Calculus One and Several Variables 10E Salas Solutions Manual

Click here to load reader

  • date post

    28-Oct-2014
  • Category

    Documents

  • view

    3.145
  • download

    407

Embed Size (px)

description

Calculus One and Several Variables 10E Salas Solutions Manual

Transcript of Calculus One and Several Variables 10E Salas Solutions Manual

P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:52SECTION1.2 1CHAPTER1SECTION1.21. rational 2. rational 3. rational4. irrational 5. rational 6. irrational7. rational 8. rational 9. rational10. rational 11.34= 0.75 12. 0.33 1.414 14. 4 =16 15. 27< 0.28571416. 2Ans: (2, )5.12(1 +x) 0(x 5)(x + 1) > 0Ans: (, 1) (5, )11. 2x2+x 1 0(2x 1)(x + 1) 0Ans: [1, 1/2]12. 3x2+ 4x 4 0(3x 2)(x + 2) 0Ans: (, 2] [2/3, )13. x(x 1)(x 2) > 0Ans: (0, 1) (2, )14. x(2x 1)(3x 5) 0Ans: (, 0] [12,53]15. x32x2+x 0x(x 1)2 0Ans: [0, )P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:524 SECTION1.316. x24x + 4 0(x 2)2 0Ans: {2}17. x3(x 2)(x + 3)2< 0Ans: (0, 2)18. x2(x 3)(x + 4)2> 0Ans: (3, )19. x2(x 2)(x + 6) > 0Ans: (, 6) (2, )20. 7x(x 4)2< 0Ans: (, 0)21. (2, 2) 22. (, 1] [1, ) 23. (, 3) (3, )24. (0, 2) 25. (32,52) 26. (32,52)27. (1, 0) (0, 1) 28. (12, 0) (0,12) 29. (32, 2) (2,52)30. (32,12) (12,52) 31. (5, 3) (3, 11) 32. (23,83)33. (58, 38) 34. (12,710) 35. (, 4) (1, )36. (, 2) (43, ) 37. |x 0| < 3 or |x| < 3 38. |x 0| < 2 or |x| < 239. |x 2| < 5 40. |x 2| < 241. |x (2)| < 5 or |x + 2| < 5 42.x b +a2 0(c) (f/g)(x) =xx + 1x 2; x > 0, x = 214.(f +g)(x) =1 x, x 12x 1, 1 < x < 22x 2, x 2(f g)(x) =1 x, x 12x 1, 1 < x < 22x, x 2(f g)(x) =_0, x < 21 2x, x 2P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:5222 SECTION1.715. 16.17. 18. y = 0 on (0, c]19. 20.21. 22.23. (f g)(x) = 2x2+ 5; dom(f g) = (, ) 24. (f g)(x) = (2x + 5)2; dom (f g) = (, )25. (f g)(x) =x2+ 5; dom(f g) = (, ) 26. (f g)(x) = x +x; dom (f g) = [0, )P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:52SECTION1.7 2327. (f g)(x) =xx 2; dom(f g) = {x : x = 0, 2}28. (f g)(x) =1x21; dom(f g) = {x = 1}29. (f g)(x) =1 cos22x = | sin 2x|; dom(f g) = (, )30. (f g)(x) =1 2 cos x; dom (f g) = [0, /3] [5/3, 2]31. (f g h) = 4 [g(h(x))] = 4 [ h(x) 1 ] = 4(x21); dom(f g h) = (, )32. (f g h)(x) = f(g(h(x))) = f(g(x2)) = f(4x2) = 4x21; dom(f g h) = (, )33. (f g h)1g(h(x))=11/[2h(x) + 1]= 2h(x) + 1 = 2x2+ 1; dom(f g h) = (, )34. (f g h)(x) = f(g(h(x))) = f(g(x2)) = f_12x2+ 1_ =1/(2x2+ 1) + 11/(2x2+ 1)= 1 + (2x2+ 1) = 2x2+ 235. Take f(x) =1xsince1 +x41 +x2= F(x) = f(g(x)) = f_1 +x21 +x4_.36. Takef(x) = ax +b since f(g(x)) = f(x2) = ax2+b = F(x)37. Take f(x) = 2 sin x since 2 sin 3x = F(x) = f(g(x)) = f(3x).38. Takef(x) =a2x, since f(g(x)) = f(x2) = _a2(x2) =a2+x2= F(x).39. Take g(x) =_1 1x4_2/3since_1 1x4_2= F(x) = f(g(x)) = [ g(x)]3.40. Takeg(x) = a2x2(x = 0), since a2x2+1a2x2= F(x) = f(g(x)) = g(x) +1g(x).41. Take g(x) = 2x31 (or (2x31)) since (2x31)2+ 1 = F(x) = f(g(x)) = [ g(x)]2+ 1.42. Takeg(x) =1xsince sin 1x= F(x) = f(g(x)) = sin(g(x)).43. (f g)(x) = f(g(x)) = _g(x) =x2= |x|;(g f)(x) = g(f(x)) = [f(x)]2= [x]2= x, x 044. (f g)(x) = f(g(x)) = 3g(x) + 1 = 3x2+ 1, (g f)(x) = g(f(x)) = (f(x))2= (3x + 1)245. (f g)(x) = f(g(x)) = 1 sin2x = cos2x; (g f)(x) = g(f(x)) = sin f(x) = sin(1 x2).P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:5224 SECTION1.746. (f g)(x) = f(g(x)) = (x 1) + 1 = x; (g f)(x) = g(f(x)) =3_(x3+ 1) 1 = x.47. (f +g)(x) = f(x) +g(x) = f(x) +c; quadg(x) = c.48. (f g)(x) = f(g(x)) = g(x) +c implies f(x) = x +c.49. (fg)(x) = f(x)g(x) = c f(x) implies g(x) = c.50. (f g)(x) = f(g(x)) = c g(x) implies f(x) = cx.51. (a) The graph of g is the graph of f shifted 3 units to the right. dom(g) = [3, a + 3], range (g) =[0, b].(b) The graph of g is the graph of f shifted 4 units to the left and scaled vertically by a factor of3. dom(g) = [4, a 4], range (g) = [0, 3b].(c) Thegraphof g isthegraphof f scaledhorizontallybyafactorof 2. dom(g) = [0, a/2],range (g) = [0, b].(d) Thegraphof g is thegraphof f scaledhorizontallybyafactor of12. dom(g) = [0, 2a],range (g) = [0, b].52. even: (fg)(x) = f(x)g(x) = (f(x))(g(x)) = f(x)g(x) = (fg)(x).53. fg is even since (fg)(x) = f(x)g(x) = f(x)g(x) = (fg)(x).54. odd: (fg)(x) = f(x)g(x) = (f(x))(g(x)) = f(x)g(x) = (fg)(x).55. (a) Iffis even, thenf(x) =_x, 1 x < 01, x < 1.(b) Iffis odd, thenf(x) =_x, 1 x < 01, x < 1.56. (a) f(x) = x2+x, (b) f(x) = x2x57. g(x) = f(x) +f[(x)] = f(x) +f(x) = g(x)58. h(x) = f(x) f[(x)] = f(x) f(x) = [f(x) f(x)] = h(x)59. f(x) =12[f(x) +f(x)]. .even+ 12[f(x) f(x)]. .oddP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:52SECTION1.8 2560.61. (a) (f g)(x) =5x2+ 16x 16(2 x)2(b) (g k)(x) = x (c) (f k g)(x) = x2462. (a) (g f)(x) =3(x24)6 x2(b) (k g)(x) = x (c) (g f k)(x) = 18(2x + 3)x2+ 18x + 2763. (a) For xeda, varying b varies they-coordinate of the vertex of the parabola.(b) For xedb, varyinga varies thex-coordinate of the vertex of the parabola(c) The graph of Fis the reection of the graph ofFin thex-axis.64. a =14, b = 491665. (a) For c > 0, the graph of cfis the graph of fscaled vertically by the factor c; for c < 0, the graphofcfis the graph offscaled vertically by the factor |c| and then reected in thex-axis.(b) For c > 1, the graph of f(cx) is the graph of fcompressed horizontally; for 0 < c < 1, the graphof f(cx) is the graph of fstretched horizontally; for 1 < c < 0, the graph of f(cx) is the graphof fstretched horizontally and reected in the y-axis; for c < 1, the graph of f(cx) is the graphof fcompressed horizontally and reected in they-axis.66. (a) The graph off(x c) is the graph off(x) shiftedc units to the right ifc > 0 and |c| units to theleft ifc < 0.(b) a changes the amplitude, b changes the period, c shifts the graph right or left |c/b| units.SECTION1.81. Let S be the set of integers for which the statement is true. Since 2(1) 21, S contains 1. Assume nowthatk S. This tells us that 2k 2k, and thus2(k + 1) = 2k + 2 2k+ 2 2k+ 2k= 2(2k) = 2k+1.(k 1)This placesk + 1 inS.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:5226 SECTION1.8We have shown that1 S and that k S implies k + 1 S.It follows thatScontains all the positive integers.2. Use 1 + 2(n + 1) = 1 + 2n + 2 3n+ 2 < 3n+ 3n= 2 3n< 3n+1.3. LetSbe the set of integers for which the statement is true. Since (1)(2) = 2 is divisible by 2, 1 S.Assume now thatk S. This tells us thatk(k + 1) is divisible by 2 and therefore(k + 1)(k + 2) = k(k + 1) + 2(k + 1)is also divisible by 2. This placesk + 1 S.We have shown that1 Sand that k Simplies k + 1 S.It follows thatScontains all the positive integers.4. Use 1 + 3 + 5 + + (2(n + 1) 1) = n2+ 2n + 1 = (n + 1)25. Use 12+ 22+ +k2+ (k + 1)2=16k(k + 1)(2k + 1) + (k + 1)2=16(k + 1)[k(2k + 1) + 6(k + 1)]=16(k + 1)(2k2+ 7k + 6)=16(k + 1)(k + 2)(2k + 3)=16(k + 1)[(k + 1) + 1][2(k + 1) + 1].6. Use13+ 23+ +n3+ (n + 1)3= (1 + 2 + +n)2+ (n + 1)3=_n(n + 1)2_2+ (n + 1)3(by example 1)=n4+ 6n3+ 13n2+ 12n + 44=_(n + 1)(n + 2)2_2= [1 + 2 + +n + (n + 1)]27. By Exercise 6 and Example 113+ 23+ + (n 1)3= [12(n 1)n]2=14(n 1)2n214n4.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:52SECTION1.8 278. By Exercise 5,12+ 22+ + (n 1)2=16(n 1)n(2n 1) 13n39. Use11 +12 +13 + +1n +1n + 1>n +1n + 1 +n_n + 1 nn + 1 n_ =n + 1.10. Use11 2 +12 3 +13 4 + +1(n + 1)(n + 2)=nn + 1 +1(n + 1)(n + 2)=n(n + 2) + 1(n + 1)(n + 2)=n + 1n + 211. LetSbe the set of integers for which the statement is true. Since32(1)+1+ 21+2= 27 + 8 = 35is divisible by 7, we see that 1 S.Assume now thatk S. This tells us that32k+1+ 2k+2is divisible by 7.It follows that32(k+1)+1+ 2(k+1)+2= 32 32k+1+ 2 2k+2= 9 32k+1+ 2 2k+2= 7 32k+1+ 2(32k+1+ 2k+2)is also divisible by 7. This placesk + 1 S.We have shown that1 S and that k S implies k + 1 S.It follows thatScontains all the positive integers.12. n 1 : True forn = 1. For the induction step, use9n+18(n + 1) 1 = 9 9n8n 9 64n + 64n = 9(9n8n 1) + 64n13. For all positive integersn 2,_1 12__1 13_ _1 1n_ =1n.To see this, let S be the set of integers n for which the formula holds. Since 1 12=12, 2 S. Supposenow thatk S. This tells us that_1 12__1 13_ _1 1k_ =1kP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:5228 REVIEWEXERCISESand therefore that_1 12__1 13_ _1 1k__1 1k + 1_ =1k_1 1k + 1_ =1k_kk + 1_ =1k + 1.This placesk + 1 Sand veries the formula forn 2.14. The product isn + 12n; usen + 12n_1 1(n + 1)2_ =n + 12n_n2+ 2n(n + 1)2_ =n + 22(n + 1)15. Fromthegure,observethataddingavertexVN+1toanN-sidedpolygonincreasesthenumberofdiagonals by (N 2) + 1 = N 1. Then use the identity12N(N 3) + (N 1) =12(N + 1)(N + 1 3).16. From the gure for Exercise 15, observe that adding a vertex (VN+1) to anN-sided polygon increasesthe angle sum by 180.17. To go fromk tok + 1, takeA = {a1, . . . , ak+1} andB = {a1, . . . , ak} . Assume thatB has 2ksubsets:B1, B2, . . . B2k. The subsets ofA are thenB1, B2, . . . , B2ktogether withB1 {ak+1} ,B2 {ak+1} , . . . , B2k {ak+1} .This gives 2(2k) = 2k+1subsets forA.18. Assumingthatwecanconstructalinesegmentoflengthk, constructarighttrianglewithsidelengths 1 and k. Then the hypotenuse is a line segment of length k + 1.19. n = 41CHAPTER1.REVIEWEXERCISES1. rational 2. rational 3. irrational4. rational 5. bounded below by 1 6. bounded above by 17. bounded; lower bound 5, upper bound 1 8. bounded; lower bound 1, upper bound149. 2x2+x 1 = (2x 1)(x + 1); x =12, 110. no real rootsP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:52REVIEWEXERCISES 2911. x210x + 25 = (x 5)2; x = 512. 9x3x = x(3x + 1)(3x 1); x = 0,13, 1313. 5x 2 < 05x < 2x 22x + 4> 2 or2x + 4< 2If2x + 4> 2x + 4 > 0 and 2 > 2x + 84 < x < 3If2x + 4< 2x + 4 < 0 and 2 > 2x 85 < x < 4Ans: (5, 4) (4, 3)22. 5x + 1 < 11 5x + 1> 1x < 6Ans: (, 6) (4, )23. d(P, Q) = _(1 2)2+ (4 (3))2= 52; midpoint:_2 + 12,4 32_ =_32,12_24.