 6, 45, and 21 is 3. The GCF of x3, x 2, and x is x. (Note that the GCF of a set of expressions in the form xn will always be the exponent of lowest degree.) And the GCF of y3, y 2, and y is y. Combine these to find the GCF of the polynomial, 3xy. Next, determine what the GCF needs to be multiplied by to obtain each term of the polynomial. We find that 3xy(2x2y2) = 6x3y3, 3xy(15xy) = 45x2y2, and 3xy(7) = 21xy. Finally, write the factored expression as the product of the GCF and the sum of the terms we needed to multiply by. (3xy)(2x 2y 2 + 15xy + 7) Analysis After factoring, we can check our work by multiplying. Use the distributive property to confirm that (3xy)(2x2y2 + 15xy + 7) = 6x3y3 + 45x2y2 + 21xy. Try It #1 Factor x(b2 − a) + 6(b2 − a) by pulling out the GCF. Factoring a Trinomial with Leading Coefﬁcient 1 Although we should always begin by looking for a GCF, pulling out the GCF is not the only way that polynomial expressions can be factored. The polynomial x 2 + 5x + 6 has a GCF of 1, but it can be written as the product of the factors (x + 2) and (x + 3). Trinomials of the form x 2 + bx + c can be factored by finding two numbers with a product of c and a sum of b. The trinomial x 2 + 10x + 16, for example, can be factored using the numbers 2 and 8 because the product of those numbers is 16 and their sum is 10. The trinomial can be rewritten as the product of (x + 2) and (x + 8). factoring a trinomial with leading coefficient 1 A trinomial of the form x 2 + bx + c can be written in factored form as (x + p)(x + q) where pq = c and p + q = b. Q & A… Can every trinomial be factored as a product of binomials? No. Some polynomials cannot be factored. These polynomials are said to be prime. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 FACTORING POLYNOMIALS 51 How To… Given a trinomial in the form x 2 + bx + c, factor it. 1. List factors of c. 2. Find p and q, a pair of factors of c with a sum of b. 3. Write the factored expression (x + p)(x + q). Example 2 Factoring a Trinomial with Leading Coefﬁcient 1 Factor x 2 + 2x − 15. Solution We have a trinomial with leading coefficient 1, b = 2, and c = −15. We need to find two numbers with a product of −15 and a sum of 2. In Table 1, we list factors until we find a pair with the desired sum. Factors of −15 1, −15 −1, 15 3, −5 −3, 5 Table 1 Sum of Factors −14 14 −2 2 Now that we have identified p and q as −3 and 5, write the factored form as (x − 3)(x + 5). Analysis We can check our work by multiplying. Use FOIL to confirm that (x − 3)(x + 5) = x2 + 2x − 15. Q & A… Does the order of the factors matter? No. Multiplication is commutative, so the order of the factors does not matter. Try It #2 Factor x 2 − 7x + 6. Factoring by Grouping Trinomials with leading coefficients other than 1 are slightly more complicated to factor. For these trinomials, we can factor by grouping by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression. The trinomial 2x 2 + 5x + 3 can be rewritten as (2x + 3)(x + 1) using this process. We begin by rewriting the original expression as 2x 2 + 2x + 3x + 3 and then factor each portion of the expression to obtain 2x(x + 1) + 3(x + 1). We then pull out the GCF of (x + 1) to find the factored expression. factor by grouping To factor a trinomial in the form ax 2 + bx + c by grouping, we find two numbers with a product of ac and a sum of b. We use these numbers to divide the x term into the sum of two terms and factor each portion of the expression separately, then factor out the GCF of the entire expression. How To… Given a trinomial in the form ax 2 + bx + c, factor by grouping. 1. List factors of ac. 2. Find p and q, a pair of factors of ac with a sum of b. 3. Rewrite the original expression as ax 2 + px + qx + c. 4. Pull out the GCF of ax 2 + px. 5. Pull out the GCF of qx + c. 6. Factor out the GCF of the expression. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 52 CHAPTER 1 PREREQUISITES Example 3 Factoring a Trinomial by Grouping Factor 5x 2 + 7x − 6 by grouping. Solution We have a trinomial with a = 5, b = 7, and c = −6. First, determine ac = −30. We need to find two numbers with a product of −30 and a sum of 7. In Table 2, we list factors until we find a pair with the desired sum. Factors of −30 1, −30 −1, 30 2, −15 −2, 15 3, −10 −3, 10 Table 2 Sum of Factors −29 29 −13 13 −7 7 So p = −3 and q = 10. 5x2 − 3x + 10x − 6 x(5x − 3) + 2(5x − 3) (5x − 3)(x + 2) Rewrite the original expression as ax2 + px + qx + c. Factor out the GCF of each part. Factor out the GCF of the expression. Analysis We can check our work by multiplying. Use FOIL to confirm that (5x − 3)(x + 2) = 5x2 + 7x − 6. Try It #3 Factor. a. 2x 2 + 9x + 9 b. 6x2 + x − 1 Factoring a Perfect Square Trinomial A perfect square trinomial is a trinomial that can be written as the square of a binomial. Recall that when a binomial is squared, the result is the square of the first term added to twice the product of the two terms and the square of the last term. a2 + 2ab + b2 = (a + b)2 and a2 − 2ab + b2 = (a − b)2 We can use this equation to factor any perfect square trinomial. perfect square trinomials A perfect square trinomial can be written as the square of a binomial: a2 + 2ab + b2 = (a + b)2 How To… Given a perfect square trinomial, factor it into the square of a binomial. 1. Confirm that the first and last term are perfect squares. 2. Confirm that the middle term is twice the product of ab. 3. Write the factored form as (a + b)2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 FACTORING POLYNOMIALS 53 Example 4 Factoring a Perfect Square Trinomial Factor 25x 2 + 20x + 4. Solution Notice that 25x 2 and 4 are perfect squares because 25x 2 = (5x)2 and 4 = 22. Then check to see if the middle term is twice the product of 5x and 2. The middle term is, indeed, twice the product: 2(5x)(2) = 20x. Therefore, the trinomial is a perfect square trinomial and can be written as (5x + 2)2. Try It #4 Factor 49x 2 − 14x + 1. Factoring a Difference of Squares A difference of squares is a perfect square subtracted from a perfect square. Recall that a difference of squares can be rewritten as factors containing the same terms but opposite signs because the middle terms cancel each other out when the two factors are multiplied. We can use this equation to factor any differences of squares. a2 − b2 = (a + b)(a − b) differences of squares A difference of squares can be rewritten as two factors containing the same terms but opposite signs. a2 − b2 = (a + b)(a − b) How To… Given a difference of squares, factor it into binomials. 1. Confirm that the first and last term are perfect squares. 2. Write the factored form as (a + b)(a − b). Example 5 Factoring a Difference of Squares Factor 9x 2 − 25. Solution Notice that 9x 2 and 25 are perfect squares because 9x 2 = (3x)2 and 25 = 52. The polynomial represents a difference of squares and can be rewritten as (3x + 5)(3x − 5). Try It #5 Factor 81y2 − 100. Q & A… Is there a formula to factor the sum of squares? No. A sum of squares cannot be factored. Factoring the Sum and Difference of Cubes Now, we will look at two new special products: the sum and difference of cubes. Although the sum of squares cannot be factored, the sum of cubes can be factored into a binomial and a trinomial. Similarly, the sum of cubes can be factored into a binomial and a trinomial, but with different signs. a3 − b3 = (a − b)(a2 + ab + b2) a3 + b3 = (a + b) (a2 − ab + b2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 54 CHAPTER 1 PREREQUISITES We can use the acronym SOAP to remember the signs when factoring the sum or difference of cubes. The first letter of each word relates to the signs: Same Opposite Always Positive. For example, consider the following
example. The sign of the first 2 is the same as the sign between x 3 − 23. The sign of the 2x term is opposite the sign between x 3 − 23. And the sign of the last term, 4, is always positive. x3 − 23 = (x − 2)(x2 + 2x + 4) sum and difference of cubes We can factor the sum of two cubes as We can factor the difference of two cubes as a3 + b3 = (a + b) (a2 − ab + b2) a3 − b3 = (a − b)(a2 + ab + b2) How To… Given a sum of cubes or difference of cubes, factor it. 1. Confirm that the first and last term are cubes, a3 + b3 or a3 − b3. 2. For a sum of cubes, write the factored form as (a + b)(a2 − ab + b2). For a difference of cubes, write the factored form as (a − b)(a2 + ab + b2). Example 6 Factoring a Sum of Cubes Factor x3 + 512. Solution Notice that x 3 and 512 are cubes because 83 = 512. Rewrite the sum of cubes as (x + 8)(x2 − 8x + 64). Analysis After writing the sum of cubes this way, we might think we should check to see if the trinomial portion can be factored further. However, the trinomial portion cannot be factored, so we do not need to check. Try It #6 Factor the sum of cubes: 216a3 + b3. Example 7 Factoring a Difference of Cubes Factor 8x 3 − 125. Solution Notice that 8x 3 and 125 are cubes because 8x 3 = (2x)3 and 125 = 53. Write the difference of cubes as (2x − 5)(4x 2 + 10x + 25). Analysis Just as with the sum of cubes, we will not be able to further factor the trinomial portion. Try It #7 Factor the difference of cubes: 1,000x 3 − 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 FACTORING POLYNOMIALS 55 Factoring Expressions with Fractional or Negative Exponents Expressions with fractional or negative exponents can be factored by pulling out a GCF. Look for the variable or exponent that is common to each term of the expression and pull out that variable or exponent raised to the lowest 3 1 __ __ power. These expressions follow the same factoring rules as those with integer exponents. For instance, 2x can + __ __ __ and being rewritten as x be factored by pulling out x 2 4 4 Example 8 Factoring an Expression with Fractional or Negative Exponents 2 1 __ _ Factor 3x (x + 2) − + 4(x + 2) . 3 3 1 _ Solution Factor out the term with the lowest value of the exponent. In this case, that would be (x + 2) − . 3 1 _ (x + 2) − (3x + 4(x + 2)) 3 1 _ (x + 2) − (3x + 4x + 8) 3 1 _ (x + 2) − (7x + 8) 3 Factor out the GCF. Simplify. Try It #8 3 1 __ _ + 7a (5a − 1) − . Factor 2(5a − 1) 4 4 Access these online resources for additional instruction and practice with factoring polynomials. • Identify GCF (http://openstaxcollege.org/l/ﬁndgcftofact) • Factor Trinomials when a Equals 1 (http://openstaxcollege.org/l/facttrinom1) • Factor Trinomials when a is not equal to 1 (http://openstaxcollege.org/l/facttrinom2) • Factor Sum or Difference of Cubes (http://openstaxcollege.org/l/sumdifcube) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 56 CHAPTER 1 PREREQUISITES 1.5 SECTION EXERCISES VERBAL 1. If the terms of a polynomial do not have a GCF, does 2. A polynomial is factorable, but it is not a perfect that mean it is not factorable? Explain. square trinomial or a difference of two squares. Can you factor the polynomial without finding the GCF? 3. How do you factor by grouping? ALGEBRAIC For the following exercises, find the greatest common factor. 4. 14x + 4xy − 18xy2 6. 30x3y − 45x2y2 + 135xy3 8. 36j4k2 − 18j3k3 + 54j2k4 5. 49mb2 − 35m2ba + 77ma2 7. 200p3m3 − 30p2m3 + 40m3 9. 6y4 − 2y3 + 3y2 − y For the following exercises, factor by grouping. 11. 2a2 + 9a − 18 10. 6x2 + 5x − 4 14. 20w2 − 47w + 24 15. 2p2 − 5p − 7 For the following exercises, factor the polynomial. 16. 7x2 + 48x − 7 17. 10h2 − 9h − 9 12. 6c2 + 41c + 63 13. 6n2 − 19n − 11 18. 2b2 − 25b − 247 19. 9d 2 −73d + 8 20. 90v2 −181v + 90 21. 12t2 + t − 13 22. 2n2 − n − 15 24. 25y2 − 196 25. 121p2 − 169 26. 4m2 − 9 23. 16x2 − 100 27. 361d2 − 81 28. 324x2 − 121 29. 144b2 − 25c2 30. 16a2 − 8a + 1 31. 49n2 + 168n + 144 32. 121x2 − 88x + 16 33. 225y2 + 120y + 16 34. m2 − 20m + 100 35. m2 − 20m + 100 36. 36q2 + 60q + 25 For the following exercises, factor the polynomials. 37. x3 + 216 38. 27y3 − 8 39. 125a3 + 343 40. b3 − 8d3 41. 64x3 − 125 42. 729q3 + 1331 43. 125r3 + 1,728s3 1 2 __ _ 44. 4x(x − 1) − + 3(x − 1) 3 3 4 1 __ __ + 7(10t + 3) 46. 3t(10t + 3) 3 3 1 6 __ __ − 2(3y − 13) 48. 9y(3y − 13) 5 5 5 1 __ _ − + 5(2d + 3) 50. 6d(2d + 3) 6 6 3 1 __ _ 45. 3c(2c + 3) − − 5(2c + 3) 4 4 3 2 __ _ 47. 14x(x + 2) − + 5(x + 2) 5 5 3 1 _ 49. 5z(2z − 9) − _ + 11(2z − 9) − 2 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.5 SECTION EXERCISES 57 REAL-WORLD APPLICATIONS For the following exercises, consider this scenario: Charlotte has appointed a chairperson to lead a city beautification project. The first act is to install statues and fountains in one of the city’s parks. The park is a rectangle with an area of 98x 2 + 105x − 27 m2, as shown in the following figure. The length and width of the park are perfect factors of the area. l × w = 98x 2 + 105x − 27 51. Factor by grouping to find the length and width of the park. 53. At the northwest corner of the park, the city is going to install a fountain. The area of the base of the fountain is 9x2 − 25 m2. Factor the area to find the lengths of the sides of the fountain. 52. A statue is to be placed in the center of the park. The area of the base of the statue is 4x 2 + 12x + 9 m2. Factor the area to find the lengths of the sides of the statue. For the following exercise, consider the following scenario: A school is installing a flagpole in the central plaza. The plaza is a square with side length 100 yd as shown in the figure below. The flagpole will take up a square plot with area x 2 − 6x + 9 yd2. Area: x 2 − 6x + 9 100 yards 100 yards 54. Find the length of the base of the flagpole by factoring. EXTENSIONS For the following exercises, factor the polynomials completely. 55. 16x 4 − 200x2 + 625 56. 81y 4 − 256 57. 16z4 − 2,401a4 3 2 __ _ 58. 5x(3x + 2) − + (12x + 8) 4 2 59. (32x 3 + 48x 2 − 162x − 243)−1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 58 CHAPTER 1 PREREQUISITES LEARNING OBJECTIVES In this section, you will: • Simplify rational expressions. • Multiply rational expressions. • Divide rational expressions. • Add and subtract rational expressions. • Simplify complex rational expressions. 1. 6 RATIONAL EXPRESSIONS A pastry shop has fixed costs of $280 per week and variable costs of $9 per box of pastries. The shop’s costs per week in terms of x, the number of boxes made, is 280 + 9x. We can divide the costs per week by the number of boxes made to determine the cost per box of pastries. 280 + 9x _ x Notice that the result is a polynomial expression divided by a second polynomial expression. In this section, we will explore quotients of polynomial expressions. Simplifying Rational Expressions The quotient of two polynomial expressions is called a rational expression. We can apply the properties of fractions to rational expressions, such as simplifying the expressions by canceling common factors from the numerator and the denominator. To do this, we first need to factor both the numerator and denominator. Let’s start with the rational expression shown. x2 + 8x + 16 ____________ x2 + 11x + 28 We can factor the numerator and denominator to rewrite the expression. (x + 4)2 ____________ (x + 4)(x + 7) Then we can simplify that expression by canceling the common factor (x + 4). x + 4 _____ x + 7 How To… Given a rational expression, simplify it. 1. Factor the numerator and denominator. 2. Cancel any common factors. Example 1 Simplifying Rational Expressions Simplify x2 − 9 __________ x2 + 4x + 3 . Solution (x + 3)(x − 3) ____________ (x + 3)(x + 1) x − 3 _____ x + 1 Factor the numerator and the denominator. Cancel common factor (x + 3). Analysis We can cancel the common factor because any expression divided by itself is equal to 1. Q & A… Can the x2 term be cancelled in Example 1? No. A factor is an expression that is multiplied by another expression. The x 2 term is not a factor of the numerator or the denominator. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.6 RATIONAL EXPRESSIONS 59 Try It #1 Simplify x − 6 _______ . x 2 − 36 Multiplying Rational Expressions Multiplication of rational expressions works the same way as multiplication of any other fractions. We multiply the numerators to find the numerator of the product, and then multiply the denominators to find the denominator of the product. Before multiplying, it is helpful to factor the numerators and denominators just as we did when simplifying rational expressions. We are often able to simplify the product of rational expressions. How To… Given two rational expressions, multiply them. 1. Factor the numerator and denominator. 2. Multiply the numerators. 3. Multiply the denominators. 4. Simplify. Example 2 Multiplying Rational Expressions Multiply the rational expressions and show the product in simplest form: x 2 + 4x − 5 __ ∙ 3x + 18 2x − 1 _______ x + 5 Solution (2x − 1) (x + 5)(x − 1) _______ ____________ ∙ (x + 5) 3(x + 6) (x + 5)(x − 1)(2x − 1) ___________________ 3(x + 6)(x + 5) , (x + 5) (x − 1)(2x − 1) ___________________ 3(x + 6) , (x + 5) (x − 1)(2x − 1) _____________ 3(x + 6) Factor the numerator and denominator. Multiply numerators and denominators. Cancel common factors to simplify. Try It #2 Multiply the rational expressions and show the product in simplest form: x 2 + 11x + 30 ____________ · x 2 + 5x + 6 x 2 + 7x + 12 ___________ x 2 + 8x + 16 Dividing Rational Expressions Division of rational expressions works the same way as division of other fractions. To divide a rational expression by another rational expression, multiply the first expression by the reciprocal of the second. Using this approach, we x2 ÷ x2 . Once the division expression has been rewritten as a mu
ltiplication would rewrite as the product 3 expression, we can multiply as we did before. 3 1 __ __ · x2 x 3 __ = x3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 60 CHAPTER 1 PREREQUISITES How To… Given two rational expressions, divide them. 1. Rewrite as the first rational expression multiplied by the reciprocal of the second. 2. Factor the numerators and denominators. 3. Multiply the numerators. 4. Multiply the denominators. 5. Simplify. Example 3 Dividing Rational Expressions Divide the rational expressions and express the quotient in simplest form: 2x2 + x − 6 __________ ÷ x2 − 1 x2 − 4 __________ x2 + 2x + 1 Solution 2x2 + x −#6 _____________ ∙ x2 − 1 x2 + 2x + 1 __ x 2 − 4 (x + 1)(x + 1) __ (x + 2)(x − 2) (2x − 3)(x + 2) __ ∙ (x + 1)(x − 1) (2x − 3)(x + 2)(x + 1)(x + 1) ___ (x + 1)(x − 1)(x + 2)(x − 2) (2x − 3)(x + 2)(x + 1)(x + 1) ___ (x + 1)(x − 1)(x + 2)(x − 2) (2x − 3)(x + 1) __ (x − 1)(x − 2) Rewrite as multiplication. Factor the numerator and denominator. Multiply numerators and denominators. Cancel common factors to simplify. Try It #3 Divide the rational expressions and express the quotient in simplest form: 9x2 − 16 __ 3x2 + 17x − 28 ÷ 3x2 − 2x − 8 __ x2 + 5x − 14 Adding and Subtracting Rational Expressions Adding and subtracting rational expressions works just like adding and subtracting numerical fractions. To add fractions, we need to find a common denominator. Let’s look at an example of fraction addition. 5 ___ 24 1 ___ 40 + = + 3 ___ 120 25 ___ 120 28 _ 120 = = 7 _ 30 We have to rewrite the fractions so they share a common denominator before we are able to add.We must do the same thing when adding or subtracting rational expressions. The easiest common denominator to use will be the least common denominator, or LCD. The LCD is the smallest multiple that the denominators have in common. To find the LCD of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, if the factored denominators were (x + 3)(x + 4) and (x + 4)(x + 5), then the LCD would be (x + 3)(x + 4)(x + 5). Once we find the LCD, we need to multiply each expression by the form of 1 that will change the denominator to the LCD. We would need to multiply the expression with a denominator of (x + 3)(x + 4) by x + 3 _ with a denominator of (x + 3)(x + 4) by . x + 3 and the expression x + 5 _ x + 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.6 RATIONAL EXPRESSIONS 61 How To… Given two rational expressions, add or subtract them. 1. Factor the numerator and denominator. 2. Find the LCD of the expressions. 3. Multiply the expressions by a form of 1 that changes the denominators to the LCD. 4. Add or subtract the numerators. 5. Simplify. Example 4 Adding Rational Expressions Add the rational expressions: 6 5 _ y _ x + Solution First, we have to find the LCD. In this case, the LCD will be xy. We then multiply each expression by the appropriate form of 1 to obtain xy as the denominator for each fraction · 5y _ xy + 6x _ xy Now that the expressions have the same denominator, we simply add the numerators to find the sum. 6x + 5y _ xy y x _ x does not change the value of the original expression because any number divided by _ y or Analysis Multiplying by itself is 1, and multiplying an expression by 1 gives the original expression. Example 5 Subtracting Rational Expressions Subtract the rational expressions: Solution 6 __ − x 2 + 4x + 4 2 _ x 2 −#4 6 _______ − (x + 2)2 2 ____________ (x + 2)(x − 2) Factor. 6 _______ (x + 2)2 ∙ x − 2 _____ x − 2 − 2 ____________ ∙ (x + 2)(x − 2) x + 2 _____ x + 2 6(x − 2) _____________ (x + 2)2(x − 2) − 2(x + 2) _____________ (x + 2)2(x − 2) 6x − 12 − (2x + 4) ________________ (x + 2)2(x − 2) 4x − 16 _____________ (x + 2)2(x − 2) 4(x − 4) _____________ (x + 2)2(x − 2) Multiply each fraction to get LCD as denominator. Multiply. Apply distributive property. Subtract. Simplify. Q & A… Do we have to use the LCD to add or subtract rational expressions? No. Any common denominator will work, but it is easiest to use the LCD. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 62 CHAPTER 1 PREREQUISITES Try It #4 Subtract the rational expressions −#3 Simplifying Complex Rational Expressions A complex rational expression is a rational expression that contains additional rational expressions in the numerator, the denominator, or both. We can simplify complex rational expressions by rewriting the numerator and denominator a _ can be simplified by rewriting the as single rational expressions and dividing. The complex rational expression 1 _ + c b a _ numerator as the fraction and combining the expressions in the denominator as 1 1 + bc _ . We can then rewrite the b a _ expression as a multiplication problem using the reciprocal of the denominator. We get ∙ 1 , which is equal b _ 1 + bc to ab _ . 1 + bc How To… Given a complex rational expression, simplify it. 1. Combine the expressions in the numerator into a single rational expression by adding or subtracting. 2. Combine the expressions in the denominator into a single rational expression by adding or subtracting. 3. Rewrite as the numerator divided by the denominator. 4. Rewrite as multiplication. 5. Multiply. 6. Simplify. Example 6 Simplifying Complex Rational Expressions Simplify Solution Begin by combining the expressions in the numerator into one expression. x 1 _ x _ x + y ∙ xy 1 _ x _ x + xy + 1 _ x x _ x to get LCD as denominator. Multiply by Add numerators. Now the numerator is a single rational expression and the denominator is a single rational expression. xy + 1 ______ x _______ x __ y We can rewrite this as division, and then multiplication. x _ y ÷ xy + 1 _ x xy + 1 y _ x _ ∙ x y(xy + 1) _ x 2 Rewrite as multiplication. Multiply. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.6 RATIONAL EXPRESSIONS 63 Try It #5 y x __ y − __ x _______ y Simplify: Q & A… Can a complex rational expression always be simplified? Yes. We can always rewrite a complex rational expression as a simplified rational expression. Access these online resources for additional instruction and practice with rational expressions. • Simplify Rational Expressions (http://openstaxcollege.org/l/simpratexpress) • Multiply and Divide Rational Expressions (http://openstaxcollege.org/l/multdivratex) • Add and Subtract Rational Expressions (http://openstaxcollege.org/l/addsubratex) • Simplify a Complex Fraction (http://openstaxcollege.org/l/complexfract) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 64 CHAPTER 1 PREREQUISITES 1.6 SECTION EXERCISES VERBAL 1. How can you use factoring to simplify rational 2. How do you use the LCD to combine two rational expressions? expressions? 3. Tell whether the following statement is true or false and explain why: You only need to find the LCD when adding or subtracting rational expressions. ALGEBRAIC For the following exercises, simplify the rational expressions. 4. x 2 − 16 __________ x 2 − 5x + 4 8. m − 12 ________ m2 − 144 5. y2 + 10y + 25 ____________ y2 + 11y + 30 6. 6a2 − 24a + 24 _____________ 6a2 − 24 7. 9b2 + 18b + 9 ____________ 3b + 3 9. 2x 2 + 7x − 4 ___________ 4x 2 + 2x − 2 10. 6x 2 + 5x − 4 _____________ 3x 2 + 19x + 20 11. a2 + 9a + 18 ___________ a2 + 3a − 18 12. 3c2 + 25c − 18 ____________ 3c2 − 23c + 14 13. 12n2 − 29n − 8 _____________ 28n2 − 5n − 3 For the following exercises, multiply the rational expressions and express the product in simplest form. 14. x 2 − x − 6 __________ · 2x 2 + x − 6 2x 2 + 7x − 15 ____________ x 2 − 9 15. c2 + 2c − 24 ___________ · c2 + 12c + 36 c2 − 10c + 24 ____________ c2 − 8c + 16 16. 2d 2 + 9d − 35 ____________ · d 2 + 10d + 21 3d 2 + 2d − 21 _____________ 3d 2 + 14d − 49 17. 10h2 − 9h − 9 _____________ · 2h2 − 19h + 24 h2 − 16h + 64 _____________ 5h2 − 37h − 24 18. 6b2 + 13b + 6 ____________ · 4b2 − 9 6b2 + 31b − 30 _____________ 18b2 − 3b − 10 19. 2d 2 + 15d + 25 _____________ · 4d 2 − 25 2d 2 − 15d + 25 _____________ 25d 2 − 1 20. 6x 2 − 5x − 50 ______________ · 15x 2 − 44x − 20 20x 2 − 7x − 6 ____________ 2x 2 + 9x + 10 21. t 2 − 1 _________ · t 2 + 4t + 3 t 2 + 2t − 15 __________ t 2 − 4t + 3 22. 2n2 − n − 15 ____________ · 6n2 + 13n − 5 12n2 − 13n + 3 _____________ 4n2 − 15n + 9 For the following exercises, divide the rational expressions. 24. 3y2 − 7y − 6 ___________ ÷ 2y2 − 3y − 9 y2 + y − 2 __________ 2y2 + y − 3 26. q2 − 9 __________ ÷ q2 + 6q + 9 q2 − 2q − 3 __________ q2 + 2q − 3 28. 16x 2 + 18x − 55 ______________ ÷ 32x 2 − 36x − 11 2x 2 + 17x + 30 _____________ 4x 2 + 25x + 6 30. 16a2 − 24a + 9 _____________ ÷ 4a2 + 17a − 15 16a2 − 9 ____________ 4a2 + 11a + 6 32. 9x2 + 3x − 20 ____________ ÷ 3x2 − 7x + 4 6x2 + 4x − 10 ____________ x2 − 2x + 1 23. 36x 2 − 25 _____________ · 6x 2 + 65x + 50 3x 2 + 32x + 20 ______________ 18x 2 + 27x + 10 25. 6p 2 + p − 12 ____________ 8p 2 + 18p + 9 ÷ 6p 2 − 11p + 4 ____________ 2p 2 + 11p − 6 27. 18d 2 + 77d − 18 ______________ ÷ 27d 2 − 15d + 2 3d 2 + 29d − 44 _____________ 9d 2 − 15d + 4 29. 144b2 − 25 _____________ ÷ 72b2 − 6b − 10 18b2 − 21b + 5 ______________ 36b2 − 18b − 10 31. 22y2 + 59y + 10 ______________ ÷ 12y2 + 28y − 5 11y2 + 46y + 8 _____________ 24y2 − 10y + 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 1.6 SECTION EXERCISES 65 For the following exercises, add and subtract the rational expressions, and then simplify. 4 __ + 33. x 10 ___ y 36. c + 2 _____ 3 − c − 4 _____ 4 34. − 12 ___ 2q 6 ___ 3p 37. y + 3 _____ y − 2 + y − 3 _____ y + 1 35. 4 _____ a + 1 + 5 _____ a − 3 38. x − 1 _____ x + 1 − 2x + 3 ______ 2x + 1 39. 3z _____ z + 1 + 2z + 5 ______ z − 2 40. 4p _____ p + 1 − p + 1 _____ 4p 41. x _____ x + 1 + y _____ y + 1 For the following exercises, simplify the rational expression. 42 45. 3 b _ a + _ 6 _______ 2b _ 3a 48. 4x _ 7 2x _ + 3 _ x _ 2 43 46 __ x − 1 _ x + 1 49. c
− 1 _ c + 1 2c _ + c + 2 __ 2c + 1 _ c + 1 REAL-WORLD APPLICATIONS 51. Brenda is placing tile on her bathroom floor. The area of the floor is 15x2 − 8x − 7 ft 2. The area of one tile is x2 − 2x + 1 ft 2. To find the number of tiles needed, simplify the rational expression: 15x2 − 8x − 7 __ . x2 − 2x + 1 44. p x _ _ − 4 8 _______ p 47. 50 ab + Area = 15x2 − 8x − 7 52. The area of Sandy’s yard is 25x2 − 625 ft 2. A patch of sod has an area of x2 − 10x + 25 ft 2. Divide the two areas and simplify to find how many pieces of sod Sandy needs to cover her yard. 53. Aaron wants to mulch his garden. His garden is x2 + 18x + 81 ft 2. One bag of mulch covers x2 − 81 ft 2. Divide the expressions and simplify to find how many bags of mulch Aaron needs to mulch his garden. EXTENSIONS For the following exercises, perform the given operations and simplify. 54. x2 + x − 6 __________ · x2 − 2x − 3 2x2 − 3x − 9 ___________ ÷ x2 − x − 2 10x2 + 27x + 18 ______________ x2 + 2x + 1 56. + 2a − 3 ______ 2a + 3 4a + 1 ______ 2a − 3 _______________ 4a2 + 9 _______ a 55. 2y2 − 3y − 20 3y2 − 10y + 3 ____________ ____________ ∙ 2y2 − y − 15 3y2 + 5y − 2 _________________________ y − 4 57. x2 + 7x + 12 ___________ ÷ x2 + x − 6 3x2 + 19x + 28 _____________ ÷ 8x2 − 4x − 24 2x2 + x − 3 ___________ 3x2 + 4x − 7 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 66 CHAPTER 1 PREREQUISITES CHAPTER 1 REVIEW Key Terms algebraic expression constants and variables combined using addition, subtraction, multiplication, and division associative property of addition the sum of three numbers may be grouped differently without affecting the result; in symbols, a + (b + c) = (a + b) + c associative property of multiplication the product of three numbers may be grouped differently without affecting the result; in symbols, a · (b · c) = (a · b) · c base in exponential notation, the expression that is being multiplied binomial a polynomial containing two terms coefficient any real number ai in a polynomial in the form an x n + ... + a2 x 2 + a1 x + a0 commutative property of addition two numbers may be added in either order without affecting the result; in symbols, a + b = b + a commutative property of multiplication two numbers may be multiplied in any order without affecting the result; in symbols, a · b = b · a constant a quantity that does not change value degree the highest power of the variable that occurs in a polynomial difference of squares the binomial that results when a binomial is multiplied by a binomial with the same terms, but the opposite sign distributive property the product of a factor times a sum is the sum of the factor times each term in the sum; in symbols, a · (b + c) = a · b + a · c equation a mathematical statement indicating that two expressions are equal exponent in exponential notation, the raised number or variable that indicates how many times the base is being multiplied exponential notation a shorthand method of writing products of the same factor factor by grouping a method for factoring a trinomial in the form ax 2 + bx + c by dividing the x term into the sum of two terms, factoring each portion of the expression separately, and then factoring out the GCF of the entire expression formula an equation expressing a relationship between constant and variable quantities greatest common factor the largest polynomial that divides evenly into each polynomial identity property of addition there is a unique number, called the additive identity, 0, which, when added to a number, results in the original number; in symbols, a + 0 = a identity property of multiplication there is a unique number, called the multiplicative identity, 1, which, when multiplied by a number, results in the original number; in symbols, a · 1 = a index the number above the radical sign indicating the nth root integers the set consisting of the natural numbers, their opposites, and 0: { … , −3, −2, −1, 0, 1, 2, 3,…} inverse property of addition for every real number a, there is a unique number, called the additive inverse (or opposite), denoted −a, which, when added to the original number, results in the additive identity, 0; in symbols, a + (−a) = 0 inverse property of multiplication for every non-zero real number a, there is a unique number, called the multiplicative 1 __ a , which, when multiplied by the original number, results in the multiplicative inverse (or reciprocal), denoted 1 __ a = 1 identity, 1; in symbols, a · irrational numbers the set of all numbers that are not rational; they cannot be written as either a terminating or repeating decimal; they cannot be expressed as a fraction of two integers leading coefficient the coefficient of the leading term leading term the term containing the highest degree least common denominator the smallest multiple that two denominators have in common Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 1 REVIEW 67 monomial a polynomial containing one term natural numbers the set of counting numbers: {1, 2, 3,…} order of operations a set of rules governing how mathematical expressions are to be evaluated, assigning priorities to operations perfect square trinomial the trinomial that results when a binomial is squared polynomial a sum of terms each consisting of a variable raised to a nonnegative integer power principal nth root the number with the same sign as a that when raised to the nth power equals a principal square root the nonnegative square root of a number a that, when multiplied by itself, equals a radical the symbol used to indicate a root radical expression an expression containing a radical symbol radicand the number under the radical symbol rational expression the quotient of two polynomial expressions rational numbers the set of all numbers of the form m __ n , where m and n are integers and n ≠ 0. Any rational number may be written as a fraction or a terminating or repeating decimal. real number line a horizontal line used to represent the real numbers. An arbitrary fixed point is chosen to represent 0; positive numbers lie to the right of 0 and negative numbers to the left. real numbers the sets of rational numbers and irrational numbers taken together scientific notation a shorthand notation for writing very large or very small numbers in the form a × 10n where 1 ≤ ∣ a ∣ < 10 and n is an integer term of a polynomial any ai x i of a polynomial in the form anx n + ... + a2 x 2 + a1 x + a0 trinomial a polynomial containing three terms variable a quantity that may change value whole numbers the set consisting of 0 plus the natural numbers: {0, 1, 2, 3,…} Key Equations Rules of Exponents For nonzero real numbers a and b and integers m and n Product rule Quotient rule Power rule Zero exponent rule Negative rule Power of a product rule Power of a quotient rule am ∙ an = am + n am ___ an = am − n (am)n = am ∙ n a0 = 1 a−n = 1 __ an (a ∙ b)n = an ∙ bn a __ ) ( b n = an __ bn perfect square trinomial (x + a)2 = (x + a)(x + a) = x2 + 2ax + a2 difference of squares difference of squares (a + b)(a − b) = a2 − b2 a2 − b2 = (a + b)(a − b) perfect square trinomial a2 + 2ab + b2 = (a + b)2 sum of cubes difference of cubes a3 + b3 = (a + b)(a2 − ab + b2) a3 − b3 = (a − b)(a2 + ab + b2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 68 CHAPTER 1 PREREQUISITES Key Concepts 1.1 Real Numbers: Algebra Essentials • Rational numbers may be written as fractions or terminating or repeating decimals. See Example 1 and Example 2. • Determine whether a number is rational or irrational by writing it as a decimal. See Example 3. • The rational numbers and irrational numbers make up the set of real numbers. See Example 4. A number can be classified as natural, whole, integer, rational, or irrational. See Example 5. • The order of operations is used to evaluate expressions. See Example 6. • The real numbers under the operations of addition and multiplication obey basic rules, known as the properties of real numbers. These are the commutative properties, the associative properties, the distributive property, the identity properties, and the inverse properties. See Example 7. • Algebraic expressions are composed of constants and variables that are combined using addition, subtraction, multiplication, and division. See Example 8. They take on a numerical value when evaluated by replacing variables with constants. See Example 9, Example 10, and Example 12. • Formulas are equations in which one quantity is represented in terms of other quantities. They may be simplified or evaluated as any mathematical expression. See Example 11 and Example 13. 1.2 Exponents and Scientiﬁc Notation • Products of exponential expressions with the same base can be simplified by adding exponents. See Example 1. • Quotients of exponential expressions with the same base can be simplified by subtracting exponents. See Example 2. • Powers of exponential expressions with the same base can be simplified by multiplying exponents. See Example 3. • An expression with exponent zero is defined as 1. See Example 4. • An expression with a negative exponent is defined as a reciprocal. See Example 5 and Example 6. • The power of a product of factors is the same as the product of the powers of the same factors. See Example 7. • The power of a quotient of factors is the same as the quotient of the powers of the same factors. See Example 8. • The rules for exponential expressions can be combined to simplify more complicated expressions. See Example 9. • Scientific notation uses powers of 10 to simplify very large or very small numbers. See Example 10 and Example 11. • Scientific notation may be used to simplify calculations with very large or very small numbers. See Example 12 and Example 13. 1.3 Radicals and Rational Expressions • The principal square root of a number a is the nonnegative number that when multiplied by itself equals a. See Example 1. • If a and b are nonnegative, th
e square root of the product ab is equal to the product of the square roots of a and b See Example 2 and Example 3. a __ is equal to the quotient of the square roots of a and b • If a and b are nonnegative, the square root of the quotient b See Example 4 and Example 5. • We can add and subtract radical expressions if they have the same radicand and the same index. See Example 6 and Example 7. • Radical expressions written in simplest form do not contain a radical in the denominator. To eliminate the square root radical from the denominator, multiply both the numerator and the denominator by the conjugate of the denominator. See Example 8 and Example 9. • The principal nth root of a is the number with the same sign as a that when raised to the nth power equals a. These roots have the same properties as square roots. See Example 10. • Radicals can be rewritten as rational exponents and rational exponents can be rewritten as radicals. See Example 11 and Example 12. • The properties of exponents apply to rational exponents. See Example 13. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 1 REVIEW 69 1.4 Polynomials • A polynomial is a sum of terms each consisting of a variable raised to a non-negative integer power. The degree is the highest power of the variable that occurs in the polynomial. The leading term is the term containing the highest degree, and the leading coefficient is the coefficient of that term. See Example 1. • We can add and subtract polynomials by combining like terms. See Example 2 and Example 3. • To multiply polynomials, use the distributive property to multiply each term in the first polynomial by each term in the second. Then add the products. See Example 4. • FOIL (First, Outer, Inner, Last) is a shortcut that can be used to multiply binomials. See Example 5. • Perfect square trinomials and difference of squares are special products. See Example 6 and Example 7. • Follow the same rules to work with polynomials containing several variables. See Example 8. 1.5 Factoring Polynomials • The greatest common factor, or GCF, can be factored out of a polynomial. Checking for a GCF should be the first step in any factoring problem. See Example 1. • Trinomials with leading coefficient 1 can be factored by finding numbers that have a product of the third term and a sum of the second term. See Example 2. • Trinomials can be factored using a process called factoring by grouping. See Example 3. • Perfect square trinomials and the difference of squares are special products and can be factored using equations. See Example 4 and Example 5. • The sum of cubes and the difference of cubes can be factored using equations. See Example 6 and Example 7. • Polynomials containing fractional and negative exponents can be factored by pulling out a GCF. See Example 8. 1.6 Rational Expressions • Rational expressions can be simplified by cancelling common factors in the numerator and denominator. See Example 1. • We can multiply rational expressions by multiplying the numerators and multiplying the denominators. See Example 2. • To divide rational expressions, multiply by the reciprocal of the second expression. See Example 3. • Adding or subtracting rational expressions requires finding a common denominator. See Example 4 and Example 5. • Complex rational expressions have fractions in the numerator or the denominator. These expressions can be simplified. See Example 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 0 CHAPTER 1 PREREQUISITES CHAPTER 1 REVIEW EXERCISES REAL NUMBERS: ALGEBRA ESSENTIALS For the following exercises, perform the given operations. 1. (5 − 3 · 2)2 − 6 2. 64 ÷ (2 · 8) + 14 ÷ 7 3. 2 · 52 + 6 ÷ 2 For the following exercises, solve the equation. 4. 5x + 9 = −11 5. 2y + 42 = 64 For the following exercises, simplify the expression. 6. 9(y + 2) ÷ 3 · 2 + 1 7. 3m(4 + 7) − m For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer. 8. 11 9. 0 5 __ 10. 6 11. √ — 11 EXPONENTS AND SCIENTIFIC NOTATION For the following exercises, simplify the expression. 12. 22 · 24 16. (xy)4 _ y3 · 2 _ x5 13. 45 __ 43 17. 4−2x 3y−3 _______ 2x 0 4 a2 b3 ) 14. ( __ −2 18. ( 2x2 _ y ) 15. 6a2 · a0 _ 2a−4 19. ( 16a3 b2 ) (4ab−1)−2 ____ 20. Write the number in standard notation: 21. Write the number in scientific notation: 16,340,000 2.1314 × 10−6 RADICALS AND RATIONAL EXPRESSIONS For the following exercises, find the principal square root. 22. √ — 121 26. √ — 162 23. √ — 196 27. √ ___ 32 _ 25 24. √ — 361 28. √ ___ 80 _ 81 25. √ — 75 29. √ _____ 49 _ 1250 30. 2 _______ 4 + √ 2 — 31 32. 12 √ — 5 − 13 √ — 5 5 33. √ — −243 34. — 3 250 √ _ −8 √ — 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 1 REVIEW 71 POLYNOMIALS For the following exercises, perform the given operations and simplify. 35. (3x3 + 2x − 1) + (4x 2 − 2x + 7) 36. (2y + 1) − (2y 2 − 2y − 5) 37. (2x 2 + 3x − 6) + (3x 2 − 4x + 9) 38. (6a2 + 3a + 10) − (6a2 −3a + 5) 39. (k + 3)(k − 6) 41. (x + 1)(x 2 + 1) 43. (a + 2b)(3a − b) 40. (2h + 1)(3h − 2) 42. (m − 2)(m2 + 2m − 3) 44. (x + y)(x − y) FACTORING POLYNOMIALS For the following exercises, find the greatest common factor. 45. 81p + 9pq − 27p2q 2 46. 12x2y + 4xy 2 −18xy 47. 88a3b + 4a2b − 144a2 For the following exercises, factor the polynomial. 48. 2x 2 − 9x − 18 51. x 2 + 10x + 25 54. 361x 2 − 121 57. 64q 3 − 27p3 1 2 _ _ 60. 4r (2r − 1) − − 5(2r − 1) 3 3 49. 8a2 + 30a − 27 52. y 2 − 6y + 9 55. p3 + 216 50. d 2 − 5d − 66 53. 4h2 − 12hk + 9k2 56. 8x3 − 125 3 1 _ _ 58. 4x(x − 1) − + 3(x − 1) 4 4 4 1 _ _ − 8(p + 3) 59. 3p(p + 3) 3 3 RATIONAL EXPRESSIONS For the following exercises, simplify the expression. 61. x2 − x − 12 ___________ x2 − 8x + 16 62. 4y2 − 25 _____________ 4y2 − 20y + 25 63. 2a2 − a − 3 ___________ · 2a2 − 6a − 8 5a2 − 19a − 4 _____________ 10a2 − 13a − 3 65. m2 + 5m + 6 ____________ ÷ 2m2 − 5m − 3 2m2 + 3m − 9 ____________ 4m2 − 4m − 3 67. 6 10 _ y _ x + 69. 1 2 _ c _ + d _________ 6c + 12d _ dc 64. d − 4 ______ · d 2 − 9 d − 3 _______ d 2 − 16 66. 4d 2 − 7d − 2 _____________ 6d 2 − 17d + 10 ÷ 8d 2 + 6d + 1 ____________ 6d 2 + 7d − 10 68. 12 __________ − a2 + 2a + 1 3 _____ a2 −1 70. 3 7 _ y _ x − _______ 2 _ x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 2 CHAPTER 1 PREREQUISITES CHAPTER 1 PRACTICE TEST For the following exercises, identify the number as rational, irrational, whole, or natural. Choose the most descriptive answer. 1. −13 2. √ — 2 For the following exercises, evaluate the equations. 3. 2(x + 3) − 12 = 18 4. y(3 + 3)2 − 26 = 10 5. Write the number in standard notation: 3.1415 × 106 6. Write the number in scientific notation: 0.0000000212. For the following exercises, simplify the expression. 7. −2 · (2 + 3 · 2)2 + 144 8. 4(x + 3) − (6x + 2) 9. 35 · 3−3 3 2 ) 10. ( _ 3 13. √ — 441 11. 8x3 ____ (2x)2 14. √ — 490 12. (16y 0)2y −2 15. √ ___ 9x _ 16 16. — 121b2 √ _______ b 1 + √ — 17. 6 √ — 24 + 7 √ — 54 − 12 √ — 6 18. — 3 −8 √ _ 625 √ — 4 19. (13q3 + 2q2 − 3) − (6q2 + 5q − 3) 20. (6p2 + 2p + 1) + (9p2 − 1) 21. (n − 2)(n2 − 4n + 4) 22. (a − 2b)(2a + b) For the following exercises, factor the polynomial. 23. 16x 2 − 81 24. y2 + 12y + 36 25. 27c 3 − 1331 3 1 __ _ 26. 3x(x − 6) − + 2(x − 6) 4 4 For the following exercises, simplify the expression. 27. 2z2 + 7z + 3 ___________ · z2 − 9 4z2 − 15z + 9 ____________ 4z2 − 1 x 2 _ x _ y + 28. 29. a 2b ___ ___ − 9a 2b ________ 3a − 2b _______ 6a Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Equations and Inequalities 2 2 Figure 1 CHAPTER OUTLINE 2.1 The Rectangular Coordinate Systems and Graphs 2.2 Linear Equations in One Variable 2.3 Models and Applications 2.4 Complex Numbers 2.5 Quadratic Equations 2.6 Other Types of Equations 2.7 Linear Inequalities and Absolute Value Inequalities Introduction For most people, the term territorial possession indicates restrictions, usually dealing with trespassing or rite of passage and takes place in some foreign location. What most Americans do not realize is that from September through December, territorial possession dominates our lifestyles while watching the NFL. In this area, territorial possession is governed by the referees who make their decisions based on what the chains reveal. If the ball is at point A (x1, y1), then it is up to the quarterback to decide which route to point B (x2, y2), the end zone, is most feasible. 73 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 4 CHAPTER 2 EQUATIONS AND INEQUALITIES LEARNING OBJECTIVES In this section you will: • Plot and identify ordered pairs in a Cartesian coordinate system. • Graph equations by plotting ordered pairs. • Find x-intercepts and y-intercepts of the graph of an equation. • Use the distance formula. • Use the midpoint formula. 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS y Schiller Avenue Mannhelm Road Bertau Avenue McLean Street Wolf Road North Avenue x Figure 1 Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Laying a rectangular coordinate grid over the map, we can see that each stop aligns with an intersection of grid lines. In this section, we will learn how to use grid lines to describe locations and changes in locations. Plotting Ordered Pairs in the Cartesian Coordinate System An old story describes how seventeenth-century philosopher/mathematician René Descartes invented the system that has become the foundation of algebra while sick in bed. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each ax
is into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis. While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the x-axis and the vertical axis the y-axis. The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the x-axis and the y-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in Figure 2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 75 y-axis Quadrant II Quadrant I x-axis Quadrant III Quadrant IV Figure 2 The center of the plane is the point at which the two axes cross. It is known as the origin, or point (0, 0). From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the x-axis and up the y-axis; decreasing, negative numbers to the left on the x-axis and down the y-axis. The axes extend to positive and negative infinity as shown by the arrowheads in Figure 3. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 3 Each point in the plane is identified by its x-coordinate, or horizontal displacement from the origin, and its y-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form (x, y). An ordered pair is also known as a coordinate pair because it consists of x- and y-coordinates. For example, we can represent the point (3, −1) in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction. See Figure 4. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 x 21 4 (3, –1) 5 Figure 4 When dividing the axes into equally spaced increments, note that the x-axis may be considered separately from the y-axis. In other words, while the x-axis may be divided and labeled according to consecutive integers, the y-axis may be divided and labeled by increments of 2, or 10, or 100. In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 6 CHAPTER 2 EQUATIONS AND INEQUALITIES Cartesian coordinate system A two-dimensional plane where the • x-axis is the horizontal axis • y-axis is the vertical axis A point in the plane is defined as an ordered pair, (x, y), such that x is determined by its horizontal distance from the origin and y is determined by its vertical distance from the origin. Example 1 Plotting Points in a Rectangular Coordinate System Plot the points (−2, 4), (3, 3), and (0, −3) in the plane. Solution To plot the point (−2, 4), begin at the origin. The x-coordinate is −2, so move two units to the left. The y-coordinate is 4, so then move four units up in the positive y direction. To plot the point (3, 3), begin again at the origin. The x-coordinate is 3, so move three units to the right. The y-coordinate is also 3, so move three units up in the positive y direction. To plot the point (0, −3), begin again at the origin. The x-coordinate is 0. This tells us not to move in either direction along the x-axis. The y-coordinate is –3, so move three units down in the negative y direction. See the graph in Figure 5. (–2, 4) –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (3, 3) 321 4 5 x (0, –3) Figure 5 Analysis Note that when either coordinate is zero, the point must be on an axis. If the x-coordinate is zero, the point is on the y-axis. If the y-coordinate is zero, the point is on the x-axis. Graphing Equations by Plotting Points We can plot a set of points to represent an equation. When such an equation contains both an x variable and a y variable, it is called an equation in two variables. Its graph is called a graph in two variables. Any graph on a twodimensional plane is a graph in two variables. Suppose we want to graph the equation y = 2x − 1. We can begin by substituting a value for x into the equation and determining the resulting value of y. Each pair of x- and y-values is an ordered pair that can be plotted. Table 1 lists values of x from −3 to 3 and the resulting values for y. x −3 −2 −1 0 1 2 3 y = 2x − 1 y = 2(−3) − 1 = −7 y = 2(−2) − 1 = −5 y = 2(−1) − 1 = −3 y = 2(0) − 1 = −1 y = 2(1) − 1 = 1 y = 2(2) − 1 = 3 y = 2(3) − 1 = 5 Table 1 (x, y) (−3, −7) (−2, −5) (−1, −3) (0, −1) (1, 1) (2, 3) (3, 5) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 77 We can plot the points in the table. The points for this particular equation form a line, so we can connect them. See Figure 6. This is not true for all equations. (4, 7) (3, 5) (2, 3) (1, 1) 321 (0, –11 –2 –3 –4 –5 –6 –7 –5 –4 –3 –2 –1 (–1, –3) (–2, –5) (–3, –7) Figure 6 Note that the x-values chosen are arbitrary, regardless of the type of equation we are graphing. Of course, some situations may require particular values of x to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph. How To… Given an equation, graph by plotting points. 1. Make a table with one column labeled x, a second column labeled with the equation, and a third column listing the resulting ordered pairs. 2. Enter x-values down the first column using positive and negative values. Selecting the x-values in numerical order will make the graphing simpler. 3. Select x-values that will yield y-values with little effort, preferably ones that can be calculated mentally. 4. Plot the ordered pairs. 5. Connect the points if they form a line. Example 2 Graphing an Equation in Two Variables by Plotting Points Graph the equation y = −x + 2 by plotting points. Solution First, we construct a table similar to Table 2. Choose x values and calculate y. x −5 −3 − = − (−5) + 2 = 7 y = − (−3) + 2 = 5 y = − (−1) + 2 = 3 y = − (0) + 2 = 2 y = − (1) + 2 = 1 y = − (3) + 2 = −1 y = − (5) + 2 = −3 Table 2 (x, y) (−5, 7) (−3, 5) (−1, 3) (0, 2) (1, 1) (3, −1) (5, −3) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 7 8 CHAPTER 2 EQUATIONS AND INEQUALITIES Now, plot the points. Connect them if they form a line. See Figure 7. (–5, 7) (–3, 5) (–1, 30, 2) (1, 1) –7 –6 –5 –4 –3 –2 –1 –1 –2 –3 321 (3, –1) 4 5 6 7 (5, –3) Figure 7 x Try It #1 1 _ x + 2. Construct a table and graph the equation by plotting points: y = 2 Graphing Equations with a Graphing Utility Most graphing calculators require similar techniques to graph an equation. The equations sometimes have to be manipulated so they are written in the style y = . The TI-84 Plus, and many other calculator makes and models, have a mode function, which allows the window (the screen for viewing the graph) to be altered so the pertinent parts of a graph can be seen. For example, the equation y = 2x − 20 has been entered in the TI-84 Plus shown in Figure 8a. In Figure 8b, the resulting graph is shown. Notice that we cannot see on the screen where the graph crosses the axes. The standard window screen on the TI-84 Plus shows −10 ≤ x ≤ 10, and −10 ≤ y ≤ 10. See Figure 8c. Plot1 Plot2 Plot3 \Y1= 2X–20 \Y2= \Y3= \Y4= \Y5= \Y6= \Y7= (a) WINDOW Xmin = −10 Xmax = 10 Xscl = 1 Ymin = −10 Ymax = 10 Yscl = 1 Xres = 1 x (b) (c) Figure 8 (a) Enter the equation. (b) This is the graph in the original window. (c) These are the original settings. By changing the window to show more of the positive x-axis and more of the negative y-axis, we have a much better view of the graph and the x- and y-intercepts. See Figure 9a and Figure 9b. WINDOW Xmin = −5 Xmax = 15 Xscl = 1 Ymin = −30 Ymax = 10 Yscl = 1 Xres = 1 y (a) (b) x Figure 9 (a) This screen shows the new window settings. (b) We can clearly view the intercepts in the new window. Example 3 Using a Graphing Utility to Graph an Equation 2 4 _ _ Use a graphing utility to graph the equation: y = − x − . 3 3 Solution Enter the equation in the y = function of the calculator. Set the window settings so that both the x- and y-intercepts are showing in the window. See Figure 105 –4 –3 –2 –1 –1 –2 321 4 5 x Figure 10 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 79 Finding x-intercepts and y-intercepts The intercepts of a graph are points at which the graph crosses the axes. The x-intercept is the point at which the graph crosses the x-axis. At this point, the y-coordinate is zero. The y-intercept is the point at which the graph crosses the y-axis. At this point, the x-coordinate is zero. To determine the x-intercept, we set y equal to zero and solve for x. Similarly, to determine the y-intercept, we set x equal to zero and solve for y. For example, lets find the intercepts of the equation y = 3x − 1. To find the x-intercept, set y = 0. To find the y-intercept, set x = 0. y = 3x − 1 0 = 3x − 1 1 = 3x = 3x − 1 y = 3(0) − 1 y = −1 (0, −1) x-intercept y-intercept We can confirm that our results make sense by observing a graph of the equation as in Figure 11. Notice that
the graph crosses the axes where we predicted it would. –5 –4 –3 –2 y = 3x − 1 321 4 5 x y 4 3 2 1 –1 –1 –2 –3 –4 Figure 11 given an equation, find the intercepts. • Find the x-intercept by setting y = 0 and solving for x. • Find the y-intercept by setting x = 0 and solving for y. Example 4 Finding the Intercepts of the Given Equation Find the intercepts of the equation y = −3x − 4. Then sketch the graph using only the intercepts. Solution Set y = 0 to find the x-intercept. Set x = 0 to find the y-intercept. y = −3x − 4 0 = −3x − 4 4 = −3x 3x − 4 y = −3(0) − 4 y = −4 (0, −4) x−intercept y−intercept Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 80 CHAPTER 2 EQUATIONS AND INEQUALITIES Plot both points, and draw a line passing through them as in Figure 12. 4 − _ , 0 3 –5 –4 –3 –2 y 3 2 1 –1 –1 –2 –3 –4 –5 –6 321 4 5 x (0, –4) Figure 12 Try It #2 3 _ x + 3. Find the intercepts of the equation and sketch the graph: y = − 4 Using the Distance Formula Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, a2 + b2 = c2, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. See Figure 13x1, y1) (x2, y2) d = c y2 − y1 = b − x1 = a x2 2 3 4 1 (x2, y1) 5 6 7 x Figure 13 The relationship of sides \|x2 − x1 \| to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, \|−3\| = 3. ) The symbols \|x2 − x1 \| indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem. \| and \|y2 − y1 \| and \|y2 − y1 It follows that the distance formula is given as c2 = a2 + b2 → c = √ — a2 + b2 We do not have to use the absolute value symbols in this definition because any number squared is positive. d 2 = (x2 − x1)2 + (y2 − y1)2 → d = √ —— (x2 − x1)2 + (y2 − y1)2 the distance formula Given endpoints (x1, y1) and (x2, y2), the distance between two points is given by d = √ —— (x2 − x1)2 + (y2 − y1)2 Example 5 Finding the Distance between Two Points Find the distance between the points (−3, −1) and (2, 3). Solution Let us first look at the graph of the two points. Connect the points to form a right triangle as in Figure 14. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 81 y 4 3 2 1 –1 –1 –2 –3 –4 –2 –3 –4 –5 (−3, −1) (2, 3) x 321 4 5 (2, −1) Figure 14 Then, calculate the length of d using the distance formula. d = √ d = √ = √ —— —— (x2 − x1)2 + (y2 − y1)2 (2 − (−3))2 + (3 − (−1))2 (5)2 + (4)2 — = √ — 25 + 16 = √ — 41 Try It #3 Find the distance between two points: (1, 4) and (11, 9). Example 6 Finding the Distance Between Two Locations Let’s return to the situation introduced at the beginning of this section. Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions. Solution The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at (1, 1). The next stop is 5 blocks to the east, so it is at (5, 1). After that, she traveled 3 blocks east and 2 blocks north to (8, 3). Lastly, she traveled 4 blocks north to (8, 7). We can label these points on the grid as in Figure 15. y Schiller Avenue (8, 7) Mannhelm Road (8, 3) Wolf Road x Bertau Avenue McLean Street North Avenue (1, 1) (5, 1) (0, 0) Figure 15 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 82 CHAPTER 2 EQUATIONS AND INEQUALITIES Next, we can calculate the distance. Note that each grid unit represents 1,000 feet. • From her starting location to her first stop at (1, 1), Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop. • Her second stop is at (5, 1). So from (1, 1) to (5, 1), Tracie drove east 4,000 feet. • Her third stop is at (8, 3). There are a number of routes from (5, 1) to (8, 3). Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet. • Tracie’s final stop is at (8, 7). This is a straight drive north from (8, 3) for a total of 4,000 feet. Next, we will add the distances listed in Table 3. From/To Number of Feet Driven (0, 0) to (1, 1) (1, 1) to (5, 1) (5, 1) to (8, 3) (8, 3) to (8, 7) Total 2,000 4,000 5,000 4,000 15,000 Table 3 The total distance Tracie drove is 15,000 feet, or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points (0, 0) and (8, 7). d = √ — (8 − 0)2 + (7 − 0)2 = √ — 64 + 49 = √ — 113 ≈ 10.63 units At 1,000 feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point (8, 7). Perhaps you have heard the saying “as the crow flies,” which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways. Using the Midpoint Formula When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula. Given the endpoints of a line segment, (x1, y1) and (x2, y2), the midpoint formula states how to find the coordinates of the midpoint M. x1 + x2 _______ , 2 A graphical view of a midpoint is shown in Figure 16. Notice that the line segments on either side of the midpoint are congruent. y1 + y2 ) _______ 2 M = ( y 0 x1 + x2 _______ , 2 y1 + y2 _______ 2 x Figure 16 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.1 THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS 83 Example 7 Finding the Midpoint of the Line Segment Find the midpoint of the line segment with the endpoints (7, −2) and (9, 5). Solution Use the formula to find the midpoint of the line segment. x1 + x2 ( _______ , 2 y1 + y2 _______ 2 ) = ( −2 + 5 7 + 9 ) _______ _____ , 2 2 3 ) = ( 8, __ 2 Try It #4 Find the midpoint of the line segment with endpoints (−2, −1) and (−8, 6). Example 8 Finding the Center of a Circle The diameter of a circle has endpoints (−1, −4) and (5, −4). Find the center of the circle. Solution The center of a circle is the center, or midpoint, of its diameter. Thus, the midpoint formula will yield the center point. x1 + x2 ( _______ , 2 y1 + y2 ) _______ 2 −4 −4 −1 + 5 ( ______ _______ , 2 2 8 4 ) = (2, −4) ) = ( __ __ , − 2 2 Access these online resources for additional instruction and practice with the Cartesian coordinate system. • Plotting Points on the Coordinate Plane (http://openstaxcollege.org/l/coordplotpnts) • Find x- and y-intercepts Based on the Graph of a Line (http://openstaxcollege.org/l/xyintsgraph) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 84 CHAPTER 2 EQUATIONS AND INEQUALITIES 2.1 SECTION EXERCISES VERBAL 1. Is it possible for a point plotted in the Cartesian coordinate system to not lie in one of the four quadrants? Explain. 3. Describe in your own words what the y-intercept of a graph is. 2. Describe the process for finding the x-intercept and the y-intercept of a graph algebraically. 4. When using the distance formula —— (x2 − x1)2 + (y2 − y1)2 , explain the correct d = √ order of operations that are to be performed to obtain the correct answer. ALGEBRAIC For each of the following exercises, find the x-intercept and the y-intercept without graphing. Write the coordinates of each intercept. 5. y = −3x + 6 7. 3x − 2y = 6 8. 4x − 3 = 2y 9. 3x + 8y = 9 6. 4y = 2x − 1 3 2 _ _ 10. 2x − y + 3 = 4 3 For each of the following exercises, solve the equation for y in terms of x. 11. 4x + 2y = 8 15. 5y + 4 = 10x 12. 3x − 2y = 6 16. 5x + 2y = 0 13. 2x = 5 − 3y 14. x − 2y = 7 For each of the following exercises, find the distance between the two points. Simplify your answers, and write the exact answer in simplest radical form for irrational answers. 17. (−4, 1) and (3, −4) 20. (−4, 3) and (10, 3) 18. (2, −5) and (7, 4) 19. (5, 0) and (5, 6) 21. Find the distance between the two points given using your calculator, and round your answer to the nearest hundredth. (19, 12) and (41, 71) For each of the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points. 22. (−5, −6) and (4, 2) 24. (−5, −3) and (−2, −8) 23. (−1, 1) and (7, −4) 25. (0, 7) and (4, −9) 26. (−43, 17) and (23, −34) GRAPHICAL For each of the following exercises, identify the information requested. 27. What are the coordinates of the origin? 28. If a point is located on the y-axis, what is the x-coordinate? 29. If a point is located on the x-axis, what is the y-coordinate? For each of the following exercises, plot the three points on the given coordinate plane. State whether the three points you plotted appear to be collinear (on the same line). 30. (4, 1)(−2, −3)(5, 0) 31. (−1, 2)(0, 4)(2, 1) y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Download the OpenStax text for free at
http://cnx.org/content/col11759/latest. SECTION 2.1 SECTION EXERCISES 85 32. (−3, 0)(−3, 4)(−3, −3) 33. Name the coordinates of the points graphed. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x A –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 B C 5 x 21 3 4 34. Name the quadrant in which the following points would be located. If the point is on an axis, name the axis. a. (−3, −4) b. (−5, 0) c. (1, −4) d. (−2, 7) e. (0, −3) For each of the following exercises, construct a table and graph the equation by plotting at least three points. 1 _ 37. 2y = x + 3 35. y = x + 2 3 36. y = −3x + 1 NUMERIC For each of the following exercises, find and plot the x- and y-intercepts, and graph the straight line based on those two points. 38. 4x − 3y = 12 41. 3y = −2x + 6 39. x − 2y = 8 40. y − 5 = 5x 42. y = x − 3 _____ 2 For each of the following exercises, use the graph in the figure below. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x 43. Find the distance between the two endpoints using the distance formula. Round to three decimal places. 44. Find the coordinates of the midpoint of the line segment connecting the two points. 45. Find the distance that (−3, 4) is from the origin. 46. Find the distance that (5, 2) is from the origin. Round to three decimal places. 47. Which point is closer to the origin? TECHNOLOGY For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu. After graphing it, use the 2nd CALC button and 1:value button, hit ENTER. At the lower part of the screen you will see “x=” and a blinking cursor. You may enter any number for x and it will display the y value for any x value you input. Use this and plug in x = 0, thus finding the y-intercept, for each of the following graphs. 48. Y1 = −2x + 5 49. Y1 = 3x − 8 ______ 4 50. Y1 = x + 5 _____ 2 For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu. After graphing it, use the 2nd CALC button and 2:zero button, hit ENTER. At the lower part of the screen you will see “left bound?” and a blinking cursor on the graph of the line. Move this cursor to the left of the x-intercept, hit ENTER. Now it says “right bound?” Move the cursor to the right of the x-intercept, hit ENTER. Now it says “guess?” Move your cursor to the left somewhere in between the left and right bound near the x-intercept. Hit ENTER. At the bottom of your screen it will display the coordinates of the x-intercept or the “zero” to the y-value. Use this to find the x-intercept. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 86 CHAPTER 2 EQUATIONS AND INEQUALITIES Note: With linear/straight line functions the zero is not really a “guess,” but it is necessary to enter a “guess” so it will search and find the exact x-intercept between your right and left boundaries. With other types of functions (more than one x-intercept), they may be irrational numbers so “guess” is more appropriate to give it the correct limits to find a very close approximation between the left and right boundaries. 51. Y1 = −8x + 6 52. Y1 = 4x − 7 53. Y1 = 3x + 5 ______ 4 Round your answer to the nearest thousandth. EXTENSIONS 54. A man drove 10 mi directly east from his home, made a left turn at an intersection, and then traveled 5 mi north to his place of work. If a road was made directly from his home to his place of work, what would its distance be to the nearest tenth of a mile? 55. If the road was made in the previous exercise, how much shorter would the man’s one-way trip be every day? 56. Given these four points: A(1, 3), B(−3, 5), C(4, 7), and D(5, −4), find the coordinates of the midpoint _ #CD . of line segments _ #AB and 57. After finding the two midpoints in the previous exercise, find the distance between the two midpoints to the nearest thousandth. 58. Given the graph of the rectangle shown and the 59. In the previous exercise, find the coordinates of the coordinates of its vertices, prove that the diagonals of the rectangle are of equal length. midpoint for each diagonal. (–6, 5) y 6 5 4 3 2 1 (10, 5) –5 –7 –4 (–6, –1) –3 –2 –1 –1 –2 –3 321 4 5 6 7 8 9 11 (10, –1) x REAL-WORLD APPLICATIONS 60. The coordinates on a map for San Francisco are (53, 17) and those for Sacramento are (123, 78). Note that coordinates represent miles. Find the distance between the cities to the nearest mile. 61. If San Jose’s coordinates are (76, −12), where the coordinates represent miles, find the distance between San Jose and San Francisco to the nearest mile. 62. A small craft in Lake Ontario sends out a distress signal. The coordinates of the boat in trouble were (49, 64). One rescue boat is at the coordinates (60, 82) and a second Coast Guard craft is at coordinates (58, 47). Assuming both rescue craft travel at the same rate, which one would get to the distressed boat the fastest? 64. If we rent a truck and pay a $75/day fee plus $.20 for every mile we travel, write a linear equation that would express the total cost y, using x to represent the number of miles we travel. Graph this function on your graphing calculator and find the total cost for one day if we travel 70 mi. 63. A man on the top of a building wants to have a guy wire extend to a point on the ground 20 ft from the building. To the nearest foot, how long will the wire have to be if the building is 50 ft tall? (20, 50) 50 (0, 0) 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 87 LEARNING OBJECTIVES In this section you will: • ole linear equations in one ariable. • ole rational equations in one ariable. • Find the slope of a line that contains two gien points. • Graph linear equations. • Use the point-slope andor slope intercept formula to write the equation of a line that satisfies certain gien properties. • Gien the equations of two lines determine whether their graphs are parallel perpendicular or neither. • Find the equation of a line parallel or perpendicular to a gien line. 2.2 LINEAR EQUATIONS IN ONE VARIABLE Figure 1 y Figure 1 x Solving Linear Equations in One Variable linear equationTh ax+b = identity equation x =x +x solution set x conditional equation x +=x − x +=x − x =− x =− Th− inconsistent equationx −=x− x −=x − x −−x =x −−x x −≠− Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 88 CHAPTER 2 EQUATIONS AND INEQUALITIES Indeed, −15 ≠ −20. There is no solution because this is an inconsistent equation. Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows. linear equation in one variable A linear equation in one variable can be written in the form ax + b = 0 where a and b are real numbers, a ≠ 0. How To… Given a linear equation in one variable, use algebra to solve it. The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = , if x is the unknown. There is no set order, as the steps used depend on what is given: 1. We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero. 2. Apply the distributive property as needed: a(b + c) = ab + ac. 3. Isolate the variable on one side of the equation. 4. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient. Example 1 Solving an Equation in One Variable Solve the following equation: 2x + 7 = 19. Solution This equation can be written in the form ax + b = 0 by subtracting 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations. 2x + 7 = 19 2x = 12 x = 6 Subtract 7 from both sides. 1 _ or divide by 2. Multiply both sides by 2 The solution is 6. Try It #1 Solve the linear equation in one variable: 2x + 1 = −9. Example 2 Solving an Equation Algebraically When the Variable Appears on Both Sides Solve the following equation: 4(x − 3) + 12 = 15 − 5(x + 6). Solution Apply standard algebraic properties. 4(x − 3) + 12 = 15 − 5(x + 6) 4x − 12 + 12 = 15 − 5x − 30 Apply the distributive property. 4x = −15 − 5x Combine like terms. x = − 9x = −15 15_ 9 5 _ x = − 3 Place x- terms on one side and simplify. 1 _ , the reciprocal of 9. Multiply both sides by 9 Analysis This problem requires the distributive property to be applied twice, and then the properties of algebra are used 5 _ to reach the final line, x = − . 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 89 Try It #2 Solve the equation in one variable: −2(3x − 1) + x = 14 − x. Solving a Rational Equation In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation. 7 2 _ _ Recall that a rational number is the ratio of two numbers, such as . A rational expression is the ratio, or quotient, or 2 3 of two polynomials. Here are three examples. x + 1 _ , x2 − 4 1 _ x − 3 , or 4 _ x2 + x − 2 Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD). Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out. Example 3 Solving a Rational Equation Solve the rational equation: 7 _ 2x − = 5 _ 3x 22 _ . 3 Solution We have three denominato
rs; 2x, 3x, and 3. The LCD must contain 2x, 3x, and 3. An LCD of 6x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6x. − 7 5 22 ) = ( (6x) ( ) (6x) _ _ _ 2x 3x 3 5 7 22 ) (6x) ) = ( ) − (6x) ( (6x) ( _ _ _ 3 3x 2x ) ( , 6 x) ) = ( ) − ( , 6x ) ( ( , 6x ) ( 7 5 22 _ _ _ , 2x , 3x , 3 3(7) − 2(5) = 22(2x) 21 − 10 = 44x 11 = 44x = x 11 _ 44 1 _ = x 4 Use the distributive property. Cancel out the common factors. Multiply remaining factors by each numerator. A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as (x + 1). Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x, x − 1, and 3x − 3. First, factor all denominators. We then have x, (x − 1), and 3(x − 1) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of (x − 1). The x in the first denominator is separate from the x in the (x − 1) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x, one factor of (x − 1), and the 3. Thus, the LCD is the following: x(x − 1)3 = 3x(x − 1) So, both sides of the equation would be multiplied by 3x(x − 1). Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 90 CHAPTER 2 EQUATIONS AND INEQUALITIES Another example is a problem with two denominators, such as x and x 2 + 2x. Once the second denominator is factored as x 2 + 2x = x(x + 2), there is a common factor of x in both denominators and the LCD is x(x + 2). Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation. We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign = = . , then If d b d b Multiply a(d) and b(c), which results in ad = bc. Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities. rational equations A rational equation contains at least one rational expression where the variable appears in at least one of the denominators. How To… Given a rational equation, solve it. 1. Factor all denominators in the equation. 2. Find and exclude values that set each denominator equal to zero. 3. Find the LCD. 4. Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left. 5. Solve the remaining equation. 6. Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator. Example 4 Solving a Rational Equation without Factoring Solve the following rational equation 2x Solution We have three denominators: x, 2, and 2x. No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2x. Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2x. 7 3 2 2x ( ) = ( _ _ _ x − 2x 2_ _ % 2 %x 2(2) − 3x = 7 ) 2x ) %2x 7 _ % 2x 4 − 3x = 7 −3x = 3 x = −1 or {−1} Distribute 2x. Denominators cancel out. The proposed solution is −1, which is not an excluded value, so the solution set contains one number, −1, or {−1} written in set notation. Try It #3 Solve the rational equation: 2 _ 3x 1 _ − = 4 1 _ . 6x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 91 Example 5 Solving a Rational Equation by Factoring the Denominator 1 1 _ _ x = Solve the following rational equation: 10 − 3 _ . 4x Solution First find the common denominator. The three denominators in factored form are x, 10 = 2 ċ 5, and 4x = 2 ċ 2 ċ x. The smallest expression that is divisible by each one of the denominators is 20x. Only x = 0 is an excluded value. Multiply the whole equation by 20x. 1 20x ( − _ x ) = ( 1 _ 10 20 = 2x − 15 ) 20x 3 ___ 4x The solution is 35 _ . 2 35 = 2x 35 ___ 2 = x Try It #4 Solve the rational equation: − 5 _ 2x + 3 _ 4x 7 _ = − . 4 Example 6 Solving Rational Equations with a Binomial in the Denominator Solve the following rational equations and state the excluded values: a. Solution . The denominators x and x − 6 have nothing in common. Therefore, the LCD is the product x(x − 6). However, for this problem, we can cross-multiply 3x = 5(x − 6) 3x = 5x − 30 −2x = −30 x = 15 Distribute. The solution is 15. The excluded values are 6 and 0. b. The LCD is 2(x − 3). Multiply both sides of the equation by 2(x − 3). 2(x − 3) ( ) = ( x _____ x − 3 2 , (x − 3) x ________ = , x − 3 5 _____ x − 3 2 ,(x − 3) 5 ________ − , x − 3 1 ) 2(x − 3) __ − 2 % 2 (x − 3) _______ % 2 2x = 10 − (x − 3) 2x = 10 − x + 3 2x = 13 − x 3x = 13 13___ 3 x = The solution is . The excluded value is 3. 13 _ 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 92 CHAPTER 2 EQUATIONS AND INEQUALITIES c. The least common denominator is 2(x − 2). Multiply both sides of the equation by x(x − 2). 2(x − 2(x − 2) _ − 2 2x = 10 − (x − 2) 2x = 12 − x 3x = 12 x = 4 The solution is 4. The excluded value is 2. Try It #5 Solve −3 _ 2x + 1 = 4 _ 3x + 1 . State the excluded values. Example 7 Solving a Rational Equation with Factored Denominators and Stating Excluded Values Solve the rational equation after factoring the denominators: . State the excluded values = 2x _ x2 − 1 Solution We must factor the denominator x2 − 1. We recognize this as the difference of squares, and factor it as (x − 1)(x + 1). Thus, the LCD that contains each denominator is (x − 1)(x + 1). Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation. (x − 1)(x + 1) ( ) (x − 1)(x + 1) 2x __ (x − 1)(x + 1(x − 1) − 1(x + 1) = 2x 2x − 2 − x − 1 = 2x Distribute the negative sign. −3 − x = 0 −3 = x The solution is −3. The excluded values are 1 and −1. Try It #6 Solve the rational equation . x2 − x − 2 Finding a Linear Equation Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = mx + b, where m = slope and b = y-intercept. Let us begin with the slope. The Slope of a Line The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run. m = y2 − y1 __ x2 − x1 If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2. The lines indicate the following slopes: m = −3, 1 _ m = 2, and m = . 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 93 y y = 2x + 1 1 y = x + 2 3 42 6 8 10 12 x y = −3x − 2 12 10 8 6 4 2 –12 –10 –8 –6 –4 –2 –2 –4 –6 –8 –10 –12 Figure 2 the slope of a line The slope of a line, m, represents the change in y over the change in x. Given two points, (x1, y1) and (x2, y2), the following formula determines the slope of a line containing these points: m = y2 − y1 __ x2 − x1 Example 8 Finding the Slope of a Line Given Two Points Find the slope of a line that passes through the points (2, −1) and (−5, 3). Solution We substitute the y-values and the x-values into the formula. m = 3 − (−1) _ −5 − 2 4 _ −7 4 _ = − 7 = 4 _ The slope is − . 7 Analysis terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result. It does not matter which point is called (x1, y1) or (x2, y2). As long as we are consistent with the order of the y Try It #7 Find the slope of the line that passes through the points (−2, 6) and (1, 4). Example 9 Identifying the Slope and y-intercept of a Line Given an Equation 3 _ Identify the slope and y-intercept, given the equation y = − x − 4. 4 3 _ Solution As the line is in y = mx + b form, the given line has a slope of m = − . The y-intercept is b = −4. 4 Analysis The y-intercept is the point at which the line crosses the y-axis. On the y-axis, x = 0. We can always identify the y-intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x = 0 and solve for y. The Point-Slope Formula Given the slope and one point on a line, we can find the equation of the line using the point-slope formula. y − y1 = m(x − x1) This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 94 CHAPTER 2 EQUATIONS AND INEQUALITIES the point-slope formula Given one point and the slope, the point-slope formula will lead to the equation of a line: y − y1 = m(x − x1) Example 10 Finding the Equation of a Line Given the Slope and One Point Write the equation of the line with slope m = −3 and passing through the point (4, 8). Write the final equation in slope-intercept form. Solution Using the point-slope formula, substitute −3 for m and the point (4, 8) for (x1, y1). y − y1 = m(x − x1) y − 8 = −3(x − 4) y − 8 = −3x + 12 y = −3x + 20 Analysis Note that any point on the line can be used to find the equat
ion. If done correctly, the same final equation will be obtained. Try It #8 Given m = 4, find the equation of the line in slope-intercept form passing through the point (2, 5). Example 11 Finding the Equation of a Line Passing Through Two Given Points Find the equation of the line passing through the points (3, 4) and (0, −3). Write the final equation in slope-intercept form. Solution First, we calculate the slope using the slope formula and two points. m = = −3 − 4_ 0 − 3 −7 _ −3 7 _ = 3 7 _ Next, we use the point-slope formula with the slope of , and either point. Let’s pick the point (3, 4) for (x1, y1). 3 7 _ (x − 3 . Distribute the 3 7 _ x − 3. In slope-intercept form, the equation is written as y = 3 Analysis To prove that either point can be used, let us use the second point (0, −3) and see if we get the same equation. 7 _ y − (−3) = (x − 0 We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 95 Standard Form of a Line Another way that we can represent the equation of a line is in standard form. Standard form is given as Ax + By = C where A, B, and C are integers. The x- and y-terms are on one side of the equal sign and the constant term is on the other side. Example 12 Finding the Equation of a Line and Writing It in Standard Form 1 , −2 ) . Write the equation in standard form. Find the equation of the line with m = −6 and passing through the point ( _ 4 Solution We begin using the point-slope formula. 1 ) y − (−2) = −6x + 2 From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right. 3 ) 2 2(y + 2) = ( −6x + _ 2 2y + 4 = −12x + 3 This equation is now written in standard form. 12x + 2y = −1 Try It #9 1 1 ) . and passing through the point ( 1, _ _ Find the equation of the line in standard form with slope m = − 3 3 Vertical and Horizontal Lines The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as x = c where c is a constant. The slope of a vertical line is undefined, and regardless of the y-value of any point on the line, the x-coordinate of the point will be c. Suppose that we want to find the equation of a line containing the following points: (−3, −5), (−3, 1), (−3, 3), and (−3, 5). First, we will find the slope. m = 5 − 3 2 _________ __ = −3 − (−3) 0 Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x-coordinates are the same and we find a vertical line through x = −3. See Figure 3. The equation of a horizontal line is given as y = c where c is a constant. The slope of a horizontal line is zero, and for any x-value of a point on the line, the y-coordinate will be c. Suppose we want to find the equation of a line that contains the following set of points: (−2, −2), (0, −2), (3, −2), and (5, −2). We can use the point-slope formula. First, we find the slope using any two points on the line. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 96 CHAPTER 2 EQUATIONS AND INEQUALITIES m = −2 − (−2)__ 0 − (−2) 0 _ = 2 = 0 Use any point for (x1, y1) in the formula, or use the y-intercept. y − (−2) = 0(x − 3) y + 2 = 0 y = −2 The graph is a horizontal line through y = −2. Notice that all of the y-coordinates are the same. See Figure 3. x = −3 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 x 21 3 4 5 y = −2 Figure 3 The line x = −3 is a vertical line. The line y = −2 is a horizontal line. Example 13 Finding the Equation of a Line Passing Through the Given Points Find the equation of the line passing through the given points: (1, −3) and (1, 4). Solution The x-coordinate of both points is 1. Therefore, we have a vertical line, x = 1. Try It #10 Find the equation of the line passing through (−5, 2) and (2, 2). Determining Whether Graphs of Lines are Parallel or Perpendicular Parallel lines have the same slope and different y-intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, m = 2. y = 2x − 3 y = 2x + 1 y = 2x + 5 21 1 –1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 4 Parallel lines All of the lines shown in the graph are parallel because they have the same slope and different y-intercepts. Lines that are perpendicular intersect to form a 90° -angle. The slope of one line is the negative reciprocal of the other. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 97 We can show that two lines are perpendicular if the product of the two slopes is −1: m1 ċ m2 = −1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 __ . 3 m1 ċ m2 = −1 1 ) = −1 3 ċ ( − _ 3 y y = 3x − 1 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 Figure 5 Perpendicular lines Example 14 Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3y = − 4x + 3 and 3x − 4y = 8. Solution The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form. First equation: Second equation: See the graph of both lines in Figure 6. 3y = −4x + 3x − 4y = 8 −4y = −3x + 5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 6 From the graph, we can see that the lines appear perpendicular, but we must compare the slopes. 4 _ m1 = − 3 3 _ m2 = 4 3 4 ) ( m1 ċ m2 = ( − ) = −1 _ _ 4 3 The slopes are negative reciprocals of each other, confirming that the lines are perpendicular. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 98 CHAPTER 2 EQUATIONS AND INEQUALITIES Try It #11 Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2y − x = 10 and 2y = x + 4. Writing the Equations of Lines Parallel or Perpendicular to a Given Line As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line. How To… Given an equation for a line, write the equation of a line parallel or perpendicular to it. 1. Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form. 2. Use the slope and the given point with the point-slope formula. 3. Simplify the line to slope-intercept form and compare the equation to the given line. Example 15 Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point Write the equation of line parallel to a 5x + 3y = 1 and passing through the point (3, 5). Solution First, we will write the equation in slope-intercept form to find the slope. 5x + 3y = 1 3y = 5x + The slope is m = − , but that really does not enter into our problem, as the only thing we need . The y-intercept is 3 3 for two lines to be parallel is the same slope. The one exception is that if the y-intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula. 5 _ y − 5 = − (x − 3 + 10 y = − 3 5 _ The equation of the line is y = − x + 10. See Figure 7. 3 y = − x + + 10 3 –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 42 6 8 10 x Figure 7 Try It #12 Find the equation of the line parallel to 5x = 7 + y and passing through the point (−1, −2). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 LINEAR EQUATIONS IN ONE VARIABLE 99 Example 16 Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point Find the equation of the line perpendicular to 5x − 3y + 4 = 0 and passes through the point (−4, 1). Solution The first step is to write the equation in slope-intercept form. 5x − 3y + 4 = 0 −3y = −5x − 4 5 _ We see that the slope is m = . This means that the slope of the line perpendicular to the given line is the negative 3 3 _ reciprocal, or − . Next, we use the point-slope formula with this new slope and the given pointx − (−4)) 5 3 12_ 12 Access these online resources for additional instruction and practice with linear equations. • Solving rational equations (http://openstaxcollege.org/l/rationaleqs) • Equation of a line given two points (http://openstaxcollege.org/l/twopointsline) • Finding the equation of a line perpendicular to another line through a given point (http://openstaxcollege.org/l/ﬁndperpline) • Finding the equation of a line parallel to another line through a given point (http://openstaxcollege.org/l/ﬁndparaline) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 100 CHAPTER 2 EQUATIONS AND INEQUALITIES 2.2 SECTION EXERCISES VERBAL 1. 2. 3. y =x + 5. x − = x + 4. x =x =− ALGEBRAIC x 6. x +=x − 7. x−= 8. x +−=x + 9. −x +=x − 10. − x= 11. x = − x + 12. = x + 13. x −+x =x + 14. x x + = − 15. x + − x − = x x- 16. = x − 17. − x+ = x+ x+ 18. x − = x − + x −x − 19. x x − += x − 20. x+ + x− = − x−x− x = + 21. x 22. 23. −− 24. − 25. −− 26. − 27. 30. 33. 36. 39. 28. 1 3 31. 34. 37. 40. 29. 32. 35. 38. 41. 42. 44. 46. − −− 43. − 45. − 47. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.2 SECTION EXERCISES 101 48. −− 49. 50. 52. (,) 51. − 53. −− GRAPHICAL 54. y =x + x − y=− NUMERIC 55. x −y = y −x = 56. x + y = y =x + 57. x = y =− 5
8. − − 59. −− 60. y =x + 61. y =x − − 62. = (,) + 63. = (,) + 64. = (,) 65. = (,) EXTENSIONS 66. y −y=mx −x x xyym 68. x −y = 70. REAL-WORLD APPLICATIONS 71. Th s 2.5 ft, fin x 67. Ax +By =C y ABCx. Th 69. − 72. fi x y p =x +y y p = x = Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 102 CHAPTER 2 EQUATIONS AND INEQUALITIES LEARNING OBJECTIVES In this section you will: • Set up a linear equation to solve a real-world application. 2.3 MODELS AND APPLICATIONS Figure 1 Credit: Kevin Dooley Josh is hoping to get an A in his college algebra class. He has scores of 75, 82, 95, 91, and 94 on his first five tests. Only the final exam remains, and the maximum of points that can be earned is 100. Is it possible for Josh to end the course with an A? A simple linear equation will give Josh his answer. Many real-world applications can be modeled by linear equations. For example, a cell phone package may include a monthly service fee plus a charge per minute of talk-time; it costs a widget manufacturer a certain amount to produce x widgets per month plus monthly operating charges; a car rental company charges a daily fee plus an amount per mile driven. These are examples of applications we come across every day that are modeled by linear equations. In this section, we will set up and use linear equations to solve such problems. Setting up a Linear Equation to Solve a Real-World Application To set up or model a linear equation to fit a real-world application, we must first determine the known quantities and define the unknown quantity as a variable. Then, we begin to interpret the words as mathematical expressions using mathematical symbols. Let us use the car rental example above. In this case, a known cost, such as $0.10/mi, is multiplied by an unknown quantity, the number of miles driven. Therefore, we can write 0.10x. This expression represents a variable cost because it changes according to the number of miles driven. If a quantity is independent of a variable, we usually just add or subtract it, according to the problem. As these amounts do not change, we call them fixed costs. Consider a car rental agency that charges $0.10/mi plus a daily fee of $50. We can use these quantities to model an equation that can be used to find the daily car rental cost C. When dealing with real-world applications, there are certain expressions that we can translate directly into math. Table 1 lists some common verbal expressions and their equivalent mathematical expressions. C = 0.10x + 50 Verbal One number exceeds another by a Twice a number One number is a more than another number One number is a less than twice another number The product of a number and a, decreased by b The quotient of a number and the number plus a is three times the number The product of three times a number and the number decreased by b is c Table 1 Translation to Math Operations x, x + a 2x x, x + a x, 2x − a ax − b x _____ x + a = 3x 3x(x − b) = c Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.3 MODELS AND APPLICATIONS 103 How To… Given a real-world problem, model a linear equation to fit it. 1. Identify known quantities. 2. Assign a variable to represent the unknown quantity. 3. If there is more than one unknown quantity, find a way to write the second unknown in terms of the first. 4. Write an equation interpreting the words as mathematical operations. 5. Solve the equation. Be sure the solution can be explained in words, including the units of measure. Example 1 Modeling a Linear Equation to Solve an Unknown Number Problem Find a linear equation to solve for the following unknown quantities: One number exceeds another number by 17 and their sum is 31. Find the two numbers. Solution Let x equal the first number. Then, as the second number exceeds the first by 17, we can write the second number as x + 17. The sum of the two numbers is 31. We usually interpret the word is as an equal sign. x + (x + 17) = 31 2x + 17 = 31 2x = 14 x = 7 x + 17 = 7 + 17 = 24 Simplify and solve. The two numbers are 7 and 24. Try It #1 Find a linear equation to solve for the following unknown quantities: One number is three more than twice another number. If the sum of the two numbers is 36, find the numbers. Example 2 Setting Up a Linear Equation to Solve a Real-World Application There are two cell phone companies that offer different packages. Company A charges a monthly service fee of $34 plus $.05/min talk-time. Company B charges a monthly service fee of $40 plus $.04/min talk-time. a. Write a linear equation that models the packages offered by both companies. b. If the average number of minutes used each month is 1,160, which company offers the better plan? c. If the average number of minutes used each month is 420, which company offers the better plan? d. How many minutes of talk-time would yield equal monthly statements from both companies? Solution a. The model for Company A can be written as A = 0.05x + 34. This includes the variable cost of 0.05x plus the monthly service charge of $34. Company B’s package charges a higher monthly fee of $40, but a lower variable cost of 0.04x. Company B’s model can be written as B = 0.04x + $40. b. If the average number of minutes used each month is 1,160, we have the following: Company A = 0.05(1,160) + 34 = 58 + 34 = 92 Company B = 0.04(1,160) + 40 = 46.4 + 40 = 86.4 So, Company B offers the lower monthly cost of $86.40 as compared with the $92 monthly cost offered by Company A when the average number of minutes used each month is 1,160. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 104 CHAPTER 2 EQUATIONS AND INEQUALITIES c. If the average number of minutes used each month is 420, we have the following: Company A = 0.05(420) + 34 = 21 + 34 = 55 Company B = 0.04(420) + 40 = 16.8 + 40 = 56.8 If the average number of minutes used each month is 420, then Company A offers a lower monthly cost of $55 compared to Company B’s monthly cost of $56.80. d. To answer the question of how many talk-time minutes would yield the same bill from both companies, we should think about the problem in terms of (x, y) coordinates: At what point are both the x-value and the y-value equal? We can find this point by setting the equations equal to each other and solving for x. 0.05x + 34 = 0.04x + 40 0.01x = 6 x = 600 Check the x-value in each equation. 0.05(600) + 34 = 64 0.04(600) + 40 = 64 Therefore, a monthly average of 600 talk-time minutes renders the plans equal. See Figure 2. y 90 80 70 60 50 40 30 0 B = 0.04x + 40 A = 0.05x + 34 100 200 300 400 500 600 700 800 900 1000 1100 1200 x Figure 2 Try It #2 Find a linear equation to model this real-world application: It costs ABC electronics company $2.50 per unit to produce a part used in a popular brand of desktop computers. The company has monthly operating expenses of $350 for utilities and $3,300 for salaries. What are the company’s monthly expenses? Using a Formula to Solve a Real-World Application Many applications are solved using known formulas. The problem is stated, a formula is identified, the known quantities are substituted into the formula, the equation is solved for the unknown, and the problem’s question is answered. Typically, these problems involve two equations representing two trips, two investments, two areas, and so on. Examples of formulas include the area of a rectangular region, A = LW; the perimeter of a rectangle, P = 2L + 2W; and the volume of a rectangular solid, V = LWH. When there are two unknowns, we find a way to write one in terms of the other because we can solve for only one variable at a time. Example 3 Solving an Application Using a Formula It takes Andrew 30 min to drive to work in the morning. He drives home using the same route, but it takes 10 min longer, and he averages 10 mi/h less than in the morning. How far does Andrew drive to work? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.3 MODELS AND APPLICATIONS 105 Solution This is a distance problem, so we can use the formula d = rt, where distance equals rate multiplied by time. Note that when rate is given in mi/h, time must be expressed in hours. Consistent units of measurement are key to obtaining a correct solution. 1 _ First, we identify the known and unknown quantities. Andrew’s morning drive to work takes 30 min, or h at 2 2 _ h, and his speed averages 10 mi/h less than the morning drive. Both trips rate r. His drive home takes 40 min, or 3 cover distance d. A table, such as Table 2, is often helpful for keeping track of information in these types of problems. Write two equations, one for each trip. To Work To Home d d d r r r − 10 Table r − 10) ( _ 3 To work To home As both equations equal the same distance, we set them equal to each other and solve for r. 2 1 ) ) = (r − 10) ( r ( _ _ 2 3 20_ 20_ 3 20_ 3 20 _ 3 r = − (−6) r = 40 We have solved for the rate of speed to work, 40 mph. Substituting 40 into the rate on the return trip yields 30 mi/h. Now we can answer the question. Substitute the rate back into either equation and solve for d. The distance between home and work is 20 mi. 1 ) d = 40 ( _ 2 = 20 Analysis Note that we could have cleared the fractions in the equation by multiplying both sides of the equation by the LCD to solve for r. 2 1 ) ) = (r − 10r − 10 3r = 4(r − 10) 3r = 4r − 40 −r = −40 r = 40 Try It #3 On Saturday morning, it took Jennifer 3.6 h to drive to her mother’s house for the weekend. On Sunday evening, due to heavy traffic, it took Jennifer 4 h to return home. Her speed was 5 mi/h slower on Sunday than on Saturday. What was her speed on Sunday? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 106 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 4 Solving a Perimeter Problem The perimeter of a rectangular outdoor patio is 54 ft. The length is 3 ft greater than the width. What are the dimensions of the patio? So
lution The perimeter formula is standard: P = 2L + 2W. We have two unknown quantities, length and width. However, we can write the length in terms of the width as L = W + 3. Substitute the perimeter value and the expression for length into the formula. It is often helpful to make a sketch and label the sides as in Figure 3. W L = W + 3 Figure 3 Now we can solve for the width and then calculate the length. P = 2L + 2W 54 = 2(W + 3) + 2W 54 = 2W + 6 + 2W 54 = 4W + 6 48 = 4W 12 = W (12 + 3) = L 15 = L The dimensions are L = 15 ft and W = 12 ft. Try It #4 Find the dimensions of a rectangle given that the perimeter is 110 cm and the length is 1 cm more than twice the width. Example 5 Solving an Area Problem The perimeter of a tablet of graph paper is 48 in.2. The length is 6 in. more than the width. Find the area of the graph paper. Solution The standard formula for area is A = LW; however, we will solve the problem using the perimeter formula. The reason we use the perimeter formula is because we know enough information about the perimeter that the formula will allow us to solve for one of the unknowns. As both perimeter and area use length and width as dimensions, they are often used together to solve a problem such as this one. We know that the length is 6 in. more than the width, so we can write length as L = W + 6. Substitute the value of the perimeter and the expression for length into the perimeter formula and find the length. P = 2L + 2W 48 = 2(W + 6) + 2W 48 = 2W + 12 + 2W 48 = 4W + 12 36 = 4W 9 = W (9 + 6) = L 15 = L Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.3 MODELS AND APPLICATIONS 107 Now, we find the area given the dimensions of L = 15 in. and W = 9 in. A = LW A = 15(9) = 135 in.2 The area is 135 in.2. Try It #5 A game room has a perimeter of 70 ft. The length is five more than twice the width. How many ft2 of new carpeting should be ordered? Example 6 Solving a Volume Problem Find the dimensions of a shipping box given that the length is twice the width, the height is 8 inches, and the volume is 1,600 in 3. Solution The formula for the volume of a box is given as V = LWH, the product of length, width, and height. We are given that L = 2W, and H = 8. The volume is 1,600 cubic inches. V = LWH 1,600 = (2W)W(8) 1,600 = 16W2 100 = W2 10 = W The dimensions are L = 20 in., W = 10 in., and H = 8 in. Analysis describing width, we can use only the positive result. Note that the square root of W 2 would result in a positive and a negative value. However, because we are Access these online resources for additional instruction and practice with models and applications of linear equations. • Problem Solving Using Linear Equations (http://openstaxcollege.org/l/lineqprobsolve) • Problem Solving Using Equations (http://openstaxcollege.org/l/equationprsolve) • Finding the Dimensions and Area Given the Perimeter (http://openstaxcollege.org/l/permareasolve) • Find the Distance Between the Cities Using the distance = rate * time formula (http://openstaxcollege.org/l/ratetimesolve) • Linear Equation Application (Write a cost equation) (http://openstaxcollege.org/l/lineqappl) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 108 CHAPTER 2 EQUATIONS AND INEQUALITIES 2.3 SECTION EXERCISES VERBAL 1. To set up a model linear equation to fit real-world applications, what should always be the first step? 2. Use your own words to describe this equation where n is a number: 5(n + 3) = 2n 3. If the total amount of money you had to invest was 4. If a man sawed a 10-ft board into two sections and $2,000 and you deposit x amount in one investment, how can you represent the remaining amount? one section was n ft long, how long would the other section be in terms of n ? 5. If Bill was traveling v mi/h, how would you represent Daemon’s speed if he was traveling 10 mi/h faster? REAL-WORLD APPLICATIONS For the following exercises, use the information to find a linear algebraic equation model to use to answer the question being asked. 7. Beth and Ann are joking that their combined ages equal Sam’s age. If Beth is twice Ann’s age and Sam is 69 yr old, what are Beth and Ann’s ages? 6. Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113? 8. Ben originally filled out 8 more applications than Henry. Then each boy filled out 3 additional applications, bringing the total to 28. How many applications did each boy originally fill out? For the following exercises, use this scenario: Two different telephone carriers offer the following plans that a person is considering. Company A has a monthly fee of $20 and charges of $.05/min for calls. Company B has a monthly fee of $5 and charges $.10/min for calls. 9. Find the model of the total cost of Company A’s plan, 10. Find the model of the total cost of Company B’s plan, using m for the minutes. using m for the minutes. 11. Find out how many minutes of calling would make the two plans equal. 12. If the person makes a monthly average of 200 min of calls, which plan should for the person choose? For the following exercises, use this scenario: A wireless carrier offers the following plans that a person is considering. The Family Plan: $90 monthly fee, unlimited talk and text on up to 5 lines, and data charges of $40 for each device for up to 2 GB of data per device. The Mobile Share Plan: $120 monthly fee for up to 10 devices, unlimited talk and text for all the lines, and data charges of $35 for each device up to a shared total of 10 GB of data. Use P for the number of devices that need data plans as part of their cost. 13. Find the model of the total cost of the Family Plan. 14. Find the model of the total cost of the Mobile Share Plan. 15. Assuming they stay under their data limit, find the number of devices that would make the two plans equal in cost. 16. If a family has 3 smart phones, which plan should they choose? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.3 SECTION EXERCISES 109 For exercises 17 and 18, use this scenario: A retired woman has $50,000 to invest but needs to make $6,000 a year from the interest to meet certain living expenses. One bond investment pays 15% annual interest. The rest of it she wants to put in a CD that pays 7%. 17. If we let x be the amount the woman invests in the 15% bond, how much will she be able to invest in the CD? 18. Set up and solve the equation for how much the woman should invest in each option to sustain a $6,000 annual return. 19. Two planes fly in opposite directions. One travels 450 mi/h and the other 550 mi/h. How long will it take before they are 4,000 mi apart? 20. Ben starts walking along a path at 4 mi/h. One and a half hours after Ben leaves, his sister Amanda begins jogging along the same path at 6 mi/h. How long will it be before Amanda catches up to Ben? 21. Fiora starts riding her bike at 20 mi/h. After a while, she slows down to 12 mi/h, and maintains that speed for the rest of the trip. The whole trip of 70 mi takes her 4.5 h. For what distance did she travel at 20 mi/h? 22. A chemistry teacher needs to mix a 30% salt solution with a 70% salt solution to make 20 qt of a 40% salt solution. How many quarts of each solution should the teacher mix to get the desired result? 23. Paul has $20,000 to invest. His intent is to earn 11% interest on his investment. He can invest part of his money at 8% interest and part at 12% interest. How much does Paul need to invest in each option to make get a total 11% return on his $20,000? For the following exercises, use this scenario: A truck rental agency offers two kinds of plans. Plan A charges $75/wk plus $.10/mi driven. Plan B charges $100/wk plus $.05/mi driven. 24. Write the model equation for the cost of renting a 25. Write the model equation for the cost of renting a truck with plan A. truck with plan B. 26. Find the number of miles that would generate the 27. If Tim knows he has to travel 300 mi, which plan same cost for both plans. should he choose? For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular. 28. A = P(1 + rt) is used to find the principal amount P deposited, earning r% interest, for t years. Use this to find what principal amount P David invested at a 3% rate for 20 yr if A = $8,000. 30. F = ma indicates that force (F) equals mass (m) times acceleration (a). Find the acceleration of a mass of 50 kg if a force of 12 N is exerted on it. mv2 _ R 29. The formula F = relates force (F), velocity (v), mass (m), and resistance (R). Find R when m = 45, v = 7, and F = 245. 31. Sum = is the formula for an infinite series 1 _ 1 − r sum. If the sum is 5, find r. For the following exercises, solve for the given variable in the formula. After obtaining a new version of the formula, you will use it to solve a question. 32. Solve for W: P = 2L + 2W 1 1_ 1 _ q = _ p + 34. Solve for f: f 36. Solve for m in the slope-intercept formula: y = mx + b 33. Use the formula from the previous question to find the width, W, of a rectangle whose length is 15 and whose perimeter is 58. 35. Use the formula from the previous question to find f when p = 8 and q = 13. 37. Use the formula from the previous question to find m when the coordinates of the point are (4, 7) and b = 12. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 110 CHAPTER 2 EQUATIONS AND INEQUALITIES 1 _ h(b1 + b2). 38. The area of a trapezoid is given by A = 2 Use the formula to find the area of a trapezoid with h = 6, b1 = 14, and b2 = 8. 1 _ h(b1 + b2) 39. Solve for h: A = 2 40. Use the formula from the previous question to find the height of a trapezoid with A = 150, b1 = 19, and b2 = 11. 41. Find the dimensions of an American football f
ield. The length is 200 ft more than the width, and the perimeter is 1,040 ft. Find the length and width. Use the perimeter formula P = 2L + 2W. 42. Distance equals rate times time, d = rt. Find the distance Tom travels if he is moving at a rate of 55 mi/h for 3.5 h. 43. Using the formula in the previous exercise, find the distance that Susan travels if she is moving at a rate of 60 mi/h for 6.75 h. 44. What is the total distance that two people travel in 3 h if one of them is riding a bike at 15 mi/h and the other is walking at 3 mi/h? 1 _ 45. If the area model for a triangle is A = bh, find the 2 area of a triangle with a height of 16 in. and a base of 11 in. 1 _ 46. Solve for h: A = bh 2 47. Use the formula from the previous question to find the height to the nearest tenth of a triangle with a base of 15 and an area of 215. 48. The volume formula for a cylinder is V = πr2 h. 49. Solve for h: V = πr2h Using the symbol π in your answer, find the volume of a cylinder with a radius, r, of 4 cm and a height of 14 cm. 50. Use the formula from the previous question to find the height of a cylinder with a radius of 8 and a volume of 16π 52. Use the formula from the previous question to find the radius of a cylinder with a height of 36 and a volume of 324π. 54. Solve the formula from the previous question for π. Notice why π is sometimes defined as the ratio of the circumference to its diameter. 51. Solve for r: V = πr2h 53. The formula for the circumference of a circle is C = 2πr. Find the circumference of a circle with a diameter of 12 in. (diameter = 2r). Use the symbol π in your final answer. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.4 COMPLEX NUMBERS 111 LEARNING OBJECTIVES In this section you will: • Add and subtract complex numbers. • Multiply complex numbers. 2.4 COMPLEX NUMBERS Figure 1 Discovered by Benoit Mandelbrot around 1980, the Mandelbrot Set is one of the most recognizable fractal images. The image is built on the theory of self-similarity and the operation of iteration. Zooming in on a fractal image brings many surprises, particularly in the high level of repetition of detail that appears as magnification increases. The equation that generates this image turns out to be rather simple. In order to better understand it, we need to become familiar with a new set of numbers. Keep in mind that the study of mathematics continuously builds upon itself. Negative integers, for example, fill a void left by the set of positive integers. The set of rational numbers, in turn, fills a void left by the set of integers. The set of real numbers fills a void left by the set of rational numbers. Not surprisingly, the set of real numbers has voids as well. In this section, we will explore a set of numbers that fills voids in the set of real numbers and find out how to work within it. Expressing Square Roots of Negative Numbers as Multiples of i We know how to find the square root of any positive real number. In a similar way, we can find the square root of any negative number. The difference is that the root is not real. If the value in the radicand is negative, the root is said to be an imaginary number. The imaginary number i is defined as the square root of −1. So, using properties of radicals, — −1 = i √ i2 = ( √ — 2 −1 ) = −1 We can write the square root of any negative number as a multiple of i. Consider the square root of −49. We use 7i and not −7i because the principal root of 49 is the positive root. √ — −49 = √ = √ = 7i — 49 ċ (−1) 49 √ −1 — — A complex number is the sum of a real number and an imaginary number. A complex number is expressed in standard form when written a + bi where a is the real part and b is the imaginary part. For example, 5 + 2i is a complex number. So, too, is 3 + 4i √ 3 . — 5 + 2i Imaginary numbers differ from real numbers in that a squared imaginary number produces a negative real number. Recall that when a positive real number is squared, the result is a positive real number and when a negative real number is squared, the result is also a positive real number. Complex numbers consist of real and imaginary numbers. Real part Imaginary part Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 112 CHAPTER 2 EQUATIONS AND INEQUALITIES imaginary and complex numbers A complex number is a number of the form a + bi where • a is the real part of the complex number. • b is the imaginary part of the complex number. If b = 0, then a + bi is a real number. If a = 0 and b is not equal to 0, the complex number is called a pure imaginary number. An imaginary number is an even root of a negative number. — How To… Given an imaginary number, express it in the standard form of a complex number. 1. Write √ 2. Express √ 3. Write √ −a as √ −1 as i. a ċ i in simplest form. a √ −1 . — — — — Example 1 Expressing an Imaginary Number in Standard Form — −9 in standard form. Express √ Solution In standard form, this is 0 + 3i. Try It #1 Express √ — −24 in standard form. √ — −9 = √ = 3i — 9 √ — −1 Plotting a Complex Number on the Complex Plane We cannot plot complex numbers on a number line as we might real numbers. However, we can still represent them graphically. To represent a complex number, we need to address the two components of the number. We use the complex plane, which is a coordinate system in which the horizontal axis represents the real component and the vertical axis represents the imaginary component. Complex numbers are the points on the plane, expressed as ordered pairs (a, b), where a represents the coordinate for the horizontal axis and b represents the coordinate for the vertical axis. Let’s consider the number −2 + 3i. The real part of the complex number is −2 and the imaginary part is 3. We plot the ordered pair (−2, 3) to represent the complex number −2 + 3i, as shown in Figure 2. −2 + 3i –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 2 complex plane In the complex plane, the horizontal axis is the real axis, and the vertical axis is the imaginary axis, as shown in Figure 3. imaginary real Figure 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.4 COMPLEX NUMBERS 113 How To… Given a complex number, represent its components on the complex plane. 1. Determine the real part and the imaginary part of the complex number. 2. Move along the horizontal axis to show the real part of the number. 3. Move parallel to the vertical axis to show the imaginary part of the number. 4. Plot the point. Example 2 Plotting a Complex Number on the Complex Plane Plot the complex number 3 − 4i on the complex plane. Solution The real part of the complex number is 3, and the imaginary part is −4. We plot the ordered pair (3, −4) as shown in Figure 4. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 4 Try It #2 Plot the complex number −4 − i on the complex plane. Adding and Subtracting Complex Numbers Just as with real numbers, we can perform arithmetic operations on complex numbers. To add or subtract complex numbers, we combine the real parts and then combine the imaginary parts. complex numbers: addition and subtraction Adding complex numbers: Subtracting complex numbers: (a + bi) + (c + di) = (a + c) + (b + d)i (a + bi) − (c + di) = (a − c) + (b − d)i How To… Given two complex numbers, find the sum or difference. 1. Identify the real and imaginary parts of each number. 2. Add or subtract the real parts. 3. Add or subtract the imaginary parts. Example 3 Adding and Subtracting Complex Numbers Add or subtract as indicated. a. (3 − 4i) + (2 + 5i) b. (−5 + 7i) − (−11 + 2i) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 114 CHAPTER 2 EQUATIONS AND INEQUALITIES Solution We add the real parts and add the imaginary parts. a. (3 − 4i) + (2 + 5i) = 3 − 4i + 2 + 5i = 3 + 2 + (−4i) + 5i = (3 + 2) + (−4 + 5)i = 5 + i b. (−5 + 7i) − (−11 + 2i) = −5 + 7i + 11 − 2i = −5 + 11 + 7i − 2i = (−5 + 11) + (7 − 2)i = 6 + 5i Try It #3 Subtract 2 + 5i from 3 − 4i. Multiplying Complex Numbers Multiplying complex numbers is much like multiplying binomials. The major difference is that we work with the real and imaginary parts separately. Multiplying a Complex Number by a Real Number Lets begin by multiplying a complex number by a real number. We distribute the real number just as we would with a binomial. Consider, for example, 3(6 + 2i): 3(6 + 2i) = (3 · 6) + (3 · 2i) = 18 + 6i Distribute. Simplify. How To… Given a complex number and a real number, multiply to find the product. 1. Use the distributive property. 2. Simplify. Example 4 Multiplying a Complex Number by a Real Number Find the product 4(2 + 5i). Solution Distribute the 4. 4(2 + 5i) = (4 ċ 2) + (4 ċ 5i) = 8 + 20i Try It #4 1 _ (5 − 2i). Find the product: 2 Multiplying Complex Numbers Together Now, let’s multiply two complex numbers. We can use either the distributive property or more specifically the FOIL method because we are dealing with binomials. Recall that FOIL is an acronym for multiplying First, Inner, Outer, and Last terms together. The difference with complex numbers is that when we get a squared term, i2, it equals −1. (a + bi)(c + di) = ac + adi + bci + bdi2 = ac + adi + bci − bd = (ac − bd) + (ad + bc)i i2 = −1 Group real terms and imaginary terms. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.4 COMPLEX NUMBERS 115 How To… Given two complex numbers, multiply to find the product. 1. Use the distributive property or the FOIL method. 2. Remember that i2 = −1. 3. Group together the real terms and the imaginary terms Example 5 Multiplying a Complex Number by a Complex Number (4 + 3i)(2 − 5i) = 4(2) − 4(5i) + 3i(2) − (3i)(5i) = 8 − 20i + 6i − 15(i2) = (8 + 15) + (−20 + 6)i = 23 − 14i Multiply: (4 + 3i)(2 − 5i). Solution Try It #5 Multiply: (3 − 4i)(2 + 3i). Dividing Complex Numbers Dividing two complex numbers is more complicated than adding, subtracting, or multip
lying because we cannot divide by an imaginary number, meaning that any fraction must have a real-number denominator to write the answer in standard form a + bi. We need to find a term by which we can multiply the numerator and the denominator that will eliminate the imaginary portion of the denominator so that we end up with a real number as the denominator. This term is called the complex conjugate of the denominator, which is found by changing the sign of the imaginary part of the complex number. In other words, the complex conjugate of a + bi is a − bi. For example, the product of a + bi and a − bi is (a + bi)(a − bi) = a2 − abi + abi − b2i2 = a2 + b2 The result is a real number. Note that complex conjugates have an opposite relationship: The complex conjugate of a + bi is a − bi, and the complex conjugate of a − bi is a + bi. Further, when a quadratic equation with real coefficients has complex solutions, the solutions are always complex conjugates of one another. Suppose we want to divide c + di by a + bi, where neither a nor b equals zero. We first write the division as a fraction, then find the complex conjugate of the denominator, and multiply. Multiply the numerator and denominator by the complex conjugate of the denominator. c + di ______ a + bi where a ≠ 0 and b ≠ 0 (c + di) _______ ċ (a + bi) (a − bi) _______ = (a − bi) (c + di)(a − bi) _____________ (a + bi)(a − bi) Apply the distributive property. Simplify, remembering that i2 = −1. = ca − cbi + adi − bdi2 __ a2 − abi + abi − b2 i2 = ca − cbi + adi − bd(−1) ___ a2 − abi + abi − b2(−1) = (ca + bd) + (ad − cb)i___ a2 + b2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 116 CHAPTER 2 EQUATIONS AND INEQUALITIES the complex conjugate The complex conjugate of a complex number a + bi is a − bi. It is found by changing the sign of the imaginary part of the complex number. The real part of the number is left unchanged. • When a complex number is multiplied by its complex conjugate, the result is a real number. • When a complex number is added to its complex conjugate, the result is a real number. Example 6 Finding Complex Conjugates Find the complex conjugate of each number. a. 2 + i √ — 5 Solution 1 _ b. − i 2 a. The number is already in the form a + bi. The complex conjugate is a − bi, or 2 − i √ 1 1 _ _ i. The complex conjugate is a − bi, or 0 + b. We can rewrite this number in the form a + bi as 0 − i. 2 2 — 5 . 1 _ i. This can be written simply as 2 Analysis Although we have seen that we can find the complex conjugate of an imaginary number, in practice we generally find the complex conjugates of only complex numbers with both a real and an imaginary component. To obtain a real number from an imaginary number, we can simply multiply by i. Try It #6 Find the complex conjugate of −3 + 4i. How To… Given two complex numbers, divide one by the other. 1. Write the division problem as a fraction. 2. Determine the complex conjugate of the denominator. 3. Multiply the numerator and denominator of the fraction by the complex conjugate of the denominator. 4. Simplify. Example 7 Dividing Complex Numbers Divide: (2 + 5i) by (4 − i). Solution We begin by writing the problem as a fraction. (2 + 5i) _ (4 − i) Then we multiply the numerator and denominator by the complex conjugate of the denominator. (2 + 5i) _ ċ (4 − i) (4 + i) _ (4 + i) To multiply two complex numbers, we expand the product as we would with polynomials (using FOIL). (4 + i) _ (4 + i) (2 + 5i) _ ċ (4 − i) = 8 + 2i + 20i + 5i2 __ 16 + 4i − 4i − i2 8 + 2i + 20i + 5(−1) __________________ 16 + 4i − 4i − (−1) 3 + 22i _______ 17 = = Because i2 = −1. Note that this expresses the quotient in standard form. = + 3 ___ 17 22 ___ i 17 Separate real and imaginary parts. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.4 COMPLEX NUMBERS 117 Simplifying Powers of i The powers of i are cyclic. Let’s look at what happens when we raise i to increasing powers. i1 = i i2 = −1 i3 = i2 ċ i = −1 ċ i = −i i4 = i3 ċ i = −i ċ i = −i2 = − (−1) = 1 i5 = i4 ċ i = 1 ċ i = i We can see that when we get to the fifth power of i, it is equal to the first power. As we continue to multiply i by increasing powers, we will see a cycle of four. Let’s examine the next four powers of i. i6 = i5 ċ i = i ċ i = i2 = −1 i7 = i6 ċ i = i2 ċ i = i3 = −i i8 = i7 ċ i = i3 ċ i = i4 = 1 i9 = i8 ċ i = i4 ċ i = i5 = i The cycle is repeated continuously: i, −1, − i, 1, every four powers. Example 8 Simplifying Powers of i Evaluate: i35. Solution Since i4 = 1, we can simplify the problem by factoring out as many factors of i4 as possible. To do so, first determine how many times 4 goes into 35: 35 = 4 ċ 8 + 3. i35 = i4 ċ 8 + 3 = i4 ċ 8 ċ i3 = (i4) ċ i3 = 18 ċ i3 = i3 = −i 8 Try It #7 Evaluate: i 18 Q & A… Can we write i35 in other helpful ways? As we saw in Example 8, we reduced i35 to i3 by dividing the exponent by 4 and using the remainder to find the simplified form. But perhaps another factorization of i35 may be more useful. Table 1 shows some other possible factorizations. Factorization of i35 i34 ċ i Reduced form Simplified form 17 (i2) ċ i (−1)17 ċ i Table 1 i33 ċ i2 i33 ċ (−1) −i33 i31 ċ i4 i31 ċ 1 i31 i19 ċ i16 4 i19 ċ (i4) i19 Each of these will eventually result in the answer we obtained above but may require several more steps than our earlier method. Access these online resources for additional instruction and practice with complex numbers. • Adding and Subtracting Complex Numbers (http://openstaxcollege.org/l/addsubcomplex) • Multiply Complex Numbers (http://openstaxcollege.org/l/multiplycomplex) • Multiplying Complex Conjugates (http://openstaxcollege.org/l/multcompconj) • Raising i to Powers (http://openstaxcollege.org/l/raisingi) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 118 CHAPTER 2 EQUATIONS AND INEQUALITIES 2.4 SECTION EXERCISES VERBAL 1. Explain how to add complex numbers. 2. What is the basic principle in multiplication of complex numbers? 3. Give an example to show that the product of two imaginary numbers is not always imaginary. 4. What is a characteristic of the plot of a real number in the complex plane? ALGEBRAIC For the following exercises, evaluate the algebraic expressions. 5. If y = x2 + x − 4, evaluate y given x = 2i. 7. If y = x2 + 3x + 5, evaluate y given x = 2 + i. 6. If y = x3 − 2, evaluate y given x = i. 8. If y = 2x2 + x − 3, evaluate y given x = 2 − 3i. 9. If y = , evaluate y given x = 5i. x + 1 _ 2 − x 10. If y = , evaluate y given x = 4i. 1 + 2x _ x + 3 GRAPHICAL For the following exercises, plot the complex numbers on the complex plane. 11. 1 − 2i 12. −2 + 3i 13. i 14. −3 − 4i NUMERIC For the following exercises, perform the indicated operation and express the result as a simplified complex number. 15. (3 + 2i) + (5 − 3i) 16. (−2 − 4i) + (1 + 6i) 17. (−5 + 3i) − (6 − i) 18. (2 − 3i) − (3 + 2i) 19. (−4 + 4i) − (−6 + 9i) 20. (2 + 3i)(4i) 21. (5 − 2i)(3i) 22. (6 − 2i)(5) 23. (−2 + 4i)(8) 24. (2 + 3i)(4 − i) 25. (−1 + 2i)(−2 + 3i) 26. (4 − 2i)(4 + 2i) 27. (3 + 4i)(3 − 4i) 28. 31. 6 + 4i_ i 3 + 4i_ 2 2 − 3i _ 4 + 3i 32. 35. √ — −9 + 3 √ — −16 36. − √ — −4 − 4 √ — −25 39. i8 40. i15 29. 6 − 2i_ 3 3 + 4i_ 2 − i 33. 37. — 2 + √ −12 __________ 2 41. i22 30. 34. −5 + 3i_ 2i 2 + 3i _ 2 − 3i 38. — 4 + √ −20 __________ 2 TECHNOLOGY For the following exercises, use a calculator to help answer the questions. 42. Evaluate (1 + i)k for k = 4, 8, and 12. Predict the 43. Evaluate (1 − i)k for k = 2, 6, and 10. Predict the value if k = 16. value if k = 14. 44. Evaluate (l + i)k − (l − i)k for k = 4, 8, and 12. Predict the value for k = 16. 45. Show that a solution of x6 + 1 = 0 is — √ 3 ____ + 2 1_ i. 2 46. Show that a solution of x8 −1 = 0 is — √ 2 ____ + 2 — √ 2 ____ i. 2 EXTENSIONS For the following exercises, evaluate the expressions, writing the result as a simplified complex number. 47. 51. 55. 4 1 __ __ + i3 i (2 + i)(4 − 2i) ____________ (1 + i) 4 + i _____ i + 3 − 4i ______ 1 − i 48. 52. 56. 1 __ − i21 1 __ i11 (1 + 3i)(2 − 4i) _____________ (1 + 2i) 49. i7(1 + i2) 53. (3 + i)2 _______ (1 + 2i)2 50. i−3 + 5i7 54. 3 + 2i ______ 2 + i + (4 + 3i) 3 + 2i ______ 1 + 2i − 2 − 3i ______ 3 + i Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 119 LEARNING OBJECTIVES In this section you will: • Solve quadratic equations by factoring. • Solve quadratic equations by the square root property. • Solve quadratic equations by using the quadratic formula. 2.5 QUADRATIC EQUATIONS Figure 1 The computer monitor on the left in Figure 1 is a 23.6-inch model and the one on the right is a 27-inch model. Proportionally, the monitors appear very similar. If there is a limited amount of space and we desire the largest monitor possible, how do we decide which one to choose? In this section, we will learn how to solve problems such as this using four different methods. Solving Quadratic Equations by Factoring An equation containing a second-degree polynomial is called a quadratic equation. For example, equations such as 2x2 + 3x − 1 = 0 and x2 − 4 = 0 are quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics. Often the easiest method of solving a quadratic equation is factoring. Factoring means finding expressions that can be multiplied together to give the expression on one side of the equation. If a quadratic equation can be factored, it is written as a product of linear terms. Solving by factoring depends on the zero-product property, which states that if a ċ b = 0, then a = 0 or b = 0, where a and b are real numbers or algebraic expressions. In other words, if the product of two numbers or two expressions equals zero, then one of the numbers or one of the expressions must equal zero because zero multiplied by anything equals zero. Multiplying the factors expands the equation to
a string of terms separated by plus or minus signs. So, in that sense, the operation of multiplication undoes the operation of factoring. For example, expand the factored expression (x − 2)(x + 3) by multiplying the two factors together. (x − 2)(x + 3) = x2 + 3x − 2x − 6 = x2 + x − 6 The product is a quadratic expression. Set equal to zero, x2 + x − 6 = 0 is a quadratic equation. If we were to factor the equation, we would get back the factors we multiplied. The process of factoring a quadratic equation depends on the leading coefficient, whether it is 1 or another integer. We will look at both situations; but first, we want to confirm that the equation is written in standard form, ax2 + bx + c = 0, where a, b, and c are real numbers, and a ≠ 0. The equation x2 + x − 6 = 0 is in standard form. We can use the zero-product property to solve quadratic equations in which we first have to factor out the greatest common factor (GCF ), and for equations that have special factoring formulas as well, such as the difference of squares, both of which we will see later in this section. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 120 CHAPTER 2 EQUATIONS AND INEQUALITIES the zero-product property and quadratic equations The zero-product property states If a ċ b = 0, then a = 0 or b = 0, where a and b are real numbers or algebraic expressions. A quadratic equation is an equation containing a second-degree polynomial; for example where a, b, and c are real numbers, and if a ≠ 0, it is in standard form. ax 2 + bx + c = 0 Solving Quadratics with a Leading Coefﬁcient of 1 In the quadratic equation x2 + x − 6 = 0, the leading coefficient, or the coefficient of x2, is 1. We have one method of factoring quadratic equations in this form. How To… Given a quadratic equation with the leading coefficient of 1, factor it. 1. Find two numbers whose product equals c and whose sum equals b. 2. Use those numbers to write two factors of the form (x + k) or (x − k), where k is one of the numbers found in step 1. Use the numbers exactly as they are. In other words, if the two numbers are 1 and −2, the factors are (x + 1)(x − 2). 3. Solve using the zero-product property by setting each factor equal to zero and solving for the variable. Example 1 Factoring and Solving a Quadratic with Leading Coefﬁcient of 1 Factor and solve the equation: x 2 + x − 6 = 0. Solution To factor x 2 + x − 6 = 0, we look for two numbers whose product equals −6 and whose sum equals 1. Begin by looking at the possible factors of −6. 1 ċ (−6) (−6) ċ 1 2 ċ (−3) 3 ċ (−2) The last pair, 3 ċ (−2) sums to 1, so these are the numbers. Note that only one pair of numbers will work. Then, write the factors. (x − 2)(x + 3) = 0 To solve this equation, we use the zero-product property. Set each factor equal to zero and solve. (x − 2)(x + 3) = 0 (x − 2) = 0 x = 2 (x + 3) = 0 x = −3 The two solutions are 2 and −3. We can see how the solutions relate to the graph in Figure 2. The solutions are the x-intercepts of x 2 + x − 6 = 0. x2 + x − 6 = 0 y (−3, 0) –4 –5 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (2, 0) 21 3 4 5 x Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 121 Try It #1 Factor and solve the quadratic equation: x 2 − 5x − 6 = 0. Example 2 Solve the Quadratic Equation by Factoring Solve the quadratic equation by factoring: x2 + 8x + 15 = 0. Solution Find two numbers whose product equals 15 and whose sum equals 8. List the factors of 15. 1 ċ 15 3 ċ 5 (−1) ċ (−15) (−3) ċ (−5) The numbers that add to 8 are 3 and 5. Then, write the factors, set each factor equal to zero, and solve. (x + 3)(x + 5) = 0 (x + 3) = 0 x = −3 (x + 5) = 0 x = −5 The solutions are −3 and −5. Try It #2 Solve the quadratic equation by factoring: x 2 − 4x − 21 = 0. Example 3 Using the Zero-Product Property to Solve a Quadratic Equation Written as the Difference of Squares Solve the difference of squares equation using the zero-product property: x 2 − 9 = 0. Solution Recognizing that the equation represents the difference of squares, we can write the two factors by taking the square root of each term, using a minus sign as the operator in one factor and a plus sign as the operator in the other. Solve using the zero-factor property. x 2 − 9 = 0 (x − 3)(x + 3) = 0 (x − 3) = 0 x = 3 (x + 3) = 0 x = −3 The solutions are 3 and −3. Try It #3 Solve by factoring: x 2 − 25 = 0. Factoring and Solving a Quadratic Equation of Higher Order When the leading coefficient is not 1, we factor a quadratic equation using the method called grouping, which requires four terms. With the equation in standard form, let’s review the grouping procedures: 1. With the quadratic in standard form, ax2 + bx + c = 0, multiply a ċ c. 2. Find two numbers whose product equals ac and whose sum equals b. 3. Rewrite the equation replacing the bx term with two terms using the numbers found in step 1 as coefficients of x. 4. Factor the first two terms and then factor the last two terms. The expressions in parentheses must be exactly the same to use grouping. 5. Factor out the expression in parentheses. 6. Set the expressions equal to zero and solve for the variable. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 122 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 4 Solving a Quadratic Equation Using Grouping Use grouping to factor and solve the quadratic equation: 4x2 + 15x + 9 = 0. Solution First, multiply ac : 4(9) = 36. Then list the factors of 36. 1 ċ 36 2 ċ 18 3 ċ 12 4 ċ 9 6 ċ 6 The only pair of factors that sums to 15 is 3 + 12. Rewrite the equation replacing the b term, 15x, with two terms using 3 and 12 as coefficients of x. Factor the first two terms, and then factor the last two terms. 4x 2 + 3x + 12x + 9 = 0 x(4x + 3) + 3(4x + 3) = 0 (4x + 3)(x + 3) = 0 Solve using the zero-product property. (4x + 3)(x + 3) = 0 (4x + 3x + 3) = 0 x = −3 3 _ and −3. See Figure 3. The solutions are − 4 2 + 15x + 9 = 0 4x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 3 Try It #4 Solve using factoring by grouping: 12x 2 + 11x + 2 = 0. Example 5 Solving a Higher Degree Quadratic Equation by Factoring Solve the equation by factoring: −3x 3 − 5x 2 − 2x = 0. Solution This equation does not look like a quadratic, as the highest power is 3, not 2. Recall that the first thing we want to do when solving any equation is to factor out the GCF, if one exists. And it does here. We can factor out −x from all of the terms and then proceed with grouping. Use grouping on the expression in parentheses. −3x 3 − 5x 2 − 2x = 0 −x(3x 2 + 5x + 2) = 0 −x(3x 2 + 3x + 2x + 2) = 0 −x[3x(x + 1) + 2(x + 1)] = 0 −x(3x + 2)(x + 1) = 0 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 123 Now, we use the zero-product property. Notice that we have three factors. −x = 0 x = 0 3x + 1 2 _ , and −1. The solutions are 0, − 3 Try It #5 Solve by factoring: x 3 + 11x 2 + 10x = 0. Using the Square Root Property When there is no linear term in the equation, another method of solving a quadratic equation is by using the square root property, in which we isolate the x 2 term and take the square root of the number on the other side of the equals sign. Keep in mind that sometimes we may have to manipulate the equation to isolate the x2 term so that the square root property can be used. the square root property With the x 2 term isolated, the square root property states that: if x 2 = k, then x = ± √ — k where k is a nonzero real number. How To… Given a quadratic equation with an x 2 term but no x term, use the square root property to solve it. 1. Isolate the x 2 term on one side of the equal sign. 2. Take the square root of both sides of the equation, putting a ± sign before the expression on the side opposite the squared term. 3. Simplify the numbers on the side with the ± sign. Example 6 Solving a Simple Quadratic Equation Using the Square Root Property Solve the quadratic using the square root property: x 2 = 8. Solution Take the square root of both sides, and then simplify the radical. Remember to use a ± sign before the radical symbol2 √ — 8 — 2 The solutions are 2 √ — 2 and −2 √ — 2 . Solving a Quadratic Equation Using the Square Root Property Example 7 Solve the quadratic equation: 4x 2 + 1 = 7. Solution First, isolate the x 2 term. Then take the square root of both sides. 4x 2 + 1 = 7 4x The solutions are and − — Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 124 CHAPTER 2 EQUATIONS AND INEQUALITIES Try It #6 Solve the quadratic equation using the square root property: 3(x − 4)2 = 15. Completing the Square Not all quadratic equations can be factored or can be solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square. We will use the example x2 + 4x + 1 = 0 to illustrate each step. 1. Given a quadratic equation that cannot be factored, and with a = 1, first add or subtract the constant term to the right sign of the equal sign. 1 _ and square it. 2. Multiply the b term by 2 x 2 + 4x = −1 1 _ (4) = 2 2 22 = 4 1 b ) 3. Add ( _ 2 2 to both sides of the equal sign and simplify the right side. We have x 2 + 4x + 4 = −1 + 4 x 2 + 4x + 4 = 3 4. The left side of the equation can now be factored as a perfect square. 5. Use the square root property and solve. x 2 + 4x + 4 = 3 (x + 2)2 = 3 √ — (x + 2)2 ± √ — 3 6. The solutions are −2 + √ — 3 and −2 − √ — 3 . Example 8 Solving a Quadr
atic by Completing the Square Solve the quadratic equation by completing the square: x 2 − 3x − 5 = 0. Solution First, move the constant term to the right side of the equal sign. x 2 − 3x = 5 1 _ of the b term and square it. Then, take 2 Add the result to both sides of the equal sign. 3 1 _ _ (−3 − 3x + ( − ) = − 3x + 4 4 2 Factor the left side as a perfect square and simplify the right side. 2 3 ) ( x − _ 2 = 29_ 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 125 Use the square root property and solve. The solutions are — 29 3 + √ _ 2 and — 29 _ 3 − √ . 2 2 ___ _________ = ± √ √ 29 3 ) ( x − _ _ 4 2 — 29 √ √ 29 _ 3 _ ± x = 2 2 — Try It #7 Solve by completing the square: x 2 − 6x = 13. Using the Quadratic Formula The fourth method of solving a quadratic equation is by using the quadratic formula, a formula that will solve all quadratic equations. Although the quadratic formula works on any quadratic equation in standard form, it is easy to make errors in substituting the values into the formula. Pay close attention when substituting, and use parentheses when inserting a negative number. We can derive the quadratic formula by completing the square. We will assume that the leading coefficient is positive; if it is negative, we can multiply the equation by −1 and obtain a positive a. Given ax 2 + bx + c = 0, a ≠ 0, we will complete the square as follows: 1. First, move the constant term to the right side of the equal sign: ax 2 + bx = −c 2. As we want the leading coefficient to equal 1, divide through by a: b 1 1 ) of the middle term, and add ( __ __ _ 3. Then, find 2 2 a b _ a x + x2 + b2 _ 4a2 = b2 c _ _ 4a2 − b2 _ 4a2 to both sides of the equal sign: = 2 4. Next, write the left side as a perfect square. Find the common denominator of the right side and write it as a single fraction: 2 ) ( x + b _ 2a = b2 − 4ac _ 4a2 5. Now, use the square root property, which gives x + b _ 2a = ± √ x + = ± b _ 2a ________ b2 − 4ac _ 4a2 b2 − 4ac __ √ 2a — 6. Finally, add − to both sides of the equation and combine the terms on the right side. Thus, b _ 2a x = — −b ± √ b2 − 4ac __ 2a the quadratic formula Written in standard form, ax2 + bx + c = 0, any quadratic equation can be solved using the quadratic formula: where a, b, and c are real numbers and a ≠ 0. x = — −b ± √ b2 − 4ac __ 2a Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 126 CHAPTER 2 EQUATIONS AND INEQUALITIES How To… Given a quadratic equation, solve it using the quadratic formula 1. Make sure the equation is in standard form: ax 2 + bx + c = 0. 2. Make note of the values of the coefficients and constant term, a, b, and c. 3. Carefully substitute the values noted in step 2 into the equation. To avoid needless errors, use parentheses around each number input into the formula. 4. Calculate and solve. Example 9 Solve the Quadratic Equation Using the Quadratic Formula Solve the quadratic equation: x 2 + 5x + 1 = 0. Solution Identify the coefficients: a = 1, b = 5, c = 1. Then use the quadratic formula. x = — −(5) ± √ (5)2 − 4(1)(1) ___ 2(1) — = −5 ± √ 25 − 4 __ 2 — = 21 __ −5 ± √ 2 Example 10 Solving a Quadratic Equation with the Quadratic Formula Use the quadratic formula to solve x 2 + x + 2 = 0. Solution First, we identify the coefficients: a = 1, b = 1, and c = 2. Substitute these values into the quadratic formula. x = — b2 − 4ac −b ± √ __ 2a −(1) ± √ — (1)2 − (4) ċ (1) ċ (2) = ___ 2 ċ 1 — = 1 − 8 −1 ± √ __ 2 — = −7 __ −1 ± √ 2 — = −1 ± i √ 7 _ 2 The solutions to the equation are — 7 −1 + i √ _ 2 and −1 − i √ _ 2 7 . — Try It #8 Solve the quadratic equation using the quadratic formula: 9x 2 + 3x − 2 = 0. The Discriminant The quadratic formula not only generates the solutions to a quadratic equation, it tells us about the nature of the solutions when we consider the discriminant, or the expression under the radical, b2 − 4ac. The discriminant tells us whether the solutions are real numbers or complex numbers, and how many solutions of each type to expect. Table 1 relates the value of the discriminant to the solutions of a quadratic equation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 QUADRATIC EQUATIONS 127 Value of Discriminant Results b2 − 4ac = 0 One rational solution (double solution) b2 − 4ac > 0, perfect square Two rational solutions b2 − 4ac > 0, not a perfect square Two irrational solutions b2 − 4ac < 0 Two complex solutions Table 1 the discriminant For ax2 + bx + c = 0, where a, b, and c are real numbers, the discriminant is the expression under the radical in the quadratic formula: b2 − 4ac. It tells us whether the solutions are real numbers or complex numbers and how many solutions of each type to expect. Example 11 Using the Discriminant to Find the Nature of the Solutions to a Quadratic Equation Use the discriminant to find the nature of the solutions to the following quadratic equations: a. x2 + 4x + 4 = 0 b. 8x2 + 14x + 3 = 0 c. 3x2 − 5x − 2 = 0 d. 3x2 − 10x + 15 = 0 Solution Calculate the discriminant b2 − 4ac for each equation and state the expected type of solutions. a. x2 + 4x + 4 = 0 b2 − 4ac = (4)2 − 4(1)(4) = 0. There will be one rational double solution. b. 8x2 + 14x + 3 = 0 b2 − 4ac = (14)2 − 4(8)(3) = 100. As 100 is a perfect square, there will be two rational solutions. c. 3x2 − 5x − 2 = 0 b2 − 4ac = (−5)2 − 4(3)(−2) = 49. As 49 is a perfect square, there will be two rational solutions. d. 3x2 −10x + 15 = 0 b2 − 4ac = (−10)2 − 4(3)(15) = −80. There will be two complex solutions. Using the Pythagorean Theorem One of the most famous formulas in mathematics is the Pythagorean Theorem. It is based on a right triangle, and states the relationship among the lengths of the sides as a2 + b2 = c2, where a and b refer to the legs of a right triangle adjacent to the 90° angle, and c refers to the hypotenuse. It has immeasurable uses in architecture, engineering, the sciences, geometry, trigonometry, and algebra, and in everyday applications. We use the Pythagorean Theorem to solve for the length of one side of a triangle when we have the lengths of the other two. Because each of the terms is squared in the theorem, when we are solving for a side of a triangle, we have a quadratic equation. We can use the methods for solving quadratic equations that we learned in this section to solve for the missing side. The Pythagorean Theorem is given as a2 + b2 = c2 where a and b refer to the legs of a right triangle adjacent to the 90° angle, and c refers to the hypotenuse, as shown in Figure 4. b c a Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 128 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 12 Finding the Length of the Missing Side of a Right Triangle Find the length of the missing side of the right triangle in Figure 5. 12 4 a Figure 5 Solution As we have measurements for side b and the hypotenuse, the missing side is a. a2 + b2 = c2 a2 + (4)2 = (12)2 a2 + 16 = 144 a2 = 128 a = √ = 8 √ — 128 2 — Try It #9 Use the Pythagorean Theorem to solve the right triangle problem: Leg a measures 4 units, leg b measures 3 units. Find the length of the hypotenuse. Access these online resources for additional instruction and practice with quadratic equations. • Solving Quadratic Equations by Factoring (http://openstaxcollege.org/l/quadreqfactor) • The Zero-Product Property (http://openstaxcollege.org/l/zeroprodprop) • Completing the Square (http://openstaxcollege.org/l/complthesqr) • Quadratic Formula with Two Rational Solutions (http://openstaxcollege.org/l/quadrformrat) • Length of a Leg of a Right Triangle (http://openstaxcollege.org/l/leglengthtri) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.5 SECTION EXERCISES 129 2.5 SECTION EXERCISES VERBAL 1. 2. 3. 5. ax+bx +c = y =ax+bx +cx 4. b−ac ALGEBRAIC 6. x +x−= 10. x −x+= 14. x +x = 18. x x = − 7. x −x += 11. x −= 15. x =x + 8. x +x −= 12. x +x −= 16. x =x 9. x +x += 13. x = 17. x +x = 19. x= 23. x += 20. x = 24. x −= 21. x −= 22. x −= 25. x−x −= 28. x + = x − 29. +z =z 26. x −x −= 30. p+p −= 27. x −x = 31. x −x −= 32. x−x += 36. x−x −= 33. x +x += 37. x −x −= 35. x −x += 34. x +x −= 38. x +x += 42. x +x += 46. ++= TECHNOLOGY 39. x +x = x − x= 43. + += 47. 40. x−x −= 44. ++= 48. += 41. x−x += 45. += 49. ++= x-2nd CALC 2:zero ,, 50. =x+x − 51. =−x+x − 52. =x+x − Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 130 CHAPTER 2 EQUATIONS AND INEQUALITIES 53. x+x −= =x+x −= 2nd CALC 5:intersection 54. x+x −= =x+x −= 2nd CALC 5:intersection 56. b a − 58. P =t−t + t t = EXTENSIONS 55. 57. 59. ax+bx +c = x 119 ft. p =−x +x − x p fi fi 2nd CALC maximum x y REAL-WORLD APPLICATIONS 60. 62. 64. A P =A−A + d =t +t t 61. 63. infl P e fl t ft P =−t +t +≤ t ≤ e fl t Th C =x + R =x −xfi x fi den. Th f 378 ft x x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 131 LEARNING OBJECTIVES In this section you will: • ole equations that inole rational eponents. • ole equations by factoring. • ole radical equations. • ole absolute alue equations. 2.6 OTHER TYPES OF EQUATIONS We have solved linear equations, rational equations, and quadratic equations using several methods. However, there are many other types of equations, and we will investigate a few more types in this section. We will look at equations involving rational exponents, polynomial equations, radical equations, absolute value equations, equations in quadratic form, and some rational equations that can be transformed into quadratics. Solving any equation, however, employs the same basic algebraic rules. We will learn some new techniques as they apply to certain equations, but the algebra never changes. Solving Equations Involving Rational Exponents Rational exponents are exponents that are fractions, where the numerator is a power and the denominator is a
root. 1 __ is another way of writing √ For example, 16 2 exponents is a useful skill, as it is highly applicable in calculus. 1 __ 16 ; 8 3 — is another way of writing 3 √ — 8 . The ability to work with rational We can solve equations in which a variable is raised to a rational exponent by raising both sides of the equation to the reciprocal of the exponent. The reason we raise the equation to the reciprocal of the exponent is because we want to eliminate the exponent on the variable term, and a number multiplied by its reciprocal equals 1. 3 1 2 ) = 1, and so on. ) = 1, 3 ( ( _ _ _ For example, 3 2 3 rational exponents A rational exponent indicates a power in the numerator and a root in the denominator. There are multiple ways of writing an expression, a variable, or a number with a rational exponentam) _ n = n √ — am = ( √ n — m a ) Example 1 Evaluating a Number Raised to a Rational Exponent 2 _ . Evaluate 8 3 Solution Whether we take the root first or the power first depends on the number. It is easy to find the cube root of 2 _ 8, so rewrite 8 3 as ( 2)2 ) Try It #1 1 _ − . Evaluate 64 3 = 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 132 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 2 Solve the Equation Including a Variable Raised to a Rational Exponent 5_ 4 = 32. Solve the equation in which a variable is raised to a rational exponent: x Solution The way to remove the exponent on x is by raising both sides of the equation to a power that is the reciprocal 5 4 _ _ . , which is of 5 4 Try It #2 3 _ = 125. Solve the equation x 2 5 _ = 32 32) 5 4 x = (2)4 The fifth root of 32 is 2. = 16 Example 3 Solving an Equation Involving Rational Exponents and Factoring 3 1 _ _ = x . Solve 3x 2 4 Solution This equation involves rational exponents as well as factoring rational exponents. Let us take this one step at a time. First, put the variable terms on one side of the equal sign and set the equation equal to zero 3x 3x 2 4 Now, it looks like we should factor the left side, but what do we factor out? We can always factor the term with the 1 2 2 _ _ _ from both terms on the left. . Then, factor out x as x lowest exponent. Rewrite 3x 4 4 2 1 − 1 ) = 0 ( 3x _ _ x 4 4 1 _ Where did x 4 come from? Remember, when we multiply two numbers with the same base, we add the exponents. 2 _ back in using the distributive property, we get the expression we had before the factoring, Therefore, if we multiply x 4 3 2 _ _ which is what should happen. We need an exponent such that when added to . Thus, the exponent on x equals 4 4 1 _ in the parentheses is . 4 Let us continue. Now we have two factors and can use the zero factor theorem. 2 1 − 1 ) = 0 ( 3x The two solutions are 0 and 1 _ . 81 Try It #3 3 _ = 8. Solve: (x + 5 81 Divide both sides by 3. 1 _ . Raise both sides to the reciprocal of 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 133 Solving Equations Using Factoring We have used factoring to solve quadratic equations, but it is a technique that we can use with many types of polynomial equations, which are equations that contain a string of terms including numerical coefficients and variables. When we are faced with an equation containing polynomials of degree higher than 2, we can often solve them by factoring. polynomial equations A polynomial of degree n is an expression of the type anx n + an − 1x n − 1 + ċ ċ ċ + a2x 2 + a1 x + a0 where n is a positive integer and an, … , a0 are real numbers and an ≠ 0. Setting the polynomial equal to zero gives a polynomial equation. The total number of solutions (real and complex) to a polynomial equation is equal to the highest exponent n. Example 4 Solving a Polynomial by Factoring Solve the polynomial by factoring: 5x 4 = 80x 2. Solution First, set the equation equal to zero. Then factor out what is common to both terms, the GCF. 5x 4 − 80x 2 = 0 5x 2(x 2 − 16) = 0 Notice that we have the difference of squares in the factor x 2 − 16, which we will continue to factor and obtain two solutions. The first term, 5x 2, generates, technically, two solutions as the exponent is 2, but they are the same solution. 5x 2 = 0 x = 0 x 2 − 16 = 0 (x − 4)(x + 4) = 0 x − 4 = 0 or x + 4 = 0 x = 4 or x = −4 The solutions are 0 (double solution ), 4, and −4. Analysis We can see the solutions on the graph in Figure 1. The x-coordinates of the points where the graph crosses the x-axis are the solutions—the x-intercepts. Notice on the graph that at 0, the graph touches the x-axis and bounces back. It does not cross the x-axis. This is typical of double solutions. y 500 400 300 200 100 –1 –100 –200 –300 –400 –500 –5 –4 –3 –2 21 3 4 5 x Figure 1 Try It #4 Solve by factoring: 12x 4 = 3x 2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 134 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 5 Solve a Polynomial by Grouping Solve a polynomial by grouping: x 3 + x 2 − 9x − 9 = 0. Solution This polynomial consists of 4 terms, which we can solve by grouping. Grouping procedures require factoring the first two terms and then factoring the last two terms. If the factors in the parentheses are identical, we can continue the process and solve, unless more factoring is suggested. x 3 + x 2 − 9x − 9 = 0 x 2(x + 1) − 9(x + 1) = 0 (x 2 − 9)(x + 1) = 0 The grouping process ends here, as we can factor x 2 − 9 using the difference of squares formula. (x 2 − 9)(x + 1) = 0 (x − 3)(x + 3)(x + 1 or or x + 3 = 0 x = −3 or or x + 1 = 0 x = −1 The solutions are 3, −3, and −1. Note that the highest exponent is 3 and we obtained 3 solutions. We can see the solutions, the x-intercepts, on the graph in Figure 2. y 25 20 15 10 5 –1 –5 –10 –15 –20 –25 –5 –4 –3 –2 21 3 4 5 x Figure 2 Analysis We looked at solving quadratic equations by factoring when the leading coefficient is 1. When the leading coefficient is not 1, we solved by grouping. Grouping requires four terms, which we obtained by splitting the linear term of quadratic equations. We can also use grouping for some polynomials of degree higher than 2, as we saw here, since there were already four terms. Solving Radical Equations Radical equations are equations that contain variables in the radicand (the expression under a radical symbol), such as — 3x + 18 = − √ √ Radical equations may have one or more radical terms, and are solved by eliminating each radical, one at a time. We have to be careful when solving radical equations, as it is not unusual to find extraneous solutions, roots that are not, in fact, solutions to the equation. These solutions are not due to a mistake in the solving method, but result from the process of raising both sides of an equation to a power. However, checking each answer in the original equation will confirm the true solutions. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 135 radical equations An equation containing terms with a variable in the radicand is called a radical equation. How To… Given a radical equation, solve it. 1. Isolate the radical expression on one side of the equal sign. Put all remaining terms on the other side. 2. If the radical is a square root, then square both sides of the equation. If it is a cube root, then raise both sides of the equation to the third power. In other words, for an nth root radical, raise both sides to the nth power. Doing so eliminates the radical symbol. 3. Solve the remaining equation. 4. If a radical term still remains, repeat steps 1–2. 5. Confirm solutions by substituting them into the original equation. Example 6 Solving an Equation with One Radical Solve √ — 15 − 2x = x. Solution The radical is already isolated on the left side of the equal side, so proceed to square both sides. — 15 − 2x = x √ 2 15 − 2x ) = (x)2 — ( √ We see that the remaining equation is a quadratic. Set it equal to zero and solve. 15 − 2x = x2 0 = x2 + 2x − 15 0 = (x + 5)(x − 3) 0 = (x + 5) or 0 = (x − 3) −5 = x or 3 = x The proposed solutions are −5 and 3. Let us check each solution back in the original equation. First, check −5. — 15 − 2x = x √ √ — — 15 − 2( − 5) = −5 25 = −5 5 ≠ −5 √ This is an extraneous solution. While no mistake was made solving the equation, we found a solution that does not satisfy the original equation. Check 3. The solution is 3. Try It #5 Solve the radical equation: √ — x + 3 = 3x − 1 √ — √ — 15 − 2x = x 15 − 2(3) = 3 9 = 3 3 = 3 √ — Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 136 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 7 Solving a Radical Equation Containing Two Radicals Solve √ — 2x + 3 + √ — x − 2 = 4. Solution As this equation contains two radicals, we isolate one radical, eliminate it, and then isolate the second radical. — √ 2x + 2x + − √ 2x + ) — ( √ Subtract √ — x − 2 from both sides. Square both sides. Use the perfect square formula to expand the right side: (a − b)2 = a2 −2ab + b2x − 2) x − 2 — 2x + 3 = (4)2 − 2(4) √ 2x + 3 = 16 − 8 √ 2x + 3 = 14 + x − 8 √ — x − 11 = −8 √ (x − 11)2 = ( − ) Combine like terms. Isolate the second radical. Square both sides. x 2 − 22x + 121 = 64(x − 2) Now that both radicals have been eliminated, set the quadratic equal to zero and solve. x 2 − 22x + 121 = 64x − 128 x 2 − 86x + 249 = 0 (x − 3)(x − 83) = 0 x − 3 = 0 or x − 83 = 0 x = 3 or x = 83 Factor and solve. The proposed solutions are 3 and 83. Check each solution in the original equation. √ √ One solution is 3. Check 83. — 2x + 2x + 3 = 4 − √ √ 2(3 √ — √ — 2x + 2x + 3 = 4 − √ √ 2(83) + 3 = 4 − √ 169 = 4 − √ 13 ≠ −383 − 2) — 81 The only solution is 3. We see that 83 is an extraneous solution. Try It #6 Solve the equation with two radicals: √ — 3x + 7 + √ — x + 2 = 1. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 137 Solving an Absolute Value Equation Next, we will learn how to solve an absolute value equat
ion. To solve an equation such as \|2x − 6\| = 8, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value bars is 8 or −8. This leads to two different equations we can solve independently. or 2x − 6 = 8 2x = 14 x = 7 2x − 6 = −8 2x = −2 x = −1 Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point. absolute value equations The absolute value of x is written as \|x\|. It has the following properties: If x ≥ 0, then \|x\| = x. If x < 0, then \|x\| = −x. For real numbers A and B, an equation of the form \|A\| = B, with B ≥ 0, will have solutions when A = B or A = −B. If B < 0, the equation \|A\| = B has no solution. An absolute value equation in the form \|ax + b\| = c has the following properties: If c < 0, \|ax + b\| = c has no solution. If c = 0, \|ax + b\| = c has one solution. If c > 0, \|ax + b\| = c has two solutions. How To… Given an absolute value equation, solve it. 1. Isolate the absolute value expression on one side of the equal sign. 2. If c > 0, write and solve two equations: ax + b = c and ax + b = − c. Example 8 Solving Absolute Value Equations Solve the following absolute value equations: b. \|3x + 4\| = −9 a. \|6x + 4\| = 8 Solution a. \|6x + 4\| = 8 Write two equations and solve each: 6x + 4 = 8 6x = 4 2 _ x = 3 2 _ and −2. The two solutions are 3 b. \|3x + 4\| = −9 c. \|3x − 5\| − 4 = 6 d. \|−5x + 10\| = 0 6x + 4 = −8 6x = −12 x = −2 There is no solution as an absolute value cannot be negative. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 138 CHAPTER 2 EQUATIONS AND INEQUALITIES c. \|3x − 5\| − 4 = 6 Isolate the absolute value expression and then write two equations. \|3x − 5\| − 4 = 6 \|3x − 5\| = 10 3x − 5 = 10 3x = 15 x = 5 5 _ There are two solutions: 5 and − . 3 d. \|−5x + 10\| = 0 3x − 5 = −10 3x = −5 5 _ x = − 3 The equation is set equal to zero, so we have to write only one equation. −5x + 10 = 0 −5x = −10 x = 2 There is one solution: 2. Try It #7 Solve the absolute value equation: \|1 − 4x\| + 8 = 13. Solving Other Types of Equations There are many other types of equations in addition to the ones we have discussed so far. We will see more of them throughout the text. Here, we will discuss equations that are in quadratic form, and rational equations that result in a quadratic. Solving Equations in Quadratic Form Equations in quadratic form are equations with three terms. The first term has a power other than 2. The middle term has an exponent that is one-half the exponent of the leading term. The third term is a constant. We can solve equations in this form as if they were quadratic. A few examples of these equations include x 4 − 5x 2 + 4 = 0, x 6 + 7x 3 − 8 = 0, 2 1 _ _ + 2 = 0. In each one, doubling the exponent of the middle term equals the exponent on the leading + 4x and x 3 3 term. We can solve these equations by substituting a variable for the middle term. quadratic form If the exponent on the middle term is one-half of the exponent on the leading term, we have an equation in quadratic form, which we can solve as if it were a quadratic. We substitute a variable for the middle term to solve equations in quadratic form. How To… Given an equation quadratic in form, solve it. 1. Identify the exponent on the leading term and determine whether it is double the exponent on the middle term. 2. If it is, substitute a variable, such as u, for the variable portion of the middle term. 3. Rewrite the equation so that it takes on the standard form of a quadratic. 4. Solve using one of the usual methods for solving a quadratic. 5. Replace the substitution variable with the original term. 6. Solve the remaining equation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 OTHER TYPES OF EQUATIONS 139 Example 9 Solving a Fourth-degree Equation in Quadratic Form Solve this fourth-degree equation: 3x 4 − 2x 2 − 1 = 0. Solution This equation fits the main criteria, that the power on the leading term is double the power on the middle term. Next, we will make a substitution for the variable term in the middle. Let u = x 2. Rewrite the equation in u. Now solve the quadratic. 3u2 − 2u − 1 = 0 3u2 − 2u − 1 = 0 (3u + 1)(u − 1) = 0 Solve each factor and replace the original term for u. 3u + 1 = 0 3u = −1 1 _ u = − 3 1 _ x2 = − 3 x = ±i √ __ 1 _ 3 __ 1 _ and ± 1. 3 The solutions are ± i √ Try It #8 Solve using substitution: x 4 − 8x 2 − 9 = 0. u − 1 = 0 u = 1 x2 = 1 x = ±1 Example 10 Solving an Equation in Quadratic Form Containing a Binomial Solve the equation in quadratic form: (x + 2)2 + 11(x + 2) − 12 = 0. Solution This equation contains a binomial in place of the single variable. The tendency is to expand what is presented. However, recognizing that it fits the criteria for being in quadratic form makes all the difference in the solving process. First, make a substitution, letting u = x + 2. Then rewrite the equation in u. Solve using the zero-factor property and then replace u with the original expression. u2 + 11u − 12 = 0 (u + 12)(u − 1) = 0 u + 12 = 0 u = −12 x + 2 = −12 x = −14 1 The second factor results in We have two solutions: −14 and −1. Try It #9 Solve: (x − 5)2 − 4(x − 5) − 21 = 0. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 140 CHAPTER 2 EQUATIONS AND INEQUALITIES Solving Rational Equations Resulting in a Quadratic Earlier, we solved rational equations. Sometimes, solving a rational equation results in a quadratic. When this happens, we continue the solution by simplifying the quadratic equation by one of the methods we have seen. It may turn out that there is no solution. Example 11 Solving a Rational Equation Leading to a Quadratic Solve the following rational equation: −4x 8 _ . x2 − 1 Solution We want all denominators in factored form to find the LCD. Two of the denominators cannot be factored further. However, x2 −1 = (x + 1)(x − 1). Then, the LCD is (x + 1)(x − 1). Next, we multiply the whole equation by the LCD. (x + 1)(x − 1) ( −4x 8 __ (x + 1)(x − 1) ) (x + 1)(x − 1) −4x(x + 1) + 4(x − 1) = −8 −4x2 − 4x + 4x − 4 = −8 −4x2 + 4 = 0 −4(x2 − 1) = 0 −4(x + 1)(x − 1) = 0 x = −1 or x = 1 In this case, either solution produces a zero in the denominator in the original equation. Thus, there is no solution. Try It #10 Solve 3x + 2 ______ x − 2 1 __ = + x −2 ______ x2 − 2x . Access these online resources for additional instruction and practice with different types of equations. • Rational Equation with No Solution (http://openstaxcollege.org/l/rateqnosoln) • Solving Equations with Rational Exponents Using Reciprocal Powers (http://openstaxcollege.org/l/ratexprecpexp) • Solving Radical Equations part 1 of 2 (http://openstaxcollege.org/l/radeqsolvepart1) • Solving Radical Equations part 2 of 2 (http://openstaxcollege.org/l/radeqsolvepart2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.6 SECTION EXERCISES 141 2.6 SECTION EXERCISES VERBAL 1. In a radical equation, what does it mean if a number 2. Explain why possible solutions must be checked in is an extraneous solution? 3 _ 3. Your friend tries to calculate the value − 9 and 2 keeps getting an ERROR message. What mistake is he or she probably making? 5. Explain how to change a rational exponent into the correct radical expression. radical equations. 4. Explain why \|2x + 5\| = −7 has no solutions. ALGEBRAIC For the following exercises, solve the rational exponent equation. Use factoring where necessary. 2 _ = 16 6. x 3 2 _ = 4 10. (x + 1) 3 3 _ = 27 7 − 5x 11 − 4x − 3x 12. (x − 1) 4 For the following exercises, solve the following polynomial equations by grouping and factoring. 13. x 3 + 2x 2 − x − 2 = 0 16. x 3 + 3x 2 − 25x − 75 = 0 19. 5x 3 + 45x = 2x 2 + 18 14. 3x 3 − 6x 2 − 27x + 54 = 0 17. m3 + m2 − m − 1 = 0 15. 4y 3 − 9y = 0 18. 2x 5 −14x 3 = 0 For the following exercises, solve the radical equation. Be sure to check all solutions to eliminate extraneous solutions. 20. √ — 3x − 1 − 2 = 0 23. √ — 3t + 5 = 7 — 21. √ x − 7 = 5 — 24. √ t + 1 + 9 = 7 22 25. √ — 12 − x = x 26. √ — 2x + 3 − √ — x + 2 = 2 27. √ — 3x + 7 + √ — x + 2 = 1 28. √ — 2x + 3 − √ — x + 1 = 1 For the following exercises, solve the equation involving absolute value. 29. \|3x − 4\| = 8 33. \|2x − 1\| − 7 = −2 30. \|2x − 3\| = −2 34. \|2x + 1\| − 2 = −3 31. \|1 − 4x\| − 1 = 5 35. \|x + 5\| = 0 32. \|4x + 1\| − 3 = 6 36. −\|2x + 1\| = −3 For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. 37. x 4 − 10x 2 + 9 = 0 40. (x + 1)2 − 8(x + 1) − 9 = 0 38. 4(t − 1)2 − 9(t − 1) = −2 41. (x − 3)2 − 4 = 0 39. (x 2 − 1)2 + (x 2 − 1) − 12 = 0 EXTENSIONS For the following exercises, solve for the unknown variable. 42. x−2 − x−1 − 12 = 0 — \|x\|2 = x 43. √ 44. t 25 − t 5 + 1 = 0 45. \|x 2 + 2x − 36\| = 12 REAL-WORLD APPLICATIONS For the following exercises, use the model for the period of a pendulum, T, such that T = 2π √ ___ L _ g , where the length of the pendulum is L and the acceleration due to gravity is g. 46. If the acceleration due to gravity is 9.8 m/s2 and the period equals 1 s, find the length to the nearest cm (100 cm = 1 m). 47. If the gravity is 32 ft/s2 and the period equals 1 s, find the length to the nearest in. (12 in. = 1 ft). Round your answer to the nearest in. For the following exercises, use a model for body surface area, BSA, such that BSA = √ kg and h = height in cm. 48. Find the height of a 72-kg female to the nearest cm whose BSA = 1.8. whose BSA = 2.1. 49. Find the weight of a 177-cm male to the nearest kg _____ wh _ 3600 , where w = weight in Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 142 CHAPTER 2 EQUATIONS AND INEQUALITIES LEARNING OBJECTIVES In this section you will: • Use interal notation. • ole linear inequalities. • ole compound linear inequalities of
both and and or type. • ole absolute alue inequalities. 2.7 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities. Figure 1 Using Interval Notation Indicating the solution to an inequality such as x ≥ 4 can be achieved in several ways. We can use a number line as shown in Figure 2. The blue ray begins at x = 4 and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4. 5 6 Figure 2 We can use set-builder notation: {x\|x ≥ 4}, which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set. 9 10 11 8 7 1 2 3 4 0 The third method is interval notation, in which solution sets are indicated with parentheses or brackets. The solutions to x ≥ 4 are represented as [4, ∞). This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses. The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval, or a set of numbers in which a solution falls, are [−2, 6), or all numbers between −2 and 6, including −2, but not including 6; (−1, 0), all real numbers between, but not including −1 and 0; and (−∞, 1], all real numbers less than and including 1. Table 1 outlines the possibilities. Set Indicated Set-Builder Notation Interval Notation All real numbers between a and b, but not including a or b {x\| a < x < b} All real numbers greater than a, but not including a All real numbers less than b, but not including b All real numbers greater than a, including a {x\| x > a} {x\| x < b} {x\| x ≥ a} (a, b) (a, ∞) (−∞, b) [a, ∞) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.7 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES 143 Set Indicated Set-Builder Notation Interval Notation All real numbers less than b, including b All real numbers between a and b, including a All real numbers between a and b, including b All real numbers between a and b, including a and b {x\| x ≤ b} {x\| a ≤ x < b} {x\| a < x ≤ b} {x\| a ≤ x ≤ b} (−∞, b] [a, b) (a, b] [a, b] All real numbers less than a or greater than b {x\| x < a and x > b} (−∞, a) ∪ (b, ∞) All real numbers {x\| x is all real numbers} (−∞, ∞) Table 1 Example 1 Using Interval Notation to Express All Real Numbers Greater Than or Equal to a Use interval notation to indicate all real numbers greater than or equal to −2. Solution Use a bracket on the left of −2 and parentheses after infinity: [−2, ∞). The bracket indicates that −2 is included in the set with all real numbers greater than −2 to infinity. Try It #1 Use interval notation to indicate all real numbers between and including −3 and 5. Example 2 Using Interval Notation to Express All Real Numbers Less Than or Equal to a or Greater Than or Equal to b Write the interval expressing all real numbers less than or equal to −1 or greater than or equal to 1. Solution We have to write two intervals for this example. The first interval must indicate all real numbers less than or equal to 1. So, this interval begins at − ∞ and ends at −1, which is written as (−∞, −1]. The second interval must show all real numbers greater than or equal to 1, which is written as [1, ∞). However, we want to combine these two sets. We accomplish this by inserting the union symbol, ∪ , between the two intervals. (−∞, −1] ∪ [1, ∞) Try It #2 Express all real numbers less than −2 or greater than or equal to 3 in interval notation. Using the Properties of Inequalities When we work with inequalities, we can usually treat them similarly to but not exactly as we treat equalities. We can use the addition property and the multiplication property to help us solve them. The one exception is when we multiply or divide by a negative number; doing so reverses the inequality symbol. properties of inequalities Addition Property If a < b, then a + c < b + c. Multiplication Property If a < b and c > 0, then ac < bc. If a < b and c < 0, then ac > bc. These properties also apply to a ≤ b, a > b, and a ≥ b. Example 3 Demonstrating the Addition Property Illustrate the addition property for inequalities by solving each of the following: a. x − 15 < 4 b. 6 ≥ x − 1 c. x + 7 > 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 144 CHAPTER 2 EQUATIONS AND INEQUALITIES Solution The addition property for inequalities states that if an inequality exists, adding or subtracting the same number on both sides does not change the inequality. a. x − 15 < 4 x − 15 + 15 < 4 + 15 b. x < 19 Try It #3 Solve: 3x−2 < 1. Add 15 to both sides. Add 1 to both sides. Subtract 7 from both sides. Example 4 Demonstrating the Multiplication Property Illustrate the multiplication property for inequalities by solving each of the following: a. 3x < 6 b. −2x − 1 ≥ 5 c. 5 − x > 10 Solution a. 3x < 6 1 1 _ _ (3x) < (6) 3 3 x < 2 b. −2x − 1 ≥ 5 −2x ≥ 6 1 1 ) ) (−2x) ≥ (6. 5 − x > 10 −x > 5 1 _ Multiply by − . 2 Reverse the inequality. (−1)(−x) > (5)( − 1) Multiply by − 1. x < − 5 Reverse the inequality. Try It #4 Solve: 4x + 7 ≥ 2x − 3. Solving Inequalities in One Variable Algebraically As the examples have shown, we can perform the same operations on both sides of an inequality, just as we do with equations; we combine like terms and perform operations. To solve, we isolate the variable. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.7 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES 145 Example 5 Solving an Inequality Algebraically Solve the inequality: 13 − 7x ≥ 10x − 4. Solution Solving this inequality is similar to solving an equation up until the last step. 13 − 7x ≥ 10x − 4 13 − 17x ≥ −4 −17x ≥ −17 x ≤ 1 Move variable terms to one side of the inequality. Isolate the variable term. Dividing both sides by −17 reverses the inequality. The solution set is given by the interval (−∞, 1], or all real numbers less than and including 1. Try It #5 1 _ Solve the inequality and write the answer using interval notation: −x + 4 < x + 1. 2 Example 6 Solving an Inequality with Fractions 5 3 2 _ _ _ x ≥ − Solve the following inequality and write the answer in interval notation: − + x. 4 3 8 Solution We begin solving in the same way we do when solving an equation. − x − 9 _ 12 12 5 17 _ _ x ≥ − 8 12 5 12 ) ( − _ _ x ≤ − 17 8 15 _ x ≤ 34 ] . 15_ The solution set is the interval ( −∞, 34 − Put variable terms on one side. Write fractions with common denominator. Multiplying by a negative number reverses the inequality. Try It #6 5 8 3 _ _ _ Solve the inequality and write the answer in interval notation: − + x ≤ x. 4 3 6 Understanding Compound Inequalities A compound inequality includes two inequalities in one statement. A statement such as 4 < x ≤ 6 means 4 < x and x ≤ 6. There are two ways to solve compound inequalities: separating them into two separate inequalities or leaving the compound inequality intact and performing operations on all three parts at the same time. We will illustrate both methods. Example 7 Solving a Compound Inequality Solve the compound inequality: 3 ≤ 2x + 2 < 6. Solution The first method is to write two separate inequalities: 3 ≤ 2x + 2 and 2x + 2 < 6. We solve them independently. 3 ≤ 2x + 2 1 ≤ 2x 1 _ ≤ x 2 and 2x + 2 < 6 2x < 4 x < 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 146 CHAPTER 2 EQUATIONS AND INEQUALITIES Then, we can rewrite the solution as a compound inequality, the same way the problem began. 1_ ≤ x < 2 2 1 In interval notation, the solution is written as [ , 2 ) . _ 2 The second method is to leave the compound inequality intact, and perform solving procedures on the three parts at the same time. 3 ≤ 2x + 2 < 6 1 ≤ 2x < 4 1 _ ≤ x < 2 2 Isolate the variable term, and subtract 2 from all three parts. Divide through all three parts by 2. 1 We get the same solution: [ , 2 ) . _ 2 Try It #7 Solve the compound inequality: 4 < 2x − 8 ≤ 10. Example 8 Solving a Compound Inequality with the Variable in All Three Parts Solve the compound inequality with variables in all three parts: 3 + x > 7x − 2 > 5x − 10. Solution Let's try the first method. Write two inequalities: and 7x − 2 > 5x − 10 2x − 2 > −10 2x > −8 3 + x > 7x − 2 3 > 6x − 2 5 > 6x 4 < x 6 5 5 ) . Notice that when we write the solution in interval or in interval notation ( −4, _ _ The solution set is −4 < x < 6 6 notation, the smaller number comes first. We read intervals from left to right, as they appear on a number line. See Figure 3. x > −4 4 Figure 3 5 6 Try It #8 Solve the compound inequality: 3y < 4 − 5y < 5 + 3y. Solving Absolute Value Inequalities As we know, the absolute value of a quantity is a positive number or zero. From the origin, a point located at (−x, 0) has an absolute value of x, as it is x units away. Consider absolute value as the distance from one point to another point. Regardless of direction, positive or negative, the distance between the two points is represented as a positive number or zero. An absolute value inequality is an equation of the form \|A\| < B, \|A\| ≤ B, \|A\| > B, or \|A\| ≥ B, Where A, and sometimes B, represents an algebraic expression dependent on a variable x. Solving the inequality means finding the set of
all x -values that satisfy the problem. Usually this set will be an interval or the union of two intervals and will include a range of values. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.7 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES 147 There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two equations. The advantage of the algebraic approach is that solutions are exact, as precise solutions are sometimes difficult to read from a graph. Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of x-values such that the distance between x and 600 is less than or equal to 200. We represent the distance between x and 600 as \|x − 600\|, and therefore, \|x − 600\| ≤ 200 or −200 ≤ x − 600 ≤ 200 −200 + 600 ≤ x − 600 + 600 ≤ 200 + 600 400 ≤ x ≤ 800 This means our returns would be between $400 and $800. To solve absolute value inequalities, just as with absolute value equations, we write two inequalities and then solve them independently. absolute value inequalities For an algebraic expression X, and k > 0, an absolute value inequality is an inequality of the form \|X\| < k is equivalent to − k < X < k \|X\| > k is equivalent to X < −k or X > k These statements also apply to \|X\| ≤ k and \|X\| ≥ k. Example 9 Determining a Number within a Prescribed Distance Describe all values x within a distance of 4 from the number 5. Solution We want the distance between x and 5 to be less than or equal to 4. We can draw a number line, such as in Figure 4, to represent the condition to be satisfied. 4 4 5 Figure 4 The distance from x to 5 can be represented using an absolute value symbol, \|x − 5\|. Write the values of x that satisfy the condition as an absolute value inequality. \|x − 5\| ≤ 4 We need to write two inequalities as there are always two solutions to an absolute value equation. x − 5 ≤ 4 x ≤ 9 and x − 5 ≥ − 4 x ≥ 1 If the solution set is x ≤ 9 and x ≥ 1, then the solution set is an interval including all real numbers between and including 1 and 9. So \|x − 5\| ≤ 4 is equivalent to [1, 9] in interval notation. Try It #9 Describe all x-values within a distance of 3 from the number 2. Example 10 Solving an Absolute Value Inequality Solve \|x − 1\| ≤ 3. Solution \|x − 1\| ≤ 3 −3 ≤ x − 1 ≤ 3 −2 ≤ x ≤ 4 [−2, 4] Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 148 CHAPTER 2 EQUATIONS AND INEQUALITIES Example 11 Using a Graphical Approach to Solve Absolute Value Inequalities 1 _ Given the equation y = − \|4x − 5\| + 3, determine the x-values for which the y-values are negative. 2 1 _ Solution We are trying to determine where y < 0, which is when − \|4x − 5\| + 3 < 0. We begin by isolating the 2 absolute value. 1 _ \|4x − 5\| < − 3 − 2 \|4x − 5\| > 6 Next, we solve for the equality \|4x − 5\| = 6. Multiply both sides by –2, and reverse the inequality. 4x − 5 = 6 or 4x − 5 = −6 4x = 11 x = 11_ 4 4x = −1 1 _ x = − 4 Now, we can examine the graph to observe where the y-values are negative. We observe where the branches are below the x-axis. Notice that it is not important exactly what the graph looks like, as long as we know that it crosses the 11 1 ___ _ horizontal axis at x = − and x = 4 4 , and that the graph opens downward. See Figure 5. y 5 4 3 2 1 –1 –2 –3 –4 –5 x = −0.25 –3 –1 –4 –5 Below x-axis –2 21 x = 2.75 x 3 4 5 Below x-axis Figure 5 Try It #10 Solve − 2\|k − 4\| ≤ − 6. Access these online resources for additional instruction and practice with linear inequalities and absolute value inequalities. • Interval Notation (http://openstaxcollege.org/l/intervalnotn) • How to Solve Linear Inequalities (http://openstaxcollege.org/l/solvelinineq) • How to Solve an Inequality (http://openstaxcollege.org/l/solveineq) • Absolute Value Equations (http://openstaxcollege.org/l/absvaleq) • Compound Inequalities (http://openstaxcollege.org/l/compndineqs) • Absolute Value Inequalities (http://openstaxcollege.org/l/absvalineqs) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 2.7 SECTION EXERCISES 149 2.7 SECTION EXERCISES VERBAL 1. 2. −x > x<− x +<x + < 3. 4. x +>x + > 5. y =\|x −\| ALGEBRAIC 6. x −≤ 7. x +≥x − 8. −x +>x − 9. x +≥x − + 10. − x≤− x 11. −x −+>x −−x 12. −x +>−x + 13. x+ − x+ ≥ 14. x− + x+ ≤ 15. −<x+≤ 17. y<−y<+y 19. x+<x+ + +< <+< 21. 23. 25. 27. 29. +  +  +<   16. x+>x−>x− 18. x−<−x+≥ 20. 22. 24. 26. 28. >  <+  < < +>   +  fi 30. \|x+\|≥− 31. \|x+\|< 32. \|x−\|> 33. \|x +\|+≤ 34. \|x−\|+≥ 35. \|−x+\|≤ 36. \|x−\|<− 37. \|x −\|>− 38. x− \| \|< Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 150 CHAPTER 2 EQUATIONS AND INEQUALITIES x- 39. 40. 41. 42. GRAPHICAL xr fi 43. \|x−\|> 44. \|x+\|≥ 45. \|x+\|≤ 46. \|x −\|< 47. \|x−\|< ftyy y- 48. x +<x − 49. x −>x + 50. x +>x + 51. x − x +> 52. x +< x + NUMERIC 53. x\|−< x < 54. x\|x≥ 55. x\|x< 56. x\| x 57. −∞ 58. ∞ 59. − 60. −∪∞ 61. − − 62. − − 63. TECHNOLOGY fts 1 Y2 =MATHNum1:abs( 2 nd CALC 5:intersection1st curve, enter, 2 nd curve, enter, guess, enter x-fi 64. \|x +\|−< 65. − \|x +\|< 66. \|x +\|−> 67. \|x −\|< 68. \|x +\|≥ EXTENSIONS 69. \|x +\|=\|x +\| 70. x−x> 71. x − ≤ x ≠ − x + REAL-WORLD APPLICATIONS 72. p=−x +x −fi xfi 73. V =T V T 74. . Th C =+x − Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 2 REVIEW 151 CHAPTER 2 REVIEW Key Terms absolute value equation an equation in which the variable appears in absolute value bars, typically with two solutions, one accounting for the positive expression and one for the negative expression area in square units, the area formula used in this section is used to find the area of any two-dimensional rectangular region: A = LW Cartesian coordinate system a grid system designed with perpendicular axes invented by René Descartes completing the square a process for solving quadratic equations in which terms are added to or subtracted from both sides of the equation in order to make one side a perfect square complex conjugate a complex number containing the same terms as another complex number, but with the opposite operator. Multiplying a complex number by its conjugate yields a real number. complex number the sum of a real number and an imaginary number; the standard form is a + bi, where a is the real part and b is the complex part. complex plane the coordinate plane in which the horizontal axis represents the real component of a complex number, and the vertical axis represents the imaginary component, labeled i. compound inequality a problem or a statement that includes two inequalities conditional equation an equation that is true for some values of the variable discriminant the expression under the radical in the quadratic formula that indicates the nature of the solutions, real or complex, rational or irrational, single or double roots. distance formula a formula that can be used to find the length of a line segment if the endpoints are known equation in two variables a mathematical statement, typically written in x and y, in which two expressions are equal equations in quadratic form equations with a power other than 2 but with a middle term with an exponent that is one- half the exponent of the leading term extraneous solutions any solutions obtained that are not valid in the original equation graph in two variables the graph of an equation in two variables, which is always shown in two variables in the two- dimensional plane identity equation an equation that is true for all values of the variable imaginary number the square root of −1: i = √ −1 . — inconsistent equation an equation producing a false result intercepts the points at which the graph of an equation crosses the x-axis and the y-axis interval an interval describes a set of numbers within which a solution falls interval notation a mathematical statement that describes a solution set and uses parentheses or brackets to indicate where an interval begins and ends linear equation an algebraic equation in which each term is either a constant or the product of a constant and the first power of a variable linear inequality similar to a linear equation except that the solutions will include sets of numbers midpoint formula a formula to find the point that divides a line segment into two parts of equal length ordered pair a pair of numbers indicating horizontal displacement and vertical displacement from the origin; also known as a coordinate pair, (x, y) origin the point where the two axes cross in the center of the plane, described by the ordered pair (0, 0) perimeter in linear units, the perimeter formula is used to find the linear measurement, or outside length and width, around a two-dimensional regular object; for a rectangle: P = 2L + 2W polynomial equation an equation containing a string of terms including numerical coefficients and variables raised to whole-number exponents Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 152 CHAPTER 2 EQUATIONS AND INEQUALITIES Pythagorean Theorem a theorem that states the relationship among the lengths of the sides of a right triangle, used to solve right triangle problems quadrant one quarter of the coordinate plane, created when the axes divide the plane into four sections quadratic equation an equation containing a second-degree polynomial; can be solved using multiple methods quadratic formula a formula that will solve all quadratic equations radical equation an equation containing at least one radical term where the variable is part of the radicand rational equation an equation consisting of a fraction of polynomials slope the change in y-values over the change in x-values solution set the set of all solutions to an equation square root property one of the methods used to solve a quadratic equation, in which the x 2 term i
s isolated so that the square root of both sides of the equation can be taken to solve for x volume in cubic units, the volume measurement includes length, width, and depth: V = LWH x-axis the common name of the horizontal axis on a coordinate plane; a number line increasing from left to right x-coordinate the first coordinate of an ordered pair, representing the horizontal displacement and direction from the origin x-intercept the point where a graph intersects the x-axis; an ordered pair with a y-coordinate of zero y-axis the common name of the vertical axis on a coordinate plane; a number line increasing from bottom to top y-coordinate the second coordinate of an ordered pair, representing the vertical displacement and direction from the origin y-intercept a point where a graph intercepts the y-axis; an ordered pair with an x-coordinate of zero zero-product property the property that formally states that multiplication by zero is zero, so that each factor of a quadratic equation can be set equal to zero to solve equations Key Equations quadratic formula Key Concepts x = — −b ± √ b2 − 4ac ______________ 2a 2.1 The Rectangular Coordinate Systems and Graphs • We can locate, or plot, points in the Cartesian coordinate system using ordered pairs, which are defined as displacement from the x-axis and displacement from the y-axis. See Example 1. • An equation can be graphed in the plane by creating a table of values and plotting points. See Example 2. • Using a graphing calculator or a computer program makes graphing equations faster and more accurate. Equations usually have to be entered in the form y = . See Example 3. • Finding the x- and y-intercepts can define the graph of a line. These are the points where the graph crosses the axes. See Example 4. • The distance formula is derived from the Pythagorean Theorem and is used to find the length of a line segment. See Example 5 and Example 6. • The midpoint formula provides a method of finding the coordinates of the midpoint dividing the sum of the x-coordinates and the sum of the y-coordinates of the endpoints by 2. See Example 7 and Example 8. 2.2 Linear Equations in One Variable • We can solve linear equations in one variable in the form ax + b = 0 using standard algebraic properties. See Example 1 and Example 2. • A rational expression is a quotient of two polynomials. We use the LCD to clear the fractions from an equation. See Example 3 and Example 4. • All solutions to a rational equation should be verified within the original equation to avoid an undefined term, or zero in the denominator. See Example 5, Example 6, and Example 7. • Given two points, we can find the slope of a line using the slope formula. See Example 8. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 2 REVIEW 153 • We can identify the slope and y-intercept of an equation in slope-intercept form. See Example 9. • We can find the equation of a line given the slope and a point. See Example 10. • We can also find the equation of a line given two points. Find the slope and use the point-slope formula. See Example 11. • The standard form of a line has no fractions. See Example 12. • Horizontal lines have a slope of zero and are defined as y = c, where c is a constant. • Vertical lines have an undefined slope (zero in the denominator), and are defined as x = c, where c is a constant. See Example 13. • Parallel lines have the same slope and different y-intercepts. See Example 14 and Example 15. • Perpendicular lines have slopes that are negative reciprocals of each other unless one is horizontal and the other is vertical. See Example 16. 2.3 Models and Applications • A linear equation can be used to solve for an unknown in a number problem. See Example 1. • Applications can be written as mathematical problems by identifying known quantities and assigning a variable to unknown quantities. See Example 2. • There are many known formulas that can be used to solve applications. Distance problems, for example, are solved using the d = rt formula. See Example 3. • Many geometry problems are solved using the perimeter formula P = 2L + 2W, the area formula A = LW, or the volume formula V = LWH. See Example 4, Example 5, and Example 6. 2.4 Complex Numbers • The square root of any negative number can be written as a multiple of i. See Example 1. • To plot a complex number, we use two number lines, crossed to form the complex plane. The horizontal axis is the real axis, and the vertical axis is the imaginary axis. See Example 2. • Complex numbers can be added and subtracted by combining the real parts and combining the imaginary parts. See Example 3. • Complex numbers can be multiplied and divided. ◦ To multiply complex numbers, distribute just as with polynomials. See Example 4 and Example 5. ◦ To divide complex numbers, multiply both numerator and denominator by the complex conjugate of the denominator to eliminate the complex number from the denominator. See Example 6 and Example 7. • The powers of i are cyclic, repeating every fourth one. See Example 8. 2.5 Quadratic Equations • Many quadratic equations can be solved by factoring when the equation has a leading coefficient of 1 or if the equation is a difference of squares. The zero-product property is then used to find solutions. See Example 1, Example 2, and Example 3. • Many quadratic equations with a leading coefficient other than 1 can be solved by factoring using the grouping method. See Example 4 and Example 5. • Another method for solving quadratics is the square root property. The variable is squared. We isolate the squared term and take the square root of both sides of the equation. The solution will yield a positive and negative solution. See Example 6 and Example 7. • Completing the square is a method of solving quadratic equations when the equation cannot be factored. See Example 8. • A highly dependable method for solving quadratic equations is the quadratic formula, based on the coefficients and the constant term in the equation. See Example 9 and Example 10. • The discriminant is used to indicate the nature of the roots that the quadratic equation will yield: real or complex, rational or irrational, and how many of each. See Example 11. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 154 CHAPTER 2 EQUATIONS AND INEQUALITIES • The Pythagorean Theorem, among the most famous theorems in history, is used to solve right-triangle problems and has applications in numerous fields. Solving for the length of one side of a right triangle requires solving a quadratic equation. See Example 12. 2.6 Other Types of Equations • Rational exponents can be rewritten several ways depending on what is most convenient for the problem. To solve, both sides of the equation are raised to a power that will render the exponent on the variable equal to 1. See Example 1, Example 2, and Example 3. • Factoring extends to higher-order polynomials when it involves factoring out the GCF or factoring by grouping. See Example 4 and Example 5. • We can solve radical equations by isolating the radical and raising both sides of the equation to a power that matches the index. See Example 6 and Example 7. • To solve absolute value equations, we need to write two equations, one for the positive value and one for the negative value. See Example 8. • Equations in quadratic form are easy to spot, as the exponent on the first term is double the exponent on the second term and the third term is a constant. We may also see a binomial in place of the single variable. We use substitution to solve. See Example 9 and Example 10. • Solving a rational equation may also lead to a quadratic equation or an equation in quadratic form. See Example 11. 2.7 Linear Inequalities and Absolute Value Inequalities • Interval notation is a method to indicate the solution set to an inequality. Highly applicable in calculus, it is a system of parentheses and brackets that indicate what numbers are included in a set and whether the endpoints are included as well. See Table 1 and Example 1 and Example 2. • Solving inequalities is similar to solving equations. The same algebraic rules apply, except for one: multiplying or dividing by a negative number reverses the inequality. See Example 3, Example 4, Example 5, and Example 6. • Compound inequalities often have three parts and can be rewritten as two independent inequalities. Solutions are given by boundary values, which are indicated as a beginning boundary or an ending boundary in the solutions to the two inequalities. See Example 7 and Example 8. • Absolute value inequalities will produce two solution sets due to the nature of absolute value. We solve by writing two equations: one equal to a positive value and one equal to a negative value. See Example 9 and Example 10. • Absolute value inequalities can also be solved by graphing. At least we can check the algebraic solutions by graphing, as we cannot depend on a visual for a precise solution. See Example 11. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 2 REVIEW 155 CHAPTER 2 REVIEW EXERCISES THE RECTANGULAR COORDINATE SYSTEMS AND GRAPHS For the following exercises, find the x-intercept and the y-intercept without graphing. 1. 4x − 3y = 12 2. 2y − 4 = 3x For the following exercises, solve for y in terms of x, putting the equation in slope–intercept form. 3. 5x = 3y − 12 4. 2x − 5y = 7 For the following exercises, find the distance between the two points. 5. (−2, 5)(4, −1) 6. (−12, −3)(−1, 5) 7. Find the distance between the two points (−71,432) and (511,218) using your calculator, and round your answer to the nearest thousandth. For the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points. 8. (−1, 5) and (4, 6) 9. (−13, 5) and (17, 18) For the following exercises, construct a table and graph the equation by plotting at least three points. 1 _
10. y = x + 4 2 11. 4x − 3y = 6 LINEAR EQUATIONS IN ONE VARIABLE For the following exercises, solve for x. 12. 5x + 2 = 7x − 8 13. 3(x + 2) − 10 = x + 4 14. 7x − 3 = 5 15. 12 − 5(x + 1) = 2x − 5 16. 2x 21 _ 4 For the following exercises, solve for x. State all x-values that are excluded from the solution set. 17. x _ x2 − x2 − 9 x ≠ 3, −3 3 1 2 _ _ _ = + 18. 4 2 x For the following exercises, find the equation of the line using the point-slope formula. 19. Passes through these two points: (−2, 1),(4, 2). 1 _ 20. Passes through the point (−3, 4) and has a slope of − . 3 21. Passes through the point (−3, 4) and is parallel to 22. Passes through these two points: (5, 1),(5, 7). 2 _ the graph y = x + 5. 3 MODELS AND APPLICATIONS For the following exercises, write and solve an equation to answer each question. 23. The number of males in the classroom is five more than three times the number of females. If the total number of students is 73, how many of each gender are in the class? 24. A man has 72 ft of fencing to put around a rectangular garden. If the length is 3 times the width, find the dimensions of his garden. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 156 CHAPTER 2 EQUATIONS AND INEQUALITIES 25. A truck rental is $25 plus $.30/mi. Find out how many miles Ken traveled if his bill was $50.20. COMPLEX NUMBERS For the following exercises, use the quadratic equation to solve. 26. x2 − 5x + 9 = 0 27. 2x2 + 3x + 7 = 0 For the following exercises, name the horizontal component and the vertical component. 28. 4 − 3i 29. −2 − i For the following exercises, perform the operations indicated. 31. (2 + 3i) − (−5 − 8i) 32. 2 √ — −75 + 3 √ — 25 34. −6i(i − 5) 37. √ — −2 ( √ — −8 − √ — 5 ) 35. (3 − 5i)2 38. 2 _ 5 − 3i 30. (9 − i) − (4 − 7i) 33. √ — −16 + 4 √ — −9 36. √ — −4 · √ — −12 39. 3 + 7i ______ i QUADRATIC EQUATIONS For the following exercises, solve the quadratic equation by factoring. 40. 2x2 − 7x − 4 = 0 41. 3x2 + 18x + 15 = 0 42. 25x2 − 9 = 0 43. 7x2 − 9x = 0 For the following exercises, solve the quadratic equation by using the square-root property. 44. x2 = 49 45. (x − 4)2 = 36 For the following exercises, solve the quadratic equation by completing the square. 46. x2 + 8x − 5 = 0 47. 4x2 + 2x − 1 = 0 For the following exercises, solve the quadratic equation by using the quadratic formula. If the solutions are not real, state No real solution. 48. 2x2 − 5x + 1 = 0 49. 15x2 − x − 2 = 0 For the following exercises, solve the quadratic equation by the method of your choice. 50. (x − 2)2 = 16 51. x2 = 10x + 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 2 REVIEW 157 OTHER TYPES OF EQUATIONS For the following exercises, solve the equations. 3 _ = 27 52. x 2 55. 3x5 − 6x3 = 0 58. \|3x − 7\| = 5 1 1 _ _ = 0 − 4x 2 4 53. x 56 59. \|2x + 3\| − 5 = 9 54. 4x3 + 8x2 − 9x − 18 = 0 57. √ — 3x + 7 + √ — x + 2 = 1 LINEAR INEQUALITIES AND ABSOLUTE VALUE INEQUALITIES For the following exercises, solve the inequality. Write your final answer in interval notation. 60. 5x − 8 ≤ 12 61. −2x + 5 > x − 7 63. \|3x + 2\| + 1 ≤ 9 64. \|5x − 1\| > 14 62. x − 1 _____ 3 + x + 2 _____ 5 3__ ≤ 5 65. \|x − 3\| < −4 For the following exercises, solve the compound inequality. Write your answer in interval notation. 66. −4 < 3x + 2 ≤ 18 67. 3y < 1 − 2y < 5 + y For the following exercises, graph as described. 68. Graph the absolute value function and graph the constant function. Observe the points of intersection and shade the x-axis representing the solution set to the inequality. Show your graph and write your final answer in interval notation. \|x + 3\| ≥ 5 69. Graph both straight lines (left-hand side being y1 and right-hand side being y2) on the same axes. Find the point of intersection and solve the inequality by observing where it is true comparing the y-values of the lines. See the interval where the inequality is true. x + 3 < 3x − 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 158 CHAPTER 2 EQUATIONS AND INEQUALITIES CHAPTER 2 PRACTICE TEST 1. Graph the following: 2y = 3x + 4. 2. Find the x- and y-intercepts for the following: 2x − 5y = 6. 3. Find the x- and y-intercepts of this equation, and sketch the graph of the line using just the intercepts plotted. 3x − 4y = 12 4. Find the exact distance between (5, −3) and (−2, 8). Find the coordinates of the midpoint of the line segment joining the two points. 5. Write the interval notation for the set of numbers 6. Solve for x: 5x + 8 = 3x − 10. represented by {x\|x ≤ 9}. 7. Solve for x: 3(2x − 5) − 3(x − 7) = 2x − 9. 9. Solve for x: 5 _____ x + 4 = 4 + 3 _____ . . Solve for x: x 2 10. The perimeter of a triangle is 30 in. The longest side is 2 less than 3 times the shortest side and the other side is 2 more than twice the shortest side. Find the length of each side. 11. Solve for x. Write the answer in simplest radical form. 12. Solve: 3x − 8 ≤ 4. x2 _ 3 1 _ − x = − 2 13. Solve: \|2x + 3\| < 5. 14. Solve: \|3x − 2\| ≥ 4. For the following exercises, find the equation of the line with the given information. 15. Passes through the points (−4, 2) and (5, −3). 16. Has an undefined slope and passes through the point (4, 3). 17. Passes through the point (2, 1) and is perpendicular 18. Add these complex numbers: (3 − 2i) + (4 − i). 2 _ x + 3. to y = − 5 — −16 . 20. Multiply: 5i(5 − 3i). — −4 + 3 √ 19. Simplify: √ 4 − i ______ . 2 + 3i 21. Divide: 23. Solve: (3x − 1)2 − 1 = 24. 25. Solve: 4x 2 − 4x − 1 = 0 27. Solve: 2 + √ — 12 − 2x = x 22. Solve this quadratic equation and write the two complex roots in a + bi form: x 2 − 4x + 7 = 0. 24. Solve: x 2 − 6x = 13. 26. Solve 28. Solve: (x − 1) 3 For the following exercises, find the real solutions of each equation by factoring. 29. 2x 3 − x 2 − 8x + 4 = 0 30. (x + 5)2 − 3(x + 5) − 4 = 0 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Functions P 1,500 1,000 500 0 Figure 1 Standard and Poor’s Index with dividends reinvested (credit "bull": modiﬁcation of work by Prayitno Hadinata; credit "graph": modiﬁcation of work by MeasuringWorth) 1970 1975 1980 1985 1990 1995 2000 2005 2010 3 y CHAPTER OUTLINE 3.1 Functions and Function Notation 3.2 Domain and Range 3.3 Rates of Change and Behavior of Graphs 3.4 Composition of Functions 3.5 Transformation of Functions 3.6 Absolute Value Functions 3.7 Inverse Functions Introduction Toward the end of the twentieth century, the values of stocks of Internet and technology companies rose dramatically. As a result, the Standard and Poor’s stock market average rose as well. Figure 1 tracks the value of that initial investment of just under $100 over the 40 years. It shows that an investment that was worth less than $500 until about 1995 skyrocketed up to about $1,100 by the beginning of 2000. That five-year period became known as the “dot-com bubble” because so many Internet startups were formed. As bubbles tend to do, though, the dot-com bubble eventually burst. Many companies grew too fast and then suddenly went out of business. The result caused the sharp decline represented on the graph beginning at the end of 2000. Notice, as we consider this example, that there is a definite relationship between the year and stock market average. For any year we choose, we can determine the corresponding value of the stock market average. In this chapter, we will explore these kinds of relationships and their properties. 159 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 160 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: • etermine whether a relation represents a function. • Find the alue of a function gien its equation or its graph. • etermine whether a function is one-to-one. • Use the ertical line test to identify functions. 3.1 FUNCTIONS AND FUNCTION NOTATION A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships. Determining Whether a Relation Represents a Function A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first. {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10)} The domain is {1, 2, 3, 4, 5}. The range is {2, 4, 6, 8, 10}. Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter x. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter y. A function f is a relation that assigns a single element in the range to each element in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, {1, 2, 3, 4, 5}, is paired with exactly one element in the range, {2, 4, 6, 8, 10}. Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as {(odd, 1), (even, 2), (odd, 3), (even, 4), (odd, 5)} Notice that each element in the domain, {even, odd} is not paired with exactly one element in the range, {1, 2, 3, 4, 5}. For example, the term “odd” corresponds to three values from the domain, {1, 3, 5} and the term “even” corresponds to two values from the range, {2, 4}. This violates the definition of a function, so this relation is not a function. Figure 1 compares relations that are functions and n
ot functions. Relation is a Function Outputs Inputs Relation is a Function Inputs Outputs Relation is NOT a Function Inputs p q Outputs x y z (a) (b) (c) Figure 1 ( a ) This relationship is a function because each input is associated with a single output. Note that input q and r both give output n. ( b ) This relationship is also a function. In this case, each input is associated with a single output. ( c ) This relationship is not a function because input q is associated with two different outputs. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 161 function A function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.” The input values make up the domain, and the output values make up the range. How To… Given a relationship between two quantities, determine whether the relationship is a function. 1. Identify the input values. 2. Identify the output values. 3. If each input value leads to only one output value, classify the relationship as a function. If any input value leads to two or more outputs, do not classify the relationship as a function. Example 1 Determining If Menu Price Lists Are Functions The coffee shop menu, shown in Figure 2 consists of items and their prices. a. Is price a function of the item? b. Is the item a function of the price? Menu Item Price Plain Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.49 Jelly Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99 Chocolate Donut . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99 Figure 2 Solution a. Let’s begin by considering the input as the items on the menu. The output values are then the prices. See Figure 2. Each item on the menu has only one price, so the price is a function of the item. b. Two items on the menu have the same price. If we consider the prices to be the input values and the items to be the output, then the same input value could have more than one output associated with it. See Figure 3. Menu Item Price Plain Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.49 Jelly Donut . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.99 Chocolate Donut . . . . . . . . . . . . . . . . . . . . . . . . . . Therefore, the item is a not a function of price. Figure 3 Example 2 Determining If Class Grade Rules Are Functions In a particular math class, the overall percent grade corresponds to a grade-point average. Is grade-point average a function of the percent grade? Is the percent grade a function of the grade-point average? Table 1 shows a possible rule for assigning grade points. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 162 CHAPTER 3 FUNCTIONS Percent grade 0-56 57-61 62-66 67-71 72-77 78-86 87-91 92-100 Grade-point average 0.0 1.0 1.5 Table 1 2.0 2.5 3.0 3.5 4.0 Solution For any percent grade earned, there is an associated grade-point average, so the grade-point average is a function of the percent grade. In other words, if we input the percent grade, the output is a specific grade-point average. In the grading system given, there is a range of percent grades that correspond to the same grade-point average. For example, students who receive a grade-point average of 3.0 could have a variety of percent grades ranging from 78 all the way to 86. Thus, percent grade is not a function of grade-point average Try It #1 Table 2[1] lists the five greatest baseball players of all time in order of rank. Player Rank Babe Ruth Willie Mays Ty Cobb Walter Johnson Hank Aaron Table 2 1 2 3 4 5 a. Is the rank a function of the player name? b. Is the player name a function of the rank? Using Function Notation Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into graphing calculators and computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions. To represent “height is a function of age,” we start by identifying the descriptive variables h for height and a for age. The letters f, g, and h are often used to represent functions just as we use x, y, and z to represent numbers and A, B, and C to represent sets. h is f of a h = f (a) f (a) We name the function f ; height is a function of age. We use parentheses to indicate the function input. We name the function f ; the expression is read as “f of a.” Remember, we can use any letter to name the function; the notation h(a) shows us that h depends on a. The value a must be put into the function h to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication. We can also give an algebraic expression as the input to a function. For example f (a + b) means “first add a and b, and the result is the input for the function f. ” The operations must be performed in this order to obtain the correct result. function notation The notation y = f (x) defines a function named f. This is read as “y is a function of x.” The letter x represents the input value, or independent variable. The letter y, or f (x), represents the output value, or dependent variable. 1 http://www.baseball-almanac.com/legendary/lisn100.shtml. Accessed 3/24/2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 163 Example 3 Using Function Notation for Days in a Month Use function notation to represent a function whose input is the name of a month and output is the number of days in that month. Solution The number of days in a month is a function of the name of the month, so if we name the function f, we write days = f (month) or d = f (m). The name of the month is the input to a “rule” that associates a specific number (the output) with each input. 31!f (January) output input rule Figure 4 For example, f (March) = 31, because March has 31 days. The notation d = f (m) reminds us that the number of days, d (the output), is dependent on the name of the month, m(the input). Analysis Note that the inputs to a function do not have to be numbers; function inputs can be names of people, labels of geometric objects, or any other element that determines some kind of output. However, most of the functions we will work with in this book will have numbers as inputs and outputs. Example 4 Interpreting Function Notation A function N = f (y) gives the number of police officers, N, in a town in year y. What does f (2005) = 300 represent? Solution When we read f (2005) = 300, we see that the input year is 2005. The value for the output, the number of police officers (N), is 300. Remember N = f (y). The statement f (2005) = 300 tells us that in the year 2005 there were 300 police officers in the town. Try It #2 Use function notation to express the weight of a pig in pounds as a function of its age in days d. Q & A… Instead of a notation such as y = f (x), could we use the same symbol for the output as for the function, such as y = y (x), meaning “y is a function of x?” Yes, this is often done, especially in applied subjects that use higher math, such as physics and engineering. However, in exploring math itself we like to maintain a distinction between a function such as f, which is a rule or procedure, and the output y we get by applying f to a particular input x. This is why we usually use notation such as y = f (x), P = W(d), and so on. Representing Functions Using Tables A common method of representing functions is in the form of a table. The table rows or columns display the corresponding input and output values. In some cases, these values represent all we know about the relationship; other times, the table provides a few select examples from a more complete relationship. Table 3 lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number of days in that month. This information represents all we know about the months and days for a given year (that is not a leap year). Note that, in this table, we define a days-in-a-month function f where D = f (m) identifies months by an integer rather than by name. Month number, m (input) 1 2 3 4 5 6 Days in month, D (output) 31 28 31 30 31 30 7 31 8 9 31 30 10 31 11 30 12 31 Table 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 164 CHAPTER 3 FUNCTIONS Table 4 defines a function Q = g (n). Remember, this notation tells us that g is the name of the function that takes the input n and gives the output Q Table 4 Table 5 below displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same input value, 5 years, has two different output values, 40 in. and 42 in. Age in years, a (input) Height in inches, h (output) 5 40 5 42 6 44 7 47 8 50 9 52 10 54 Table 5 How To… Given a table of input and output values, determine whether the table represents a function. 1. Identify the input and output values. 2. Check to see if each input value is paired with only one output value. If so, the table represents a function. Example 5 Identifying Tables that Represent Functions Which table, Table 6, Table 7, or Table 8, represents a function (if any)? Input Output Input Output Input Output 2 5 8 1 3 6 −3 0 4 5 1 5 Table 6 Table 7 0 2 4 1 5 5 Table 8 Solution Table 6 and Table 7 define functions. In both, each input value corresponds to exactly one output value. Table 8 does not define a function because the input value of 5 corresponds to two different output values. When a table represents a function, corresponding input and output values can also be specified us
ing function notation. The function represented by Table 6 can be represented by writing Similarly, the statements f (2) = 1, f (5) = 3, and f (8) = 6 g (−3) = 5, g (0) = 1, and g (4) = 5 represent the function in table Table 7. Table 8 cannot be expressed in a similar way because it does not represent a function. Try It #3 Does Table 9 represent a function? Input Output 1 2 3 10 100 1000 Table 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 165 Finding Input and Output Values of a Function When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value. When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the function’s formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value. Evaluation of Functions in Algebraic Forms When we have a function in formula form, it is usually a simple matter to evaluate the function. For example, the function f (x) = 5 − 3x 2 can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5. How To… Given the formula for a function, evaluate. 1. Replace the input variable in the formula with the value provided. 2. Calculate the result. Example 6 Evaluating Functions at Speciﬁc Values Evaluate f(x) = x 2 + 3x − 4 at: a. 2 b. a c. a + h d. f (a + h) − f (a) __ h Solution Replace the x in the function with each specified value. a. Because the input value is a number, 2, we can use simple algebra to simplify. f (2) = 22 + 3(2. In this case, the input value is a letter so we cannot simplify the answer any further. c. With an input value of a + h, we must use the distributive property. f (a) = a2 + 3a − 4 f (a + h) = (a + h)2 + 3(a + h) − 4 = a2 + 2ah + h2 + 3a + 3h −4 d. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that and we know that f (a + h) = a2 + 2ah + h2 + 3a + 3h − 4 f(a) = a2 + 3a − 4 Now we combine the results and simplify. f (a + h) − f(a) __ = h (a2 + 2ah + h2 + 3a + 3h − 4) − (a2 + 3a − 4) ____ h 2ah + h2 + 3h ___________ h h(2a + h + 3) ___________ h Factor out h. = = = 2a + h + 3 Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 166 CHAPTER 3 FUNCTIONS Example 7 Evaluating Functions Given the function h(p) = p2 + 2p, evaluate h(4). Solution To evaluate h(4), we substitute the value 4 for the input variable p in the given function. h(p) = p2 + 2p h(4) = (4)2 +2 (4) = 16 + 8 = 24 Therefore, for an input of 4, we have an output of 24. Try It #4 Given the function g(m) = √ — m − 4 . Evaluate g(5). Example 8 Solving Functions Given the function h(p) = p2 + 2p, solve for h(p) = 3. Solution h(p) = 3 p2 + 2p = 3 Substitute the original function h(p) = p2 + 2p. p2 + 2p − 3 = 0 Subtract 3 from each side. (p + 3)(p − 1) = 0 Factor. If (p + 3)(p − 1) = 0, either (p + 3) = 0 or (p − 1) = 0 (or both of them equal 0). We will set each factor equal to 0 and solve for p in each case. (p + 3) = 0, p = −3 (p − 1) = 0, p = 1 This gives us two solutions. The output h(p) = 3 when the input is either p = 1 or p = −3. We can also verify by graphing as in Figure 5. The graph verifies that h(1) = h(−3) = 3 and h(4) = 24. h(p) 35 30 25 20 15 10 5 1 2 3 4 5 p p –3 –2 h(p) 3 0 0 0 1 3 4 24 Figure 5 Try It #5 Given the function g(m) = √ — m − 4 , solve g(m) = 2. Evaluating Functions Expressed in Formulas Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation 2n + 6p = 12 expresses a functional relationship between n and p. We can rewrite it to decide if p is a function of n. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 167 How To… Given a function in equation form, write its algebraic formula. 1. Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves only the input variable. 2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity. Example 9 Finding an Equation of a Function Express the relationship 2n + 6p = 12 as a function p = f (n), if possible. Solution To express the relationship in this form, we need to be able to write the relationship where p is a function of n, which means writing it as p = [expression involving n]. 2n + 6p = 12 6p = 12 − 2n Subtract 2n from both sides. p = 12 − 2n _______ 6 Divide both sides by 6 and simplify. p = 12 __ − 6 2n ___ 6 Therefore, p as a function of n is written as 1 __ p = 2 − n 3 1 __ p = f (n) = 2 − n 3 Example 10 Expressing the Equation of a Circle as a Function Does the equation x 2 + y 2 = 1 represent a function with x as input and y as output? If so, express the relationship as a function y = f (x). Solution First we subtract x 2 from both sides. We now try to solve for y in this equation − x2 = + √ — 1 − x2 and − √ — 1 − x2 We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function y = f (x). If we graph both functions on a graphing calculator, we will get the upper and lower semicircles. Try It #6 If x − 8y 3 = 0, express y as a function of x. Q & A… Are there relationships expressed by an equation that do represent a function but that still cannot be represented by an algebraic formula? Yes, this can happen. For example, given the equation x = y + 2y, if we want to express y as a function of x, there is no simple algebraic formula involving only x that equals y. However, each x does determine a unique value for y, and there are mathematical procedures by which y can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for y as a function of x, even though the formula cannot be written explicitly. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 168 CHAPTER 3 FUNCTIONS Evaluating a Function Given in Tabular Form As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppy’s memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours. The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See Table 10.[2] Pet Puppy Adult dog Cat Goldfish Beta fish Memory span in hours 0.008 0.083 16 2160 3600 Table 10 At times, evaluating a function in table form may be more useful than using equations. Here let us call the function P. The domain of the function is the type of pet and the range is a real number representing the number of hours the pet’s memory span lasts. We can evaluate the function P at the input value of “goldfish.” We would write P(goldfish) = 2160. Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the pertinent row of the table. The tabular form for function P seems ideally suited to this function, more so than writing it in paragraph or function form. How To… Given a function represented by a table, identify specific output and input values. 1. Find the given input in the row (or column) of input values. 2. Identify the corresponding output value paired with that input value. 3. Find the given output values in the row (or column) of output values, noting every time that output value appears. 4. Identify the input value(s) corresponding to the given output value. Example 11 Evaluating and Solving a Tabular Function Using Table 11, a. Evaluate g(3) b. Solve g(n) = 6. n g (n Table 11 Solution a. Evaluating g (3) means determining the output value of the function g for the input value of n = 3. The table output value corresponding to n = 3 is 7, so g (3) = 7. b. Solving g (n) = 6 means identifying the input values, n, that produce an output value of 6. Table 11 shows two solutions: 2 and 4. When we input 2 into the function g, our output is 6. When we input 4 into the function g, our output is also 6. 2 http://www.kgbanswers.com/how-long-is-a-dogs-memory-span/4221590. Accessed 3/24/2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 169 Try It #7 Using Table 11, evaluate g(1). Finding Function Values from a Graph Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value( s ). Example 12 Reading Function Values from a Graph Given the graph in Figure 6, a. Evaluate f (2). b. Solve f (x) = 4. f (x) 7 6 5 4 3 2 1 –1 –1 2 – – 3 –5 –4 –3 –2 21 3 4 5 Figure 6 Solution a. To evaluate f (2), locate the point on the curve where x = 2, then read the y-coordinate of that
point. The point has coordinates (2, 1), so f (2) = 1. See Figure 7. f (x) 7 6 5 4 3 2 1 –1 –1 2 – – 3 –5 –4 –3 –2 (2, 1) f (2) = 1 21 3 4 5 Figure 7 b. To solve f (x) = 4, we find the output value 4 on the vertical axis. Moving horizontally along the line y = 4, we locate two points of the curve with output value 4: (−1, 4) and (3, 4). These points represent the two solutions to f (x) = 4: −1 or 3. This means f (−1) = 4 and f (3) = 4, or when the input is −1 or 3, the output is 4. See Figure 8. f (x) 7 6 5 4 3 2 1 (−1, 4) (3, 4) 21 3 4 5 –5 –4 –3 –2 –1 –1 2 – – 3 Figure 8 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 170 CHAPTER 3 FUNCTIONS Try It #8 Using Figure 7, solve f (x) = 1. Determining Whether a Function is One-to-One Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in Figure 1 at the beginning of this chapter, the stock price was $1,000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1,000. However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in Table 12. Letter grade A B C D Grade-point average 4.0 3.0 2.0 1.0 Table 12 This grading system represents a one-to-one function, because each letter input yields one particular grade-point average output and each grade-point average corresponds to one input letter. To visualize this concept, let’s look again at the two simple functions sketched in Figure 1(a) and Figure 1(b). The function in part ( a) shows a relationship that is not a one-to-one function because inputs q and r both give output n. The function in part (b ) shows a relationship that is a one-to-one function because each input is associated with a single output. one-to-one function A one-to-one function is a function in which each output value corresponds to exactly one input value. There are no repeated x- or y-values. Example 13 Determining Whether a Relationship Is a One-to-One Function Is the area of a circle a function of its radius? If yes, is the function one-to-one? Solution A circle of radius r has a unique area measure given by A = πr 2, so for any input, r, there is only one output, A. The area is a function of radius r. If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure A is given by the formula A = πr2. Because areas and radii are positive numbers, there is exactly one solution: r = √ ___ A __ π So the area of a circle is a one-to-one function of the circle’s radius. Try It #9 a. Is a balance a function of the bank account number? b. Is a bank account number a function of the balance? c. Is a balance a one-to-one function of the bank account number? Try It #10 a. If each percent grade earned in a course translates to one letter grade, is the letter grade a function of the percent grade? b. If so, is the function one-to-one? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 171 Using the Vertical Line Test As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many input-output pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis. The most common graphs name the input value x and the output value y, and we say y is a function of x, or y = f (x) when the function is named f. The graph of the function is the set of all points (x, y) in the plane that satisfies the equation y = f (x). If the function is defined for only a few input values, then the graph of the function consists of only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in Figure 9 tell us that f (0) = 2 and f (6) = 1. However, the set of all points (x, y) satisfying y = f (x) is a curve. The curve shown includes (0, 2) and (6, 1) because the curve passes through those points. y Figure 9 x The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See Figure 10. Function Not a Function Not a Function Figure 10 How To… Given a graph, use the vertical line test to determine if the graph represents a function. 1. Inspect the graph to see if any vertical line drawn would intersect the curve more than once. 2. If there is any such line, determine that the graph does not represent a function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 172 CHAPTER 3 FUNCTIONS Example 14 Applying the Vertical Line Test Which of the graphs in Figure 11 represent( s ) a function y = f (x)? f (x) (a) y 5 4 3 2 1 f (x) x 4 5 6 7 8 9 10 11 12 x x (b) Figure 11 (c) Solution If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts ( a) and (b) of Figure 11. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as shown in Figure 12. Try It #11 Does the graph in Figure 13 represent a function? y Figure 12 y Figure 13 x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 173 Using the Horizontal Line Test Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. How To… Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function. 1. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once. 2. If there is any such line, determine that the function is not one-to-one. Example 15 Applying the Horizontal Line Test Consider the functions shown in Figure 11(a) and Figure 11(b). Are either of the functions one-to-one? Solution The function in Figure 11(a) is not one-to-one. The horizontal line shown in Figure 14 intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.) f (x) Figure 14 x The function in Figure 11(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once. Try It #12 Is the graph shown here one-to-one? y x Identifying Basic Toolkit Functions In this text, we will be exploring functions—the shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our “toolkit functions,” which form a set of basic named functions for which Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 174 CHAPTER 3 FUNCTIONS we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use x as the input variable and y = f (x) as the output variable. We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in Table 13. Name Function Toolkit Functions Graph f (x) Constant f (x) = c, where c is a constant Identity f (x) = x Absolute value f (x) = .#x . Quadratic f (x) = x2 Cubic f (x) = x3 f (x) f (x) f (x) f (x) x x x x x x –2 0 2 x –2 0 2 x –2 0 2 x –2 –1 0 1 2 f(x) 2 2 2 f(x) –2 0 2 f(x) 2 0 2 f(x) 4 1 0 1 4 x –1 –0.5 0 0.5 1 f(x) –1 –0.125 0 0.125 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 FUNCTIONS AND FUNCTION NOTATION 175 Reciprocal f (x) = 1 __ x Reciprocal squared 1 _ f (x) = x 2 Square root f (x) = √ — x Cube root f (x) = 3 √ — x f (x) f (x) f (x) f (x) x –2 –1 –0.5 0.5 1 2 x –2 –1 –0.5 0.5 1 2 x 0 1 4 f(x) –0.5 –1 –2 2 1 0.5 f(x) 0.25 1 4 4 1 0.25 f(x) 0 1 2 x –1 f(x) –1 –0.125 –0.5 0 0.125 1 0 0.5 1 x x x x Table 13 Access the following online resources for additional instruction and practice with functions. • Determine if a Relation is a Function (http://openstaxcollege.org/l/relationfunction) • Vertical Line Test (http://openstaxcollege.org/l/vertlinetest) • Introduction to Functions (http://openstaxcollege.org/l/introtofunction) • Vertical Line Test of Graph (http://openstaxcollege.org/l/vertlinegraph) • One-to-one Functions (http://openstaxcollege.org/l/onetoone) • Graphs as One-to-one Functions (http://openstaxcollege.org/l/graphonetoone) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 176 CHAPTER 3 FUNCTIONS 3.1 SECTION EXERCISES VERBAL 1. What is the difference between a relation and 2. What is the difference between the in
put and the a function? output of a function? 3. Why does the vertical line test tell us whether the 4. How can you determine if a relation is a one-to-one graph of a relation represents a function? function? 5. Why does the horizontal line test tell us whether the graph of a function is one-to-one? ALGEBRAIC For the following exercises, determine whether the relation represents a function. 6. {(a, b), (c, d), (a, c)} 7. {(a, b),(b, c),(c, c)} For the following exercises, determine whether the relation represents y as a function of x. 9. y = x 2 8. 5x + 2y = 10 10. x = y 2 11. 3x 2 + y = 14 1 __ 14. y = x 3x + 5 ______ 7x − 1 17. y = 20. x = y 3 23. x = ± √ — 1 − y 26. y 3 = x 2 12. 2x + y 2 = 6 15. x = 3y + 5 _ 7y − 1 18. x 2 + y 2 = 9 21. y = x 3 24. y = ± √ — 1 − x 13. y = −2x 2 + 40x 16. x = √ — 1 − y 2 19. 2xy =#1 22. y = √ — 1 − x 2 25. y 2 = x 2 For the following exercises, evaluate the function f at the indicated values f (−3), f (2), f (−a), −f (a), f (a + h). 28. f (x) = −5x 2 + 2x − 1 29. f (x) = √ — 2 − x + 5 27. f (x) = 2x − 5 30. f (x) = 6x − 1 ______ 5x + 2 32. Given the function g(x) =#5 − x 2, simplify 33. Given the function g(x) =#x 2 + 2x, simplify 31. f (x) = ∣#x − 1 ∣ − ∣#x + 1 ∣ g(x + h) − g(x) __ h g(x) − g(a ≠#0 34. Given the function k(t) = 2t − 1: 35. Given the function f (x) =#8 − 3x: a. Evaluate k(2). b. Solve k(t) =#7. a. Evaluate f (−2). b. Solve f (x) = −1. 36. Given the function p(c) = c 2 + c: 37. Given the function f (x) =#x 2 − 3x a. Evaluate p(−3). b. Solve p(c) = 2. 38. Given the function f (x) =# √ — x + 2 : a. Evaluate f (7). b. Solve f (x) =#4 a. Evaluate f (5). b. Solve f (x) =#4 39. Consider the relationship 3r + 2t = 18. a. Write the relationship as a function r = f (t). b. Evaluate f (−3). c. Solve f (t) = 2. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 SECTION EXERCISES 177 GRAPHICAL For the following exercises, use the vertical line test to determine which graphs show relations that are functions. 40. y 41. y 42. y 43. y 46. y 49. y x x x x 44. y 47. y 50. y x x x x 45. y 48. y 51. y x x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 178 CHAPTER 3 FUNCTIONS 52. Given the following graph 53. Given the following graph 54. Given the following graph a. Evaluate f (−1). b. Solve for f (x) = 3. a. Evaluate f (0). b. Solve for f (x) = −3. a. Evaluate f (4). b. Solve for f (x) = 1. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, determine if the given graph is a one-to-one function. 57. 56. 55. y y 21 3 4 5 x –5 –4 –3 –2 59. 21 3 4 5 x !π 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 !1 !2 !3 !4 !5 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 58. –5 –4 –3 –2 NUMERIC 21 3 4 5 x –5 –4 –3 –2 x π For the following exercises, determine whether the relation represents a function. 60. {(−1, −1),(−2, −2),(−3, −3)} 61. {(3, 4),(4, 5),(5, 6)} 62. {(2, 5),(7, 11),(15, 8),(7, 9)} For the following exercises, determine if the relation represented in table form represents y as a function of x. 63. x y 5 3 10 8 15 14 64. x y 5 3 10 8 15 8 65. x y 5 3 10 8 10 14 For the following exercises, use the function f represented in Table 14 below. x f (x) 0 74 1 28 2 1 3 53 4 56 Table 14 5 3 6 36 7 45 8 14 9 47 66. Evaluate f (3). 67. Solve f (x) = 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.1 SECTION EXERCISES 179 For the following exercises, evaluate the function f at the values f (−2), f (−1), f (0), f (1), and f (2). 68. f (x) = 4 − 2x 69. f (x) = 8 − 3x 70. f (x) = 8x 2 − 7x + 3 71. f (x) = 3 + √ — x + 3 72. f (x) = x − 2 _ x + 3 73. f (x) = 3x For the following exercises, evaluate the expressions, given functions f, g, and h: f (x) = 3x − 2 g(x) = 5 − x2 h(x) = −2x2 + 3x − 1 74. 3f (1) − 4g(−2) TECHNOLOGY 7 75. f ( ) − h(−2) _ 3 For the following exercises, graph y = x2 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 76. [−0.1, 0.1] 77. [−10, 10] 78. [−100, 100] For the following exercises, graph y = x3 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 79. [−0.1, 0.1] 81. [−100, 100] 80. [−10, 10] For the following exercises, graph y = √ viewing window. Show each graph. — x on the given viewing window. Determine the corresponding range for each 82. [0, 0.01] 83. [0, 100] 84. [0, 10,000] For the following exercises, graph y = viewing window. Show each graph. 85. [−0.001, 0.001] — 3 √ x on the given viewing window. Determine the corresponding range for each 86. [−1,000, 1,000] 87. [−1,000,000, 1,000,000] REAL-WORLD APPLICATIONS 88. The amount of garbage, G, produced by a city with population p is given by G = f (p). G is measured in tons per week, and p is measured in thousands of people. a. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function f. b. Explain the meaning of the statement f (5) = 2. 89. The number of cubic yards of dirt, D, needed to cover a garden with area a square feet is given by D = g(a). a. A garden with area 5,000 ft2 requires 50 yd3 of dirt. Express this information in terms of the function g. b. Explain the meaning of the statement g(100) = 1. 90. Let f (t) be the number of ducks in a lake t years after 1990. Explain the meaning of each statement: a. f (5) = 30 b. f (10) = 40 91. Let h(t) be the height above ground, in feet, of a rocket t seconds after launching. Explain the meaning of each statement: a. h(1) = 200 b. h(2) = 350 92. Show that the function f (x) = 3(x − 5)2 + 7 is not one-to-one. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 180 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Find the domain of a function identified by an equation using interal notation. • Gien the graph of a function find the domain and range using interal notation. • Graph piecewise-defined functions. 3.2 DOMAIN AND RANGE If you’re in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all time—I am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 1 shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these. Market Share of Horror Moveis, by Year Top-Five Grossing Horror Movies for years 2000–201 fl 350 300 250 200 150 100 50 0 8% 7% 6% 5% 4% 3% 2% 1% 0% I am Legend (2007) Hannibal (2001) The Ring (2002) The Grudge (2004) The Conjuring (2013) 2005 Figure 1 Based on data compiled by www.the-numbers.com.[3] 2002 2000 2001 2003 2004 2006 2007 2008 200 9 2010 2011 2012 2013 Finding the Domain of a Function Deﬁned by an Equation In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0. We can visualize the domain as a “holding area” that contains “raw materials” for a “function machine” and the range as another “holding area” for the machine’s products. See Figure 2. Domain a b c Function machine Figure 2 Range x y z We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [when the set includes the endpoint and a parenthesis (to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write (0, 100]. We will discuss interval notation in greater detail later. 3 The Numbers: Where Data and the Movie Business Meet. “Box Office History for Horror Movies.” http://www.the-numbers.com/market/genre/Horror. Accessed 3/24/2014 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 181 Let’s turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the function’s equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative. Before we begin, let us review the conventions of interval notation: • The smallest number from the interval is written first. • The largest number in the interval is written second, following a comma. • • Parentheses, (or), are used to signify that an endpoint value is not included, called exclusive
. Brackets, [or], are used to indicate that an endpoint value is included, called inclusive. See Figure 3 for a summary of interval notation. Inequality Interval Notation Graph on Number Line a, ∞) (−∞, a) [a, ∞) (−∞, a] (a, b) [a, b) (a, b] [a, b Figure 3 Description x is greater than a x is less than a x is greater than or equal to a x is less than or equal to a x is strictly between a and b x is between a and b, to include a x is between a and b, to include b x is between a and b, to include a and b Example 1 Finding the Domain of a Function as a Set of Ordered Pairs Find the domain of the following function: {(2, 10), (3, 10), (4, 20), (5, 30), (6, 40)}. Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs. {2, 3, 4, 5, 6} Try It #1 Find the domain of the function: {(−5, 4), (0, 0), (5, −4), (10, −8), (15, −12)} How To… Given a function written in equation form, find the domain. 1. Identify the input values. 2. Identify any restrictions on the input and exclude those values from the domain. 3. Write the domain in interval form, if possible. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 182 CHAPTER 3 FUNCTIONS Example 2 Finding the Domain of a Function Find the domain of the function f (x) = x2 − 1. Solution The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In interval form, the domain of f is (−∞, ∞). Try It #2 Find the domain of the function: f (x) = 5 − x + x 3. How To… Given a function written in an equation form that includes a fraction, find the domain. 1. Identify the input values. 2. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for x. If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve. 3. Write the domain in interval form, making sure to exclude any restricted values from the domain. Example 3 Finding the Domain of a Function Involving a Denominator Find the domain of the function f (x) = x + 1 _____ . 2 − x Solution When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x. 2 − x = 0 −x = −2 x = 2 Now, we will exclude 2 from the domain. The answers are all real numbers where x < 2 or x > 2 as shown in Figure 4. We can use a symbol known as the union, ∪, to combine the two sets. In interval notation, we write the solution: (−∞, 2) ∪ (2, ∞). –3 –2 –1 0 x < 2 or x > 2 1 2 3 ↓ ↓ (−∞, 2) ∪ (2, ∞) Figure 4 Try It #3 Find the domain of the function: f (x) = 1 + 4x ______ 2x − 1 . How To… Given a function written in equation form including an even root, find the domain. 1. Identify the input values. 2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 3. The solution(s) are the domain of the function. If possible, write the answer in interval form. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 183 Example 4 Finding the Domain of a Function with an Even Root Find the domain of the function f (x) = √ — 7 − x . Solution When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 7 − x ≥ 0 −x ≥ −7 x ≤ 7 Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to 7, or (−∞, 7]. Try It #4 Find the domain of the function f (x) = √ — 5 + 2x . Q & A… Can there be functions in which the domain and range do not intersect at all? Yes. For example, the function f (x) = − # 1 _ — x √ negative real numbers as its range. As a more extreme example, a function’s inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common. has the set of all positive real numbers as its domain but the set of all Using Notations to Specify Domain and Range In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, {x \| 10 ≤ x < 30} describes the behavior of x in set-builder notation. The braces { } are read as “the set of,” and the vertical bar \| is read as “such that,” so we would read {x \| 10 ≤ x < 30} as “the set of x-values such that 10 is less than or equal to x, and x is less than 30.” Figure 5 compares inequality notation, set-builder notation, and interval notation. 5 5 5 5 5 Inequality Notation 5 < h ≤ 10 Set-builder Notation Interval Notation {h \| 5 < h ≤ 10} (5, 10] 5 ≤ h < 10 {h \| 5 ≤ h < 10} [5, 10) 5 < h < 10 {h \| 5 < h < 10} (5, 10) h < 10 h ≥ 10 {h \| h < 10} (−∞, 10) {h \| h ≥ 10} [10, ∞) All real numbers / (−∞, ∞) Figure 5 10 10 10 10 10 10 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 184 CHAPTER 3 FUNCTIONS To combine two intervals using inequality notation or set-builder notation, we use the word “or.” As we saw in earlier examples, we use the union symbol, ∪, to combine two unconnected intervals. For example, the union of the sets {2, 3, 5} and {4, 6} is the set {2, 3, 4, 5, 6}. It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending order of numerical value. If the original two sets have some elements in common, those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is {x \| \| x \| ≥ 3} = (−∞, −3] ∪ [3, ∞) set-builder notation and interval notation Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form {x \| statement about x} which is read as, “the set of all x such that the statement about x is true.” For example, {x \| 4 < x ≤ 12} Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example, (4, 12] How To… Given a line graph, describe the set of values using interval notation. 1. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. 2. At the left end of each interval, use [with each end value to be included in the set (solid dot) or (for each excluded end value (open dot). 3. At the right end of each interval, use] with each end value to be included in the set (filled dot) or) for each excluded end value (open dot). 4. Use the union symbol ∪ to combine all intervals into one set. Example 5 Describing Sets on the Real-Number Line Describe the intervals of values shown in Figure 6 using inequality notation, set-builder notation, and interval notation. –2 –1 0 1 2 3 4 5 6 7 Figure 6 Solution To describe the values, x, included in the intervals shown, we would say, “x is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.” Inequality 1 ≤ x ≤ 3 or x > 5 Set-builder notation { x \|1 ≤ x ≤ 3 or x > 5 } Interval notation [1, 3] ∪ (5, ∞) Remember that, when writing or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 185 Try It #5 Given this figure, specify the graphed set in a. words b. set-builder notation c. interval notation Finding Domain and Range from Graphs –5 –4 –3 –2 –1 0 1 2 3 4 5 Figure 7 Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure 8. Domain y 7 6 5 4 3 2 1 –6 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 –6 –7 – 8 – 9 Range 21 3 4 5 6 x Figure 8 We can observe that the graph extends horizontally from −5 to the right without bound, so the domain is [−5, ∞). The vertical extent of the graph is all range values 5 and below, so the range is (−∞, 5]. Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range. Example 6 Finding Domain and Range from a Graph Find the domain and range of the function f whose graph is shown in Figure 9. –5 –4 –3 –2 –1 f y 1 –1 –2 – 3 – 4 –5 1 2 3 4 5 x Figure 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 186 CHAPTER 3 FUNCTIONS Solution We can observe that the horizontal extent of the graph is −3 to 1, so the domain of f is (−3, 1]. The vertical extent of the graph is 0 to −4, so the range is [−4, 0]. See Figure 10. y Domain 1 –5 –4 –3
–2 –1 f –1 –2 – 3 – 4 –5 1 2 3 4 5 x Range Figure 10 Example 7 Finding Domain and Range from a Graph of Oil Production Find the domain and range of the function f whose graph is shown in Figure 11. Alaska Crude Oil Production Th 2200 2000 1800 1600 1400 1200 1000 800 600 400 200 0 1975 1980 1985 1990 1995 2000 2005 Figure 11 (credit: modiﬁcation of work by the U.S. Energy Information Administration) [4] Solution The input quantity along the horizontal axis is “years,” which we represent with the variable t for time. The output quantity is “thousands of barrels of oil per day,” which we represent with the variable b for barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as 1973 ≤ t ≤ 2008 and the range as approximately 180 ≤ b ≤ 2010. In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines. Try It #6 Given Figure 12, identify the domain and range using interval notation. World Population Increase 100 90 80 70 60 50 40 30 20 10 0 1950 1960 1970 1980 1990 2000 Year Figure 12 Q & A… Can a function’s domain and range be the same? Yes. For example, the domain and range of the cube root function are both the set of all real numbers. 4 http://www.eia.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=MCRFPAK2&f=A. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 187 Finding Domains and Ranges of the Toolkit Functions We will now return to our set of toolkit functions to determine the domain and range of each. f (x) f (x) = c Domain: (−∞, ∞) Range: [c, c] x For the constant function f(x) = c, the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant c, so the range is the set {c} that contains this single element. In interval notation, this is written as [c, c], the interval that both begins and ends with c. Figure 13 f (x) Figure 14 f (x) Figure 15 f (x) Figure 16 f (x) Figure 17 For the identity function f(x) = x, there is no restriction on x. Both the domain and range are the set of all real numbers. Domain: (−∞, ∞) Range: (−∞, ∞) x Domain: (−∞, ∞) Range: [0, ∞) x For the absolute value function f(x) = ∣ x ∣ , there is no restriction on x. However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. Domain: (−∞, ∞) Range: [0, ∞) x For the quadratic function f(x) = x 2, the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers. Domain: (−∞, ∞) Range: (−∞, ∞) x For the cubic function f(x) = x 3, the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 188 CHAPTER 3 FUNCTIONS Domain: (−∞, 0) ∪ (0, ∞) Range: (−∞, 0) ∪ (0, ∞) x For the reciprocal function f(x) = 1 _ x , we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also write {x \| x ≠ 0}, the set of all real numbers that are not zero. Domain: (−∞, 0) ∪ (0, ∞) Range: (0, ∞) x For the reciprocal squared function f(x) = 1 __ x 2 , we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers. Domain: [0, ∞) Range: [0, ∞) x Domain: (−∞, ∞) Range: (−∞, ∞) x — x , we cannot take the For the square root function f(x) = √ square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive number x is defined to be positive, even though the square of the negative number − √ — x also gives us x. — 3 √ x the domain and range For the cube root function f(x) = include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function). f (x) Figure 18 f (x) Figure 19 f (x) Figure 20 f (x) Figure 21 How To… Given the formula for a function, determine the domain and range. 1. Exclude from the domain any input values that result in division by zero. 2. Exclude from the domain any input values that have nonreal (or undefined) number outputs. 3. Use the valid input values to determine the range of the output values. 4. Look at the function graph and table values to confirm the actual function behavior. Example 8 Finding the Domain and Range Using Toolkit Functions Find the domain and range of f (x) = 2x3 − x. Solution There are no restrictions on the domain, as any real number may be cubed and then subtracted from the result. The domain is (−∞, ∞) and the range is also (−∞, ∞). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 189 Example 9 Finding the Domain and Range Find the domain and range of f (x) = 2 ____ x + 1 . Solution We cannot evaluate the function at −1 because division by zero is undefined. The domain is (−∞, −1) ∪ (−1, ∞). Because the function is never zero, we exclude 0 from the range. The range is (−∞, 0) ∪ (0, ∞). Example 10 Finding the Domain and Range Find the domain and range of f (x) = 2 √ — x + 4 . Solution We cannot take the square root of a negative number, so the value inside the radical must be nonnegative. The domain of f (x) is [−4, ∞). x + 4 ≥ 0 when x ≥ −4 We then find the range. We know that f (−4) = 0, and the function value increases as x increases without any upper limit. We conclude that the range of f is [0, ∞). Analysis Figure 22 represents the function f . f (x) 5 4 3 2 1 f –5 –4 –3 –2 –1 –1 21 3 4 5 x Figure 22 Try It #7 Find the domain and range of f (x) = − √ — 2 − x . Graphing Piecewise-Deﬁned Functions Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function f (x) = \|x\|. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0. If we input 0, or a positive value, the output is the same as the input. f (x) = x if x ≥ 0 If we input a negative value, the output is the opposite of the input. f (x) = −x if x < 0 Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function. A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain. We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain “boundaries.” For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total income S would be 0.1S if S ≤ $10,000 and $1000 + 0.2(S − $10,000) if S > $10,000. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 190 CHAPTER 3 FUNCTIONS piecewise function A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this: f (x) = { formula 1 formula 2 formula 3 if x is in domain 1 if x is in domain 2 if x is in domain 3 In piecewise notation, the absolute value function is x −x { \|x\| = if x ≥ 0 if x < 0 How To… Given a piecewise function, write the formula and identify the domain for each interval. 1. Identify the intervals for which different rules apply. 2. Determine formulas that describe how to calculate an output from an input in each interval. 3. Use braces and if-statements to write the function. Example 11 Writing a Piecewise Function A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of 10 or more people. Write a function relating the number of people, n, to the cost, C. Solution Two di fferent formulas will be needed. For n-values under 10, C = 5n. For values of n that are 10 or greater, C = 50. { C(n) = 5n if 0 < n < 10 50 if n ≥ 10 Analysis The function is represented in Figure 23. The graph is a diagonal line from n = 0 to n = 10 and a constant after that. In this example, the two formulas agree at the meeting point where n = 10, but not all piecewise functions have this property. C (n) 50 40 30 20 10 C (n) – 25 – 20 – 15 – 10 – 5 –10 5 10 15 20 25 n Figure 23 Example 12 Working with a Piecewise Function A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer. { C(g) = 25 if 0 < g < 2 25 + 10(g − 2) if g ≥ 2 Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data. Solution To find the cost of using 1.5 gigabytes of data, C(1.5), we first look to see which part of the domain our input falls in. Because 1.5 is less th
an 2, we use the first formula. C(1.5) = $25 To fi nd the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula. C(4) = 25 + 10(4 − 2) = $45 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 DOMAIN AND RANGE 191 Analysis The function is represented in Figure 24. We can see where the function changes from a constant to a shifted and stretched identity at g = 2. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain. C(g) 50 40 30 20 10 C (g10 1 2 3 4 5 g Figure 24 How To… Given a piecewise function, sketch a graph. 1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain. 2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function. Example 13 Graphing a Piecewise Function Sketch a graph of the function. f (x) = { x ≤ 1 x2 if 3 x if 1 < x ≤ 2 x > 2 if Solution Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to inequality. Figure 25 shows the three components of the piecewise function graphed on separate coordinate systems. f (x) 5 4 3 2 1 –1 (ax) 5 4 3 2 1 –1 (b) f (x) 5 4 3 2 1 –1 (c Figure 25 ( a) f (x ) = x 2 if x ≤ 1; ( b) f (x ) = 3 if 1 < x ≤ 2; ( c) f (x ) = x if x > 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 192 CHAPTER 3 FUNCTIONS Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure 26. f (x) 5 4 3 2 1 – Figure 26 Analysis Note that the graph does pass the vertical line test even at x = 1 and x = 2 because the points (1, 3) and (2, 2) are not part of the graph of the function, though (1, 1) and (2, 3) are. Try It #8 Graph the following piecewise function. f (x) = { x3 −2 — √ x x < −1 if if −1 < x < 4 x > 4 if Q & A… Can more than one formula from a piecewise function be applied to a value in the domain? No. Each value corresponds to one equation in a piecewise formula. Access these online resources for additional instruction and practice with domain and range. • Domain and Range of Square Root Functions (http://openstaxcollege.org/l/domainsqroot) • Determining Domain and Range (http://openstaxcollege.org/l/determinedomain) • Find Domain and Range Given the Graph (http://openstaxcollege.org/l/drgraph) • Find Domain and Range Given a Table (http://openstaxcollege.org/l/drtable) • Find Domain and Range Given Points on a Coordinate Plane (http://openstaxcollege.org/l/drcoordinate) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 SECTION EXERCISES 193 3.2 SECTION EXERCISES VERBAL 1. Why does the domain differ for different functions? 2. How do we determine the domain of a function defined by an equation? 3. Explain why the domain of f (x) = from the domain of f (x is different 4. When describing sets of numbers using interval notation, when do you use a parenthesis and when do you use a bracket? 5. How do you graph a piecewise function? ALGEBRAIC For the following exercises, find the domain of each function using interval notation. 6. f (x) = −2x(x − 1)(x − 2) 7. f (x) = 5 − 2x2 8. f (x) = 3 √ — x − 2 9. f (x) = 3 − √ — 6 − 2x 10. f (x) = √ — 4 − 3x 11. f (x) = √ — x 2 + 4 12. f (x) = # 3 √ — 1 − 2x 13. f (x) = # 3 √ — x − 1 15. f (x) = #3x + 1 ______ 4x + 2 18. f (x) = # 1 ________ x2 − x − 6 16. f (x) = # — √ x + 4 _______ x − 4 19. f (x) = # 2x3 − 250 __________ x2 − 2x − 15 14. f (x) = # 9 _____ x − 6 17. f (x) = # x − 3 __________ x2 + 9x − 22 20. f (x) = # 5 _ — x − 3 √ 21. f (x) = 2x + 1_ 5 − x √ — 22. f (x √ — 23. f (x √ — 24. f (x) = #x __ x 25. f (x) = x2 − 9x_ x2 − 81 26. Find the domain of the function f (x) = √ — 2x 3 − 50x by: a. using algebra. b. graphing the function in the radicand and determining intervals on the x-axis for which the radicand is nonnegative. GRAPHICAL For the following exercises, write the domain and range of each function using interval notation. 27. y 28. y 29. –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 42 6 8 10 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 194 30. 33. 36. CHAPTER 3 FUNCTIONS y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 y –1 –1 –2 –3 –4 –5 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –5 –4 –3 –2 –5 –4 –3 –2 –10 –8 –6 –4 31. 21 3 4 5 x –5 –4 –3 –2 34. 21 3 4 5 x –5 –4 –3 –2 37. 42 6 8 10 x –10 –8 –6 –4 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 5 4 3 2 1 y –1 –1 –2 –3 –4 –5 10 8 6 4 2 –2 –2 –4 –6 –8 –10 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 32. 21 3 4 5 x –5 –4 –3 –2 35. 21 3 4 5 x –6 –5 –4 –3 –2 1 6 –6, – ) ( ) , –6 1 6 ( – 42 6 8 10 x 21 , ( 1 6 ) 21 3 4 5 6 x For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation. x + 1 if x < −2 { 38. f (x) =# −2x − 3 if x ≥ −2 { 39. f (x) =# 2x − 1 if x < 1 if x ≥ 1 1 + x { 40. f (x) =# x + 1 if x < 0 x − 1 if x > 0 { 41. f (x) =# 3 √ if x < 0 x if x ≥ 0 — { 42. f (x) =# if x < 0 x2 1 − x if x > 0 { 43. f (x) =# if x < 0 x2 x + 2 if x ≥ 0 { 44. f (x) =# x + 1 if x < 1 if x ≥ 1 x3 { 45. f (x) =# \| x \| if x < 2 if x ≥ 2 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.2 SECTION EXERCISES 195 NUMERIC For the following exercises, given each function f, evaluate f (−3), f (−2), f (−1), and f (0). { 46. f (x) =# x + 1 if x < −2 −2x − 3 if x ≥ −2 { 47. f (x) =# 1 if x ≤ −3 0 if x > −3 { 48. f (x) =# −2x 2 + 3 if x ≤ −1 if x > −1 5x − 7 For the following exercises, given each function f, evaluate f (−1), f (0), f (2), and f (4). { 49. f (x) =# 7x + 3 if x < 0 7x + 6 if x ≥ 0 { 50. f (x) =# if \| if x ≥ 2 51. f (x) =# { 5x if 3 x 2 if x < 0 if 0 ≤ x ≤ 3 x > 3 For the following exercises, write the domain for the piecewise function in interval notation. { 52. f (x) =# x + 1 if x < −2 −2x − 3 if x ≥ −2 { 53. f (x) =# x2 − 2 if x < 1 −x2 + 2 if x > 1 { 54. f (x) =# 2x − 3 if x < 0 if x ≥ 2 −3x2 TECHNOLOGY 55. Graph y = 1 __ x 2 on the viewing window [−0.5, −0.1] and [0.1, 0.5]. Determine the corresponding range for the viewing window. Show the graphs. 1 _ x on the viewing window [−0.5, −0.1] and [0.1, 0.5]. Determine the corresponding range for the 56. Graph y = viewing window. Show the graphs. EXTENSION 57. Suppose the range of a function f is [−5, 8]. What is the range of \| f (x) \|? 58. Create a function in which the range is all nonnegative real numbers. 59. Create a function in which the domain is x > 2. REAL-WORLD APPLICATIONS 60. The height h of a projectile is a function of the time t it is in the air. The height in feet for t seconds is given by the function h(t) = −16t 2 + 96t. What is the domain of the function? What does the domain mean in the context of the problem? 61. The cost in dollars of making x items is given by the function C(x) = 10x + 500. a. The fixed cost is determined when zero items are produced. Find the fixed cost for this item. b. What is the cost of making 25 items? c. Suppose the maximum cost allowed is $1500. What are the domain and range of the cost function, C(x)? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 196 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Find the aerage rate of change of a function. • Find the difference quotient of a function. • Use a graph to determine where a function is increasing decreasing or constant. 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS Gasoline costs have experienced some wild fluctuations over the last several decades. Table 1[5] lists the average cost, in dollars, of a gallon of gasoline for the years 2005–2012. The cost of gasoline can be considered as a function of year. y C(y) 2005 2.31 2006 2.62 2007 2.84 2008 3.30 2009 2.41 2010 2.84 2011 3.58 2012 3.68 Table 1 If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year. In this section, we will investigate changes such as these. Finding the Average Rate of Change of a Function The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in Table 1 did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value. Average rate of change = Change in output __ Change in input ∆y_ ∆x y2 − y1 _ x2 − x1 f (x2) − f (x1) __ − x1 = = = x2 The Greek letter ∆ (delta) signifies the change in a quantity; we read the ratio as “delta-y over delta-x” or “the change in y divided by the change in x.” Occasionally we write ∆ f instead of ∆y, which still represents the change in the function’s output value resulting from a change to its input value. It does not mean we are changing the function into some other function. In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was ∆y _ ∆x = $1.37 _ 7
years ≈ 0.196 dollars per year On average, the price of gas increased by about 19.6¢ each year. Other examples of rates of change include: • A population of rats increasing by 40 rats per week • A car traveling 68 miles per hour (distance traveled changes by 68 miles each hour as time passes) • A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon) • The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage • The amount of money in a college account decreasing by $4,000 per quarter 5 http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 197 rate of change A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are “output units per input units.” The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values. ∆y _ ∆x = f (x2) − f (x1) __ x2 − x1 How To… Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values x1 and x2. 1. Calculate the difference y2 − y1 = ∆y. 2. Calculate the difference x2 − x1 = ∆x. ∆y _ . 3. Find the ratio ∆x Example 1 Computing an Average Rate of Change Using the data in Table 1, find the average rate of change of the price of gasoline between 2007 and 2009. Solution In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is ∆y ___ ∆x = y2 − y1 _ x2 − x1 $2.41 − $2.84 __ 2009 − 2007 −$0.43 ______ 2 years = = Analysis Note that a decrease is expressed by a negative change or “negative increase.” A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases. = −$0.22 per year Try It #1 Using the data in Table 1 at the beginning of this section, find the average rate of change between 2005 and 2010. Example 2 Computing Average Rate of Change from a Graph Given the function g (t) shown in Figure 1, find the average rate of change on the interval [−1, 2]. g(t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 198 CHAPTER 3 FUNCTIONS Solution At t = −1, Figure 2 shows g (−1) = 4. At t = 2, the graph shows g (2) = 1. g(t) ∆g(t) = –3 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 ∆t = 3 t Figure 2 The horizontal change ∆ t = 3 is shown by the red arrow, and the vertical change ∆ g(t) = −3 is shown by the turquoise arrow. The average rate of change is shown by the slope of the red line segment. The output changes by –3 while the input changes by 3, giving an average rate of change of 1 − 4 _______ = 2 − (−1) −3 ___ 3 = −1 Analysis Note that the order we choose is very important. If, for example, we use answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as ( x1 , y1 , we will not get the correct ). ) and ( x2 , y2 y2 − y1 _ x1 − x2 Example 3 Computing Average Rate of Change from a Table After picking up a friend who lives 10 miles away and leaving on a trip, Anna records her distance from home over time. The values are shown in Table 2. Find her average speed over the first 6 hours. t (hours) D(t) (miles) 0 10 1 55 2 90 3 153 4 214 5 240 6 282 7 300 Table 2 Solution Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours. The average speed is 47 miles per hour. 292 − 10 _______ = 6 − 0 282___ 6 = 47 Analysis Because the speed is not constant, the average speed depends on the interval chosen. For the interval [2, 3], the average speed is 63 miles per hour. Computing Average Rate of Change for a Function Expressed as a Formula Example 4 Compute the average rate of change of f (x) = x 2 − #1 __ on the interval [2, 4]. x Solution We can start by computing the function values at each endpoint of the interval. 1 __ f (2) = 22 − 2 1 __ f(4) = 42 − 4 1 __ = 4 − 2 7 __ = 2 1 __ = 16 − 4 = 63__ 4 Now we compute the average rate of change. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 199 Average rate of change = f (4) − f (2)_ 4 − 2 7 63 __ __ − 4 2 _ 4 − 2 49 __ 4 _ 2 49__ 8 = = = Try It #2 Find the average rate of change of f (x) = x − 2 √ — x on the interval [1, 9]. Example 5 Finding the Average Rate of Change of a Force The electrostatic force F, measured in newtons, between two charged particles can be related to the distance between 2 _ the particles d, in centimeters, by the formula F(d) = d 2 . Find the average rate of change of force if the distance between the particles is increased from 2 cm to 6 cm. Solution We are computing the average rate of change of F (d) = 2 __ d 2 on the interval [2, 6]. Average rate of change = = F(6) − F(2) __ 6 − 2 2 2 _ _ 62 − 22 _ 6 − 2 2 −# #2 __ __ 36 4 _ 4 − #16 __ 36 _ 4 = − #1 __ 9 = = Simplify. Combine numerator terms. Simplify. The average rate of change is − #1 __ newton per centimeter. 9 Example 6 Finding an Average Rate of Change as an Expression Find the average rate of change of g(t) = t2 + 3t + 1 on the interval [0, a]. The answer will be an expression involving a in simplest form. Solution We use the average rate of change formula. g(a) − g(0)_ a − 0 Average rate of change = Evaluate. = (a2 + 3a + 1) − (02 + 3(0) + 1) ___ a − 0 Simplify. = a2 + 3a + 1 − 1 _____________ a a(a + 3) _______ a = a + 3 = Simplify and factor. Divide by the common factor a. This result tells us the average rate of change in terms of a between t = 0 and any other point t = a. For example, on the interval [0, 5], the average rate of change would be 5 + 3 = 8. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 200 CHAPTER 3 FUNCTIONS Try It #3 Find the average rate of change of f (x) = x2 + 2x − 8 on the interval [5, a] in simplest forms in terms of a. Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3 shows examples of increasing and decreasing intervals on a function. Increasing –5 –4 –3 –2 f(x) 20 16 12 8 4 –1 –4 –8 –12 –16 –20 Decreasing 21 3 4 5 x Increasing f(b) > f(a) where b > a f(b) < f(a) where b > a f(b) > f(a) where b > a Figure 3 The function f(x ) = x 3 − 12x is increasing on (−∞, −2) ∪ (2, ∞) and is decreasing on (−2, 2). While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form is “local minima.” Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The singular form is “extremum.”) Often, the term local is replaced by the term relative. In this text, we will use the term local. Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the function’s entire domain. For the function whose graph is shown in Figure 4, the local maximum is 16, and it occurs at x = −2. The local minimum is −16 and it occurs at x = 2. Local maximum = 16 occurs at x = –2 f(x) Increasing (–2, 16) 20 16 12 8 4 Decreasing –5 –4 –3 –2 –1 –4 –8 –12 –16 –20 21 3 4 5 x Increasing (2, –16) Local minimum = –16 occurs at x = 2 Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 201 To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 5 illustrates these ideas for a local maximum. f(x) f(b) Local maximum Increasing function Decreasing function a b c Figure 5 Deﬁnition of a local maximum x These observations lead us to a formal definition of local extrema. local minima and local maxima A function f is an increasing function on an open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a. A function f is a decreasing function on an open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a. A function f has a local maximum at x = b if there exists an interval (a, c) with a < b < c such that, for any x in the interval (a, c), f (x) ≤ f (b). Likewise, f has a local minimum at x = b if there exists an interval (a, c) with a < b < c such that, for any x in the interval (a, c), f (x) ≥ f (b). Example 7 Finding Increasing and Decreasing Intervals on a Grap
h Given the function p(t) in Figure 6, identify the intervals on which the function appears to be increasing. p 5 4 3 2 1 –1 1 2 3 4 5 6 t –1 –2 Figure 6 Solution We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from t = 1 to t = 3 and from t = 4 on. In interval notation, we would say the function appears to be increasing on the interval (1, 3) and the interval (4, ∞). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 202 CHAPTER 3 FUNCTIONS Analysis Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at t = 1, t = 3, and t = 4. These points are the local extrema (two minima and a maximum). Example 8 Finding Local Extrema from a Graph x 2 __ __ + Graph the function f (x) = . Then use the graph to estimate the local extrema of the function and to determine 3 x the intervals on which the function is increasing. Solution Using technology, we find that the graph of the function looks like that in Figure 7. It appears there is a low point, or local minimum, between x = 2 and x = 3, and a mirror-image high point, or local maximum, somewhere between x = −3 and x = −2. f(x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 7 Analysis Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 8 provides screen images from two different technologies, showing the estimate for the local maximum and minimum. y 6 4 2 0 –2 2.4494898, 1.6329932 2 4 6 x Maximum X = –2.449491 Y = –1.632993 (a) (b) Figure 8 Based on these estimates, the function is increasing on the interval (−∞, −2.449) and (2.449, ∞). Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing — approximation algorithms used by each. (The exact location of the extrema is at ± √ 6 , but determining this requires calculus.) Try It #4 Graph the function f (x) = x3 − 6x2 − 15x + 20 to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 203 Example 9 Finding Local Maxima and Minima from a Graph For the function f whose graph is shown in Figure 9, find all local maxima and minima. y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 –5 –4 –3 –2 21 3 4 5 x f Figure 9 Solution Observe the graph of f. The graph attains a local maximum at x = 1 because it is the highest point in an open interval around x = 1. The local maximum is the y-coordinate at x = 1, which is 2. The graph attains a local minimum at x = −1 because it is the lowest point in an open interval around x = −1. The local minimum is the y-coordinate at x = −1, which is −2. Analyzing the Toolkit Functions for Increasing or Decreasing Intervals We will now return to our toolkit functions and discuss their graphical behavior in Figure 10, Figure 11, and Figure 12. Function Increasing/Decreasing Example Constant Function f(x) = c Neither increasing nor decreasing Identity Function f(x) = x Increasing Quadratic Function f(x) = x2 Increasing on (0, ∞) Decreasing on (−∞, 0) Minimum at x = 0 Figure 10 y y y x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 204 CHAPTER 3 FUNCTIONS Function Increasing/Decreasing Example Cubic Function f(x) = x3 Increasing Reciprocal f(x) = 1 __ x Decreasing (−∞, 0) ∪ (0, ∞) y y y Reciprocal Squared f(x) = 1 _ x2 Increasing on (−∞, 0) Decreasing on (0, ∞) Figure 11 Function Increasing/Decreasing Example y Cube Root — x f(x) = 3 √ Increasing y y Square Root f(x) = √ — x Increasing on (0, ∞) Absolute Value f(x) = ∣ x ∣ Increasing on (0, ∞) Decreasing on (−∞, 0) Figure 12 x x x x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 RATES OF CHANGE AND BEHAVIOR OF GRAPHS 205 Use A Graph to Locate the Absolute Maximum and Absolute Minimum There is a difference between locating the highest and lowest points on a graph in a region around an open interval (locally) and locating the highest and lowest points on the graph for the entire domain. The y-coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively. To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function. See Figure 13. y 5 4 3 2 1 f Absolute maximum is f (2) = 2 –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 21 3 4 5 x Absolute minimum is f (0) = −2 Figure 13 Not every function has an absolute maximum or minimum value. The toolkit function f (x) = x3 is one such function. absolute maxima and minima The absolute maximum of f at x = c is f (c) where f (c) ≥ f (x) for all x in the domain of f. The absolute minimum of f at x = d is f (d) where f (d) ≤ f (x) for all x in the domain of f. Example 10 Finding Absolute Maxima and Minima from a Graph For the function f shown in Figure 14, find all absolute maxima and minima. y 20 16 12 8 4 –1 –4 –8 –12 –16 –20 –5 –4 –3 –2 f 21 3 4 5 x Figure 14 Solution Observe the graph of f. The graph attains an absolute maximum in two locations, x = −2 and x = 2, because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at x = −2 and x = 2, which is 16. The graph attains an absolute minimum at x = 3, because it is the lowest point on the domain of the function’s graph. The absolute minimum is the y-coordinate at x = 3, which is −10. Access this online resource for additional instruction and practice with rates of change. • Average Rate of Change (http://openstaxcollege.org/l/aroc) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 206 CHAPTER 3 FUNCTIONS 3.3 SECTION EXERCISES VERBAL 1. 3. d diff 2. fab bc fac 4. y=x ALGEBRAIC fifi bh 5. f x=x 6. hx=−x − 8. g x=x −− 11. __ − k t=t+ t 9. y= x 7. 10. qx=x− + t − t t + pt= − 12. f x=x −b 13. g x=x − 14. px=x++h 15. kx=x−+h 16. f x=x +xx+h 17. gx=x−xx+h 18. at= +h t+ 19. bx= +h x+ 20. jx=x +h 21. rt=t +h 22. =+ 23. =+ + 26. f(x=x−xx, x+h 29. = + 24. 27. = = , + + 25. 28. = = + + + 30. = + 31. = + + = = GRAPHICAL fFigure 15 32. x=x= 33. x=x= y – x Figure 15 6 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.3 SECTION EXERCISES 207 y – – – – – – – – – – 35. x – – – – y – – – – – – 36 34. 37. 38 – – – – – – – – – – – – Figure 16 – – – 39. 40. f Figure 17. 41. 42 Figure 16 NUMERIC Figure 17 43. Table 3 ab Year Sales (millions of dollars) Table 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 208 CHAPTER 3 FUNCTIONS 44. Table 4 ab Year Population (thousands) Table 4 TECHNOLOGY 45. f x=x −x + 47. 49. — t+ gt=t √ mx=x +x −x −x+ EXTENSION 51. ThfFigure 18 46. 48. 50. hx=x +x +x +x − kt=t −t nx=x −x +x −x+ a. b. c. d. = = Figure 18 52. f x= c x f c − " __ 53. f(x)= " __ .b x fb __ − REAL-WORLD APPLICATIONS 54. 56. dt=ttdt t=t= 55. o fi 57. ThFigure 19 Time (days) Figure 19 t t=t= Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 209 LEARNING OBJECTIVES In this section, you will: • Combine functions using algebraic operations and find their domains. • Create a new function by composition of functions and find its domain. • aluate composite functions. • ecompose a composite function into its component functions. 3.4 COMPOSITION OF FUNCTIONS Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day. Using descriptive variables, we can notate these two functions. The function C(T) gives the cost C of heating a house for a given average daily temperature in T degrees Celsius. The function T(d) gives the average daily temperature on day d of the year. For any given day, Cost = C(T(d)) means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature T(d). For example, we could evaluate T(5) to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write C(T(5)). Cost for the temperature C(T(5)) Temperature on day 5 By combining these two relationships into one function, we have performed function composition, which is the focus of this section. Combining Functions Using Algebraic Operations Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function. Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If w(y) is the wife’s income and h(y) is the husband’s income in year y, and we want T to represent the total income, then we can define a new function. T(y) = h(y) + w(y) If this holds true for every year, then we can focus on the relation between the functions without
reference to a year and write T = h + w Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 210 CHAPTER 3 FUNCTIONS For two functions f (x) and g(x) with real number outputs, we define new functions f + g, f − g, fg, and relations f _ g by the (f + g)(x) = f (x) + g(x) (f − g)(x) = f (x) − g(x) (fg)(x) = f (x)g(x) f g ) (x) = (# _ f (x)_ g(x) where g(x) ≠ 0 Example 1 Performing Algebraic Operations on Functions g _ ) (x), given f (x) = x − 1 and g(x) = x 2 − 1. Are they the same Find and simplify the functions (g − f )(x) and (# f function? Solution Begin by writing the general form, and then substitute the given functions. (g − f )(x) = g(x) − f (x) (g − f )(x) = x 2 − 1 − (x − 1) (g − f )(x) = x 2 − x (g − f )(x) = x(x − 1) x2 − 1_ x − 1 g(x)_ f (x) g _ ) (x) = (# f g _ ) (x) = (# f g _ ) (x) = (# f g _ ) (x) = x + 1 (# f (x + 1)(x − 1) ____________ x − 1 where x ≠ 1 where x ≠ 1 where x ≠ 1 No, the functions are not the same. g _ ) (x), the condition x ≠ 1 is necessary because when x = 1, the denominator is equal to 0, which makes Note: For (# f the function undefined. Try It #1 Find and simplify the functions (fg)(x) and (f − g)(x). f (x) = x − 1 and g(x) = x 2 − 1 Are they the same function? Create a Function by Composition of Functions Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation: ( f#∘#g)(x) = f (g(x)) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 211 We read the left-hand side as “f composed with g at x,” and the right-hand side as “ f of g of x.” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol ∘ is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases f (g(x)) ≠ f (x)g(x). It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function g takes the input x first and yields an output g(x). Then the function f takes g(x) as an input and yields an output f (g(x)). g(x), the output of g is the input of f (f ° g)(x) = f(g(x)) x is the input of g In general, f#∘#g and g#∘#f are different functions. In other words, in many cases f ( g(x)) ≠ g(f (x)) for all x. We will also see that sometimes two functions can be composed only in one specific order. For example, if f (x) = x2 and g(x) = x + 2, then but f (g(x)) = f (x + 2) = (x + 2)2 = x2 + 4x + 4 g(f (x)) = g(x2) = x2 + 2 These expressions are not equal for all values of x, so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value x = − #1 _ . 2 Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs. composition of functions When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input x and functions f and g, this action defines a composite function, which we write as f#∘#g such that (f#∘#g)(x) = f (g(x)) The domain of the composite function f#∘#g is all x such that x is in the domain of g and g(x) is in the domain of f. It is important to realize that the product of functions fg is not the same as the function composition f (g(x)), because, in general, f (x)g(x) ≠ f (g(x)). Example 2 Determining whether Composition of Functions is Commutative Using the functions provided, find f (g(x)) and g(f (x)). Determine whether the composition of the functions is commutative. f (x) = 2x + 1 g(x) = 3 − x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 212 CHAPTER 3 FUNCTIONS Solution Let’s begin by substituting g(x) into f (x). Now we can substitute f (x) into g(x). f (g(x)) = 2(3 − x) + 1 = 6 − 2x + 1 = 7 − 2x g(f (x)) = 3 − (2x + 1) = 3 − 2x − 1 = − 2x + 2 We find that g(f (x)) ≠ f (g(x)), so the operation of function composition is not commutative. Example 3 Interpreting Composite Functions The function c(s) gives the number of calories burned completing s sit-ups, and s(t) gives the number of sit-ups a person can complete in t minutes. Interpret c(s(3)). Solution The inside expression in the composition is s(3). Because the input to the s-function is time, t = 3 represents 3 minutes, and s(3) is the number of sit-ups completed in 3 minutes. Using s(3) as the input to the function c(s) gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups). Example 4 Investigating the Order of Function Composition Suppose f (x) gives miles that can be driven in x hours and g(y) gives the gallons of gas used in driving y miles. Which of these expressions is meaningful: f (g(y)) or g(f (x))? Solution The function y = f (x) is a function whose output is the number of miles driven corresponding to the number of hours driven. number of miles = f (number of hours) The function g(y) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means: number of gallons = g (number of miles) The expression g(y) takes miles as the input and a number of gallons as the output. The function f (x) requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression f (g(y)) is meaningless. The expression f (x) takes hours as input and a number of miles driven as the output. The function g(y) requires a number of miles as the input. Using f (x) (miles driven) as an input value for g(y), where gallons of gas depends on miles driven, does make sense. The expression g(f (x)) makes sense, and will yield the number of gallons of gas used, g, driving a certain number of miles, f (x), in x hours. Q & A… Are there any situations where f (g(y)) and g(f (x)) would both be meaningful or useful expressions ? Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order. Try It #2 The gravitational force on a planet a distance r from the sun is given by the function G(r). The acceleration of a planet subjected to any force F is given by the function a(F). Form a meaningful composition of these two functions, and explain what it means. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 213 Evaluating Composite Functions Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function. Evaluating Composite Functions Using Tables When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function. Example 5 Using a Table to Evaluate a Composite Function Using Table 1, evaluate f (g(3)) and g(f (3)). x 1 2 3 4 f (x) 6 8 3 1 Table 1 g(x) 3 5 2 7 Solution To evaluate f (g(3)), we start from the inside with the input value 3. We then evaluate the inside expression g(3) using the table that defines the function g: g(3) = 2. We can then use that result as the input to the function f, so g(3) is replaced by 2 and we get f (2). Then, using the table that defines the function f, we find that f (2) = 8. g(3) = 2 f (g(3)) = f (2) = 8 To evaluate g(f (3)), we first evaluate the inside expression f (3) using the first table: f (3) = 3. Then, using the table for g, we can evaluate g(f (3)) = g(3) = 2 Table 2 shows the composite functions f#∘#g and g#∘#f as tables. x 3 g(x) 2 f (g(x)) 8 Table 2 f (x) 3 g(f (x)) 2 Try It #3 Using Table 1, evaluate f (g(1)) and g(f (4)). Evaluating Composite Functions Using Graphs When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read
the input and output values, but this time, from the x- and y-axes of the graphs. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 214 CHAPTER 3 FUNCTIONS How To… Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs. 1. Locate the given input to the inner function on the x-axis of its graph. 2. Read off the output of the inner function from the y-axis of its graph. 3. Locate the inner function output on the x-axis of the graph of the outer function. 4. Read the output of the outer function from the y-axis of its graph. This is the output of the composite function. Example 6 Using a Graph to Evaluate a Composite Function Using Figure 1, evaluate f (g(1)). g(x) f(x) 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 321 1 –2 –3 –4 –5 321 4 5 6 7 x (a) (b) Figure 1 Solution To evaluate f (g(1)), we start with the inside evaluation. See Figure 2. g(x) f (x) 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 (1, 3) 321 4 5 6 7 x g(1) = 3 Figure 2 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 (3, 6) 321 4 5 6 7 x f (3) = 6 We evaluate g(1) using the graph of g(x), finding the input of 1 on the x-axis and finding the output value of the graph at that input. Here, g(1) = 3. We use this value as the input to the function f. f (g(1)) = f (3) We can then evaluate the composite function by looking to the graph of f (x), finding the input of 3 on the x-axis and reading the output value of the graph at this input. Here, f (3) = 6, so f ( g(1)) = 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 215 Analysis Figure 3 shows how we can mark the graphs with arrows to trace the path from the input value to the output value. g(x) f(x) –10 –8 –6 –4 10 8 6 4 2 –2–2 –4 –6 –8 –10 10 8 6 4 2 –2–2 –4 –6 –8 –10 42 6 8 10 x 42 6 8 10 x –10 –8 –6 –4 Figure 3 Try It #4 Using Figure 1, evaluate g(f (2)). Evaluating Composite Functions Using Formulas When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression. While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition f ( g(x)). To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like f (t) = t 2 − t, we substitute the value inside the parentheses into the formula wherever we see the input variable. How To… Given a formula for a composite function, evaluate the function. 1. Evaluate the inside function using the input value or variable provided. 2. Use the resulting output as the input to the outside function. Example 7 Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input Given f (t) = t2 − t and h(x) = 3x + 2, evaluate f (h(1)). Solution Because the inside expression is h(1), we start by evaluating h(x) at 1. Then f (h(1)) = f (5), so we evaluate f (t) at an input of 5. h(1) = 3(1) + 2 h(1) = 5 f (h(1)) = f (5) f (h(1)) = 52 − 5 f (h(1)) = 20 Analysis specific numerical values. It makes no difference what the input variables t and x were called in this problem because we evaluated for Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 216 CHAPTER 3 FUNCTIONS Try It #5 Given f (t) = t 2 − t and h(x) = 3x + 2, evaluate a. h(f (2)) b. h(f (−2)) Finding the Domain of a Composite Function As we discussed previously, the domain of a composite function such as f#∘#g is dependent on the domain of g and the domain of f. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as f ∘ g. Let us assume we know the domains of the functions f and g separately. If we write the composite function for an input x as f (g(x)), we can see right away that x must be a member of the domain of g in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that g(x) must be a member of the domain of f, otherwise the second function evaluation in f (g(x)) cannot be completed, and the expression is still undefined. Thus the domain of f ∘ g consists of only those inputs in the domain of g that produce outputs from g belonging to the domain of f. Note that the domain of f composed with g is the set of all x such that x is in the domain of g and g(x) is in the domain of f. domain of a composite function The domain of a composite function f (g(x)) is the set of those inputs x in the domain of g for which g(x) is in the domain of f. How To… Given a function composition f (g(x)), determine its domain. 1. Find the domain of g. 2. Find the domain of f. 3. Find those inputs x in the domain of g for which g(x) is in the domain of f. That is, exclude those inputs x from the domain of g for which g(x) is not in the domain of f. The resulting set is the domain of f#∘#g. Example 8 Finding the Domain of a Composite Function Find the domain of (f#∘#g)(x) where f (x) = and g(x) = 5 ____ x − 1 4 _____ 3x − 2 2 __ Solution The domain of g(x) consists of all real numbers except x = , since that input value would cause us to divide by 3 0. Likewise, the domain of f consists of all real numbers except 1. So we need to exclude from the domain of g(x) that value of x for which g(x) = 1. 4 _____ 3x − 2 = 1 4 = 3x − 2 6 = 3x x = 2 2 __ So the domain of f#∘#g is the set of all real numbers except and 2. This means that 3 We can write this in interval notation as 2 __ or x ≠ 2 x ≠ 3 2 2 __ __ , 2 ) ∪ (2, ∞) ) ∪ (# (#− ∞, 3 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 217 Example 9 Finding the Domain of a Composite Function Involving Radicals Find the domain of (f#∘#g)(x) where f (x) = √ — x + 2 and g(x) = √ — 3 − x Solution Because we cannot take the square root of a negative number, the domain of g is (−∞, 3]. Now we check the domain of the composite function (f#∘#g)(x) = √ — 3 − x + 2 or (f#∘#g)(x) = √ — 5 − x The domain of this function is (−∞, 5]. To find the domain of f#∘#g, we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since (−∞, 3] is a proper subset of the domain of f#∘#g. This means the domain of f#∘#g is the same as the domain of g, namely, (−∞, 3]. Analysis This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of f#∘#g can contain values that are not in the domain of f, though they must be in the domain of g. Try It #6 Find the domain of (f#∘#g)(x) where f (x) = 1 ____ x − 2 and g(x) = √ — x + 4 Decomposing a Composite Function into its Component Functions In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient. Example 10 Decomposing a Function Write f (x) = √ 5 − x 2 as the composition of two functions. — Solution We are looking for two functions, g and h, so f (x) = g(h(x)). To do this, we look for a function inside a function in the formula for f (x). As one possibility, we might notice that the expression 5 − x2 is the inside of the square root. We could then decompose the function as h(x) = 5 − x2 and g(x) = √ — x We can check our answer by recomposing the functions. g(h(x)) = g(5 − x2) = √ — 5 − x2 Try It #7 Write f (x) = 4 __ — 4 + x2 3 − √ as the composition of two functions. Access these online resources for additional instruction and practice with composite functions. • Composite Functions (http://openstaxcollege.org/l/compfunction) • Composite Function Notation Application (http://openstaxcollege.org/l/compfuncnot) • Composite Functions Using Graphs (http://openstaxcollege.org/l/compfuncgraph) • Decompose Functions (http://openstaxcollege.org/l/decompfunction) • Composite Function Values (http://openstaxcollege.org/l/compfuncvalue) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 218 CHAPTER 3 FUNCTIONS 3.4 SECTION EXERCISES VERBAL 1. 2. f∘g f g 3. 4. f∘g ALGEBRAIC 5. fx=x+xgx=−xf+g 6. fx=−x+xgx=f+g f−gfg, f g f−gfg, f g 7. fx=x +xgx= f+g x f−gfg, f g 9. fx=x gx=√ f g f−gfg, 8. fx= gx= x− f g f+gf−gfg, −x ,fi — x− f+g 10. fx=√ — g x gx=x− f 11. fx=x+gx=x− a. fg b. fgx c gfx d. g"∘"gx e. f"∘"f − o fif gxgfx 12. fx=x +gx=√ — x+ 13. fx=√ — x +gx=x + 14. fx=xgx=x+ 15. fx= √ — x gx= x+ x 16. fx= x − x + gx= 17. fx= gx= " ___ x− x + o fif ghx 18. fx=x +gx=x−hx=√ — x x hx=x+ 19. fx=x +gx= x ,gx=x− 20. fx= a. f"∘"gx b. f"∘"gx c. g"∘"f x d. g"∘"f x e. ( f g ) x — −x gx=− " 21. fx=√ a. g"∘"f x b. g"∘"f x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 SECTION EXERCISES 219 22. fx= gx= − x x +x a. g"∘"f x b. g"∘"f 23. px = " mx=x − — x √ a. px mx b. pmx c. mpx 24. qx= hx=x − — x √ x gx=√ 25. fx= — x− f"∘"gx a. qx hx b. qhx c. hqx , fif xgxhx=f x 26. hx=x+ 27. hx=x− 30. hx=+ √ — x √ 31. hx= _______ x− 28. hx= x − 32. hx= x −− 29. hx= x+ √ 33. hx= _______ x− x+ 34. hx = ( +x −x ) 35. hx=√ — x+ 36. hx=x− 37. hx= √ — x− 38. hx=x + 39. hx= x− 40. hx=( ) x− 41. hx=√ x− x+ GRAPHICAL fFigure 4 g,Figure 5 f(x) – – – x (x) – – – x Figure 4 Figure 5 42. fg 46. ff 43. fg 47. ff 44. gf 48. gg 45. gf 49. gg Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 220 CHAPTER 3 FUNCTIONS f x,Figure 6 gx,Figure 7hx,Figure 8 fx f x f x gx hx Figure 6 Figure 7 Figure 8 50. gf 51. gf 52. fg 53. fg 54. fh 55. hf 56. fgh 57. fgf− NUMERIC fgTable 3 x fx
gx Table 3 58. fg 62. ff 59. fg 63. ff 60. gf 64. gg 61. gf 65. gg fgTable 4 − − x f (x) g(x) − − − Table 4 − − − 66. f"∘"g 69. g"∘"f 67. f"∘"g 70. g"∘"g 68. g"∘"f 71. f"∘"f o fif ggf 72. fx=x+gx=−x 73. fx=x+gx=−x 74. fx=√ — x+gx=−x 75. fx= gx=x+ _ x +" f x=x+gx=x +fi 76. fg 77. fgx 78. g(f− 79. g"∘"g x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 SECTION EXERCISES 221 EXTENSIONS f x=x+x= √ — x − 80. f"∘"gxg"∘"fx 81. f"∘"gg"∘"f 82. g"∘"fx 83. f"∘"gx __ 84. fx= x a. f"∘"fx b. f"∘"f xf F x=x+f x=xgx=x + 85. g"∘"fx=Fx 86. f"∘"gx=Fx , fif x=x+x ≥ gx=√ — x − 87. f"∘"gg"∘"f 88. g"∘"faf"∘"ga 89. f"∘"gg"∘"f REAL-WORLD APPLICATIONS 90. ThDp p. Th Cxx a. DC b. CD c. DCx= d. CDp= 92. ff xs. Thff Px x 94. rt=t+ t 96. Thr ___ "V " . π √ VrV= ftt Vt=+t a. rVt b. exact 10 in 91. ThAd d mi em. Th ftt m(t) a. Am b. mA c. Amt= d. mAd= 93. — t+ rt= t= √ 95. ft 97. Th NT=T −T+<T< T Tt=t+t a. NTt b. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 222 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: • ecreate the library of functions. • Graph and describe functions using ertical and horiontalshifts. • Graph and describe functions using reflections about the x-ais and the y-ais. • Graph and describe functions using compressions and stretches. • Combine transformations. • Gien a graph or description write the equation as a transformation of a basic function. • Perform transformations on a gien graph. 3.5 TRANSFORMATION OF FUNCTIONS Figure 1 (credit: "Misko"/Flickr) We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror? Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or processes in the real world. In this section, we will take a look at several kinds of transformations. Graphing Functions Using Vertical and Horizontal Shifts Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, and equations. One method we can employ is to adapt the basic graphs of the toolkit functions to build new models for a given scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to solve. Identifying Vertical Shifts One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function g (x) = f (x) + k, the function f (x) is shifted vertically k units. See Figure 2 for an example. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 223 f(x) f(x)+1 f(x) 3 4 5 21 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 2 Vertical shift by k = 1 of the cube root function f ( x) = 3 √ — x . To help you visualize the concept of a vertical shi ft , consider that y = f (x). Therefore, f (x) + k is equivalent to y + k. Every unit of y is replaced by y + k, so the y-value increases or decreases depending on the value of k. The result is a shift upward or downward. vertical shift Given a function f (x), a new function g(x) = f (x) + k, where k is a constant, is a vertical shift of the function f (x). All the output values change by k units. If k is positive, the graph will shift up. If k is negative, the graph will shift down. Example 1 Adding a Constant to a Function To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. Figure 3 shows the area of open vents V (in square feet) throughout the day in hours after midnight, t. During the summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open vents by 20 square feet throughout the day and night. Sketch a graph of this new function. V 300 250 200 150 100 50 – –4 50– 4 8 12 16 20 24 28 t Figure 3 Solution We can sketch a graph of this new function by adding 20 to each of the output values of the original function. This will have the effect of shifting the graph vertically up, as shown in Figure 4. V 300 250 240 200 150 100 50 20 – –4 50– Up 20 4 8 12 16 20 24 28 t Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 224 CHAPTER 3 FUNCTIONS Notice that in Figure 4, for each input value, the output value has increased by 20, so if we call the new function S(t), we could write S(t) = V(t) + 20 This notation tells us that, for any value of t, S(t) can be found by evaluating the function V at the same input and then adding 20 to the result. This defines S as a transformation of the function V, in this case a vertical shift up 20 units. Notice that, with a vertical shift, the input values stay the same and only the output values change. See Table 1. t V(t) S(t) 0 0 20 8 0 20 17 220 240 19 0 20 24 0 20 10 220 240 Table 1 How To… Given a tabular function, create a new row to represent a vertical shift. 1. Identify the output row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down. Example 2 Shifting a Tabular Function Vertically A function f (x) is given in Table 2. Create a table for the function g(x) = f (x) − 3. x f (x) 2 1 4 3 Table 2 6 7 8 11 Solution The formula g (x) = f (x) − 3 tells us that we can find the output values of g by subtracting 3 from the output values of f. For example: Given Given transformation f (2) = 1 g(x) = f (x) − 3 g(2) = f (2) − 3 = 1 − 3 = −2 Subtracting 3 from each f (x) value, we can complete a table of values for g(x) as shown in Table 3. x f (x) g(x) 2 1 −2 4 3 0 Table 3 6 7 4 8 11 8 Analysis As with the earlier vertical shift, notice the input values stay the same and only the output values change. Try It #1 The function h(t) = –4.9t2 + 30t gives the height h of a ball (in meters) thrown upward from the ground after t seconds. Suppose the ball was instead thrown from the top of a 10-m building. Relate this new height function b(t) to h(t), and then find a formula for b(t). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 225 Identifying Horizontal Shifts We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift, shown in Figure 5. f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 f(x+1) f(x) 21 3 4 5 x Figure 5 Horizontal shift of the function f (x) = 3 √ — x . Note that h = +1 shifts the graph to the left, that is, towards negative values of x. For example, if f (x) = x2, then g(x) = (x − 2)2 is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in f. horizontal shift Given a function f, a new function g(x) = f (x − h), where h is a constant, is a horizontal shift of the function f. If h is positive, the graph will shift right. If h is negative, the graph will shift left. Example 3 Adding a Constant to an Input Returning to our building airflow example from Figure 3, suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function. Solution We can set V(t) to be the original program and F (t) to be the revised program. V(t) = the original venting plan F(t) = starting 2 hrs sooner In the new graph, at each time, the airflow is the same as the original function V was 2 hours later. For example, in the original function V, the airflow starts to change at 8 a.m., whereas for the function F, the airflow starts to change at 6 a.m. The comparable function values are V(8) = F(6). See Figure 6. Notice also that the vents first opened to 220 ft2 at 10 a.m. under the original plan, while under the new plan the vents reach 220 ft2 at 8 a.m., so V(10) = F(8). In both cases, we see that, because F (t) starts 2 hours sooner, h = −2. That means that the same output values are reached when F (t) = V(t − (−2)) = V(t + 2). V 300 250 240 200 150 100 50 –2 0 50– – –4 Left 2 4 8 12 16 20 24 28 t Figure 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 226 CHAPTER 3 FUNCTIONS Analysis Note that V(t + 2) has the effect of shifting the graph to the left. Horizontal changes or “inside changes” affect the domain of a function (the input) instead of the range and often seem counterintuitive. The new function F(t) uses the same outputs as V(t), but matches those outputs to inputs 2 hours earlier than those of V(t). Said another way, we must add 2 hours to the input of V to find the corresponding output for F : F(t) = V(t + 2). How To… Given a tabular function, create a new row to represent a horizontal shift. 1. Identify the input row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each input cell. Example 4 Shifting a Tabular Function Horizontally A function f (x) is given in Table 4. Create a table for the function g (x) = f (x − 3). x f (x) 2 1 4 3 Table 4 6 7 8 11 Solution The formula g(x) = f (x − 3) tells us t
hat the output values of g are the same as the output value of f when the input value is 3 less than the original value. For example, we know that f (2) = 1. To get the same output from the function g, we will need an input value that is 3 larger. We input a value that is 3 larger for g (x) because the function takes 3 away before evaluating the function f. We continue with the other values to create Table 5. g (5) = f (5 − 3) = f (2) = 1 x x − 3 f (x) g(x) 5 2 1 1 7 4 3 3 Table 5 9 6 7 7 11 8 11 11 The result is that the function g (x) has been shifted to the right by 3. Notice the output values for g (x) remain the same as the output values for f (x), but the corresponding input values, x, have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11. Analysis Figure 7 represents both of the functions. We can see the horizontal shift in each point. y 12 10 8 6 4 2 –2–2 –4 –6 –8 –10 –12 –12 –10 –8 –6 –4 42 6 8 10 12 x f (x) g (x) = f(x – 3) Figure 7 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 227 Example 5 Identifying a Horizontal Shift of a Toolkit Function Figure 8 represents a transformation of the toolkit function f (x) = x2. Relate this new function g (x) to f (x), and then find a formula for g (x). f (x) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 x Figure 8 Solution Notice that the graph is identical in shape to the f (x) = x2 function, but the x-values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so g(x) = f (x − 2) Notice how we must input the value x = 2 to get the output value y = 0; the x-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the f (x) function to write a formula for g (x) by evaluating f (x − 2). f(x) = x2 g(x) = f (x − 2) g (x) = f (x − 2) = (x − 2)2 Analysis To determine whether the shift is +2 or −2, consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, f (0) = 0. In our shifted function, g (2) = 0. To obtain the output value of 0 from the function f, we need to decide whether a plus or a minus sign will work to satisfy g (2) = f (x − 2) = f (0) = 0. For this to work, we will need to subtract 2 units from our input values. Example 6 Interpreting Horizontal versus Vertical Shifts The function G (m) gives the number of gallons of gas required to drive m miles. Interpret G (m) + 10 and G (m + 10). Solution G (m) + 10 can be interpreted as adding 10 to the output, gallons. This is the gas required to drive m miles, plus another 10 gallons of gas. The graph would indicate a vertical shift. G (m + 10) can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to drive 10 miles more than m miles. The graph would indicate a horizontal shift. Try It #2 Given the function f (x) = √ Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units? — x , graph the original function f (x) and the transformation g (x) = f (x + 2) on the same axes. Combining Vertical and Horizontal Shifts Now that we have two transformations, we can combine them. Vertical shifts are outside changes that affect the output ( y-) axis values and shift the function up or down. Horizontal shifts are inside changes that affect the input (x-) axis values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and right or left. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 228 CHAPTER 3 FUNCTIONS How To… Given a function and both a vertical and a horizontal shift, sketch the graph. 1. Identify the vertical and horizontal shifts from the formula. 2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant and down for a negative constant. 3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant and right for a negative constant. 4. Apply the shifts to the graph in either order. Example 7 Graphing Combined Vertical and Horizontal Shifts Given f (x) = . x ., sketch a graph of h (x) = f (x + 1) − 3. Solution The function f is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of h has transformed f in two ways: f (x + 1) is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in f (x + 1) − 3 is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated in Figure 9. Let us follow one point of the graph of f (x) = . x .. • The point (0, 0) is transformed first by shifting left 1 unit: (0, 0) → (−1, 0) • The point (−1, 0) is transformed next by shifting down 3 units: (−1, 0) → (−1, −3) y = \|x + 1\| y = \|x\| y = \|x + 1\| – 3 21 1 –1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 10 shows the graph of h. Figure 9 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 y = \|x + 1\| – 3 21 3 4 5 x Figure 10 Try It #3 Given f (x) = . x ., sketch a graph of h(x) = f (x − 2) + 4. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 229 Example 8 Identifying Combined Vertical and Horizontal Shifts Write a formula for the graph shown in Figure 11, which is a transformation of the toolkit square root function. h(x) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 x Figure 11 Solution The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as Using the formula for the square root function, we can write h(x(x) = f (x − 1) + 2 Analysis Note that this transformation has changed the domain and range of the function. This new graph has domain [1, ∞) and range [2, ∞). Try It #4 1 __ Write a formula for a transformation of the toolkit reciprocal function f (x) = that shifts the function’s graph one x unit to the right and one unit up. Graphing Functions Using Reﬂections about the Axes Another transformation that can be applied to a function is a reflection over the x- or y-axis. A vertical reflection reflects a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The reflections are shown in Figure 12. y Horizontal reflection f(x) Original function f(–x) x Vertical reflection –f(x) Figure 12 Vertical and horizontal reﬂections of a function. Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 230 CHAPTER 3 FUNCTIONS reflections Given a function f (x), a new function g(x) = −f (x) is a vertical reflection of the function f (x), sometimes called a reflection about (or over, or through) the x-axis. Given a function f (x), a new function g(x) = f (−x) is a horizontal reflection of the function f (x), sometimes called a reflection about the y-axis. How To… Given a function, reflect the graph both vertically and horizontally. 1. Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis. 2. Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph about the y-axis. Example 9 Reﬂecting a Graph Horizontally and Vertically Reflect the graph of s(t) = √ — t a. vertically and b. horizontally. Solution a. Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis as shown in Figure 13. s(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 t –5 –4 –3 –2 v(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 13 Vertical reﬂection of the square root function Because each output value is the opposite of the original output value, we can write V(t) = −s(t) or V(t) = − √ — t Notice that this is an outside change, or vertical shift, that affects the output s(t) values, so the negative sign belongs outside of the function. b. Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in Figure 14. s(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 t –5 –4 –3 –2 H(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 14 Horizontal reﬂection of the square root function Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 231 Because each input value is the opposite of the original input value, we can write H(t) = s(−t) or H(t) = √ — −t Notice that this is an inside change or horizontal change that affects the input values, so the negative sign is on the inside of the function. Note that these transformations can affect the domain and range of the functions. While the original square root function has domain [0, ∞) and range [0, ∞), the vertical reflection gives the V(t) function the range (−∞, 0] and the horizontal reflection gives the H(t) function the domain (−∞, 0]. Try It #5 Reflect the graph of f (x) = \|x − 1\| a. vertically and b. horizontally. Example 10 Reﬂecting a Tabular Function Horizontally and Vertically A function f (x) is given as Table 6. Create a table for the functions below. a. g(x) = −f (x) b. h(x) = f (−x) x f (x) 2 1 4 3 Table 6 6 7 8 11 Solution a. For g(x), the negative sign outside the function indicates a vertical reflection, so the x-values stay the same and each output value will be the opposite of the original output value. See Table 7. x g (x) 2 –1 6 –7 8 –11 4 –3 Table 7 b. For h(x), the negative sign inside the
function indicates a horizontal reflection, so each input value will be the opposite of the original input value and the h(x) values stay the same as the f (x) values. See Table 8. x h(x) –2 1 –4 3 Table 8 –6 7 –8 11 Try It #6 A function f (x) is given as Table 9. Create a table for the functions below. x f(x) −2 5 0 10 Table 9 2 15 4 20 a. g(x) = −f (x) b. h(x) = f (−x) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 232 CHAPTER 3 FUNCTIONS Example 11 Applying a Learning Model Equation A common model for learning has an equation similar to k(t) = −2−t + 1, where k is the percentage of mastery that can be achieved after t practice sessions. This is a transformation of the function f (t) = 2t shown in Figure 15. Sketch a graph of k(t). f (t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Solution This equation combines three transformations into one equation. Figure 15 • A horizontal reflection: f (−t) = 2−t • A vertical reflection: −f (−t) = −2−t • A vertical shift: −f (−t) + 1 = −2−t + 1 We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two points through each of the three transformations. We will choose the points (0, 1) and (1, 2). 1. First, we apply a horizontal reflection: (0, 1) (−1, 2). 2. Then, we apply a vertical reflection: (0, −1) (1, −2). 3. Finally, we apply a vertical shift: (0, 0) (1, 1). This means that the original points, (0,1) and (1,2) become (0,0) and (1,1) after we apply the transformations. In Figure 16, the first graph results from a horizontal reflection. The second results from a vertical reflection. The third results from a vertical shift up 1 unit. f (t) f (t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (a) 1 2 3 4 5 t –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t –5 –4 –3 –2 (b) Figure 16 k(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (c) 21 3 4 5 t Analysis As a model for learning, this function would be limited to a domain of t ≥ 0, with corresponding range [0, 1). Try It #7 Given the toolkit function f (x) = x 2, graph g(x) = −f (x) and h(x) = f (−x). Take note of any surprising behavior for these functions. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 233 Determining Even and Odd Functions Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions f (x) = x2 or f (x) = ⎪x⎪ will result in the original graph. We say that these types of graphs are symmetric about the y-axis. A function whose graph is symmetric about the y-axis is called an even function. 1 _ If the graphs of f (x) = x3 or f (x) = x were reflected over both axes, the result would be the original graph, as shown in Figure 17. y y f (x) = x3 f(x) = x3 y y f (−x) f(−xf (−x) −f (−x5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5 –5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5 21 21 3 3 4 4 5 5 x x 21 21 3 3 4 4 5 5 x x –5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5 (a) (a) (b) (b) (c) (c) Figure 17 (a) The cubic toolkit function (b) Horizontal reﬂection of the cubic toolkit function (c) Horizontal and vertical reﬂections reproduce the original cubic function. We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called an odd function. Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, f (x) = 2x is neither even nor odd. Also, the only function that is both even and odd is the constant function f (x) = 0. even and odd functions A function is called an even function if for every input x: f (x) = f (−x) The graph of an even function is symmetric about the y-axis. A function is called an odd function if for every input x: f (x) = −f (−x) The graph of an odd function is symmetric about the origin. How To… Given the formula for a function, determine if the function is even, odd, or neither. 1. Determine whether the function satisfies f (x) = f (−x). If it does, it is even. 2. Determine whether the function satisfies f (x) = −f (−x). If it does, it is odd. 3. If the function does not satisfy either rule, it is neither even nor odd. Example 12 Determining whether a Function Is Even, Odd, or Neither Is the function f (x) = x 3 + 2x even, odd, or neither? Solution Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return us to the original function. Let’s begin with the rule for even functions. f (−x) = (−x)3 + 2(−x) = −x 3 − 2x This does not return us to the original function, so this function is not even. We can now test the rule for odd functions. Because −f (−x) = f (x), this is an odd function. −f (−x) = −(−x 3 − 2x) = x 3 + 2x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 234 CHAPTER 3 FUNCTIONS Analysis Consider the graph of f in Figure 18 . Notice that the graph is symmetric about the origin. For every point (x, y) on the graph, the corresponding point (−x, −y) is also on the graph. For example, (1, 3) is on the graph of f, and the corresponding point (−1, −3) is also on the graph. f (x) f 5 4 3 2 1 –1 –2 –3 –4 –5 (−1, −3) –1 –2 –3 –4 –5 (1, 3) 21 3 4 5 x Figure 18 Try It #8 Is the function f (s) = s 4 + 3s 2 + 7 even, odd, or neither? Graphing Functions Using Stretches and Compressions Adding a constant to the inputs or outputs of a function changed the position of a graph with respect to the axes, but it did not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity. We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically. Vertical Stretches and Compressions When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression. Figure 19 shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression. y Vertical stretch Vertical compression 2 f(x) f(x) 0.5 f(x) x Figure 19 Vertical stretch and compression vertical stretches and compressions Given a function f (x), a new function g(x) = af (x), where a is a constant, is a vertical stretch or vertical compression of the function f (x). • If a > 1, then the graph will be stretched. • If 0 < a < 1, then the graph will be compressed. • If a < 0, then there will be combination of a vertical stretch or compression with a vertical reflection. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 235 How To… Given a function, graph its vertical stretch. 1. Identify the value of a. 2. Multiply all range values by a. 3. If a > 1, the graph is stretched by a factor of a. If 0 < a < 1, the graph is compressed by a factor of a. If a < 0, the graph is either stretched or compressed and also reflected about the x-axis. Example 13 Graphing a Vertical Stretch A function P(t) models the population of fruit flies. The graph is shown in Figure 20. P(t) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 t Figure 20 A scientist is comparing this population to another population, Q, whose growth follows the same pattern, but is twice as large. Sketch a graph of this population. Solution Because the population is always twice as large, the new population’s output values are always twice the original function’s output values. Graphically, this is shown in Figure 21. If we choose four reference points, (0, 1), (3, 3), (6, 2) and (7, 0) we will multiply all of the outputs by 2. The following shows where the new points for the new graph will be located. (0, 1) → (0, 2) (3, 3) → (3, 6) (6, 2) → (6, 4) (7, 0) → (7, 0) Q(t) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 t Figure 21 Symbolically, the relationship is written as Q(t) = 2P(t) This means that for any input t, the value of the function Q is twice the value of the function P. Notice that the effect on the graph is a vertical stretching of the graph, where every point doubles its distance from the horizontal axis. The input values, t, stay the same while the output values are twice as large as before. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 236 CHAPTER 3 FUNCTIONS How To… Given a tabular function and assuming that the transformation is a vertical stretch or compression, create a table for a vertical compression. 1. Determine the value of a. 2. Multiply all of the output values by a. Example 14 Finding a Vertical Compression of a Tabular Function A function f is given as Table 10. Create a table for the function g(x) = #1 __ f (x). 2 x f (x) 2 1 6 7 8 11 4 3 Table 10 Solution The formula g(x) =# #1 __ f (x) tells us that the output values of g are half of the output values of f with the same 2 inputs. For example, we know that f (4) = 3. Then We do the same for the other values to produce Table 11. (3) =# #3 f (4) =# #1 g(4) =# #1 __ __ __ 2 2 2 x g(x) 2 1 __ 2 4 3 __ 2 Table 11 6 7 __ 2 8 11 __ 2 1 __ Analysis The result is that the function g(x) has been compressed vertically by . Each output value is divided in half, 2 so the graph is half the original height. Try It #9 A function f is given as Table 12. Create a table for the function g(x) =# #3 __ f (x). 4 x f(x) 2 12 4 16 Table 12 6 20 8 0 Example 15 Recognizing a Vertical Stretch The graph in Figure 22 is a transformation of the toolkit function f (x) = x3. Relate this new function g(x) to f (x), and then find a formula for g(x). g(x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 22 Download the OpenStax text for free at http://cnx.org/content/col11
759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 237 Solution When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that g(2) = 2. With the basic cubic function at the same input, f (2) = 23 = 8. the outputs of the function f because g(2) =# #1 1 __ __ f (2). From this we Based on that, it appears that the outputs of g are 4 4 can fairly safely conclude that g(x) =# #1 __ f (x). 4 We can write a formula for g by using the definition of the function f. f (x) =# #1 g(x) =# #1 __ __ x3 4 4 Try It #10 Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units. Horizontal Stretches and Compressions Now we consider changes to the inside of a function. When we multiply a function’s input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function. Horizontal compression y = x2 y = (0.5 x)2 Horizontal stretch y = (2 x)2 y 10 5 –4 –3 –2 –1 1 2 3 4 5 x –1 Figure 23 Given a function y = f (x), the form y = f (bx) results in a horizontal stretch or compression. Consider the function y = x2. Observe Figure 23. The graph of y = (0.5x)2 is a horizontal stretch of the graph of the function y = x 2 by a factor of 2. The graph of y = (2x)2 is a horizontal compression of the graph of the function y = x 2 by a factor of 2. horizontal stretches and compressions Given a function f (x), a new function g(x) = f (bx), where b is a constant, is a horizontal stretch or horizontal compression of the function f (x). 1 __ • If b > 1, then the graph will be compressed by . b 1 __ • If 0 < b < 1, then the graph will be stretched by . b • If b < 0, then there will be combination of a horizontal stretch or compression with a horizontal reflection. How To… Given a description of a function, sketch a horizontal compression or stretch. 1. Write a formula to represent the function. 2. Set g(x) = f (bx) where b > 1 for a compression or 0 < b < 1 for a stretch. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 238 CHAPTER 3 FUNCTIONS Example 16 Graphing a Horizontal Compression Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, R, will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population. Solution Symbolically, we could write R(1) = P(2), R(2) = P(4), and in general, R(t) = P(2t). See Figure 24 for a graphical comparison of the original population and the compressed population. f (x) 6 5 4 3 2 1 – –1 1– Original population, P(t) Transformed population, R(t) f (x1 1a) (b) Figure 24 (a) Original population graph (b) Compressed population graph Analysis Note that the effect on the graph is a horizontal compression where all input values are half of their original distance from the vertical axis. Example 17 Finding a Horizontal Stretch for a Tabular Function 1 __ x ) . A function f (x) is given as Table 13. Create a table for the function g(x) = f (# 2 x f (x) 2 1 6 7 8 11 4 3 Table 13 1 __ x ) tells us that the output values for g are the same as the output values for Solution The formula g(x) = f (# 2 the function f at an input half the size. Notice that we do not have enough information to determine g(2) because 1 __ ċ 2 ) = f (1), and we do not have a value for f (1) in our table. Our input values to g will need to be twice as g(2) = f (# 2 large to get inputs for f that we can evaluate. For example, we can determine g(4). We do the same for the other values to produce Table 14. 1 __ ċ 4 ) = f (2) = 1 g(4) = f (# 2 x g(x) 4 1 8 3 12 7 16 11 Table 14 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 239 Figure 25 shows the graphs of both of these sets of points. –20 –16 –12 –8 y 12 10 8 6 4 2 –4 –2 –4 –6 –8 –10 –12 (a) 84 12 16 20 x –20 –16 –12 –8 Figure 25 y 12 10 8 6 4 2 –4 –2 –4 –6 –8 –10 –12 (b) 84 12 16 20 x Analysis Because each input value has been doubled, the result is that the function g(x) has been stretched horizontally by a factor of 2. Example 18 Recognizing a Horizontal Compression on a Graph Relate the function g(x) to f (x) in Figure 26. f (x) 6 5 4 3 2 1 – –1 1 Figure 26 Solution The graph of g(x) looks like the graph of f (x) horizontally compressed. Because f (x) ends at (6, 4) and g(x) ends at 1 1 __ __ ) = 2. We might also notice that g(2) = f (6) , because 6 (# (2, 4), we can see that the x-values have been compressed by 3 3 1 __ and g(1) = f (3). Either way, we can describe this relationship as g(x) = f (3x). This is a horizontal compression by . 3 Analysis Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or 1 1 __ __ x ) . in our function: f (# compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of 4 4 This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. Try It #11 Write a formula for the toolkit square root function horizontally stretched by a factor of 3. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 240 CHAPTER 3 FUNCTIONS Performing a Sequence of Transformations When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first. When we see an expression such as 2f (x) + 3, which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of f (x), we first multiply by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition. Horizontal transformations are a little trickier to think about. When we write g(x) = f (2x + 3), for example, we have to think about how the inputs to the function g relate to the inputs to the function f . Suppose we know f (7) = 12. What input to g would produce that output? In other words, what value of x will allow g(x) = f (2x + 3) = 12? We would need 2x + 3 = 7. To solve for x, we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression. This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function. Let’s work through an example. We can factor out a 2. p_ f (bx + px) = (2x + 4)2 f (x) = (2(x + 2))2 Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally. combining transformations When combining vertical transformations written in the form af (x) + k, first vertically stretch by a and then vertically shift by k. When combining horizontal transformations written in the form f (bx + h), first horizontally shift by h and 1 __ then horizontally stretch by . b 1_ When combining horizontal transformations written in the form f (b(x + h)), first horizontally stretch by b and then horizontally shift by h. Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are performed first. Example 19 Finding a Triple Transformation of a Tabular Function Given Table 15 for the function f (x), create a table of values for the function g(x) = 2f (3x) + 1. x f (x) 6 10 12 14 Table 15 18 15 24 17 Solution There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal 1 1 __ __ . See Table 16. , which means we multiply each x-value by transformations, f (3x) is a horizontal compression by 3 3 x f (3x) 2 10 4 14 Table 16 6 15 8 17 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 241 Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation. See Table 17. x 2f (3x) 2 20 4 28 Table 17 6 30 8 34 Finally, we can apply the vertical shift, which will add 1 to all the output values. See Table 18. x g(x) = 2f (3x) + 1 4 29 6 31 8 35 2 21 Table 18 Example 20 Finding a Triple Transformation of a Graph 1 __ Use the graph of f (x) in Figure 27 to sketch a graph of k(x) = f (# x + 1 ) − 3. 2 f (x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 27 Solution To simplify, let’s start by factoring out the inside of the function. 1 1 __ __ (x + 2 (# f (# 2 2 1 __ on the inside of the function. By factoring the inside, we can first horizontally stretch by 2, as indicated by the 2 Remember that twice the size of 0 is still 0, so the point (0, 2) remains at (0, 2) while the point (2, 0) will stretch to (4, 0). See Figure 28. f (x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 28 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 242 CHAPTER 3 FUNCTIONS Next, we horizontally shift left by 2 units, as indicated by x + 2. See Figure 29. f (x) –6 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4
–5 21 3 4 5 6 x Figure 29 Last, we vertically shift down by 3 to complete our sketch, as indicated by the −3 on the outside of the function. See Figure 30. f (x) –6 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 6 x Figure 30 Access this online resource for additional instruction and practice with transformation of functions. • Function Transformations (http://openstaxcollege.org/l/functrans) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 SECTION EXERCISES 243 3.5 SECTION EXERCISES VERBAL 1. 2. x y 3. When examining the formula of a function that 4. When examining the formula of a function that is is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression? the result of multiple transformations, how can you tell a reflection with respect to the x-axis from a reflection with respect to the y-axis? 5. How can you determine whether a function is odd or even from the formula of the function? ALGEBRAIC ft 6. fx) =√ — x hift 8. fx=! ! __ hift x 7. fx=xhift 9. fx= __ x hift f 10. y = f x − 11. y = f x + 12. y = f x + 13. y = f x − 14. y = f x + 15. y = f x + 16. y = f x − 17. y = f x − 18. y = f x − + 19. y = f x + − s — — 20. f x = x + − 21. gx=x+− 22. ax=√ −x+ 23. kx = −√ x − GRAPHICAL f x=xFigure 31 f x 24. gx = x + 25. hx = x − 26. wx = x − 27. f t = t + − 28. hx = \|x − \| + – – – – 29. kx = x − − 30. mt = + √ — Figure 31 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 244 CHAPTER 3 FUNCTIONS NUMERIC 31. 34. 37. 33. 36. 32. 35. y f – – – – –– – – – – x y x f – – – – –– – – – – y f x – – – – –– – – – – – – – – – – – – 38 –– – – – –– – – – – – – – – – –– – – – – fi 39. y x f – – – – –– – – – – 40. y f – – – – –– – – – – x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 SECTION EXERCISES 245 41. 44. y f – – – – –– – – – – 42. x y f x – – – – –– – – – – 43. y f – – – – –– – – – – x y – – – – –– – – – – f x 45. fx=x 48. fx=x− 46. gx=√x — 49. gx=x 47. hx=__ x +x 50. hx=x−x f 51. gx=−fx 52. gx=f−x 53. gx=fx 54. gx=fx 55. gx=fx 56. gx=fx 59. gx=f−x 60. gx=−fx 57. __ x ) gx=f (! 58. __ x ) gx=f (! g 61. Thfx=∣ x ∣y __ 63. __ Thfx= x __ hift 3 uni 62. 64. Thfx=√x x — Thfx=__ x hift 65. Thfx=x __ hift 66. Thfx=x hift Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 246 CHAPTER 3 FUNCTIONS Th 67. gx=x+− 68. gx=x+− 70. kx=−√x − — 71. mx=! !__ x 73. px= ! (! !__ x ) − 74. qx= (! !__ x ) + 69. hx=−∣!x−∣+ 72. nx= !__ x− 75. — ax=√−x+ 78. ()=(+)+ y f x 81. ()= () y f x x x ()=( )+ y 77. ()=(+) 76. 79. f ()=( ) y f 82. ()=( ) y f y f 80. ()= () y f x x x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.6 ABSOLUTE VALUE FUNCTIONS 247 LEARNING OBJECTIVES In this section you will: • Graph an absolute value function. • Solve an absolute value equation. 3.6 ABSOLUTE VALUE FUNCTIONS Figure 1 Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: “s58y”/Flickr) Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will continue our investigation of absolute value functions. Understanding Absolute Value Recall that in its basic form f (x) = \| x \|, the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems. absolute value function The absolute value function can be defined as a piecewise function x { f (x) = ∣ x ∣ #= −x if x ≥ 0 if x < 0 Example 1 Using Absolute Value to Determine Resistance Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often ±1%, ±5%, or ±10%. Suppose we have a resistor rated at 680 ohms, ±5%. Use the absolute value function to express the range of possible values of the actual resistance. Solution We can find that 5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance R in ohms, \| R − 680 \| ≤ 34 Try It #1 Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 248 CHAPTER 3 FUNCTIONS Graphing an Absolute Value Function The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure 2. –5 –4 –3 –2 y = \|x\| 21 1–1 –2 –3 –4 –5 Figure 2 Figure 3 shows the graph of y = 2\| x − 3\|#+ 4. The graph of y = \| x\| has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at (3, 4) for this transformed function. y = 2\|x − 3\| Vertical stretch y = \|x\| y 10 \|x − 3\| + 4 Up 4 y = \|x − 3\| Right 3 –6 –5 –4 –3 –2 –1–1 1 2 3 4 5 6 x Figure 3 Example 2 Writing an Equation for an Absolute Value Function Given a Graph Write an equation for the function graphed in Figure 4. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 4 Solution The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.6 ABSOLUTE VALUE FUNCTIONS 249 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x (3 , –2) Figure 5 We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 6. Ratio 2/1 y 5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 Ratio 1/1 x 21 3 4 5 (3 , −2) From this information we can write the equation Figure 6 f (x) = 2\| x − 3 \|#− 2, treating the stretch as a vertical stretch, or f (x) = \| 2(x − 3) \|#− 2, treating the stretch as a horizontal compression. Analysis Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. Q & A… If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it? Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for x and f (x). Now substituting in the point (1, 2) f (x) = a\| x − 3 \|#− 2 2 = a\| 1 − 3 \|#− 2 4 = 2a a = 2 Try It #2 Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 250 CHAPTER 3 FUNCTIONS Q & A… Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis? Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero. No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (see Figure 7). y 5 4 3 2 1 –1 –1 –2 –3 –4 (a) –5 –4 –3 –2 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 (b) 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 (c) 21 3 4 5 x Figure 7 (a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points. Solving an Absolute Value Equation In Other Types of Equations, we touched on the concepts of absolute value equations. Now that we understand a little more about their graphs, we can take another look at these types of equations. Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as 8 = \| 2x − 6\| , we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or −8. This leads to two different equations we can solve independently. 2x − 6 = 8 or 2x − 6 = −8 2x = −2 x = −1 2x =#14 x = 7 Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point. An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example, \| x \|#= 4, \| 2x − 1 \|#= 3 \| 5x + 2 \|#− 4 = 9 solutions to absolute value equations For real numbers
A and B, an equation of the form \| A \|#= B, with B ≥ 0, will have solutions when A = B or A = −B. If B <#0, the equation \| A \|#= B has no solution. How To… Given the formula for an absolute value function, find the horizontal intercepts of its graph. 1. Isolate the absolute value term. 2. Use \| A \|#= B to write A = B or −A = B, assuming B > 0. 3. Solve for x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.6 ABSOLUTE VALUE FUNCTIONS 251 Example 3 Finding the Zeros of an Absolute Value Function For the function f (x) = \| 4x + 1 \|#− 7 , find the values of x such that f (x) = 0. Solution 0 = \| 4x +#1 \|#−#7 7 = \| 4x +#1 \| 7 =#4x +#1 or −7 =#4x +#1 6 = 4x −8 =#4x −8 ___ 4 6 __ x = = 1.5 4 The function outputs 0 when x = 1.5 or x = −2. See Figure 8. = −2 x = Substitute 0 for f (x). Isolate the absolute value on one side of the equation. Break into two separate equations and solve. y 10 8 6 4 2 –0.5 –2 –4 –6 –8 –10 (−2, 0) –1 –1.5 –2 –2.5 x where f(x) = 0 f x (1.5, 0) 0.5 1 1.5 2 2.5 x where f(x) = 0 Try It #3 For the function f (x) = \| 2x − 1 \|#− 3, find the values of x such that f (x) = 0. Figure 8 Q & A… Should we always expect two answers when solving . A .#= B? No. We may find one, two, or even no answers. For example, there is no solution to 2 + \| 3x − 5 \|#= 1. Access these online resources for additional instruction and practice with absolute value. • Graphing Absolute Value Functions (http://openstaxcollege.org/l/graphabsvalue) • Graphing Absolute Value Functions 2 (http://openstaxcollege.org/l/graphabsvalue2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 252 CHAPTER 3 FUNCTIONS 3.6 SECTION EXERCISES VERBAL 1. How do you solve an absolute value equation? 3. When solving an absolute value function, the isolated absolute value term is equal to a negative number. What does that tell you about the graph of the absolute value function? ALGEBRAIC 2. How can you tell whether an absolute value function has two x-intercepts without graphing the function? 4. How can you use the graph of an absolute value function to determine the x-values for which the function values are negative? 5. Describe all numbers x that are at a distance of 4 from the number 8. Express this using absolute value notation. 6. Describe all numbers x that are at a distance of 1__ 2 from the number −4. Express this using absolute value notation. 7. Describe the situation in which the distance that point x is from 10 is at least 15 units. Express this using absolute value notation. 8. Find all function values f (x) such that the distance from f (x) to the value 8 is less than 0.03 units. Express this using absolute value notation. For the following exercises, find the x- and y-intercepts of the graphs of each function. 9. f (x)#= 4 ∣ x − 3 ∣ + 4 10. f (x)#= −3 ∣ x − 2 ∣ − 1 11. f (x)#= −2 ∣ x + 1 ∣ + 6 12. f (x)#= −5 ∣ x + 2 ∣ + 15 13. f (x)#= 2 ∣ x − 1 ∣ − 6 14. f (x)#= ∣ −2x + 1 ∣ − 13 15. f (x)#= − ∣ x − 9 ∣ + 16 GRAPHICAL For the following exercises, graph the absolute value function. Plot at least five points by hand for each graph. 16. y = \| x − 1 \| 17. y = \| x + 1#\|# 18. y = \| x \|#+ 1 For the following exercises, graph the given functions by hand. 19. y = \| x \|#− 2 20. y = −\| x#\|# 21. y = −\| x \| −#2 22. y = −\| x − 3 \|#−#2 23. f (x) = −\| x − 1 \|#−#2 24. f (x) = −\| x + 3 \|#+ 4 25. f (x) = 2\| x + 3 \|#+ 1 26. f (x) = 3\| x − 2 \|#+ 3 27. f (x) = \| 2x − 4 \|#− 3 28. f (x) = \| 3x + 9 \|#+ 2 29. f (x) = −\| x − 1 \|#− 3 30. f (x) = −\| x + 4 \| −3 1 __ \| x + 4 \|#− 3 31. f (x) = 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.6 SECTION EXERCISES 253 TECHNOLOGY 32. Use a graphing utility to graph f (x) = 10\| x − 2\| on the viewing window [0, 4]. Identify the corresponding range. Show the graph. 33. Use a graphing utility to graph f (x) = −100\| x\|#+ 100 on the viewing window [−5, 5]. Identify the corresponding range. Show the graph. For the following exercises, graph each function using a graphing utility. Specify the viewing window. 35. f (x) = 4 × 109 ∣ x − (5 × 109) ∣ #+ 2 × 109 34. f (x) = −0.1\| 0.1(0.2 − x)\|#+ 0.3 EXTENSIONS For the following exercises, solve the inequality. 36. If possible, find all values of a such that there are no x-intercepts for f (x) = 2\| x + 1\|#+ a. 37. If possible, find all values of a such that there are no y-intercepts for f (x) = 2\| x + 1\|#+ a. REAL-WORLD APPLICATIONS 38. Cities A and B are on the same east-west line. Assume that city A is located at the origin. If the distance from city A to city B is at least 100 miles and x represents the distance from city B to city A, express this using absolute value notation. 39. The true proportion p of people who give a favorable rating to Congress is 8% with a margin of error of 1.5%. Describe this statement using an absolute value equation. 40. Students who score within 18 points of the number 82 will pass a particular test. Write this statement using absolute value notation and use the variable x for the score. 41. A machinist must produce a bearing that is within 0.01 inches of the correct diameter of 5.0 inches. Using x as the diameter of the bearing, write this statement using absolute value notation. 42. The tolerance for a ball bearing is 0.01. If the true diameter of the bearing is to be 2.0 inches and the measured value of the diameter is x inches, express the tolerance using absolute value notation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 254 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: • etermine whether a function is one-to-one. • erify the inerse using composition. • Find the inerse of a function. • etermine the domain and range of an inerse function and restrict the domain of a function to mae it one-to-one. • Use the graph of a one-to-one function to graph its inerse function on the same aes. • Use a graph or table to ealuate a function and its inerse. 3.7 INVERSE FUNCTIONS A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating. If some physical machines can run in two directions, we might ask whether some of the function “machines” we have been studying can also run backwards. Figure 1 provides a visual representation of this question. In this section, we will consider the reverse nature of functions. x y f ? y x Figure 1 Can a function “machine” operate in reverse? Verifying That Two Functions Are Inverse Functions Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula and substitutes 75 for F to calculate 5 __ C = (F − 32) 9 5 __ (75 − 32) ≈ 24°C. 9 Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from Figure 2 for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit. Mon Tue Web Thu 26°C \| 19°C 29°C \| 19°C 30°C \| 20°C 26°C \| 18°C Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 255 At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for F after substituting a value for C. For example, to convert 26 degrees Celsius, she could write 26 =# #5 __ (F − 32) 9 9 __ = F − 32 26 ċ 5 9 __ + 32 ≈ 79 F = 26 ċ 5 After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature. The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function. Given a function f (x), we represent its inverse as f −1(x), read as “ f inverse of x.” The raised −1 is part of the notation. It is not an exponent; it does not imply a power of −1. In other words, f −1(x) does not mean is the reciprocal of f and not the inverse. The “exponent-like” notation comes from an analogy between function composition and multiplication: just as a−1 a = 1 (1 is the identity element for multiplication) for any nonzero number a, so f −1 ∘ f equals the identity function, that is, 1 ___ f (x) 1 ___ f (x) because (f −1 ∘ f )(x) = f −1( f (x)) = f −1 (y) = x This holds for all x in the domain of f. Informally, this means that inverse functions “undo” each other. However, just as zero does not have a reciprocal, some functions do not have inverses. Given a function f (x), we can verify whether some other function g (x) is the inverse of f (x) by checking whether either g ( f (x)) = x or f (g (x)) = x is true. We can test whichever equation is more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.) For example, y = 4x and y =# #1 __ x are inverse functions. 4 and (f −1 ∘ f )(x) = f −1(4x) =# #1 __ (4x) = x 4 1 1 __ __ (f ∘ f −1)(x) = f (# x ) = x x ) = 4 (# 4 4 A few coordinate pairs from the graph of the function y = 4x are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs 1 __ x are (−8, −2), (0, 0), and (8, 2). If we in
terchange the input and output of each from the graph of the function y = 4 coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function. inverse function For any one-to-one function f (x) = y, a function f −1 (x) is an inverse function of f if f −1(y) = x. This can also be written as f −1( f (x)) = x for all x in the domain of f . It also follows that f ( f −1(x)) = x for all x in the domain of f −1 if f −1 is the inverse of f . The notation f −1 is read “ f inverse.” Like any other function, we can use any variable name as the input for f −1, so we will often write f −1(x), which we read as “ f inverse of x.” Keep in mind that and not all functions have inverses. f −1(x) ≠ 1 ___ f (x) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 256 CHAPTER 3 FUNCTIONS Example 1 Identifying an Inverse Function for a Given Input-Output Pair If for a particular one-to-one function f (2) = 4 and f (5) = 12, what are the corresponding input and output values for the inverse function? Solution The inverse function reverses the input and output quantities, so if f (2) = 4, then f −1 (4) = 2; f (5) = 12, then f −1 (12) = 5. Alternatively, if we want to name the inverse function g, then g (4) = 2 and g (12) = 5. Analysis Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table 1. (x, f (x)) (2, 4) (5, 12) (x, g (x)) (4, 2) (12, 5) Table 1 Try It #1 Given that h−1(6) = 2, what are the corresponding input and output values of the original function h? How To… Given two functions f (x) and g (x), test whether the functions are inverses of each other. 1. Determine whether f (g(x)) = x or g(f (x)) = x. 2. If either statement is true, then both are true, and g = f −1 and f = g −1. If either statement is false, then both are false, and g ≠ f −1 and f ≠ g−1. Example 2 Testing Inverse Relationships Algebraically If f (x) = 1 ____ x + 2 and g (x) =# #1 __ − 2, is g = f −1? x Solution so 1 _ g (f (x)) = − 2 1 _ ) (# 1 and f = g −1 This is enough to answer yes to the question, but we can also verify the other formula. f (g (x)) = 1 _ 1 __ − 2 + 2 x 1 _ 1 __ x = x = Analysis Notice the inverse operations are in reverse order of the operations from the original function. Try It #2 If f (x) = x3 − 4 and g (x) = # 3 √ — x − 4 , is g = f −1? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 257 Example 3 Determining Inverse Relationships for Power Functions If f (x) = x3 (the cube function) and g (x) =# #1 __ x, is g = f −1? 3 x3 __ Solution 27 f (g (x)) = ≠ x No, the functions are not inverses. Analysis The correct inverse to the cube is, of course, the cube root not a multiplier. 3 √ — x = x1/3 that is, the one-third is an exponent, Try It #3 If f (x) = (x − 1)3 and g (x) = 3 √ — x + 1, is g = f −1? Finding Domain and Range of Inverse Functions The outputs of the function f are the inputs to f −1, so the range of f is also the domain of f −1. Likewise, because the inputs to f are the outputs of f −1, the domain of f is the range of f −1. We can visualize the situation as in Figure 3. Domain of f Range of f f (x) a b Range of f –1 f –1(x) Domain of f –1 Figure 3 Domain and range of a function and its inverse When a function has no inverse function, it is possible to create a new function where that new function on a limited x is f −1(x) = x2, because a square “undoes” domain does have an inverse function. For example, the inverse of f (x) = √ a square root; but the square is only the inverse of the square root on the domain [0, ∞), since that is the range of f (x) = √ x . — — We can look at this problem from the other side, starting with the square (toolkit quadratic) function f (x) = x2. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and −3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the “inverse” is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function. In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is one-to-one. For example, we can make a restricted version of the square function f (x) = x2 with its range limited to [0, ∞), which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function). If f (x) = (x − 1)2 on [1, ∞), then the inverse function is f −1(x) = √ x + 1. — • The domain of f = range of f −1 = [1, ∞). • The domain of f −1 = range of f = [0, ∞). Q & A… Is it possible for a function to have more than one inverse? No. If two supposedly different functions, say, g and h, both meet the definition of being inverses of another function f, then you can prove that g = h. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 258 CHAPTER 3 FUNCTIONS domain and range of inverse functions The range of a function f (x) is the domain of the inverse function f −1(x). The domain of f (x) is the range of f −1(x). How To… Given a function, find the domain and range of its inverse. 1. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse. 2. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function. Example 4 Finding the Inverses of Toolkit Functions Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table 2. We restrict the domain in such a fashion that the function assumes all y-values exactly once. Constant f (x) = c Identity f (x) = x Quadratic f (x) = x 2 Cubic f (x) = x 3 Reciprocal f (x) = 1_ x Reciprocal squared Cube root Square root Absolute value f (x) = 1 _ x2 f (x) = 3 √ — x f (x) = √ — x f (x) = .x. Table 2 Solution The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse. The absolute value function can be restricted to the domain [0, ∞), where it is equal to the identity function. The reciprocal-squared function can be restricted to the domain (0, ∞). Analysis We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 4. They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse. f (x) f (x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 (a) 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 (b) 321 4 5 x Figure 4 (a ) Absolute value (b ) Reciprocal squared Try It #4 The domain of function f is (1, ∞) and the range of function f is (−∞, −2). Find the domain and range of the inverse function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 259 Finding and Evaluating Inverse Functions Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases. Inverting Tabular Functions Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range. Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Example 5 A function f (t) is given in Table 3, showing distance in miles that a car has traveled in t minutes. Find and interpret f −1(70). Interpreting the Inverse of a Tabular Function t (minutes) f (t) (miles) 30 20 50 40 70 60 90 70 Table 3 Solution The inverse function takes an output of f and returns an input for f. So in the expression f −1(70), 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function f, 90 minutes, so f −1(70) = 90. The interpretation of this is that, to drive 70 miles, it took 90 minutes. Alternatively, recall that the definition of the inverse was that if f (a) = b, then f −1(b) = a. By this definition, if we are given f −1(70) = a, then we are looking for a value a so that f (a) = 70. In this case, we are looking for a t so that f (t) = 70, which is when t = 90. Try It #5 Using Table 4, find and interpret a. f (60), and b. f −1(60). t (minutes) f (t) (miles) 30 20 60 50 70 60 90 70 50 40 Table 4 Evaluating the Inverse of a Function, Given a Graph of the Original Function We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the origin
al function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original function’s graph. How To… Given the graph of a function, evaluate its inverse at specific points. 1. Find the desired input on the y-axis of the given graph. 2. Read the inverse function’s output from the x-axis of the given graph. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 260 CHAPTER 3 FUNCTIONS Example 6 Evaluating a Function and Its Inverse from a Graph at Speciﬁc Points A function g(x) is given in Figure 5. Find g(3) and g−1(3). g(x) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 x Figure 5 Solution To evaluate g(3), we find 3 on the x-axis and find the corresponding output value on the y-axis. The point (3, 1) tells us that g(3) = 1. To evaluate g −1(3), recall that by definition g −1(3) means the value of x for which g(x) = 3. By looking for the output value 3 on the vertical axis, we find the point (5, 3) on the graph, which means g(5) = 3, so by definition, g −1(3) = 5. See Figure 6. g(x) 6 5 4 3 2 1 – –1 1– (5, 3) (3, 1) 1 2 3 4 5 6 7 x Figure 6 Try It #6 Using the graph in Figure 6, a. find g−1(1), and b. estimate g−1(4). Finding Inverses of Functions Represented by Formulas Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formula—for example, y as a function of x—we can often find the inverse function by solving to obtain x as a function of y. How To… Given a function represented by a formula, find the inverse. 1. Make sure f is a one-to-one function. 2. Solve for x. 3. Interchange x and y. Example 7 Inverting the Fahrenheit-to-Celsius Function Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature. 5 __ C = (F − 32) 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 261 Solution 5 __ (F − 32) C = 9 9 __ = F − 32 C ċ 5 9 __ C + 32 F = 5 By solving in general, we have uncovered the inverse function. If then 5 __ C = h(F) = (F − 32), 9 9 __ F = h−1(C) = C + 32. 5 In this case, we introduced a function h to represent the conversion because the input and output variables are descriptive, and writing C−1 could get confusing. Try It #7 1 __ (x − 5) Solve for x in terms of y given y = 3 Example 8 Solving to Find an Inverse Function Find the inverse of the function f (x) = 2 ____ x − 3 + 4. Solution y = + 4 Set up an equation. 2 ____ x − 3 2 _____ x − 3 2 ____ y − 4 2 ____ = Subtract 4 from both sides. Multiply both sides by x − 3 and divide by y − 4. x = + 3 Add 3 to both sides. So f −1 (y) = + 3 or f −1 (x) = 2 _____ y − 4 2 ____ x − 4 + 3. Analysis The domain and range of f exclude the values 3 and 4, respectively. f and f −1 are equal at two points but are not the same function, as we can see by creating Table 5 . x f (x) 1 3 2 2 Table 5 5 5 f −1(y) y Example 9 Solving to Find an Inverse with Radicals Find the inverse of the function f (x) = 2 + √ — x − 4 . Solution y − 2)2 = x − 4 x = (y − 2)2 + 4 So f −1 (x) = (x − 2)2 + 4. The domain of f is [4, ∞). Notice that the range of f is [2, ∞), so this means that the domain of the inverse function f −1 is also [2, ∞). Analysis The formula we found for f −1(x) looks like it would be valid for all real x. However, f −1 itself must have an inverse (namely, f) so we have to restrict the domain of f −1 to [2, ∞) in order to make f −1 a one-to-one function. This domain of f −1 is exactly the range of f. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 262 CHAPTER 3 FUNCTIONS Try It #8 What is the inverse of the function f (x) = 2 − √ — x ? State the domains of both the function and the inverse function. Finding Inverse Functions and Their Graphs Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function f (x) = x2 restricted to the domain [0, ∞), on which this function is one-to-one, and graph it as in Figure 7. f (x) 5 4 3 2 1 –1–1 –2 –5 –4 –3 –2 21 3 4 5 x Figure 7 Quadratic function with domain restricted to [0, ∞). Restricting the domain to [0, ∞) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain. We already know that the inverse of the toolkit quadratic function is the square root function, that is, f −1(x) = √ What happens if we graph both f and f −1 on the same set of axes, using the x-axis for the input to both f and f −1 ? We notice a distinct relationship: The graph of f −1(x) is the graph of f (x) reflected about the diagonal line y = x, which we will call the identity line, shown in Figure 8. — x . f (x) y = x f –1(x) 21 1–1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 8 Square and square-root functions on the non-negative domain This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes. Example 10 Given the graph of f (x) in Figure 9, sketch a graph of f −1(x). Finding the Inverse of a Function Using Reﬂection about the Identity Line Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 263 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 9 Solution This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of (0, ∞) and range of (−∞, ∞), so the inverse will have a domain of (−∞, ∞) and range of (0, ∞). If we reflect this graph over the line y = x, the point (1, 0) reflects to (0, 1) and the point (4, 2) reflects to (2, 4). Sketching the inverse on the same axes as the original graph gives Figure 10. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 f –1(x) y = x f (x) 21 3 4 5 x Figure 10 The function and its inverse, showing reﬂection about the identity line Try It #9 Draw graphs of the functions f and f −1 from Example 8. Q & A… Is there any function that is equal to its own inverse? Yes. If f = f −1, then f (f (x)) = x, and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because Any function f (x) = c − x, where c is a constant, is also equal to its own inverse. 1 _ = x 1 _ x Access these online resources for additional instruction and practice with inverse functions. • Inverse Functions (http://openstaxcollege.org/l/inversefunction) • One-to-one Functions (http://openstaxcollege.org/l/onetoone) • Inverse Function Values Using Graph (http://openstaxcollege.org/l/inversfuncgraph) • Restricting the Domain and Finding the Inverse (http://openstaxcollege.org/l/restrictdomain) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 264 CHAPTER 3 FUNCTIONS 3.7 SECTION EXERCISES VERBAL 1. n eff 2. fx=x 3. 4. 5. ALGEBRAIC 6. fx=a−xa 7. 8. 9. 10. , fi 11. fx=x+ 12. fx=x+ 13. fx=−x 14. fx=x + 15. fx=x + 16. fx=−x 17. fx=−x 21. fx=√x−+ — 25. fx= _____ x− 26. fx= x_____+ x+ 18. 22. 27. + fx=√x — − fx=+ √ x fx= x_____− x+ 19. fx=√−x — 23. 28. — fx=−√ x x______+ −x fx= 20. 24. 29. — fx= fx=+ √x− _____ x+ x______+ fx= −x 30. fx=x + x−∞ 31. fx=x + x+ −∞ 32. fx=x −x+ ∞ fif . Thn fif 33. fx=x+ 34. fx=x− 35. fx=x − 36. fx= x − gx= x − x a. fgxg fx b. fxgx f xgx 37. fx=√ x− g x=x + — 38. fx=−x+ g x= x− − GRAPHICAL 39. fx=√x — 40. — fx=√ x+ 41. fx=−x+ 42. fx=x − Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 SECTION EXERCISES 265 44. x – – – – y – – – – – – 45 43. 46 – – – – – – – – – – 47. y 48. y 49. y x x 50. y x x fFigure 11 – – – – 51. f 52. fx= 53. f − 54. f −x Figure 11 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 266 CHAPTER 3 FUNCTIONS Figure 12 55. f − 56. ff − 57. f f 58. f f NUMERIC y –– – – –– – – – – f x Figure 12 f 59. f=f − 61. f −−=−f− 60. f=f − 62. f −−=−f− Table 6 x f (x) Table 6 63. f 65. f − 64. fx= 66. f −x= 67. fTable 7 f −x x f (x) Table 7 TECHNOLOGY , fi. Th fx= 68. fx=x − 69. _____ x− 70. fx= _ x− REAL-WORLD APPLICATIONS 71. xy __ x+ fx= 72. ThC Cr=πr rCrπ 73. Th tdt=t td t Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 267 CHAPTER 3 REVIEW Key Terms absolute maximum the greatest value of a function over an interval absolute minimum the lowest value of a function over an interval average rate of change the difference in the output values of a function found for two values of the input divided by the difference between the inputs composite function the new function formed by function composition, when the output of one function is used as the input of another decreasing function a function is decreasing in some open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a dependent variable an output variable domain the set of all possible input values for a relation even function a function whose graph is unchanged by horizontal reflection, f (x) = f (−x), and is symmetric about the y-axis function a relation in which each input value yields a unique output value horizontal compression a transformation that compresses a function’s graph horizontally, by multiplying the input by a constant b > 1 horizontal line test a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once horizontal reflection a transformation that reflects a function’s graph across the y-axis by multiplying the input by −1 horizontal shift a transformat
ion that shifts a function’s graph left or right by adding a positive or negative constant to the input horizontal stretch a transformation that stretches a function’s graph horizontally by multiplying the input by a constant 0 < b < 1 increasing function a function is increasing in some open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a independent variable an input variable input each object or value in a domain that relates to another object or value by a relationship known as a function interval notation a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion inverse function for any one-to-one function f (x), the inverse is a function f −1(x) such that f −1(f (x)) = x for all x in the domain of f; this also implies that f ( f −1(x)) = x for all x in the domain of f −1 local extrema collectively, all of a function’s local maxima and minima local maximum a value of the input where a function changes from increasing to decreasing as the input value increases. local minimum a value of the input where a function changes from decreasing to increasing as the input value increases. odd function a function whose graph is unchanged by combined horizontal and vertical reflection, f (x) = − f (−x), and is symmetric about the origin one-to-one function a function for which each value of the output is associated with a unique input value output each object or value in the range that is produced when an input value is entered into a function piecewise function a function in which more than one formula is used to define the output range the set of output values that result from the input values in a relation rate of change the change of an output quantity relative to the change of the input quantity relation a set of ordered pairs Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 268 CHAPTER 3 FUNCTIONS set-builder notation a method of describing a set by a rule that all of its members obey; it takes the form {x . statement about x} vertical compression a function transformation that compresses the function’s graph vertically by multiplying the output by a constant 0 < a < 1 vertical line test a method of testing whether a graph represents a function by determining whether a vertical line intersects the graph no more than once vertical reflection a transformation that reflects a function’s graph across the x-axis by multiplying the output by −1 vertical shift a transformation that shifts a function’s graph up or down by adding a positive or negative constant to the output vertical stretch a transformation that stretches a function’s graph vertically by multiplying the output by a constant a > 1 Key Equations Constant function f (x) = c, where c is a constant Identity function f (x) = x Absolute value function f (x) = .x. Reciprocal squared function f (x) = Quadratic function Cubic function Reciprocal function Square root function Cube root function Average rate of change f (x) = x2 f (x) = x3 1_ f (x) = x 1 _ x2 f (x) = √ — x f (x) = ∆y _ ∆x = — x 3 √ f (x2) − f (x1) _ x2 − x1 Composite function (f ∘#g)(x) = f (g(x)) Vertical shift g(x) = f (x) + k (up for k > 0) Horizontal shift g(x) = f (x − h) (right for h > 0) Vertical reflection g(x) = −f (x) Horizontal reflection g(x) = f (−x) Vertical stretch g(x) = af (x) (a > 0) Vertical compression g(x) = af (x) (0 < a < 1) Horizontal stretch g(x) = f (bx) (0 < b < 1) Horizontal compression g(x) = f (bx) (b > 1) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 269 Key Concepts 3.1 Functions and Function Notation • A relation is a set of ordered pairs. A function is a specific type of relation in which each domain value, or input, leads to exactly one range value, or output. See Example 1 and Example 2. • Function notation is a shorthand method for relating the input to the output in the form y = f (x). See Example 3 and Example 4. • In tabular form, a function can be represented by rows or columns that relate to input and output values. See Example 5. • To evaluate a function, we determine an output value for a corresponding input value. Algebraic forms of a function can be evaluated by replacing the input variable with a given value. See Example 6 and Example 7. • To solve for a specific function value, we determine the input values that yield the specific output value. See Example 8. • An algebraic form of a function can be written from an equation. See Example 9 and Example 10. • Input and output values of a function can be identified from a table. See Example 11. • Relating input values to output values on a graph is another way to evaluate a function. See Example 12. • A function is one-to-one if each output value corresponds to only one input value. See Example 13. • A graph represents a function if any vertical line drawn on the graph intersects the graph at no more than one point. See Example 14. • The graph of a one-to-one function passes the horizontal line test. See Example 15. 3.2 Domain and Range • The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number. • The domain of a function can be determined by listing the input values of a set of ordered pairs. See Example 1. • The domain of a function can also be determined by identifying the input values of a function written as an equation. See Example 2, Example 3, and Example 4. • Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation. See Example 5. • For many functions, the domain and range can be determined from a graph. See Example 6 and Example 7. • An understanding of toolkit functions can be used to find the domain and range of related functions. See Example 8, Example 9, and Example 10. • A piecewise function is described by more than one formula. See Example 11 and Example 12. • A piecewise function can be graphed using each algebraic formula on its assigned subdomain. See Example 13. 3.3 Rates of Change and Behavior of Graphs • A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is determined using only the beginning and ending data. See Example 1. • Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example 2. • Comparing pairs of input and output values in a table can also be used to find the average rate of change. See Example 3. • An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See Example 4 and Example 5. • The average rate of change can sometimes be determined as an expression. See Example 6. • A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See Example 7. • A local maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 270 CHAPTER 3 FUNCTIONS • A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values. • Minima and maxima are also called extrema. • We can find local extrema from a graph. See Example 8 and Example 9. • The highest and lowest points on a graph indicate the maxima and minima. See Example 10. 3.4 Composition of Functions • We can perform algebraic operations on functions. See Example 1. • When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function. • The function produced by combining two functions is a composite function. See Example 2 and Example 3. • The order of function composition must be considered when interpreting the meaning of composite functions. See Example 4. • A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function. • A composite function can be evaluated from a table. See Example 5. • A composite function can be evaluated from a graph. See Example 6. • A composite function can be evaluated from a formula. See Example 7. • The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See Example 8 and Example 9. • Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions. • Functions can often be decomposed in more than one way. See Example 10. 3.5 Transformation of Functions • A function can be shifted vertically by adding a constant to the output. See Example 1 and Example 2. • A function can be shifted horizontally by adding a constant to the input. See Example 3, Example 4, and Example 5. • Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal shifts. See Example 6. • Vertical and horizontal shifts are often combined. See Example 7 and Example 8. • A vertical reflection reflects a graph about the x-axis. A graph can be reflected vertically by multiplying the output by –1. • A horizontal reflection reflects a graph about the y-axis. A graph can be reflected horizontally by multiplying the input by –1. • A graph can be reflected both vertically and horizontally. The order in
which the reflections are applied does not affect the final graph. See Example 9. • A function presented in tabular form can also be reflected by multiplying the values in the input and output rows or columns accordingly. See Example 10. • A function presented as an equation can be reflected by applying transformations one at a time. See Example 11. • Even functions are symmetric about the y-axis, whereas odd functions are symmetric about the origin. • Even functions satisfy the condition f (x) = f (−x). • Odd functions satisfy the condition f (x) = −f (−x). • A function can be odd, even, or neither. See Example 12. • A function can be compressed or stretched vertically by multiplying the output by a constant. See Example 13, Example 14, and Example 15. • A function can be compressed or stretched horizontally by multiplying the input by a constant. See Example 16, Example 17, and Example 18. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 271 • The order in which different transformations are applied does affect the final function. Both vertical and horizontal transformations must be applied in the order given. However, a vertical transformation may be combined with a horizontal transformation in any order. See Example 19 and Example 20. 3.6 Absolute Value Functions • Applied problems, such as ranges of possible values, can also be solved using the absolute value function. See Example 1. • The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction. See Example 2. • In an absolute value equation, an unknown variable is the input of an absolute value function. • If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable. See Example 3. 3.7 Inverse Functions • If g(x) is the inverse of f (x), then g(f (x)) = f (g(x)) = x. See Example 1, Example 2, and Example 3. • Each of the toolkit functions has an inverse. See Example 4. • For a function to have an inverse, it must be one-to-one (pass the horizontal line test). • A function that is not one-to-one over its entire domain may be one-to-one on part of its domain. • For a tabular function, exchange the input and output rows to obtain the inverse. See Example 5. • The inverse of a function can be determined at specific points on its graph. See Example 6. • To find the inverse of a formula, solve the equation y = f (x) for x as a function of y. Then exchange the labels x and y. See Example 7, Example 8, and Example 9. • The graph of an inverse function is the reflection of the graph of the original function across the line y = x. See Example 10. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 272 CHAPTER 3 FUNCTIONS CHAPTER 3 REVIEW EXERCISES FUNCTIONS AND FUNCTION NOTATION For the following exercises, determine whether the relation is a function. 1. {(a, b), (c, d), (e, d)} 2. {(5, 2), (6, 1), (6, 2), (4, 8)} 3. y 2 + 4 = x, for x the independent variable and y the dependent variable 4. Is the graph in Figure 1 a function? y 25 20 15 10 5 –5 –5 –10 –15 –20 –25 –25 –20 –15 –10 f 5 10 15 20 25 x Figure 1 For the following exercises, evaluate the function at the indicated values: f (−3); f (2); f (−a); −f (a); f (a + h). 5. f (x) = −2x 2 + 3x 6. f (x) = 2.3x − 1. For the following exercises, determine whether the functions are one-to-one. 7. f (x) = −3x + 5 8. f (x) = .x − 3. For the following exercises, use the vertical line test to determine if the relation whose graph is provided is a function. 9. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 10. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 11. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, graph the functions. 12. f (x) = .x + 1. 13. f (x) = x 2 − 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 273 For the following exercises, use Figure 2 to approximate the values. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 14. f (2) 15. f (−2) 21 3 4 5 x 16. If f (x) = −2, then solve for x. 17. If f (x) = 1, then solve for x. Figure 2 For the following exercises, use the function h(t) = −16t 2 + 80t to find the values. 18. h(2) − h(1) __ 2 − 1 19. h(a) − h(1) __ a − 1 DOMAIN AND RANGE For the following exercises, find the domain of each function, expressing answers using interval notation. 20. f (x) = 2 _ 3x + 2 21. f (x) = x − 3 ___________ x 2 − 4x − 12 23. Graph this piecewise function: f (x) = { x + 1 −2x − 3 x < −2 x ≥ −2 — x − 6 _ 22. f (x) = # √ x − 4 √ — RATES OF CHANGE AND BEHAVIOR OF GRAPHS For the following exercises, find the average rate of change of the functions from x = 1 to x = 2. 24. f (x) = 4x − 3 25. f (x) = 10x 2 + x 26. f (x) = − #2 _ x 2 For the following exercises, use the graphs to determine the intervals on which the functions are increasing, decreasing, or constant. 27. y 28. –5 –4 –3 –2 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 29. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x 30. Find the local minimum of the function graphed in Exercise 27. 31. Find the local extrema for the function graphed in Exercise 28. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 274 CHAPTER 3 FUNCTIONS 32. For the graph in Figure 3, the domain of the function is [−3, 3]. The range is [−10, 10]. Find the absolute minimum of the function on this interval. 33. Find the absolute maximum of the function graphed in Figure 3. –5 –4 –3 –2 y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x Figure 3 COMPOSITION OF FUNCTIONS For the following exercises, find (f ∘#g)(x) and (g ∘ f)(x) for each pair of functions. 34. f (x) = 4 − x, g(x) = −4x 35. f (x) = 3x + 2, g(x) = 5 − 6x 36. f (x) = x 2 + 2x, g(x) = 5x + 1 37. f (x(x) = x 38. f (x) = , g(x _____ 2 For the following exercises, find (f ∘#g) and the domain for (f ∘#g)(x) for each pair of functions. 39. f (x __ , g(x) = x 42. f (x) = 1 _ x2 − 1 , g(x) = √ — x + 1 40. f (x) = , g(x(x) = √ 41. f (x) = — x For the following exercises, express each function H as a composition of two functions f and g where H(x) = (f ∘#g)(x). 43. H(x) = √ _______ 2x − 1 ______ 3x + 4 44. H(x) = 1 _ (3x2 − 4)−3 TRANSFORMATION OF FUNCTIONS For the following exercises, sketch a graph of the given function. 45. f (x) = (x − 3)2 46. f (x) = (x + 4)3 48. f (x) = −x3 49. f (x) = 3 √ — −x 51. f (x) = 4[.x − 2. − 6] 52. f (x) = −(x + 2)2 − 1 47. f (x) = √ — x + 5 50. f (x) = 5 √ — −x − 4 For the following exercises, sketch the graph of the function g if the graph of the function f is shown in Figure 4. y 5 4 3 2 1 –1 –1 –2 –5 –4 –3 –2 21 3 4 5 x Figure 4 53. g(x) = f (x − 1) 54. g(x) = 3f (x) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 275 For the following exercises, write the equation for the standard function represented by each of the graphs below. 55. y 56. y –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, determine whether each function below is even, odd, or neither. 57. f (x) = 3x4 58. g(x) = √ — x 1 _ x + 3x 59. h(x) = For the following exercises, analyze the graph and determine whether the graphed function is even, odd, or neither. 60. y 61. y 62. y –25 –20 –15 –10 25 20 15 10 5 –5 –5 –10 –15 –20 –25 5 10 15 20 25 x –10 –8 –6 –4 25 20 15 10 5 –2 –5 –10 –15 –20 –25 42 6 8 10 x –10 –8 –6 –4 ABSOLUTE VALUE FUNCTIONS For the following exercises, write an equation for the transformation of f (x) = \| x \|. 63. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 64. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 65. 21 3 4 5 x –5 –4 –3 –2 10 8 6 4 2 –2 –2 –4 –6 –8 –10 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 42 6 8 10 x 21 3 4 5 x For the following exercises, graph the absolute value function. 66. f (x) = \| x − 5 \| 67. f (x) = −\| x − 3 \| 68. f (x) = \| 2x − 4 \| Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 276 CHAPTER 3 FUNCTIONS INVERSE FUNCTIONS For the following exercises, find f −1(x) for each function. 69. f (x) = 9 + 10x 70. f (x) = x _ x + 2 For the following exercise, find a domain on which the function f is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain. 71. f (x) = x 2 + 1 72. Given f (x) = x 3 − 5 and g(x. Find f (g(x)) and g(f (x)). b. What does the answer tell us about the relationship between f (x) and g(x)? For the following exercises, use a graphing utility to determine whether each function is one-to-one. 1 _ x 73. f (x) = 76. If f (1) = 4, find f −1(4). 74. f (x) = −3x 2 + x 75. If f (5) = 2, find f −1(2). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 PRACTICE TEST 277 CHAPTER 3 PRACTICE TEST For the following exercises, determine whether each of the following relations is a function. 1. y = 2x + 8 2. {(2, 1), (3, 2), (−1, 1), (0, −2)} For the following exercises, evaluate the function f (x) = −3x 2 + 2x at the given input. 3. f (−2) 4. f (a) 5. Show that the function f (x) = −2(x − 1)2 + 3 is not one-to-one. 6. Write the domain of the function f (x) = √ — 3 − x in interval notation. 7. Given f (x) = 2x 2 − 5x, find f (a + 1) − f (1). 8. Graph the function f (x) = { x + 1 if −2 < x < 3 x ≥ 3 −x if 9. Find the average rate of change of the function f (x) = 3 − 2x 2 + x by finding f (b) − f (a) _ b − a . For the following exercises, use the functions f (x) = 3 − 2x 2 + x and g(x) = √ — x to find the composite functions. 10. ( g ∘ f )(x) 11. ( g ∘ f )(1) 12. Express H(x) = 3 √ — 5x 2 − 3x as a composition of two functions, f and g, where ( f ∘#g )(x) = H(x). For the following exercises, graph the functions by translating, stretching, and/or compressing a toolkit function. 13. f (x) = √ — x + 6 − 1 14. f (x) = 1 _ x + 2 − 1 For the following exercises, dete
rmine whether the functions are even, odd, or neither. 5 _ 15. f (x) = − x2 + 9x 6 1_ 17. f (x) = x 5 _ 16. f (x) = − x 3 + 9x 5 18. Graph the absolute value function f (x) = −2\| x − 1 \| + 3. For the following exercises, find the inverse of the function. 19. f (x) = 3x − 5 20. f (x) = 4 _____ x + 7 For the following exercises, use the graph of g shown in Figure 1. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21. On what intervals is the function increasing? 22. On what intervals is the function decreasing? 21 3 4 5 x 23. Approximate the local minimum of the function. Express the answer as an ordered pair. 24. Approximate the local maximum of the function. Express the answer as an ordered pair. Figure 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 278 CHAPTER 3 FUNCTIONS For the following exercises, use the graph of the piecewise function shown in Figure 2. 25. Find f (2). 26. Find f (−2). 27. Write an equation for the piecewise function. y 5 4 3 2 1 f 21 3 4 5 x –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 Figure 2 For the following exercises, use the values listed in Table 1. x F (x 11 6 13 7 15 8 17 Table 1 28. Find F (6). 29. Solve the equation F (x) = 5. 30. Is the graph increasing or decreasing on its domain? 31. Is the function represented by the graph one-to-one? 32. Find F −1(15). 33. Given f (x) = −2x + 11, find f −1(x). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Linear Functions 4 Figure 1 A bamboo forest in China (credit: “JFXie”/Flickr) CHAPTER OUTLINE 4.1 Linear Functions 4.2 Modeling with Linear Functions 4.3 Fitting Linear Models to Data Introduction Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour.[6] In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function. Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data. 6 http://www.guinnessworldrecords.com/records-3000/fastest-growing-plant/ 279 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 280 CHAPTER 4 LINEAR FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Represent a linear function. • Determine whether a linear function is increasing, decreasing, or constant. • Interpret slope as a rate of change. • Write and interpret an equation for a linear function. • Graph linear functions. • Determine whether lines are parallel or perpendicular. • Write the equation of a line parallel or perpendicular to a given line. 4.1 LINEAR FUNCTIONS Figure 1 Shanghai MagLev Train (credit: “kanegen”/Flickr) Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train (Figure 1). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.[7] Suppose a maglev train travels a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time. Representing Linear Functions The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method. Representing a Linear Function in Word Form Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship. • The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed. The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station. 7 http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 281 Representing a Linear Function in Function Notation Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept form of a line, where x is the input value, m is the rate of change, and b is the initial value of the dependent variable. Equation form y = mx + b Function notation f (x) = mx + b In the example of the train, we might use the notation D(t) in which the total distance D is a function of the time t. The rate, m, is 83 meters per second. The initial value of the dependent variable b is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train. Representing a Linear Function in Tabular Form D(t) = 83t + 250 A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 2. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time. 1 second 1 second 1 second t D(t) 0 250 1 333 2 416 3 499 83 meters 83 meters 83 meters Figure 2 Tabular representation of the function D showing selected input and output values Q & A… Can the input in the previous example be any real number? No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers. Representing a Linear Function in Graphical Form Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, D(t) = 83t + 250, to draw a graph, represented in Figure 3. Notice the graph is a line. When we plot a linear function, the graph is always a line. The rate of change, which is constant, determines the slant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph that the y-intercept in the train example we just saw is (0, 250) and represents the distance of the train from the station when it began moving at a constant speed 500 400 300 200 100 0 1 2 3 4 Time (s) 5 Figure 3 The graph of D(t) = 83t + 250. Graphs of linear functions are lines because the rate of change is constant. Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line f (x) = 2x + 1. Ask yourself what numbers can be input to the function. In other words, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 282 CHAPTER 4 LINEAR FUNCTIONS linear function A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line f (x) = mx + b where b is the initial or starting value of the function (when input, x = 0), and m is the constant rate of change, or slope of the function. The y-intercept is at (0, b). Example 1 Using a Linear Function to Find the Pressure on a Diver The pressure, P, in pounds per square inch (PSI) on the diver in Figure 4 depends upon her depth below the water surface, d, in feet. This relationship may be modeled by the equation, P(d) = 0.434d + 14.696. Restate this function in words. Figure 4 (credit: Ilse Reijs and Jan-Noud Hutten) Solution To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths. Analysis The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases. Determining Whether a Linear Function Is Increasing, Decreasing, or Constant The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with
a positive slope slants upward from left to right as in Figure 5(a). For a decreasing function, the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 5(b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 5(c). Increasing function Decreasing function Constant function f (x) f (x) f (x) f f f x x x (a) (b) Figure 5 (c) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 283 increasing and decreasing functions The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function. • f (x) = mx + b is an increasing function if m > 0. • f (x) = mx + b is an decreasing function if m < 0. • f (x) = mx + b is a constant function if m = 0. Example 2 Deciding Whether a Function Is Increasing, Decreasing, or Constant Some recent studies suggest that a teenager sends an average of 60 texts per day.[8] For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant. a. The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent. b. A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month. c. A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month. Solution Analyze each function. a. The function can be represented as f (x) = 60x where x is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day. b. The function can be represented as f (x) = 500 − 60x where x is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after x days. c. The cost function can be represented as f (x) = 50 because the number of days does not affect the total cost. The slope is 0 so the function is constant. Interpreting Slope as a Rate of Change In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Recall that given two values for the input, x1 and x2, and two corresponding values for the output, y1 and y2—which can be represented by a set of points, (x1, y1) and (x2, y2)—we can calculate the slope m. m = change in output (rise) __ change in input (run) = ∆y _ ∆x = y2 − y1 _ x2 − x1 Note in function notation two corresponding values for the output y1 and y2 for the function f, y1 = f (x1) and y2 = f (x2), so we could equivalently write Figure 6 indicates how the slope of the line between the points, (x1, y1) and (x2, y2), is calculated. Recall that the slope measures steepness, or slant. The greater the absolute value of the slope, the steeper the line is. m = f (x2) − f (x1) _ x2 − x1 8 http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 284 1. CHAPTER 4 LINEAR FUNCTIONS x2 – x1 y2 – y1 y 10 x2 , y2) (x1 , y1) 0 –1 321 4 ∆y ∆x = y2 – y1 x2 – x1 m = x Figure 6 The slope of a function is calculated by the change in y divided by the change in x. It does not matter which coordinate is used as the (x2, y2) and which is the (x1, y1), as long as each calculation is started with the elements from the same coordinate pair. Q & A… Are the units for slope always units for the output __ ? units for the input Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input. calculate slope The slope, or rate of change, of a function m can be calculated according to the following: m = change in output (rise) __ change in input (run) = ∆y _ ∆x = y2 − y1 _ x2 − x1 where x1 and x2 are input values, y1 and y2 are output values. How To… Given two points from a linear function, calculate and interpret the slope. 1. Determine the units for output and input values. 2. Calculate the change of output values and change of input values. 3. Interpret the slope as the change in output values per unit of the input value. Example 3 Finding the Slope of a Linear Function If f (x) is a linear function, and (3, −2) and (8, 1) are points on the line, find the slope. Is this function increasing or decreasing? Solution The coordinate pairs are (3, −2) and (8, 1). To find the rate of change, we divide the change in output by the change in input. m = change in output __ = change in input 1 − (−2) ________ 8 − 3 3 __ = 5 We could also write the slope as m = 0.6. The function is increasing because m > 0. Analysis As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used. Note that if we had reversed them, we would have obtained the same slope. m = (−2) − (1) _________ 3 − 8 = −3 ___ −5 3 _ = 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 285 Try It #1 If f (x) is a linear function, and (2, 3) and (0, 4) are points on the line, find the slope. Is this function increasing or decreasing? Example 4 Finding the Population Change from a Linear Function The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012. Solution The rate of change relates the change in population to the change in time. The population increased by 27,800 − 23,400 = 4,400 people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years. 4,400 people __ 4 years = 1,100 people__ year So the population increased by 1,100 people per year. Analysis Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable. Try It #2 The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012. Writing and Interpreting an Equation for a Linear Function Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form. Now we can choose which method to use to write equations for linear functions based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f in Figure 7. y f 10 8 6 4 2 –2 –2 –4 (0, 7) (4, 4) 42 6 8 10 x Figure 7 –10 –8 –6 –4 We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose (0, 7) and (4, 4). We can use these points to calculate the slope. m = y2 − y1 _ x2 − x1 = 4 − 7 _____ 4 − 0 = − #3 _ 4 Now we can substitute the slope and the coordinates of one of the points into the point-slope form. y − y1 = m(x − x1) y − 4 = − #3 _ (x − 4) 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 286 CHAPTER 4 LINEAR FUNCTIONS If we want to rewrite the equation in the slope-intercept form, we would find y − 4 = − #3 _ (x − 4) 4 y − 4 = − #3 _ x + 7 4 If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, b = 7. We now have the initial value b and the slope m so we can substitute m and b into the slope-intercept form of a line. f (x) = mx + b ↑ ↑ − #3 _ 7 4 f (x) = − #3 _ x + 7 4 So the function is f (x) = − #3 x + 7, and the linear equation would be y = − #3 _ _ x + 7. 4 4 How To… Given the graph of a linear function, write an equation to represent the function. 1. Identify two points on the line. 2. Use the two points to calculate the slope. 3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection. 4. Substitute the slope and y-intercept into the slope-intercept form of a line equation. Example 5 Writing an Equation for a Linear Function Write an equation for a linear function given a graph of f shown in Figure 8. y f 642 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 8 Solution Identify two points on the line, such as (0, 2) and (−2, −4). Use the points to calculate the slope. m = y2 − y1 _ x2 − x1 −4 − 2 _______ −2 − 0 −6 ___ −2 = = Substitute the slope and the coordinates of one of the points into the point-slope form. = 3 y − y1 = m(x − x1) y − (−4) = 3(x − (−2)) y + 4 = 3(x + 2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 287 We can use algebra to rewrite the equation in the slope-intercept form. y + 4 = 3(x + 2) y + 4 = 3x + 6 y = 3x + 2 Analysis This makes sense because we can see from Figure 9 that the line crosses the y-axis at the point (0, 2), which is the y-intercept, so b = 2. y 10 8 6 4 2 –10 –2 –4 –6 –8 (−2, −4) –2 –4 –6 –8 –10 (0, 2) 642 8 10 x Example 6 Writing an Equation for a Linear Cost Function Figure 9 Suppose Ben starts a company in wh
ich he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function C where C(x) is the cost for x items produced in a given month. Solution The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by C(x) = 1250 + 37.5x. Analysis If Ben produces 100 items in a month, his monthly cost is found by substitution 100 for x. So his monthly cost would be $5,000. C(100) = 1,250 + 37.5(100) = 5,000 Example 7 Writing an Equation for a Linear Function Given Two Points If f is a linear function, with f (3) = −2 , and f (8) = 1 , find an equation for the function in slope-intercept form. Solution We can write the given points using coordinates. We can then use the points to calculate the slope. f (3) = −2 → (3, −2) f (8) = 1 → (8, 1) m = y2 − y1 _ x2 − x1 1 − (−2) ________ 8 − 3 = Substitute the slope and the coordinates of one of the points into the point-slope form. = #3 _ 5 y − y1 = m(x − x1) y − (−2) =# #3 _ (x − 3) 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 288 CHAPTER 4 LINEAR FUNCTIONS We can use algebra to rewrite the equation in the slope-intercept form. y + 2 =# #3 _ (x − 3) 5 x −# #9 y + 2 =# #3 _ _ 5 5 19_ y = #3 _ x − 5 5 Try It #3 If f (x) is a linear function, with f (2) = −11, and f (4) = −25, find an equation for the function in slope-intercept form. Modeling Real-World Problems with Linear Functions In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems. How To… Given a linear function f and the initial value and rate of change, evaluate f (c). 1. Determine the initial value and the rate of change (slope). 2. Substitute the values into f (x) = mx + b. 3. Evaluate the function at x = c. Example 8 Using a Linear Function to Determine the Number of Songs in a Music Collection Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of months. How many songs will he own in a year? Solution The initial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means that b = 200. The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line. f (x) = mx + b ↑ ↑ 15 200 N(t) = 15t + 200 Figure 10 We can write the formula N(t) = 15t + 200. With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at t = 12. N(12) = 15(12) + 200 = 180 + 200 = 380 Marcus will have 380 songs in 12 months. Analysis Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 289 Example 9 Using a Linear Function to Calculate Salary Based on Commission Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I(n), and interpret the meaning of the components of the equation. Solution The given information gives us two input-output pairs: (3,760) and (5,920). We start by finding the rate of change. m = 920 − 760 ________ 5 − 3 $160 _______ 2 policies = $80 per policy = Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week. We can then solve for the initial value. I(n) = 80n + b 760 = 80(3) + b when n = 3, I(3) = 760 760 − 80(3) = b 520 = b The value of b is the starting value for the function and represents Ilya’s income when n = 0, or when no new policies are sold. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold. We can now write the final equation. I(n) = 80n + 520 Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold. Example 10 Using Tabular Form to Write an Equation for a Linear Function Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation. Number of weeks, w 0 2 4 6 Number of rats, P(w) 1,000 1,080 1,160 1,240 Table 1 Solution We can see from the table that the initial value for the number of rats is 1,000, so b = 1,000. Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week. P(w) = 40w + 1000 If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240) m = 1240 − 1080 __________ 6 − 2 = 160___ 4 = 40 Q & A… Is the initial value always provided in a table of values like Table 1? No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f (x) = mx + b, and solve for b. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 290 CHAPTER 4 LINEAR FUNCTIONS Try It #4 A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2 shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x), where x is the number of months since the start of the experiment. x 0 2 4 8 12 H(x) 12.5 13.5 14.5 16.5 18.5 Table 2 Graphing Linear Functions Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third method is by using transformations of the identity function f (x) = x. Graphing a Function by Plotting Points To fi nd points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs.We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, f (x) = 2x, we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. How To… Given a linear function, graph by plotting points. 1. Choose a minimum of two input values. 2. Evaluate the function at each input value. 3. Use the resulting output values to identify coordinate pairs. 4. Plot the coordinate pairs on a grid. 5. Draw a line through the points. Example 11 Graphing by Plotting Points Graph f (x) = − #2 __ x + 5 by plotting points. 3 Solution Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6. Evaluate the function at each input value, and use the output value to identify coordinate pairs0) = − #2 __ (0) + 5 = 5 ⇒ (0, 5) 3 f (3) = − #2 __ (3) + 5 = 3 ⇒ (3, 3) 3 f (6) = − #2 __ (6) + 5 = 1 ⇒ (6, 1) 3 Plot the coordinate pairs and draw a line through the points. Figure 11 represents the graph of the function 2 __ x + 5. f (x) = − 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 291 f(x) (0, 5) f 6 5 4 3 2 1 (3, 3) (6, 1) – – __ x + 5. Figure 11 The graph of the linear function f (x) = − # 3 Analysis The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function. Try It #5 Graph f (x) = − #3 _ x + 6 by plotting points. 4 Graphing a Function Using y-intercept and Slope Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set x = 0 in the equation. The other characteristic of the linear function is its slope. Let’s consider the following function. f (x) =# #1 _ x + 1 2 1 _ The slope is . Because the slo
pe is positive, we know the graph will slant upward from left to right. The y-intercept is 2 the point on the graph when x = 0. The graph crosses the y-axis at (0, 1). Now we know the slope and the y-intercept. rise _ We can begin graphing by plotting the point (0, 1). We know that the slope is rise over run, m = run . From our example, we have m =# #1 _ , which means that the rise is 1 and the run is 2. So starting from our y-intercept (0, 1), we can 2 rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 12. y 5 4 3 2 (0, 1) –2 –1 f y-intercept ←Rise = 1 ↑Run = 2 1 2 4 3 Figure 12 x 5 6 7 graphical interpretation of a linear function In the equation f (x) = mx + b • b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis. • m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope: m = change in output (rise) ___ = change in input (run) ∆y _ ∆x = y2 − y1 _ x2 − x1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 292 CHAPTER 4 LINEAR FUNCTIONS Q & A… Do all linear functions have y-intercepts? Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.) How To… Given the equation for a linear function, graph the function using the y-intercept and slope. 1. Evaluate the function at an input value of zero to find the y-intercept. 2. Identify the slope as the rate of change of the input value. 3. Plot the point represented by the y-intercept. 4. Use rise _ run to determine at least two more points on the line. 5. Sketch the line that passes through the points. Example 12 Graphing by Using the y-intercept and Slope Graph f (x) = − #2 _ x + 5 using the y-intercept and slope. 3 Solution Evaluate the function at x = 0 to find the y-intercept. The output value when x = 0 is 5, so the graph will cross the y-axis at (0, 5). According to the equation for the function, the slope of the line is − #2 _ . This tells us that for each vertical decrease in 3 the “rise” of −2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in Figure 13. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a line through the points. f f(x) 6 5 4 3 2 1 – – __ x + 5 and shows how to calculate the rise over run for the slope. Figure 13 Graph of f (x ) = − 3 Analysis The graph slants downward from left to right, which means it has a negative slope as expected. Try It #6 Find a point on the graph we drew in Example 12 that has a negative x-value. Graphing a Function Using Transformations Another option for graphing is to use a transformation of the identity function f (x) = x. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression. Vertical Stretch or Compression In the equation f (x) = mx, the m is acting as the vertical stretch or compression of the identity function. When m is negative, there is also a vertical reflection of the graph. Notice in Figure 14 that multiplying the equation of f (x) = x by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if 0 < m < 1. This means the larger the absolute value of m, the steeper the slope. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 293 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 1 2 3 4 5 6 x Figure 14 Vertical stretches and compressions and reﬂections on the function f (x) = x. Vertical Shift In f (x) = mx + b, the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Notice in Figure 15 that adding a value of b to the equation of f (x) = x shifts the graph of f a total of b units up if b is positive and .b. units down if b is negative(x) = f(x) = f(x) = f(x) = f(x) = x 42 6 8 10 y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 Figure 15 This graph illustrates vertical shifts of the function f (x) = x. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. How To… Given the equation of a linear function, use transformations to graph the linear function in the form f (x) = mx + b. 1. Graph f (x) = x. 2. Vertically stretch or compress the graph by a factor m. 3. Shift the graph up or down b units. Example 13 Graphing by Using Transformations Graph f (x) = #1 __ x − 3 using transformations. 2 1 Solution The equation for the function shows that m =# #1 __ __ . The so the identity function is vertically compressed by 2 2 equation for the function also shows that b = −3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 16. Then show the vertical shift as in Figure 17. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 294 1. CHAPTER 4 LINEAR FUNCTIONS y = x y = x1 2 x 5 6 7 21 3 4 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –7 –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –7 –6 –5 –4 –3 –2 y = x1 2 y = x − 3 1 2 x 21 3 4 5 6 7 Figure 16 The function, y = x, compressed by a factor of 1 __ . 2 1 __ x , shifted down 3 units. Figure 17 The function y =# 2 Try It #7 Graph f (x) = 4 + 2x, using transformations. Q & A… In Example 15, could we have sketched the graph by reversing the order of the transformations? No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2. f (2) =# #1 __ (22 Writing the Equation for a Function from the Graph of a Line Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 18. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept. y f 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 18 Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (−2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be m = rise run =# #4 _ _ = 2 2 Substituting the slope and y-intercept into the slope-intercept form of a line gives y = 2x + 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 295 How To… Given a graph of linear function, find the equation to describe the function. 1. Identify the y-intercept of an equation. 2. Choose two points to determine the slope. 3. Substitute the y-intercept and slope into the slope-intercept form of a line. Example 14 Matching Linear Functions to Their Graphs Match each equation of the linear functions with one of the lines in Figure 19. a. f (x) = 2x + 3 b. g(x) = 2x − 3 c. h(x) = −2x + 3 d. j(x) =# #1 __ 2 x + 3 –7 –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 I II III 21 3 4 5 6 7 x IV Figure 19 Solution Analyze the information for each function. a. This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g has the same slope, but a different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f must be represented by line I. b. This function also has a slope of 2, but a y-intercept of −3. It must pass through the point (0, −3) and slant upward from left to right. It must be represented by line III. c. This function has a slope of −2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right. 1 _ and a y-intercept of 3. It must pass through the point (0, 3) and slant upward d. This function has a slope of 2 from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II. Now we can re-label the lines as in Figure 20. j(x) = x + 3 1 2 –7 –6 –5 –4 –3 –2 g(x) = 2 x − 3 21 1 –1 –2 –3 –4 –5 f (x) = 2 x + 3 h(x) = −2 x + 3 Figure 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 296 CHAPTER 4 LINEAR FUNCTIONS Finding the x-intercept of a Line So far, we have been finding the y-intercept of a function: the point at which the graph of the function crosses the y-axis. Recall that a function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero. To fi nd the x-intercept, set a function f (x) equal to zero and solve for the value of x. For example, consider the function shown. Set the function equal to 0 and solve for x. f (x) = 3x − 6 0 = 3x − 6 6 = 3x 2 = x x = 2 The graph of the function crosses the x-axis at the point (2, 0). Q & A… Do all linear functions have x-intercepts? No. However, linear functions of the form y = c, where c is a nonzero real number
, are the only examples of linear functions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 21. y y = 5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 21 x-intercept The x-intercept of the function is value of x when f (x) = 0. It can be solved by the equation 0 = mx + b. Example 15 Finding an x-intercept Find the x-intercept of f (x) =# #1 _ x − 3. 2 Solution Set the function equal to zero to solve for x. 0 =# #1 _ x − 3 2 3 =# #1 __ x 2 6 = x x = 6 The graph crosses the x-axis at the point (6, 0). Analysis A graph of the function is shown in Figure 22. We can see that the x-intercept is (6, 0) as we expected. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 297 –10 –8 –6 –4 y 4 2 –2 –2 –4 –6 –8 –10 642 8 10 x Figure 22 Try It #8 Find the x-intercept of f (x) = #1 _ x − 4. 4 Describing Horizontal and Vertical Lines There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 23, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 in the equation f (x) = mx + b, the equation simplifies to f (x) = b. In other words, the value of the function is a constant. This graph represents the function f (x) = 24 −5 –4 –3 –2 –1 21 3 4 5 x Figure 23 A horizontal line representing the function f (x) = 2. A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined. m = change of output __ change of input ←#Non-zero real number ← 0 Figure 24 Example of how a line has a vertical slope. 0 in the denominator of the slope. Notice that a vertical line, such as the one in Figure 25, has an x-intercept, but no y-intercept unless it’s the line x = 0. This graph represents the line x = 2. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x x 2 2 y −4 −2 2 0 2 2 2 4 Figure 25 The vertical line, x = 2, which does not represent a function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 298 CHAPTER 4 LINEAR FUNCTIONS horizontal and vertical lines Lines can be horizontal or vertical. A horizontal line is a line defined by an equation in the form f (x) = b. A vertical line is a line defined by an equation in the form x = a. Example 16 Writing the Equation of a Horizontal Line Write the equation of the line graphed in Figure 26. –10 –8 –6 –4 y –2 –2 –4 –6 –8 –10 642 8 10 x f Figure 26 Solution For any x-value, the y-value is −4, so the equation is y = −4. Example 17 Writing the Equation of a Vertical Line Write the equation of the line graphed in Figure 27. y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 f 42 6 8 10 x Figure 27 Solution The constant x-value is 7, so the equation is x = 7. Determining Whether Lines are Parallel or Perpendicular The two lines in Figure 28 are parallel lines: they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the y-intercept of the other, they would become coincident1 –1 –2 –3 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 28 Parallel lines We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 299 f (x) = −2x + 6} f (x) = −2x − 4 parallel f (x) = 3x + 2} f (x) = 2x + 2 not parallel Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 29 are perpendicular. –6 –5 –4 –3 –2 y 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –7 –8 21 3 4 5 6 x Figure 29 Perpendicular lines Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if m1 and m2 are negative reciprocals of one another, they can be multiplied together to yield −1. m1m2 = −1 1 1 _ _ To fi nd the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is is 8. To , and the reciprocal of 8 8 find the negative reciprocal, first find the reciprocal and then change the sign. As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular. The product of the slopes is −1. f (x) =# #1 1 _ _ is −4 x + 2 negative reciprocal of 4 4 1 _ f (x) = −4x + 3 negative reciprocal of −4 is 4 1 −4 ( ) = −1 _ 4 parallel and perpendicular lines Two lines are parallel lines if they do not intersect. The slopes of the lines are the same. f (x) = m1x + b1 and g(x) = m2x + b2 are parallel if m1 = m2. If and only if b1 = b2 and m1 = m2, we say the lines coincide. Coincident lines are the same line. Two lines are perpendicular lines if they intersect at right angles. f (x) = m1x + b1 and g(x) = m2x + b2 are perpendicular if and only if m1m2 = −1, and so m2 = − # 1 ___ m1 Example 18 Identifying Parallel and Perpendicular Lines Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines. f (x) = 2x + 3 g(x) =# #1 _ x − 4 2 h(x) = −2x + 2 j(x) = 2x − 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 0 CHAPTER 4 LINEAR FUNCTIONS Solution Parallel lines have the same slope. Because the functions f (x) = 2x + 3 and j(x) = 2x − 6 each have a slope 1 _ of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and are negative 2 reciprocals, the equations, g(x) =# #1 _ x − 4 and h(x) = −2x + 2 represent perpendicular lines. 2 Analysis A graph of the lines is shown in Figure 30 . h(x) = −2x + 2 y f (x) = 2x + 3 j(x) = 2x − 6 g(x) = x − 4 1 2 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 30 The graph shows that the lines f (x) = 2x + 3 and j(x) = 2x − 6 are parallel, and the lines g(x) =# #1 _ 2 h(x) = −2x + 2 are perpendicular. x − 4 and Writing the Equation of a Line Parallel or Perpendicular to a Given Line If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line. Writing Equations of Parallel Lines Suppose for example, we are given the equation shown. f (x) = 3x + 1 We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line with a slope of 3 will be parallel to f (x). So the lines formed by all of the following functions will be parallel to f (x). g(x) = 3x + 6 h(x) = 3x + 1 p(x) = 3x +# #2 __ 3 Suppose then we want to write the equation of a line that is parallel to f and passes through the point (1, 7). This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value for b will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form. y − y1 = m(x − x1) y − 7 = 3(x − 1) y − 7 = 3x − 3 y = 3x + 4 So g(x) = 3x + 4 is parallel to f (x) = 3x + 1 and passes through the point (1, 7). How To… Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point. 1. Find the slope of the function. 2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line. 3. Simplify. Example 19 Finding a Line Parallel to a Given Line Find a line parallel to the graph of f (x) = 3x + 6 that passes through the point (3, 0). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 301 Solution The slope of the given line is 3. If we choose the slope-intercept form, we can substitute m = 3, x = 3, and f (x) = 0 into the slope-intercept form to find the y-intercept. g(x) = 3x + b 0 = 3(3) + b b = −9 The line parallel to f (x) that passes through (3, 0) is g(x) = 3x − 9. Analysis We can confirm that the two lines are parallel by graphing them. Figure 31 shows that the two lines will never intersect. y 5 4 3 2 1 Right 1 –6 –3 –4 –5 y = 3x + 6 –2 –1 –1 –2 –3 –4 –5 Up 3 Up 3 1 2 4 5 6 3 Right 1 x y = 3x − 9 Figure 31 Writing Equations of Perpendicular Lines We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function: f (x) = 2x + 4 . Any function with a slope of − #1 The slope of the line is 2, and its negative reciprocal is − #1 __ __ will be perpendicular to 2 2 f (x). So the lines formed by all of the following functions will be perpendicular to f (x). g(x) = − #1 __ x + 4 2 h(x) = − #1 __ x + 2 2 x − #1 p(x) = − #1 __ __ 2 2 As before, we can narrow down our choices for a particular perpendicular line if we know that it passe
s through a given point. Suppose then we want to write the equation of a line that is perpendicular to f (x) and passes through the point (4, 0). We already know that the slope is − #1 __ . Now we can use the point to find the y-intercept by substituting 2 the given values into the slope-intercept form of a line and solving for b. g(x) = mx + b 0 = − #1 __ (4) + b 2 0 = −2 + b 2 = b b = 2 The equation for the function with a slope of − #1 __ and a y-intercept of 2 is 2 g(x) = − #1 __ x + 2. 2 So g(x) = − #1 __ x + 2 is perpendicular to f (x) = 2x + 4 and passes through the point (4, 0). Be aware that perpendicular 2 lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature. Q & A… A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not −1. Doesn’t this fact contradict the definition of perpendicular lines? No. For two perpendicular linear functions, the product of their slopes is −1. However, a vertical line is not a function so the definition is not contradicted. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 2 CHAPTER 4 LINEAR FUNCTIONS How To… Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line. 1. Find the slope of the function. 2. Determine the negative reciprocal of the slope. 3. Substitute the new slope and the values for x and y from the coordinate pair provided into g(x) = mx + b. 4. Solve for b. 5. Write the equation for the line. Example 20 Finding the Equation of a Perpendicular Line Find the equation of a line perpendicular to f (x) = 3x + 3 that passes through the point (3, 0). Solution The original line has slope m = 3, so the slope of the perpendicular line will be its negative reciprocal, or − #1 __ . Using this slope and the given point, we can find the equation for the line. 3 g(x) = − #1 __ x + b 3 0 = − #1 __ (3 The line perpendicular to f (x) that passes through (3, 0) is g(x) = − #1 __ x + 1. 3 Analysis A graph of the two lines is shown in Figure 321 –1 –2 –3 –6 –5 –4 –3 –2 y f (x) = 3 x + 6 g(x) = − x + 1 1 3 x 21 3 4 5 6 Note that that if we graph perpendicular lines on a graphing calculator using standard zoom, the lines may not appear to be perpendicular. Adjusting the window will make it possible to zoom in further to see the intersection more closely. Figure 32 Try It #9 Given the function h(x) = 2x − 4, write an equation for the line passing through (0, 0) that is a. parallel to h(x) b. perpendicular to h(x) How To… Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point. 1. Determine the slope of the line passing through the points. 2. Find the negative reciprocal of the slope. 3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values. 4. Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 303 Example 21 Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point A line passes through the points (−2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5). Solution From the two points of the given line, we can calculate the slope of that line. Find the negative reciprocal of the slope. m1 = 5 − 6 _______ 4 − (−2) = −1___ 6 = − #1 __ 6 m2 = −1_ − #1 __ 6 = −1 ( − #6 __ ) 1 = 6 We can then solve for the y-intercept of the line passing through the point (4, 5). g(x) = 6x + b 5 = 6(4) + b 5 = 24 + b −19 = b b = −19 The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is y = 6x − 19 Try It #10 A line passes through the points, (−2, −15) and (2,−3). Find the equation of a perpendicular line that passes through the point, (6, 4). Access this online resource for additional instruction and practice with linear functions. • Linear Functions (http://openstaxcollege.org/l/linearfunctions) • Finding Input of Function from the Output and Graph (http://openstaxcollege.org/l/ﬁndinginput) • Graphing Functions Using Tables (http://openstaxcollege.org/l/graphwithtable) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 4 CHAPTER 4 LINEAR FUNCTIONS 4.1 SECTION EXERCISES VERBAL 1. Terry is skiing down a steep hill. Terry’s elevation, E(t), in feet after t seconds is given by E(t) = 3000 − 70t. Write a complete sentence describing Terry’s starting elevation and how it is changing over time. 2. Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour? 3. A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours. 4. If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y-intercepts. 5. If a horizontal line has the equation f (x) = a and a vertical line has the equation x = a, what is the point of intersection? Explain why what you found is the point of intersection. ALGEBRAIC For the following exercises, determine whether the equation of the curve can be written as a linear function. 6. y =# #1 __ x + 6 4 9. 3x + 5y = 15 12. −2x 2 + 3y2 = 6 7. y = 3x − 5 10. 3x 2 + 5y = 15 13. − #x − 3 ______ = 2y 5 8. y = 3x 2 − 2 11. 3x + 5y 2 = 15 For the following exercises, determine whether each function is increasing or decreasing. 14. f (x) = 4x + 3 15. g(x) = 5x + 6 16. a(x) = 5 − 2x 17. b(x) = 8 − 3x 20. j(x) =# #1 __ x − 3 2 23. m(x) = − #3 __ x + 3 8 18. h(x) = −2x + 4 21. p(x) =# #1 __ x − 5 4 19. k(x) = −4x + 1 22. n(x) = − #1 __ x − 2 3 For the following exercises, find the slope of the line that passes through the two given points. 24. (2, 4) and (4, 10) 27. (8, −2) and (4, 6) 25. (1, 5) and (4, 11) 28. (6, 11) and (−4, 3) 26. (−1, 4) and (5, 2) For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible. 29. f (−5) = −4, and f (5) = 2 30. f (−1) = 4 and f (5) = 1 31. Passes through (2, 4) and (4, 10) 33. Passes through (−1, 4) and (5, 2) 32. Passes through (1, 5) and (4, 11) 34. Passes through (−2, 8) and (4, 6) 35. x-intercept at (−2, 0) and y-intercept at (0, −3) 36. x-intercept at (−5, 0) and y-intercept at (0, 4) For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither. 37. 4x − 7y = 10 7x + 4y = 1 39. 3y + 4x = 12 −6y = 8x + 1 40. 6x − 9y = 10 38. 3y + x = 12 −y = 8x + 1 3x + 2y = 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 SECTION EXERCISES 305 For the following exercises, find the x- and y-intercepts of each equation 41. f (x) = −x + 2 44. k(x) = −5x + 1 42. g(x) = 2x + 4 45. −2x + 5y = 20 43. h(x) = 3x − 5 46. 7x + 2y = 56 For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 47. Line 1: Passes through (0, 6) and (3, −24) Line 2: Passes through (−1, 19) and (8, −71) 48. Line 1: Passes through (−8, −55) and (10, 89) Line 2: Passes through (9, −44) and (4, −14) 49. Line 1: Passes through (2, 3) and (4, −1) Line 2: Passes through (6, 3) and (8, 5) 50. Line 1: Passes through (1, 7) and (5, 5) Line 2: Passes through (−1, −3) and (1, 1) 51. Line 1: Passes through (2, 5) and (5, −1) Line 2: Passes through (−3, 7) and (3, −5) For the following exercises, write an equation for the line described. 52. Write an equation for a line parallel to f (x) = −5x − 3 53. Write an equation for a line parallel to g (x) = 3x − 1 and passing through the point (2, −12). and passing through the point (4, 9). 54. Write an equation for a line perpendicular to 55. Write an equation for a line perpendicular to h (t) = −2t + 4 and passing through the point (−4, −1). p (t) = 3t + 4 and passing through the point (3, 1). GRAPHICAL For the following exercises, find the slope of the lines graphed. 57. 321 4 5 6 x –6 –5 –4 –3 –2 56. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 321 4 5 6 x For the following exercises, write an equation for the lines graphed. 58. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 59. 321 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 60. 321 4 5 6 x –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 321 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 6 61. CHAPTER 4 LINEAR FUNCTIONS y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 62. 321 4 5 x –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 63. 321 4 5 6 x –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 321 4 5 6 x For the following exercises, match the given linear equation with its graph in Figure 33. B 5 A 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 21 3 4 5 E F Figure 33 C D 64. f (x) = −x − 1 65. f (x) = −2x − 1 66. f (x) = − #1 __ x − 1 2 67. f (x) = 2 68. f (x) = 2 + x 69. f (x) = 3x + 2 For the following exercises, sketch a line with the given features. 70. An x-intercept of (−4, 0) and y-intercept of (0, −2) 71. An x-intercept of (−2, 0) and y-intercept of (0, 4) 72. A y-intercept of (0, 7) and slope − #3 __ 2 2 __ 73. A y-intercept of (0, 3) and slope 5 74. Passing through the points (−6, −2) and (6, −6) 75. Passing through the points (−3, −4) and (3, 0) For the following exercises, sketch the graph of each equation. 76. f (x) = −2x − 1 77. g(x) = −3x + 2 79. k(x) =# #2 __ x − 3 3 82. x = 3 80. f (t) = 3 + 2t 83. x = −2 For the following exercises, write the equation of the line shown in the graph. 85. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6
86. 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 87. 21 3 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –2 –3 –4 –5 –6 78. h(x) =# #1 __ x + 2 3 81. p(t) = −2 + 3t 84. r(x) = 4 88. 21 3 4 5 6 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 21 3 4 5 x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 SECTION EXERCISES 307 NUMERIC For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data. x g(x) x k(x) 89. 92. 95. 5 13 x 2 f (x) −4 5 15 0 5 −10 −25 −40 10 10 28 4 16 20 58 6 36 25 73 8 56 90. 93. 96. x h(x) x g(x) x k(x) 0 5 5 30 10 105 15 230 2 0 6 6 −19 −44 −69 4 0 6 2 31 6 106 8 231 91. 94. 0 x f (x) −5 x h (x) 2 13 5 20 4 23 10 45 8 43 15 70 10 53 TECHNOLOGY For the following exercises, use a calculator or graphing technology to complete the task. 97. If f is a linear function, f (0.1) = 11.5, and f (0.4) = −5.9, find an equation for the function. 98. Graph the function f on a domain of [−10, 10] : f (x) = 0.02x − 0.01. Enter the function in a graphing utility. For the viewing window, set the minimum value of x to be −10 and the maximum value of x to be 10. 99. Graph the function f on a domain of [−10, 10] : f (x) = 2,500x + 4,000 100. Table 3 shows the input, w, and output, k, for a linear function k. a. Fill in the missing values of the table. b. Write the linear function k, round to 3 decimal places. w −10 5.5 67.5 b k 30 −26 a −44 Table 3 101. Table 4 shows the input, p, and output, q, for a linear function q. a. Fill in the missing values of the table. b. Write the linear function k. p q 0.5 400 0.8 700 12 a b 1,000,000 Table 4 1 __ 102. Graph the linear function f on a domain of [−10, 10] for the function whose slope is and y-intercept is 8 31 __ . 16 Label the points for the input values of −10 and 10. 103. Graph the linear function f on a domain of [−0.1, 0.1] for the function whose slope is 75 and y-intercept is −22.5. Label the points for the input values of −0.1 and 0.1. 104. Graph the linear function f where f (x) = ax + b on the same set of axes on a domain of [−4, 4] for the following values of a and b. a. a = 2; b = 3 b. a = 2; b = 4 c. a = 2; b = −4 d. a = 2; b = −5 EXTENSIONS 105. Find the value of x if a linear function goes through the following points and has the following slope: (x, 2), (−4, 6), m = 3 106. Find the value of y if a linear function goes through the following points and has the following slope: (10, y), (25, 100), m = −5 107. Find the equation of the line that passes through the following points: (a, b) and (a, b + 1) 108. Find the equation of the line that passes through the following points: (2a, b) and (a, b + 1) 109. Find the equation of the line that passes through the following points: (a, 0) and (c, d) 110. Find the equation of the line parallel to the line g(x) = −0.01x + 2.01 through the point (1, 2). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 8 CHAPTER 4 LINEAR FUNCTIONS 111. Find the equation of the line perpendicular to the line g(x) = −0.01x + 2.01 through the point (1, 2). For the following exercises, use the functions f (x) = −0.1x + 200 and g(x) = 20x + 0.1. 112. Find the point of intersection of the lines f and g. 113. Where is f (x) greater than g (x)? Where is g (x) greater than f (x)? REAL-WORLD APPLICATIONS 114. At noon, a barista notices that she has $20 in her tip jar. If she makes an average of $0.50 from each customer, how much will she have in her tip jar if she serves n more customers during her shift? 115. A gym membership with two personal training sessions costs $125, while gym membership with five personal training sessions costs $260. What is cost per session? 117. A phone company charges for service according to the formula: C(n) = 24 + 0.1n, where n is the number of minutes talked, and C(n) is the monthly charge, in dollars. Find and interpret the rate of change and initial value. 119. A city’s population in the year 1960 was 287,500. In 1989 the population was 275,900. Compute the rate of growth of the population and make a statement about the population rate of change in people per year. 121. Suppose that average annual income (in dollars) for the years 1990 through 1999 is given by the linear function: I(x) = 1,054x + 23,286, where x is the number of years after 1990. Which of the following interprets the slope in the context of the problem? a. As of 1990, average annual income was $23,286. b. In the ten-year period from 1990–1999, average annual income increased by a total of $1,054. c. Each year in the decade of the 1990s, average annual income increased by $1,054. d. Average annual income rose to a level of $23,286 by the end of 1999. 116. A clothing business finds there is a linear relationship between the number of shirts, n, it can sell and the price, p, it can charge per shirt. In particular, historical data shows that 1,000 shirts can be sold at a price of $30, while 3,000 shirts can be sold at a price of $22. Find a linear equation in the form p(n) = mn + b that gives the price p they can charge for n shirts. 118. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 30 oz of beans. When she plants 34 stalks, each plant produces 28 oz of beans. Find a linear relationship in the form y = mn + b that gives the yield when n stalks are planted. 120. A town’s population has been growing linearly. In 2003, the population was 45,000, and the population has been growing by 1,700 people each year. Write an equation, P(t), for the population t years after 2003. 122. When temperature is 0 degrees Celsius, the Fahrenheit temperature is 32. When the Celsius temperature is 100, the corresponding Fahrenheit temperature is 212. Express the Fahrenheit temperature as a linear function of C, the Celsius temperature, F (C). a. Find the rate of change of Fahrenheit temperature for each unit change temperature of Celsius. b. Find and interpret F (28). c. Find and interpret F (−40). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 MODELING WITH LINEAR FUNCTIONS 309 LEARNING OBJECTIVES In this section, you will: • Build linear models from verbal descriptions. • Model a set of data with a linear function. 4.2 MODELING WITH LINEAR FUNCTIONS Figure 1 (credit: EEK Photography/Flickr) Emily is a college student who plans to spend a summer in Seattle. She has saved $3,500 for her trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent her situation? What would be the x-intercept, and what can she learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models. Building Linear Models from Verbal Descriptions When building linear models to solve problems involving quantities with a constant rate of change, we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them: 1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system. 2. Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value. 3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret. 4. Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem. 5. When needed, write a formula for the function. 6. Solve or evaluate the function using the formula. 7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically. 8. Clearly convey your result using appropriate units, and answer in full sentences when necessary. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 0 CHAPTER 4 LINEAR FUNCTIONS Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units. Output: M, money remaining, in dollars Input: t, time, in weeks So, the amount of money remaining depends on the number of weeks: M(t) We can also identify the initial value and the rate of change. Initial Value: She saved $3,500, so $3,500 is the initial value for M. Rate of Change: She anticipates spending $400 each week, so −$400 per week is the rate of change, or slope. Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week. The rate of change is constant, so we can start with the linear model M(t) = mt + b. Then we can substitute the intercept and slope provided. M(t) = mt + b ↑ ↑ −400 3500 M(t) = − 400t + 3500 To find the x-intercept, we set the output to zero, and solve for the input. 0 = −400t + 3500 t = 3500 ____ 400 = 8.75 The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks. When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—al
most no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is 0 ≤ t ≤ 8.75. In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model. Using a Given Intercept to Build a Model Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is −$250 per month. We can then use the slopeintercept form and the given information to develop a linear model. Now we can set the function equal to 0, and solve for x to find the x-intercept. f (x) = mx + b = −250x + 1000 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 MODELING WITH LINEAR FUNCTIONS 311 0 = −250x + 1000 1000 = 250x 4 = x x = 4 The x-intercept is the number of months it takes her to reach a balance of $0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan. Using a Given Input and Output to Build a Model Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output. How To… Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem. 1. Identify the input and output values. 2. Convert the data to two coordinate pairs. 3. Find the slope. 4. Write the linear model. 5. Use the model to make a prediction by evaluating the function at a given x-value. 6. Use the model to identify an x-value that results in a given y-value. 7. Answer the question posed. Example 1 Using a Linear Model to Investigate a Town’s Population A town’s population has been growing linearly. In 2004 the population was 6,200. By 2009 the population had grown to 8,100. Assume this trend continues. a. Predict the population in 2013. b. Identify the year in which the population will reach 15,000. Solution The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y-intercept would correspond to the year 0, more than 2,000 years ago! To make computation a little nicer, we will define our input as the number of years since 2004: Input: t, years since 2004 Output: P(t), the town’s population To predict the population in 2013 (t = 9), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope. To determine the rate of change, we will use the change in output per change in input. m = change in output __ change in input The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to t = 0, giving the point (0, 6200). Notice that through our clever choice of variable definition, we have “given” ourselves the y-intercept of the function. The year 2009 would correspond to t = 5, giving the point (5, 8100). The two coordinate pairs are (0, 6200) and (5, 8100). Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 2 CHAPTER 4 LINEAR FUNCTIONS m = 8100 − 6200 __________ 5 − 0 = 1900 ____ 5 = 380 people per year We already know the y-intercept of the line, so we can immediately write the equation: P(t) = 380t + 6200 To predict the population in 2013, we evaluate our function at t = 9. P(9) = 380(9) + 6,200 = 9,620 If the trend continues, our model predicts a population of 9,620 in 2013. To fi nd when the population will reach 15,000, we can set P(t) = 15000 and solve for t. 15000 = 380t + 6200 8800 = 380t t ≈ 23.158 Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027. Try It #1 A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs $0.25 to produce each doughnut. a. Write a linear model to repre sent the cost C of the company as a function of x, the number of doughnuts produced. b. Find and interpret the y-intercept. Try It #2 A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues. a. Predict the population in 2014. b. Identify the year in which the population will reach 54,000. Using a Diagram to Model a Problem It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful. Example 2 Using a Diagram to Model Distance Walked Anna and Emanuel start at the same intersection. Anna walks east at 4 miles per hour while Emanuel walks south at 3 miles per hour. They are communicating with a two-way radio that has a range of 2 miles. How long after they start walking will they fall out of radio contact? Solution In essence, we can partially answer this question by saying they will fall out of radio contact when they are 2 miles apart, which leads us to ask a new question: “How long will it take them to be 2 miles apart?” Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 MODELING WITH LINEAR FUNCTIONS 313 In this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be 2 miles apart. We can see that time will be our input variable, so we’ll define our input and output variables. Input: t, time in hours. Output: A(t), distance in miles, and E(t), distance in miles Because it is not obvious how to define our output variable, we’ll start by drawing a picture such as Figure 2. Anna walking east, 4 miles/hour Distance between them Emanuel walking south, 3 miles/hour Figure 2 Initial Value: They both start at the same intersection so when t = 0, the distance traveled by each person should also be 0. Thus the initial value for each is 0. Rate of Change: Anna is walking 4 miles per hour and Emanuel is walking 3 miles per hour, which are both rates of change. The slope for A is 4 and the slope for E is 3. Using those values, we can write formulas for the distance each person has walked. A(t) = 4t E(t) = 3t For this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the “starting point” at the intersection where they both started. Then we can use the variable, A, which we introduced above, to represent Anna’s position, and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, E, to represent Emanuel’s position, measured from the starting point in the southward direction. Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure. We can then define a third variable, D, to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful, as we can see from Figure 3. Recall that we need to know how long it takes for D, the distance between them, to equal 2 miles. Notice that for any given input t, the outputs A(t), E(t), and D(t) represent distances. E A D Figure 3 Figure 3 shows us that we can use the Pythagorean Theorem because we have drawn a right angle. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 4 CHAPTER 4 LINEAR FUNCTIONS Using the Pythagorean Theorem, we get: D(t)2 = A(t)2 + E(t)2 = (4t)2 + (3t)2 = 16t2 + 9t2 = 25t2 D(t) = ± √ 25t2 — Solve for D(t) using the square root. In this scenario we are considering only positive values of t, so our distance D(t) will always be positive. We can simplify this answer to D(t) = 5t. This means that the distance between Anna and Emanuel is also a linear function. Because D is a linear function, we can now answer the question of when the distance between them will reach 2 miles. We will set the output D(t) = 2 and solve for t. = ±5 \| t \| D(t) = 2 5t = 2 t =# #2 __ = 0.4 5 They will fall out of radio contact in 0.4 hours, or 24 minutes. Q & A…
Should I draw diagrams when given information based on a geometric shape? Yes. Sketch the figure and label the quantities and unknowns on the sketch. Example 3 Using a Diagram to Model Distance Between Cities There is a straight road leading from the town of Westborough to Agritown 30 miles east and 10 miles north. Partway down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough. If the town of Eastborough is located 20 miles directly east of the town of Westborough, how far is the road junction from Westborough? Solution It might help here to draw a picture of the situation. See Figure 4. It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates (30, 10), and Eastborough at (20, 0). Agritown (30, 10) (0, 0) Westborough (20, 0) Eastborough 20 miles Figure 4 Using this point along with the origin, we can find the slope of the line from Westborough to Agritown: m = 10 − 0 =# #1 __ ______ 30 − 0 3 Now we can write an equation to describe the road from Westborough to Agritown. W(x) =# #1 __ x 3 From this, we can determine the perpendicular road to Eastborough will have slope m = −3. Because the town of Eastborough is at the point (20, 0), we can find the equation: E(x) = −3x + b 0 = −3(20) + b b = 60 E(x) = −3x + 60 Substitute (20, 0) into the equation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 MODELING WITH LINEAR FUNCTIONS 315 We can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal, 1 __ x = −3x + 60 3 10 __ x = 60 3 10x = 180 x = 18 y = W(18) = #1 __ (18) 3 = 6 Substitute this back into W(x). The roads intersect at the point (18, 6). Using the distance formula, we can now find the distance from Westborough to the junction. distance#=# √ —— (x2 − x1)2 + (y2 − y1)2 — (18 − 0)2 + (6 − 0)2 = √ ≈ 18.974 miles Analysis One nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points. Try It #3 There is a straight road leading from the town of Timpson to Ashburn 60 miles east and 12 miles north. Partway down the road, it junctions with a second road, perpendicular to the first, leading to the town of Garrison. If the town of Garrison is located 22 miles directly east of the town of Timpson, how far is the road junction from Timpson? Modeling a Set of Data with Linear Functions Real-world situations including two or more linear functions may be modeled with a system of linear equations. Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure 5. Exactly one solution x g Infinitely many solutions y f (a) y y x g f x No solutions (c) (b) Figure 5 How To… Given a situation that represents a system of linear equations, write the system of equations and identify the solution. 1. Identify the input and output of each linear model. 2. Identify the slope and y-intercept of each linear model. 3. Find the solution by setting the two linear functions equal to one another and solving for x, or find the point of intersection on a graph. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 6 CHAPTER 4 LINEAR FUNCTIONS Example 4 Building a System of Linear Models to Choose a Truck Rental Company Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile[9]. When will Keep on Trucking, Inc. be the better choice for Jamal? Solution The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions in Table 1. Input Outputs Initial Value Rate of Change d, distance driven in miles K(d): cost, in dollars, for renting from Keep on Trucking M(d) cost, in dollars, for renting from Move It Your Way Up-front fee: K(0) = 20 and M(0) = 16 K(d) = $0.59/mile and P(d) = $0.63/mile Table 1 A linear function is of the form f (x) = mx + b. Using the rates of change and initial charges, we can write the equations K(d) = 0.59d + 20 M(d) = 0.63d + 16 Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when K(d) < M(d). The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the K(d) function is smaller. These graphs are sketched in Figure 6, with K(d) in red. $ 120 110 100 90 80 70 60 50 40 30 20 10 0 K(d) = 0.59d + 20 (100, 80) M(d) = 0.63d + 16 0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 d Figure 6 To find the intersection, we set the equations equal and solve: K(d) = M(d) 0.59d + 20 = 0.63d + 16 4 = 0.04d 100 = d d = 100 This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that K(d) is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is d > 100. Access this online resources for additional instruction and practice with linear function models. • Interpreting a Linear Function (http://openstaxcollege.org/l/interpretlinear) 9 Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 SECTION EXERCISES 317 4.2 SECTION EXERCISES VERBAL 1. 2. 3. 4. ALGEBRAIC 5. yx=f x=+x f(x 6. x __ f(x=– x f(x 7. y 8. __ f(x=– x f(x xg x=f(x=x f x 9. 10. 11. 12. 13. 14. Ptt P P 16. P y 17. 18. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 31 8 CHAPTER 4 LINEAR FUNCTIONS : Th. Th s fi 19. 20. W, t W 21. W 22. W xy 23. 24. : Thffl ffl 25. C, t 26. C 27. C 28. C xy 29. 30. GRAPHICAL Figure 7fiy tt y Figure 7 t 31. yyt 32. y 33. x 34. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 SECTION EXERCISES 319 Figure 8fiy tt y t Figure 8 35. yyt 36. y 37. x 38. NUMERIC fl Table 2 Year Mississippi Table 2 Hawaii 39. 40. 41. ft r? (Th fl Table 3 Year Indiana Table 3 Alabama 42. 43. 44. ft r? (Th Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 0 CHAPTER 4 LINEAR FUNCTIONS REAL-WORLD APPLICATIONS 45. a. 46. a. b. b. c. d. e. P t c. d. e. Pt f. f. 47. ys a fl a. x 48. ys a fl a. x b. y c. b. y c. 49. a. 50. e 285. Th a. P P b. b. 51. Th a. Rt b. c. 52. a. Rt b. c. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.2 SECTION EXERCISES 321 53. o diff s. Th . Th plus 54. o diff nies. Th Th 55. 56. 57. 58. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 32 2 CHAPTER 4 LINEAR FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Draw and interpret scatter plots. • Use a graphing utility to ﬁnd the line of best ﬁt. • Distinguish between linear and nonlinear relations. • Fit a regression line to a set of data and use the linear model to make predictions. 4.3 FITTING LINEAR MODELS TO DATA A professor is attempting to identify trends among final exam scores. His class has a mixture of students, so he wonders if there is any relationship between age and final exam scores. One way for him to analyze the scores is by creating a diagram that relates the age of each student to the exam score received. In this section, we will examine one such diagram known as a scatter plot. Drawing and Interpreting Scatter Plots A scatter plot is a graph of plotted points that may show a relationship between two sets of data. If the relationship is from a linear model, or a model that is nearly linear, the professor can draw conclusions using his knowledge of linear functions. Figure 1 shows a sample scatter plot. Final Exam Score vs. Age 100 90 80 70 60 50 40 30 20 10 10 20 30 40 50 Age Figure 1 A scatter plot of age and ﬁnal exam score variables. Notice this scatter plot does not indicate a linear relationship. The points do not appear to follow a trend. In other words, there does not appear to be a relationship between the age of the student and the score on the final exam. Example 1 Using a Scatter Plot to Investigate Cricket Chirps Table 1 shows the number of cricket chirps in 15 seconds, for several different air temperatures, in degrees Fahrenheit[10]. Plot this data, and determine whether the data appears to be linearly related. Chirps Temperature 44 80.5 35 70.5 20.4 57 33 66 31 68 35 72 18.5 52 37 73.5 26 53 Table 1 Cricket Chirps vs Air Temperature Solution Plotting this data, as depicted in Figure 2 suggests that there may be a trend. We can see from the trend in the data that the number of chirps increases as the temperature increases. The trend appears to be roughly linear, though certainly not perfectly so. 10 Selected data from http://classic.globe.gov/fsl/scientistsblog/2007/10/. Retrieved Aug 3, 2010. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.3 FITTING LINEAR MODELS TO DATA 323 Cricket Chirps vs. Temperature ) 90 80 70 60 50 40 0 0 10 20 30 40 50 Cricket Chirps in 15 Seconds Figure 2 Finding the Line of Best Fit Once we recognize a need for a linear function to model that data, the natural follow-up question is “what is that linear function?” One way to approximate our linear function is to sketch the line that seems to best fit the data. Then we can extend the line until we can verify the y-in

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