Calculo purcell 9 ed solucionario

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Transcript of Calculo purcell 9 ed solucionario
 1. CHAPTER00.1 Concepts Review 1. rational numbersPreliminaries 1 2 1 1 1 1 8. = 3 5 2 3 5 2 1 5 3 3 5 2 15 15 2. dense1 2 1 2 1 2 1 = = 3 5 2 15 3 5 15 1 6 1 1 5 1 = = = 3 15 15 3 15 93. If not Q then not P. 4. theorems2Problem Set 0.1 1. 4 2(8 11) + 6 = 4 2(3) + 6 = 4 + 6 + 6 = 16 2. 3 2 4 ( 7 12 ) = 3[ 2 4(5) ] = 3[ 2 + 20] = 3(22) = 66 3.4[5(3 + 12 4) + 2(13 7)] = 4[5(5) + 2(6)] = 4[25 + 12] = 4(37) = 1484.5 [ 1(7 + 12 16) + 4] + 2 = 5 [ 1(3) + 4] + 2 = 5 ( 3 + 4 ) + 2 = 5 (1) + 2 = 5 + 2 = 75.6.7.5 1 65 7 58 = = 7 13 91 91 91 3 3 1 3 3 1 + = + 4 7 21 6 3 21 6 42 6 7 43 = + = 42 42 42 42 1 1 1 1 1 1 1 3 4 1 = + + 3 2 4 3 6 3 2 12 6 1 1 1 1 = + 3 2 12 6 1 1 4 = + 3 24 24 1 3 1 = = 3 24 24Instructors Resource Manual22 14 2 14 2 14 6 = = 9. 21 5 1 21 14 21 14 3 3 2=14 3 2 9 6 = = 21 7 3 49 492 2 35 33 5 7 = 7 7 = 7 = 33 = 11 10. 6 2 1 7 1 6 1 7 7 7 7 7 11 12 11 4 7 21 = 7 7 = 7 = 7 11. 11 + 12 11 + 4 15 15 7 21 7 7 7 1 3 7 4 6 7 5 + + 5 12. 2 4 8 = 8 8 8 = 8 = 1 3 7 4 6 7 3 3 + + 2 4 8 8 8 8 813. 1 1 1 2 3 2 1 =1 =1 = = 1 3 3 3 3 3 1+ 2 214. 2 +15.(3 5 1+ 25+ 33 3 = 2+ 2 5 7 2 2 2 6 14 6 20 = 2+ = + = 7 7 7 7= 2+)() ( 5) ( 3)5 3 =22=53= 2Section 0.11 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
2. 16.(5 3) = ( 5) 222( 5 )( 3 ) + ( 3 )227.= 5 2 15 + 3 = 8 2 1517. (3x 4)( x + 1) = 3 x 2 + 3 x 4 x 4= 3x2 x 4 18. (2 x 3)2 = (2 x 3)(2 x 3)124 2 + x + 2x x x + 2 12 4( x + 2) 2x = + + x( x + 2) x( x + 2) x( x + 2) 12 + 4 x + 8 + 2 x 6 x + 20 = = x( x + 2) x( x + 2) 2(3 x + 10) = x( x + 2) 2+= 4 x2 6 x 6 x + 9 = 4 x 2 12 x + 919.28.(3x 9)(2 x + 1) = 6 x 2 + 3 x 18 x 92 y + 2(3 y 1) (3 y + 1)(3 y 1) 2(3 y + 1) 2y = + 2(3 y + 1)(3 y 1) 2(3 y + 1)(3 y 1) =2= 6 x 15 x 920. (4 x 11)(3x 7) = 12 x 2 28 x 33 x + 77=21. (3t 2 t + 1) 2 = (3t 2 t + 1)(3t 2 t + 1) 32322= 9t 3t + 3t 3t + t t + 3t t + 16y + 2 + 2y 8y + 2 = 2(3 y + 1)(3 y 1) 2(3 y + 1)(3 y 1)== 12 x 2 61x + 7742 y + 6 y 2 9 y2 12(4 y + 1) 4y +1 = 2(3 y + 1)(3 y 1) (3 y + 1)(3 y 1)= 9t 4 6t 3 + 7t 2 2t + 100 = 0b.0 is undefined. 0c.0 =0 17d.3 is undefined. 0e.05 = 0f. 170 = 129. a. 22. (2t + 3)3 = (2t + 3)(2t + 3)(2t + 3) = (4t 2 + 12t + 9)(2t + 3) = 8t 3 + 12t 2 + 24t 2 + 36t + 18t + 27 = 8t 3 + 36t 2 + 54t + 2723.x 2 4 ( x 2)( x + 2) = = x+2, x 2 x2 x224.x 2 x 6 ( x 3)( x + 2) = = x+2, x 3 x3 ( x 3)25.t 2 4t 21 (t + 3)(t 7) = = t 7 , t 3 t +3 t +326.22x 2x 322x 2x + x=2 x(1 x) 2x( x 2 x + 1) 2 x( x 1) = x( x 1)( x 1) 2 = x 1Section 0.10 = a , then 0 = 0 a , but this is meaningless 0 because a could be any real number. No 0 single value satisfies = a . 030. If31..083 12 1.000 96 40 36 4Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 3. 32..285714 7 2.000000 14 60 56 40 35 50 49 10 7 30 28 233..142857 21 3.000000 21 90 84 60 42 180 168 120 105 150 147 334..294117... 17 5.000000... 0.2941176470588235 34 160 153 70 68 20 17 30 17 130 119 11Instructors Resource Manual35.3.6 3 11.0 9 20 18 236..846153 13 11.000000 10 4 60 52 80 78 20 13 70 65 50 39 1137. x = 0.123123123... 1000 x = 123.123123... x = 0.123123... 999 x = 123 123 41 x= = 999 333 38. x = 0.217171717 1000 x = 217.171717... 10 x = 2.171717... 990 x = 215 215 43 x= = 990 198 39. x = 2.56565656... 100 x = 256.565656... x = 2.565656... 99 x = 254 254 x= 99 40. x = 3.929292 100 x = 392.929292... x = 3.929292... 99 x = 389 389 x= 99Section 0.13 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 4. 41. x = 0.199999... 100 x = 19.99999...52.10 x = 1.99999... 90 x = 18 18 1 x= = 90 554. 55.10 x = 3.99999... 90 x = 36 36 2 x= = 90 556.43. Those rational numbers that can be expressed by a terminating decimal followed by zeros. 1 p 1 = p , so we only need to look at . If q q q q = 2n 5m , then nm1 1 1 = = (0.5)n (0.2)m . The product q 2 5 of any number of terminating decimals is also a nmterminating decimal, so (0.5) and (0.2) , and hence their product, decimal. Thus1 , is a terminating qp has a terminating decimal qexpansion. 45. Answers will vary. Possible answer: 0.000001, 1 0.0000010819... 1246. Smallest positive integer: 1; There is no smallest positive rational or irrational number. 47. Answers will vary. Possible answer: 3.14159101001... 48. There is no real number between 0.9999(repeating 9's) and 1. 0.9999 and 1 represent the same real number. 49. Irrational 50. Answers will vary. Possible answers: and , 2 and 2 51. ( 3 + 1)3 20.392304854Section 0.12 3)4 0.010205144353. 4 1.123 3 1.09 0.0002830738842. x = 0.399999 100 x = 39.99999...44.(( 3.1415 )1/ 2 0.5641979034 8.92 + 1 3 0.000691744752 4 (6 2 2) 3.66159180757. Let a and b be real numbers with a < b . Let n be a natural number that satisfies 1 / n < b a . Let S = {k : k n > b} . Since a nonempty set of integers that is bounded below contains a least element, there is a k 0 S such that k 0 / n > b but(k 0 1) / n b . Thenk0 1 k0 1 1 = >b > a n n n n k 0 1 k 0 1 Thus, a < n b . If n < b , then choose r=k 0 1 n. Otherwise, choose r =k0 2 n.1 0 but x . x xe.b.b. The statement, converse, and contrapositive are all false. 69. a.True; x + ( x ) < x + 1 + ( x ) : 0 < 12b. The statement, converse, and contrapositive are all true. 68. a.True; Let y be any positive number. Take y x = . Then 0 < x < y . 2d. True; 1/ n can be made arbitrarily close to 0.b. If a < b then a < b. If a b then a b. 67. a. x 2 .If a triangle is a right triangle, then 21 2 . Then x = 2d. True; Let x be any number. Takeb. If I take off next week, then I finished my research paper. If I do not take off next week, then I did not finish my research paper. 65. a.False; Take x =Some natural number is larger than its square. The original statement is true.Prove the contrapositive. Suppose n is even. Then there is an integer k such that n = 2k . Then n 2 = (2k )2 = 4k 2 = 2(2k 2 ) .Thus n 2 is even. Parts (a) and (b) prove that n is odd if and74.only if n 2 is odd. 75. a. b.243 = 3 3 3 3 3 124 = 4 31 = 2 2 31 or 22 31Some natural number is not rational. The original statement is true.Instructors Resource ManualSection 0.15 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6. 5100 = 2 2550 = 2 2 1275c.82. a.= 2 2 3 425 = 2 2 3 5 85 = 2 2 3 5 5 17 or 22 3 52 17c.76. For example, let A = b c 2 d 3 ; thenA2 = b 2 c 4 d 6 , so the square of the number is the product of primes which occur an even number of times. 77.p p2 ;2 = ; 2q 2 = p 2 ; Since the prime 2 q q 2 must occur an even number of factors of p p times, 2q2 would not be valid and = 2 q must be irrational. 3=p p2 ; 3= ; 3q 2 = p 2 ; Since the prime q q2factors of p 2 must occur an even number of times, 3q 2 would not be valid ande.f. 83. a.p = 3 qx = 2.4444...; 10 x = 24.4444... x = 2.4444... 9 x = 22 22 x= 92 3 n = 1: x = 0, n = 2: x = , n = 3: x = , 3 2 5 n = 4: x = 4 3 The upper bound is . 2 2Answers will vary. Possible answer: An example is S = {x : x 2 < 5, x a rational number}. Here the least upper bound is 5, which is real but irrational.must be irrational. 79. Let a, b, p, and q be natural numbers, sob. 2d. 12=78.2a bp a p aq + bp are rational. + = This q b q bq sum is the quotient of natural numbers, so it is also rational.andb. True0.2 Concepts Review 1. [1,5); (, 2] 2. b > 0; b < 0p 80. Assume a is irrational, 0 is rational, and q p r qr is = is rational. Then a = q s ps rational, which is a contradiction. a81. a. 9 = 3; rationalb.3 0.375 = ; rational 8c.(1 + 3)2 = 1 + 2 3 + 3 = 4 + 2 3; irrational4. 1 x 5Problem Set 0.2 1. a.b.(3 2)(5 2) = 15 4 = 30; rationald.3. (b) and (c)c.d.6Section 0.2Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 7. 3 < 1 6 x 4 9. 4 < 6 x 3e.2 1 1 2 > x ; , 3 2 2 3f.2. a. c.(2, 7) (, 2]b. d.[3, 4)[1, 3]10.3. x 7 < 2 x 5 2 < x;( 2, )4 < 5 3x < 7 1 < 3x < 2 1 2 2 1 > x > ; , 3 3 3 34. 3x 5 < 4 x 61 < x; (1, ) 11. x2 + 2x 12 < 0; x=5.7 x 2 9x + 3 5 2 x= 1 13(7. 4 < 3 x + 2 < 5 6 < 3 x < 3 2 < x < 1; (2, 1))() x 1 + 13 x 1 13 < 0; 5 5 x ; , 2 2 6. 5 x 3 > 6 x 4 1 > x;(,1)2 (2)2 4(1)(12) 2 52 = 2(1) 2( 1 13, 1 + 13)12. x 2 5 x 6 > 0 ( x + 1)( x 6) > 0; (, 1) (6, )13. 2x2 + 5x 3 > 0; (2x 1)(x + 3) > 0; 1 (, 3) , 2 8. 3 < 4 x 9 < 11 6 < 4 x < 20 3 3 < x < 5; ,5 2 2 14. 3 (4 x + 3)( x 2) < 0; , 2 4 15.Instructors Resource Manual4 x2 5x 6 < 0x+4 0; [4, 3) x3Section 0.27 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 8. 16.3x 2 2 0; , (1, ) x 1 3 3 >2 x+520.3 2 > 0 x+517.2 5 < 0 x 2 5x < 0; x 2 ( , 0) , 5 18.3 2( x + 5) >0 x+52 0; 5, 2 x+5 21. ( x + 2)( x 1)( x 3) > 0; (2,1) (3,8)3 1 22. (2 x + 3)(3x 1)( x 2) < 0; , , 2 2 3 3 23. (2 x  3)( x 1)2 ( x  3) 0; , [3, ) 2 24. (2 x 3)( x 1) 2 ( x 3) > 0;19.( ,1) 1, 3 ( 3, ) 21 4 3x 2 1 4 0 3x 2 1 4(3 x 2) 0 3x 2 9 12 x 2 3 0; , , 3x 2 3 4 25.x3 5 x 2 6 x < 0 x( x 2 5 x 6) < 0 x( x + 1)( x 6) < 0; (, 1) (0, 6)26. x3 x 2 x + 1 > 0 ( x 2 1)( x 1) > 0 ( x + 1)( x 1) 2 > 0; (1,1) (1, )27. a. c. 8Section 0.2False. False.b.True.Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9. 28. a.True.c.False.29. a.b.True.33. a.( x + 1)( x 2 + 2 x 7) x 2 1x3 + 3 x 2 5 x 7 x 2 1 x3 + 2 x 2 5 x 6 0 ( x + 3)( x + 1)( x 2) 0 [3, 1] [2, ) Let a < b , so ab < b 2 . Also, a 2 < ab .Thus, a 2 < ab < b 2 and a 2 < b 2 . Let a 2 < b 2 , so a b Then 0 < ( a b ) = a 2 2ab + b 2 2x4 2 x2 8b.< b 2 2ab + b 2 = 2b ( b a )x4 2 x2 8 0Since b > 0 , we can divide by 2b to get ba > 0.( x 2 4)( x 2 + 2) 0 ( x 2 + 2)( x + 2)( x 2) 0b. We can divide or multiply an inequality by any positive number. a 1 1 a < b 1 and 2x + 1 < 3 3x > 6 and 2x < 2 x > 2 and x < 1; (2, 1)( x 2 4)( x 2 1) < 0 ( x + 2)( x + 1)( x 1)( x 2) < 0 (2, 1) (1, 2)34. a.32. a.3x + 7 > 1 and 2x + 1 < 4 5 x > 2 and x < ; 21 1 , 2.01 1.99 2 x 7 > 1 or 2 x + 1 < 3b.x > 4 or x < 1 (,1) (4, )2.99 2 x 7 1 or 2 x + 1 < 3 2 x 8 or 2 x < 22.99 5.02 4.98 8 or 2 x < 2b.1.99 1 x< 1.99 2.01b. 3x + 7 > 1 and 2x + 1 > 4 3x > 6 and 2x > 5 5 x > 2 and x > ; ( 2, ) 2 c.( x 2 + 1)2 7( x 2 + 1) + 10 < 02 x 7 1 or 2 x + 1 > 33.012 x 8 or 2 x > 2x 4 or x > 1 (, )35.x 2 5; x 2 5 or x 2 5 x 3 or x 7 (, 3] [7, )Instructors Resource ManualSection 0.29 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 10. 36.x + 2 < 1; 1 < x + 2 < 143.3 < x < 1 (3, 1)37.4 x + 5 10; 10 4 x + 5 10 15 4 x 5 38.15 5 15 5 x ; , 4 4 4 42 x 1 > 2;2x 1 < 2 or 2x 1 > 2 2x < 1 or 2x > 3; 1 3 1 3 x < or x > , , , 2 2 2 2 39.40.2x 5 7 7 2x 2x 5 7 or 5 7 7 7 2x 2x 2 or 12 7 7 x 7 or x 42; (, 7] [42, ) x +1 < 1 4 x 1 < + 1 < 1 4 x 2 < < 0; 4 8 < x < 0; (8, 0)41. 5 x 6 > 1; 5 x 6 < 1 or 5 x 6 > 1 5 x < 5 or 5 x > 7 7 7 x < 1 or x > ;(,1) , 5 5 42.2 x 7 > 3;2x 7 < 3 or 2x 7 > 3 2x < 4 or 2x > 10 x < 2 or x > 5; (, 2) (5, )44.1 3 > 6; x 1 1 3 < 6 or 3 > 6 x x 1 1 + 3 < 0 or 9 > 0 x x 1 + 3x 1 9x < 0 or > 0; x x 1 1 , 0 0, 3 9 5 > 1; x 5 5 2 + < 1 or 2 + > 1 x x 5 5 3 + < 0 or 1 + > 0 x x 3x + 5 x+5 < 0 or > 0; x x 5 ( , 5) , 0 (0, ) 3 2+45. x 2 3x 4 0; x=3 (3)2 4(1)(4) 3 5 = = 1, 4 2(1) 2( x + 1)( x 4) = 0; (, 1] [4, )4 (4)2 4(1)(4) =2 2(1) ( x 2)( x 2) 0; x = 246. x 2 4 x + 4 0; x =47. 3x2 + 17x 6 > 0; x=17 (17) 2 4(3)(6) 17 19 1 = = 6, 2(3) 6 31 (3x 1)(x + 6) > 0; ( , 6) , 3 48. 14 x 2 + 11x 15 0; 11 (11) 2 4(14)(15) 11 31 = 2(14) 28 3 5 x= , 2 7 3 5 3 5 x + x 0; , 2 7 2 7x=49. x 3 < 0.5 5 x 3 < 5(0.5) 5 x 15 < 2.5 50. x + 2 < 0.3 4 x + 2 < 4(0.3) 4 x + 18 < 1.2 10Section 0.2Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 11. 51.x2 053. 3x 15 < 3( x 5) < (3x 7)( x 5) > 0; 3 x5 < x5 0, > 0. 2 x +2 x +3x + (2) 4 + 4 + 7 = 15 1 and x 2 + 1 1 so 1. 2 x +11==x +9 x +2 2 + = 2 2 2 x + 9 x + 9 x + 9 x2 + 9 1 1 Since x 2 + 9 9, 2 x +9 9 x +2 x +2 9 x2 + 9 x +2 x2 9 x2 + 9b a< b.of absolute values. c.x +9 2a b a b a b Use Property 4b.2x2a b = a + (b) a + b = a + b65. a.x269.1 3 1 2 1 1 x + x + x+ 2 4 8 16 1 3 1 2 1 1 x4 + x + x + x + 2 4 8 16 1 1 1 1 1+ + + + since x 1. 2 4 8 16 1 1 1 1 1.9375 < 2. So x 4 + x3 + x 2 + x + 2 4 8 16 x4 +Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13. x < x270. a.77.x x2 < 0 x(1 x) < 0 x < 0 or x > 11 11 R 602x 0 a 2 + b 2 > 4c 455. Label the points C, P, Q, and R as shown in the figure below. Let d = OP , h = OR , and a = PR . Triangles OPR and CQR aresimilar because each contains a right angle and they share angle QRC . For an angle of 30 ,a 1 d 3 and = h = 2a . Using a = h 2 h 256. The equations of the two circles are ( x R)2 + ( y R)2 = R 2 ( x r )2 + ( y r )2 = r 2Let ( a, a ) denote the point where the two circles touch. This point must satisfy (a R)2 + (a R)2 = R 2 R2 2 2 a = 1 R 2 (a R)2 = 2 Since a < R , a = 1 R. 2 At the same time, the point where the two circles touch must satisfy (a r )2 + (a r )2 = r 2 2 a = 1 r 2 2 Since a > r , a = 1 + r. 2 Equating the two expressions for a yields 2 2 1 R = 1 + r 2 2 22 1 2 R= r= 2 1+ 2 r=1 2 + 2 1 2 R 2 2 1 + 1 2 2 1 2R1 2 r = (3 2 2) R 0.1716 R 1property of similar triangles, QC / RC = 3 / 2 , 2 3 4 = a = 2+ a2 2 3 By the Pythagorean Theorem, we have d = h 2 a 2 = 3a = 2 3 + 4 7.464Instructors Resource ManualSection 0.319 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 20. 57. Refer to figure 15 in the text. Given ine l1 with slope m, draw ABC with vertical and horizontal sides m, 1. Line l2 is obtained from l1 by rotating it around the point A by 90 counterclockwise. Triangle ABC is rotated into triangle AED . We read off 1 1 slope of l2 = = . m m60. See the figure below. The angle at T is a right angle, so the Pythagorean Theorem gives ( PM + r )2 = ( PT )2 + r 2 ( PM )2 + 2rPM + r 2 = ( PT )2 + r 2 PM ( PM + 2r ) = ( PT )2 PM + 2r = PN so this gives ( PM )( PN ) = ( PT ) 258. 2 ( x 1)2 + ( y 1)2 = ( x 3) 2 + ( y 4)2 4( x 2 2 x + 1 + y 2 2 y + 1) = x 2 6 x + 9 + y 2 8 y + 163x 2 2 x + 3 y 2 = 9 + 16 4 4; 2 17 x + y2 = ; 3 3 1 17 1 2 2 2 x x+ + y = + 3 9 3 9 3x 2 2 x + 3 y 2 = 17; x 2 21 52 2 x + y = 3 9 B = (6)2 + (8)2 = 100 = 10 52 1 center: , 0 ; radius: 3 3 59. Let a, b, and c be the lengths of the sides of the right triangle, with c the length of the hypotenuse. Then the Pythagorean Theoremsays that a 2 + b 2 = c 2 Thus,a 2 b 2 c 2 + = or 8 8 8 261. The lengths A, B, and C are the same as the corresponding distances between the centers of the circles: A = (2)2 + (8)2 = 68 8.221 a 1 b 1 c + = 2 2 2 2 2 2C = (8)2 + (0)2 = 64 = 8 Each circle has radius 2, so the part of the belt around the wheels is 2(2 a ) + 2(2 b ) + 2(2 c ) = 2[3  (a + b + c)] = 2(2 ) = 4 Since a + b + c = , the sum of the angles of a triangle. The length of the belt is 8.2 + 10 + 8 + 4 38.8 units.221 x is the area of a semicircle with 2 2 diameter x, so the circles on the legs of the triangle have total area equal to the area of the semicircle on the hypotenuse.From a 2 + b 2 = c 2 , 3 2 3 2 3 2 a + b = c 4 4 4 3 2 x is the area of an equilateral triangle 4 with sides of length x, so the equilateral triangles on the legs of the right triangle have total area equal to the area of the equilateral triangle on the hypotenuse of the right triangle.20Section 0.362 As in Problems 50 and 61, the curved portions of the belt have total length 2 r. The lengths of the straight portions will be the same as the lengths of the sides. The belt will have length 2 r + d1 + d 2 + + d n .Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 21. 63. A = 3, B = 4, C = 6 3(3) + 4(2) + (6) 7 d= = 5 (3) 2 + (4)2 64. A = 2, B = 2, C = 4 d=2(4) 2(1) + 4) 2(2) + (2)2=14 8=7 2 265. A = 12, B = 5, C = 1 12(2) 5(1) + 1 18 d= = 13 (12) 2 + (5) 2 66. A = 2, B = 1, C = 5 d=2(3) 1(1) 5 2(2) + (1)2=2 5=2 5 567. 2 x + 4(0) = 5 5 x= 2 d=2( 5 ) + 4(0) 7 = 2 (2)2 + (4) 22 20=5 568. 7(0) 5 y = 1 1 y= 5 1 7(0) 5 6 7 7 74 5 d= = = 2 2 74 74 (7) + (5) 2 3 5 3 = ; m = ; passes through 1+ 2 3 5 2 + 1 3 2 1 1 , = , 2 2 2 2 1 3 1 y = x+ 2 5 2 3 4 y = x+ 5 569. m =Instructors Resource Manual04 1 = 2; m = ; passes through 20 2 0+2 4+0 , = (1, 2) 2 2 1 y 2 = ( x 1) 2 1 3 y = x+ 2 2 60 1 m= = 3; m = ; passes through 42 3 2+4 0+6 , = (3, 3) 2 2 1 y 3 = ( x 3) 3 1 y = x+4 3 1 3 1 x+ = x+4 2 2 3 5 5 x= 6 2 x=3 1 3 y = (3) + = 3 2 2 center = (3, 3)70. m =71. Let the origin be at the vertex as shown in the figure below. The center of the circle is then ( 4 r , r ) , so it has equation ( x (4 r ))2 + ( y r )2 = r 2 . Along the side oflength 5, the ycoordinate is always3 4timesthe xcoordinate. Thus, we need to find the value of r for which there is exactly one x23 solution to ( x 4 + r ) 2 + x r = r 2 . 4 Solving for x in this equation gives 16 x = 16 r 24 r 2 + 7r 6 . There is 25 ()exactly one solution when r 2 + 7 r 6 = 0, that is, when r = 1 or r = 6 . The root r = 6 is extraneous. Thus, the largest circle that can be inscribed in this triangle has radius r = 1.Section 0.321 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 22. 72. The line tangent to the circle at ( a, b ) will beThe slope of PS is 1 [ y1 + y4 ( y1 + y2 )] y y 2 2 = 4 . The slope of 1 x4 x2 x1 + x4 ( x1 + x2 ) ] [ 2 1 [ y3 + y4 ( y2 + y3 )] y y 2 . Thus QR is 2 = 4 1 x4 x2 [ x3 + x4 ( x2 + x3 )] 2 PS and QR are parallel. The slopes of SR and y y PQ are both 3 1 , so PQRS is a x3 x1 parallelogram.perpendicular to the line through ( a, b ) and the center of the circle, which is ( 0, 0 ) . The line through ( a, b ) and ( 0, 0 ) has slope 0b b a r2 = ; ax + by = r 2 y = x + 0a a b b a so ax + by = r 2 has slope and is b perpendicular to the line through ( a, b ) and m=( 0, 0 ) ,so it is tangent to the circle at ( a, b ) .73. 12a + 0b = 36 a=3 32 + b 2 = 36 b = 3 3 3x 3 3 y = 36 x 3 y = 12 3x + 3 3 y = 36 x + 3 y = 1274. Use the formula given for problems 6366, for ( x, y ) = ( 0, 0 ) .77. x 2 + ( y 6) 2 = 25; passes through (3, 2) tangent line: 3x 4y = 1 The dirt hits the wall at y = 8.A = m, B = 1, C = B b;(0, 0) d=m(0) 1(0) + B b m2 + (1) 2=Bb m2 + 175. The midpoint of the side from (0, 0) to (a, 0) is 0+a 0+0 a , = , 0 2 2 2 The midpoint of the side from (0, 0) to (b, c) is 0+b 0+c b c , = , 2 2 2 2 c0 c m1 = = ba ba c 0 c m2 = 2 = ; m1 = m2 ba ba 20.4 Concepts Review 1. yaxis 2.( 4, 2 )3. 8; 2, 1, 4 4. line; parabolaProblem Set 0.4 1. y = x2 + 1; yintercept = 1; y = (1 + x)(1 x); xintercepts = 1, 1 Symmetric with respect to the yaxis276. See the figure below. The midpoints of the sides are x2 + x3 y2 + y3 x + x y + y2 P 1 2 , 1 , , , Q 2 2 2 2 x + x y + y4 R 3 4 , 3 , and 2 2 x + x y + y4 S 1 4 , 1 . 2 2 22Section 0.4Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 23. 2. x = y 2 + 1; y intercepts = 1,1; xintercept = 1 . Symmetric with respect to the xaxis.3. x = 4y2 1; xintercept = 1 Symmetric with respect to the xaxis4.y = 4 x 2 1; y intercept = 1 1 1 y = (2 x + 1)(2 x 1); xintercepts = , 2 2 Symmetric with respect to the yaxis.5. x2 + y = 0; y = x2 xintercept = 0, yintercept = 0 Symmetric with respect to the yaxis6. y = x 2 2 x; y intercept = 0 y = x(2 x); xintercepts = 0, 27 7. 7x2 + 3y = 0; 3y = 7x2; y = x 2 3 xintercept = 0, yintercept = 0 Symmetric with respect to the yaxis8. y = 3x 2 2 x + 2; y intercept = 2Instructors Resource ManualSection 0.423 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 24. 9. x2 + y2 = 4 xintercepts = 2, 2; yintercepts = 2, 2 Symmetric with respect to the xaxis, yaxis, and origin10. 3x 2 + 4 y 2 = 12; yintercepts = 3, 3 xintercepts = 2, 2 Symmetric with respect to the xaxis, yaxis, and origin11. y = x2 2x + 2: yintercept = 2 2 4+8 22 3 xintercepts = = = 1 3 2 224Section 0.412. 4 x 2 + 3 y 2 = 12; y intercepts = 2, 2xintercepts = 3, 3 Symmetric with respect to the xaxis, yaxis, and origin13. x2 y2 = 4 xintercept = 2, 2 Symmetric with respect to the xaxis, yaxis, and origin14. x 2 + ( y 1)2 = 9; y intercepts = 2, 4xintercepts = 2 2, 2 2 Symmetric with respect to the yaxisInstructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 25. 15. 4(x 1)2 + y2 = 36; yintercepts = 32 = 4 2 xintercepts = 2, 4 Symmetric with respect to the xaxis18. x 4 + y 4 = 1; y intercepts = 1,1 xintercepts = 1,1 Symmetric with respect to the xaxis, yaxis, and origin16. x 2 4 x + 3 y 2 = 219. x4 + y4 = 16; yintercepts = 2, 2 xintercepts = 2, 2 Symmetric with respect to the yaxis, xaxis and originxintercepts = 2 2 Symmetric with respect to the xaxis17. x2 + 9(y + 2)2 = 36; yintercepts = 4, 0 xintercept = 0 Symmetric with respect to the yaxisInstructors Resource Manual20. y = x3 x; yintercepts = 0; y = x(x2 1) = x(x + 1)(x 1); xintercepts = 1, 0, 1 Symmetric with respect to the originSection 0.425 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 26. 21. y =1 2; yintercept = 1x +1 Symmetric with respect to the yaxis224. 4 ( x 5 ) + 9( y + 2) 2 = 36; xintercept = 525. y = (x 1)(x 2)(x 3); yintercept = 6 xintercepts = 1, 2, 3 22. y =x; y intercept = 0 x +1 xintercept = 0 Symmetric with respect to the origin 226. y = x2(x 1)(x 2); yintercept = 0 xintercepts = 0, 1, 223. 2 x 2 4 x + 3 y 2 + 12 y = 2 2( x 2 2 x + 1) + 3( y 2 + 4 y + 4) = 2 + 2 + 12 2( x 1)2 + 3( y + 2)2 = 12yintercepts = 2 30 3xintercept = 1 27. y = x 2 ( x 1)2 ; yintercept = 0 xintercepts = 0, 126Section 0.4Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 27. 28. y = x 4 ( x 1)4 ( x + 1)4 ; y intercept = 0 xintercepts = 1, 0,1 Symmetric with respect to the yaxisIntersection points: (0, 1) and (3, 4)32. 2 x + 3 = ( x 1) 2 29.x + y = 1; yintercepts = 1, 1;xintercepts = 1, 1 Symmetric with respect to the xaxis, yaxis and origin2 x + 3 = x2 + 2 x 1 x2 + 4 = 0 No points of intersection33. 2 x + 3 = 2( x 4)2 30.x + y = 4; yintercepts = 4, 4;xintercepts = 4, 4 Symmetric with respect to the xaxis, yaxis and origin2 x + 3 = 2 x 2 + 16 x 32 2 x 2 18 x + 35 = 0 x=18 324 280 18 2 11 9 11 = = ; 4 4 2 9 11 Intersection points: 2 , 6 + 11 , 9 + 11 2 , 6 11 31. x + 1 = ( x + 1)2 x + 1 = x2 + 2 x + 1 x 2 + 3x = 0 x( x + 3) = 0 x = 0, 3Instructors Resource ManualSection 0.427 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 28. 37. y = 3x + 134. 2 x + 3 = 3x 2 3 x + 12x 2 + 2 x + (3 x + 1) 2 = 153x 2 x + 9 = 0 No points of intersectionx 2 + 2 x + 9 x 2 + 6 x + 1 = 15 10 x 2 + 8 x 14 = 0 2(5 x 2 + 4 x 7) = 0 2 39 1.65, 0.85 5 Intersection points: 2 39 1 3 39 , and 5 5 2 + 39 1 + 3 39 , 5 5 [ or roughly (1.65, 3.95) and (0.85, 3.55) ] x=35. x 2 + x 2 = 4 x2 = 2 x= 2()(Intersection points: 2, 2 ,2, 2)38. x 2 + (4 x + 3) 2 = 81 x 2 + 16 x 2 + 24 x + 9 = 81 17 x 2 + 24 x 72 = 0 12 38 2.88, 1.47 17 Intersection points: 12 38 3 24 38 , and 17 17 12 + 38 3 + 24 38 , 17 17 [ or roughly ( 2.88, 8.52 ) , (1.47,8.88 ) ] x=36. 2 x 2 + 3( x 1)2 = 12 2 x 2 + 3 x 2 6 x + 3 = 12 5x2 6 x 9 = 0 6 36 + 180 6 6 6 3 3 6 = = 10 10 5 Intersection points: 3 3 6 2 3 6 3 + 3 6 2 + 3 6 , 5 , 5 , 5 5 x=28Section 0.4Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 29. 39. a.y = x 2 ; (2)(b.ax3 + bx 2 + cx + d , with a > 0 : (1)c.ax3 + bx 2 + cx + d , with a < 0 : (3)d.y = ax3 , with a > 0 : (4)40. x 2 + y 2 = 13;(2, 3), (2,3), (2, 3), (2,3) 22222d3 = (2 2) + (3 + 3) = 6 Three such distances.())()()()d1 = (2 2) 2 + 1 + 21 1 + 13 21 13)21 + 13)f (2u ) = 3(2u ) 2 = 12u 2 ; f ( x + h) = 3( x + h)2)2=f (0) = 1 02 = 1 f (k ) = 1 k 2 f (5) = 1 (5) 2 = 24f.1 15 1 1 f =1 =1 = 16 16 4 4g.f (1 + h ) = 1 (1 + h ) = 2h h 2h.f (1 + h ) f (1) = 2h h 2 0 = 2h h 2i.( 2 21)2c.e.)f (2) = 1 (2)2 = 3f ( 2 + h ) f ( 2) = 1 ( 2 + h ) + 322= 2 21 9.17 d 4 = (2 2)2 + 1 21 (1 + 13) (f (1) = 1 12 = 0d.)22(= 16 + 21 13)2222= 4h h 2= 50 + 2 273 9.11()d5 = (2 2)2 + 1 21 1 13 (2b.d3 = (2 + 2)2 + 1 + 21 1 21 21 + 212.1. a.= 50 + 2 273 9.11= 16 +( 2 13 )0.5 Concepts Review2((=Problem Set 0.5d 2 = (2 2)2 + 1 + 21 1 13 = 0+2= 2 13 7.21 Four such distances ( d 2 = d 4 and d1 = d5 ).2= 50 2 273 4.12()4. even; odd; yaxis; origin21 , 2, 1 + 13 , 2, 1 13= 16 +13 + 133. asymptote41. x2 + 2x + y2 2y = 20; 2, 1 + 21 ,((1. domain; ranged 2 = (2 + 2) + (3 3) = 52 = 2 13= 16 += 0+22d1 = (2 + 2) + (3 + 3) = 4( 2, 1 )d6 = (2 2)2 + 1 + 13 1 13 13 21)222. a. b.F ( 2) = ( 2)3 + 3( 2) = 2 2 + 3 2 =5 2= 50 2 273 4.123c.Instructors Resource ManualF (1) = 13 + 3 1 = 41 1 1 1 3 49 F = + 3 = + = 4 4 4 64 4 64Section 0.529 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 30. d.F (1 + h ) = (1 + h ) + 3 (1 + h ) 3f. ( x 2 + x) == 1 + 3h + 3h 2 + h3 + 3 + 3h = 4 + 6h + 3h 2 + h3e.F (1 + h ) 1 = 3 + 6h + 3h + hf.=F ( 2 + h) F ( 2)232= 15h + 6h + h3. a.b.G (0) =d.e.f.4. a.b.0.25 3 1f ( x) =c.f (3 + 2) =31 = 1 0 1 3G( y ) =G ( x) =c.1 1 = x 1 x +17. a.21 x = 1 1 x21 2 0.841 3.293(12.26) 2 + 9 12.26 3 1.1991t 1 21 = 2d. (u + 1) =e.( x2 ) =+(1) 2=21 2=x2=x2 + y2 = 1c.x = 2 y +1x2 = 2 y + 1 1.06(u + 1) + (u + 1) 2( x2 ) + ( x2 )2Section 0.5tu +1; undefinedb. xy + y + x = 1 y(x + 1) = 1 x 1 x 1 x y= ; f ( x) = x +1 x +1t2 t3 4 1 23 3y = 1 x 2 ; not a function=2t + ( t ) 2( 3)2 + 9f ( 3) =y 2 = 1 x 21 x21 + 12 (t ) =is noty 2 1 1 G = x2 (1) ==3+ 2 3 0.250.79 3b. f(12.26) =1 2.75 2.658(0.79) 2 + 9f(0.79) =1 G (1.01) = = 100 1.01 1 21=1=21 G (0.999) = = 1000 0.999 1c.301f (0.25) =b.6. a. c.x2 + xdefined= ( 2 + h ) + 3 ( 2 + h ) 23 3 ( 2 ) = 8 + 12h + 6h 2 + h3 + 6 + 3h 14x2 + xx 4 + 2 x3 + 2 x 2 + x35. a.( x 2 + x) + ( x 2 + x) 2=y=u 2 + 3u + 2x2 + x4 xu +1d.x2 1 x2 1 ; f ( x) = 2 2y y+1 xy + x = y x = y xy x = y(1 x) x x ; f ( x) = y= 1 x 1 x x=Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 31. 8. The graphs on the left are not graphs of functions, the graphs on the right are graphs of functions. 9.f (a + h) f (a) [2(a + h) 2 1] (2a 2 1) = h h 4ah + 2h 2 = = 4a + 2h hx 2 9 0; x 2 9; x 3Domain: {x d.F (a + h) F (a ) 4(a + h) 4a = h h4a3 + 12a 2 h + 12ah 2 + 4h3 4a3 h 12a 2 h + 12ah 2 + 4h3 h14. a.b.= 12a 2 + 12ah + 4h 2f ( x) =4 x2=4 x2 ( x 3)( x + 2)x2 x 6 Domain: {x : x 2, 3} G ( y ) = ( y + 1) 1Domain: { y = x 4 x + hx 2h + 4 h 3h = 2 h( x 4 x + hx 2h + 4) 3 = 2 x 4 x + hx 2h + 4: y > 1}c. (u ) = 2u + 3 (all real numbers) Domain:d.F (t ) = t 2 / 3 4 (all real numbers) Domain:2a+h a + h+ 4: y 5}1 0; y > 1 y +13 3 g ( x + h) g ( x) x + h 2 x 2 = h h 3x 6 3x 3h + 6G ( a + h) G ( a ) = hH ( y ) = 625 y 4Domain: { y =12.: x 3}3=11. ( x) = x 2 9625 y 4 0; 625 y 4 ; y 5 310.c.15. f(x) = 4; f(x) = 4; even functiona a+4h2a + 4a + ah + 4h a 2 ah 4a a 2 + 8a + ah + 4h + 16 h 4h= = =13. a.h(a 2 + 8a + ah + 4h + 16) 4 a 2 + 8a + ah + 4h + 16F ( z) = 2 z + 32z + 3 0; z Domain: z b.16. f(x) = 3x; f(x) = 3x; odd functiong (v ) =3 23 :z 21 4v 14v 1 = 0; v = Domain: v 1 4 1 :v 4Instructors Resource ManualSection 0.531 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 32. 17. F(x) = 2x + 1; F(x) = 2x + 1; neither20. g (u ) =18. F ( x) = 3x 2; F ( x) = 3x 2; neither 21. g ( x) =19. g ( x) = 3x 2 + 2 x 1; g ( x) = 3 x 2 2 x 1 ; neither32Section 0.522. ( z ) =u3 u3 ; g ( u ) = ; odd function 8 8x 2x 1; g ( x) =x 2x 1; odd2z +1 2 z + 1 ; ( z ) = ; neither z 1 z 1Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 33. 23.f ( w) = w 1; f ( w) = w 1; neither2224. h( x ) = x + 4; h( x) = x + 4; even function26. F (t ) = t + 3 ; F ( t ) = t + 3 ; neither27. g ( x) =x x ; g ( x ) = ; neither 2 228. G ( x) = 2 x 1 ; G ( x) = 2 x + 1 ; neither25.f ( x) = 2 x ; f ( x) = 2 x = 2 x ; evenfunction1 if t 0 29. g (t ) = t + 1 if 0 < t < 2 2 t 1 if t 2Instructors Resource ManualneitherSection 0.533 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 34. x 2 + 4 if x 1 30. h( x ) = if x > 1 3x neither35. Let y denote the length of the other leg. Then x2 + y 2 = h2 y 2 = h2 x 2 y = h2 x 2 L ( x ) = h2 x 236. The area is 1 1 A ( x ) = base height = x h 2 x 2 2 2 37. a.31. T(x) = 5000 + 805x Domain: {x integers: 0 x 100}T ( x) 5000 u ( x) = = + 805 x x Domain: {x integers: 0 < x 100} P ( x) = 6 x (400 + 5 x( x 4))32. a.= 6 x 400 5 x( x 4)E(x) = 24 + 0.40xb. 120 = 24 + 0.40x 0.40x = 96; x = 240 mi 38. The volume of the cylinder is r 2 h, where h is the height of the cylinder. From the figure, 2 2 2 h 2 h 2 = 3r ; r + = (2r ) ; 2 4 h = 12r 2 = 2r 3. V (r ) = r 2 (2r 3) = 2r 3 3P(200) 190 ; P (1000 ) 610b.c. ABC breaks even when P(x) = 0; 6 x 400 5 x( x 4) = 0; x 390 33. E ( x) = x x 2 y 0.539. The area of the two semicircular ends is 0.51x0.51 exceeds its square by the maximum amount. 234. Each side has lengthp . The height of the 3d 2 . 41 d . 2 2 d 2 d d 2 1 d d A(d ) = +d + = 4 4 2 2 The length of each parallel side is2d d 2 4 Since the track is one mile long, d < 1, so 1 1 d < . Domain: d : 0 < d < =3p . 6 1 p 3p 3 p2 A( p ) = = 2 3 6 36 triangle is34Section 0.5Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 35. 40. a.1 3 A(1) = 1(1) + (1)(2 1) = 2 242. a.f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y)b.f ( x + y ) = ( x + y )2 = x 2 + 2 xy + y 2 f ( x) + f ( y )c.f(x + y) = 2(x + y) + 1 = 2x + 2y + 1 f(x) + f(y)d. f(x + y) = 3(x + y) = 3x 3y = f(x) + f(y)b.1 A(2) = 2(1) + (2)(3 1) = 4 2c.A(0) = 0d.1 1 A(c) = c(1) + (c)(c + 1 1) = c 2 + c 2 2e.43. For any x, x + 0 = x, so f(x) = f(x + 0) = f(x) + f(0), hence f(0) = 0. Let m be the value of f(1). For p in N, p = p 1 = 1 + 1 + ... + 1, so f(p) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1) = pf(1) = pm. 1 1 1 1 1 = p = + + ... + , so p p p p 1 1 1 m = f (1) = f + + ... + p p p 1 1 1 1 = f + f + ... + f = pf , p p p p 1 m hence f = . Any rational number can p pbe written asf.Domain: {c Range: { y 41. a. b.: c 0} : y 0}B (0) = 0 1 1 1 1 1 B = B (1) = = 2 6 12 2 2p with p, q in N. q1 1 1 p 1 = p = + + ... + , q q q q q p 1 1 1 so f = f + + ... + q q q q 1 1 1 = f + f + ... + f q q q 1 m p = pf = p = m q q qc.Instructors Resource ManualSection 0.535 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 36. 44. The player has run 10t feet after t seconds. He reaches first base when t = 9, second base when t = 18, third base when t = 27, and home plate when t = 36. The player is 10t 90 feet from first base when 9 t 18, hence46. a.xf(x)46.190230.4118213.765119.95790017.3369217.7388if 18 < t 2730.4521if 27 < t 3644.4378b.902 + (10t 90)2 feet from home plate. The player is 10t 180 feet from second base when 18 t 27, thus he is 90 (10t 180) = 270 10t feet from third baseand 902 + (270 10t ) 2 feet from home plate. The player is 10t 270 feet from third base when 27 t 36, thus he is 90 (10t 270) = 360 10t feet from home plate.a.b.45. a.f(1.38) 76.8204 f(4.12) 6.750810t 902 + (10t 90)2 s= 902 + (270 10t ) 2 360 10t if 0 t 9180 180 10t s = 902 + (10t 90) 2 2 2 90 + (270 10t ) if 0 t 9if 9 < t 18or 27 < t 3647.if 9 < t 18 if 18 < t 27f(1.38) 0.2994 f(4.12) 3.6852 xf(x)44.0533.1538a.22.375b. f(x) = 0 when x 1.1, 1.7, 4.3 f(x) 0 on [1.1, 1.7] [4.3, 5]11.801.2510.221.12532.38464b.3.5548.a. 36Section 0.5Range: {y R: 22 y 13}f(x) = g(x) at x 0.6, 3.0, 4.6 Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 37. b. f(x) g(x) on [0.6, 3.0] [4.6, 5] c.f ( x) g ( x) = x3 5 x 2 + x + 8 2 x 2 + 8 x + 1= x3 7 x 2 + 9 x + 9b. On 6, 3) , g increases from 13 g ( 6 ) = 4.3333 to . On ( 2, 6 , g 3 26 2.8889 . On decreased from to 9( 3, 2 ) the maximum occurs aroundLargest value f (2) g (2) = 45x = 0.1451 with value 0.6748 . Thus, the range is ( , 0.6748 2.8889, ) . 49. c.x 2 + x 6 = 0; (x + 3)(x 2) = 0 Vertical asymptotes at x = 3, x = 2d. Horizontal asymptote at y = 30.6 Concepts Review 1. ( x 2 + 1)3 a.xintercept: 3x 4 = 0; x =4 33 0 42 = yintercept: 2 0 +06 32. f(g(x)) 3. 2; left 4. a quotient of two polynomial functionsb. c.x 2 + x 6 = 0; (x + 3)(x 2) = 0 Vertical asymptotes at x = 3, x = 2Problem Set 0.6 1. a.( f + g )(2) = (2 + 3) + 22 = 9d. Horizontal asymptote at y = 0 b.( f g )(1) = f (12 ) = 1 + 3 = 4e.( g f )(1) = g (1 + 3) = 4 2 = 16f.( g f )(8) = g (8 + 3) = (5) 2 = 252. a.xintercepts: 3x 2 4 = 0; x = ( g f )(3) =d.a.( f g )(0) = (0 + 3)(02 ) = 0c.50.4 2 3 = 3 332 9 3 = = 3+3 6 2( f g )(2) = (22 + 2) 12 + 12b.( f g )(1) =c.2 yintercept: 31 2 1 g 2 (3) = = = 3 + 3 3 9 2 1+ 3 2Instructors Resource Manual=2 2 28 =6 = 2+3 5 52 4=42Section 0.637 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 38. 2d.e.f.3. a.(f 2 1 1 3 g )(1) = f = + = 1+ 3 2 2 45.= x2 + 2 x 32 2 = 2+3 5 2 ( g g )(3) = g = 3+32 2 3 = 10 = 1 +3 3 5 36. g 3 (x) = (x 2 +1) 3 = (x 4 + 2x 2 + 1)(x 2 + 1) 61 tb.c.( )(r ) = (r + 1) =d. 3 ( z ) = ( z 3 + 1) 3e.( )(5t) = [(5t) 3 +1] 13b.2= g[(x 2 + 1) 2 + 1] = g( x 4 + 2x 2 + 2) = ( x 4 + 2x 2 + 2) 2 + 1 = x 8 + 4x 6 + 8x 4 + 8x 2 + 57. g(3.141) 1.188 8. g(2.03) 0.000205r 3 +11/ 39. g 2 ( ) g ( ) 4.7891 5t1/ 32 = (11 7 ) 11 7 10. [ g 3 () g ()]1/ 3 = [(6 11)3 (6 11)]1/ 3 7.8071 5t1 (( ) )(t ) = ( ) t 3 1 1 1 = + 1 1 = 3 + 1 t t t t11. a. b. 12. a.4. a.4= x + 3x + 3x + 1 ( g g g )( x) = ( g g )( x 2 + 1)3f. f ) ( x) = g x 2 4 = 1 + x 2 4 = 1 + x2 41 1 1 ( )(r ) = = + 1 = 3 + 1 r r r = 125t3 + 1 2(g( g f )(1) = g (12 + 1) =( + )(t ) = t 3 + 1 +g ) ( x) = f ( 1 + x ) = 1 + x 4(fg ( x) = x , f ( x) = x + 7g (x) = x15 , f (x) = x 2 + x 2f ( x) =x2 x2 1 x Domain: ( , 1] [1, )3, g ( x) = x 2 + x + 1( f g )( x) =b.4 2 f 4 ( x) + g 4 ( x) = x 2 1 + x 16 = ( x 2 1)2 + x4 Domain: ( , 0 ) (0, ) 2c.2 2 g )( x) = f = 1 = x x Domain: [2, 0) (0, 2]d.( g f )( x) = g x 2 1 = (f4f ( x) =13. p = f1 , g (x) = x 3 + 3 x x g h if f(x) =1/ x , g ( x) = x ,h( x ) = x 2 + 1 p= fg h if f ( x) = 1/ x , g(x) = x + 1,h( x) = x 24 1 x214. p = f g h l if f ( x) = 1/ x , g ( x) = x , 2 h(x) = x + 1, l( x) = x2 2x 1 Domain: ( , 1) (1, )38Section 0.6Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 39. 15. Translate the graph of g ( x) = x to the right 2 units and down 3 units.17. Translate the graph of y = x 2 to the right 2 units and down 4 units.18. Translate the graph of y = x 3 to the left 1 unit and down 3 units.16. Translate the graph of h( x) = x to the left 3units and down 4 units. 19. ( f + g )( x) =x3 + x 220. ( f + g )( x) = x + xInstructors Resource ManualSection 0.639 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 40. 21. F (t ) =t t24. a.tF(x) F(x) is odd because F(x) F(x) = [F(x) F(x)]b. F(x) + F(x) is even because F(x) + F((x)) = F(x) + F(x) = F(x) + F(x) c.22. G (t ) = t tF ( x ) F ( x) F ( x ) + F ( x) is odd and is 2 2 even. F ( x ) F ( x) F ( x) + F ( x) 2 F ( x) + = = F ( x) 2 2 225. Not every polynomial of even degree is an even function. For example f ( x) = x 2 + x is neither even nor odd. Not every polynomial of odd degree is an odd function. For example g ( x) = x 3 + x 2 is neither even nor odd. 26. a.Neitherb. c.b. Odd; (f + g)(x) = f(x) + g(x) = f(x) g(x) = (f + g)(x) if f and g are both odd functions. c.Even; ( f g )( x) = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = ( f g )( x) if f and g are both even functions.d. Even; ( f g )( x) = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = ( f g )( x) if f and g are both odd functions. e.40PFe.Even; (f + g)(x) = f(x) + g(x) = f(x) + g(x) = (f + g)(x) if f and g are both even functions.RFd.23. a.PFRFf.Neither27. a.P = 29 3(2 + t ) + (2 + t )2 = t + t + 27b. When t = 15, P = 15 + 15 + 27 6.773 28. R(t) = (120 + 2t + 3t2 )(6000 + 700t ) = 2100 t3 + 19, 400t 2 + 96, 000t + 720, 000400t 29. D(t ) = 2 2 (400t ) + [300(t 1)] if 0 < t < 1 if t 1if 0 < t < 1 400t D(t ) = 2 250, 000t 180, 000t + 90, 000 if t 1 30. D(2.5) 1097 miOdd; ( f g )( x) = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = [ f ( x)][ g ( x)] = ( f g )( x) if f is an even function and g is an odd function.Section 0.6Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 41. 31.(ax+ab ) + b cx (ax+ab ) a cx36. ax + b a f ( f ( x)) = f = cx a c =a 2 x + ab + bcx ab=x(a 2 + bc)=x acx + bc acx + a 2 a 2 + bc 2 If a + bc = 0 , f(f(x)) is undefined, while if x = a , f(x) is undefined. c x 3 3 x 3 x +1 ( f ( f ( x))) = f f 32. f = f x 3 x +1 +1 x +1 x 3 3x 3 2 x 6 x 3 = f = f = f x 3 + x +1 2x 2 x 1 x3 3 x 3 3x + 3 4 x 1 = xx 3 = = =x x 3+ x 1 4 +1 x 1If x = 1, f(x) is undefined, while if x = 1, f(f(x)) is undefined. 33. a.b.1 f = x34. a. b.1 x1=1 1 x1 f1 ( f 2 ( x)) = ; x f1 ( f3 ( x)) = 1 x; 1 ; 1 x x 1 f1 ( f5 ( x)) = ; x x ; f1 ( f6 ( x)) = x 1f1 ( f 4 ( x)) =f 2 ( f1 ( x)) = f 2 ( f 2 ( x)) =x x 1 x 1 x 1= x;1 ; 1 x 1 f 2 ( f 4 ( x)) = = 1 x;f 2 ( f 6 ( x)) =x =x x x +1=1 x x 1x ; x 1=1 x 1 xx 1 ; xf3 ( f1 ( x)) = 1 x; 1 x 1 f f ( x) = f x = =1x 1/ x 1 / x 1x 1 x x 1 1 x=x 1 x 1 x1 x 1 ; = x x f3 ( f3 ( x)) = 1 (1 x) = x;f3 ( f 2 ( x)) = 1 1 x = ; 1 x x 1 x 1 1 = ; f3 ( f5 ( x)) = 1 x x x 1 f3 ( f 6 ( x)) = 1 = ; x 1 1 x f3 ( f 4 ( x)) = 1 1=xxf ( f ( x)) = f ( x /( x 1)) ==1 xf 2 ( f3 ( x)) =f 2 ( f 5 ( x )) =f (1 / x) =1 ; x 11 1 x x f ( f ( x)) = f = x 1=c.1 xf1 ( f1 ( x)) = x;x /( x 1)x x( x 1) + 1 x35. ( f1 ( f 2 f3 ))( x) = f1 (( f 2 f3 )( x)) = f1 ( f 2 ( f3 ( x))) (( f1 f 2 ) f3 )( x) = ( f1 f 2 )( f3 ( x)) = f1 ( f 2 ( f3 ( x))) = ( f1 ( f 2 f3 ))( x)x 1 x 11 ; 1 x 1 x ; f 4 ( f 2 ( x)) = = 1 1 x x 1 f 4 ( f1 ( x)) =f 4 ( f3 ( x)) = f 4 ( f 4 ( x)) = f 4 ( f5 ( x)) =f 4 ( f 6 ( x)) =Instructors Resource Manual1 1 = ; 1 (1 x) x 1 1 1 1 x1 1x 1 x=1 x x 1 = ; x 1 x 1=x = x; x ( x 1)1 x 1 = = 1 x; x 1 x 1 x 1 x Section 0.641 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 42. a.x 1 f5 ( f1 ( x)) = ; x 1 1 f5 ( f 2 ( x)) = x = 1 x;b.=f 5 ( f 5 ( x)) =x 1 1 x x 1 xf 5 ( f 6 ( x)) =x 1 x 1 x x 1=1 (1 x) = x; 1f3f4f5f6f3 )f4 )f5 )f6 )= ( f 4 f 4 ) ( f5 f 6 ) = f5 f 2 = f3 c.x ( x 1) 1 = ; x xIf Ff 6 = f 1 , then F = f 6 .d. If G f 3 G = f5. e.1 ; 1 xf 6 = f 1 , then G f 4 = f 1 soIf f 2 f 5 H = f 5 , then f 6 H = f 5 so H = f3.37.f 6 ( f3 ( x)) =x 1 1 x ; = 1 x 1 xf 6 ( f 4 ( x)) =1 1 x 1 1 1 x=1 1 = ; 1 (1 x) xf 6 ( f 5 ( x)) =x 1 x x 1 1 x=x 1 = 1 x; x 1 xf 6 ( f 6 ( x )) =x x 1 x 1 x 1=x =x x ( x 1)38.f1f2f3f4f5f6f1f1f2f3f4f5f6f2f2f1f4f3f6f5f3f3f5f1f6f2f4f4f4f6f2f5f1f3f5f5f3f6f1f4f2f642f1 f 2 = (((( f 2x ; x 1 =f3 ) f3 ) f3 ) f3 )= ((((( f 1 f 2 ) f 3 ) f 4 ) f 5 ) f 6 )x 1 x 1 = = ; x 1 1 x1f3= f1 f 3 = f 31 1 1 x 1 1 x1 xf3= (( f3 f3 ) f3 )f 5 ( f 4 ( x)) =f 6 ( f 2 ( x)) =f3= ((( f1 f3 ) f3 ) f3 )1 x 1 x f5 ( f3 ( x)) = = ; 1 x x 11 xf3= (((( f 31 xf 6 ( f1 ( x)) =f3f6f4f5f2f3f1Section 0.639.Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 43. Problem Set 0.740.1. a.b. 60 = 3 180 d. 4 240 = 180 3e.37 370 = 18 180 f.b. 45 = 180 4c.41. a. 30 = 180 6 10 = 180 182. a.7 180 = 210 6 180 = 135 b. c.1 180 = 60 3 d.4 180 = 240 3 e.f.c.3 43 180 = 30 18 42. 3. a.35 180 = 350 18 33.3 0.5812 180 b.c. 66.6 1.1624 180 d. 240.11 4.1907 180 e.0.7 Concepts Review 46 0.8029 180 369 6.4403 180 f. 11 0.1920 180 1. ( , ); [1, 1] 2. 2 ; 2 ; 3. odd; even4 x 4. r = (4) + 3 = 5; cos = = 5 r 22Instructors Resource ManualSection 0.743 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 44. 4. a. 180 3.141 180 Thusb. 180 6. 28 359. 8 c. 180 286.5 5. 00 d. 180 0. 001 0 .057 e. 180 0.1 5.73 f. 180 36. 0 2062.6 5. a.56. 4 tan34. 2 68.37 sin 34.1b.cos= cos=and by the Pythagorean Identity, sin3=3 . 2sin (0.361) 0.35326. a. b. 7. a.234.1sin(1.56) 248.3 cos(0.34 ) sin 2 (2.51) + cos(0.51) 1.2828 56. 3 tan34. 2 46.097 sin 56.1Referring to Figure 2, it is clear that sin3b.sin 35 0. 0789 sin 26 + cos 26 Identity, cos 2= 1 sin 29. a.=123 1 = 1 = . 6 4 21 = 1 cos()b.sec() =1 3 sec = = 2 4 cos 3 4d. Section 0.72 sin 6 = 3 tan = 3 6 cos 6 c.6= 0 . The rest of the values are 2 obtained using the same kind of reasoning in the second quadrant.and cos8. Referring to Figure 2, it is clear that sin 0 = 0 and cos 0 = 1 . If the angle is / 6 , then the triangle in the figure below is 1 1 equilateral. Thus, PQ = OP = . This 2 2 1 implies that sin = . By the Pythagorean 6 244tan (0.452) 0.4855d.63 . The results 2=2 were derived in the text. 4 4 2 If the angle is / 3 then the triangle in the 1 figure below is equilateral. Thus cos = 3 2 sin5.34 tan 21.3 0.8845 sin 3.1+ cot 23.5c.1 csc = =1 2 sin ( )(2)Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 45. e.f.10. a.b.( ) ( )b. cos 3t = cos(2t + t ) = cos 2t cos t sin 2t sin t cos 4 cot = =1 4 sin 4= (2 cos 2 t 1) cos t 2sin 2 t cos t = 2 cos3 t cos t 2(1 cos 2 t ) cos t( ) ( ) sin 4 tan = = 1 4 cos 4( ) ( )= 2 cos3 t cos t 2 cos t + 2 cos3 t= 4 cos3 t 3cos t c. sin 3 tan = = 3 3 cos 3= 2(2sin x cos x)(2 cos 2 x 1) = 2(4sin x cos3 x 2sin x cos x) = 8sin x cos3 x 4sin x cos x1 sec = =2 3 cos 3( )d.( ) ( )c. 3 cos 3 cot = = 3 3 sin 3d.1 csc = = 2 4 sin e. 3 sin 6 tan = = 6 cos 3 6f. 1 cos = 3 213. a. b.(4)( ) ( )c.d. 11. a.(1 + sin z )(1 sin z ) = 1 sin 2 z 1= cos 2 z =sin 4 x = sin[2(2 x)] = 2sin 2 x cos 2 x(1 + cos )(1 cos ) = 1 cos 2 = sin 2 sin u cos u + = sin 2 u + cos 2 u = 1 csc u sec u (1 cos 2 x)(1 + cot 2 x) = (sin 2 x)(csc2 x) 2 1 = sin x 2 = 1 sin x 1 sin t (csc t sin t ) = sin t sin t sin t 2 2 = 1 sin t = cos t 1 csc 2 t csc 2 t= cos 2 t = 2sec zb.(sec t 1)(sec t + 1) = sec 2 t 1 = tan 2 tc.sec t sin t tan t = =d.12. a.=14. a.cot 2 t csc 2 t=cos 2 t sin 2 t 1 sin 2 t1 sec 2 ty = sin 2x1 sin 2 t cos t cos t1 sin 2 t cos 2 t = = cos t cos t cos tsec2 t 1 sec 2 t sin 2 v +=tan 2 t sec 2 t1 2=sin 2 t cos 2 t 1 cos 2 t= sin 2 t= sin 2 v + cos 2 v = 1sec vInstructors Resource ManualSection 0.745 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 46. b.y = 2 sin tb. y = 2 cos tc. y = cos x 4 c.y = cos 3td.y = sec td. y = cos t + 315. a.y = csc t46Section 0.7x 2 Period = 4 , amplitude = 316. y = 3 cosInstructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 47. 17. y = 2 sin 2x Period = , amplitude = 221. y = 21 + 7 sin( 2 x + 3) Period = , amplitude = 7, shift: 21 units up, 3 units left 2 22. y = 3cos x 1 2 18. y = tan x Period = Period = 2 , amplitude = 3, shifts: units 2right and 1 unit down.19. y = 2 +1 cot(2 x) 6Period = 2, shift: 2 units up 23. y = tan 2 x 3 units right Period = , shift: 6 220. y = 3 + sec( x ) Period = 2 , shift: 3 units up, units rightInstructors Resource ManualSection 0.747 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 48. 24. a. and g.: y = sin x + = cos x = cos( x) 2 b. and e.: y = cos x + = sin( x + ) 2 = sin( x ) c. and f.: y = cos x = sin x 2 = sin( x + ) d. and h.: y = sin x = cos( x + ) 2 = cos( x )t sin (t) = t sin t; even25. a. b. c.1 csc( t ) = = csc t; odd sin( t )( )=32. a.sin(cos(t)) = sin(cos t); evenf.x + sin(x) = x sin x = (x + sin x); odd cot(t) + sin(t) = cot t sin t = (cot t + sin t); odd26. a.b.sin 3 (t ) = sin 3 t ; oddc.sec( t) =1 = sec t; even cos(t )sin 4 ( t ) = sin 4 t ; evend. e.cos(sin(t)) = cos(sin t) = cos(sin t); evenf.( x )2 + sin( x ) = x 2 sin x; neither 227. cos 2 1 1 = cos = = 3 3 4 2 228. sin 222 1 1 = sin = = 6 6 4 2 32sin(x y) = sin x cos(y) + cos x sin(y) = sin x cos y cos x sin yb.cos(x y) = cos x cos(y) sin x sin (y) = cos x cos y + sin x sin yc.tan( x y ) =2e.32 2 4=sin(t ) = sin t = sin t ; evend.1 cos 4 1 2 1 cos 2 8 31. sin = = = 8 2 2 2 2sin ( t ) = sin t ; even 2()1 + cos 6 1 + 2 1 + cos 2 12 = = = 30. cos 12 2 2 2 2+ 3 = 4 2tan x + tan( y ) 1 tan x tan( y ) tan x tan y 1 + tan x tan ytan t + tan tan t + 0 = 1 tan t tan 1 (tan t )(0) = tan t33. tan(t + ) =34. cos( x ) = cos x cos( ) sin x sin( ) = cos x 0 sin x = cos x 35. s = rt = (2.5 ft)( 2 rad) = 5 ft, so the tire goes 5 feet per revolution, or 1 revolutions 5 per foot. ft 1 rev mi 1 hr 60 5280 5 ft hr 60 min mi 336 rev/min 36. s = rt = (2 ft)(150 rev)( 2 rad/rev) 1885 ft 37. r1t1 = r2 t2 ; 6(2)t1 = 8(2)(21) t1 = 28 rev/sec 38. y = sin and x = cos y sin m= = = tan x cos 31 1 3 29. sin 6 = sin 6 = 2 = 8 48Section 0.7Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 49. 39. a.b.tan = 3 = 3 3x + 3 y = 6 3 y = 3x + 6 y=3 3 x + 2; m = 3 3 3 3tan = =44. Divide the polygon into n isosceles triangles by drawing lines from the center of the circle to the corners of the polygon. If the base of each triangle is on the perimeter of the polygon, then 2 . the angle opposite each base has measure n Bisect this angle to divide the triangle into two right triangles (See figure).5 640. m1 = tan 1 and m2 = tan 2 tan 2 + tan(1 ) tan = tan( 2 1 ) = 1 tan 2 tan(1 ) =tan 2 tan 1 m m1 = 2 1 + tan 2 tan 1 1 + m1m2 b h = so b = 2r sin and cos = so n 2r n n r h = r cos . n P = nb = 2rn sin n 1 A = n bh = nr 2 cos sin n n 2 sin32 1 = 1 + 3(2 ) 7 0.141941. a.tan =b.tan =1 1 2 1+( 1 ) (1) 2= 3 1.8925 c.2x 6y = 12 2x + y = 0 6y = 2x + 12y = 2x 1 y= x2 3 1 m1 = , m2 = 2 3 2 1 3 = 7; 1.7127 tan = 1 + 1 (2) 3()42. Recall that the area of the circle is r 2 . The measure of the vertex angle of the circle is 2 . Observe that the ratios of the vertex angles must equal the ratios of the areas. Thus, t A = , so 2 r 2 1 A = r 2t . 2 43. A =45. The base of the triangle is the side opposite the t angle t. Then the base has length 2r sin 2 (similar to Problem 44). The radius of the t semicircle is r sin and the height of the 2 t triangle is r cos . 2 A=1 t t t 2r sin r cos + r sin 2 2 2 2 22t t r 2 t = r 2 sin cos + sin 2 2 2 2 21 (2)(5) 2 = 25cm 2 2Instructors Resource ManualSection 0.749 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 50. x x x x 46. cos cos cos cos 2 4 8 16 1 3 1 1 3 1 = cos x + cos x cos x + cos x 2 2 4 4 16 16 1 3 1 3 1 = cos x + cos x cos x + cos x 4 4 4 16 16 1 3 3 3 1 = cos x cos x + cos x cos x 4 4 16 4 16 3 1 1 1 + cos x cos x + cos x cos x 16 4 16 4 1 1 15 9 1 13 11 = cos + cos x + cos x + cos x 4 2 16 16 2 16 16 1 7 1 1 5 3 + cos x + cos x + cos x + cos x 2 16 16 2 16 16 1 15 13 11 9 = cos x + cos x + cos x + cos x 8 16 16 16 16 7 5 3 1 + cos x + cos x + cos x + cos x 16 16 16 16 49. As t increases, the point on the rim of the wheel will move around the circle of radius 2. a.x(2) 1.902 y (2) 0.618 x(6) 1.176 y (6) 1.618 x(10) = 0 y (10) = 2 x(0) = 0 y (0) = 2b. x(t ) = 2 sin t , y (t ) = 2 cos t 5 5 c.The point is at (2, 0) when is , when t = 5t= 2; that5 . 22 . When 10 you add functions that have the same frequency, the sum has the same frequency.50. Both functions have frequency 47. The temperature function is 2 7 T (t ) = 80 + 25 sin 12 t 2 . The normal high temperature for November 15th is then T (10.5) = 67.5 F.a.y (t ) = 3sin( t / 5) 5cos( t / 5)+2sin(( t / 5) 3)48. The water level function is 2 F (t ) = 8.5 + 3.5 sin (t 9) . 12 The water level at 5:30 P.M. is then F (17.5) 5.12 ft .b.50Section 0.7y (t ) = 3cos( t / 5 2) + cos( t / 5) + cos(( t / 5) 3)Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 51. a.C sin(t + ) = (C cos )sin t + (C sin ) cos t. Thus A = C cos and B = C sin .b.51.A2 + B 2 = (C cos )2 + (C sin ) 2 = C 2 (cos 2 ) + C 2 (sin 2 ) = C 2Also, c.B C sin = = tan A C cos A1 sin(t + 1 ) + A2 sin(t + 2 ) + A3 (sin t + 3 ) = A1 (sin t cos 1 + cos t sin 1 ) + A2 (sin t cos 2 + cos t sin 2 ) + A3 (sin t cos 3 + cos t sin 3 ) = ( A1 cos 1 + A2 cos 2 + A3 cos 3 ) sin t + ( A1 sin 1 + A2 sin 2 + A3 sin 3 ) cos t = C sin (t + )where C and can be computed from A = A1 cos 1 + A2 cos 2 + A3 cos 3 B = A1 sin 1 + A2 sin 2 + A3 sin 3 as in part (b).d.Written response. Answers will vary.52. ( a.), (b.), and (c.) all look similar to this:d.53. a.b.c. e.The windows in (a)(c) are not helpful because the function oscillates too much over the domain plotted. Plots in (d) or (e) show the behavior of the function. Instructors Resource ManualThe plot in (a) shows the long term behavior of the function, but not the short term behavior, whereas the plot in (c) shows the short term behavior, but not the long term behavior. The plot in (b) shows a little of each. Section 0.751 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 52. 54. a.h( x ) = ( f g ) ( x ) 3 cos(100 x) + 2 100 = 2 1 2 cos (100 x) + 1 100 j ( x) = ( g f )( x) =56. 2 f ( x ) = ( x 2n ) , 0.0625, 1 1 x 2n , 2n + 4 4 otherwisewhere n is an integer. y1 3x + 2 cos 100 100 x2 + 1 0.5b.0.252c.11x20.8 Chapter Review Concepts Test 1. False: 2. True: 1 4 x x + 1 : x n, n + 4 55. f ( x ) = 4 x x + 7 : x n + 1 , n + 1 3 3 4 where n is an integer.()p and q must be integers. p1 p2 p1q2 p2 q1 = ; since q1 q2 q1q2 p1 , q1 , p2 , and q2 are integers, so are p1q2 p2 q1 and q1q2 .If the numbers are opposites ( and ) then the sum is 0, which is rational.4. True:Between any two distinct real numbers there are both a rational and an irrational number.5. False:0.999... is equal to 1.6. True:( am ) = ( an )7. False:(a * b) * c = abc ; a *(b * c) = ab8. True:(3. False:Since x y z and x z , x = y = z)y2111x9. True:52Section 0.8nm= a mn cx would 2 be a positive number less than x .If x was not 0, then =Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 53. 10. True:y x = ( x y ) so ( x y )( y x) = ( x y )(1)( x y )20. True:If r > 1, then 1 r < 0. Thus, since 1 + r 1 r ,= (1)( x y ) . 21 1 . 1 r 1+ r( x y ) 2 0 for all x and y, so ( x y ) 2 0.11. True: 12. True:If r > 1, r = r , and 1 r = 1 r , so 1 1 1 . = 1 r 1 r 1+ r[ a, b] and [b, c ]If r < 1, r = r and 1 r = 1 + r ,a 1 1 > 1; < b b aa < b < 0; a < b;so share point b incommon. 13. True:14. True:21. True:If (a, b) and (c, d) share a point then c < b so they share the infinitely many points between b and c.1 1 1 = . 1 r 1 r 1+ rIf x and y are the same sign, then x y = x y . x y x+ y when x and y are the same sign, so x y x + y . If x and y have opposite signs then either x y = x ( y ) = x + yx 2 = x = x if x < 0.16. False:17. True:For example, if x = 3 , then x = ( 3) = 3 = 3 which does(x > 0, y < 0) or x y = x y = x + ynot equal x.15. False:(x < 0, y > 0). In either case x y = x+ y .For example, take x = 1 and y = 2 . 4x < y x < y 4If either x = 0 or y = 0, the inequality is easily seen to be true.4422. True:x = x and y = y , so x < y18. True:444x + y = ( x + y )4x2 =23. True:= x + ( y ) = x + y19. True:If r = 0, then 1 1 1 = = = 1. 1+ r 1 r 1 r For any r, 1 + r 1 r . Since r < 1, 1 r > 0 so1 1 ; 1+ r 1 rIf y is positive, then x =( y)x3 == y.(3 y )3=y24. True:For example x 2 0 has solution [0].25. True:x 2 + ax + y 2 + y = 0 x 2 + ax +If 1 < r < 0, then r = r and2a2 1 a2 1 + y2 + y + = + 4 4 4 4 2a a2 + 1 1 x+ + y+ = 2 2 4 is a circle for all values of a.1 r = 1 + r , so1 1 1 = . 1+ r 1 r 1 r 1 r = 1 r , soy satisfiesFor every real number y, whether it is positive, zero, or negative, the cube root x = 3 y satisfiesalso, 1 < r < 1.If 0 < r < 1, then r = r and226. False:If a = b = 0 and c < 0 , the equation does not represent a circle.1 1 1 . = 1+ r 1 r 1 rInstructors Resource ManualSection 0.853 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 54. 27. True;28. True:3 ( x a) 4 3 3a y = x + b; 4 4 If x = a + 4: 3 3a y = (a + 4) +b 4 4 3a 3a = +3 +b = b+3 4 4 y b =40. True:41. True:The equation is (3 + 2m) x + (6m 2) y + 4 2m = 0 which is the equation of a straight line unless 3 + 2m and 6m 2 are both 0, and there is no real number m such that 3 + 2m = 0 and 6m 2 = 0. f ( x) = ( x 2 + 4 x + 3) = ( x + 3)( x + 1)31. True:32. True:If ab > 0, a and b have the same sign, so (a, b) is in either the first or third quadrant.The domain does not include n + where n is an integer. 243. True:The domain is ( , ) and the range is [6, ) .44. False:The range is ( , ) .45. False:The range ( , ) .46. True:If f(x) and g(x) are even functions, f(x) + g(x) is even. f(x) + g(x) = f(x) + g(x)47. True:If f(x) and g(x) are odd functions, f(x) + g(x) = f(x) g(x) = [f(x) + g(x)], so f(x) + g(x) is odd48. False:If f(x) and g(x) are odd functions, f(x)g(x) = f(x)[g(x)] = f(x)g(x), so f(x)g(x) is even. If f(x) is even and g(x) is odd, f(x)g(x) = f(x)[g(x)] = f(x)g(x), so f(x)g(x) is odd.50. False:If f(x) is even and g(x) is odd, f(g(x)) = f(g(x)) = f(g(x)); while if f(x) is odd and g(x) is even, f(g(x)) = f(g(x)); so f(g(x)) is even.51. False:30. True:42. False:49. True:29. True:If the points are on the same line, they have equal slope. Then the reciprocals of the slopes are also equal.If f(x) and g(x) are odd functions, f ( g ( x)) = f(g(x)) = f(g(x)), so f(g(x)) is odd.Let x = / 2. If > 0 , then x > 0 and x < . If ab = 0, a or b is 0, so (a, b) lies on the xaxis or the yaxis. If a = b = 0, (a, b) is the origin. y1 = y2 , so ( x1 , y1 ) and ( x2 , y2 )are on the same horizontal line. 33. True:( x 2 + 4 x + 3) 0 on 3 x 1 .d = [(a + b) (a b)]2 + (a a) 2 = (2b) 2 = 2b34. False:The equation of a vertical line cannot be written in pointslope form.35. True:This is the general linear equation.36. True:Two nonvertical lines are parallel if and only if they have the same slope.37. False:The slopes of perpendicular lines are negative reciprocals.38. True:If a and b are rational and ( a, 0 ) , ( 0, b ) are the intercepts, the slope is 39. False:b which is rational. af ( x) =2( x)3 + ( x)ax + y = c y = ax + c ax y = c y = ax c (a )( a) 1. (unless a = 1 )5452. True:Section 0.8=( x)2 + 1=2 x3 x x2 + 12 x3 + x x2 + 1Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 55. 53. True:f (t ) = =(sin(t )) 2 + cos( t ) tan( t ) csc( t )( sin t )2 + cos t (sin t )2 + cos t = tan t ( csc t ) tan t csc t54. False:f(x) = c has domain ( , ) , yet the range has only one value, c.b.g (1.8) =57. True:(ff )( x) = ( x 2 )3 = x 6(f2=4 25 2(n 2 n + 1)2 ; (1)2 (1) + 1 = 1; 2 (2) 2 (2) + 1 = 9; g )( x) = ( x 3 ) 2 = x 6(g21 1 1 25 n + ; 1 + = 2; 2 + = ; 2 4 n 1 1 2 + 2 1.8 = 0.9 = 1 256. True:1n1. a.f(x) = c has domain ( , ) and the only value of the range is c.55. False:Sample Test Problemsg )( x) = ( x 3 ) 2 = x 658. False:f ( x) g ( x) = x x = x 2 359. False:60. True:61. True:52 (2)2 (2) + 1 = 49 c.43 / n ; 43 /1 = 64; 43 / 2 = 8; 4 3 / 2 =d.n1 1 ; 1 = 1; n 121 = 2 2f The domain of excludes any g values where g = 0.f(a) = 0 Let F(x) = f(x + h), then F(a h) = f(a h + h) = f(a) = 0 cos x sin x cos( x) cot( x) = sin( x) cos x = = cot x sin x2. a.1 1 2 = = ; 2 2 21 1 1 1 1 + + 1 + m n m n 63. False:1cot x =b.The domain of the tangent function excludes all n + where n is an 2 integer. The cosine function is periodic, so cos s = cos t does not necessarily imply s = t; e.g., cos 0 = cos 2 = 1 , but 0 2 .Instructors Resource Manual=c.1 1 + m n = 1 1 1 + m n mn + n + m = mn n + m 1+2 x 2 x 2 x + 1 x x 2 x + 1 ( x 2)( x + 1) = 3 2 3 2 x +1 x 2 x +1 x 2 =62. False:1 82( x 2) x 3 ( x 2) 2( x + 1) x4 x 8(t 3 1) (t 1)(t 2 + t + 1) 2 = = t + t +1 t 1 t 13. Let a, b, c, and d be integers. a+ c a c ad + bc b d which is rational. = + = 2 2b 2d 2bdSection 0.855 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 56. 4. x = 4.1282828 1000 x = 4128.282828 10 x = 41.28282813. 21t 2 44t + 12 3; 21t 2 44t + 15 0; t=990 x = 4087 4087 x= 99044 442 4(21)(15) 44 26 3 5 = = , 2(21) 42 7 3 3 5 3 5 t t 0; , 7 3 7 35. Answers will vary. Possible answer: 13 0.50990... 502x 1 1 > 0; , ( 2, ) x2 2 14.2 3 8.15 104 1.32 545.39 6. 3.247.( 2.0)2.515. ( x + 4)(2 x 1) 2 ( x 3) 0;[4,3] 3 2.0 2.6616. 3x 4 < 6; 6 < 3 x 4 < 6; 2 < 3x < 10; 8. sin2( 2.45 ) + cos ( 2.40 ) 1.00 0.0495 29. 1 3 x > 0 3x < 1 1 x< 3 1 , 3 10. 6 x + 3 > 2 x 5 4 x > 8 x > 2; ( 2, )11. 3 2 x 4 x + 1 2 x + 7 3 2 x 4 x + 1 and 4 x + 1 2 x + 7 6 x 2 and 2 x 6 1 1 x and x 3; , 3 3 3 12. 2 x 2 + 5 x 3 < 0;(2 x 1)( x + 3) < 0; 1 1 3 < x < ; 3, 2 217.2 10 2 10 < x < ; , 3 3 3 33 2 1 x 3 20 1 x 3 2(1 x) 0 1 x 2x +1 0; 1 x 1 , (1, ) 2 18. 12 3 x x (12 3 x)2 x 2 144 72 x + 9 x 2 x 2 8 x 2 72 x + 144 0 8( x 3)( x 6) 0 (,3] [6, )19. For example, if x = 2, (2) = 2 2 x x for any x < 020. If x = x, then x = x. x056Section 0.8Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 57. 2 + 10 0 + 4 , 27. center = = (6, 2) 2 2 1 1 radius = (10 2) 2 + (4 0) 2 = 64 + 16 2 2 =2 521. t 5 = (5 t) = 5 t If 5 t = 5 t, then 5 t 0. t 5 22. t a = (a t ) = a tIf a t = a t , then a t 0. tacircle: ( x 6)2 + ( y 2) 2 = 2023. If x 2, then28. x 2 + y 2 8 x + 6 y = 0 x 2 8 x + 16 + y 2 + 6 y + 9 = 16 + 90 2 x 2 + 3 x + 2 2 x 2 + 3 x + 2 8 + 6 + 2 = 16( x 4) 2 + ( y + 3) 2 = 25;11 . Thus also x + 2 2 so 2 x +2 2 22 x2 + 3x + 2 x2 + 2= 2 x2 + 3x + 21 16 2 2 x +2 1=824. a.The distance between x and 5 is 3.b. The distance between x and 1 is less than or equal to 2. c.The distance between x and a is greater than b.center = ( 4, 3) , radius = 5 x2 2 x + y 2 + 2 y = 229.x2 2 x + 1 + y2 + 2 y + 1 = 2 + 1 + 1 ( x 1) 2 + ( y + 1) 2 = 4 center = (1, 1) x 2 + 6 x + y 2 4 y = 7 x 2 + 6 x + 9 + y 2 4 y + 4 = 7 + 9 + 4 ( x + 3)2 + ( y 2)2 = 6 center = (3, 2) d = (3 1) 2 + (2 + 1)2 = 16 + 9 = 525.30. a.d ( A, B ) = (1 + 2) 2 + (2 6)23x + 2 y = 6 2 y = 3 x + 6 3 y = x+3 2 3 m= 2 3 y 2 = ( x 3) 2 3 13 y = x+ 2 2= 9 + 16 = 5 d ( B, C ) = (5 1)2 + (5 2)2 = 16 + 9 = 5 d ( A, C ) = (5 + 2)2 + (5 6) 2 = 49 + 1 = 50 = 5 2 ( AB) + ( BC )2 = ( AC ) 2 , so ABC is a right triangle. 21+ 7 2 + 8 , 26. midpoint: = ( 4,5 ) 2 2 d = (4 3)2 + (5 + 6)2 = 1 + 121 = 122Instructors Resource ManualSection 0.857 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 58. b.2 m= ; 33x 2 y = 5 2 y = 3 x + 5 3 5 y = x ; 2 2 3 m= 2 3 y 1 = ( x + 2) 2 3 y = x+4 2c.b.3 x + 4y = 9 4y = 3x + 9; 4 3 9 y = x+ ; m = 3 4 4 4 y 1 = ( x + 2) 3 4 11 y = x+ 3 32 ( x 1) 3 2 5 y = x 3 3 y +1 =c.y=9d. x = 2 e.contains (2, 1) and (0, 3); m =3 1 ; 0+2y=x+3 3 +1 4 11 3 8 4 = ; m2 = = = ; 52 3 11 5 6 3 11 + 1 12 4 m3 = = = 11 2 9 3 m1 = m2 = m3 , so the points lie on the same line.32. m1 = d. x = 333. The figure is a cubic with respect to y.The equation is (b) x = y 3 . 34. The figure is a quadratic, opening downward, with a negative yintercept. The equation is (c) y = ax 2 + bx + c. with a < 0, b > 0, and c < 0.35. 31. a.583 1 2 m= = ; 7+2 9 2 y 1 = ( x + 2) 9 2 13 y = x+ 9 9Section 0.8Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 59. 36.x2 2 x + y 2 = 3 x2 2 x + 1 + y 2 = 4 ( x 1)2 + y 2 = 440. 4 x y = 2 y = 4 x 2; 1 4 contains ( a, 0 ) , ( 0, b ) ; m=ab =8 2 ab = 16 16 b= a 1 b0 b = = ; 0a 4 a a = 4b 16 a = 4 a37.a 2 = 64 a =8 b=41. a.b.38.c.16 1 = 2; y = x + 2 8 4 f (1) =1 1 1 = 1+1 1 21 1 1 f = =4 2 1 +1 1 2 2f(1) does not exist. 1 1 1 1 = t 1+1 t 1 t t 1d.e.39. y = x2 2x + 4 and y x = 4; x + 4 = x2 2 x + 4 x 2 3x = 0 x( x 3) = 0 points of intersection: (0, 4) and (3, 7)f (t 1) =1 1 t 1 f = = t 1 +1 1 1+ t t t t42. a.b.c.g (2) =2 +1 3 = 2 21 g = 21 2+1 1 2=32 + h +1 + 22 1 g ( 2 + h ) g ( 2) = 2+ h h h h 2 h + 6 3h 6 2( h + 2) 1 2( h + 2) = = = h h 2(h + 2)43. a.: x 1, 1}b.Instructors Resource Manual{x {x : x 2}Section 0.859 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 60. 44. a.b.f ( x) =3( x) ( x) + 1 2=3x x +1 2; odd46.g ( x) = sin( x) + cos( x)= sin x + cos x = sin x + cos x; evenc.h( x) = ( x)3 + sin( x) = x3 sin x ; oddd.k ( x) =45. a.( x )2 + 1 x + ( x) 4=x2 + 1 x + x4; even47. V(x) = x(32 2x)(24 2x) Domain [0, 12]2f (x) = x 148. a.1 13 ( f + g )(2) = 2 + (22 + 1) = 2 2 b.15 3 ( f g )(2) = (5) = 2 2c.(fg )(2) = f (5) = 5 d.(g13 3 3 f )(2) = g = + 1 = 2 2 4 e.1 f 3 (1) = 1 + = 0 1 1 24 = 5 5 2b.x g(x) = 2 x +132f.49. a. c.60 x2 h(x) = 6 xSection 0.83 f 2 (2) + g 2 (2) = + (5) 2 2 9 109 = + 25 = 4 4 y=1 2 x 4if 0 x 2 if x > 2Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 61. b.y=1 ( x + 2)2 453. a. b.sin (t) = sin t = 0.8 sin 2 t + cos2 t = 1 cos 2 t = 1 (0.8)2 = 0.36 cos t = 0. 6c. d.c.tan t =e. cos t = sin t = 0.8 2 f.1 y = 1 + ( x + 2) 2 4sin 2t = 2 sin t cos t = 2(0.8)(0.6) = 0.96sin( + t ) = sin t = 0.8sin t 0.8 4 = = 1.333 cos t 0.6 354. sin 3t = sin(2t + t ) = sin 2t cos t + cos 2t sin t = 2sin t cos 2 t + (1 2sin 2 t ) sin t = 2sin t (1 sin 2 t ) + sin t 2sin 3 t= 2sin t 2sin 3 t + sin t 2sin 3 t = 3sin t 4sin 3 t 55. s = rt rev rad 1 min = 9 20 2 (1 sec) = 6 min rev 60 sec 18.85 in. 50. a. b. c.(,16] fReview and Preview Problemsg = 16 x 4 ; domain [2, 2]g f = ( 16 x ) 4 = (16 x) 2 ; domain (,16] (note: the simplification ( 16 x ) 4 = (16 x) 2 is only true given the restricted domain)51.f ( x) = x , g ( x) = 1 + x, h( x) = x 2 , k(x) = sin x, F ( x) = 1 + sin 2 x = f g h k52. a.sin(570) = sin(210) = 1 2b. 9 cos = cos = 0 2 2c.3 13 cos = cos = 6 6 2Instructors Resource Manual1. a) b) 2. a) b)0 < 2 x < 4; 0 < x < 2 6 < x < 16 13 < 2 x < 14; 6.5 < x < 74 < x / 2 < 7; 14 < x < 83. x 7 = 3 or x 7 = 3 x = 10 or x=4 4. x + 3 = 2 or x = 1 orx + 3 = 2 x = 55. x 7 = 3 or x 7 = 3 x = 10 or x=4 6. x 7 = d or x 7 = d x = 7 + d or x = 7 d 7. a)x 7 < 3 and x 7 > 3 x < 10 and x>4 4 < x < 10Review and Preview61 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 62. b)c)d)8. a)x 7 3 and x 7 3 x 10 and x4 4 x 10 x 7 1 and x 8 and 6 x8b)g ( 0.9 ) = 0.0357143 g ( 0.99 ) = 0.0033557 g ( 0.999 ) = 0.000333556x 7 1 x6g (1.001) = 0.000333111 g (1.01) = 0.00331126 g (1.1) = 0.03125x 7 < 0.1 and x 7 > 0.1 x < 7.1 and x > 6.9 6.9 < x < 7.1 x 2 < 1 and x 2 > 1 x < 3 and x >1 1< x < 3b)c)12. a)x 2 < 0.1 and x 2 > 0.1 x < 2.1 and x > 1.9 1.9 < x < 2.1d)g ( 2) =x 2 1 or x 2 1 x 3 or x 1x 2 < 0.01 and x 2 > 0.01 x < 2.01 and x > 1.99 1.99 < x < 2.019. a) b) 10. a) 11. a)x 1 0; x 1 2 x 2 x 1 0; x 1, 0.5 x0b)g ( 0 ) = 1b)1 = 1 1 0.1 = 1 F ( 0.1) = 0.1 0.01 F ( 0.01) = = 1 0.01 0.001 F ( 0.001) = = 1 0.001 0.001 F ( 0.001) = =1 0.001 0.01 F ( 0.01) = =1 0.01 0.01 F ( 0.1) = =1 0.01 1 F (1) = = 1 1 F ( 1) =G ( 1) = 0.841471 G ( 0.1) = 0.998334 G ( 0.01) = 0.999983x00 1 f ( 0) = =1 0 1 0.81 1 f ( 0.9 ) = = 1.9 0.9 1 0.9801 1 = 1.99 f ( 0.99 ) = 0.99 1 0.998001 1 = 1.999 f ( 0.999 ) = .999 1 1.002001 1 = 2.001 f (1.001) = 1.001 1 1.0201 1 = 2.01 f (1.01) = 1.01 1 1.21 1 = 2.1 f (1.1) = 1.1 1 4 1 =3 f ( 2) = 2 11 5G ( 0.001) = 0.99999983 G ( 0.001) = 0.99999983 G ( 0.01) = 0.999983 G ( 0.1) = 0.998334 G (1) = 0.84147113. x 5 < 0.1 and x 5 > 0.1 x < 5.1 and x > 4.9 4.9 < x < 5.1 14. x 5 < and x 5 > x < 5 + and x > 5 5 < x < 5+ 15. a.True.b. False: Choose a = 0.c.True.d. True16. sin ( c + h ) = sin c cos h + cos c sin h 62Review and PreviewInstructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 63. 1CHAPTERLimits 9.1.1 Concepts Reviewx3 4 x 2 + x + 6 x 1 x +1 lim( x + 1)( x 2 5 x + 6) x 1 x +11. L; c= lim2. 6= lim ( x 2 5 x + 6) x 1 23. L; right= (1) 5(1) + 64. lim f ( x) = M= 12x cProblem Set 1.1x2 = lim( x 2 + 2 x 1) = 1 x 01. lim( x 5) = 2x 0x 32. lim (1 2t ) = 311.t 13. 4.lim ( x 2 + 2 x 1) = (2) 2 + 2(2) 1 = 1= t t = 2tlim ( x 2 + 2t 1) = (2) 2 + 2t 1 = 3 + 2tx 2(2) ( ( 1)6. lim t 2 x 2 = t 112.) ( ( 1) 1) = 05. lim t 2 1 = t 12)x2 4 ( x 2)( x + 2) = lim x2 x 2 x2 x2 = lim( x + 2)x2 9 x 3 x 3 ( x 3)( x + 3) = lim x 3 x3 = lim( x + 3) limx 3 x2 = 1 x27. lim=3+3=6 13.x2lim(t + 4)(t 2) 4 (3t 6) 2t 2= lim=2+2=4 8.x2 t 2 ( x + t )( x t ) = lim xt x + t x t x+t = lim ( x t ) limx tx 2(x 4 + 2 x3 x 210. lim(t 2) 2 t + 4 9(t 2) 2t 2t 2 + 4t 21 t 7 t+7 (t + 7)(t 3) = lim t 7 t+7 = lim (t 3)t+4 9= limlimt 2=2+4 6 = 9 9t 7= 7 3 = 1014.(t 7)3 t 7limt 7+= lim t 7+= limt 7+(t 7) t 7 t 7 t 7= 77 = 0Instructors Resource ManualSection 1.163 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 64. 15. limx 4 18 x 2 + 81x 3( x 3)2= limx 3( x 3) 2 ( x + 3) 2= limx 3( x 3)( x 2 9) 2 ( x 3)= lim( x + 3)2 = (3 + 3) 22limt 02( x sin x ) 2 / x 2x21.x 31.0.02513140.12.775 1068(3u + 4)(u 1)30.012.77775 1010(u 1) 20.0012.77778 10141. 0.10.0251314 2.775 1060.012.77775 10100.0012.77778 1014= 36 (3u + 4)(2u 2)316. lim1 cos t =0 2t(u 1) 2u 1= limu 1= lim 8(3u + 4)(u 1) = 8[3(1) + 4](1 1) = 0 u 117.(2 + h) 2 4 4 + 4h + h 2 4 = lim h0 h0 h h limh 2 + 4h = lim(h + 4) = 4 h 0 h 0 h= limlim( x sin x) 2 x2x 018.( x + h) 2 x 2 x 2 + 2 xh + h 2 x 2 = lim h0 h 0 h h limh 2 + 2 xh = lim(h + 2 x) = 2 x h 0 h 0 h2(1 cos x ) / xx22.=0 21.x19.0.002495840.01 0.0010.0000249996 2.5 1071.0.2113220.1sin x 2x0.2113220.1= lim0.00249584 0.0000249996 2.5 1071.0.4207350.10.4991670.010.4999920.010.0010.499999920.0011.0.4207350.10.4991670.010.4999920.0010.499999920.01 0.0012(t 1) /(sin(t 1))t23.=01.12.10351.012.010031.0012.0010.22984901.18840.02497920.91.903170.002499980.991.990030.9991.9991 cos t 2tt0.1x2x 03.56519sin x = 0.5 x 0 2 x1.(1 cos x) 22.lim20.lim0.00024999998t 1 =2 1) 2lim1. 0.10.02497920.010.002499980.001640.229849t 1 sin(t0.00024999998Section 1.1Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 65. x sin( x 3) 3 x 3x24.4.1. + 4 0.1 +0.1585293.12.limx 4(1 + sin( x 3 / 2)) /( x )x 1. + 0.6741170.1 + 0.01 + 0.00500.001 + (x ) 2 4(tan x 1)20.119210.001 +0.1 +1 + sin ( x 32 ) xx =00.0000210862 2.12072 107 20.536908 20.01 +0.0500 0.0050lim0.001993390.01 + 20.45970.0005= 0.250.1 + 21. + 20.001 + 0.2505(2 2sin u ) / 3u0.00050.01 + 0.2550081. + 20.05000.1 + 0.300668u28.0.45971. + 0.249540.001 + 4x sin( x 3) 3 =0 lim x 3 x3 25.0.245009 0.01 + 40.0000166666 1.66667 1072.99940.1 + 40.001665832.990.201002 1. + 40.1585292.940.001 +0.0000166666 1.66667 1073.0010.0320244 0.01 +0.001665833.01( x / 4) 2 /(tan x 1) 2x27.0.00226446 20.001 + 20.0000213564 2.12342 1072 2sin u lim =0 u 3u 229. a.t(1 cot t ) /(1 / t )1.0.3579070.10.8966640.010.9899670.0010.999d.1.1.64209e.0.11.09666f.0.011.009970.0011.001lim f ( x) = 2x 326.limt 01 cot t 1 tb. f(3) = 1 c.g.= 1 h.i.Instructors Resource Manualf(1) does not exist. lim f ( x) =x 15 2f(1) = 2 lim f(x) does not exist. x1lim f ( x) = 2x 1lim f ( x) = 1x 1+lim f ( x ) = +x 15 2Section 1.165 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 66. lim f ( x) does not exist.b.lim f ( x) does not exist.b.f(3) = 1c.f(1) = 2c.f(1) = 1d.30. a.d.x 3lim f ( x) = 2x 1e.lim f ( x) = 2x 1+34.f(1) = 1f.x 1lim f ( x) does not exist.g. h. i. 31. a. b. c. d. e. f. 32. a. b. c.x 1lim f ( x) = 1x 1lim f ( x) does not exist.x 1+lim f ( x ) = 2x 1+a.f(3) = 2 f(3) is undefined.x 1b. g(1) does not exist.lim f ( x) = 2c.x 3lim f ( x) = 4x 3+d.lim f ( x) does not exist.x 3lim g ( x) = 035.lim g ( x ) = 1x2lim g ( x ) = 1x 2+f ( x) = x [ x ] lim f ( x) does not exist.x 3+lim f ( x) = 2x 1lim f ( x) = 2x 1+lim f ( x) = 2x 1d. f (1) = 2 e.lim f ( x) = 0f.f (1) = 0x 1a. b.33. c.d.a.66f(0) = 0 lim f ( x) does not exist.x 0lim f ( x ) = 1x 0 lim f ( x) =x 1 21 2lim f ( x) = 0x 0Section 1.1Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 67. f ( x) =36.41. lim f ( x) exists for a = 1, 0, 1.x xxa42. The changed values will not change lim f ( x) at xaany a. As x approaches a, the limit is still a 2 . 43. a.x 1limx 1limx 1 x 1x 1b.f (0) does not exist.a.lim f ( x) does not exist.b.x 0limx 1limc.does not exist.x 1 x 1 x 1= 1 and lim+x 1x 1=1= 1x2 x 1 1 x 1x 1x 1= 3lim f ( x ) = 1c.x 0 d. 1 1 lim does not exist. x 1 x 1 x 1 d.lim f ( x) = 1x 1244. a.x2 1 37. lim does not exist. x 1 x 1 limx 1x2 1 x2 1 =2 = 2 and lim x 1 x 1+ x 1x 0= lim( x + 2 2)( x + 2 + 2)= limx 0x 0x( x + 2 + 2)x 0= limx 01x1 does not exist. x 1/ xlim x (1)2 = = = 4 0+2 + 2 2 2 x+2+ 2=0b) 01c)=01/ xx 0+45. a) 1x( x + 2 + 2)1+x 0+x( x + 2 + 2) x+22x x =0lim x(1)d.= limb.limc.x 039. a.b.x+2 2 x38. limlim x 1+d)1146. a) Does not exist c)lim f ( x) does not exist.1b) 0 d) 0.556x 1lim f ( x) = 047. lim x does not exist since x 0x 040.x is not definedfor x < 0. 48.lim x x = 1x 0+49. limx 0x =0 x50. lim x = 1 x 0sin 2 x 1 = x 0 4 x 251. limInstructors Resource ManualSection 1.267 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 68. 52. limx 07. If x is within 0.001 of 2, then 2x is within 0.002 of 4.sin 5 x 5 = 3x 31 53. lim cos does not exist. x 0 x 1 54. lim x cos = 0 x 0 xx3 155. lim56. limx 057.=62x + 2 2x 1x sin 2 x sin( x 2 )limx 258. lim+x 18. If x is within 0.0005 of 2, then x2 is within 0.002 of 4.=2x2 x 2 = 3 x22 1/( x 1)1+ 2=059. lim x ; The computer gives a value of 0, but x 0limx 09. If x is within 0.0019 of 2, then 0.002 of 4.8 x is withinx does not exist.1.2 Concepts Review 1. L ; L + 2. 0 < x a < ; f ( x) L < 10. If x is within 0.001 of 2, then 3.8 is within 0.002 xof 4.34. ma + bProblem Set 1.2 1. 0 < t a < f (t ) M < 2. 0 < u b < g (u ) L < 11. 0 < x 0 < (2 x 1) (1) < 2x 1+ 1 < 2x < 3. 0 < z d < h( z ) P < 2 x 0. ) x x=39. For x < 0, x = x, thus limx 0 x x= limx 0 x = 1 x40. For x > 0, x = x, thus limx 0 +45.x x= limx 0 +x =1 x41. As x 0 , 1 + cos x 2 while sin x 0 . 1 + cos x lim = sin x x 0limx 2x 2 = lim = 2, x 3 x 1 3 x2x 2 lim = lim = 2, x x 3 x 1 3 xHorizontal asymptote y = 2 2x 2x lim = , lim = ; x 3+ x 3 x 3 x 3 Vertical asymptote x = 342. 1 sin x 1 for all x, and 1 sin x lim = 0, so lim = 0. x x x xInstructors Resource ManualSection 1.581 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 82. 46.lim3 2x 9 3= 0, lim249. f ( x ) = 2 x + 3 = 0;x 9 x x Horizontal asymptote y = 0 3 3 lim = , lim = , 9 x2 + 9 x2 x 3 x 3 3 3 lim = , lim = ; 2 2 x 3+ 9 x x 3 9 x Vertical asymptotes x = 3, x = 350.14x 2 x2f ( x) = 3x + 4 4x + 3 x2 + 1, thusWe say that lim f ( x) = if to each x c +negative number M there corresponds a > 0 such that 0 < x c < f(x) < M.14= 0, lim, thus 4x + 3 lim [ f ( x) (3 x + 4)] = lim x x x 2 + 1 4+ 3 x x2 = lim =0. 1+ 1 x 2 x The oblique asymptote is y = 3x + 4.= 0; x 2 x2 + 7 +7 Horizontal asymptote y = 0 2 Since 2x + 7 > 0 for all x, g(x) has no vertical asymptotes. limx 11 lim [ f ( x) (2 x + 3)] = lim =0 3 x x x 1 The oblique asymptote is y = 2x + 3.51. a.47.1 3b. We say that lim f ( x) = if to each x c positive number M there corresponds a > 0 such that 0 < c x < f(x) > M. We say that lim f ( x) = if to each52. a.x positive number M there corresponds an N > 0 such that N < x f(x) > M. b. We say that lim f ( x ) = if to each x positive number M there corresponds an N < 0 such that x < N f(x) > M. 53. Let > 0 be given. Since lim f ( x ) = A, there is x 48.limx limx 2x 2x +5= lim2x x2 + 5x = lim2 1+x 5 x2=2 1+5 x22 1 =a corresponding number M1 such that= 2, 2 1x > M1 f ( x) A < . Similarly, there is a 2= 2Since x 2 + 5 > 0 for all x, g(x) has no vertical asymptotes.number M2 such that x > M 2 g ( x) B < . 2 Let M = max{M1 , M 2 } , then x > M f ( x) + g ( x) ( A + B) = f ( x) A + g ( x) B f ( x) A + g ( x) B= 2 2 Thus, lim [ f ( x) + g ( x)] = A + B 0. Thus, there is at least one number c between 0 and 1 such that x 3 + 3x 2 = 0. 53. Because the function is continuous on [ 0,2 ] and (cos 0)03 + 6sin 5 0 3 = 3 < 0,Cost $0.60(cos 2)(2)3 + 6sin 5 (2) 3 = 83 3 > 0, there is at least one number c between 0 and 2 such0.48that (cos t )t 3 + 6sin 5 t 3 = 0.0.720.3654. Let f ( x ) = x 7 x + 14 x 8 . f(x) is continuous at all values of x. f(0) = 8, f(5) = 12 Because 0 is between 8 and 12, there is at least one number c between 0 and 5 such that 30.24 0.12 1 3 5 2 4 6 Length of call in minutes50. The function is continuous on the intervals [0, 200], (200,300], (300, 400], 2f ( x ) = x 3 7 x 2 + 14 x 8 = 0 .This equation has three solutions (x = 1,2,4)Cost $ 80 60 4055. Let f ( x ) = x cos x. . f(x) is continuous at all20100 200 300 400 500 Miles Driven51. The function is continuous on the intervals (0, 0.25], (0.25, 0.375], (0.375, 0.5], values of x 0. f(0) = 1, f(/2) = / 2 Because 0 is between 1 and / 2 , there is at least one number c between 0 and /2 such that f ( x ) = x cos x = 0. The interval [0.6,0.7] contains the solution.Cost $ 4 3 2 10.250.5 0.75 Miles Driven156. Let f ( x) = x5 + 4 x3 7 x + 14 f(x) is continuous at all values of x. f(2) = 36, f(0) = 14 Because 0 is between 36 and 14, there is at least one number c between 2 and 0 such that f ( x) = x5 + 4 x3 7 x + 14 = 0.86Section 1.6Instructors Resource Manual 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 87. 57. Suppose that f is continuous at c, so lim f ( x) = f (c). Let x = c + t, so t = x c, then x cas x c , t 0 and the statement lim f ( x) = f (c) becomes lim f (t + c ) = f (c). x ct 0Suppose that lim f (t + c) = f (c) and let x = t + t 0c, so t = x c. Since c is fixed, t 0 means that x c and the statement lim f (t + c) = f (c) t 0becomes lim f ( x) = f (c) , so f is continuous at x cc. 58. Since f(x) is continuous at c, lim f ( x) = f (c) > 0. Choose = f ( c ) , then x cthere exists a > 0 such that 0 < x c < f ( x) f (c) < . Thus, f ( x ) f ( c ) > = f ( c ) , or f ( x ) > 0 . Since also f ( c ) > 0 , f ( x ) > 0 for all x in (c , c + ).59. Let g(x) = x f(x). Then, g(0) = 0 f(0) = f(0) 0 and g(1) = 1 f(1) 0 since 0 f(x) 1 on [0, 1] . If g(0) = 0, then f(0) = 0 and c = 0 is a fixed point of f. If g(1) = 0, then f(1) = 1 and c = 1 is a fixed point of f. If neither g(0) = 0 nor g(1) = 0, then g(0) < 0 and g(1) > 0 so there is some c in [0, 1] such that g(c) = 0. If g(c) = 0 then c f(c) = 0 or f(c) = c and c is a fixed point of f. 60. F