Calculus One and Several Variables 10E Salas Solutions Manual

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P1: PBU/OVY P2: PBU/OVY QC: PBU/OVY T1: PBU JWDD027-01 JWDD027-Salas-v1 November 25, 2006 15:52 SECTION 1.2 1 CHAPTER 1 SECTION 1.2 1. rational 2. rational 3. rational 4. irrational 5. rational 6. irrational 7. rational 8. rational 9. rational 10. rational 11. 3 4 =0.75 12. 0.33 < 1 3 13. 2 > 1.414 14. 4= 16 15. 2 7 < 0.285714 16. π< 22 7 17. |6| =6 18. |− 4| =4 19. |− 3 7| = 10 20. |− 5|−|8| = 3 21. |− 5| + |− 8| = 13 22. |2 π| = π 2 23. |5 5| =5 5 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. bounded, lower bound 0, upper bound 4 42. bounded above by 0 43. not bounded 44. bounded above by 4 45. not bounded 46. bounded; lower bound 0, upper bound 1 47. bounded above, upper bound 2 48. 2 < 3 π< 2 π 3 < 3 π 49. x 0 =2,x 1 =2.75,x 2 =2.58264,x 3 =2.57133,x 4 =2.57128,x 5 =2.57128; bounded; lower bound 2, upper bound 3 (the smallest upper bound =2.57128 ···); x n =2.5712815907 (10 decimal places)
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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. d(P, Q) = _(1 (3)2+ (6 (4))2= 226; midpoint:_3 12, 4 + 62_ = (2, 1)25. x = 2 26. y = 327. Theline l : 2x 3y = 6 hasslope m = 2/3. Therefore, anequationforthelinethrough(2, 3)perpendicular to l is: y + 3 = 32 (x 2) or 3x + 2y = 0P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:5230 REVIEWEXERCISES28. The line l : 3x + 4y = 12 has slope m = 3/4. Therefore, an equation for the line through (2, 3)parallel to l is: y + 3 = 34 (x 2) or 3x + 4y = 629.x 2y = 43x + 4y = 35x = 5 x = 1; (1,32).30.4x y = 23x + 2y = 011x = 4 x =411; (411,611).31. Solve the equations simultaneously:2x2= 8x 62x28x + 6 = 02(x 1)(x 3) = 0the line and the parabola intersect at (1, 2) and (3, 18).32. Thelinetangenttothecircleatthepoint(2, 1)isperpendiculartotheradiusatthatpoint. Thecenter of the circle is at (1, 3). The slope of the line through (1, 3) and (2, 1) is 2/3. Therefore anequation for the line tangent to the circle at (2, 1) isy 1 =32(x 2) or 3x 2y = 4.33. domain: (, ); range: (, 4] 34. domain: (, ); range: (, )35. domain: [4, ); range: [0, ) 36. domain: [12,12]; range; [0,12]37. domain: (, ); range: [1, ) 38. domain: (, ); range: [0, )39. domain: (, ); range: [0, ) 40. domain: (, ); range: (, )1 2 3 4x12345y-2 -1 1 2x-11234yP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:52REVIEWEXERCISES 3141. x =76,116 42. x =13,23,43,5343. x =3244. x = 0,13,23, ,43,53, 245. 46.-6 -3 3 6x-11y-6 -3 3 6x-11y47. 48.-6 -3 3 6x-3-113y-6 -3 3 6x-11y49. (f +g)(x) = (3x + 2) + (x21) = x2+ 3x + 1, dom(f +g) = (, ).(f g)(x) = (3x + 2) (x21) = 3 + 3x x2, dom(f g) = (, ).(f g)(x) = (3x + 2)(x21) = 3x3+ 2x23x 2, dom(f g) = (, )._fg_(x) =3x + 2x21, dom(f/g) = (, 1) (1, 1) (1, ).50. (f +g)(x) = x24 +x + 1/x = x2+x + 1/x 4, dom(f +g) = (, 0) (0, ).(f g)(x) = x2x 1/x 4, dom(f g) = (, 0) (0, ).(f g)(x) = (x24)(x + 1/x) = x33x 4/x, dom(f g) = (, 0) (0, )._fg_(x) =x24x + 1/x=x34xx2+ 1, dom(f/g) = (, 0) (0, ).51. (f +g)(x) = cos2x + sin 2x, dom(f +g) = [0, 2].(f g)(x) = cos2x sin 2x, dom(f g) = [0, 2].(f g)(x) = cos2x(sin 2x) = 2 cos3xsin x, dom(f g) = [0, 2]._fg_(x) =cos2xsin 2x=12cotx, dom(f/g) : x (0, 2), x =12, ,32.52. (f g)(x) = (x + 1)22(x + 1) = x21, dom(f g) = (, ).(g f)(x) = x22x + 1 = (x 1)2, dom(f g) = (, ).53. (f g)(x) = _(x25) + 1 =x24, dom(f g) = (, 2] [2, ).(g f)(x) = _x + 1_25 = x 4, dom(g f) = [1, ).54. (f g)(x) =_1 sin22x = | cos 2x|, dom(f g) = (, ).(g f)(x) = sin 21 x2, dom(g f) = [1, 1].P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-01 JWDD027-Salas-v1 November25,2006 15:5232 REVIEWEXERCISES55. (a) y = kx.(b) Ifb = ka, then b = ka = k(b). Hence,Q is a point onl.(c) If > 0,P, Q are on the same side of the origin; if < 0,P, Q are on opposite sides of the origin.56. (a) Set = a2. Thena = 2 and the quadratic equation can be written asx22x +b = 0 or (x )2(2b) = 0.if 2b > 0, set 2= (2b) and we have (x )22= 0;if 2b = 0, we have (x )2= 0;if 2b < 0, set 2= (2b) and we have (x )2+2= 0.(b) x = +orx = .(c) x = .(d) (x )2+2> 0 for all x.57. Since |a| = |a b +b| |a b| +|b| by the given inequality, we have |a| |b| |a b|.58. (a) P=12D(b) A =18D2P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53SECTION2.1 33CHAPTER2SECTION2.11. (a) 2 (b) 1 (c) does not exist (d) 32. (a) 4 (b) 4 (c) 4 (d)23. (a) does not exist (b) 3 (c) does not exist (d) 34. (a) 1 (b)does not exist (c) does not exist (d)15. (a) does not exist (b)does not exist (c) does not exist (d)16. (a) 1 (b) 2 (c) does not exist (d) 27. (a) 2 (b)2 (c) 2 (d) 18. (a) 2 (b)2 (c) 2 (d)29. (a) 0 (b)0 (c) 0 (d)010. (a) does not exist (b)does not exist (c) does not exist (d)111. c = 0, 6 12. c = 3 13. 1 14. 315. 12 16. 5 17. 1 18. does not exist19.3220. does not exist 21. does not exist 22. does not exist23. limx32x 6x 3= limx32 = 224. limx3x26x + 9x 3= limx3(x 3)2x 3= limx3(x 3) = 025. limx3x 3x26x + 9= limx3x 3(x 3)2= limx31x 3; does not exist26. limx2x23x + 2x 2= limx2(x 1)(x 2)x 2= limx2(x 1) = 127. limx2x 2x23x + 2= limx2x 2(x 1)(x 2)= limx21x 1= 128. does not exist 29. does not exist30. limx1_x + 1x_ = 2 31. limx02x 5x2x= limx0(2 5x) = 2P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5334 SECTION2.132. limx3x 36 2x= limx3x 32(3 x)= limx3_12_ = 1233. limx1x21x 1= limx1(x 1)(x + 1)x 1= limx1(x + 1) = 234. limx1x31x 1= limx1(x 1)(x2+ x + 1)(x 1)= limx1x2+ x + 1 = 335. 0 36. does not exist 37. 138. 3 39. 16 40. 041. does not exist 42. 2 43. 444. 0 45. does not exist 46. 247.limx1x2+ 1 2x 1=limx1(x2+ 1 2 )(x2+ 1 +2 )(x 1)(x2+ 1 +2 )=limx1x21(x 1)(x2+ 1 +2 )= limx1x + 1x2+ 1 +2=222=1248.limx5x2+ 5 30x 5=limx5(x2+ 5 30 )(x2+ 5 +30 )(x 5)(x2+ 5 +30 )=limx5x225(x 5)(x2+ 5 +30 )= limx5x + 5x2+ 5 +30=10230=53049.limx1x212x + 2 2=limx1x212x + 2 22x + 2 + 22x + 2 + 2=limx1(x 1)(x + 1)_2x + 2 + 2_2x + 2 4= limx1(x 1)(x + 1)_2x + 2 + 2_2(x 1)=limx1(x + 1)_2x + 2 + 2_2=2 42= 450. 0 51. 2 52. 0.16753.3254.11255. (i) 5 (ii) does not exist56. (i) 0 (ii)2357. (i)54(ii) 0 58. (i)1732(ii)1259. c = 1 60. c =5261. c = 2P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53SECTION2.2 3562. c = 3 63. limx0f(x) = 1;limx0g(x) = 064. limx0f(x) = 1;limx0g(x) = 0SECTION2.21.122. limx0x2(1 + x)2x= limx012 x(1 + x) = 03. limx0x(1 + x)2x2= limx01 + x2x; does not exist4.435. limx1x41x 1= limx1(x3+ x2+ x + 1) = 46. does not exist 7. does not exist8. limx1x21x22x + 1= limx1(x 1)(x + 1)(x 1)2= limx1x + 1x 1; does not exist9. 1 10. does not exist 11. does not exist12. 1 13. 0 14. 015. limx2+f(x) =limx2+(x2x) = 2 16. limx1f(x) =limx11 = 117. 1 18. 9 19. 1 20. 1021. 122. 2and 323. =12 = .05 24. =15= 0.125. 2 = .02 26. = 5= 0.5 27. = 1.75 28. = .2247429. for = 0.5 take = 0.24; for = 0.25 take = 0.130. for = 0.5 take = 0.06; for = 0.25 take = 0.0331. for = 0.5 take = 0.75; for = 0.25 take = 0.4332. for = 0.5 take = 0.125; for = 0.1 take = 0.02833. for = 0.25 take = 0.23; for = 0.1 take = 0.1434. for = 0.5 take = 0.25; for = 0.1 take = 0.0535. Since|(2x 5) 3| = |2x 8| = 2|x 4|,we can take =12:if 0 < |x 4| 0 such that() if c < x < c + then |f(x) L| < .Suppose now that 0 < |h| < . Thenc < c +|h| < c + and, by (),|f(c +|h|) L| < .Thus limh0f(c +|h|) = L.Conversely we now assume thatlimh0f(c +|h|) = L. Then for > 0 there exists> 0 such that() if 0 < |h| < then |f(c +|h|) L| < .Suppose now thatc < x < c + . Then0 < x c < so that, by (),|f(c + (x c)) L| = |f(x) L| < .Thus limxc+f(x) = L.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5340 SECTION2.258. Suppose that limxcf(x) = L and let> 0. Then there exists> 0 such thatif 0 < |x c| < , then |f(x) L| 0. Then there exists> 0 such thatif 0 < |x c| < , then |f(x) L 0| 0 such that if 0 < |x c| < thenL f(x) |L f(x)| = |f(x) L| < LTherefore, f(x) > L L = 0 for all x (c , c + ); take = .(b) Let= L and repeat the argument in part (a).60. Counterexample: Setf(x) =_x ifx = 11 ifx = 1.Then f(1) = 1 > 0 and limx1f(x) =limx1x = 1. By Exercise 51 (b), f(x) < 0for all x = 1in an interval of the form(1 , 1 + ), > 0.61. (a) Let limxcf(x) = L and limxcg(x) = M, and let> 0. There exist positive numbers 1 and 2such that|f(x) L| < /2 if 0 < |x c| < 1and|g(x) M| < /2 if 0 < |x c| < 2Let = min (1, 2). ThenM L = M g(x) + g(x) f(x) + f(x) L [M g(x)] + [f(x) L] /2 /2 = for allx such that 0 < |x c| < . Since is arbitrary, it follows that M L.(b) No. For example, if f(x) = x2and g(x) = |x| on (1, 1), then f(x) < g(x) on (1, 1) exceptat x = 0, butlimx0x2= limx0|x| = 0.62. Let= 1. Since limxcf(x) = L, there exists a > 0 such that if 0 < |x c| < , then|f(x) L| < = 1. Thus,1 < f(x) L < 1 or L 1 < f(x) < L + 1Take B = max {|L 1|, |L + 1| }. Then|f(x)| < B for all x such that 0 < |x c| < .P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53SECTION2.3 41SECTION2.31. (a) 3 (b) 4 (c) 2 (d) 0 (e) does not exist (f)132. (a) 5 (b) 8 (c) does not exist (d) 0 (e) 4/5 (f) 93. limx4_1x 14__1x 4_ = limx4_4 x4x__1x 4_ = limx414x= 116; Theorem 2.3.2 does not applysincelimx41x 4does not exist.4. Theorem 2.3.10 does not apply since limx3(x2+ x 12) = 0.limx3x2+ x 12x 3= limx3(x + 4)(x 3)x 3= limx3(x + 4) = 75. 3 6. 49 7. 38. 9 9. 5 10. 211. does not exist 12. 232013. 114. 0 15. does not exist16. limh0h_1 1h_ = limh0(h 1) = 1 17. limh0h_1 +1h_ = limh0(h + 1) = 118. limx2x 2x24= limx21x + 2=1419. limx2x24x 2= limx2x + 21= 420. limx2(x2x 6)2x + 2=limx2(x 3)2(x + 2)2x + 2=limx2(x + 3)2(x + 2) = 021. limx4x 2x 4= limx4x 2x 4x + 2x + 2= limx4x 4(x 4)(x + 2)=1422. limx1x 1x 1= limx1(x 1)(x + 1)x 1= limx1(x + 1) = 223. limx1x2x 6(x + 2)2= limx1(x + 2)(x 3)(x + 2)2= limx1x 3x + 2= 2324. limx2(x2x 6)(x + 2)2=limx2(x 3)(x + 2)(x + 2)2=limx2x 3x + 2; does not exist25. limh01 1/h21 1/h= limh0h21h2h= limh0(h + 1)(h 1)h(h 1)= limh0h + 1h; does not exist26. limh01 1/h21 + 1/h2= limh0h21h2+ 1= 1P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5342 SECTION2.327. limh01 1/h1 + 1/h= limh0h 1h + 1= 128. limh01 + 1/h1 + 1/h2= limh0h2+ hh2+ 1= 029. limt1t2+ 6t + 5t2+ 3t + 2=limt1(t + 1)(t + 5)(t + 1)(t + 2)=limt1t + 5t + 2= 430. limx2+x24x 2=limx2+x + 2x 2x 2=limx2+x + 2x 2; does not exist31. limt0t + a/tt + b/t= limt0t2+ at2+ b=ab32. limx1x21x31= limx1(x 1)(x + 1)(x 1)(x2+ x + 1)= limx1x + 1x2+ x + 1=2333. limx1x51x41= limx1(x 1)(x4+ x3+ x2+ x + 1)(x 1)(x3+ x2+ x + 1)= limx1x4+ x3+ x2+ x + 1x3+ x2+ x + 1=5434. limh0h2_1 +1h_ = limh0(h2+ h) = 035. limh0h_1 +1h2_ = limh0h2+ 1h; does not exist36. limx4_3xx + 4 +8x + 4_ =limx43x + 8x + 4 ; does not exist37. limx4_2xx + 4 +8x + 4_ =limx42x + 8x + 4=limx42 = 238. limx4_2xx + 4 8x + 4_ =limx42x 8x + 4 ; does not exist39. (a) limx4_1x 14_ = limx44 x4x= 0(b) limx4__1x 14__1x 4__ = limx4__4 x4x__1x 4__ = limx4_14x_ = 116(c) limx4__1x 14_(x 2)_ = limx4(4 x)(x 2)4x= 0(d) limx4__1x 14__1x 4_2_ = limx44 x4x(x 4)2= limx414x(4 x); does not exist40. (a) limx3x2+ x + 12x 3; does not exist(b) limx3x2+ x 12x 3= limx3(x + 4)(x 3)x 3= limx3(x + 4) = 7P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53SECTION2.3 43(c) limx3(x2+ x 12)2x 3= limx3(x + 4)2(x 3)2x 3= limx3(x + 4)2(x 3) = 0(d) limx3x2+ x 12(x 3)2= limx3(x + 4)(x 3)(x 3)2= limx3x + 4x 3; does not exist41. (a) limx4f(x) f(4)x 4= limx4(x24x) (0)x 4= limx4x = 4(b) limx1f(x) f(1)x 1= limx1x24x + 3x 1= limx1(x 1)(x 3)x 1= limx1(x 3) = 2(c) limx3f(x) f(1)x 3= limx3x24x + 3x 3= limx3(x 1)(x 3)x 3= limx3(x 1) = 2(d) limx3f(x) f(2)x 3= limx3x24x + 4x 3; does not exist42. (a) limx3f(x) f(3)x 3= limx3x327x 3= limx3(x2+ 3x + 9) = 27(b) limx3f(x) f(2)x 3= limx3x38x 3= limx3(x 2)(x2+ 2x + 4)x 3; does not exist(c) limx3f(x) f(3)x 2= limx3x327x 2= 0(d) limx1f(x) f(1)x 1= limx1x31x 1= limx1(x2+ x + 1) = 343. f(x) = 1/x, g(x) = 1/x withc = 044. Set, for instance,f(x) =_0, x < c1, x > c,g(x) =_1, x < c0, x > c45. True. Let limxc[f(x) + g(x)] = L. If limxcg(x) = Mexists, then limxcf(x) = limxc[f(x) + g(x) g(x)] =L Malso exists. This contradicts the fact that limxcf(x) does not exist.46. False, because limxcg(x) = limxc[f(x) + g(x) f(x)] = limxc[f(x) + g(x)] limxcf(x). exists.47. True. If limxc_f(x) = L exists, then limxc_f(x)_f(x) = L2also exists.48. False. Set f(x) = 1 for all x, c = 0.49. False; for example setf(x) = x and c = 050. False; for example, neither limit need exist: setf(x) =1(x 3)2and g(x) =2(x 3)2.51. False; for example, setf(x) = 1 x2, g(x) = 1 + x2, andc = 0.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5344 SECTION2.352. (a) If f(x) g(x) then|f(x) g(x)| = f(x) g(x) and12 {[f(x) + g(x)] +|f(x) g(x)|} =12 {f(x) + g(x) + f(x) g(x)}=12 2f(x) = f(x) = max {f(x),g(x)}.If f(x) g(x) then|f(x) g(x)| = [f(x) g(x)] = g(x) f(x) and12 {[f(x) + g(x)] +|f(x) g(x)|} =12 {f(x) + g(x) + g(x) f(x)}=12 2g(x) = g(x) = max {f(x),g(x)}.(b) min {f(x),g(x)} =12 {[f(x) + g(x)] |f(x) g(x)|}53. If limxcf(x) = L and limxcg(x) = L, thenlimxch(x) = limxc12{[f(x) + g(x)] |f(x) g(x)|}= limxc12[f(x) + g(x)] limxc|f(x) g(x)|=12(L + L) 12(L L) = L.A similar argument works forH.54. (a) Let> 0. Since limxcf(x) = L, there exists 1> 0 such thatif 0 < |x c| < 1then () |f(x) L| < Let 2 = min {|xic| : 1 i n, xi = c}. Then() f(x) = g(x) if 0 < |x c| < 2Now, choose = min {1,2}. Then, if 0 < |x c| < |f(x) L| 0 such that|f(x) f(c)| < f(c) for allx (c , c + ).This implies that f(x) > 0 for all x (c , c + ).(b) Apply the result in part (a) to h(x) = f(x).(c) Apply the result in part (a) to h(x) = g(x) f(x).51. Suppose that g does not have a non-removable discontinuity at c. Then either g is continuous at c or ithas a removable discontinuity atc. In either case, limg(x) asx c exists. Since g(x) = f(x) exceptat a nite set of pointsx1, x2, . . . , xn, limf(x) exists asx c by Exercise 54, Section 2.3.52. (a) Choose any point c and let> 0. Since f is continuous at c, there exists > 0 such thatif |x c| < then |f(x) f(c)| < .Now, since| |f(x)| |f(c)| | |f(x) f(c)|,it follows that| |f(x)| |f(c)| | 0.Let > 0andlet = min{/B,p}. If |x c| < thenx (c p,c + p) and|f(x) f(c)| B|x c| < B B

B= Thus,fis continuous atc.54. Choose any point c (a, b), and let> 0. Now choosing = , we haveif |x c| < , x (a, b), then |f(x) f(c)| |x c| < = .Thus, f is continuous at c, and it follows that f is continuous on (a, b.)55. limh0[f(c + h) f(c)] = limh0_f(c + h) f(c)h h_ = limh0_f(c + h) f(c)h_ limh0h = L 0 = 0.Thereforefis continuous atc by Exercise 47.56. Let fe(x) =12 [f(x) + f(x)] and fo(x) =12 [f(x) f(x)]. Then feis an even function, foisan odd function, each function is continuous on (, ), and f= fe + fo.57.5258.1259. limx0f(x) = 1; limx0+f(x) = 1; fis discontinuous for allk.60. 2SECTION2.51. limx0sin 3xx= limx03_sin 3x3x_ = 3(1) = 3 2. limx02xsin x= 2 limx0sin xx= 2(1) = 23. limx03xsin 5x= limx035_5xsin 5x_ =35(1) =354. limx0sin 3x2x=32limx0sin 3x3x=325. limx0sin 4xsin 2x= limx04x2x sin 4x4x2xsin 2x= 2(1)(1) = 26. limx0sin 3x5x=35limx0sin 3x3x=35P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53SECTION2.5 517. limx0sin x2x= limx0x_sin x2x2_ = limx0x limx0sin x2x2= 0(1) = 08. limx0sin x2x2= 1 9. limx0sin xx2= limx0(sin x)/xx; does not exist10. limx0sin2x2x2= limx0_sin x2__sin x2x2_ = 0(1) = 011. limx0sin23x5x2= limx095_sin 3x3x_2=95(1) =9512. limx0tan23x4x2= limx094 1cos23x sin23x(3x)2=94(1)(1) =9413. limx02xtan 3x= limx02xcos 3xsin 3x= limx023_3xsin 3x_cos 3x =23(1)(1) =2314. limx04xcot 3x= limx04x sin 3xcos 3x= 0 15. limx0xcsc x = limx0xsin x= 116. limx0cos x 12x= 12limx01 cos xx= 017. limx0x21 cos 2x= limx0x21 cos 2x _1 + cos 2x1 + cos 2x_ = limx0x2(1 + cos 2x)sin22x= limx014_2xsin 2x_2(1 + cos 2x) =14(1)(2) =1218. limx0x22xsin 3x= limx0_x 23__3xsin 3x_ =_23_(1) = 2319. limx01 sec22xx2= limx0tan22xx2= limx0sin22xx2cos22x= limx0_4_sin 2x2x_21cos22x_ = 420. limx012x csc x=12limx0sin xx=1221. limx02x2+ xsin x= limx0(2x + 1)xsin x= 122. limx01 cos 4x9x2= limx0_1 cos 4x9x2__1 + cos 4x1 + cos 4x_ = limx01 cos24x9x2(1 + cos 4x)= limx0sin24x9x2(1 + cos 4x)=169limx0sin24x(4x)211 + cos 4x=169 1 12=8923. limx0tan 3x2x2+ 5x= limx01x(2x + 5)sin 3xcos 3x= limx032x + 5_sin 3x3x_1cos 3x=35(1)(1) =3524. limx0x2(1 + cot23x) = limx0_x2+x2cos23xsin23x_ = limx0_x2+(3x)2sin23x cos23x9_ = 0 + (1)19=19P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5352 SECTION2.525. limx0sec x 1xsec x= limx01cos x 1x_1cos x_= limx01 cos xx= 026. limx/41 cos xx=1 cos(/4)/4=4 2227.2228. limx0sin2xx(1 cos x)= limx0x1 cos x sin2xx2; does not exist29. limx/2cos xx /2= limh0cos (h + /2)h= limh0sin hh= 1h = x /2 cos (h + /2) = cos hcos /2 sin hsin /230. limxsin xx =limx sin(x )x = 131. limx/4sin (x + /4) 1x /4= limh0sin (h + /2) 1h= limh0cos h 1h= 0h = x /432. limx/6sin[x + (/3)] 1x (/6)= limx/6sin[x (/6) + (/2)] 1x (/6)= limx/6cos[x (/6)] 1x (/6)= 033. Equivalently we will show thatlimh0cos (c + h) = cos c. The identitycos (c + h) = cos c cos h sin c sin hgiveslimh0cos (c + h) = cos c_limh0cos h_sin c_limh0sin h_ = (cos c)(1) (sin c)(0) = cos c.34.ab35. 0 36.ab37. 138. Let> 0. There exists 1> 0 such thatif 0 < |x| < 1, then |f(x) L| < .Take = 1/|a| :if 0 < |x| < , then 0 < |ax| = |a| |x| < 1and |f(ax) L| < .P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53SECTION2.5 5339. f(x) = sin x; a = /4limh0f(a + h) f(a)h= limh0sin_4+ h_sin_4_h= limh0sin (/4) cos h + cos (/4) sin h sin (/4)h= limh0sin (/4) (1 cos h) + cos (/4) sin hh= sin (/4)limh01 cos hh+ cos (/4)limh0sin hh= cos (/4) =22tangent line: y 22=22_x 4_40. f(x) = cos x; a = /3limh0f(a + h) f(a)h= limh0cos_3+ h_cos_3_h= limh0cos (/3) cos h sin (/3) sin h cos (/3)h= limh0cos (/3) (1 cos h) sin(/3) sin hh= cos (/3)limh01 cos hhsin (/3)limh0sin hh= sin (/3) = 32tangent line: y 12= 32_x 3_41. f(x) = cos 2x; a = /6limh0f(a + h) f(a)h= limh0cos 2_6+ h_cos_26_h= limh0cos (2h + /3) cos (/3)h= limh0cos (/3) cos 2h sin (/3) sin 2h cos (/3)h= cos (/3)limh02 1 cos 2hhsin (/3)limh02 sin 2hh= cos (/3) 2 0 sin (/3) 2 1 = 3tangent line: y 12= 3_x 6_42. f(x) = sin 3x; a = /2limh0f(a + h) f(a)h=limh0sin 3_2+ h_sin_32_h= limh0sin (3h + 3/2) sin (3/2)h=limh0sin (3/2) cos 3h + cos (3/2) cos 3h sin (3/2)h=limh01 cos 3hh= 3 limh01 cos 3h3h= 3 0 = 0tangent line: y (1) = 0 [x (3/2)] or y = 143. For x = 0, |x sin(1/x)| = |x| | sin(1/x)| |x|. Thus,|x| |x sin(1/x)| |x|Since limx0(|x|) = limx0|x| = 0, the result follows by the pinching theorem.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5354 SECTION2.544. Since 0 cos2[1/(x )] 1 for all x = , we have |(x ) cos2[1/(x )]| |x |. Thus,|x | |(x ) cos[1/(x )]| |x |Since limx(|x |) =limx|x | = 0, the result follows by the pinching theorem.45. Forx close to 1(radian), 0 < sin x 1. Thus,0 < |x 1| sin x |x 1|and the result follows by the pinching theorem.46. Clearly, |f(x)| 1 for all x. Therefore,|xf(x)| |x| which implies |x| xf(x) |x| for allx.The result follows by the pinching theorem.47. Suppose that there is a numberBsuch that |f(x)| Bfor allx = 0. Then |xf(x)| B|x| andB|x| xf(x) B|x|The result follows by the pinching theorem.48. We havef(x) = x f(x)xandf(x)x B, x = 0.Therefore, the result follows from Exercise 47.49. Suppose that there is a numberBsuch thatf(x) Lx c B for x = c. Then0 |f(x) L| =(x c) f(x) Lx c B|x c|By the pinching theorem, limxc|f(x) L| = 0 which implies limxcf(x) = L.50. Let > 0. Let> 0 be such that for | c| < p,if |x c| < then |f(x)| 0.By the intermediate-value theorem there is ac in [1, 2] such thatf(c) = 0.2. Set f(x) = x4x 1. Then f is continuous on [1, 1] and f(1) = 1 > 0, f(1) = 1 < 0.By the intermediate-value theorem there is ac in [1, 1] such thatf(c) = 0.3. Set f(x) = sin x + 2 cos x x2Then fis continuous on [0, 2] and f(0) = 2 > 0, f(2) = 1 24< 0.By the intermediate-value theorem there is ac in [0, 2] such thatf(c) = 0.4. Set f(x) = 2 tan x x. Then f is continuous on [0, 4] and f(0) = 0 < 1, f(4) = 2 4> 1.By the intermediate-value theorem there is ac in [0, 4] such thatf(c) = 1.5. Set f(x) = x22 +12x. Then f is continuous on [14, 1] and f(14) =116> 0, f(1) = 12< 0.By the intermediate-value theorem there is ac in [14, 1] such thatf(c) = 0.6. Set f(x) = x53+ x13 . Then f is continuous on [1, 1] and f(1) = 2 < 1, f(1) = 2 > 1.By the intermediate-value theorem there is ac in [1, 1] such thatf(c) = 1.7. Set f(x) = x3x + 2. Then f is continuous on [1, 2] and f(1) = 1 3 < 0, f(2) = 6 > 0.By the intermediate-value theorem there is ac in [1, 2] such thatf(c) = 0.i.e.c3=c + 2.8. Set f(x) =x23x 2. Thenf iscontinuouson[3, 5] andf(3) = 2 < 0, f(5) =10 + 2 > 0.By the intermediate-value theorem there is ac in [3, 5] such thatf(c) = 0.9. f(x) is continuous on [0, 1]; f(0) = 0 < 1 andf(1) = 4 > 1.By the intermediate value theorem there is ac in (0, 1) such thatf(c) = 1.10. f(x) is continuous on [2, 3]; f(2) =12> 0 andf(3) = 12< 0,By the intermediate value theorem there is ac in (2, 3) (hence in (1, 4)) such thatf(c) = 0.11. Setf(x) = x34x + 2. Thenf(x) is continuous on [3, 3].Checking the integer values on this interval,f(3) = 13 < 0, f(2) = 2 > 0, f(0) = 2 > 0, f(1) = 1 < 0, andf(2) = 2 > 0.By the intermediate value theorem there are roots in (3, 2), (0, 1) and (1, 2).12. Setf(x) = x2. Thenf(x) is continuous on [1, 2], f(1) = 1 < 2 andf(2) = 4 > 2.By the intermediate value theorem there is ac in (1, 2) such thatf(c) = 2.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5356 SECTION2.613. 14. 15. 16. Impossible17. 18. 19. 20.21. Impossible 22. 23. 24. Impossible25. If f(0) = 0or if f(1) = 1, we have a xed point. If f(0) = 0andf(1) = 1, then set g(x) = x f(x).giscontinuouson[ 0, 1 ] andg(0) < 0 < g(1). Thereforethereexistsanumber c (0, 1)suchthatg(c) = c f(c) = 0.26. Seth(x) = f(x) g(x). Then h is continuous on [a, b], andh(a) = f(a) g(a) < 0, h(b) = f(b) g(b) > 0. By the intermediate value theorem there exists a number c (a, b) such that h(c) = 0. Thus,f(c) = g(c).27. (a)If f(0) = 1 orif f(1) = 0, weredone.Suppose f(0) = 1 and f(1) = 0.Thediagonalfrom(0, 1) to (1, 0) has equation y = 1 x, 0 x 1. Set g(x) = (1 x) f(x). g is continuous on[0, 1] and g(0) = 1 f(0) > 0; g(1) = f(1) < 0. Thereforethereexistsanumber c (0, 1)suchthatg(c) = (c 1) f(c) = 0.(b) Suppose g(0) = 0 and g(1) = 1. Set h(x) = g(x) f(x); h is continuous on [0, 1]. If h(0) = 0orif h(1) = 0, weredone. Ifnot, then h(0) < 0 and h(1) > 0. Thereforethereexistsanumberc (0, 1) such that h(c) = 0. The same argument works for the other case.28. Letf(x) = x3and repeat the argument given in Exercise 37.29. The cubic polynomialP(x) = x3+ ax2+ bx + c is continuous on (, ).. WritingPasP(x) = x3_1 +ax +bx2+cx3_x = 0P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53SECTION2.6 57it follows that P(x) < 0 for large negative values ofx and P(x) > 0 for large positive values ofx.Thus there exists a negative numberNsuch that P(x) < 0 for x < N, and a positive numberMsuch that P(x) > 0 for x > M. By the intermediate-value theorem, Phas a zero in [N, M].30. (a) LetSbe the set of positive integers for which the statement is true. Then 1 Sby hypothesis. Now assume thatk S. Thenak< bkandak+1= (a)ak< (a)bk< (b)bk= bk+1.Therefore,k + 1 SandSis the set of positive integers.(b) Clearly0is theunique nthroot of 0. Chooseanypositivenumber xandlet f(t) = tnx.Sincef(0) = x < 0andf(t) ast , thereexistsanumberc > 0suchthatf(c) = 0.The numberc is annth root ofx. The uniqueness follows from part (a).31. ThefunctionTiscontinuouson[4000, 4500]]; and T(4000)= 98.0995, T(4500)= 97.9478. Thus,bytheintermediate-valuetheorem,thereisanelevationhbetween4000and4500meterssuchthatT(h) = 98.32. ThinkoftheequatorasbeingacircleandchooseareferencepointPandapositivedirection.Forexample, choose P to be0longitudeandlet eastwardbethepositivedirection. Usingradianmeasure, letx, 0 x 2denotethecoordinateof apointxradiansfromP. Then, xandx + arediametricallyoppositepointsontheequator. LetT(x)bethetemperatureatthepointx, andlet f(x) = T(x) T(x + ). If f(0) = 0, thenthetemperaturesatthepoints0andareequal. Iff(x) = 0, thenf(0) = T(0) T() andf() = T() T(2) = t() T(0) have opposite sign. Thus,there exists a pointc (0, ) at whichf(c) = 0, andT(c) = T(c + ).33. LetA(r)denotetheareaofacirclewithradiusr, r [0, 10]. Then A(r) = r2iscontinuouson[0, 10], and A(0) = 0 and A(10) = 100 = 314. Since 0 < 250 < 314 it follows from the intermediatevalue theorem that there exists a numberc (0, 10) such that A(c) = 250.34. Letxandybethedimensionsof arectanglein R. Then, 2x + 2y = Pandy =P2 x. TheareafunctionA(x) = xy = x_P2 x_ =P2x x2, x [0, P/2]is continuous. Therefore,A has a maximum value on [0, P/2]. SinceA(x) =P2x x2=P216 _x P4_2it is clear that the rectangle with maximum area has dimensionsx = y =P4 .35. Inscribe a rectangle in a circle of radius R. Introduce a coordinate system with the origin at the centerof the circle and the sides of the rectangle parallel to the coordinate axes. Then the area of the rectangleP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5358 SECTION2.6is given byA(x) = 4x_R2x2, x [0, R].SinceA is continuous on [0, R], A has a maximum value.36. f(0) = 4, f(1) = 2. Thus, fhas a zero in (0, 1) atr = 0.771.37. f(3) = 9, f(2) = 5; f(0) = 3, f(1) = 1; f(1) = 1, f(2) = 1 Thus, f hasazeroin(3, 2, ) in (0, 1) and in (1, 2).r1 = 2.4909, r2 = 0.6566, and r3 = 1.834338. f(2) = 25, f(1) = 1; f(0) = 1, f(1) = 1; f(1) = 1, f(2) = 27 Thus, f has a zero in(2, 1, ) in (0, 1) and in (1, 2).r1 = 1.3888, r2 = 0.3345, and r3 = 1.214639. f(2) = 5.6814, f(1) = 1.1829; f(0) = 0.5, f(1) = 0.1829; f(1) = 0.1829, f(2) = 6.681Thus,fhas a zero in (2, 1), in (0, 1) and in (1, 2).r1 = 1.3482, r2 = 0.2620, and r3 = 1.081640. fsatises the hypothesis of the intermediate-value theorem: c= 0.8341. fisnotcontinuousatx = 1. Thereforefdoesnotsatisfythehypothesisoftheintermediate-valuetheorem. However,f(3) + f(2)2=12= f(c) for c= 0.16342. fis not continuous atx = /2, /2, 3/2;12 [f() + f(2)] = 0; f(x) = 0 for allx [, 2].43. fsatises the hypothesis of the intermediate-value theorem:f(/2) + f(2)2=12= f(c) for c= 2.38, 4.16, 5.2544. fis bounded.max (f) = 6 [f(0) = 6]min (f)= 0.376 [f(1.46)= 0.376]P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53PROJECT2.6 5945. fis bounded.max (f) = 1 [f(1) = 1]min (f) = 1 [f(1) = 1]46. fis unbounded47. fis bounded.no maxfis not dened atx = 0min (f)= 0.3540PROJECT2.61. (a)2 12n< 0.001 = 2n> 1000 = n > 9.The minimum number of iterations required isn = 10.2 144910241.415034.(b)2 12n< 0.0001 = 2n> 10, 000 = n > 13.The minimum number of iterations required isn = 14.2 12n< 0.00001 = 2n> 100, 000 = n > 16.The minimum number of iterations required isn = 17.2. f(x) = x3+ x 9; f(1) = 7 andf(2) = 1. Therefore,f(c) = 0 for somec (1, 2).(a) A minimum of 7 iterations are required;c 2451281.9140625.Accurate to 7 decimal places, the root isc1.9201751.(b)2 12n< 0.001 = 2n> 1, 000 = n > 9.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5360 REVIEWEXERCISESThe minimum number of iterations required isn = 10.2 12n< 0.0001 = 2n> 10, 000 = n > 13.The minimum number of iterations required isn = 14.3. f(x) = sinx + x + 3; f(3) 0.142 and f(2) 0.091. Therefore, f(c) = 0 for some c (3, 2).(a) A minimum of 7 iterations are required;c 279128 2.17969.Accurate to 6 decimal places, the root isc2.179727.(b)2 12n< 0.00001 = 2n> 100, 000 = n > 16.The minimum number of iterations required isn = 17.2 12n< 0.000001 = 2n> 1, 000, 000 = n > 19.The minimum number of iterations required isn = 20.4. For f(x) = x22, the rst three iterations are:c1 =43, c2 =53 1.6667, c3 =32= 1.5For f(x) = x3+ x 9, the rst three iterations are:c1 =158= 1.875, c2 =3116= 1.9375, c3 =6132= 1.90625.For f(x) = sinx + x + 3, the rst three iterations are:c1 2.39056, c2 2.19528, c3 2.09764.CHAPTER2.REVIEWEXERCISES1. limx3x23x + 3=3233 + 3= 1.2. limx2x2+ 4x2+ 2x + 1=22+ 422+ 2(2) + 1=89.3. limx3(x 3)2x + 3=06= 0.4. limx3x29x25x + 6= limx3(x 3)(x + 3)(x 2)(x 3)= limx3x + 3x 2= 6.5. limx2+x 2|x 2|=limx2+x 2x 2= 1.6. limx2|x|x 2=limx2xx 2= 12.7. limx0_1x 1 xx_ = limx0xx= 1.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53REVIEWEXERCISES 618. limx3+x 3|x 3|=limx3+x 3x 3=limx3+1x 3; does not exist.9. limx1+|x 1|x=limx1+x 1x= 0 and limx1|x 1|x= limx1x + 1x= 0. limx1|x 1|x= 0.10. limx1+x31|x31|=limx1+x31x31= 1 and limx1x31|x31|= limx1x31x3+ 1= 1; does not exist.11. limx1x 1x 1= limx1x 1(x 1)(x + 1)= limx11x + 1=12.12. limx3+x22x 3x 3=limx3+(x 3)(x + 1)x 3=limx3+x 3(x + 1) = 0.13. limx3+x22x 3x 3=limx3+x + 1x 3, does not exist.14. limx4x + 5 3x 4= limx4(x + 5 3)(x + 5 + 3)(x 4)(x + 5 + 3)= limx4x 4(x 4)(x + 5 + 3)=16.15. limx2x38x43x24= limx2(x 2)(x2+ 2x + 4)(x 2)(x + 2)(x2+ 1)= limx2x2+ 2x + 4(x + 2)(x2+ 1)=35.16. limx05xsin 2x= limx0522xsin 2x=52.17. limx0tan22x3x2= limx043 cos22x_sin 2x2x_2=43.18. limx0xcsc 4x = limx04x4 sin 4x=14.19. limx0x23xtan x= limx0x(x 3) cos xsin x= 320. limx/2cos x2x = limx/2sin(/2 x)2(/2 x)= 12.21. limx0sin 3x5x24x= limx0sin 3x3x(53x 43)= 34.22. limx05x21 cos 2x= limx05x22 sin2x=52.23. limxx + sin x=limxx + sin(x + )= 1.24. limx2x24x38= limx2(x 2)(x + 2)(x 2)(x2+ 2x + 4)= limx2x + 2x2+ 2x + 4=13.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5362 REVIEWEXERCISES25. limx2x 2|x24|=limx2x 2(x 2)(x + 2)=limx21x + 2= 14.26. limx21 2/x1 4/x2= limx2x22xx24= limx2xx + 2=12.27. limx1+x23x + 2x 1=limx1+(x 2)(x 1)x 1=limx1+(x 2)x 1 = 0.28. limx31 9/x21 + 3/x=1 9/91 + 3/3= 0.29. limx2f(x) = 3(2) 22= 2.30. limx2+f(x) = (2)23 = 1 and limx2f(x) = 3 2 = 1. Hence limx2f(x) = 1.31. (a) False (b) False (c) True (d) False (e) True32. Since limx3x22x 3 = 0, we must have limx3(2x23ax + x a 1) = 0.limx3(2x23ax + x a 1) = 18 9a + 3 a 1 = 0; 10a + 20 = 0; a = 2.limx32x25x 3x22x 3= limx3(2x + 1)(x 3)(x + 1)(x 3)=74.33. (a)-3 -2 -1 1 2 3x-6-4-2246y(b) (i) 1 (ii) 0 (iii) does not exist (iv) 6 (v) 4 (vi) does not exist(c) (i)fis continuous from the left at 1; fis not continuous from the right at 1.(ii)fis not continuous from the left at 2; fis continuous from the right at 2.34. (a)limx0cos_1 cos x2x_ = cos_limx01 cos x2x_ = cos 0 = 1(b)limx0cos_ sin x2x_ = cos_limx0 sin x2x_ = cos12 = 0P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:53REVIEWEXERCISES 6335. Forfto be continuous at 2, we must havelimx2(2x21) = 7 = A =limx2+(x32Bx) = 8 4B.These equations imply that A = 7 andB =14.36. Forfto continuous atx = 1, we must havelimx1(Ax + B) = B A = 2 or A = B + 2.Forfto be discontinuous atx = 2, we must havelimx2+(2Bx A) = 4B A = f(2) = 4.A = B + 2 and 4B A = 4 implythat B = 2. Necessaryandsucientconditionsfor f tobecontinuous at x = 1 and discontinuous at x = 2 are: A = B + 2 with B = 2.37. limx3x22x 15x + 3=limx3(x 5)(x + 3)x + 3= 8; set f(3) = 8.38. limx3x + 1 2x 3= limx3(x + 1 2)(x + 1 + 2)(x 3)(x + 1 + 2)= limx3x 3(x 3)(x + 1 + 2)=14; set f(3) =14.39. limx0sin xx= limx0 sin xx= ; set f(0) = .40.1 cos xx2=1 cos xx2 1 + cosx1 + cosx=1 cos2xx211 + cosx=sin2xx211 + cosx.Therefore, limx01 cosxx2= limx0sin2xx211 + cosx=12; set f(0) =12.41. (a) False; need f continuous (b)True; intermediate-value theorem(c) False; need f continuous on [a, b] (d)False; consider f(x) = x2on [1.1].42. Setf(x) = x33x 4. Thenf(2) = 2 andf(3) = 14. By the intermediate-value theorem,f(x) haszero in [2, 3].43. Setf(x) = 2 cos x x + 1. Thenf(1) = 2 cos 1 > 0andf(2) = 2 cos 2 1 < 0. Bytheintermediate-value theorem,f(x) as zero in [1, 2].44. Let> 0.|(5x 4) 6| = |5x 10| = 5|x 2|.If 0 < |x 2| < = /5, then|(5x 4) 6| < . Therefore, limx2(5x 4) = 6.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-02 JWDD027-Salas-v1 November25,2006 15:5364 REVIEWEXERCISES45. If 0 < |x (4)| = |x + 4| < 1, then 2x + 5 < 0 and|2x + 5| = 2x 5.Let> 0.|2x + 5| 3 = | 2x 5 3| = | 2x 8| = 2|x + 4|.If = min {1, /2}, then |2x + 5| 3 < . Therefore, limx4|2x + 5| = 3.46. Let> 0.x 5 2 =x 5 2 x 5 + 2x 5 + 2 =x 9x 5 + 2 < |x 9|.If 0 < |x 9| < = , then x 5 2 < . Therefore, limx9x 5 = 2.47. Suppose that limx0f(x)x= L exists. Thenlimx0f(x) = limx0f(x) xx= limx0f(x)xx = Llimx0x = 0.48. Supposethat limxcg(x) = l andthat f iscontinuousat l.Let> 0.Bythecontinuityof f atl, there is a number 1> 0 such that |f(y) f(l)| 0, x = 0(b) No. By the continuity of f (from the right) at 0, f(1/n) = 0 for each postive integer impliesthat f(0) = 0.50. (a) No. Suppose that f and g are everywhere continuous. If f(c) = g(c), then there is an interval(c p, c + p) on which f(c) = g(c).(b) No. Suppose that f andg are everywhere continuous. If f(x) = g(x) on[a, b], thenf(a) = g(a)and f(b) = g(b). Therefore, by the continuity of fand g, there exist postive numbers p and q suchthat f(x) = g(x) on (a p, b + q).(c) Yes. For example, set f(x) = 0 for all x and g(x) =_x(1 x), 0 x 10, otherwise.f and g are everywhere continuous and f(x) = g(x) only on (0, 1).P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.1 65CHAPTER3SECTION3.11. f

(x) = limh0f(x +h) f(x)h= limh0[2 3(x +h)] [2 3x]h= limh03hh= limh03 = 32. f

(x) = limh0f(x +h) f(x)h= limh0k kh= limh00 = 03. f

(x) = limh0f(x +h) f(x)h= limh0[5(x +h) (x +h)2] (5x x2)h= limh05h 2xh h2h= limh0(5 2x h) = 5 2x4. f

(x) = limh0f(x +h) f(x)h= limh0[2(x +h)3+ 1] [2x3+ 1]h= limh02(x3+ 3x2h + 3xh2+h3) 2x3h= limh06x2h + 6xh2+ 2h3h= limh0(6x2+ 6xh + 2h2) = 6x25. f

(x) = limh0f(x +h) f(x)h= limh0(x +h)4x4h= limh0(x4+ 4x3h + 6x2h2+ 4xh3+h4) x4h= limh0(4x3+ 6x2h + 4xh2+h3) = 4x36. f

(x) = limh0f(x +h) f(x)h= limh01x +h + 3 1x + 3hlimh0(x + 3) (x +h + 3)h(x +h + 3)(x + 3)= limh0hh(x +h + 3)(x + 3)limh01(x +h + 3)(x + 3)=1(x + 3)27. f

(x) = limh0f(x +h) f(x)h= limh0x +h 1 x 1h= limh0(x +h 1) (x 1)h(x +h 1 +x 1 )= limh01x +h 1 +x 1=12x 18. f

(x) = limh0f(x +h) f(x)h= limh0[(x +h)34(x +h)] [x34x]h= limh03x2h + 3xh2+h34hh= limh0(3x2+ 3xh +h24) = 3x24P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5366 SECTION3.19. f

(x) = limh0f(x +h) f(x)h= limh01(x +h)2 1x2h= limh0x2(x2+ 2hx +h2)hx2(x +h)2= limh02x hx2(x +h)2= 2x310. f

(x) = limh0f(x +h) f(x)h= limh01x +h1xhlimh0x x +hhxx +h= limh0_x x +h_ _x +x +h_hxx +h_x +x +h_ = limh0x (x +h)hxx +h_x +x +h_limh0hhxx +h_x +x +h_=12xx11. f(x) = x24x; c = 3:dierence quotient:f(3 +h) f(3)h=(3 +h)24(3 +h) (3)h=9 + 6h +h212 4h + 3h=2h +h2h= 2 +hTherefore, f

(3) = limh0f(3 +h) f(3)h= limh0(2 +h) = 212. f(x) = 7x x2; c = 2:dierence quotient:f(2 +h) f(2)h=7(2 +h) (2 +h)2(10)h=14 + 7h 4 4h h210h=3h h2h= 3 hTherefore, f

(2) = limh0f(2 +h) f(2)h= limh0(3 h) = 313. f(x) = 2x3+ 1; c = 1:dierence quotient:f(1 +h) f(1)h=2(1 +h)3+ 1 (1)h=2_1 + 3h 3h2+h3+ 2h=6h 6h2+ 2h3h= 6 6h + 22Therefore, f

(1) = limh0f(1 +h) f(1)h= limh0(6 6h + 2h2) = 614. f(x) = 5 x4; c = 1:dierence quotient:f(1 +h) f(1)h=5 (1 +h)4(4)hP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.1 67=5 1 4h 6h24h3h44h= 4h 6h24h3h4h= 4 6h 4h2h3Therefore, f

(3) = limh0f(1 +h) f(1)h= limh0_4 6h 4h2h3_ = 415. f(x) =8x + 4; c = 2:dierence quotient:f(2 +h) f(2)h=8(2 +h) + 4 4h=8h + 2 4h=8 4h 8h(h + 2)=4h + 2Therefore, f

(2) = limh0f(2 +h) f(2)h= limh04h + 2= 216. f(x) =6 x; c = 2:dierence quotient:f(2 +h) f(2)h=_6 (2 +h) (2)h=4 h 2h=4 h 2h4 h + 24 h + 2=14 h + 2Therefore, f

(2) = limh0f(2 +h) f(2)h= limh014 h + 2= 1417. f(4) = 4. Slope of tangent at (4, 4) isf

(4) = 3. Tangent y 4 = 3(x 4).18. f(4) = 2. Slope of tangent at (4, 2) isf

(4) =124=14. Tangenty 2 =14(x 4).19. f(2) =14. Slope of tangent at (2, 14) is14. Tangent: y 14=14(x + 2).20. f(2) = 3. Slope of tangent at (2, 3) isf

(2) = 3(2)2= 12. Tangent: y + 3 = 12(x 2) ;21. (a) fis not continuous atc = 1 and c = 1; fhas a removable discontinuity atc = 1 and a jumpdiscontinuity atc = 1.(b) fis continuous but not dierentiable atc = 0 and c = 3.22. (a) fis not continuous atc = 2; fhas a jump discontinuity at 2(b) fis continuous but not dierentiable atc = 2 and c = 3.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5368 SECTION3.123. atx = 1 24. atx =5225. atx = 026. atx = 2, 2 27. atx = 1 28. atx = 229. f

(1) = 4limh0f(1 +h) f(1)h=limh04(1 +h) 4h= 4limh0+f(1 +h) f(1)h=limh0+2(1 +h)2+ 2 4h= 430. f

(1) = 6limh0f(1 +h) f(1)h=limh03(1 +h)23h= 6limh0+f(1 +h) f(1)h=limh0+[2(1 +h)3+ 1] 3h= 631. f

(1) does not existlimh0f(1 +h) f(1)h=limh0h 0h= 1limh0+f(1 +h) f(1)h=limh0+h20h= 032. f

(3) does not exist becausefis not continuous atx = 3.33. 34. 35.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.1 6936. 37. 38.39. Since f(1) = 1 and limx1+f(x) = 2, f is not continuous at 1. Therefore, by (3.1.3), f is not dierentiableat 1.40. (a) f

(x) =_2(x + 1), x < 02(x 1), x > 0.(b) limh0f(0 +h) f(0)h=limh0(h + 1)21h=limh0(h + 2) = 2,limh0+f(0 +h) f(0)h=limh0+(h 1)21h=limh0+(h 2) = 2.41. Continuity atx = 1 : limx1f(x) = 1 =limx1+f(x) = A+B. ThusA+B = 1.Dierentiability atx = 1:limh0f(1 +h) f(1)h=limh0(1 +h)31h= 3 =limh0+f(1 +h) f(1)h= ATherefore,A = 3, = B = 2.42. Continuity atx = 2 = 4B + 2A = 2; dierentiability atx = 2 = 4B +A = 4;A = 2, B =3243. f(x) = c, c any constant. 44. f(x) =_1, x = 00, x = 0.45. f(x) = |x + 1|; or f(x) =_0, x = 11, x = 1.46. f(x) = |x21|47. f(x) = 2x + 5 48. f(x) = |x|49. (a) limx2+f(x) =limx2f(x) = f(2) = 2 Thus,fis continuous atx = 2.(b) limh0f(2 +h) f(2)h=limh0(2 +h)2(2 +h) 2h= 3limh0+f(2 +h) f(2)h=limh0+2(2 +h) 2 2h= 2 fis not dierentiable at 2.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5370 SECTION3.150.f

(x) =limh0(x +h)x +h xxh=limh0x_x +h x_+hx +hh=limh0xhh_x +h +x_ + limh0x +h=12x +x =32x51. (a) fis not continuous at 0 since limx0f(x) = 1 and limx0+f(x) = 0. Therefore fis not dierentiableat 0.(b)1x1y52. (a) limh0f(h) 0h=limh0hh= 1, h rationallimh00h= 0, h irrational.Therefore, limh0f(0 +h) f(0)hdoes not exist.(b) limh0g(h) 0h=limh0h2h= 0, h rationallimh00h= 0, h irrationalTherefore,g is dierentiable at 0 andg

(0) = 0.53. (a) If the tangent line is horizontal, then the normal line is vertical: x = c.(b) If the tangent line has slopef

(c) = 0, then the normal line has slope 1f

(c):y f(c) = 1f

(c)(x c).(c) If the tangent line is vertical, then the normal line is horizontal: y = f(c).54. The center of the circle.55. The normal line has slope 4 : y 2 = 4(x 4).P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.1 7156.xy2 42457. f(x) = x2, f

(x) = 2xThe tangent line at (3, 9) has equation y 9 = 6(x 3) andx-intercept x =32.The normal line at (3, 9) has equation y 9 = 16(x 3) andx-intercept x = 57.s = 57 32=1112.58. (a) The slope of the normal atPis y/x.(b) The slope of the tangent atPis x/y.(c)_1 (x +h)21 x2h=_1 (x +h)21 x2h_1 (x +h)2+1 x2_1 (x +h)2+1 x2=1 (x +h)2(1 x2)h__1 (x +h)2+1 x2_=2xh h2h__1 (x +h)2+1 x2_=2x h__1 (x +h)2+1 x2_limh0f(x +h) f(x)h=limh02x h_1 (x +h)2+1 x2= 2x21 x2= xywhere y =1 x2.59. (a) Since | sin(1/x)| 1 it follows thatx f(x) x and x2 g(x) x2Thus limx0f(x) = f(0) = 0 and limx0g(x) = g(0) = 0, which implies thatfandgare continuous at 0.(b) limh0h sin(1/h) 0h= limh0sin(1/h) does not exist.(c) limh0h2sin(1/h) 0h= limh0h sin(1/h) = 0. Thusg is dierentiable at 0 andg

(0) = 0.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5372 SECTION3.160. f

(2) = limh0f(2 +h) f(2)h= limh0[(2 +h)3+ 1] (23+ 1)h= limh012h + 6h2+h3h= 12f

(2) = limx2f(x) f(2)x 2= limx2(x3+ 1) 9x 2= limx2(x 2)(x2+ 2x + 4)x 2= limx2(x2+ 2x + 4) = 1261. f

(1) = limh0f(1 +h) f(1)h= limh0[(1 +h)23(1 +h)] (2)]h= limh0h +h2h= 1f

(1) = limx1f(x) f(1)x 1= limx1(x23x) (2)x 1= limx1(x 2)(x 1)x 1= limx1(x 2) = 162. f

(3) = limh0f(3 +h) f(3)h= limh0_1 + (3 +h) 2h= limh04 +h 2h= limh0hh4 +h + 2=14f

(3) = limx3f(x) f(3)x 3= limx11 +x 2x 3= limx3x 3(x 3)(1 +x + 2)= limx311 +x + 2=1463. f

(1) = limh0f(1 +h) f(1)h= limh0(1 +h)1/3+ 1h= limh0(1 +h)1/3+ 1h (1 +h)2/3(1 +h)1/3+ 1(1 +h)2/3(1 +h)1/3+ 1= limh0hh[1 +h)2/3(1 +h)1/3+ 1]=13f

(1) =limx1f(x) f(1)x (1)=limx1x1/3+ 1x + 1=limx1x1/3+ 1x + 1x2/3x1/3+ 1x2/3x1/3+ 1=limx1x + 1(x + 1)(x2/3x1/3+ 1)=1364. f

(0) = limh0f(0 +h) f(0)h= limh01h + 2 12h= limh0h2h(h + 2)= limh012(h + 2)= 14f

(0) = limx0f(x) f(0)x= limx01x + 2 12x= limx0x2x(x + 2)= limx012(x + 2)= 1465. (a) D =(2 +h)5/225/2h1 h 1(b) f

(2)= 7.07166. (a) D =(2 +h)2/322/3h1 h 1P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.1 73(b) f

(2)= 0.52967. (a) f

(x) =525x 4; f

(3) =5211(b)f

(x) = 2x + 16x36x5; f

(2) = 68(c) f

(x) = 3(3 2x)2 + 3x)222 + 3x; f

(1) = 1368. (a) f

(1) does not exist (b)f

(2) = 0 (c) f

(3) does not exist69. (c) f

(x) = 10x 21x2(d)f

(x) = 0 atx = 0,102170. (c) f

(x) = 3x2+ 2x 4 (d)f

(x) = 0 atx= 1.5352, 0.868571. (a) f

(32) = 114 ; tangent line: y 218= 114_x 32_, normal line: y 218=411_x 32_(c) (1.453, 1.547)72. (a) Setf(x) = x3. Thenf(x +h) f(x)h=(x +h)3x3h=x3+ 3x2h + 3xh2+h3x3h=3x2h + 3xh2+h3h= 3x2+ 3xh +h2limh0f(x +h) f(x)h= limh0(3x2+ 3xh +h2) = 3x2.(b) LetSbe the set of integers for which the statement is true. We have shown that 1, 2, 3 S.Assume thatk S. This tells us that iff(x) = xk, then f

(x) = limh0(x +h)kxkh= kxk1.Setf(x) = xk+1. Then, by the hint,f

(x) = limh0(x +h)k+1xk+1h=limh0x(x +h)kx xk+h(x +h)kh=limh0x_(x +h)kxkh_+ limh0(x +h)k= x kxk1+xk= (k + 1)xk.Therefore,k + 1 S. ThusScontains the set of positive integers.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5374 SECTION3.2SECTION3.21. F

(x) = 1 2. F

(x) = 2 3. F

(x) = 55x418x24. F

(x) = 6x35. F

(x) = 2ax +b 6. F

(x) = x3x2+x 17. F

(x) = 2x38. F

(x) =x3(2x) (x2+ 2)3x2x6= x2+ 6x49. G

(x) = (x21)(1) + (x 3)(2x) = 3x26x 1 10. F

(x) = 1 +1x211. G

(x) =(1 x)(3x2) x3(1)(1 x)2=3x22x3(1 x)212. F

(x) =(cx d)a (ax b)c(cx d)2=bc ad(cx d)213. G

(x) =(2x + 3)(2x) (x21)(2)(2x + 3)2=2(x2+ 3x + 1)(2x + 3)214. G

(x) =(x + 1)(28x3) (7x4+ 11)(1)(x + 1)2=21x4+ 28x311(x + 1)215. G

(x) = (x32x)(2) + (2x + 5)(3x22) = 8x3+ 15x28x 1016. G

(x) =(x21)(3x2+ 3) (x3+ 3x)2x(x21)2=x46x23(x21)217. G

(x) =(x 2)(1/x2) (6 1/x)(1)(x 2)2= 2(3x2x + 1)x2(x 2)218. G

(x) =x2(4x3) (1 +x4)2xx4=2(x41)x319. G

(x) = (9x88x9)_1 1x2_+_x + 1x_(72x772x8) = 80x9+ 81x864x7+ 63x620. G

(x) =_1x2__1 +1x2_+_1 + 1x__2x3_ = 1x2 2x3 3x421. f

(x) = (x 2)2; f

(0) = 14, f

(1) = 122. f

(x) = 3x3+ 2x; f

(0) = 0, f

(1) = 523. f

(x) =(1 +x2)(2x) (1 x2)(2x)(1 +x2)2=4x(1 +x2)2; f

(0) = 0, f

(1) = 124. f

(x) =(x2+ 2x + 1)(4x + 1) (2x2+x + 1)(2x + 2)(x2+ 2x + 1)2=(x + 1)(4x + 1) 2(2x2+x + 1)(x + 1)3;f

(0) = 1, f

(1) =14P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.2 7525. f

(x) =(cx +d)a (ax +b)c(cx +d)2=ad bc(cx +d)2, f

(0) =ad bcd2, f

(1) =ad bc(c +d)226. f

(x) =(cx2+bx +a)(2ax +b) (ax2+bx +c)(2cx +b)(cx2+bx +a)2; f

(0) =b(a c)a2, f

(1) =2(a c)a +b +c27. f

(x) = xh

(x) +h(x); f

(0) = 0h

(0) +h(0) = 0(2) + 3 = 328. f

(x) = 6xh(x) + 3x2h

(x) 5; f

(0) = 529. f

(x) = h

(x) +h

(x)[h(x)]2, f

(0) = h

(0) +h

(0)[h(0)]2= 2 +232=20930. f

(x) = h

(x) +h(x) xh

(x)h2(x); f

(0) = h

(0) +h(0)h2(0)= 2 + 13=7331. f

(x) =(x + 2)(1) x(1)(x + 2)2=2(x + 2)2,slope of tangent at (4, 2) :f

(4) =12; equation for tangent:y 2 =12(x + 4)32. f

(x) = (x32x + 1)(4) + (4x 5)(3x22) = 16x315x216x + 14slope of tangent at (2, 15) :f

(2) = 50; equation for tangent:y 15 = 50(x 2)33. f

(x) = (x23)(5 3x2) + (5x x3)(2x);slope of tangent at (1, 8) :f

(1) = (2)(2) + (4)(2) = 4; equation for tangent:y + 8 = 4(x 1)34. f

(x) = 2x + 10x2;slope of tangent at (2, 9) :f

(2) = 32; equation for tangent:y 9 = 32(x + 2)35. f

(x) = (x 2)(2x 1) + (x2x 11)(1) = 3(x 3)(x + 1);f

(x) = 0 atx = 1, 3; (1, 27), (3, 5)36. f

(x) = 2x + 16x2=2(x3+ 8)x2; f

(x) = 0 atx = 2; (2, 12)37. f

(x) =(x2+ 1)(5) 5x(2x)(x2+ 1)2=5(1 x2)(x2+ 1)2, f

(x) = 0 atx = 1; (1, 5/2), (1, 5/2)38. f

(x) = (x + 2)(2x 2) + (x22x 8)(1) = 3x212 = 3(x24);f

(x) = 0 atx = 2; (2, 0), (2, 32)39. f(x) = x48x2+ 3; f

(x) = 4x316x = 4x(x24) = 4x(x 2)(x + 2)(a) f

(x) = 0 atx = 0, 2(b) f

(x) > 0 on (2, 0) (2, )(c) f

(x) < 0 on (, 2) (0, 2)P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5376 SECTION3.240. f(x) = 3x44x32; f

(x) = 12x312x2= 12x2(x 1)(a) f

(x) = 0 atx = 0, 1.(b) f

(x) > 0 on (1, ).(c) f

(x) < 0 on (, 0) (0, 1)41. f(x) = x +4x2; f

(x) = 1 8x3=x38x3=(x 2)(x2+ 2x + 4)x3(a) f

(x) = 0 atx = 2.(b) f

(x) > 0 on (, 0) (2, )(c) f

(x) < 0 on (0, 2)42. f(x) =x22x + 4x2+ 4; f

(x) =(x2+ 4)(2x 2) (x22x + 4)(2x)(x2+ 4)2=2(x24)(x2+ 4)2(a) f

(x) = 0 atx = 2, 2.(b) f

(x) > 0 on (, 2) (2, )(c) f

(x) < 0 on (2, 2)43. slope of line 4; slope of tangentf

(x) = 2x; 2x = 4 atx = 2; (2, 10)44. slope of line 3/5; slope of tangentf

(x) = 3x23;perpendicular when 3x23 = 53; x = 23; _23,4627_,_23, 4627_45. f(x) = x3+x2+x +C 46. f(x) = x4x2+ 4x +C47. f(x) =2x33 3x22+ 1x +C 48. f(x) =x55+x42+x +C49. We wantfto be continuous atx = 2. That is, we wantlimx2f(x) = f(2) =limx2+f(x).This gives(1) 8A+ 2B + 2 = 4B A.We also wantlimx2f

(x) =limx2+f

(x).This gives(2) 12A+B = 4B.Equations (1) and (2) together imply thatA = 2 andB = 8.50. First we need, limx1f(x) = limx1+f(x), = A+B = B A+ 4 or A+B = 2.Next we need, limx1f

(x) = limx1+f

(x) = 2A = 5B +Aor 3A+ 5B = 0.Solving these equations gives A = 5,B = 3.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.2 7751. slope of tangent at (5, 5) isf

(5) = 4tangenty 5 = 4(x 5) intersectsx-axis at (254, 0)normaly 5 =14(x 5) intersectsx-axis at (15, 0)area of triangle is12(5)(15 +254 ) =425852. slope of tangent at (2, 5) isf

(2) = 4tangenty 5 = 4(x 2) intersectsx-axis at (134, 0)normaly 5 =14(x 2) intersectsx-axis at (18, 0)area of triangle is12(5)(18 +134 ) =425853. If the point (1, 3) lies on the graph, we havef(1) = 3 and thus() A+B +C = 3.If the line 4x +y = 8 (slope 4) is tangent to the graph at (2, 0), thenf(2) = 0 andf

(2) = 4. Thus,() 4A+ 2B +C = 0 and 4A+B = 4.Solving the equations in () and (), we nd thatA = 1, B = 0, C = 4.54. First,f(1) = 0 andf

(1) = 3 = A+B +C +D = 0 and 3A+ 2B +C = 3Next,f(2) = 9 andf

(2) = 18 = 8A+ 4B + 2C +D = 9 and 12A+ 4B +C = 18Solving these equations givesA = 3, B = 6,C = 6,D = 3.55. Let f(x) = ax2+bx +c. Then f

(x) = 2ax +b and f

(x) = 0 at x = b/2a.56. The derivative ofp is the quadraticp

(x) = 3ax2+ 2bx +c. Its discriminant isD = (2b)24(3a)(c) = 4b212ac(a)p has two horizontal tangents ip

has two real roots iD > 0.(b)p has exactly one horizontal tangent ip has only one real root iD = 0.(c)p has no horizontal tangent ip has no real roots iD < 0.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5378 SECTION3.257. Let f(x) = x3x. The secant line through (1, f(1)) = (1, 0) and (2, f(2)) = (2, 6) has slopem =6 02 (1)= 2. Now, f

(x) = 3x21 and 3c21 = 2 implies c = 1, 1.58. msec =f(3) f(1)3 1=34 122=18f

(x) =(x + 1)(1) x(1)(x + 1)2=1(x + 1)2; f

(c) =18= c = 1 2259. Let f(x) = 1/x, x > 0. Then f

(x) = 1/x2. An equation for the tangent line to the graph offatthe point (a, f(a)), a > 0, is y = (1/a2)x + 2/a. The y-intercept is 2/a and the x-intercept is 2a.The area of the triangle formed by this line and the coordinate axes is: A =12 (2/a)(2a) = 2 squareunits.60. Let (x, y) be the point on the graph that the tangent line passes through. f

(x) = 3x2, sox38 =3x2(x 2). Thusx = 2 orx = 1. The lines arey 8 = 12(x 2) andy + 1 = 3(x + 1).61. Let (x, y) be the point on the graph that the tangent line passes through. f

(x) = 3x21, sox3x 2 = (3x21)(x + 2). Thusx = 0 orx = 3. The lines arey = x andy + 24 = 26(x + 3).62. (a) f(c) = c3; f

(x) = 3x2andf

(c) = 3c2. Tangent line:y c3= 3c2(x c) ory = 3c2x 2c3.(b) We solve the equation 3c2x 2c3= x3:x33c2x + 2c3= 0 = (x c)(x2+cx 2c2) = 0 = (x c)2(x + 2c) = 0Thus, the tangent line atx = c, c = 0 intersects the graph atx = 2c.63. Sincefandf +gare dierentiable, g = (f +g) fis dierentiable. The functions f(x) = |x| andg(x) = |x| are not dierentiable at x = 0 yet their sumf(x) +g(x) 0 is dierentiable for allx.64. No. Iffandfg are dierentiable, theng =fgfwill be dierentiable wheref(x) = 0.65. Since_fg_(x) =f(x)g(x)= f(x) 1g(x),it follows from the product and reciprocal rules that_fg_

(x) =_f 1g_

(x) = f(x)_g

(x)[g(x)]2_+f

(x) 1g(x)=g(x)f

(x) f(x)g

(x)[g(x)]2.66.(fgh)

(x) = [(fg)(x) h(x)]

= (fg)(x)h

(x) +h(x)[(fg)(x)]

= f(x)g(x)h

(x) +h(x)[f(x)g

(x) +g(x)f

(x)]= f

(x)g(x)h(x) +f(x)g

(x)h(x) +f(x)g(x)h

(x)67. F

(x) = 2x_1 + 1x_(2x3x + 1) + (x2+ 1)_1x2_(2x3x + 1) + (x2+ 1)_1 + 1x_(6x21)P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.2 7968. G

(x) =12x_11 + 2x_(x2+x 1) +x_2(1 + 2x)2_(x2+x 1) +x_11 + 2x_(2x + 1)69. g(x) = [f(x)]2= f(x) f(x)g

(x) = f(x)f

(x) +f(x)f

(x) = 2f(x)f

(x)70. g

(x) = 2(x32x2+x + 2)(3x24x + 1)71. (a) f

(x) = 0 atx = 0, 2 (b)f

(x) > 0 on (, 2) (0, )(c) f

(x) < 0 on (2, 1) (1, 0)72. (a) f

(x) = 0 atx = 0, 1,52(b)f

(x) > 0 on (, 0) _1,52_ (5/2, )(c) f

(x) < 0 on (0, 1)73. (a) f

(x) = 0 for allx = 0 (b)f

(x) > 0 on (0, )(c) f

(x) < 0 on (, 0)74. (a) f

(x) = 0 atx = 34= 1.587 (b)f

(x) > 0 on _34, 0_(c) f

(x) < 0 on _, 34_ (0, )75. (a)sin(0 + 0.001) sin 00.001= 0.99999sin(0 0.001) sin 00.001= 0.99999sin[(/6) + 0.001] sin(/6)0.001= 0.86578sin[(/6) 0.001] sin(/6)0.001= 0.86628sin[(/4) + 0.001] sin(/4)0.001= 0.70675sin[(/4) 0.001] sin(/4)0.001= 0.70746sin[(/3) + 0.001] sin(/3)0.001= 0.49957sin[(/3) 0.001] sin(/3)0.001= 0.50043sin[(/2) + 0.001] sin(/2)0.001= 0.0005sin[(/2) 0.001] sin(/2)0.001= 0.0005(b) cos 0 = 1, cos(/6)= 0.866025, cos(/4)= 0.707107, cos(/3) = 0.5, cos(/2) = 0(c) If f(x) = sin x then f

(x) = cos x.76. (a)x = 2, 0,54(b)x1 = 2.732, x2 = 0.618,x3 = 0.732, x4 = 1.618P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5380 SECTION3.3SECTION3.31.dydx= 12x32x 2.dydx= 2x 8x53.dydx= 1 +1x24.dydx=(1 x)2 2x(1)(1 x)2=2(1 x)25.dydx=(1 +x2)(1) x(2x)(1 +x2)2=1 x2(1 +x2)26. y = x3x22x;dydx= 3x22x 27.dydx=(1 x)2x x2(1)(1 x)2=x(2 x)(1 x)28. y =2x x23 + 3x;dydx=(3 + 3x)(2 2x) (2x x2)(3)(3 + 3x)2=2 2x x23(1 +x)29.dydx=(x31)3x2(x3+ 1)3x2(x31)2=6x2(x31)210.dydx=(1 +x)2x x2(1)(1 +x)2=x2+ 2x(1 +x)211.ddx (2x 5) = 2 12.ddx (5x + 2) = 513.ddx [(3x2x1)(2x + 5)] = (3x2x1)2 + (2x + 5)(6x +x2) = 18x2+ 30x + 5x214.ddx_(2x2+ 3x1)(2x 3x2) = (2x2+ 3x1)(2 + 6x3) + (2x 3x2)(4x 3x2) = 12x2+ 27x415.ddt_t42t31_ =(2t31)4t3t4(6t2)(2t31)2=2t3(t32)(2t31)216.ddt_2t3+ 1t4_ =ddt_2t+1t4_ = 2t2 4t5= 2(t3+ 2)t517.ddu_2u1 2u_ =(1 2u)2 2u(2)(1 2u)2=2(1 2u)218.ddu_u2u3+ 1_ =(u3+ 1)(2u) u2(3u2)(u3+ 1)2=u(2 u3)(u3+ 1)219.ddu_uu 1 uu + 1_ =(u 1)(1) u(u 1)2 (u + 1)(1) u(u + 1)2= 1(u 1)2 1(u + 1)2= 2(1 +u2)(u21)220.ddu_u2(1 u2)(1 u3) =ddu[u2u4u5+u7] = 2u 4u35u4+ 7u6P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.3 8121.ddx_x3+x2+x + 1x3x2+x 1_ =(x3x2+x 1)(3x2+ 2x + 1) (x3+x2+x + 1)(3x22x + 1)(x3x2+x 1)2= 2(x4+ 2x2+ 1)(x2+ 1)2(x 1)2=2(x 1)222.ddx_x3+x2+x 1x3x2+x + 1_ =(x3x2+x + 1)(3x2+ 2x + 1) (x3+x2+x 1)(3x22x + 1)(x3x2+x + 1)2=2x4+ 8x2+ 2(x3x2+x + 12)23.dydx= (x + 1)ddx [(x + 2)(x + 3)] + (x + 2)(x + 3)ddx (x + 1) = (x + 1)(2x + 5) + (x + 2)(x + 3)Atx = 2,dydx= (3)(9) + (4)(5) = 47.24.dydx= (x + 1)(x2+ 2)(3x2) + (x + 1)(x3+ 3)(2x) + (x2+ 2)(x3+ 3)(1)Atx = 2,dydx= 3(6)(12) + 3(11)(4) + (6)(11) = 414.25.dydx=(x + 2)ddx[(x 1)(x 2)] (x 1)(x 2)(1)(x + 2)2=(x + 2)(2x 3) (x 1)(x 2)(x + 2)2Atx = 2,dydx=4(1) 1(0)16=14.26. y =x4x22x2+ 2;dydx=(x2+ 2)(4x32x) (x4x22)(2x)(x2+ 2)2Atx = 2,dydx=6(28) (10)436=32927. f

(x) = 21x230x4, f

(x) = 42x 120x328. f

(x) = 10x424x3+ 2, f

(x) = 40x372x229. f

(x) = 1 + 3x2, f

(x) = 6x330. f

(x) = 2x + 2x3, f

(x) = 2 6x431. f(x) = 2x22x23, f

(x) = 4x + 4x3, f

(x) = 4 12x432. f(x) = 4x 9x1, f

(x) = 4 + 9x2, f

(x) = 18x333.dydx= x2+x + 1d2ydx2= 2x + 1d3ydx3= 234.dydx= 10 + 50xd2ydx2= 50d3ydx3= 035.dydx= 8x 20d2ydx2= 8d3ydx3= 0P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5382 SECTION3.336.dydx=12 x212 x + 1d2ydx2= x 12d3ydx3= 137.dydx= 3x2+ 3x4d2ydx2= 6x 12x5d3ydx3= 6 + 60x638.dydx= 3x22x2d2ydx2= 6x + 4x3d3ydx3= 6 12x439.ddx_xddx (x x2)_ =ddx [x(1 2x)] =ddx [x 2x2] = 1 4x40.d2dx2_(x23x)ddx (x +x1)_ =d2dx2_(x23x)(1 x2)=d2dx2[x23x 1 + 3x1]=ddx (2x 3 3x2) = 2 + 6x341.d4dx4 [3x x4] =d3dx3 [3 4x3] =d2dx2 [12x2] =ddx [24x] = 2442.d5dx5 [ax4+bx3+cx2+dx +e] =d4dx4 [4ax3+ 3bx2+ 2cx +d]=d3dx3 [12ax2+ 6bx + 2c]=d2dx2 [24ax + 6b] =ddx [24a] = 043.d2dx2_(1 + 2x)d2dx2 (5 x3)_ =d2dx2[(1 + 2x)(6x)] =d2dx2_6x 12x2 = 2444.d3dx3_1xd2dx2 [x45x2]_ =d3dx3_1x (12x210)_=d3dx3 [12x 10x1] =d2dx2 [12 + 10x2]=ddx[20x3] = 60x445. y = x4x33+ 2x2+C 46. y =x22+1x2+ 3x +C47. y = x51x4+C 48. y =2x63+53x3 2x +C49. Let p(x) = ax2+bx +c. Then p

(x) = 2ax +b and p

(x) = 2a. Nowp

(1) = 2a = 4 =a = 2; p

(1) = 2(2)(1) +b = 2 =b = 6;p(1) = 2(1)26(1) +c = 3 =c = 7Thus p(x) = 2x26x + 7.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.3 8350. p(x) = ax3+bx2+cx +d p

(1) = 6 = a = 1p

(x) = 3ax2+ 2bx +c p

(1) = 2 = b = 2p

(x) = 6ax + 2b p

(1) = 3 = c = 4p

(x) = 6a p(1) = 0 = d = 3Therefore,p(x) = x3+ 2x2+ 4x + 3.51. (a) If k = n, f(n)(x) = n! (b)If k > n, f(k)(x) = 0.(c) If k < n, f(k)(x) = n(n 1)(n 2) (n k + 1)xnk.52. (a)dndxn= n! an(b)dkdxk= 0 ifk > n.53. Let f(x) =_x2x 00 x 0(a) f

+(0) =limh0+f(0 +h) f(0)h=limh0+h20h= 0,f

(0) =limh0f(0 +h) f(0)h=limh00h= 0Therefore,fis dierentiable at 0 andf

(0) = 0.(b) f

(x) =_2x x 00 x 0(d)(c) f

+(0) =limh0+f

(0 +h) f

(0)h=limh0+2h 0h= 2,f

(0) =limh0f

(0 +h) f

(0)h=limh00h= 0Since f

+(0) = f

(0), f

(0) does not exist.54. Let g(x) =_x3x 00 x < 0(a) g

+(0) =limh0+g(0 +h) g(0)h=limh0+h30h= 0,g

(0) =limh0g(0 +h) g(0)h=limh00h= 0Therefore,g is dierentiable at 0 andg

(0) = 0.g

(x) =_3x2x 00 x < 0g

+(0) =limh0+g

(0 +h) g, (0)h=limh0+3h20h= 0,g

(0) =limh0g

(0 +h) g

(0)h=limh00h= 0Therefore,g

is dierentiable at 0 andg

(0) = 0.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5384 SECTION3.3(b) g

(x) =_3x2x 00 x < 0(d)g

(x) =_6x x 00 x < 0(c) g

+(0) =limh0+g

(0 +h) g

(0)h=limh0+6h 0h= 6g

(0) =limh0g

(0 +h) g

(0)h=limh00h= 0Since g

+(0) = g

(0), g

(0) does not exist.55. It suces to give a single counterexample. For instance, iff(x) = g(x) = x, then (fg)(x) = x2so that (fg)

(x) = 2 butf(x)g

(x) +f

(x)g(x) = x 0 + 0 x = 0.56.ddx[f(x)g

(x) f

(x)g(x)] = [f(x)g

(x) +f

(x)g

(x)] [f

(x)g

(x) +f

(x)g(x)]= f(x)g

(x) f

(x)g(x)57. f

(x) = 6x; (a) x = 0 (b) x > 0 (c) x < 058. f

(x) = 12x2; (a) x = 0 (b) allx = 0 (c) none59. f

(x) = 12x2+ 12x 24; (a) x = 2, 1 (b) x < 2, x > 1 (c) 2 < x < 160. f

(x) = 12x2+ 18x 12; (a) x = 2,12(b) x < 2, x >12(c) 2 < x 0.70. (a) f(x) = x4x35x2x 2; f

(x) = 4x33x210x 1.(b)xyff-20-101020-2 -1 1 2 3(c) The graph is falling when f

(x) < 0;the graph is rising when f

(x) > 0.71. (a) f(x) =12 x33x2+ 3x + 3; f

(x) =32 x26x + 3(b) (c) The line tangent to the graph is horizontal; thegraph turns from rising to falling, or fromfalling to rising.(d) x1 = 0.586, x2 = 3.414P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.4 8772. (a) f(x) =12 x33x2+ 4x + 1; f

(x) =32 x26x + 4.(b)(c) The line tangent to the graph is horizontal; thegraph turns from rising to falling, or fromfalling to rising.(d) x1 = 0.845, x2 = 3.155SECTION3.41. A = r2,dAdr= 2r. Whenr = 2,dAdr= 4. 2. V= s3,dVds= 3s2. Whens = 4,dVds= 48.3. A =12z2,dAdz= z. Whenz = 4,dAdz= 4. 4.dydx= x2. Whenx = 1,dydx= 1.5. y =1x(1 +x),dydx= (2x + 1)x2(1 +x)2. Atx = 2,dydx= 536.6.dydx= 3x224x + 45 = 3(x 3)(x 5);dydx= 0 atx = 3, 5.7. V=43r3,dVdr= 4r2= the surface area of the ball.8. S = 4r2;dSdr= 8r anddSdr= 8r0atr = r0.dSdr= 1 = r0 =18.9. y = 2x2+x 1,dydx= 4x + 1;dydx= 4 atx =34. Thereforex0 =34.10. (a) A = _d2_2=4d2; A

=2d (b)A = _ C2_2=C24;dAdC=C211. (a) w = s2, V= s3=_w2_3=24w3,dVdw=324w2.(b) z2= s2+w2= 3s2, z = s3. V= s3=_z3_3=39z3,dVdz=33z2.12. A = bh = (constant) = h =Ab ;dhdb= Ab2= bhb2= hb13. (a)dAd=12r2(b)dAdr= r(c) =2Ar2soddr= 4Ar3= 4r3_12r2_ = 2r14. (a)dAdh= 2r (b)dAdr= 2(2r +h)(c) h =A2r r;dhdr= A2r2 1 = 2r(r +h) 2r22r2= 2r +hrP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5388 SECTION3.515. y = ax2+bx +c, z = bx2+ax +c.dydx= 2ax +b,dzdx= 2bx +a.dydx=dzdxi 2ax +b = 2bx +a. Witha = b, this occurs only atx =12.16. Setk(x) = f(x)g(x)h(x). Thenk

(x) = f(x)g(x)h

(x) +f(x)g

(x)h(x) +f

(x)g(x)h(x) andk

(1) = (0)(2)(0) + (0)(1)(2) + (1)(2)(2) = 4SECTION3.51. y = x4+ 2x2+ 1, y

= 4x3+ 4x = 4x(x2+ 1)y = (x2+ 1)2, y

= 2(x2+ 1)(2x) = 4x(x2+ 1)2. y = x62x3+ 1, y

= 6x56x2= 6x2(x31)y = (x31)2, y

= 2(x31)(3x2) = 6x2(x31)3. y = 8x3+ 12x2+ 6x + 1, y

= 24x2+ 24x + 6 = 6(2x + 1)2y = (2x + 1)3, y

= 3(2x + 1)2(2) = 6(2x + 1)24. y = x6+ 3x4+ 3x2+ 1, f

(x) = 6x5+ 12x3+ 6x = 6x(x2+ 1)2y = (x2+ 1)3, y

= 3(x2+ 1)2(2x) = 6x(x2+ 1)25. y = x2+ 2 +x2, y

= 2x 2x3= 2x(1 x4)y = (x +x1)2, y

= 2(x +x1)(1 x2) = 2x(1 +x2)(1 x2) = 2x(1 x4)6. y = 9x412x3+ 4x2, y

= 36x336x2+ 8x = 4x(3x 2)(3x 1)y = (3x22x)2, y

= 2(3x22x)(6x 2) = 4x(3x 2)(3x 1)7. f

(x) = 1(1 2x)2ddx (1 2x) = 2(1 2x)28. f

(x) = 5(1 + 2x)4ddx (1 + 2x) = 10(1 + 2x)49. f

(x) = 20(x5x10)19ddx (x5x10) = 20(x5x10)19(5x410x9)10. f

(x) = 3(x2+x2)2ddx (x2+x2) = 6(x2+x2)2(x x3)11. f

(x) = 4_x 1x_3ddx_x 1x_ = 4_x 1x_3_1 +1x2_12. f(t) = (1 +t)4; f

(t) = 4(1 +t)5ddt(1 +t) = 4(1 +t)513. f

(x) = 4(x x3+x5)3ddx (x x3+x5) = 4(x x3+x5)3(1 3x2+ 5x4)14. f

(t)= 3(t t2)2ddt(t t2) = 3(t t2)2(1 2t)P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.5 8915. f

(t)= 4(t1+t2)3ddt(t1+t2) = 4(t1+t2)3(t22t3)16. f

(x) = 3_4x + 35x 2_2ddx_4x + 35x 2_ = 3_4x + 35x 2_2_(5x 2)4 (4x + 3)5(5x 2)2_ = 69(4x + 3)2(5x 2)417. f

(x) = 4_3xx2+ 1_3ddx_3xx2+ 1_ = 4_3xx2+ 1_3_(x2+ 1)3 3x(2x)(x2+ 1)2_ =324x3(1 x2)(x2+ 1)518. f

(x) = 3_(2x + 1)2+ (x + 1)22 ddx_(2x + 1)2+ (x + 1)2= 3_(2x + 1)2+ (x + 1)22[2(2x + 1)(2) + 2(x + 1)(1)]= 6_(2x + 1)2+ (x + 1)22(5x + 3)19. f

(x) = _x33+x22+x1_2ddx_x33+x22+x1_ = _x33+x22+x_2(x2+x + 1)20. f

(x) = 2[(6x +x5)1+x]ddx [(6x +x5)1+x] = 2[(6x +x5)1+x][1 (6x +x5)2(6 + 5x4)]21.dydx=dydududx=2u(1 +u2)2(2)Atx = 0, we haveu = 1 and thusdydx= 44= 1.22.dydx=dydududx= (1 u2) 4(3x + 1)3(3)Atx = 0, we haveu = 1 and thusdydx= 0.23.dydx=dydududx=(1 4u)2 2u(4)(1 4u)24(5x2+ 1)3(10x) =2(1 4u)240x(5x2+ 1)3Atx = 0, we haveu = 1 and thusdydx=29(0) = 0.24.dydx=dydududx= (3u21) _2(1 +x)2_Atx = 0, we haveu = 1 and thusdydx= 2(2) = 4.25.dydt=dydududxdxdt=(1 +u2)(7) (1 7u)(2u)(1 +u2)2(2x)(2)=7u22u 7(1 +u2)2(4x) =4x(7x4+ 12x22)(x4+ 2x2+ 2)2=4(2t 5)[7(2t 5)4+ 12(2t 5)22][(2t 5)4+ 2(2t 5)2+ 2]226.dydt=dydududxdxdt= 2u_(1 +x2)(7) (1 7x)(2x)(1 +x2)2_(5)= 10u 7x22x 7(1 +x2)2=10[1 7(5t + 2)][7(5t + 2)22(5t + 2) 7][1 + (5t + 2)2]2P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5390 SECTION3.527.dydx=dydsdsdtdtdx= 2(s + 3) 12t 3 (2x)Atx = 2, we havet = 4 so thats = 1 and thusdydx= 2(4)12 1(4) = 16.28.dydx=dydsdsdtdtdx=2(1 s)2_1 +1t2_12xAtx = 2, we havet =2 ands =2/2. Thusdydx=2(1 2/2)2_1 + 12_122=332 4.29. (f g)

(0) = f

(g(0))g

(0) = f

(2)g

(0) = (1)(1) = 130. (f g)

(1) = f

(g(1))g

(1) = f

(1)g

(1) = (1)(0) = 031. (f g)

(2) = f

(g(2))g

(2) = f

(2)g

(2) = (1)(1) = 132. (g f)

(0) = g

(f(0))f

(0) = g

(1)f

(0) = (0)(2) = 033. (g f)

(1) = g

(f(1))f

(1) = g

(0)f

(1) = (1)(1) = 134. (g f)

(2) = g

(f(2))f

(2) = g

(1)f

(2) = (0)(1) = 035. (f h)

(0) = f

(h(0))h

(0) = f

(1)h

(0) = (1)(2) = 236. (f h g)

(1) = f

(h(g(1)))h

(g(1))g

(1) = f

(2)h

(1)g

(1) = (1)(1)(0) = 037. (g f h)

(2) = g

(f(h(2)))f

(h(2))h

(2) = g

(1)f

(0)h

(2) = (0)(2)(2) = 038. (g h f)

(0) = g

(h(f(0)))h

(f(0))f

(0) = g

(2)h

(1)f

(0) = (1)(1)(2) = 239. f

(x) = 4(x3+x)3(3x2+ 1)f

(x) = 3(4)(x3+x)2(3x2+ 1)2+ 4(x3+x)3(6x) = 12(x3+x)2[(3x2+ 1)2+ 2x(x3+x)]40. f

(x) = 10(x25x + 2)9(2x 5)f

(x) = 9(10)(x25x + 2)8(2x 5)2+ 10(x25x + 2)9(2)= (10)(x25x + 2)8_9(2x 5)2+ 2(x25x + 2)41. f

(x) = 3_x1 x_21(1 x)2=3x2(1 x)4f

(x) =6x(1 x)43x2(4)(1 x)3(1)(1 x)8=6x(1 +x)(1 x)542. f

(x) =12x2+ 1(2x) =xx2+ 1P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.5 91f

(x) =x2+ 1(1) xxx2+ 1_x2+ 1_2=1(x2+ 1)3/243. 2xf

(x2+ 1) 44. f

_x 1x + 1_ddx_x 1x + 1_ =2(x + 1)2f

_x 1x + 1_45. 2f(x)f

(x) 46.[f(x) + 1]f

(x) [f(x) 1]f

(x)[f(x) + 1]2=2f

(x)[f(x) + 1]247. f

(x) = 4x(1 +x2)3; (a) x = 0 (b) x < 0 (c) x > 048. f

(x) = 2(1 x2)(2x) = 4x(1 x2); (a) x = 1, 0, 1 (b) 1 < x < 0, x > 1(c) x < 1, 0 < x < 149. f

(x) =1 x2(1 +x2)2; (a) x = 1 (b) 1 < x < 1 (c) x < 1, x > 150. f

(x) = (1 x2)3+x(3)(1 x2)2(2x) = (1 x2)2(1 7x2);(a) x = 1, x = 177 (b) 177 < x 151.n!(1 x)n+152.(1)n+1n!(1 +x)n+153. n! bn54.(1)nn! abn(bx +c)n+155. y = (x2+ 1)3+C 56. y =(x21)22+C57. y = (x32)2+C 58. y =(x3+ 2)33+C59. L

(x) =1x2+ 1 2x =2xx2+ 160. H

(x) = 2f(x)f

(x) 2g(x)g

(x) = 2f(x)g(x) 2g(x)f(x) = 061. T

(x) = 2f(x) f

(x) + 2g(x) g

(x) = 2f(x) g(x) 2g(x) f(x) = 062. (a) Supposefis even: [f(x)]

= [f(x)]

= f

(x)(1) = f

(x); thus f

(x) = f

(x).(b) Supposefis odd: [f(x)]

= [f(x)]

= f

(x)(1) = f

(x); thus f

(x) = f

(x).63. SupposeP(x) = (x a)2q(x), whereq(a) = 0. ThenP

(x) = 2(x a)q(x) + (x a)2q

(x) and P

(x) = 2q(x) + 4(x a)q

(x) + (x a)2q

(x),and it follows thatP(a) = P

(a) = 0, andP

(a) = 0.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5392 SECTION3.5Now suppose thatP(a) = P

(a) = 0 andP

(a) = 0.P(a) = 0 = P(x) = (x a)g(x) for some polynomialg.ThenP

(x) = g(x) + (x a)g

(x) andP

(a) = 0 = g(a) = 0 and sog(x) = (x a)q(x) for some polynomialq.Therefore,P(x) = (x a)2q(x). Finally, P

(a) = 0 implies q(a) = 0.64. Suppose P(x) = (x a)3q(x), whereq(a) = 0. ThenP

(x) = 3(x a)2q(x) + (x a)3q

(x)P

(x) = 6(x a)q(x) + 6(x a)2q

(x) + (x a)3q

(x)P

(x) = 6q(x) + 18(x a)q

(x) + 9(x a)2q

(x) + (x a)3q

(x)and it follows thatP(a) = P

(a) = P

(a) = 0, P

(a) = 0.Now suppose that P(a) = P

(a) = P

(a) = 0 and P

(a) = 0.P(a) = 0 = P(x) = (x a)g(x) for some polynomialg.Then P

(x) = g(x) + (x a)g

(x) andP

(a) = 0 = g(a) = 0 and so g(x) = (x a)h(x) for some polynomialh.Therefore,P(x) = (x a)2h(x). NowP

(x) = 2h(x) + 4(x a)h

(x) + (x a)2h

(x) andP

(a) = 0 = h(a) = 0 and so h(x) = (x a)q(x) for some polynomialq.Therefore,P(x) = (x a)3q(x). Finally, P

(a) = 0 implies q(a) = 0.65. LetPbe a polynomial function of degreen. The numbera is a root ofPof multiplicityk, (k < n) ifand only if P(a) = P

(a) = = P(k1)(a) = 0 and P(k)(a) = 0.66. A =34x2, where x =233h. NowdAdh=dAdxdxdh=32x 233=233h;dAdh= 4 whenh = 2367. V=43 r3anddrdt= 2 cm/sec. By the chain rule,dVdt=dVdrdrdt= 4r2drdt= 8r2.At the instant the radius is 10 centimeters, the volume is increasing at the ratedVdt= 8(10)2= 800 cm3/sec.68. V=43 r3, S = 4r2, anddVdt= 200.dSdt=dSdr drdVdVdt= 8r 14r2 200P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.5 93=400r= 80 whenr = 5The surface area is increasing 80 cm2/sec. at the instant the radius is 5 centimeters.69. (a)dFdt=dFdr drdt=2kr3 (49 9.8t) =2k(49t 4.9t2)3(49 9.8t), 0 t 10(b)dFdt (3) =2k(102.9)3(19.6);dFdt (7) = 2k(102.9)3(19.6).70. (a) f(9) = 2, f

(9) = 112; tangentT: y = 112x 54.(b)2 4 6 8 10x-2-1yfT(c) (7.4, 10.8)71. (a) f(1) =12, f

(1) = 12; tangentT: y = 12x + 1.(b)1 2x1yfT(c) (0.8, 1.2)72.ddx_x2d4dx4_x2+ 1_4_ = 288_35x5+ 20x3+x_73. (a)ddx_f_1x__ = f

(1/x)x2(b)ddx_f_x21x2+ 1__ =4xf

_(x21)/(x2+ 1)(1 +x2)2(c)ddx_f(x)1 +f(x)_ =f

(x)[1 +f(x)]274. (a)ddx [u1 (u2(x))] = u

1 (u2(x)) u

2(x) (b)ddx [u1 (u2(u3(x)))] = u

1 [u2(u3(x)] u

2 (u3(x))u

3(x)(c)ddx [u1 (u2[u3(u4(x))])] = u

1 [u2(u3[u4(x)])] u

2 (u3[u4(x)]) u

3[u4(x)]u

4(x)75.d2dx2[f(g(x))] = [g

(x)]2f

[g(x)] +f

[g(x)]g

(x)P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5394 SECTION3.6SECTION3.61.dydx= 3 sin x 4 sec xtan x 2.dydx= 2x sec x +x2sec x tan x3.dydx= 3x2csc x x3csc xcot x 4.dydx= 2 sin x cos x5.dydt= 2 cos t sin t 6.dydt= 6t tan t + 3t2sec2t7.dydu= 4 sin3uddu (sin u) = 4 sin3ucosuddu (u) = 2u1/2sin3u cosu8.dydu= csc u22u2csc u2cot u29.dydx= sec2x2ddx (x2) = 2xsec2x210.dydx= 12xsinx 11.dydx= 4[x + cot x]3[1 csc2x]12.dydx= 3(x2sec 2x)2(2x 2 sec 2x tan 2x) = 6(x2sec 2x)2(x sec 2x tan 2x)13.dydx= cos x,d2ydx2= sin x 14.dydx= sin x,d2ydx2= cos x15.dydx=(1 + sin x)(sin x) cos x(cos x)(1 + sin x)2= sin x (sin2x + cos2x)(1 + sin x)2= (1 + sin x)1d2ydx2= (1 + sin x)2ddx (1 + sin x) = cos x(1 + sin x)216.dydx= 3 tan2(2x) sec2(2x)(2) = 6 tan2(2x) sec2(2x)d2ydx2= 6(2) tan(2x) sec2(2x) sec2(2x)(2) + 6 tan2(2x)(2) sec(2x)[sec(2x) tan(2x)](2)= 242tan(2x) sec2(2x)[sec2(2x) + tan2(2x)]17.dydu= 3 cos22uddu (cos 2u) = 6 cos22usin 2ud2ydu2= 6[cos22uddu (sin 2u) + sin 2uddu (cos22u)]= 6[2 cos32u + sin 2u(4 cos 2usin 2u)] = 12 cos 2u[2 sin22u cos22u]18.dydt= 5 sin4(3t) cos(3t)(3) = 15 sin4(3t) cos(3t)d2ydt2= 15(4) sin3(3t) 3 cos2(3t) + 15 sin4(3t)[3 sin(3t)] = 45 sin3(3t)[4 cos2(3t) sin2(3t)]19.dydt= 2 sec22t,d2ydt2= 4 sec 2tddt (sec 2t) = 8 sec22t tan 2tP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.6 9520.dydu= 4 csc24u;d2ydu2= 4(2) csc(4u)[csc(4u) cot(4u)(4)] = 32 csc2(4u) cot(4u)21.dydx= x2(3 cos 3x) + 2xsin 3xd2ydx2= [x2(9 sin 3x) + 2x(3 cos 3x)] + [2x(3 cos 3x) + 2(sin 3x)]= (2 9x2) sin 3x + 12xcos 3x22.dydx=(1 cos x) cos x sin x([ sin x])(1 cos x)2=cos x 1(1 cos x)2=11 cos xd2ydx2= sin x(1 cos x)223. y = sin2x + cos2x = 1 sodydx=d2ydx2= 024. y = sec2x tan2x = 1 sodydx=d2ydx2= 025.d4dx4 (sin x) =d3dx3 (cos x) =d2dx2 (sin x) =ddx (cos x) = sin x26.d4dx4 (cos x) =d3dx3 ( sin x) =d2dx2 ( cos x) =ddx (sin x) = cos x27.ddt_t2d2dt2 (t cos 3t)_ =ddt_t2ddt (cos 3t 3t sin 3t)_=ddt[t2(3 sin 3t 3 sin 3t 9t cos 3t)]=ddt[6t2sin 3t 9t3cos 3t]= (18t2cos 3t 12t sin 3t) + (27t3sin 3t 27t2cos 3t)= (27t312t) sin 3t 45t2cos 3t28.ddt_t ddt (cos t2)_ =ddt_t sin t2(2t) =ddt_2t2sin t2= 4t sin t22t2cos t2(2t) = 4t(sin t2+t2cos t2)29.ddx [f(sin 3x)] = f

(sin 3x)ddx (sin 3x) = 3 cos 3xf

(sin 3x)30.ddx [sin f(3x)] = cos[f(3x)]f

(3x)(3) = 3f

(3x) cos[f(3x)]31.dydx= cos x; slope of tangent at (0, 0) is 1; tangent: y = x.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5396 SECTION3.632.dydx= sec2x; slope of tangent at (/6,3/3) is sec2(/6) = 4/3;tangent: y 133 =43_x 16_33.dydx= csc2x; slope of tangent at (6,3) is 4, an equation fortangent: y 3 = 4_x 6_.34.dydx= sin x; slope of tangent at (0, 1) is 0; tangent: y = 1.35.dydx= sec xtan x, slope of tangent at (4,2) is 2, an equation fortangent isy 2 =2_x 4_.36.dydx= csc x cot x, slope of tangent at (/3, 23/3) is 2/3;tangent: y 233 = 23_x 13_.37.dydx= sin x; x = 38.dydx= cos x; x =12, x =3239.dydx= cos x 3 sin x;dydx= 0 gives tan x =13; x =6,7640.dydx= sin x 3 cos x;dydx= 0 gives tan x = 3; x =23,5341.dydx= 2 sin xcos x = sin 2x; x =2, ,3242.dydx= 2 sin x cos x = sin 2x; x =2, ,3243.dydx= sec2x 2;dydx= 0 gives sec x = 2; x =4,34,54,7444.dydx= 3 csc2x + 4;dydx= 0 gives csc x = 23; x =3,23,43,5345.dydx= 2 sec xtan x + sec2x; since sec x is never zero,dydx= 0 gives2 tan x + sec x = 0 so that sin x = 1/2; x =76,11646.dydx= csc2x + 2 csc x cot x; since csc x is never zero,dydx= 0 gives2 cot x csc x = 0 so that cos x = 1/2; x =3,53P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.6 9747. f(x) = x + 2 cosx, f

(x) = 1 2 sinx.(a) xf

(x) = 0 : 1 2 sinx = 0, sinx = 1/2, x = /6, 5/6.f

:0 65 62 - + +(b) f

(x) > 0 on (0, /6) (5/6, 2), (c) f

(x) < 0 on (/6, 5/6).48. f(x) = x 2 sinx, f

(x) = 1 2 cosx.(a) f

(x) = 0 : 1 2 cosx = 0, cosx =2/2, x = /4, 7/4.f

:0 47 42+ (b) f

(x) > 0 on (/4, 7/4), (c) f

(x) < 0 on (0, /4) (7/4, 2).49. f(x) = sinx + cosx, f

(x) = cosx sinx.(a) f

(x) = 0 : cosx sinx = 0, cosx = sinx, x = /4, 5/4.f

:0 45 42 - + +(b) f

(x) > 0 on (0, /4) (5/4, 2), (c) f

(x) < 0 on (/4, 5/4).50. f(x) = sinx cosx, f

(x) = cosx + sinx.(a) f

(x) = 0 : cosx + sinx = 0, cosx = sinx, x = 3/4, 7/4.f

:0 3 47+ - +42 (b) f

(x) > 0 on (0, 3/4) (7/4, 2), (c) f

(x) < 0 on (3/4, 7/4).51. (a)dydt=dydududxdxdt= (2u)(sec xtan x) = 2 sec2t tan t(b) y = sec2t 1,dydt= 2 sec t (sec t tan t) = 2 sec2t tan t52. (a)dydt=dydududxdxdt= 3_12(1 +u)_2_12_( sin x)(2) = 3_12(1 + cos 2t)_2( sin 2t)= 3(cos4t)(2 sin t cos t) = 6 cos5t sin t(b) y =_12(1 + cos 2t)_3= cos6t;dydt= 6 cos5t( sin t) = 6 cos5t sin t53. (a)dydt=dydududxdxdt= 4_12(1 u)_3_12_(sin x)(2) = 4_12(1 cos 2t)_3sin 2t= 4 sin6t (2 sin t cos t) = 8 sin7t cos tP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:5398 SECTION3.6(b) y =_12(1 cos 2t)_4= sin8t,dydt= 8 sin7t cos t54. (a)dydt=dydududxdxdt= (2u)( csc x cot x)(3) = (2 csc(3t)( csc(3t) cot(3t)3= 6 csc2(3t) cot(3t)(b) y = 1 csc2(3t);dydt= 2 csc(3t)[csc(3t) cot(3t)](3) = 6 csc2(3t) cot(3t)55.dndxn (cos x) =_(1)(n+1)/2sin x, n odd(1)n/2cos x, n even56. (a)ddx (cot x) =ddx_cos xsin x_ =sin x( sin x) cos x(cos x)sin2x=1sin2x= csc2x(b)ddx (sec x) =ddx_1cos x_ =1cos2x( sin x) =1cos x sin xcos x= sec x tan x(c)ddx (csc x) =ddx_1sin x_ =1sin2x(cos x) =1sin x cos xsin x= csc x cot x57.ddx (cos x) =ddx_sin_2x__ = cos_2x_ = sin x58. Dierentiating both sides, 2 cos 2x = 2(cos2x sin2x). Thus cos 2x = cos2x sin2x.59. f

(0) = limh0sin(0 +h) sin 0h= limh0sin hh= limx0sin xx60. f

(0) = limh0cos(0 +h) cos 0h= limh0cos h 1h= limx0cos x 1x61. f(x) = 2 sin x + 3 cos x +C 62. f(x) = tan x + cot x +C63. f(x) = sin 2x + sec x +C 64. f(x) = cos 3x3+ csc 2x2+C65. f(x) = sin(x2) + cos 2x +C 66. f(x) =tan(x3)3+ sec 2x +C67. (a) f

(x) = sin(1/x) +x cos(1/x)(1/x2)= sin(1/x) (1/x) cos(1/x)g

(x) = 2x sin(1/x) +x2cos(1/x)(1/x2)= 2x sin(1/x) cos(1/x)(b) limx0g

(x) = limx0[2x sin(1/x) cos(1/x)] = limx0cos(1/x) does not existP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.6 9968. (a) fmust be continuous at 0:limx0+f(x) =limx0+cos x = 1, limx0f(x) =limx0(ax +b) = b; thusb = 1Dierentiable at 0 :limh0+f(h) f(0)h=limh0+cos h 1h= 0,limh0f(h) f(0)h=limh0+ah + 1 1h= aTherefore,fis dierentiable at 0 ifa = 0 andb = 1.(b)1x1yf69. (a) Continuity:limx2/3sin x =32, limx2/3+(ax +b) =2a3+b; thus2a3+b =32Dierentiability:limx2/3cos x = 12, limx2/3+(a) = a; thus a = 12Therefore,fis dierentiable at 2/3 ifa = 12and b =123 +13(b)23x1gy70. (a) Continuity:limx/3(1 +a cosx) = 1 +12 a, limx/3+[b + sin (x/2)] = b +12which implies b =12 +12 a.Dierentiability:limx/3(a sinx) = 123 a, limx/3+[12cos (x/2)] =143 which implies a = 12.Therefore, f is dierentiable at /3 if a = 12and b =14.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53100 SECTION3.6(b)3x1fy71. Let y(t) = A sin t +Bcos t. Theny

(t) = A cos t Bsin t and y

(t) = 2A sin t 2Bcos tThus,d2ydt2+2y = 0.72. (a) = a sin(t +);

= a cos(t +);

= a2sin(t +)Thus, satises the equation.(b) = a sin(t +0)= a sin(t) cos 0a cos(t) sin 0= A sin(t) +Bcos(t) whereA = a sin 0, B = a cos 073. A =12 c2sin x;dAdx=12 c2cos x74. c =a2+b22ab cos x;dcdx=12a2+b22ab cos x(2ab sin x) =ab sin xa2+b22ab cos x75. (a) f(4p)(x) = k4pcos kx, f(4p+1)(x) = k4p+1sin kx, f(4p+2)(x) = k4p+2cos kx,f(4p+3)(x) = k4p+3sin kx, p = 0, 1, 2, . . .(b) m = k2, k = 1, 2, 3, . . .76. f(x) = A cos 2 x +Bsin 2 x; f

(x) = A2 sin 2 x +B2 cos 2 xf(0) = 2 = A = 2; f

(0) = 3 = B = 32= 32277. fhas horizontal at the points withx-coordinate12,32, = 3.39, = 6.0378. f(x) = sin x sin2x has horizontal tangents at:_6,14_,_2, 0_,_56,14_,_32, 2_79. f(x) = sinx, f

(x) = cosx; f(0) = 0, f

(0) = 1. Therefore an equation for the tangent line T isy = x.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.7 101The graphs of f and T are:2x-1-1fT2y| sinx x| < 0.01 on (0.39, 0.39).80. f(x) = tanx, f

(x) = sec2x; f(/4) = 1, f

(/4) = 2. Therefore an equation for the tangent lineT is y = 2x + 1 12.The graphs of f and T are:4 2x-11yfT| tanx (2x + 1 12)| < 0.01 on (0.712, 0.852).SECTION3.71. x2+y2= 42x + 2ydydx= 0dydx= xy2. x3+y33xy = 03x2+ 3y2dydx 3_y +x dydx_ = 0dydx=y x2y2x3. 4x2+ 9y2= 368x + 18ydydx= 0dydx= 4x9y4.x +y = 412x +12ydydx= 0dydx= yx5. x4+ 4x3y +y4= 14x3+ 12x2y + 4x3dydx + 4y3dydx= 0dydx= x3+ 3x2yx3+y3P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53102 SECTION3.76.x2x2y +xy2+y2= 12x 2xy x2dydx +y2+ 2xydydx + 2ydydx= 0dydx=2xy 2x y22xy + 2y x27. (x y)2y = 02(x y)_1 dydx_dydx= 0dydx=2(x y)2(x y) + 18. (y + 3x)24x = 02(y + 3x)_dydx + 3_4 = 0dydx= 3 +2y + 3x9. sin (x +y) = xycos (x +y)_1 +dydx_ = xdydx +ydydx=y cos (x +y)cos (x +y) x10. tan xy = xy; sec2(xy)_y +xdydx_ = y +xdydx;dydx= yx11. y2+ 2xy = 162ydydx + 2xdydx + 2y = 0(x +y)dydx +y = 0.Dierentiating a second time, we have(x +y)d2ydx2+dydx_2 +dydx_ = 0.Substitutingdydx=yx +y, we have(x +y)d2ydx2 y(x +y)_2x +yx +y_ = 0,d2ydx2=2xy +y2(x +y)3=16(x +y)3.12. x22xy + 4y2= 32x 2y 2xdydx + 8ydydx= 0x y + (4y x)dydx= 0.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.7 103Dierentiating a second time, we have1 dydx +_4dydx 1_ dydx + (4y x)d2ydx2= 0.Substitutingdydx=y x4y x, we haved2ydx2= 3(x22xy + 4y2)(4y x)3=9(4y x)3.13. y2+xy x2= 92ydydx +xdydx +y 2x = 0.Dierentiating a second time, we have_2_dydx_2+ 2yd2ydx2_+_xd2ydx2+dydx_+dydx 2 = 0(2y +x)d2ydx2+ 2__dydx_2+dydx 1_ = 0.Substitutingdydx=2x y2y +x, we have(2y +x)d2ydx2+ 2_(2x y)2+ (2x y)(2y +x) (2y +x)2(2y +x)2_ = 0d2ydx2=10(y2+xy x2)(2y +x)3=90(2y +x)3.14. x23xy = 182x 3y 3xdydx= 0.Dierentiating a second time, we have2 3dydx 3dydx 3xd2ydx2= 0Substitutingdydx=2x 3y3x, we haved2ydx2= 6(x 3y)9x2= 6(x23xy)9x3= 6(18)9x3= 12x315. 4 tan y = x34 sec2ydydx= 3x2dydx=34 x2cos2yd2ydx2=32 x cos2y + 34 x2_2 cos y(sin y)dydx_=32 x cos2y 98 x4sin y cos3yP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53104 SECTION3.716.sin2x + cos2y = 12 sin x cos x 2 cos y sin ydydx= 0sin 2x sin 2ydydx= 0dydx=sin 2xsin 2yd2ydx2=sin 2y(cos 2x)2 sin 2x(cos 2y)2dydxsin22ySubstitutingdydx=sin 2xsin 2yand using the double angle formulas, we nd thatd2ydx2=8_cos2xcos2y(sin2y sin2x) sin2xsin2y(cos2y cos2x)sin32y= 0since sin2x = sin2y and cos2x = cos2y from the original equation.17. x24y2= 9, 2x 8ydydx= 0.At (5, 2), we getdydx=58. Then,2 8_yd2ydx2+_dydx_2_ = 0.At (5, 2) we get2 8_2d2ydx2+ 2564_ = 0 so thatd2ydx2= 9128.18. x2+ 4xy +y3+ 5 = 0, 2x + 4y + 4xdydx + 3y2dydx= 0.At (2, 1), we get 4 4 + 8dydx + 3dydx= 0 sodydx= 0. Dierentiating again,2 + 4dydx + 4dydx + 4xd2ydx2+ 6y_dydx_2+ 3y2d2ydx2= 0At (2, 1) we get 2 + 11d2ydx2= 0 sod2ydx2= 21119. cos (x + 2y) = 0 sin (x + 2y)_1 + 2dydx_ = 0.At (/6,/6), we getdydx= 1/2. Then,cos (x + 2y)_1 + 2dydx_2sin (x + 2y)_2d2ydx2_ = 0.At (/6,/6), we getcos 2(0)2sin 2_2d2ydx2_ = 0 so thatd2ydx2= 0.P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.7 10520. x = sin2y 1 = 2 sin y cos ydydx= sin 2ydydx.At (1/2, /4), we getdydx= 1. Dierentiating again,0 = 2 cos 2y_dydx_2+ sin 2yd2ydx2.At (1/2, /4), we getd2ydx2= 0.21.2x + 3y = 52 + 3dydx= 0slope of tangent at (2, 3) : 2/3tangent: y 3 = 23(x + 2)normal: y 3 =32(x + 2)22.9x2+ 4y2= 7218x + 8ydydx= 0slope of tangent at (2, 3) : 32tangent: y 3 = 32(x 2)normal: y 3 =23(x 2)23.x2+xy + 2y2= 282x +xdydx +y + 4ydydx= 0slope of tangent at (2, 3) : 1/2tangent: y + 3 = 12(x + 2)normal: y + 3 = 2(x + 2)24.x3axy + 3ay2= 3a33x2ay axdydx + 6aydydx= 0slope of tangent at (a, a) : 25tangent: y a = 25(x a)normal: y a =52(x a)25.x = cos ydydx1 = sin ydydxslope of tangent at_12, 3_ :23tangent: y 3= 23_x 12_normal: y 3=32_x 12_26.tan xy = xsec2(xy)_y +xdydx_ = 1slope of tangent at (1, /4) :2 4tangent: y 4=2 4(x 1)normal: y 4= 42 (x 1)27.dydx=12(x3+ 1)1/2ddx (x3+ 1) =32x2(x3+ 1)1/228.dydx=13(x + 1)2/329.dydx=14(2x2+ 1)3/4ddx (2x2+ 1) = x(2x2+ 1)3/430.dydx=13(x + 1)2/3(x + 2)2/3+ (x + 1)1/3_23_(x + 2)1/3=3x + 43(x + 1)2/3(x + 2)1/331.dydx=_2 x2_x3 x2_+_3 x2_x2 x2_ =x(2x25)2 x23 x232.dydx=32(x4x + 1)1/2(4x31) =32x4x + 1 (4x31)P1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53106 SECTION3.733.ddx_x +1x_ =ddx (x1/2+x1/2) =12x1/2 12x3/2=12x3/2(x 1)34.ddx__3x + 12x + 5_ =12_3x + 12x + 5_1/2_(2x + 5)3 (3x + 1)2(2x + 5)2_ =132(2x + 5)2_2x + 53x + 135.ddx_xx2+ 1_ =ddx_x(x2+ 1)1/2_= x_12(x2+ 1)3/2(2x)_+ (x2+ 1)1/2= (x2+ 1)3/236.ddx_x2+ 1x_ =x_12_(x2+ 1)1/2(2x) (x2+ 1)1/2x2=x2x2+ 1 _x2+ 1x2=1x2x2+ 137. (a) (b) (c)38. y = (a2+x2)1/2;dydx=12(a2+x2)1/2(2x) = x(a2+x2)1/2;d2ydx2= (a2+x2)1/2 12x(a2+x2)3/2(2x) =a2(a2+x2)3/239. y = (a +bx)1/3;dydx=b3(a +bx)2/3;d2ydx2= 2b29(a +bx)5/340. y = x(a2x2)1/2;dydx= (a2x2)1/2+12 x(a2x2)1/2(2x) = (a2x2)1/2x2(a2x2)1/2d2ydx2=12(a2x2)1/2(2x) 2x(a2x2)1/2x2_12_(a2x2)3/2(2x) =x(2x23a2)(a2x2)3/241. y =x tanx;dydx=12x tanx +x sec2x_12x_ =12x tanx + 12sec2xd2ydx2=2x sec2x(1/2x) tanx(1/x)4x+ secx secx tanx(1/2x)=x sec2x tanx + 2x sec2xtanx4xx42. y =x sinx;dydx=12x sinx +x cosx_12x_ =12x sinx + 12cosxP1:PBU/OVY P2:PBU/OVY QC:PBU/OVY T1:PBUJWDD027-03 JWDD027-Salas-v1 November25,2006 15:53SECTION3.7 107d2ydx2=2xcosx(1/2x) sinx(1/x)4x 12sinx(1/2x)=xcosx sinx4x3/2 sinx4x43. Dierentiation of x2+y2= r2gives 2x + 2ydydx= 0 so that the slope of the normal line is1dy/dx=yx(x = 0).Let(x0,y0)beapointonthecircle. Clearly, if x0 = 0, thenormal line, x = 0, passesthroughtheorigin. Ifx0 = 0, the normal line isy y0 =y0x0(x x0), which simplies t