(Solution Manual) Brief Calculus-An Applied Approach

Larson, 9

Math & Statistics

Solution Manual

9th Edition

• $15.00 $50.00

---

Description

Solution Manual for Brief Calculus-An Applied Approach, The downloadable file consists of 759 pages, and the total size of the file is 53.14 MB. Purchase now only for highly discounted price.

Author(s): Larson

---

---

Study Note Preview

Solutions to Tech Tutors 757 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. Solutions to TECH TUTORS Chapter 1 Section 1.3 (page 29) The lines appear perpendicular in the setting 99 x−≤ ≤ and 66 . y−≤ ≤ Section 1.6 (page 61) Most calculators set in connect ed mode will join the two branches of the graph with a nearly vertical line near 2. x= This line is not part of the graph. Chapter 4 Section 4.5 (page 287) Answers will vary. Chapter 6 Section 6.3 (page 395) 1.46265 Chapter 8 Section 8.4 (page 537) Answers will vary. Section 8.5 (page 546) Answers will vary. Chapter 11 Section 11.3 (page 676) ()2ln 1 yx x=+ CONTENTS Chapter 1 Functions, Graphs, and Limits ............................................................... 1 Chapter 2 Differentiation....................................................................................... 66 Chapter 3 Applications of the Derivative ........................................................... 145 Chapter 4 Exponential and Logarithmic Functions ........................................... 247 Chapter 5 Integration and Its Applications......................................................... 309 Chapter 6 Techniques of Integration .................................................................. 367 Chapter 7 Functions of Several Variables .......................................................... 416 Chapter 8 Trigonometric Functions.................................................................... 497 Chapter 9 Probability and Calculus .................................................................... 548 Chapter 10 Series and Taylor Polynomials .......................................................... 580 Chapter 11 Differential Equations ........................................................................ 646 Appendix A Precalculu s Review ............................................................................ 683 Appendix B Alternate Introduction to the Fundamental Theorem of Calculus.......................................................................................... 700 Checkpoints ............................................................................................................. 704 Tech Tutors ............................................................................................................. 757 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. APPENDICES Appendix A A Precalculus Review ........................................................................ 684 A.1 The Real Number Line and Order ..................................................... 684 A.2 Absolute Value and Distance on the Real Number Line.................. 686 A.3 Exponents and Radicals ..................................................................... 689 A.4 Factoring Po lynomials........................................................................ 691 A.5 Fractions and Rationalization............................................................. 696 Appendix B Alternative Introduction to the Fundamental Theorem of Calculus .......................................................................................... 700 684 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. APPENDIX A A Precalculus Review Section A.1 The Real Number Line and Order 1. Because 1 40.25 ,= it is rational. 2. Because 3678 13678 ,−= − it is rational. 3. Because π is irrational, 3 2πis irrational. 4. Because 2is irrational, 32 1− is irrational. 5. Because 4.3451 has a repeating decimal expansion, it is rational. 6. 22 7is rational. 7. Because 364 4,= it is rational. 8. Because 0.8177 has a repeating decimal expansion, it is rational. 9. Because 60 is not the cube of a rational number, 360is irrational. 10. Because e is irrational, 2 e is irrational. 11. 12 551 2 0 51 2x x x−> > > (a) Yes, if 3, x= then 15 53 x== is greater than 12 5. (b) No, if 3, x=− then 15 53 x=− =− is not greater than 12 5. (c) Yes, if 5 2, x= then 52 5 21 0x== is greater than 12 24 51 0.= 12. 13 33 23 0 3 2xx x x x x+< +< +< <− (a) No, if 0, x= then x is not less than 3.2− (b) No, if 4, x= then x is not less than 3.2− (c) Yes, if 4, x=− then x is less than 3.2− 13. 2024 02 8 21 0x x x−<< <−< << (a) Yes, if 4, x= then 2 10. x<< (b) No, if 10, x= then x is not less than 10. (c) No, if 0, x= then x is not greater than 2. 14. 3112 23 2 51 51 o r 15x x x xx−−< ≤ −< − ≤ −< −≤ − >≥ ≤< (a) No, if 0, x= then x is not greater than or equal to 1. (b) Yes, if 1, x=then 1 5. x≤< (c) No, if 5, x= then x is not less than 5. 15. 57 55 75 12x x x−≥ −+≥+ ≥ 16. ()()11 22 3 223 23x x x> > > 17. ()()11 22 1 241 2 41 21 221 21 21xx xx x x x x x+< +−− <−− <− <− <− 18. ()()11 2227 3 27 7 3 7 24 24 2x x x x x+< +−<− <− <− <− x 10 12 14 16 12 0 −1 −2x21− −6 −4 −2 x −1 −2 1 0 2 x 3 2 Section A.1 The Real Number Line and Order 685 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 19. () ()11 5542 3 1 42 43 3 143 55 55 1xx x xx x x x x−<− −− −<− − − −< − −− > −− > 20. 42 1 41 2 11 5xx xx x x x−≤ + −−−≤ +−− −≤ 21. 42 34 43 2 33 43 12 7 12 7 22 2 17 22x x x x x−< − < −+ < − + < + −< < −<< −<< 22. 03 5 03 33 53 32x x x≤+< −≤+−<− −≤ < 23. 3 1 44 3 1 44 3 1 44 3 1 441 11 11x x x x>+> −>+−>− −> > − −< < − 24. () ()113 31 3 3 13 33 33x x x x−< − < ⎛⎞−<− <⎜⎟⎝⎠ −< −< >> − 25. () ()()523 66 6 523 32 3 0 53 0 1153 055 6xx xx xx x x x+> ⎛⎞ ⎛⎞+>⎜⎟ ⎜⎟⎝⎠ ⎝⎠ +> > > > 26. ()523 66 523 32 3 0 30xx xx xx x−> ⎛⎞−>⎜⎟⎝⎠ −> > 27. () ( )2 226 26 0 23 20xx xx xx−< −−< +−< Zeros of the polynomial ()() 23 2xx+− are 3 2x=−and 2. x= By testing the intervals ()()33 22,, , 2 ,−∞ − − and() 2, ,∞the solution set is 3 22. x−<< 28. () ( )2 221 9 3 29 4 0 21 40xx xx xx+<− −+ < −− < Zeros of the polynomial ()( ) 21 4xx−− are 1 2x=and 4. x= By testing the intervals ()()11 22,,, 4 ,−∞ and () 4, ,∞the solution set is 1 24. x<< 29. Let E represent the earnings per share, in dollars. Then 4.1 4.25. E≤≤ 30. Let p represent the daily oil pr oduction, in millions of barrels. Then 2 2.4. p<< 31. Let p represent the percent of Americans who conduct banking transactions online. Then 40. p≤ 32. Let I represent the net income, in millions of dollars. Then 239. I≥ 33. 20A= and 220 200. rA=−= Let T be the target heart rate. Then ()()()() 0.60 200 0.90 200 120 180T T≤≤ ≤≤ So, the target heart rate for a 20-year-old is between 120 beats per minute and 180 beats per minute. 34. C=0.35 2500 13,000 0.35 10,500 30,000m m m+< < < So, the number of miles driven must be less than 30,000. 2 0 −2x −6 −4 −2 x −5 −2024x21−27 x −3 − 2 −1 0 1 2 3 −10x4 4 2131−− x −3 −2 −1 0 1 2 3 x 68 4x −10 0 10 40 30 20 50 x −1 −2 0123 2− x −1 0 4 5 3 2 1 1 2 686 Appendix A A Precalculus Review © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 35. 115.95R x= and 95 750Cx=+ and because , R C> you can write 115.95 95 750 20.95 750 75035.799520.95 36.xx x x x>+ > >= ≥… So, this product will return a profit if 36 units. x≥ 36. Revenue: 4.50 R x= Cost: 2.75 220 Cx=+ Profit: () 4.50 2.75 220 1.75 220PR C xx x=− =−+ =− So, 60 1.75 220 270 280 1.75 490 160 280x x x≤− ≤ ≤≤ ≤≤ So, the daily donut sales vary between 160 dozen donuts per day and 280 dozen donuts per day. 37. (a) False. Because , ab< 22 .ab−> − (b) True. Because , ab< 22 . ab+<+ (c) True. Because , ab< 66 .ab< (d) False, if 0, ab> then 11.ab> 38. (a) True. Because , ab< 44 .ab−<− (b) False. Because , ab< ab−> − and 44 .ab−>− (c) True. Because , ab< 33 .ab−> − (d) True. Because , ab<11 44. ab< Section A.2 Absolute Value a nd Distance on the Real Number Line 1. (a) 126 75 51 d=− = (b) 75 126 51 d=− = − (c) 126 75 51d=− = 2. (a) () 126 75 51 d=− −− = (b) () 75 126 51 d=− −− = (c) () 126 75 51 d=− −− =− 3. (a) () 9.34 5.65 14.99 d=− −= (b) 5.65 9.34 14.99 d=− − =− (c) () 9.34 5.65 14.99 d=− −= 4. (a) 2.05 4.25 6.3 d=− − = (b) () 4.25 2.05 6.3 d=− −= (c) 2.05 4.25 6.3 d=− − =− 5. (a) 16 128 112 57 5 7 5d=− = (b) 16 128 112 75 5 75d=− = − (c) 16 128 112 57 5 7 5d=− = 6. (a) 18 61 23 51 5 3d=− − = (b) ()61 18 23 15 5 3d=− −= (c) 18 61 23 51 5 3d=−− = − 7. 2 x≤ 8. 3 x< 9. 2 x> 10. 3 x≥ 11. 53x−≤ 12. 43x+< 13. 22x−> 14. 22 2x−> 15. 53x−< 16. 25x−> 17. 2 ya−≤ 18. yc h−< Section A.2 Absolute Value and Distance on the Real Number Line 687 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 19. 44 x−< < 20. ()()()111 22262 6 62 6 33x x x−< < −< < −< < 21. ()32 22 32 6x x x<− ⎛⎞<−⎜⎟⎝⎠ <− or ()32 22 32 6x x x> ⎛⎞>⎜⎟⎝⎠ > 22. 31 2 31 2 33 4x x x<− −< <− or 31 2 31 2 33 4x x x> > > 23. 25 2 25 55 25 37x x x−< − < −+ < − + < + << 24. 31 4 31 1 4 1 35 35 33 5 3x x x x x+≤ − +−≤ −− ≤− −≤ ≤− or 31 4 31 1 4 1 33 33 33 1x x x x x+≥ +−≥− ≥ ≥ ≥ 25. () ()352 325 22 31 0 33 1 03 7x x x x x−≤− −≤− −≤ − −+≤ −+ ≤− or () ()352 325 22 31 0 331 03 13x x x x x−≥ −≥ −≥ −+≥ + ≥ 26. 52 15 51 2 11 51 62 4 62 4 222 32x x x x x−< +< −− < +− < − −< < −< < −< < 27. 10 4 10 10 4 10 14 14x x x x−<− −− < − − −< − > or 10 4 10 10 4 10 6 6x x x x−> −− >− −> − < 28. 25 20 25 25 20 25 45 45x x x x−≤− −− ≤ −− −≤ − ≥ or 25 20 25 25 20 25 5 5x x x x−≥ −− ≥ − −≥ − ≤ 29. () () ()111 222192 1 19 92 9 19 10 2 8 10 2 8 54 45x x x x x x−<− < −−<− −<− −< −< − −−> − −> − − >> << 30. () () ()() ()211 13 211 1 1 113 2203 232 3 3 03 620 11 162022 2 30 03x x x x x x x x−< − < −−<− −<− −< − < ⎛⎞−<− <⎜⎟⎝⎠ −< − < −−> −− > − >> << 31. bx ab ba xaaba ab x ab−≤−≤ −+≤−+≤+ −≤≤+ 0x −6−4−22 4 6 x −3 − 2 −1 0 1 2 3 x 6 0 −6 0 x −6 −4 −2 2 4 6 012345678x x −3 − 2 −1 0 1 2 3 5 3 − 10 −10 0x−71 3x 10 14 6 x 0 10 45 40 30 20 50 5 6 4 2x −1 1 0 3 x 2 x a − ba + b a x −3 − 2 −1 1 0 2 3 688 Appendix A A Precalculus Review © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 32. 2 2 2 2 22 2xa b xaa ba x ab x ab abx−≤ − −+≤ − + ≤− −≤ −≤ or 2 2 2 2 22 2xab xaaba x ab x ab abx−≥ −+≥+ ≥+ +≥ +≥ 33. () () () ()3224 342 4 4 24 83 8 83 8 118833xabb xabb bx ab ba xaa ba abx ab−−< < −⎛⎞−< <⎜⎟⎝⎠ −< −< −+< −+< + −< <+ 34. () () ()5 2 5 2 5 2 5112 5 2 25 2 52 5 2 5xab xaa b a xab xab xab xab x ab−< − −− < − − −<− − −⎛⎞−< − − −⎜⎟⎝⎠ >+ ⎛⎞>+⎜⎟⎝⎠ >+ or ()5 2 5 2 5 2 2 5xab xba xab x ab−> −>− <− <− 35. Midpoint 82 4162+== 36. Midpoint 7.3 12.7102+== 37. Midpoint 6.85 9.351.252−+== 38. Midpoint () 4.6 1.32.952−+ −== − 39. Midpoint 3 11 24 4 1 22 8−+== = 40. Midpoint 55 62 5 23+ = = 41. 33.15 2p−≤ 42. 200,000 25,000 25,000 200,000 25,000 175,000 225,000x x x−≤ −≤ − ≤ ≤≤ The low production level is 175,000 barrels of oil, and the high production level is 275,000 barrels of oil. 43. 20 0.75 0.75 20 0.75 19.25 20.75x x x−≤ −≤ − ≤ ≤≤ The lowest and highest acceptable weights for a 20-ounce cereal box are 19.25 ounces and 20.75 ounces. 44. 57.517.5 57.5117.5 7.5 57.5 7.5 50 65w w w w−≤ −−≤ ≤ −≤− ≤ ≤≤ The guidelines specify that the weights for male collies lie between 50 pounds and 65 pounds 45. (a) ()4750 500 0.05 4750 237.50 4750 237.50E E−≤ = −≤ (b) $5116.37 is not within 5% of the specified budgeted amount; at variance. 46. (a) ()15,000 500 0.05 15,000 750 15,000 750I I−≤ = −≤ (b) $14,695.00 is within $500 of the specified budgeted amount; not at variance. 47. (a) ()20,000 500 0.05 20,000 1000 20,000 1000E E−≤ = −≤ (b) $22,718.35 is not within $500 of the specified budgeted amount; at variance. 48. (a) ()7500 500 0.05 7500 375 7500 375T T−≤ = −≤ (b) $8691.00 is not within 5% of the specified budgeted amount; at variance.x a − 8b 3a + 8b 3a 3 x a 2 5 a − 2 5 b 2 5 a + 2 5 b 2 5 x a − ba + b 2a 22 Section A.3 Exponents and Radicals 689 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 49. Let r be the percent of defective units. 0.0005 0.0001 0.0005 0.0001 0.0004 0.0006r r−≤ ≤+ ≤≤ Number of defective units 150,000 xr== () () 0.0004 150,000 150,000 0.0006 150,000 60 90r x≤≤ ≤≤ Cost of refunds 195.99 Cx== () () 60 195.99 195.99 90 195.99 11,759.40 17,639.10x C≤≤ ≤≤ The total cost of refunds shoul d be between $11,759.40 and $17,639.10. Section A.3 Exponents and Radicals 1. () ( )323 22 7 5 4−= −= − 2. 263 61233== 3. ()()3 11 8242 4−== 4. ()2 2777552 5−== 5. 1 113 11 3 4 3431 3 1 3− −++== = 6. () ()2 244 2 334 3 3 399 3−−= − = − = 7. () () ( ) ()233 2 4 2 3 4 4 8 12 32 44−− −= − − =+= 8. () ( )35 3 5 27 135−= −= − 9. () () ( ) ()0 006 10 6 10 6 1 60 6 1 5−⎡ ⎤ = − = − =⎣⎦ 10. () ()()3311 16416 4 41 4−=== −− −− 11. 3 3 227 729 9== 12. ()()()3 3311 1 1 99 3 2 7== = 13. 12 1142 4−== 14. () ()34 33 34411 1 11616 8 2 16−== = = 15. () ()() ()25 25 2 2511 1 1324 32 2 32−−== = = −− − 16. ()323 210 10 100== 17. ()60500 1.01 908.3483 ≈ 18. ()12010,0000.1079 1.1≈ 19. 354 3.7798−≈ − 20. 6325 2.6221≈ 21. ()()324 2 3 1 2 21 2 3 14 1462 62 162 168 3 4yy y y y y y−−− − − −− −= ⎛⎞=⎜⎟⎝⎠ ⎛⎞=⎜⎟⎝⎠ = 22. ()34 3 433 3zz z z−− +== 23. ()22410 10x x= 24. ()23641 6x x= 25. 2 23 5 3777xx xx+ −== 26. ()3 31 2 72 721 xxxx x− −− −== = 690 Appendix A A Precalculus Review © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 27. () ()() ()3 32 5 210 55 22 4xyx yx y xy+ −+=+ =+ + 28. 3 3 2312 4 64 93 2 7s ss s⎛⎞ ⎛⎞==⎜⎟ ⎜⎟⎝⎠ ⎝⎠ 29. ()()()12 11 2 1 2 12 123 333xx xxx xxx+−== = 30. ()()333 22 3 2x xx== 31. 84 24 2 2 2=⋅ = = 32. 33 3 33 316 8 2 8 2 2 2 27 27 3 27⋅=== 33. ()()33 3 3 53 23 2 2 3 54 27 2 27 2 3 2x xx x x xx=== 34. ()4 3 23 23 2 433 3xyx y x y== 35. ()()94 5 93 3 1 2 3 3 31 1 2 3 32 3144 8 18 21 8 21 8xyz x yz yz xy z y z xz z yy−− − −−= = = 36. ()()58 48 4 4 24 4 232 16 2 22 22xyz yz x y yz xy yxyz−− −= = = 37. ()()()3231 2 3 4 3 2 2xx x x x xx−= − = +− 38. ()42 2 286 2 43xx x x−= − 39. ()3 52 12 12 3 122122 1xxx x xx−− ++= + = 40. ()3 32 32 32 3 325155 1xxx x xx−− −−= − = 41. () () ()()( ) ()() ()() ()32 12 12 12 2 12316 13 1 1 2 31 2 31 1 2xx x x xx xx x xx x+− +=+ + − =+ + − =+ −+ 42. () () () ()( ) ()() ()() ()52 32 32 32 2 322 1 41 21 1 2 21 2 21 2 1xx x x xx xx x xx x−−−=− − − =− − − =− −+ 43. () ()63 6 3 22 2 4 351 5 33 153 1513x xx x x xx x x xx⎛⎞ +=+⎜⎟ ⎝⎠ =+ =+ 44. () () () ()() ()() ()() () ()() () ()23 2 22 2 2 2 211 1 111 11 12 1 21 1xx x xxx xx x x x x+− − − −=+ − − ++ −= + −= + 45. 4x−is defined when 4. x≥ Therefore, the domain is [) 4, .∞ 46. 52 x− is defined when 5 2. x≤ Therefore, the domain is (5 2,.⎤−∞⎦ 47. 23 x+is defined for all real numbers. Therefore, the domain is (),.−∞ ∞ 48. 241x+is defined for all real numbers. Therefore, the domain is (),.−∞ ∞ 49. 31 4x−is defined for all real numbers except 4. x= Therefore, the domain is ()() ,4 4 , .−∞ ∪ ∞ 50. 31 4x+is defined when 4. x≠− Therefore, the domain is ()() ,4 4 , .−∞ − ∪ − ∞ 51. 2 1x x+ − The numerator is defined when 2. x≥− The denominator is defined when 1. x≠ Therefore, the domain is [)() 2, 1 1, .−∪ ∞ Section A.4 Factoring Polynomials 691 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 52. 1 23x+is defined when 3,2x>− and 64 x− is defined when 3.2x≤ Therefore, the domain is 33,.22⎛⎤−⎜⎥⎝⎦ 53. 1200.06510,000 1 $19,121.8412A⎛⎞=+ ≈⎜⎟⎝⎠ 54. 10000.057000 1 $8027.61365A⎛⎞=+ ≈⎜⎟⎝⎠ 55. 600.0555000 1 $11,345.464A⎛⎞=+ ≈⎜⎟⎝⎠ 56. 1800.078000 1 $22,791.5712A⎛⎞=+ ≈⎜⎟⎝⎠ 57. 41 1222 232 32 8 22 22.22 seconds2 2LTπππ π ππ=== = == ≈ 58. ()() () () ()() () ()23 2111 1 1 11 1 1 1n nA Pr Pr Pr Pr Pr r r r−=+ +++++ ++ ⎡ ⎤ = + ++++ +++⎣ ⎦" " Section A.4 Factoring Polynomials 1. 6, 7, ab== − and 1 c= 74 9 2 4 7 5 12 12x±− ±== So, 75112x+== or 75 1.12 6x−== 2. 8, 2, ab== − and 1 c=− 24 3 2 2 6 16 16x±+ ±== So, 26 1 16 2x+== or 26 1.16 4x−== − 3. 4, 12, ab== − and 9 c= 12 144 144 12 3 88 2x±−== = 4. 9, 12, ab== and 4 c= 12 144 144 12 2 18 18 3x−± − −== = − 5. 1,a= 4, b= and 1 c= 41 6 44 2 32322y−± − −±== = − ± 6. 1, 5, ab== and 2 c=− 52 5 853 3 22y−± + −±== 7. 2, 3, ab== and 4 c=− 39 3 234 1 44x−± + −±== 8. 3, 8, ab==− and 4 c=− 86 4 4 8 8 4 7 66x±+ ±== 9. ()2 281 6 4 xx x−+=− 10. ()2 210 25 5 xx x++ = + 11. ()2 244 1 2 1xx x++ = + 12. ()2 291 2 4 3 2xx x−+ =− 13. ()()234 1 3 11xx xx−+= − − 14. ()()221 2 1 1xx x x−−= + − 15. ()()235 2 3 2 1xx x x−+= − − 16. ()()241 9 1 2 4 3 4xx x x++ =++ 17. ()2 2244 2 x xy y x y−+= − 18. ()( )2222 x xy y x y x y−− =− + 19. ()() ()() ()42 2 281 9 9 93 3yy y yy y−=+ − =+ + − 20. ()()()()()42 2216 4 4 4 2 2 xx xx x x−= + − = ++− 21. ()()33 3 282 22 4xx xx x−= − =− ++ 692 Appendix A A Precalculus Review © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible website, in whole or in part. 22. ()( )33 3 264 4 4 4 16 yy y y y−=−=− ++ 23. ()( )33 3 264 4 4 4 16 yy y y y+=+=+ −+ 24. ()( )33 3 2125 5 5 5 25 zz z z z+= + = + − + 25. ()( )33 2 2x yx y x x y y−=− ++ 26. () () () () () ()( ) ()32 32 22 2 22 22 2xa b xa bxa xa bb xab x x aa x ba bb xab x x ab a a bb⎡⎤ −+= ⎡ − + ⎤ −− −+⎣⎦ ⎣⎦ ⎡⎤ =− +⋅ − +−++⎣⎦ ⎡⎤ =− +⋅ − ++ ++⎣⎦ 27. ()() ()() () () ()32 2 244 4 4 41 411xx x x x x xxxx x−− + = − − − =− −=− + − 28. ()() ()() () () () () ()32 2 2 211 1 11 111 11xxx x x x xxxxx xx−− + = − −− =− − =− + − =+ − 29. ()() ()()32 2 2234 6 2 3 2 2 3 23 2xxx x x x xx−+ − = − + − =− + 30. ()() ()()32 2 255 2 5 5 55 55xxx x x x xx−− + = − −− =− − 31. ()() ()()32 2 224 2 2 2 2 22 1xx x x x x xx−− + = − − − =− − 32. ()() ()() () () ()32 2 27 4 28 7 4 7 74 722xxx x x x xxxxx−− + = − −− =− −=− + − 33. ()() () () ()42 2 2 215 16 16 1 44 1xx x x xxx−− = − + =− + + 34. () () () () ()42 22 224 92 5 21 2 5 55 2 1xx xx xx x−− =+− =− + + 35. ()250 50 0, 5xx xx x−= −= = 36. ()2 3 223 0 230 0,xx xx x−= −= = 37. () ()290 33 0 3, 3x xx x−= +− = =− 38. () ()225 0 55 0 5, 5x xx x−= +− = =− 39. ()()230 33 0 3x xx x−= +− = =± 40. ()() ()()280 88 0 22 22 0 22x xx xx x−= +− = +− = =± 41. () ()2 239 0 69 9 0 60 0, 6x xx xx x−− = −+ − = −= = 42. () ()2 213 6 0 13 6 16 16 7, 5x x x xx+−= += += ± =−± =− 43. () ()220 21 0 2, 1xx xx x+−= +− = =− 44. () ()256 0 32 0 3, 2xx xx x++ = ++ = =−− Section A.4 Factoring Polynomials 693 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 45. () ()256 0 23 0 2, 3xx xx x−+ = −− = = 46. () ()220 0 54 0 5, 4xx xx x+− = +− = =− 47. () ( )2 2 335 2 0 32 10 ,1xx xx x++ = ++ = =− − 48. () ( )2 1 221 0 21 10 ,1xx xx x−−= +− = =− 49. 3 3 364 0 64 64 4x x x+= =− =−= − 50. 3 3 3216 0 216 216 6x x x−= = == 51. 4 4 416 0 16 16 2x x x−= = =± =± 52. 4 4 4625 0 625 625 5x x x−= = =± =± 53. () () ()() () () ()32 2 244 0 14 1 0 14 0 122 0 1, 2xx x xx x xx xx x x−−+ = −− −= −− = −−+ = =± 54. () () ()()32 2 2 1 226 3 0 21 3 210 21 30xx x xx x xx x+++ = ++ += ++ = =− [ Note: 230 x+= has no real roots.] 55. Because () ()242 2 , xx x−= + − the roots are 2. x=± By testing points inside and outside the interval []2, 2 ,− we find that the expression is defined when 2 x≤− or 2.x≥ So, the domain is (][) ,2 2 , .−∞− ∪ ∞ 56. Because () ()242 2 ,x xx −= + − the roots are 2. x=±By testing points inside and outside the interval []2, 2 ,− we find that the expression is defined when 22 . x−≤≤ So, the domain is []2, 2 .− 57. Because () ()271 2 3 4 , xx x x−+= − − the roots are 3x=and 4. x= By testing points inside and outside the interval []3, 4 , we find that the expression is defined when 3 x≤or 4.x≥ So, the domain is (][) ,3 4 , .−∞∪∞ 58. Because () ()281 5 3 5 , xx x x−+= − − the roots are 3x=and 5. x= By testing the intervals ()() ,3, 3 ,5,−∞ and ()5, ,∞we find that the expression is defined when 3 x≤or 5.x≥ So, the domain is (][) ,3 5 , .−∞∪∞ 59. Because () ( )256 1 5 11 ,xx xx++ = + + the roots are 1 5x=−and 1. x=−By testing the intervals ()()1 5,1 , 1 , ,−∞− − − and ()1 5,,−∞ we find that the expression is defined when 1 x≤− or 1 5. x≥− So, the domain is (])1 5,1 , .⎡ −∞− ∪− ∞⎣ 60. Because () ( )231 0 3 3 1 3 ,xx x x−+ = −− the roots are 1 3x=and 3. x= By testing the intervals ()1 3,,−∞()1 3,3, and ()3, ,∞we find that the expression is defined when 1 3x≤or 3.x≥ So, the domain is ([)1 3,3 , .⎤−∞∪∞⎦ 61. So, the factorization is ()()32 236 2 1 4 2 . xxx x xx−− − = + − − −1 1 −3 −6 −2 −1 4 2 1 −4 −2 0 694 Appendix A A Precalculus Review © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 62. So, the factorization is ()( ) () () ()32 222 1 3 2 112 .xx x xxx xxx−− + = + − + =+ − − 63. So, the factorization is ()( )32 222 1 1 2 3 1 .xx x x x x−−+ =+ −+ 64. So, the factorization is ()()43 2 3216 96 256 256 41 2 4 8 6 4 .xxx x xx x x−+−+= −−+ − 65. Possible rational zeros: 8, 4, 2, 1±±±± Use synthetic division for 1. x=− So, ()() () () ()32 210 8 0 12 8 0 124 0 1, 2, 4.xx x xxx xx x x−− − = +− − = ++− = =− − 66. Possible rational zeros: 6, 3, 2, 1±±±± Use synthetic division for 1. x=− So, ()() () () ()3 276 0 16 0 132 0 1, 3, 2.xx xx x xxx x−− = +− − = +−+ = =− − 67. Possible rational roots: 1, 2, 3, 6±±±± Use synthetic division for 1. x= So, ()() () () ()32 261 1 6 0 15 6 0 123 0 1, 2, 3.xx x xxx xx x x−+− = −− + = −−− = = 68. Possible rational roots: 1, 2, 3, 6±±±± Use synthetic division for 2. x= So, ()() () () ()32 225 6 0 24 3 0 213 0 2, 1, 3.xxx xx x xx x x+− − = −+ + = −++ = =−− 69. Possible rational zeros: 3 1211 223366 , 3 , 2 , 1 ,,,,,±±±±±±±±± Use synthetic division for 3. x= So, ()() () ( ) ( )32 2 21 3261 11 9 6 0 36 7 2 0 33 22 1 0 3, , .xxx xx x xx x x−− − = −+ + = −++ = =−− −1 1 −2 −1 2 −1 3 −2 1 −3 2 0 −1 2 −1 −2 1 −2 3 −1 2 −3 1 0 4 1 −16 96 −256 256 4 −48 192 −256 1 −12 48 −64 0 −1 1 −1 −10 −8 −1 2 8 1 −2 −8 0 1 1 −6 11 −6 1 −5 6 1 −5 6 0 2 1 2 −5 −6 2 8 6 1 4 3 0 3 6 −11 −19 −6 18 21 6 6 7 2 0 −1 1 0 −7 −6 −1 1 6 1 −1 −6 0 Section A.4 Factoring Polynomials 695 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 70. Possible rational roots: 1 21, 2, 4, ,±±±± 11 36,,±±11 91 8,,±±2244 3939,,,±±±± Use synthetic division for 1 2. x= So, ()() ()()() () () ()32 2 21 2 1 2 12 2318 9 8 4 0 18 8 0 29 4 0 21 32 320 ,.xxx xx xx xx x x−− + = −− = −− = −+− = =± 71. Possible rational roots: 1, 2, 4±±± Use synthetic division for 4. x= So, ()()32 233 4 0 41 0 .xxx xx x−− − = −+ + = Because 21 xx++ has no real solutions, 4 x=is the only real solution. 72. Possible rational zeros: 3 1 226, 3, , 2, 1,±±±±±± Use synthetic division for 3. x= So, ()() () ( ) ()32 2 1 221 3 6 0 32 5 2 0 32 1 2 0 3, , 2.xx x xx x xx x x−− − = −+ + = −+ + = =−− 73. Possible rational zeros: 11 42,, 1 , 2±±±± Use synthetic division for 1. x=− So, ()() () ( ) ()32 2 1 441 15 2 0 14 7 2 0 14 1 2 0 1, , 2.xx x xx x xx x x++ − = ++ − = +− + = =− − 74. Possible rational zeros: 12 33,, 1 , 2 , 3 , 6±±±±±± Use synthetic division for 3. x=− So, ()() () ( ) ()32 2 2 3341 3 6 0 33 5 2 0 33 2 1 0 3, , 1xx x xx x xx x x+−+ = +− + = +− − = =− 75. The equation is a second-degr ee equation, so it has two solutions. 2 2 20.0003 1200 0 0.0003 1200 4,000,000 2000x x x x−= = = =± There cannot be a negative number of units, so 2000 x=− should be rejected. A production level of 2000 units will minimize the average cost. 76. () () ()2 2 2200 2000 3800 1000 0 200 2000 4800 0 200 10 24 0 200 4 6xx xx xx xx−+ −> >−+ >− + >− − The roots are 4, 6. x= By testing points inside and outside the interval []4, 6 , we find that the expression is defined when 4 6. x<< 1 2 18 −9 −8 4 9 0 −4 18 0 −8 0 4 1 −3 −3 −4 4 4 4 1 1 1 0 3 2 −1 −13 −6 6 15 6 2 5 2 0 −1 4 11 5 −2 −4 −7 2 4 7 −2 0 −3 3 4 −13 6 −9 15 −6 3 −5 2 0 696 Appendix A A Precalculus Review © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. Section A.5 Fractions and Rationalization 1. ()() () ()2 243 71 2 31 8 6 3 4,36xx xx xx x x xxx−− −+=+− + − −=≠+ 2. ()() () ( )2 261 76 11 10 10 1xx xx xx x x−− −+=++ + + 3. ()() () ( )2 252 31 0 26 2 3 2 5,223xx xx xx x x xxx+− +−=−− + − +=≠+ 4. ()() () ()2 234 3 31 3 1 2 42 1 7 3 34,37xx xx xx x x xxx++ ++=−− − + +=≠ −− 5. 33 22 2xx xx x++=−−− 6. () 51 021 0 51 0 21 0 21 21 21 3 21xx xx xx x x x+− + ++−=−− − =− 7. 2 233 3xxx xx x x−=− −= 8. 3 22 2 3 223 23 32xxx xx x x+= + += 9. () () ( )() () ( ) () ( ) () ( )2 223 4 5 3 25 33 4 3 3 4 3 3 4 68 5 1 5 33 4 59 8 33 4xx x x x x xx xx xx x xx xx xx+−+= +− + −+ −+ ++ −=−+ −+=−+ 10. () () () () () () () () ()32 31 3 1 31 2 31 2 31 2 36 31 31 2 7 31 2x x xx x x x x xx xx xx+ −−= −− + −+ −+ +− +=−+ =−+ 11. () ()() () () () ()2 22 21 2 1 42 2 2 2 2 22 22 4x xx x xx x x xx x x+−= −⋅−− − + −+ −+=−+ =−− 12. () () () ()() () () () () () ()2 23 55 33 3 33 33 53 33 35 33xx x xx x xx xx xx xx xx xx−+= ++− + +− +− +−=+− −+=+− 13. () () () ()() () () () () () () ()211 22 2 1 2 11 21 21 1 21 1 21xx xx x x x x x x xx xx xx xx xx−= −+− + + − + −=−+− +− −−=+− =+− Section A.5 Fractions and Rationalization 697 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 14. () ()() () () () () () () () ()23 23 2 3 2 13 1 3 1 3 1 26 32 31 58 31x xx xx x x x x x xx xx x xx− −−+=++− + − + − + −+ −=−+ −=−+ 15. ()() () () () () () ()2 2 22 2 2 2 2 2 221 1 21 1 1 11 21 1 1 2 1 2 1 2 1xx xx xx xx xx xx xx xx xx xx x x−+ + −+ =+ ++ −− += + −+= + −−= + −−=+ 16. () ()() () () () () () () ()32 32 1 33 1 22 2 2 2 2 36 36 1 22 61 22xx x x xx xx xx xx x xx−++ +++ =+ − +− +− −+ ++=+− +=+− 17. () () ()() () ()32 12 32 32 3221 2 1111 2 1x xx xxxx x x+ −−+=+ ++++ += + 18. ()() () () () ()2 2 2222 2 22 2 2 1 22 42 2 2xxx x xx x x x x xx x x−−+ −−=⋅ + −+−= () () () ( )24 2 2 25 2 2xx x x xx x−⎡+ −⎤⎣⎦= −−= 19. () ()21 21 21 2 1 1 21 21 22 1 21 3 21tt t tt t tt tt t t t−−+ + − ++=− ⋅ + + −− += + −= + 20. ()2 2 2 22 2 22 22 22 22 2211 1 11 1 11 1 1 1 1x x x x xx x x xx xx xx xx+−+ + ++=− ⋅ + ⋅ ++ −+ + = + =− + 698 Appendix A A Precalculus Review © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 21. () ()3 22 2 23 2 22 33 2 221 1 1 21 1 1 11 22 1 1 1xxx x x xx x x xx xx x x x⎛⎞+− ÷ +⎜⎟ + ⎝⎠ ⎛⎞+⎜⎟=−⎜⎟ + ++⎝⎠ +−=⋅+ + () () ()3 22 2 3222 11 2 1xx xx xx x+= ++ + = + 22. () () () ()() ()3 33 3 33 3 33 333 12 3233 3311 21 21 31 1 21 21 21 3 2 21 1 21xxx x x x x xx xx x xx x⎛⎞+− ÷ +⎜⎟ + ⎝⎠ ⎛⎞+⎜⎟=−⋅⎜⎟ + ++⎝⎠ +− −== ++ + 23. () () () ()12 1222 2 2 1222 2 2 2222 22 2 2xx x x x xx x xx− −+− + ⎡⎤++ −⎣⎦= = + 24. () () () ()12 12 12 22 211 11 1 1xx x xx x xx xx− −+⎡ − + ⎤ +− + ⎣⎦= −= + 25. () () () ()321 1 21 1 1 21 11 1 21xx xx x x x x xx xx xx+− + + ⎛+ ⎞=−⎜⎟+ ++⎝⎠ = + 26. ()() ()() () () () ()2132 232 2 22 23 23 222 22 2322 2 232221 31 31 21 31 31 23 1 31 3 31xx x x x x xxx xx xx x xx−− − ⎛⎞−⎜⎟=−⎜⎟⎜⎟−−⎝⎠ −− = − −= − 27. () () ()() () () () ()2 32 12 2 32 32 2 32 322 23 23 22 3 23 23 36 23 32 23xx xx xx x xx xx x xx x−+ ++ + −=+ ++ += + += + 28. () () ()()() () ()32 1222 2 32 3222 2 3223 23 3 32 3 23 23 61 8 23x xx x x xx xx x−+ ++ + −=+ ++ −+= + 29. 22 1 0 2 1 0 1 0 10 5 10 10 10=⋅= = 30. 33 2 1 3 2 1 2 1 21 7 21 21 21=⋅= = 31. 44 1 4 1 1 11 1xx x x x x xx x−−=⋅=− −− − 32. 75 7 55 7 77 7yy y yy y yy y++=⋅=+ ++ + 33. () () () () ()2 22 2 2 249 3 49 3 9 99 9 49 3 9 33 49 9,33xx x xx x xx xx xxx−− −=⋅ − −− −−=+− −=≠+ Section A.5 Fractions and Rationalization 699 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 34. () () () () () ()2 22 2 2 2 2 210 2 10 2 6 66 6 10 2 6 6 10 2 6 23 10 6,23xx xx xx xx xx xx x xx xx x xx xxxx++ −−=⋅ −− −− −− +− −=−− +− −=+− −−=≠ −− 35. ()55 1 4 2 14 2 14 2 14 2 51 4 2 14 4 14 2 2+=⋅ −− + + =− += 36. ()13 13 6 10 61 0 61 0 61 0 13 6 10 36 10 61 0 2−=⋅ ++ − − =− −= 37. 11 6 5 65 65 65 65 65 65−=⋅ ++ − −=− =− 38. () ()23 23 23 23 23 23 32xx x x−=⋅ ++ − − =− =− 39. () ()22 2 22 2 22 2 2xx xx xx xx xx xx xx−−=⋅ +− +− −− −− =−− =−− 40. () () () ()10 10 5 55 5 10 5 5 25 25xx xx xx xx xx xx xx xx−+=⋅ +++ + − + −+ =−+ =− − + =+ − 41. ()22 22 22 22 22 22 1,0 22xx x xx x x xx x x+− +− ++=⋅ ++ +−= ++ =≠ ++ 42. ()11 11 11 11 11 11 1,0 11xx x xx x x xx x x+−+ − + +=⋅ ++ +−= ++ =≠ ++ 43. 10,000,P= 0.075 r= 60 N= ()600.075 1210,000 $200.38 110.075 12 1M⎡⎤=≈⎢⎥⎛⎞⎢⎥−⎜⎟⎢⎥⎜⎟ +⎝⎠⎣⎦ 44. (a) ()2 2 2900,0006 6 900,000 6 900,000 6 150,000Cxx x xx x x x x=+ =+ += + = (b) When 240: x= ( )26 240 150,000 $5190.00240C+ == When 387: x= ( )26 387 150,000 $4647.58387C+ =≈ When 480: x= ( )26 480 150,000 $4755.00480C+ == The retailer should place an order of 387 units, because this cost is the lowest. 700 Appendix B Alternative Introduction to the Theorem of Fundamental Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. APPENDIX B Alternative Introduction to the Fundamental Theorem of Calculus 1. The subintervals are [][][][] 0, 1 , 1, 2 , 2, 3 , 3, 4 , and []4, 5 . The midpoints of each subinterval are 3571 2222,,,, a n d 9 2. Because the width of each rectangle is 1, th e sum of the areas of the five rectangles is ()()()()()135791111122222 3517.5 square units.2S fffff⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=++++⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ == 2. The subintervals are [][][] 0, 0.5 , 0.5, 1 , 1, 1.5 , and [] 1.5, 2 . The midpoints of each subinterval are 0.25, 0.75, 1.25, and 1.75. Because the width of each rectangle is 0.5, th e sum of the areas of the four rectangles is ()()()()()()()() 0.5 0.25 0.5 0.75 0.5 1.25 0.5 1.75 5.375 square units.S ffff=+++ = 3. Left Riemann sum: 0.518 Right Riemann sum: 0.768 4. Left Riemann sum: 3.685 Right Riemann sum: 4.038 5. Left Riemann sum: 0.746 Right Riemann sum: 0.646 6. Left Riemann sum: 0.760 Right Riemann sum: 0.635 7. Left Riemann sum: 0.859 Right Riemann sum: 0.659 8. Left Riemann sum: 1.166 Right Riemann sum: 1.270 9. (a) (b) 20 2xnn−∆= = Endpoints: ()24 200 0 0 2 22 2201 2 1n nn n nnnn nn<+<+< <+ < ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞<<< < −<⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠" " (c) Because yx= is increasing, ()()1 iifm fx−= on [] 1, .iix x− () ( ) ( ) () 1 11 122 2211nn n i ii isn f x x f i inn nn− == =⎡⎤ ⎡⎤⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=∆ = −= − ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎢⎥ ⎢⎥⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠⎣⎦ ⎣⎦∑∑ ∑ (d) ()()[] 1 on , ii i ifMf xx x− = () () 11 122 22nn n i ii iSn f x x f i inn nn== =⎡⎤ ⎡⎤⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=∆ = = ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎢⎥ ⎢⎥⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠⎣⎦ ⎣⎦∑∑ ∑ 32 xy Appendix B Alternative Introduction to the Fundamental Theorem of Calculus 701 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. (e) (f ) () 122 2 2lim 1 lim 2n nnininn n →∞ →∞=⎡⎤ + ⎛⎞ ⎛⎞−= =⎜⎟ ⎜⎟⎢⎥⎝⎠ ⎝⎠⎣⎦∑ 122 2lim lim 2 2n nniinn n →∞ →∞=⎡⎤⎛⎞ ⎛⎞ ⎡ ⎤=+ = ⎜⎟ ⎜⎟⎢⎥ ⎢⎥⎝⎠ ⎝⎠ ⎣ ⎦⎣⎦∑ 10. (a) (b) 31 2xnn−∆= = Endpoints: ()24 211 1 1 3 22 2211 1 12 1 1 1n nn n nnnn nn<+ <+ < <+ < ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞<+ <+ < <+ − <+⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠" " (c) Because yx=is increasing, ()()1 iifm fx−= on 1, .iix x−⎡⎤⎣⎦ () ( ) ( ) () 1 11 122 2211 11nn n i ii isn f x x f i inn nn− == =⎡⎤ ⎡⎤ ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=∆ = + −= + − ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥ ⎢⎥⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ ⎣⎦ ⎣⎦∑∑ ∑ (d) () ( ) iifMf x= on 1,i ix x−⎡⎤⎣⎦ () () 11 122 2211nn n i ii iSn f x x f i inn nn== =⎡⎤ ⎡⎤⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞=∆ = + = + ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟⎢⎥ ⎢⎥⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠⎣⎦ ⎣⎦∑∑ ∑ (e) (f ) ()() ()1 11 22 2 2lim 1 1 lim 2 22 4 2lim 2 lim 4 4 1 22 2 2lim 1 lim 2 lim n nni nn n nni nnnin nnn n n n nn n nninnn n n→∞ →∞= →∞ →∞ →∞ →∞= →∞⎡⎤⎛+ ⎞ ⎡⎤ ⎛⎞ ⎛⎞ ⎛⎞+− = + − ⎢⎥⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎢⎥⎝⎠ ⎝⎠ ⎝⎠ ⎢⎥ ⎣⎦ ⎝⎠⎣⎦ +⎡⎤ ⎡ ⎤= + −= −=⎢⎥ ⎢ ⎥⎣⎦ ⎣ ⎦ ⎡+ ⎤ ⎡⎤⎛⎞ ⎛⎞ ⎛⎞+=+ ⎢⎥ ⎜⎟ ⎜⎟ ⎜⎟⎢⎥⎝⎠ ⎝⎠ ⎝⎠⎣⎦ ⎣⎦ =∑ ∑ ()21 22l i m 4 4 nn nn →∞⎡+ ⎤ ⎡⎤+= + =⎢⎥ ⎢⎥⎣⎦ ⎣⎦ n 5 10 50 100 Left sum, LS 1.6 1.8 1.96 1.98 Right sum, RS 2.4 2.2 2.04 2.02 n 5 10 50 100 Left sum, LS 3.6 3.8 3.96 3.98 Right sum, RS 4.4 4.2 4.04 4.02 x 241234y702 Appendix B Alternative Introduction to the Theorem of Fundamental Calculus © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 11. 5 03dx∫ 12. ()2 042 xdx−∫ 13. () () ()40 4 44 044 4 xdx x dx x dx −−−=+ + −∫∫ ∫ 14. 22 0xdx∫ 15. ()22 24 xdx −−∫ 16. 1 211 1dxx−+∫ 17. 2 01x dx+∫ 18. ()1 22 11 x dx −+∫ 19. 12 A= 20. 8A a= 21. 8A= 22. 4 A= 23. 14 A= 24. 25 2A= 25. 1 A= 26. 2A a= 3 3x Rectangley 44 x Triangley1 −11 x Trian gleyTrapezoidy x −1 1234 −3369 x5 a −ayx1234 −11 3 4 5y 2 x −11 2 3456 −1123456y a a −axy Appendix B Alternative Introduction to the Fundamental Theorem of Calculus 703 © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. 27. 9 2Aπ= 28. 2 1 2A rπ= 29. () ()123 1 13 2 1Sn n Snn=+++ + −+ =+−++++" " Adding, ()()()() () ()2 1 1 1 times 21 1 2Sn n n n Sn n nnS=++++++ =+ +=" 30. (a) 2 1 21 1 11ixd x xnn cin−∆== =+∫ () ()2 1 01 1 11 lim 11lim 1 11lim 1 1 11lim 2 11 3lim 122 2n ixi nn nn nii n nxd x f c x inn inn nnnnn n∆→= ∞ →∞= →∞== →∞ →∞=∆ ⎛⎞ ⎛ ⎞=+ ⎜⎟ ⎜ ⎟⎝⎠ ⎝ ⎠ ⎡⎤=+⎢⎥⎣⎦ ⎡ +⎤=+ ⋅⎢ ⎥ ⎣ ⎦ ⎡⎤=+ + =⎢⎥⎣⎦∑ ∫ ∑ ∑∑ (b) 4 0 3 40 4 4 ixd x xnn cin−∆== =∫ () ()4 0 01 1 1 2 1 2 3 l i m 44lim 3 41 2lim 48lim 1 48lim 2 24lim 24 24n ixi n ni n ni n ni n nxd x f c x inn inn in nn n n∆→= →∞= →∞= →∞= →∞ →∞=∆ ⎛⎞= ⎜⎟⎝⎠ = = ⎛+ ⎞=⋅⎜⎟ ⎝⎠ ⎛⎞=+ =⎜⎟⎝⎠∑ ∫ ∑ ∑ ∑ 31. () ()5 11n i ifxx f x d x =∆>∑ ∫ 32. () ()5 11n i ifxx f x d x =∆<∑ ∫ 4 3 −3x Semicircley xr −ry704 Solutions to Checkpoints © 2013 Cengage Learning. All Rights Reserved. May not be scanned, copi ed or duplicated, or posted to a publicly accessible webs ite, in whole or in part. Solutions to CHECKPOINTS Chapter 1 Section 1.1 1. 2. 3. Let ()() 11,2 , 1xy=− and () ( )22,2 , 4 .xy= () () () ( ) () ()22 21 21 2 2 2222 4 1 43 25 5dx xy y=− + − =⎡ − − ⎤ +−⎣⎦ =+ == 4. () ( ) () ( ) () ( )2 2 1 2 2 2 22 352 5 1 93 6 4 5 62 3 1 1 64 2 0 65 35 16 4 6 5d d d=− + ⎡ − − ⎤ = + =⎣⎦ =− + ⎡ − − − ⎤ =+ =⎣⎦ =− + − − = + = Because ()()2222 2 12 3 45 20 65 , dd d+= + == you can apply the converse of the Pythagorean Theorem to conclude that the triangle must be a right triangle. 5. () 10, 10 and () 25, 30 () ()2230 10 25 10 400 225 625 25 ydd=− + − =+ == 6. Let ()() 11,6 , 2xy=− and ()() 22,2 , 8 .xy= ()12 12Midpoint ,22 62 28,22 2, 5x xy y++⎛⎞=⎜⎟⎝⎠ −+ +⎛⎞=⎜⎟⎝⎠ =− 7. Midpoint ()2007 2009 76 99, 2008, 87.522++⎛⎞==⎜⎟⎝⎠ For 2008, you would estimate the annual sales to have been $87.5 billion. 8. Original point Translated point () () () ()1, 0 3, 2 3, 6 1, 4 ( )() () ( ) () ( ) () ( )12 , 04 1 , 4 32 , 24 1 , 2 32 , 64 1 , 2 12 , 44 1 , 0−−= − − −− = − −− = −− = − −4 −3 −2 −1 1 2 3 4 4 3 2 1 −3 −4 −1 y x (−1, −2) (−3, 2) (3, 1) (0, −2) (4, −2) 2000 2002 2004 2006 200827002750280028502900 tE Year −2 1 4 5 6 7 8 −1 −2 −3 −4 −5 1 2 3 4 5 x y d 2 d 3 d 1 (5, 5) (6, −3)(2, −1)