Solution Manual for Calculus and Its Applications, 2nd Edition

INSTRUCTOR’S SOLUTIONS MANUAL CALCULUS AND ITS APPLICATIONS BRIEF VERSION TWELFTH EDITION AND CALCULUS AND ITS APPLICATIONS SECOND EDITION Marvin L. Bittinger Indiana University Purdue University Indianapolis David J. Ellenbogen Community College of Vermont Scott A. Surgent Arizona State University The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright © 2020, 2015 by Pearson Education, Inc. Publishing as Pearson, 501 Boylston Street, Boston, MA 02116. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-13-518244-4 ISBN-10: 0-13-518244-1 Contents Chapter R: Functions, Graphs, and Models …………………………………………………… 1 Chapter R Test …………………………………………………………………………………… 86 Chapter 1: Differentiation …………………………………………………………………………. 90 Chapter 1 Test…………………………………………………………………………………… 235 Chapter 2: Exponential and Logarithmic Functions ……………………………………. 239 Chapter 2 Test ………………………………………………………………………………….. 314 Chapter 3: Applications of Differentiation ………………………………………………… 318 Chapter 3 Test ………………………………………………………………………………….. 510 Chapter 4: Integration …………………………………………………………………………….. 527 Chapter 4 Test ………………………………………………………………………………….. 658 Chapter 5: Applications of Integration ……………………………………………………… 664 Chapter 5 Test ………………………………………………………………………………….. 763 Chapter 6: Functions of Several Variables ………………………………………………… 772 Chapter 6 Test ………………………………………………………………………………….. 895 Chapter 7: Trigonometric Functions ………………………………………………………… 901 Chapter 7 Test ………………………………………………………………………………….. 944 Chapter 8: Differential Equations …………………………………………………………….. 949 Chapter 8 Test ………………………………………………………………………………… 1015 Chapter 9: Sequences and Series ……………………………………………………………. 1021 Chapter 9 Test ………………………………………………………………………………… 1104 Chapter 10: Probability Distributions ……………………………………………………… 1109 Chapter 10 Test ………………………………………………………………………………. 1160 Chapter 11: Systems and Matrices (online at bit.ly/2ScEtHt) ……………………… 1165 Chapter 11 Test……………………………………………………………………………….. 1263 Chapter 12: Combinatorics and Probability (online at bit.ly/2HyJiH0) ………… 1267 Chapter 12 Test……………………………………………………………………………….. 1349 iii Copyright © 2020 Pearson Education, Inc. iv Copyright © 2016 Pearson Education, Inc. Chapter R Functions, Graphs, and Models Exercise Set R.1 1. 2. 3. 4. x y 1 3 0 0 2 6 Graph y  x  1 . Graph y  x  4 . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y 2 2 0 4 3 7  x, y  2, 2 0, 4 3, 7 1 Graph y   x . 4  x, y  1,3 0, 0 2, 6 5. 5 Graph y   x  3 . 3 6. Graph y  2 x4. 3 We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. y x  x, y  3 6 0 4 3 2 Graph y  3x . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. Copyright © 2020 Pearson Education, Inc. 3, 6 0, 4 3, 2 2 7. Chapter R Functions, Graphs, and Models Graph x  y  5 . We solve for y first. x y 5 y  5 x subtract x from both sides y  x  5 commutative property Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y 1 6 0 5 2 3 Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y 2 0 2 1 6 2  x, y  2, 0 2,1 6, 2  x, y  1, 6 0,5 2,3 11. Graph 2 x  5 y  10 . 8. Graph x  y  4 . 12. Graph 5 x  6 y  12 . We solve for y first. 5 x  6 y  12 9. subtract 5 x from both sides 6 y  12  5 x 1 y divide both sides by  6 12  5 x  6 5 y  2  x 6 5 y  x2 6 We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. Graph 6 x  3 y  9 . 10. Graph 8 y  2 x  4 . We solve for y first. 8 y  2x  4 8 y  2x  4 2 4 x 8 8 1 1 y  x 4 2 y add 2x to both sides divide both sides by 8 x y 6 7 0 2 6 3 Copyright © 2020 Pearson Education, Inc.  x, y  6, 7 0, 2 6,3 Exercise Set R.1 3 16. Graph x  2  y 2 . Since x is expressed in terms of y we first choose values for y and then compute x. Then we plot the points that are found and connect them with a smooth curve. 13. Graph y  x 2  5 . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y  x, y  2 1 1 4 0 5 1 4 2 1 2, 1 1, 4 0, 5 1, 4 2, 1 x y  x, y  2 2 1 1 2 0 1 1 2 2 2, 2 1, 1 2, 0 1,1 2, 2 17. Graph y  5 . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. 14. Graph y  x 2  3 . x y 2 5 1 5 0 5 1 5 2 5 15. Graph x  y 2  2 . Copyright © 2020 Pearson Education, Inc.  x, y  2,5 1,5 0,5 1,5 2,5 4 Chapter R Functions, Graphs, and Models 18. Graph y  2 . 22. Graph y  1  x3 . First we solve for y. y  1  x3 y  x3  1 Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. 19. Graph y  7  x 2 . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. x y 2 3 1 6 0 7 1 6 2 3  x, y  2,3 1, 6 0, 7 1,6 2,3 23. y  x, y  2 9 1 2 0 1 1 0 2 7 2, 9 1, 2 0, 1 1,0 2, 7 A  0.5t 4  3.45t 3  96.65t 2  347.7t , 0t6 We substitute t  2 A  0.5  2  3.45  2  96.65  2  347.7  2 4 20. Graph y  5  x 2 . 3 2  344.4 Approximately 344.4 milligrams of ibuprofen will remain in the blood stream 2 hours after 400 mg have been swallowed. 24. 21. Graph y  7  x3 . x R  0.006 x  15.714 We substitute 1954 in for x to get R  0.006 1954  15.714  3.99 According to this model, the world record for the mile in 1954 is approximately 3.99 minutes. Likewise, we substitute 2000 in for x to get R  0.006  2000  15.714  3.71 According to the model, the world record for the mile in 2000 will be approximately 3.71 minutes. Copyright © 2020 Pearson Education, Inc. Exercise Set R.1 5 Finally, we substitute 2025 in for x to get R  0.006  2025  15.714  3.56 According to the model, the world record for the mile in 2025 will be approximately 3.56 minutes. 27. a. N  0.0319t  3.081 We substitute t  13 N  0.0319(13)  3.081  3.46 In 2020, the number of female athletes will be approximately 3.46 million. b. N  0.0319t  3.081 We substitute N  3.6 3.6  0.0319t  3.081 t  17.7 In the year 2007 + 17 = 2024, the number of female athletes will be approximately 3.6 million. 25. a. A  0.002 x 2  0.924 x  0.152 We substitute x  21.3 A  0.002(21.3) 2  0.924(21.3)  0.152  18.62 The optimum angle, in degrees, to tilt the solar panel in Honolulu is 18.62. b. A  0.002 x 2  0.924 x  0.152 We substitute x  39.1 A  0.002(39.1) 2  0.924(39.1)  0.152  32.919 The optimum angle, in degrees, to tilt the solar panel in Kansas City is 32.919. 28. a. N  0.0011t 2  0.0412t  3.032 We substitute t  13 N  0.0011(13) 2  0.0412(13)  3.032  3.38 According to this model, in 2020 there will be approximately 3.38 million female athletes. This model predicts a slightly lower level than the model in exercise 25. c. A  0.002 x 2  0.924 x  0.152 We substitute x  53.5 A  0.002(53.5)2  0.924(53.5)  0.152  43.558 The optimum angle, in degrees, to tilt the solar panel in Edmonton is 43.558. b. N  0.0011t 2  0.0412t  3.032 We substitute t  30 N  0.0011(30) 2  0.0412(30)  3.032  3.28 According to this model, in 2037 there will be approximately 3.28 million female athletes. 26. a. S  0.00173t 2  3.477t  0.924 We substitute t  10 S  0.00173(10) 2  3.477(10)  0.924  35.5 In 2003, the sea had risen 35.5 mm over the 1993 level. b. S  0.00173t 2  3.477t  0.924 We substitute t  27 S  0.00173(27) 2  3.477(27)  0.924  93.5 In 2020, the sea will rise approximately 93.5 mm over the 1993 level. c. S  0.00173t 2  3.477t  0.924 We substitute S  1000 2 1000  0.00173t  3.477t  0.924 t  93.5 In 2024, the sea level will have risen 1 meter above the 1993 level. c. Answer will vary. 29. v t   10.9t We substitute 2.5 in for t to get v 2.5  10.9  2.5  27.25 White was traveling at 27.25 miles per hour when he reentered the half pipe. 30. s t   16t 2 s t   28 28  16t 2 28 2 t 16 28  t2 16 1.3228  t Copyright © 2020 Pearson Education, Inc. 6 Chapter R Functions, Graphs, and Models Danny took approximately 1.32 seconds to hit the ramp. 31. a) A  P 1  i  t 1  100, 000 1.028 nt  100, 000 1  0.00000196347   100, 000 1.00000196347  8760  100, 000 1.02839563811  102,839.56 At the end of 1 year, the investment is worth $102,839.56. 21  100, 000 1  0.014 2  100, 000 1.014 32. a) A  P 1  i  t 2 A  300, 000 1  0.022 1 A  100, 000 1.028196  102,819.60 At the end of 1 year, the investment is worth $102,819.60. nt  0.028  A  100, 000 1    4   100, 000 1  0.07   i b) A  P 1    n nt 21  306, 636.30 At the end of 1 year, the investment is worth $306,636.30. 4 4  i c) A  P 1    n  102,829.537  102,829.54 At the end of 1 year, the investment is worth $102,829.54. nt  0.028  A  100, 000 1    365   300, 000 1.022  306, 600.00 At the end of 1 year, the investment is worth $306,600.00.  0.022  A  300, 000 1    2  41  100, 000 1.0282953744  i d) A  P 1    n 87601  102,839.563811  0.028  A  100, 000 1    2   100, 000 1.07  nt 8760  102,800 At the end of 1 year, the investment is worth $102,800.  i c) A  P 1    n  i A  P 1    n  0.028  A  100, 000 1   8760  A  100, 000 1  0.028  i b) A  P 1    n e) There are 24  365  8760 hours in one year. nt  0.022  A  300, 000 1    4   300, 000 1.0055 41 4  306, 654.65 At the end of 1 year, the investment is worth $306,654.65. 3651  100, 000 1  0.00076712329 365  100, 000 1.00076712329 365  100, 000 1.02839458002  102,839.458002  102,839.46 At the end of 1 year, the investment is worth $102,839.46.  i d) A  P 1    n nt  0.022  A  300, 000 1    365  3651  300, 000 1.000060273973 365  306, 672.93 At the end of 1 year, the investment is worth $306,672.93. Copyright © 2020 Pearson Education, Inc. Exercise Set R.1 7 e) There are 24  365  8760 hours in one year.  i A  P 1    n nt  30, 000 1.000109589 87601  30, 000 1.127489438  300, 000 1.000002511416 8760  306, 673.13 At the end of 1 year, the investment is worth $306,673.13. 33. a) A  P 1  i  A  30, 000 1  0.04 3  33,824.68 At the end of 3 year, the investment is worth $33,824.68. e) There are 24  365  8760 hours in one year. 3  33, 745.92 At the end of 3 year, the investment is worth $33,745.92. nt  0.04  A  30, 000 1    2  23  30, 000 1.000004566210046 26,280  30, 000 1.127496541  33,824.89624  33,824.90 At the end of 3 year, the investment is worth $33,824.90. t 6 A  1000 1  0.05 4  30, 000 1.1262  33, 784.87 At the end of 3 year, the investment is worth $33,784.87. nt  1215.51 At the end of 4 year, the investment is worth $1512.51.  i b) A  P 1    n nt  0.05  A  1000 1    2  43 2 4  1000 1.050625 8  30, 000 1.01 12  30, 000 1.12682503  33,804.7509  33,804.75 At the end of 3 year, the investment is worth $33,804.75. nt 87603 34. a) A  P 1  i   30, 000 1.02  0.04  A  30, 000 1    4  nt  0.04  A  30, 000 1   8760   30, 000 1.04  i d) A  P 1    n  33,824.68315  i A  P 1    n t  i c) A  P 1    n 3653 1095  0.022  A  300, 000 1   8760   i b) A  P 1    n  0.04  A  30, 000 1   365   1218.40 At the end of 4 year, the investment is worth $1218.40.  i c) A  P 1    n nt  0.05  A  1000 1    4  4 4  1000 1.0125 16  1219.89 At the end of 4 year, the investment is worth $1219.89. Copyright © 2020 Pearson Education, Inc. 8 d) Chapter R Functions, Graphs, and Models  i A  P 1    n  1  r n  1 37. W  P   r   We substitute 3000 for P, 0.0305 3.05%  0.0305 for r, and 18 for n. nt  0.05  A  1000 1   365  3654  1000 1.000136986 1460 e)  1221.39 At the end of 4 year, the investment is worth $1221.39. There are 24  365  8760 hours in one year.  i A  P 1    n nt  0.05  A  1000 1   8760  87604  1000 1.000005708 35,040  1221.40 At the end of 4 year, the investment is worth $1221.40. 35. Using the formula: n  r  r    1    12  12   M  P n   r   1    1   12   We substitute 18,000 for P , 0.046 4.6%  0.046 for r, and 36 3 12  36 for n. Then we use a calculator to perform the computation.  0.046  0.046  36   1    12  12    M  18, 000   0.046  36  1   1   12      536.25 The monthly payment on the loan will be approximately $536.25. 36. 30 years  30 12  360 months P  100, 000; r  0.024  0.024  0.024  360   1    12  12   M  100, 000    0.024  360  1   1   12      389.94 The monthly payment on the loan will be approximately $389.94.  1  0.030518  1  W  3000  0.0305    70,561.01 Rounded to the nearest cent, the annuity will be worth $70,561.01 after 18 years. 38. We substitute 50,000 for W, 0.0725  1   4 %  0.0425  in for r, and 20 for n. Then 4 we proceed to solve for P.  1  0.042520  1 50, 000  P   0.0425   50, 000 P  1  0.042520  1   0.0425   1635.99  P You will need to invest $1635.99 annually to reach a goal of $50,000 after 20 years. 39. a) Locate 5.8 on the vertical axis and then think of horizontal lines extending across the graph from this point. The years for which the graph lies above this line are the years for which the unemployment rate was at or above 5.8%. This time period was 2008 – 2014. b) Locate 7 on the vertical axis and then think of a horizontal line extending across the graph from this point. The years for which the graph lies below this line are the years for which unemployment was below 7%. Those time periods are 2006 – 2008 and 2014 – 2016. c) Locate the highest point on the graph and extend a line vertically to the horizontal axis. The year which the unemployment rate was the highest was 2010 at 9.6%. d) Locate the lowest point on the graph and extend a line vertically to the horizontal axis. In this case there are two points that are exactly at 4.6%. The years when the unemployment rate was the lowest are 2006 and 2007. Copyright © 2020 Pearson Education, Inc. Exercise Set R.1 9 We substitute 0.053 5.3%  0.053 for r, and 40. Answers will vary.  1  r n  1 41. a) Using the formula W  P   we r   substitute 1200 for P, 0.04  4%  0.04 for r and 35 for n.  1  0.0435  1  W  1200  0.04    88, 382.67 Sally will have approximately $88,382.67 in her account when she retires. b) Sally invested $1200 per year for 35 years. Therefore, the total amount of her original payments is: $120035  $42, 000 . Since the total amount in the account was $88,382.67, the interest earned over the 35 years is: $88,382.67  $42, 000  $46,382.67 Therefore, $42,000 was the total amount of Sally’s payments and $46,382.67 was the total amount of her interest. 42. a) Using the formula: n  r  r   1      12  12   M  P n   r   1    1   12   We substitute 206,780.16 for P , 0.04 4%  0.04 for r, and 15 80  65  15 for n. Then we use a calculator to perform the computation.  0.04  0.04 180   1    12  12   M  206, 780.16    0.04 180  1   1   12      653.76 Sally should take a monthly payment of $653.76. b) Sally received 180 payments of $653.76, therefore she received a total of $653.76180  $117, 676.80 during the 15 years. Of that 42,000 was what she originally contributed, leaving $75,676.80 in interest. 43. Using the formula: n  r Y  1    1  n 12 (monthly) for n. Then we use a calculator to perform the computation. 12  0.053  Y  1   1  12   0.0543  5.43% The annual yield of 5.3% compounded monthly would be approximately 5.43%. 44. Using the formula: n  r Y  1    1  n We substitute 0.041  4.1%  0.041 for r, and 4 (quarterly) for n. Then we use a calculator to perform the computation. 4  0.041  Y  1   1  4   0.0416  4.16% The annual yield of 4.1% compounded quarterly would be approximately 4.16%. 45. Using the formula: n  r Y  1    1  n We substitute 0.0375 3.75%  0.0375 for r, and 52 (weekly) for n. Then we use a calculator to perform the computation.  0.0375  Y  1    52  52 1  0.0382  3.82% The annual yield of 3.75% compounded weekly would be approximately 3.82%. 46. Using the formula: n  r Y  1    1  n We substitute 0.04  4%  0.04 for r, and 360 (daily) for n. Then we use a calculator to perform the computation.  0.04  Y  1   360  360 1  0.0408  4.08% The annual yield of 4% compounded daily would be approximately 4.08%. Copyright © 2020 Pearson Education, Inc. 10 Chapter R Functions, Graphs, and Models  0.0297  Y  1    52  47. a. Using the formula: n  r Y  1    1  n We substitute 0.025  2.5%  0.025 for r, and 1 (annually) for n. Then we use a calculator to perform the computation. 52 1  0.0301  3.01% The annual yield for Foothill Bank will be approximately 3.01% b. Foothill Bank has a higher yield. 49. Using the formula: n  r Y  1    1  n 1  0.025  Y  1   1  1   0.025  2.5% The annual yield for Western Bank will be 2.5%. Using the formula: We substitute 0.022  2.2%  0.022 for Y, and 12 (monthly) for n. Then we use a calculator to perform the computation. 12 r   0.022  1    1  12  n  r Y  1    1  n We substitute 0.0243 2.43%  0.0243 for r, and 12 (monthly) for n. Then we use a calculator to perform the computation. r  0.02179  2.179% Mesalands would need to pay at least 2.179% to exceed an annual yield of 2.2%. 50. Using the formula: n  r Y  1    1  n 12  0.0243  Y  1   1  12   0.0246  2.46% The annual yield for Commonwealth Savings will be approximately 2.46% b. Western Bank has a higher yield. We substitute 0.0375 3.75%  0.0375 for Y, and 4 (quarterly) for n. Then we use a calculator to perform the computation. 4  r 0.0375  1    1  4 48. a. Using the formula: n  r Y  1    1  n We substitute 0.03 3%  0.03 for r, and 1 (annually) for n. Then we use a calculator to perform the computation. r  0.03698  3.698% Shea Savings would need to pay at least 3.698% to exceed an annual yield of 3.75%. 51. Graph y  x3  x 2 The resulting graph is: 1  0.03  Y  1   1  1   0.03  3% The annual yield for Sierra Savings will be 3%. Using the formula: n  r Y  1    1  n We substitute 0.0297 2.97%  0.0297 for r, and 52 (weekly) for n. Then we use a calculator to perform the computation. Copyright © 2020 Pearson Education, Inc. Exercise Set R.1 11 52. Graph y  2 x The resulting graph is Copyright © 2020 Pearson Education, Inc. 12 Chapter R Functions, Graphs, and Models Exercise Set R.2 13. The correspondence is a function because every book has a unique Library of Congress number. 1. 14. The correspondence is a function because every book has a unique ISBN. The correspondence is a function because each member of the domain corresponds to only one member of the range. 2. The correspondence is a function because each member of the domain corresponds to only one member of the range. 3. The correspondence is a function because each member of the domain corresponds to only one member of the range. 4. The correspondence is not a function because one member of the domain, 6, corresponds to two members of the range, – 6 and – 7. 5. The correspondence is a function because each member of the domain corresponds to only one member of the range, even though two members of the domain, Quarter Pounder with Cheese ® and Big N’ Tasty with Cheese ® correspond to $3.20. 6. 7. The correspondence is a function because each member of the domain corresponds to only one member of the range. The correspondence is a function because each number doubled is exactly one number. 8. The correspondence is a function because each number that is two less than the number is a number. 9. The correspondence is a function because each square root of a number is exactly one positive number. 10. The correspondence is a function because each cubed root of a number is exactly one number. 11. The correspondence is not a function because there may be more than one number less than or equal to a positive number. 15. This correspondence is a function, because all people have a specific birthday. 16. This correspondence is a function, because all people have a specific weight when they step on a scale. 17. The correspondence is not a function because more than one person is born on any specific date. 18. The correspondence is not a function because more multiple people may have the same weight. 19. This correspondence is a function, because a rectangles area is determined by the specific length and width. 20. This correspondence is a function, because a rectangles perimeter is determined by the specific length and width. 21. The correspondence is not a function because there may be more than one length and width to obtain the given area. 22. The correspondence is not a function because there may be more than one length and width to obtain the given perimeter. 23. a) f  x  4x  3 f 5.1  4 5.1  3  17.4 f 5.01  4 5.01  3  17.04 f 5.001  4 5.001  3  17.004 f 5  4 5  3  17 x f  x 5.1 17.4 12. The correspondence is not a function because there is more than one odd integer less than or equal to the given number. Copyright © 2020 Pearson Education, Inc. 5.01 17.04 5.001 17.004 5 17 Exercise Set R.2 13 b) f  x   4 x  3 26. f  4  4  4  3  13 g  x  x2  4 g  3  3  4  13 2 f 3  4 3  3  9 g 0   0   4  4 2 f  2  4  2  3  11 g 1  1  4  5 2 f  k   4  k   3  4k  3 g 7   7   4  53 2 f  x  h   4  x  h   3  4 x  4h  3 24. a) g  a  h   a  h   4  a 2  2ah  h 2  4 2 f  x   3x  2 x f  x 4.1 14.3 2 4.01 14.03 4.001 14.003 4 14 b) f  x   3x  2 f 5  3 5  2  17 27. f 1  3  1  2  1 f  k   3  k   2  3k  2 f  x  h   3  x  h   2  3x  3h  2 25. f  x  f  4  f 0   g  x  x  3 2 g  1   1  3  1  3  2 2 f a   g 0  0  3  0  3  3 2 g 1  1  3  1  3  2  g  a  h   g a  a  2ah  h  4  a  4  h h 2 2ah  h  h  2a  h 2 2 1  x  32 1 2  2  4  3 1 0  3 1 a   3  2 1 2  1 49 2  1 9 7  1 3 1 a  32 2 f  x  h  g 5  5  3  25  3  22 2 g a  h   a  h   3  a  2ah  h  3 2 2 2 2  2  g  a  h   g  a   a  h   3   a   3  h h 2 a  2ah  h 2  3   a 2  3  h 2 2ah  h  h h  2a  h   h  2a  h 1   x  h   3 2 1  x  h  32 f  x  h  f  x h 1 1  2    x h x 3    32  h   x  32  x  h  32   x  h  32  x  32  x  h  32  x  32  2     h 2 x  6 x  9  x  2hx  6 x  h 2  6h  9 h  x  h  3  x  3 2 2hx  h 2  6h h  x  h  3  x  3 2 2 h 2 x  h  6 h  x  h  3  x  3 2 2 x  h  6  x  h  32  x  32 Copyright © 2020 Pearson Education, Inc. 2 , h0 2   14 28. Chapter R Functions, Graphs, and Models f  x  f 3  33. Graph f  x   2 x  5 . 1  x  5 2 1 3  5 f 1  f k   2  1 2 1  1  5 2 2   1 6 2 First, we choose some values for x and compute the values for f  x  , in order to form the 1 4  ordered pairs that we will plot on the graph. f  1  2  1  5  7 1 36 f 0  2 0  5  5 f 1  2 1  5  3 1 f  2  2  2  5  1  k  5 2 f  x  h  1  x  h   5  2  29. a. Graph f  x   4 x  2 . b. 1  x  h  52 x f  x 1 7 0 5 1 3  x, f  x  1, 7 0, 5 1, 3 2, 1 2 1 Next we plot the input – output pairs from the table and, in this case, draw the line to complete the graph. 30. a. Graph g  x   3x  4 . b. 34. Graph f  x   3x  1 . 31. a. Graph h  x   x 2  x . b. 35. Graph g  x   4 x . First, we choose some values for x and compute the values for g  x  , in order to form the 32. a. Graph k  x   x 2  3 x . b. ordered pairs that we will plot on the graph. g  1  4  1  4 g 0  4 0  0 g 1  4 1  4 . Copyright © 2020 Pearson Education, Inc. Exercise Set R.2 15 x g  x 1 4 0 0  x, g  x  1, 4 0, 0 0,  4 4 1 Next we plot the input – output pairs from the table and, in this case, draw the line to complete the graph. 38. Graph f  x   x 2  4 . 36. Graph g  x   2 x . 39. Graph f  x   6  x 2 . First, we choose some values for x and compute the values for f  x  , in order to form the ordered pairs that we will plot on the graph. f  2  6  2  2 2 f  1  6  1  5 2 37. Graph f  x   x 2  2 . f 0  6  0   6 2 First, we choose some values for x and compute the values for f  x  , in order to form the ordered pairs that we will plot on the graph. Choosing some values for x and evaluating the function, we have: f  2   2  2  2 2 f  1   1  2  1 2 f 0  0  2  2 2 f 1  1  2  1 2 f  2   2  2  2 2 x f  x 2 2 1 1 0 2 1 1  x, f  x  2, 2 1, 1 0, 2 1, 1 2, 2 f 1  6  1  5 2 f  2  6   2   2 2 x f  x 2 2 1 5 0 6 1 5  x, f  x  2, 2 1,5 0, 6 1,5 2, 2 2 2 Next we plot the input – output pairs from the table and, in this case, draw the curve to complete the graph. 2 2 Next we plot the input – output pairs from the table and, in this case, draw the curve to complete the graph. Copyright © 2020 Pearson Education, Inc. 16 Chapter R Functions, Graphs, and Models 40. Graph g  x    x 2  1 . 42. Graph g  x   41. Graph g  x   x3 . First, we choose some values for x and compute the values for g  x  , in order to form the ordered pairs that we will plot on the graph. g 2   2  8 3 g  1   1  1 3 g  0   0   0 3 1 3 x . 2 43. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 44. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 45. The graph is not that of a function. A vertical line can intersect the graph more than once. g 1  1  1 3 g  2   2  8 3 x f  x 2 8 1 1 0 0 1 1 2 8  x, f  x  2, 8 1, 1 0, 0 1,1 2,8 Next we plot the input – output pairs from the table above and, in this case, draw the curve to complete the graph. 46. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 47. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 48. The graph is not that of a function. A vertical line can intersect the graph more than once. 49. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 50. The graph is not that of a function. A vertical line can intersect the graph more than once. 51. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 52. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 53. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 54. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. Copyright © 2020 Pearson Education, Inc. Exercise Set R.2 17 b) The graph is not that of a function. A vertical line can intersect the graph more than once. 55. The graph is not that of a function. A vertical line can intersect the graph more than once. 56. The graph is not that of a function. A vertical line can intersect the graph more than once. 59. f  x  h  f  x  h 57. Graph x  y 2  2 . a) First, we choose some values for y (since x is expressed in terms of y) and compute the values for x , in order to form the ordered pairs that we will plot on the graph.  2 2 For y  1; x  1  2  1 2 For y  2; x  2  2  2 2 2 1 1 2 0 1 1 2 2 Next we plot the input – output pairs from the table and, in this case, draw the curve to complete the graph. 2 h 2 2 xh  h  3h combining like terms h h  2 x  h  3 Factoring  h  2 x  h  3, h  0 For y  0; x  0  2  2 2 2  2  x, y  2, 2 1, 1 2, 0 1,1 2, 2 h x  2 xh  h  3 x  3h   x  3 x   For y  1; x   1  2  1 y  x  h2  3  x  h    x 2  3x  2 For y  2; x   2  2  2 x f  x   x2  3x 60. f  x  x2  4x f  x  h  f  x h   x  h2  4  x  h   x 2  4 x  h x  2 xh  h  4 x  4h   x 2  2 2  4 x  h 2 xh  h 2  4h  h h 2 x  h  4  h  2 x  h  4, h  0 61. To find f  1 we need to locate which piece b) The graph is not that of a function. A vertical line can intersect the graph more than once. 58. a) Graph x  y 2  3 . defines the function on the domain that contains x  1 . When x  1 , the function is defined by f  x   2 x  1; for x  0 ; therefore, f  1  2  1  1  2  1  3 . To find f 1 we need to locate which piece defines the function on the domain that contains x  1 . When x  1 , the function is defined by f  x   x 2  3; for 0  x  4 ; therefore, f 1  1  3  1  3  2 . 2 Copyright © 2020 Pearson Education, Inc. 18 Chapter R Functions, Graphs, and Models 62. When x  3 , the function is defined by f  x   2 x  1; for x  0 ; therefore, Note that for f  x   1 . f  3  2  3  1  6  1  7 . When x  3 , the function is defined by f  x   x 2  3; for 0  x  4 ; therefore, f 3  3  3  9  3  6 . 2 f  0   1 f 1  1 f  2  1 The solid dot indicates that 0, 1 is part of the graph. 63. To find f 0 we need to locate which piece defines the function on the domain that contains x  0. When x  0 , the function is defined by f  x   17; for x  0 ; therefore, f 0  17 . To find f 10 we need to locate which piece defines the function on the domain that contains x  10 . When x  10 , the function is defined 1 by f  x   x  1; for x  4 ; therefore, 2 1 f 10  10  1  5  1  6 . 2 64. When x  5 , the function is defined by f  x   2 x  1; for x  0 ; therefore, f  5  2  5  1  10  1  11 . When x  5 , the function is defined by 1 f  x   x  1; for x  4 ; therefore, 2 1 5 7 f 5  5  1   1   3.5 . 2 2 2  1 for x  0 65. Graph f  x    .  1 for x  0 First, we graph f  x   1 for inputs less than 0. We note for any x-value less than 0, the graph is the horizontal line y  1 . Note that for f  x   1 f 2  1  2, for x  3 66. Graph f  x    .  2, for x  3 6, for x  2 67. Graph f  x    2 .  x , for x  2 First, we graph f  x   6 for x  2 . This graph consists of only one point, 2, 6 . The solid dot indicates that 2, 6 is part of the graph. Next, we graph f  x   x 2 for inputs x  2 . Note that for f  x   x 2 f  3   3  9 2 f  1   1  1 2 f  0  0   0 2 f 1  1  1 2 f 1  1 The solution is continued on the next page. The open circle indicates that 0,1 is not part of f  2   2   4 2 the graph. Next, we graph f  x   1 for inputs greater than or equal to 0. We note for any x-value less than 0, the graph is the horizontal line y  1 . Copyright © 2020 Pearson Education, Inc. Exercise Set R.2 19 x f  x 3 9 1 1 0 0 1 1  x, f  x  3,9 1,1 0, 0 1,1 2, 4 2 4 Since the input x  2 is not defined on this part of the graph, the point 2, 4 is not part of Next, we graph g  x   x  2 for inputs x  0 . Creating the input – output table, we have: g  x x  x, g  x  1,3 2, 4 2 4 3,5 3 5 The open circle indicates that 0, 2 is not part 1 3 of the graph. the graph. The open circle indicates that 2, 4 is not part of the graph. 5, for x  1 68. Graph f  x    3 .  x , for x  1  x, for x  0  for x  0 . 69. Graph g  x   4,  x  2, for x  0  First, we graph g  x    x for inputs x  0 . Creating the input – output table, we have: g  x x  x, g  x  3,3 2, 2 2 2 1,1 1 1 The open circle indicates that 0, 0 is not part 3 3 of the graph. Next, we graph g  x   4 for x  0 . This part of the graph consists of a single point. The solid dot indicates that 0, 4 is part of the graph. 2 x  3,  70. Graph g  x   5,  x  2,  for x  1 for x  1 .  12 x  1,  71. Graph g  x    4,  x  3,  for x  2 for x  1 for x  2 . for x  2 First, we graph g  x   12 x  1 for inputs x  2 . Creating the input – output table, we have: x g  x 2 2 0 1 1  12  x, g  x  2, 2 0, 1 1,  12  The open circle indicates that 2, 0 is not part of the graph. Next, we graph g  x    4 for x  2 . This part of the graph consists of a single point. The solid dot indicates that  2,  4 is part of the graph. Next, we graph g  x   x  3 for inputs x  2 . Copyright © 2020 Pearson Education, Inc. 20 Chapter R Functions, Graphs, and Models Choosing some values for x and evaluating the function, we have: f  x x  x, f  x  Creating the input – output table, we have: g  x x  x, g  x  3, 0 4,1 4 1 5 5, 2 2 The open circle indicates that 2, 1 is not part 3 0 of the graph. 3 6 1 2 0 3 1 2 3 6 3, 6 1, 2 0, 3 1, 2 3, 6 Since the input x  2 is not defined on this part of the graph, the point 2,1 is not part of the graph. The open circle indicates that 2,1 is not part of the graph.  x2 , for x  0  72. Graph g  x   3, for x  0 . 2 x  3, for x  0  for x  3  6, 74. Graph f  x    2 .   x  5, for x  3 for x  2 7, . 73. Graph f  x    2  x  3, for x  2 First, we graph f  x   7 for x  2 . This graph consists of only one point, 2, 7 . The solid dot indicates that 2, 7 is part of the graph. Next, we graph f  x   x 2  3 for inputs x  2 . Note that for f  x   x 2  3 4t 75.  0.03  A t   P 1    4  We substitute 500 in for P and 2 in for t :  0.03  A t   500 1    4  4 2  500 1.0075 8 f 3  3  3  6 2  500 1.061598848 f  1   1  3  2 2  530.7994239  530.80 The investment will be worth approximately $530.80 after 2 years. f 0  0  3  3 2 f 1  1  3  2 2 f 3  3  3  6 2 We create the input – output table. Copyright © 2020 Pearson Education, Inc. Exercise Set R.2 76. 21  0.03  A t   P 1    4  80. a) Yes, the table represents a function. Each event is assigned exactly one scale of impact number. b) The inputs are the events; the outputs are the scale of impact numbers. 4t  0.03  A t   800 1    4  43  800 1.0075 12  875.05 The investment will be worth approximately $875.05 after 3 years. hw 77. s  3600 a) We substitute 170 for h and 70 for w. 17070 s  1.818 3600 The patient’s approximate surface area is 1.818m2 b) We substitute 170 for h and 100 for w. 170100 s  2.173 3600 The patient’s approximate surface area is 2.173m2 c) We substitute 170 for h and 50 for w. 17050 s  1.537 3600 The patient’s approximate surface area is 1.537 m2 78. s 81. Solve the equation for y. 2 x  y  16  4  3 y  2 x y  16  4  3 y 4 y  20 y5 We sketch a graph of the equation. No vertical line meets the graph more than once. Thus, the equation represents a function. 82. First we solve the equation for y. 2 y 2  3x  4 x  5 2 y2  x  5 subtract 3 x from both sides x5 y2  divide both sides by 2 2 x5 take the square root of both sides y 2 We sketch a graph of the equation. hw 3600 a) s  15070  1.708 3600 The patient’s approximate surface area is 1.708m2 . b) s  18070  1.871 3600 The patient’s approximate surface area is 1.871m2 . 79. a) Yes, each term has a specific monthly payment. b) Yes, each month payment has a specific term. c) No, the interest rate has multiple monthly payments. d) Yes, each monthly payment has a specific interest rate. We can see that a vertical line will intersect the graph more than once; therefore, this is not a function. 83. First we solve the equation for y. 4 y   64 x 4 y   4 x 2 3 2 3 3 3 3 3 y2  x y x Copyright © 2020 Pearson Education, Inc. 22 Chapter R Functions, Graphs, and Models Next, we sketch a graph of the equation. Now, we are able to look at the table. We will have to enter the appropriate values of x. We can see that a vertical line will intersect the graph more than once; therefore, this is not a function. 84. First, we solve the equation for y. 3 y   72x 3 2 2 From the table we conclude: f 3  2 9 y 3  72 x y3  8x f  2   0 y  3 8x y  23 x  y  0 Note: since y must be non-negative to satisfy the original equation, we only graph the points for which y is non-negative. Next, we sketch a graph of the equation: No vertical line meets the graph more than once. Thus, the equation represents a function. 85. Answers will vary. The vertical line test works, because you are locating the values to which each input corresponds. For a relation to be a function, each input must correspond with exactly one output. Therefore, if a vertical line intersects the graph in more than one point, that particular input corresponds to more than one output and the graph is not a function. 86. f  x  x  2  x  1  5 f  0   2 f  4  2 87. f  x   x3  2 x 2  4 x  13 88. f  x  3 2 x 4 We begin by setting up the table: Next, we will type in the equation into the graphing editor. First, we set up the table to allow us to ask the calculator to compute specific values. Next, we type the equation into the graphing editor. Copyright © 2020 Pearson Education, Inc. Exercise Set R.2 23 Now, we are able to look at the table: 89. Each graph is shown below. In order to graph f  x   x3  2 x 2  4 x  13 , we use the window at the top of the next column. After entering the function into the graphing editor, we get: 90. Answers will vary. Some ordered pairs are 2.978723,1.0617  , 1.382979, 2.8438301 , 0,3.1622777  , 1.7021277, 2.6651006 , and 2.6595745,1.7107494. 91. a) f  x   5( x  2) After entering the function into the graphing editor, we get: In order to graph f  x   standard window: 3 2 x 4 g  x  5x  2 b) , we use the c) No, the graph is not the same function. 92. a) f  x   ( x  4)2 g  x  x2  4 After entering the function into the graphing editor, we get: In order to graph f  x   x  2  x  1  5 , we use the standard window. b) c) No, the graph is not the same function. Copyright © 2020 Pearson Education, Inc. 24 93. Chapter R Functions, Graphs, and Models f  x   3x  6 g  x   3( x  h) The functions represent the same function, so set them equal to each other to find the value for h. 3 x  6  3( x  h) 3 x  6  3x  3h 6  3h 2h 94. f  x   ( x  3) 2 g  x  x2  6 x  h The functions represent the same function, so set them equal to each other to find the value for h. ( x  3) 2  x 2  6 x  h x2  6 x  9  x2  6 x  h 9h Copyright © 2020 Pearson Education, Inc. Exercise Set R.3 Exercise Set R.3 1.  2, 4 3. 0,5 4.  1, 2 5. 9, 5 6.  9, 4 7.  x, x  h  8.  x, x  h  9.  4, 1  2,3 10. , 0  3,  11.  2, 2 13. 19.  4, 3  0,5 20. , 2  1, 4 1,3 2. 12. 25 5,5 6, 20 14.  4, 1 15. 3,  16. , 2 17. 2,3 18.  10, 4 21. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, 3. Thus, f 1  3 . b) The domain is the set of all x-values of the points on the graph. The domain is 3, 1,1,3,5 . c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. One such point exists, 3, 2 . Thus the x-value for which f  x   2 is x  3. d) The range is the set of all y-values of the points on the graph. The range is 2, 0, 2,3, 4 . 22. a) f 1  1 . b) The domain is 4, 3, 2, 1, 0,1, 2 . c) The point on the graph with the second coordinate 2 is 2, 2 . Thus the x-value for which f  x   2 is x  2 . d) The range is 2, 1, 0,1, 2,3, 4 . 23. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, 4. Thus, f 1  4 . b) The domain is the set of all x-values of the points on the graph. The domain is 5, 3,1, 2,3, 4,5 . Copyright © 2020 Pearson Education, Inc. 26 Chapter R Functions, Graphs, and Models c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. Three such point exists, 5, 2 ; 3, 2 ; and 4, 2 . Thus the xvalues for which f  x   2 are 5, 3, 4 . d) The range is the set of all y-values of the points on the graph. The range is 3, 2, 4,5 24. a) f 1  2 . b) The domain is 6, 4, 2, 0,1,3, 4 . c) The points on the graph with the second coordinate 2 are 1, 2 and 3, 2 . Thus the x-values for which f  x   2 are 1,3 . d) The range is 5, 2, 0, 2,5 . 25. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, -1. Thus, f 1  1 . b) The domain is the set of all x-values of the points on the graph. These extend from -2 to 4. Thus, the domain is  x | 2  x  4 , or in interval notation  2, 4 . c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. One such point exists, 3, 2 . Thus the x-value for which f  x   2 is x  3. d) The range is the set of all y-values of the points on the graph. These extend from -3 to 3. Thus, the range is  y | 3  y  3 , or, in interval notation  3,3 . 26. a) 27. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, -2. Thus, f 1  2 . b) The domain is the set of all x-values of the points on the graph. These extend from -4 to 2. Thus, the domain is  x | 4  x  2 , or, in interval notation  4, 2 . c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. One such point exists, 2, 2 . Thus the x-value for which f  x   2 are x  2. d) The range is the set of all y-values of the points on the graph. These extend from -3 to 3. Thus, the range is  y | 3  y  3 , or in interval notation  3,3 . 28. a) f 1  2.25 . b) The domain is  x | 4  x  3 or  4,3 . c) The point on the graph with the second coordinate 2 appears to be 0, 2 . Thus the x-value for which f  x   2 is x  0. . d) The range is  y | 5  y  4 , or,  5, 4 . 29. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, 3. Thus, f 1  3 . b) The domain is the set of all x-values of the points on the graph. These extend from -3 to 3. Thus, the domain is  x | 3  x  3 , or, in interval notation  3,3 . f 1  2.5 . b) The domain is  x | 3  x  5 or  3,5 . c) The point on the graph with the second coordinate 2 appears to be 2.25, 2 . Thus the x-value for which f  x   2 is x  2.25. d) The range is  y |1  y  4 , or, 1, 4 . c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. Two such point exists, 1.4, 2 and 1.4, 2 . Thus the x-values for which f  x   2 are 1.4,1.4 . Copyright © 2020 Pearson Education, Inc. Exercise Set R.3 27 d) The range is the set of all y-values of the points on the graph. These extend from -5 to 4. Thus, the range is  y | 5  y  4 , or in 33. interval notation  5, 4 . 30. a) f 1  2 . b) The domain is  x | 5  x  4 or  5, 4 . adding x to both sides 2 x Thus, 2 is not in the domain of f, while all other real numbers are. The domain of f is x | x is a real number and x  2 ; or, in c) The points on the graph with the second coordinate 2 are all the points with the xvalue in the set  x |1  x  4 . Thus the xvalues for which f  x   2 are interval notation,   , 2   2,    x |1  x  4 , or 1, 4 . d) The range is  y | 3  y  2 , or  3, 2 . 31. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, 1. Thus, f 1  1 . 34. b) The domain is the set of all x-values of the points on the graph. These extend from -5 to 5. However, the open circle at the point 5, 2 indicates that 5 is not in the domain. 35. dividing both sides by 2 x0 The domain of f is x | x is a real number and x  0 ; or, in interval values for which f  x   2 are x | 3  x  5 , or 3,5 . d) The range is the set of all y-values of the points on the graph. The range is 2, 1, 0,1, 2 f 1  2 . b) The domain is  x | 4  x  4 or  4, 4 . c) The points on the graph with the second coordinate 2 are all the points with the xvalue in the set  x | 0  x  2 . Thus the xvalues for which f  x   2 are x | 0  x  2 , or 0, 2. d) The range is 1, 2,3, 4 . f  x  2 x Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which 2 x  0 . We find them by solving the inequality. 2x  0 setting the radicand  0 c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. We notice all the points with xvalues in the set  x | 3  x  5 Thus the x- 32. a) 2 x3 Set the denominator equal to 0 and solve. x3  0 x  3 The domain is x | x is a real number and x  3 , or, in f  x  interval notation,  , 3   3,   Thus, the domain is  x | 5  x  5 , or in interval notation  5,5 . 6 2 x Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain of f. 2 x  0 setting the denominator equal to 0 f  x  notation, 0,   36. f  x  x  2 Solve x  2  0 x2 The domain of f is x | x is a real number and x  2 ; or, in interval notation,  2,   Copyright © 2020 Pearson Education, Inc. 28 37. Chapter R Functions, Graphs, and Models f  x  x2  2 x  3 42. We can calculate the function value for all values of x, so the domain is the set of all real numbers  We can calculate the function value for all values of x, so the domain is the set of all real numbers  38. 43. f  x  x2  3 We can calculate the function value for all values of x, so the domain is the set of all real numbers  39. f  x   3x  7 x2 6 x  12 Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain of f. 6 x  12  0 setting the denominator equal to 0 adding 12 to both sides 6 x  12 dividing both sides by 6 x2 Thus, 2 is not in the domain of f, while all other real numbers are. The domain of f is x | x is a real number and x  2 ; or, in f  x  3x  1 7  2x Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain of f. 7  2x  0 setting the denominator equal to 0 f  x  adding 2 x to both sides 7  2x 7 dividing both sides by 2 x 2 7 Thus, is not in the domain of f, while all 2 other real numbers are. The domain of f is 7   x | x is a real number and x   ; or, in 2  7 7   interval notation,   ,    ,    2 2  interval notation,   , 2   2,   44. 40. 8 3x  6 Solve 3x  6  0 x2. The domain of f is x | x is a real number and x  2 ; or, in f  x  interval notation,   , 2   2,   41. f  x  x  4 2x  1 9  2x Solve 9  2 x  0 9 x 2 9 Thus, is not in the domain of f, while all 2 other real numbers are. The domain of f is 9   x | x is a real number and x   ; or, in 2  9 9   interval notation,   ,    ,    2 2  f  x  We can calculate the function value for all values of x, so the domain is the set of all real numbers  Copyright © 2020 Pearson Education, Inc. Exercise Set R.3 45. 29 g  x  4  5x Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which 4  5 x  0 . We find them by solving the inequality. 4  5x  0 setting the radicand  0 subtracting 4 from both sides 5 x  4 4 x dividing both sides by 5 5 The domain of g is 4   x | x is a real number and x    ; or, in 5   4  interval notation,   ,    5  49. g  x  2x 2 x  25 Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain of g. setting the denominator equal to 0 x 2  25  0 x 2  25 adding 25 to both sides x   25 taking the square root or both sides x  5 Thus, 5 and 5 are not in the domain of g, while all other real numbers are. The domain of g is  x | x is a real number and x  5, x  5 ; or, in interval notation,  , 5  5,5  5,  46. g  x   2  3x Solve 2  3x  0 2  3x 2 x 3 The domain of g is 2   x | x is a real number and x   ; or, in 3  2  interval notation,   ,   3 47. g  x  x2  2 x  1 50. interval notation,  , 6  6, 6  6,   51. g  x   4 x3  5 x 2  2 x We can calculate the function value for all values of x, so the domain is the set of all real numbers  2 x 2  36 x  6 The domain of g is x | x is a real number and x  6, x  6 ; or, in We can calculate the function value for all values of x, so the domain is the set of all real numbers  48. x 1 x  36 2 Solve x  36  0 g  x  52. 1 x2  9 We can calculate the function value for all values of x, so the domain is the set of all real numbers  g  x  x x 1 We can calculate the function value for all values of x, so the domain is the set of all real numbers  g  x  2 Copyright © 2020 Pearson Education, Inc. 30 53. Chapter R Functions, Graphs, and Models g  x  x  1  6  2 x Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which x  1  0 and 6  2 x  0 . We find them by solving the inequality. x 1  0 setting the radicand  0 x  1 subtracting 1 from both sides 6  2x  0 setting the radicand  0 2 x  6 subtracting 6 from both sides x  3 dividing both sides by -2 Thus, the domain of g is the union of these two inequalities  x | x  1 and x  3 ; or, in interval notation,  1,3 56. The graph crosses the line y  1 at every integer value of x. Therefore, the set of x-values for which g  x   1 is x | 5, 4, 3, 2, 1, 0,1, 2,3, 4,5 . 57. First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. We notice the point with x-value is 1. 58. First, we locate –4 on the vertical axis and then we look horizontally to find any points on the graph for which the second coordinate is greater than –4. We notice all the points with x-values in the set  x | 4  x  2 Thus the x-values for which G  x   4 are x | 4  x  2 , or  4, 2 . 54. g  x   2 x  3  12  5 x Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which 2 x  3  0 and 12  5 x  0 . We find them by solving the inequality. 2x  3  0 setting the radicand  0 2 x  3 subtracting 3 from both sides 3 dividing both sides by 2 x 2 12  5 x  0 setting the radicand  0 subtracting 12 from both sides 5 x  12 12 x dividing both sides by -5 5 Thus, the domain of g is the union of these two 3 12   inequalities  x | x   and x   ; or, in 2 5   3 12  interval notation,   ,   2 5 59. First, we locate 0 on the vertical axis and then we look horizontally to find any points on the graph for which the second coordinate is 0. We notice all the points with x-values in the set x | x  3 and 4 Thus the x-values for which W  x   0 are 3 and 4 . 60. First, we locate 0 on the vertical axis and then we look horizontally to find any points on the graph for which the second coordinate is greater than or equal to 0. We notice all the points with x-values in the set  x | 6  x  4 Thus the x- values for which k  x   0 are x | 6  x  4 , or  6, 4 . 61. The domain is the number of hours that Karen will work, which is 0 to 80 or 0,80 . The range of the function is how much Karen will earn, which is a minimum of P(0)  40(0)  0 and a maximum of P(80)  40(80)  3200 . Therefore the range of the function will be  0,3200 . 55. The graph of f lies on or below the x-axis when f  x   0 , so we scan the graph from left to right looking for the values of x for which the graph lies on or below the x axis. Those values extend from -1 to 2. So the set of x-values for which f  x   0 is  x | 1  x  2 , or, in interval notation,  1, 2 . 62. The domain is the amount of money Marcus will spend, which is 0 to 200 or  0, 200 . The range of the function is how much Marcus will pay in taxes, which is a minimum of T (0)  0.05(0)  0 and a maximum of T (200)  0.05(200)  10 . Therefore the range of the function will be 0,10 . Copyright © 2020 Pearson Education, Inc. Exercise Set R.3 31 63. a) The domain in interval notation is  25,102 . b) The range in interval notation is  0, 450 . c) Answers will vary. Since we are looking for the greatest increase in the incidence of breast cancer, we are looking for the steepest portion of the graph. It appears that the greatest increase occurs from age 50 to age 60. 64. a) The graph extends from x  0 to x  92.3 , so the domain, in interval notation, of the function N is  0,92.3 b) The graph extends from N  x   0 to N  x   6 million. Therefore, the range, in interval notation, of the function N is 0, 6, 000, 000 . c) Answers will vary. We would target the 50 year old to 60 year old age group, because that is the age group that has the most number of hearing-impaired Americans. 65. a) f (4)  5  3.50(3)  15.50 . The charge for the first mile is $5, and then the charge for the next 3 miles is $3.50 each, giving a fare of $15.50 for a 4-mile trip. b) f (4.25)  5  3.50(4)  19 . The 0.25 mile is considered part of a fourth additional mile after the first mile is charged at the $5 rate, so the total fare is $19.00 for a trip of at least 4 miles up to and including 5 miles. c) The fares will be $5, $8.50, $12, and so on, in increments of $3.50, up to $36.50. Thus the range is {5, 8.5, 12, 15.5, 19, 22.5, 26, 29.5, 33, 36.5}. 66. a) S (95)  20 . It cost $20 to ship an order totaling $95. b) S (102)  20  5  15 . It cost $15 to ship an order totaling $102, since $2 is considered a part of an extra $20. c) Orders totaling up to and including $120 will be shipped for $15, those totaling up to and including $140 will be shipped for $10, those totaling up to and including $160 will be shipped for $5. Thus, and order totaling $160.01 or more will be shipped for free. d) The shipping charges are reduced in increments of $5. Thus, the range of S is {0, 5, 10, 15, 20}. 67. The domain of the function is the union of where the numerator and denominator are defined. Thus it would be where x is greater than equal to 0 and 5  x is greater than 0. x0 5 x  0 5 x The domain is the union of these sets which is 0,5 . 68. The domain of the function is the union of where the numerator and denominator are defined. Thus it would be where 3  x is greater than equal to 0 and x does not equal 0. x0 3 x  0 3 x The domain is the union of these sets which is , 0  0,3 . 69. The domain of the function is where the denominator is defined. Thus it would be where   x x 2  9 does not equal 0.   x x2  9  0 x( x  3)( x  3)  0 0, 3,3  x The domain is the union of these sets which is , 3  3, 0  0,3  3,  . 70. The domain of the function is where the denominator is defined. Thus it would be where   x  x  x  12  0 x x 2  x  12 does not equal 0. 2 x( x  4)( x  3)  0 0, 4, 3  x The domain is the union of these sets which is , 3  3, 0  0, 4  4,  . 71. Answers may vary. One example is 1 . f  x  2 x x 4   72. Answers may vary. One example is 1 f  x  . x( x  1)(7) Copyright © 2020 Pearson Education, Inc.