[PDF] solutions-Adams-Calculus-A-Complete-Course-8th-Edition-[konkur

INSTRUCTORíS SOLUTIONS MANUAL to accompany ADAMS / ESSEX CALCULUS: A COMPLETE COURSE; CALCULUS: SINGLE VARIABLE; and CALCULUS: SEVERAL VARIABLES Eighth Edition Prepared by Robert A. Adams University of British Columbia Christopher Essex University of Windsor Ontario Toronto Copyright © 2014 Pearson Canada Inc., Toronto, Ontario. Pearson Canada. All rights reserved. This work is protected by Canadian copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination or sale of any part of this work (including on the Internet) will destroy the integrity of the work and is not permitted. The copyright holder grants permission to instructors who have adopted Calculus: A Complete Course, Eighth Edition, by Adams/Essex to post this material online only if the use of the website is restricted by access codes to students in the instructorís class that is using the textbook and provided the reproduced material bears this copyright notice.

FOREWORDThese solutions are provided for the benefit of instructors using the textbooks:Calculus: A Complete Course (8th Edition),Single-Variable Calculus (8th Edition),andCalculus of Several Variables (8th Edition)by R. A. Adams and Chris Essex, published by Pearson Education Canada. For the most part,the solutions are detailed, especially in exercises on core material and techniques. Occasion-ally some details are omitted - for example, in exercises on applications of integration, theevaluation of the integrals encountered is not always given with the same degree of detail asthe evaluation of integrals found in those exercises dealing specifically with techniques of in-tegration.Instructors may wish to make these solutions available to their students. However, studentsshould use such solutions with caution. It is always more beneficial for them to attempt ex-ercises and problems on their own, before they look at solutions done by others. If they ex-amine solutions as "study material" prior to attempting the exercises, they can lose much ofthe benefit that follows from diligent attempts to develop their own analytical powers. Whenthey have tried unsuccessfully to solve a problem, then looking at a solution can give them a"hint" for a second attempt. SeparateStudent Solutions Manualsfor the books are availablefor students. They contain the solutions to the even-numbered exercises only.Apr, 2012.R. A. Adamsadms@math.ubc.caChris Essexessex@uwo.cawww.konkur.in

CONTENTSSolutions for Chapter P1Solutions for Chapter 123Solutions for Chapter 240Solutions for Chapter 382Solutions for Chapter 4109Solutions for Chapter 5179Solutions for Chapter 6215Solutions for Chapter 7270Solutions for Chapter 8318Solutions for Chapter 9353Solutions for Chapter 10393Solutions for Chapter 11421Solutions for Chapter 12450Solutions for Chapter 13494Solutions for Chapter 14541Solutions for Chapter 15583Solutions for Chapter 16614Solutions for Chapter 17641Solutions for Chapter 18648Solutions for Chapter 18 extended669Solutions for Appendices699NOTE: Chapter 18 extended is only needed by users ofCalculus of Several Variables (Eighth Edition)www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.1 (PAGE 10)CHAPTER P. PRELIMINARIESSection P.1 Real Numbers and the Real Line(page 10)1.29D0:22222222 D0:22.111D0:09090909 D0:093.IfxD0:121212, then 100xD12:121212 D12Cx.Thus 99xD12 andxD12=99D4=33.4.IfxD3:277777, then 10x32D0:77777and100x320D7C.10x32/, or 90xD295. ThusxD295=90D59=18.5.1=7D0:142857142857 D0:1428572=7D0:285714285714 D0:2857143=7D0:428571428571 D0:4285714=7D0:571428571428 D0:571428note the same cyclic order of the repeating digits5=7D0:714285714285 D0:7142856=7D0:857142857142 D0:8571426.Two different decimal expansions can represent the samenumber. For instance, both 0:999999 D0:9 and1:000000 D1:0 represent the number 1.7.x0 andx5 dene the interval [0;5].8.x<2 andx 3 dene the interval [3;2/.9.x>5 orx<6 denes the union.1;6/[.5;1/.10.x 1 denes the interval.1;1].11.x>2 denes the interval.2;1/.12.x<4 orx2 denes the interval.1;1/,that is, thewhole real line.13.If2x>4, thenx<2. Solution:.1;2/14.If 3xC58, then 3x853 andx1. Solution:.1;1]15.If 5x373x, then 8x10 andx5=4. Solution:.1;5=4]16.If6x43x42, then 6x6x8. Thus 147xandx2. Solution:.1;2]17.If 3.2x/ <2.3Cx/, then 0<5xandx>0. Solution:.0;1/18.Ifx2<9, thenjxj<3 and32, then 1>3.2x/D63x, so 3x>5andx>5=3. This case has solutionsx>2.Solution:.1;5=3/[.2;1/.20.Given:.xC1/=x2.CASE I. Ifx>0, thenxC12x, sox1.CASE II. Ifx<0, thenxC12x, sox1. (notpossible)Solution:.0;1].21.Given:x22x0. Thenx.x2/0. This is onlypossible ifx0 andx2. Solution: [0;2].22.Given 6x25x 1, then.2x1/.3x1/0, soeitherx1=2 andx1=3, orx1=3 andx1=2.The latter combination is not possible. The solution set is[1=3;1=2].23.Givenx3>4x, we havex.x24/ >0. This is possibleifx<0 andx2<4, or ifx>0 andx2>4. Thepossibilities are, therefore,20, thenx22xC8, so thatx22x80, or.x4/.xC2/0. This ispossible forx>0 only ifx4.CASE II. Ifx<0, then we must have.x4/.xC2/0,which is possible forx<0 only ifx 2.Solution: [2;0/[[4;1/.26.Given:3x1<2xC1.CASE I. Ifx>1 then.x1/.xC1/ >0, so that3.xC1/ <2.x1/. Thusx<5. There are no solutionsin this case.CASE II. If12.x1/. Thusx>5. In this case allnumbers in.1;1/are solutions.CASE III. Ifx<1, then.x1/.xC1/ >0, so that3.xC1/ <2.x1/. Thusx<5. All numbersx<5are solutions.Solutions:.1;5/[.1;1/.27.Ifjxj D3 thenxD 3.28.Ifjx3j D7, thenx3D 7, soxD 4 orxD10.29.Ifj2tC5j D4, then 2tC5D 4, sotD 9=2 ortD 1=2.30.Ifj1tj D1, then 1tD 1, sotD0 ortD2.1www.konkur.in

SECTION P.1 (PAGE 10)ADAMS and ESSEX: CALCULUS 831.Ifj83sj D9, t?e? 83sD 9, s? 3sD 1 ?? 17, a?dsD 1=3 ??sD17=3.32.Ifs21D1, t?e?s21D 1, s?sD0 ??sD4.33.Ifjxj<2, t?e?x?s ??.2;2/.34.Ifjxj 2, t?e?x?s ?? [2;2].35.Ifjs1j 2, t?e? 12s1C2, s?s?s ?? [1;3].36.IfjtC2j<1, t?e?21jx3jsays t?at t?e d?sta?cef??mxt?1 ?s g?eate? t?a? t?e d?sta?ce f??mxt? 3, s?xmust be t? t?e ??g?t ?f t?e ????t ?alfway betwee?1a?d 3. T?usx>1.42.jx3j<2jxj ,x26xC9D.x3/2<4x2,3x2C6x9>0,3.xC3/.x1/ >0. T??s??equal?ty ??lds ?fx<3 ??x>1.43.jaj Da?f a?d ??ly ?fa0. It ?s false ?fa<0.44.T?e equat???jx1j D1x??lds ?fjx1j D .x1/,t?at ?s, ?fx1<0, ??, equ?vale?tly, ?fx<1.45.T?e t??a?gle ??equal?tyjxCyj jxj C jyj?m?l?es t?atjxj jxCyj jyj:A??ly t??s ??equal?ty w?t?xDaba?dyDbt? getjabj jaj jbj:S?m?la?ly,jabj D jbaj jbj jaj. S??cejaj jbj?s equal t? e?t?e?jaj jbj??jbj jaj, de?e?d??g ?? t?es?zes ?faa?db, we ?avejabj jaj jbj:Section P.2 Cartesian Coordinates in thePlane (page 16)1.F??mA.0;3/t?B.4;0/,∞xD40D4 a?d∞yD03D 3.jABj Dp42C.3/2D5.2.F??mA.1;2/t?B.4;10/,∞xD4.1/D5 a?d∞yD 102D 12.jABj Dp52C.12/2D13.3.F??mA.3;2/t?B.1;2/,∞xD 13D 4 a?d∞yD 22D 4.jABj Dp.4/2C.4/2D4p2.4.F??mA.0:5;3/t?B.2;3/,∞xD20:5D1:5 a?d∞yD33D0.jABj D1:5.5.Sta?t??g ????t:.2;3/. I?c?eme?ts∞xD4,∞yD 7.New ??s?t??? ?s.2C4;3C.7//, t?at ?s,.2;4/.6.A???val ????t:.2;2/. I?c?eme?ts∞xD 5,∞yD1.Sta?t??g ????t was.2.5/;21/, t?at ?s,.3;3/.7.x2Cy2D1 ?e??ese?ts a c??cle ?f ?ad?us 1 ce?t?ed at t?e???g??.8.x2Cy2D2 ?e??ese?ts a c??cle ?f ?ad?usp2 ce?t?ed att?e ???g??.9.x2Cy21 ?e??ese?ts ????ts ??s?de a?d ?? t?e c??cle ?f?ad?us 1 ce?t?ed at t?e ???g??.10.x2Cy2D0 ?e??ese?ts t?e ???g??.11.yx2?e??ese?ts all ????ts ly??g ?? ?? ab?ve t?e?a?ab?layDx2.12.y

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.2 (PAGE 16)24.T?e l??e t???ug?.2;0/a?d.0;2/?as sl??emD.20/=.0C2/D1 a?d equat???yD2Cx.25.IfmD 2 a?dbDp2, t?e? t?e l??e ?as equat???yD 2xCp2.26.IfmD 1=2 a?dbD 3, t?e? t?e l??e ?as equat???yD .1=2/x3, ??xC2yD 6.27.3xC4yD12 ?asx??te?ce?taD12=3D4 a?dy??te?ce?tbD12=4D3. Its sl??e ?sb=aD 3=4.yx3xC4yD12F?g. P.2.2728.xC2yD 4 ?asx??te?ce?taD 4 a?dy??te?ce?tbD 4=2D 2. Its sl??e ?sb=aD2=.4/D 1=2.yxxC2yD 4F?g. P.2.2829.p2xp3yD2 ?asx??te?ce?taD2=p2Dp2a?dy??te?ce?tbD 2=p3. Its sl??e ?sb=aD2=p6Dp2=3.yxp2xp3yD2F?g. P.2.2930.1:5x2yD3 ?asx??te?ce?taD 3=1:5D2 a?dy??te?ce?tbD 3=.2/D3=2. Its sl??e ?sb=aD3=4.yx1:5x2yD3F?g. P.2.3031.l??e t???ug?.2;1/?a?allel t?yDxC2 ?syDx1; l??e?e??e?d?cula? t?yDxC2 ?syD xC3.32.l??e t???ug?.2;2/?a?allel t? 2xCyD4 ?s2xCyD 2; l??e ?e??e?d?cula? t? 2xCyD4 ?sx2yD 6.33.We ?ave3xC4yD 62x3yD13H)6xC8yD 126x9yD39:Subt?act??g t?ese equat???s g?ves 17yD 51, s?yD 3a?dxD.139/=2D2. T?e ??te?sect??? ????t ?s.2;3/.34.We ?ave2xCyD85x7yD1H)14xC7yD565x7yD1:Add??g t?ese equat???s g?ves 19xD57, s?xD3 a?dyD82xD2. T?e ??te?sect??? ????t ?s.3;2/.35.Ifa6D0 a?db6D0, t?e?.x=a/C.y=b/D1 ?e??ese?tsa st?a?g?t l??e t?at ?s ?e?t?e? ????z??tal ??? ve?t?cal, a?dd?es ??t ?ass t???ug? t?e ???g??. Putt??gyD0 we getx=aD1, s? t?ex??te?ce?t ?f t??s l??e ?sxDa; ?utt??gxD0 g?vesy=bD1, s? t?ey??te?ce?t ?syDb.36.T?e l??e.x=2/.y=3/D1 ?asx??te?ce?taD2, a?dy??te?ce?tbD 3.yx3x2y3D12F?g. P.2.3637.T?e l??e t???ug?.2;1/a?d.3;1/?as sl??emD.11/=.32/D 2 a?d equat???yD12.x2/D52x. Itsy??te?ce?t ?s 5.3www.konkur.in

SECTION P.2 (PAGE 16)ADAMS and ESSEX: CALCULUS 838.T?e l??e t???ug?.2;5/a?d.k;1/?asx??te?ce?t 3, s?als? ?asses t???ug?.3;0/. Its sl??emsat?ses10k3DmD053C2D 1:T?usk3D 1, a?d s?kD2.39.+DAxCB. If+D5;000 w?e?xD10;000 a?d+D6;000 w?e?xD15;000, t?e?10;000ACBD5;00015;000ACBD6;000Subt?act??g t?ese equat???s g?ves 5;000AD1;000, s?AD1=5. F??m t?e ?st equat???, 2;000CBD5;000,s?BD3;000. T?e c?st ?f ????t??g 100,000 ?am??lets ?s$100;000=5C3;000D$23;000.40.40δa?d40δ?s t?e same tem?e?atu?e ?? b?t? t?eFa??e??e?t a?d Cels?us scales.+504030201010203040F504030 2010 10 20 30 405060 70 80.40;40/+D59.F32/+DFF?g. P.2.4041.AD.2;1/;BD.6;4/;+D.5;3/jABj Dp.62/2C.41/2Dp25D5jA+j Dp.52/2C.31/2Dp25D5jB+j Dp.65/2C.4C3/2Dp50D5p2:S??cejABj D jA+j, t??a?gleAB+?s ?s?sceles.42.AD.0;0/;BD.1;p3/;+D.2;0/jABj Dq.10/2C.p30/2Dp4D2jA+j Dp.20/2C.00/2Dp4D2jB+j Dq.21/2C.0p3/2Dp4D2:S??cejABj D jA+j D jB+j, t??a?gleAB+?s equ?late?al.43.AD.2;1/;BD.1;3/;+D.3;2/jABj Dp.12/2C.3C1/2Dp17jA+j Dp.32/2C.2C1/2Dp34Dp2p17jB+j Dp.31/2C.23/2Dp17:S??cejABj D jB+ja?djA+j Dp2jABj, t??a?gleAB+?s a? ?s?sceles ??g?ta?gled t??a?gle w?t? ??g?t a?gle atB. T?usAB+=?s a squa?e ?f=?s d?s?laced f??m+by t?e same am?u?tA?s f??mB, t?at ?s, by ??c?eme?ts1xD21D1 a?d1yD 13D 4. T?us=D.3C1;2C.4//D.2;2/.44.IfMD.xm;ym/?s t?e m?d????t ?fP1P2, t?e? t?e d?s?laceme?t ?fMf??mP1equals t?e d?s?laceme?t ?fP2f??mM:xmx1Dx2xm;ymy1Dy2ym:T?usxmD.x1Cx2/=2 a?dymD.y1Cy2/=2.45.IfQD.xq;yq/?s t?e ????t ??P1P2t?at ?s tw? t???ds ?ft?e way f??mP1t?P2, t?e? t?e d?s?laceme?t ?fQf??mP1equals tw?ce t?e d?s?laceme?t ?fP2f??mQ:xqx1D2.x2xq/;yqy1D2.y2yq/:T?usxqD.x1C2x2/=3 a?dyqD.y1C2y2/=3.46.Let t?e c???d??ates ?fPbe.x;0/a?d t??se ?fQbe.X;2X/. If t?e m?d????t ?fPQ?s.2;1/, t?e?.xCX/=2D2; .02X/=2D1:T?e sec??d equat??? ?m?l?es t?atXD 1, a?d t?e sec??d t?e? ?m?l?es t?atxD5. T?usP?s.5;0/.47.p.x2/2Cy2D4 says t?at t?e d?sta?ce ?f.x;y/f??m.2;0/?s 4, s? t?e equat??? ?e??ese?ts a c??cle ?f ?ad?us 4ce?t?ed at.2;0/.48.p.x2/2Cy2Dpx2C.y2/2says t?at.x;y/?sequ?d?sta?t f??m.2;0/a?d.0;2/. T?us.x;y/mustl?e ?? t?e l??e t?at ?s t?e ??g?t b?sect?? ?f t?e l??e f??m.2;0/t?.0;2/. A s?m?le? equat??? f?? t??s l??e ?sxDy.49.T?e l??e 2xCkyD3 ?as sl??emD 2=k. T??s l??e?s ?e??e?d?cula? t? 4xCyD1, w??c? ?as sl??e4,???v?dedmD1=4, t?at ?s, ???v?dedkD 8. T?e l??e ?s?a?allel t? 4xCyD1 ?fmD 4, t?at ?s, ?fkD1=2.50.F?? a?y value ?fk, t?e c???d??ates ?f t?e ????t ?f ??te?sect??? ?fxC2yD3 a?d 2x3yD 1 w?ll als? sat?sfyt?e equat???.xC2y3/Ck.2x3yC1/D04www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.3 (PAGE 22)because they cause both expressions in parentheses to be0. The equation above is linear inxandy, and so rep-resents a straight line for any choice ofk. This line willpass through.1;2/provided 1C43Ck.26C1/D0,that is, ifkD2=3. Therefore, the line through the pointof intersection of the two given lines and through thepoint.1;2/has equationxC2y3C23.2x3yC1/D0;or, on simplification,xD1.Section P.3 Gra⎷hs o{ Quadratic Equations(⎷a}e 22?1.x2Cy2D162.x2C.y2/2D4, orx2Cy24yD03..xC2/2Cy2D9, orx2Cy2C4yD54..x3/2C.yC4/2D25, orx2Cy26xC8yD0.5.x2Cy22xD3x22xC1Cy2D4.x1/2Cy2D4centre:.1;0/; radius 2.6.x2Cy2C4yD0x2Cy2C4yC4D4x2C.yC2/2D4centre:.0;2/; radius 2.7.x2Cy22xC4yD4x22xC1Cy2C4yC4D9.x1/2C.yC2/2D9centre:.1;2/; radius 3.8.x2Cy22xyC1D0x22xC1Cy2yC14D14.x1/2Cy122D14centre:.1;1=2/; radius 1=2.9.x2Cy2>1 represents all points lying outside the circleof radius 1 centred at the origin.10.x2Cy2<4 represents the open disk consisting of allpoints lying inside the circle of radius 2 centred at theorigin.11..xC1/2Cy24 represents the closed disk consisting ofall points lying inside or on the circle of radius 2 centredat the point.1;0/.12.x2C.y2/24 represents the closed disk consisting ofall points lying inside or on the circle of radius 2 centredat the point.0;2/.13.Together,x2Cy2>1 andx2Cy2<4 represent annulus(washer-shaped region) consisting of all points that areoutside the circle of radius 1 centred at the origin andinside the circle of radius 2 centred at the origin.14.Together,x2Cy24 and.xC2/2Cy24 represent theregion consisting of all points that are inside or on boththe circle of radius 2 centred at the origin and the circleof radius 2 centred at.2;0/.15.Together,x2Cy2<2xandx2Cy2<2y(or, equivalently,.x1/2Cy2<1 andx2C.y1/2<1) represent theregion consisting of all points that are inside both thecircle of radius 1 centred at.1;0/and the circle of radius1 centred at.0;1/.16.x2Cy24xC2y>4 can be rewritten.x2/2C.yC1/2>9. This equation, taken together withxCy>1, represents all points that lie both outside thecircle of radius 3 centred at.2;1/and above the linexCyD1.17.The interior of the circle with centre.1;2/and radiusp6 is given by.xC1/2C.y2/2<6, orx2Cy2C2x4y<1.18.The exterior of the circle with centre.2;3/and ra-dius 4 is given by.x2/2C.yC3/2>16, orx2Cy24xC6y>3.19.x2Cy2<2;x120.x2Cy2>4; .x1/2C.y3/2<1021.The parabola with focus.0;4/and directrixyD 4 hasequationx2D16y.22.The parabola with focus.0;1=2/and directrixyD1=2has equationx2D 2y.23.The parabola with focus.2;0/and directrixxD 2 hasequationy2D8x.24.The parabola with focus.1;0/and directrixxD1 hasequationy2D 4x.25.yDx2=2 has focus.0;1=2/and directrixyD 1=2.yx.0;1=2/yD1=2yDx2=2Fig. P.3.2526.yD x2has focus.0;1=4/and directrixyD1=4.5www.konkur.in

SECTION P.3 (PAGE 22)ADAMS and ESSEX: CALCULUS 8yxyD1/4(0,1/4)yDx2Fig. P.3.2627.xD y2/4 has focus(1,0)and directrixxD1.yxxD1(1,0)xDy2/4Fig. P.3.2728.xDy2/16 has focus(4,0)and directrixxD 4.yx(4,0)xDy2/16xD4Fig. P.3.2829.yx(3,3)4(4,2)3yDx2Version (b)Version (c)Version (d)Version (a)Fig. P.3.29a) has equationyDx23.b) has equationyD(x4)2oryDx28xC16.c) has equationyD(x3)2C3 oryDx26xC12.d) has equationyD(x4)22, oryDx28xC14.3?.a) IfyDmxis shifted to the right by amountx1, theequationyDm(xx1)results. If(a,b)satises thisequation, thenbDm(ax1), and sox1Da(b/m).Thus the shifted equation isyDm(xaC(b/m))Dm(xa)Cb.b) IfyDmxis shifted vertically by amounty1,the equationyDmxCy1results. If(a,b)satises this equation, thenbDmaCy1, andsoy1Dbma. Thus the shifted equation isyDmxCbmaDm(xa)Cb, the sameequation obtained in part (a).3∞.yDp(x/3)C132.4yD?xC133.yDp(3x/2)C134.(y/2)D?4xC135.yD1x2shifted down 1, left 1 givesyD (xC1)2.3?.x2Cy2D5 shifted up 2, left 4 gives(xC4)2C(y2)2D5.37.yD(x1)21 shifted down 1, right 1 givesyD(x2)22.38.yD?xshifted down 2, left 4 givesyD?xC42.?www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.3 (PAGE 22)39.yDx2C3,yD3xC1( Subtracting these equationsgivesx23xC2D0, or.x1/.x2/D0( ThusxD1 orxD2( The corresponding values ofyare 4 and 7( Theintersection points are.1;4/and.2;7/(40.yDx26,yD4xx2( Subtracting these equationsgives2x24x6D0, or 2.x3/.xC1/D0( ThusxD3orxD 1( The corresponding values ofyare 3 and5(The intersection points are.3;3/and.1;5/(41.x2Cy2D25, 3xC4yD0( The second equation says thatyD 3x=4( Substituting this into the rst equation gives25x2=16D25, soxD 4( IfxD4, then the secondequation givesyD 3, ifxD 4, thenyD3( Theintersection points are.4;3/and.4;3/( Note thathaving found values forx, we substituted them into thelinear equation rather than the quadratic equation to ndthe corresponding values ofy( Had we substituted intothe quadratic equation we would have got more solutions(four points in all), but two of them would have failed tosatisfy 3xC4yD12( When solving systems of nonlinearequations you should always verify that the solutions yound do satisfy the given equations(42.2x2C2y2D5,xyD1( The second equation says thatyD1=x( Substituting this into the rst equation gives2x2C.2=x2/D5, or 2x45x2C2D0( This equationfactors to.2x21/.x22/D0, so its solutions arexD 1=p2 andxD p2( The corresponding valuesofyare given byyD1=x( Therefore, the intersectionpoints are.1=p2;p2/,.1=p2;p2/,.p2;1=p2/, and.p2;1=p2/(43..x2=4/Cy2D1 is an ellipse with major axis between.2;0/and.2;0/and minor axis between.0;1/and.0;1/(yxx24Cy2D1Fig( P(3(4344.9x2C16y2D144 is an ellipse with major axis between.4;0/and.4;0/and minor axis between.0;3/and.0;3/(yx9x2C16y2D144Fig( P(3(4445..x3/29C.yC2/24D1 is an ellipse with centre at.3;2/, major axis between.0;2/and.6;2/andminor axis between.3;4/and.3;0/(yx.3;2/.x3/29C.yC2/24D1Fig( P(3(4546..x1/2C.yC1/24D4 is an ellipse with centre at.1;1/, major axis between.1;5/and.1;3/and minoraxis between.1;1/and.3;1/(yx.x1/2C.yC1/24D4.1;1/Fig( P(3(4647..x2=4/y2D1 is a hyperbola with centre at the origin and passing through.2;0/( Its asymptotes areyD x=2(7www.konkur.in

SECTION P.3 (PAGE 22)ADAMS and ESSEX: CALCULUS 8yxx24y2D1yDx/2yDx/2F?g. P.3.4748.x2y2D 1 ?s a ?ecta?gula? ?y?e?b?la w?t? ce?t?e att?e ???g?? a?d ?ass??g t???ug?(0,1). Its asym?t?tesa?eyD x.yxx2y2D1yDxyDxF?g. P.3.4849.xyD 4 ?s a ?ecta?gula? ?y?e?b?la w?t? ce?t?e att?e ???g?? a?d ?ass??g t???ug?(2,2)a?d(2,2). Itsasym?t?tes a?e t?e c???d??ate axes.yxxyD4F?g. P.3.495?.(x1)(yC2)D1 ?s a ?ecta?gula? ?y?e?b?la w?t? ce?t?eat(1,2)a?d ?ass??g t???ug?(2,1)a?d(0,3). Itsasym?t?tes a?exD1 a?dyD 2.yxxD1yD 2(x1)(yC2)D1F?g. P.3.505∞.a) ?e?lac??gxw?t?x?e?laces a g?a?? w?t? ?ts ?eect??? ac??ss t?eyax?s.b) ?e?lac??gyw?t?y?e?laces a g?a?? w?t? ?ts ?eect??? ac??ss t?exax?s.52.?e?lac??gxw?t?xa?dyw?t?y?eects t?e g?a?? ??b?t? axes. T??s ?s equ?vale?t t? ??tat??g t?e g?a?? 180ab?ut t?e ???g??.53.jxj C jyj D1.I? t?e ?st quad?a?t t?e equat??? ?sxCyD1.I? t?e sec??d quad?a?t t?e equat??? ?sxCyD1.I? t?e t???d quad?a?t t?e equat??? ?sxyD1.I? t?e f?u?t? quad?a?t t?e equat??? ?sxyD1.yx1jxj C jyj D1111F?g. P.3.53Section P.4 Functions and Their Graphs(page 32)∞.f(x)D1Cx2; d?ma??R, ?a?ge [1,1)2.f(x)D1px; d?ma?? [0,1),?a?ge(1,1]3.G(x)Dp82x; d?ma??(1,4], ?a?ge [0,1)4.F(x)D1/(x1); d?ma??(1,1)[(1,1),?a?ge(1,0)[(0,1)8www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.4 (PAGE 32)5.?.t/Dtp2t; domain.1;2/? ran}eR. (The equationyD?.t/can be squared and rewritten ast2Cy2t2y2D?? a quadratic equation inthavin} realsolutions {or every real value o{y. Thus the ran}e o{?contains all real numbers.?6.g.x/D∞∞px2; domain.2;3/[.3;1/?ran}e.1;?/[.?;1/.The equationyDg.x/can be solved{orxD2.∞.∞=y//2so has a real solution ⎷rovidedy6D?.7.yxyxyxyx}ra⎷h (i?}ra⎷h (iii?}ra⎷h (iv?}ra⎷h (ii?Fi}. P.4.7Gra⎷h (ii? is the }ra⎷h o{ a {unction because verticallines can meet the }ra⎷h only once. Gra⎷hs (i?? (iii??and (iv? do not have this ⎷ro⎷erty? so are not }ra⎷hs o{{unctions.8.yxyxyxyx}ra⎷h (a?}ra⎷h (b?}ra⎷h (d?}ra⎷h (c?Fi}. P.4.8a? is the }ra⎷h o{x.∞x/2? which is ⎷ositive {orx>?.b? is the }ra⎷h o{x2x3Dx2.∞x/? which is ⎷ositivei{x<∞.c? is the }ra⎷h o{xx4? which is ⎷ositive i{ ?

SECTION P.4 (PAGE 32)ADAMS and ESSEX: CALCULUS 810.x f.x/Dx2=3? ??:5 ?:?299?∞∞∞:5 ∞:3∞?42∞:5874yxyDx2=3Fi}. P.4.∞?11.f.x/Dx2C∞ is even:f.x/Df.x/12.f.x/Dx3Cxis odd:f.x/D f.x/13.f.x/Dxx2∞is odd:f.x/D f.x/14.f.x/D∞x2∞is even:f.x/Df.x/15.f.x/D∞x2is odd about.2;?/:f.2x/D f.2Cx/16.f.x/D∞xC4is odd about.4;?/:f.4x/D f.4Cx/17.f.x/Dx2?xis even aboutxD3:f.3x/Df.3Cx/18.f.x/Dx32 is odd about.?;2/:f.x/C2D .f.x/C2/19.f.x/D jx3j D jxj3is even:f.x/Df.x/20.f.x/D jxC∞jis even aboutxD ∞:f.∞x/Df.∞Cx/21.f.x/Dp2xhas no symmetry.22.f.x/Dp.x∞/2is even aboutxD∞:f.∞x/Df.∞Cx/23.yxyDx224.yxyD∞x225.yxyD.x∞/226.yxyD.x∞/2C∞27.yxyD∞x328.yxyD.xC2/310www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.4 (PAGE 32)29.yxyDpxC130.yxyDpxC131.yxyDjxj32.yxyDjxj133.yxyDjx2j34.yxyD1Cjx2j35.yxyD2xC2xD236.yxxD2yD12x37.yxyDxxC1xD1yD138.yxxD1yD1yDx1x11www.konkur.in

SECTION P.4 (PAGE 32)ADAMS and ESSEX: CALCULUS 839.yxyDf(x)C2(∞,3)2(2,2)yxyDf(x)(∞,∞)2Fi}. P.4.39(a? Fi}. P.4.39(b?40.yxyDf(x)C2(∞,3)2(2,2)yx∞yDf(x)∞(2,∞)∞Fi}. P.4.4?(a? Fi}. P.4.4?(b?41.yxyDf(xC2)(∞,∞)242.yx(2,∞)∞ 3yDf(x∞)43.yx2yDf(x)(∞,∞)44.yxyDf(x)(∞,∞)245.yx(3,∞)2 4yDf(4x)46.yx(∞,∞)yD∞f(∞x)(∞,∞)47.Ran}e is a⎷⎷roximately [?.∞8,?.?8].y∞.??.8?.??.4?.2?.2?.4?.??.8x54 3 2 ∞ ∞ 2 3 4yD ?.∞8yD?.?8yDxC2x2C2xC3Fi}. P.4.4748.Ran}e is a⎷⎷roximately(1,?.∞7].y7?5432∞x54 3 2 ∞ ∞ 2 3 4yD?.∞7yDx∞x2CxFi}. P.4.4812www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.5 (PAGE 38)49.y112345x54 3 2 1 1 2 3 4yDx46x3C9x21Fig. P.4.49Apparent symmetry aboutxD1:5.This can be conrmed by calculatingf.3x/, whichturns out to be equal tof.x/.50.y112x54 3 2 1 1 2 3 4yD32xCx222xCx2Fig. P.4.50Apparent symmetry aboutxD1.This can be conrmed by calculatingf.2x/, whichturns out to be equal tof.x/.51.y211234x3 2 1 1 2 3 456yDx1x2yD xC3yDx1Fig. P.4.51Apparent symmetry about.2;1/, and about the linesyDx1 andyD3x.These can be conrmed by noting thatf.x/D1C1x2,so the graph is that of 1=xshifted right 2 units and upone.52.y2112345x7 654 3 2 1 1 2yD2x2C3xx2C4xC5Fig. P.4.52Apparent symmetry about.2;2/.This can be conrmed by calculating shifting the graphright by 2 (replacexwithx2) and then down 2 (subtract 2). The result is5x=.1Cx2/, which is odd.53.Iffis both even and odd thef.x/Df.x/D f.x/,sof.x/D0 identically.Section P.5 Combining Functions to MakeNew Functions (page 38)1.f.x/Dx,g.x/Dpx1.D.f/DR,D.g/D[1;1/.D.fCg/DD.fg/DD.fg/DD.g=f/D[1;1/,D.f=g/D.1;1/..fCg/.x/DxCpx1.fg/.x/Dxpx1.fg/.x/Dxpx1.f=g/.x/Dx=px1.g=f/.x/D.p1x/=x2.f.x/Dp1x,g.x/Dp1Cx.D.f/D.1;1],D.g/D[1;1/.D.fCg/DD.fg/DD.fg/D[1;1],D.f=g/D.1;1],D.g=f/D[1;1/..fCg/.x/Dp1xCp1Cx.fg/.x/Dp1xp1Cx.fg/.x/Dp1x2.f=g/.x/Dp.1x/=.1Cx/.g=f/.x/Dp.1Cx/=.1x/13www.konkur.in

SECTION P.5 (PAGE 38)ADAMS and ESSEX: CALCULUS 83.yDxyD x2yDxx2yx4.y211x2 11yD xyDx3yDx3x5.yxyDxC jxjyD jxjyDxD jxjyDx6.y11234x2 1 1 2 3 45yD jxjyD jx2jyD jxj C jx2j7.f.x/DxC5,g.x/Dx23(fg.0/Df.3/D2;g.f.0//Dg.5/D22f.g.x//Df.x23/Dx2C2gf.x/Dg.f.x//Dg.xC5/D.xC5/23ff.5/Df.0/D5;g.g.2//Dg.1/D 2f.f.x//Df.xC5/DxC10gg.x/Dg.g.x//D.x23/238.f.x/D2=x,g.x/Dx=.1x/(ff.x/D2=.2=x/DxID.ff/D fx.x6D0gfg.x/D2=.x=.1x//D2.1x/=xID.fg/D fx.x6D0;1ggf.x/D.2=x/=.1.2=x//D2=.x2/ID.gf/D fx.x6D0;2ggg.x/D.x=.1x//=.1.x=.1x///Dx=.12x/ID.gg/D fx.x6D1=2;1g9.f.x/D1=.1x/,g.x/Dpx1(ff.x/D1=.1.1=.1x///D.x1/=xID.ff/D fx.x6D0;1gfg.x/D1=.1px1/ID.fg/D fx.x1;x6D2ggf.x/Dp.1=.1x//1Dpx=.1x/ID.gf/D[0;1/gg.x/Dqpx11ID.gg/D[2;1/10.f.x/D.xC1/=.x1/D1C2=.x1/,g.x/Dsgn.x/(ff.x/D1C2=.1C.2=.x1/1//DxID.ff/D fx.x6D1gfg.x/DsgnxC1sgnx1D0ID.fg/D.1;0/gf.x/DsgnxC1x1Dn1 ifx<1 orx>11 if1

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.5 (PAGE 38)yxyD?xyD2C?xyD2C?xC3yD1/(2C?xC3)Fig. P.5.17∞8.yxyD2xyD2x1yD12xyD?12xyD1?12xyD1?12x1Fig. P.5.18∞9.yx82(1,2)yD2f(x)2?.yx2yD(1/2)f(x)2∞.yxyDf(2x)(1/2,1)122.yxyDf(x/3)6323.yxyD1Cf(x/2)(2,2)24.yxyD2f((x1)/2)1525.yxyDf(x)(1,1)22?.yxyDg(x)(1,1)2∞5www.konkur.in

SECTION P.5 (PAGE 38)ADAMS and ESSEX: CALCULUS 827.F.x/DAxCB(a?FδF.x/DF.x/)A.AxCB/CBDAxCB)A[.A∞/xCB]D?Thus? eitherAD? orAD∞ andBD?.(b?FδF.x/Dx)A.AxCB/CBDx).A2∞/xC.AC∞/BD?Thus? eitherAD ∞ orAD∞ andBD?28.bxc D? {or ?x<∞;dxe D? {or∞x

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.6 (PAGE 45)7.x3C1D.xC1/.x2xC1/. One root is1. The othertwo are the solutions ofx2xC1D0, namelyxD1p142D12p32?:We havex3C1D.xC1/ x12p32?! x12Cp32?!:8.x41D.x21/.x2C1/D.x1/.xC1/.x?/.xC?/.The roots are 1,1,?, and?.9.x63x4C3x21D.x21/3D.x1/3.xC1/3. Theroots are 1 and1, each with multiplicity 3.10.x5x416xC16D.x1/.x416/D.x1/.x24/.x4C4/D.x1/.x2/.xC2/.x2?/.xC2?/:The roots are 1, 2,2, 2?, and2?.11.x5Cx3C8x2C8D.x2C1/.x3C8/D.xC2/.x?/.xC?/.x22xC4/Three of the ve roots are2,?and?. The remaining two are solutions ofx22xC4D0, namelyxD2p4162D1p3?. We havex5Cx3C8x2C8D.xC2/.x?/.xC?/.xaCp3?/.xap3?/:12.x94x7x6C4x4Dx4.x5x24x3C4/Dx4.x31/.x24/Dx4.x1/.x2/.xC2/.x2CxC1/:Seven of the nine roots are: 0 (with multiplicity 4),1, 2, and2. The other two roots are solutions ofx2CxC1D0, namelyxD1p142D 12p32?:The required factorization ofx94x7x6C4x4isx4.x1/.x2/.xC2/ x12Cp32?! x12p32?!:13.The denominator isx2C2xC2D.xC1/2C1 which isnever 0. Thus the rational function is dened for all realnumbers.14.The denominator isx3xDx.x1/.xC1/whichis zero ifxD0, 1, or1. Thus the rational function isdened for all real numbers except 0, 1, and1.15.The denominator isx3Cx2Dx2.xC1/which is zeroonly ifxD0 orxD 1. Thus the rational function isdened for all real numbers except 0 and1.16.The denominator isx2Cx1, which is a quadraticpolynomial whose roots can be found with the quadraticformula. They arexD.1p1C4/=2. Hence thegiven rational function is dened for all real numbersexcept.1p5/=2 and.1Cp5/=2.17.x31x22Dx32xC2x1x22Dx.x22/C2x1x22DxC2x1x22:18.x2x2C5xC3Dx2C5xC35x3x2C5xC3D1C5x3x2C5xC3:19.x3x2C2xC3Dx3C2x2C3x2x23xx2C2xC3Dx.x2C2xC3/2x23xx2C2xC3Dx2.x2C2xC3/4x6C3xx2C2xC3Dx2CxC6x2C2xC3:20.x4Cx2x3Cx2C1Dx.x3Cx2C1/x3xCx2x3Cx2C1DxC.x3Cx2C1/Cx2C1xCx2x3Cx2C1Dx1C2x2xC1x3Cx2C1:21.As in Example 6, we wanta4D4, soa2D2andaDp2,bD p2aD 2. ThusP.x/D.x22xC2/.x2C2xC2/.22.Following the method of Example 6, we calculate.x2bxCa2/.x2CbxCa2/Dx4Ca4C.2a2b2/x2Dx2Cx2C1providedaD1 andb2D1C2a2D3, sobDp3. ThusP.x/D.x2p3xC1/.x2Cp3xC1/.23.LetP.x/Da?x?Ca?1x?1C Ca1xCa0, where?1. By the Factor Theorem,x1 is a factor ofP.x/if and only ifP.1/D0, that is, if and only ifa?Ca?1C Ca1Ca0D0.24.LetP.x/Da?x?Ca?1x?1C Ca1xCa0, where?1. By the Factor Theorem,xC1 is a factor ofP.x/if and only ifP.1/D0, that is, if and only ifa0a1Ca2a3CC.1/?a?D0. This condition saysthat the sum of the coefcients of even powers is equalto the sum of coefcients of odd powers.17www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.7 (PAGE 57)12.ta?xc?txta?xCc?txDs??xc?sxc?sxs??xs??xc?sxCc?sxs??xDs??2xc?s2xc?sxs??xs??2xCc?s2xc?sxs??xDs??2xc?s2x13.c?s4xs??4xD.c?s2xs??2x/.c?s2xCs??2x/Dc?s2xs??2xDc?s.2x/14..1c?sx/.1Cc?sx/D1c?s2xDs??2x?m?l?es1c?sxs??xDs??x1Cc?sx. N?w1c?sxs??xD1c?s2x2s??2x2D112s??2x22s??x2c?sx2Ds??x2c?sx2Dta?x215.1c?sx1Cc?sxD2s??2x22c?s2x2Dta?2x216.c?sxs??xc?sxCs??xD.c?sxs??x/2.c?sxCs??x/.c?sxs??x/Dc?s2x2s??xc?sxCs??2xc?s2xs??2xD1s??.2x/c?s.2x/Dsec.2x/ta?.2x/17.s??3xDs??.2xCx/Ds??2xc?sxCc?s2xs??xD2s??xc?s2xCs??x.12s??2x/D2s??x.1s??2x/Cs??x2s??3xD3s??x4s??3x18.c?s3xDc?s.2xCx/Dc?s2xc?sxs??2xs??xD.2c?s2x1/c?sx2s??2xc?sxD2c?s3xc?sx2.1c?s2x/c?sxD4c?s3x3c?sx19.c?s2x?as ?e???d.yx2=21yDc?s.2x/F?g. P.7.1920.s??x2?as ?e???d 4.yx211F?g. P.7.2021.s??x?as ?e???d 2.yx243111yDs??.x/F?g. P.7.2122.c?sx2?as ?e???d 4.yx13511F?g. P.7.2223.y32112xyD2c?sx319www.konkur.in

SECTION P.7 (PAGE 57)ADAMS and ESSEX: CALCULUS 824.y112xyD1Csin425.sinxD35;2

INSTRUCTOR'S SOLUTIONS MANUALSECTION P.7 (PAGE 57)39.bcDcosA)cDbsecA40.sinADac41.sinADacDpc2b2c42.sinADacDapa2Cb243.aD4,bD3,AD4sinBDbsinAaD341p2D34p244.Given thataD2;bD2;cD3:CbAcBaSincea2Db2Cc22bccosA,cosADa2b2c22bcD4492.2/.3/D34(45.aD2,bD3,cD4b2Da2Cc22accosBThus cosBD4C169224D1116sinBDs1112162Dp25612116Dp1351646.Given thataD2;bD3;CD4:c2Da2Cb22abcosCD4C92.2/.3/cos4D1312p2(Hence,cDs1312p22:12479(47.cD3,AD4,BD3impliesCD512asinADcsinC)aD1p23sin512aD3p21sin712D3p22p21Cp3(by #5)D61Cp348.Given thataD2;bD3;CD35( Thenc2D4C92.2/.3/cos35, hencec1:78050(49.aD4,BD40,CD70ThusAD70(bsin40D4sin70sobD4sin40sin70D2:73650.IfaD1;bDp2;AD30, thensinBbDsinAaD12(Thus sinBDp22D1p2,BD4or34, andCD4C6D712orCD34C6D12(Thus, cosCDcos712Dcos4C3D1p32p2orcosCDcos12Dcos34D1Cp32p2(Hence,c2Da2Cb22abcosCD1C22p2cosCD3.1p3/or 3.1Cp3/D2Cp3 or 2p3:Hence,cDp2Cp3 orp2p3(=6p211CA B0B00Fig( P(7(5051.Let?be the height of the pole andxbe the distancefromCto the base of the pole(Then?Dxtan50and?D.xC10/tan35Thusxtan50Dxtan35C10tan35soxD10tan35tan50tan35?D10tan50tan35tan50tan3516:98The pole is about 16(98 metres high(52.See the following diagram( Since tan40D?=a, thereforeaD?=tan40( Similarly,bD?=tan70(SinceaCbD2 km, therefore,?tan40C?tan70D2?D2.tan40tan70/tan70Ctan401:286 km:21www.konkur.in

SECTION P.7 (PAGE 57)ADAMS and ESSEX: CALCULUS 8habBABall???7040F?g. P.7.5253(A?ea4ABCD12jBCjhDah2Dacs??B2Dabs??C2By symmet?y, a?ea4ABCals?D12bcs??AbCAhBcPF?g. P.7.5354(F??m Exe?c?se 53, a?eaD12acs??B. By C?s??e Law,c?sBDa2Cc2b22ac. T?us,s??BDs1a2Cc2b22ac2Dpa4b4c4C2a2b2C2b2c2C2a2c22ac:He?ce, A?eaDpa4b4c4C2a2b2C2b2c2C2a2c24squa?e u??ts. S??ce,s.sa/.sb/.sc/DbCcCa2bCca2abCc2aCbc2D116.bCc/2a2a2.bc/2D116a2.bCc/2C.bc/2a4.b2c2/2D1162a2b2C2a2c2a4b4c4C2b2c2T?usps.sa/.sb/.sc/= A?ea ?f t??a?gle.22www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION 1.1 (PAGE 63)CHAPTER 1. LIMITS AND CONTINUITYSection 1.1 Examples of Velocity, GrowthRate, and Area (page 63)1.Average velocity =1x1tD(tCh)2t2hm/s.2.hAvg. vel. over [2,2Ch]15.00000.14.10000.014.01000.0014.00100.00014.00013.Guess velocity isvD4 m/s attD2 s.4.Average volocity on [2,2Ch] is(2Ch)24(2Ch)2D4C4hCh24hD4hCh2hD4Ch.Ashapproaches 0 this average velocity approaches 4m/s5.xD3t212tC1 m at timets.Average velocity over interval [1,2] is(322122C1)(312121C1)21D 3m/s.Average velocity over interval [2,3] is(332123C1)(322122C1)32D3 m/s.Average velocity over interval [1,3] is(332123C1)(312121C1)31D0 m/s.6.Average velocity over [t,tCh] is3(tCh)212(tCh)C1(3t212tC1)(tCh)tD6thC3h212hhD6tC3h12 m/s.This average velocity approaches 6t12 m/s ashapproaches 0.AttD1 the velocity is 6112D 6 m/s.AttD2 the velocity is 6212D0 m/s.AttD3 the velocity is 6312D6 m/s.7.AttD1 the velocity isvD 6<0 so the particle ismoving to the left.AttD2 the velocity isvD0 so the particle is stationary.AttD3 the velocity isvD6>0 so the particle ismoving to the right.8.Average velocity over [tk,tCk] is3(tCk)212(tCk)C1[3(tk)212(tk)C1](tCk)(tk)D12k3t2C6tkC3k212t12kC13t2C6tk3k2C12t12kC1D12tk24k2kD6t12 m/s,which is the velocity at timetfrom Exercise 7.9.y12t1 2 3 45yD2C1πsin(πt)Fig. 1.1.9AttD1 the height isyD2 ft and the weight ismoving downward.10.Average velocity over [1,1Ch] is2C1πsinπ(1Ch)2C1πsinπhDsin(πCπh)πhDsinπcos(πh)Ccosπsin(πh)πhD sin(πh)πh.hAvg. vel. on [1,1Ch]1.000000.1000 0.9836316430.0100 0.9998355150.0010 0.99999835511.The velocity attD1 is aboutvD 1 ft/s. The “"indicates that the weight is moving downward.23www.konkur.in

SECTION 1.1 (PAGE 63)ADAMS and ESSEX: CALCULUS 812.We sketc?ed a ta?ge?t l??e t? t?e g?a?? ?? ?age 55 ??t?e text attD20. T?e l??e a??ea?ed t? ?ass t???ug?t?e ????ts.10;0/a?d.50;1/. O? day 20 t?e b??mass ?sg??w??g at ab?ut.10/=.5010/D0:025 mm2/d.13.T?e cu?ve ?s stee?est, a?d t?e?ef??e t?e b??mass ?s g??w??g m?st ?a??dly, at ab?ut day 45.14.a)???t255075100125150175yea?2008 2009 2010 2011 2012F?g. 1.1.14b) Ave?age ?ate ?f ??c?ease ?? ???ts betwee? 2010 a?d2012 ?s1746220122010D1122D56 (t??usa?d$/y?).c) D?aw??g a ta?ge?t l??e t? t?e g?a?? ?? (a) attD2010 a?d measu???g ?ts sl??e, we ?d t?att?e ?ate ?f ??c?ease ?f ???ts ?? 2010 ?s ab?ut 43t??usa?d$/yea?.Section 1.2 Limits of Functions (page 71)1.F??m ??s?ect??g t?e g?a??yx111yDf.x/F?g. 1.2.1we see t?atl?mxω1f.x/D1;l?mxω0f.x/D0;l?mxω1f.x/D1:2.F??m ??s?ect??g t?e g?a??yx1 2 31yDg.x/F?g. 1.2.2we see t?atl?mxω1g.x/d?es ??t ex?st(left l?m?t ?s 1, ??g?t l?m?t ?s 0)l?mxω2g.x/D1;l?mxω3g.x/D0:3.l?mxω1g.x/D14.l?mxω1Cg.x/D05.l?mxω3Cg.x/D06.l?mxω3g.x/D07.l?mxω4.x24xC1/D424.4/C1D18.l?mxω23.1x/.2x/D3.1/.22/D09.l?mxω3xC3xC6D3C33C6D2310.l?mtω4t24tD.4/24C4D211.l?mxω1x21xC1D1211C1D02D012.l?mxω1x21xC1Dl?mxω1.x1/D 213.l?mxω3x26xC9x29Dl?mxω3.x3/2.x3/.xC3/Dl?mxω3x3xC3D06D014.l?mxω2x2C2xx24Dl?mxω2xx2D24D1215.l?mhω214h2d?es ??t ex?st; de??m??at?? a????ac?es 0but ?ume?at?? d?es ??t a????ac? 0.16.l?mhω03hC4h2h2h3Dl?mhω03C4hhh2d?es ??t ex?st; de??m??at?? a????ac?es 0 but ?ume?at?? d?es ??t a????ac? 0.24www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION 1.2 (PAGE 71)17.l?mxω9px3x9Dl?mxω9.px3/.pxC3/.x9/.pxC3/Dl?mxω9x9.x9/.pxC3/Dl?mxω91pxC3D1618.l?mhω0p4Ch2hDl?mhω04Ch4h.p4ChC2/Dl?mhω01p4ChC2D1419.l?mxω.x/2xD022D020.l?mxω2jx2j D j 4j D421.l?mxω0jx2jx2Dj 2j2D 122.l?mxω2jx2jx2Dl?mxω21;?fx>21;?fx<2.He?ce, l?mxω2jx2jx2d?es ??t ex?st.23.l?mtω1t21t22tC1l?mtω1.t1/.tC1/.t1/2Dl?mtω1tC1t1d?es ??t ex?st(de??m??at??ω0, ?ume?at??ω2.)24.l?mxω2p44xCx2x2Dl?mxω2jx2jx2d?es ??t ex?st.25.l?mtω0tp4Ctp4tDl?mtω0t.p4CtCp4t/.4Ct/.4t/Dl?mtω0p4CtCp4t2D226.l?mxω1x21pxC32Dl?mxω1.x1/.xC1/.pxC3C2/.xC3/4Dl?mxω1.xC1/.pxC3C2/D.2/.p4C2/D827.l?mtω0t2C3t.tC2/2.t2/2Dl?mtω0t.tC3/t2C4tC4.t24tC4/Dl?mtω0tC38D3828.l?msω0.sC1/2.s1/2sDl?msω04ssD429.l?myω1y4pyC3y21Dl?myω1.py1/.py3/.py1/.pyC1/.yC1/D24D1230.l?mxω1x3C1xC1Dl?mxω1.xC1/.x2xC1/xC1D331.l?mxω2x416x38Dl?mxω2.x2/.xC2/.x2C4/.x2/.x2C2xC4/D.4/.8/4C4C4D8332.l?mxω8x2=34x1=32Dl?mxω8.x1=32/.x1=3C2/.x1=32/Dl?mxω8.x1=3C2/D433.l?mxω21x24x24Dl?mxω2xC24.x2/.xC2/Dl?mxω21xC2D1434.l?mxω21x21x24Dl?mxω2xC21.x2/.xC2/Dl?mxω2xC1.x2/.xC2/d?es ??t ex?st.35.l?mxω0p2Cx2p2x2x2Dl?mxω0.2Cx2/.2x2/x2.p2Cx2Cp2x2/Dl?mxω02x2x2p2Cx2/Cp2x2D2p2Cp2D1p236.l?mxω0j3x1j j3xC1jxDl?mxω0.3x1/2.3xC1/2x.j3x1j C j3xC1j/Dl?mxω012xx.j3x1j C j3xC1j/D121C1D 637.f.x/Dx2l?mhω0f.xCh/f.x/hDl?mhω0.xCh/2x2hDl?mhω02hxCh2hDl?mhω02xChD2x25www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION 1.2 (PAGE 71)61.f.x/D(x1 ifx 1x2C1 if10limxω1f.x/Dlimxω1x1D 11D 262.limxω1Cf.x/Dlimxω1Cx2C1D1C1D263.limxω0Cf.x/Dlimxω0C.xC/2D264.limxω0f.x/Dlimxω0x2C1D165.If limxω4f.x/D2 and limxω4g.x/D 3, thena) limxω4g.x/C3D 3C3D0b) limxω4xf.x/D42D8c) limxω4g.x/2D.3/2D9d) limxω4g.x/f.x/1D321D 366.If limxωa f.x/D4 and limxωag.x/D 2, thena) limxωaf.x/Cg.x/D4C.2/D2b) limxωaf.x/g.x/D4.2/D8c) limxωa4g.x/D4.2/D8d) limxωaf.x/g.x/D42D 267.If limxω2f.x/5x2D3, thenlimxω2f.x/5Dlimxω2f.x/5x2.x2/D3.22/D0:Thus limxω2f.x/D5.68.If limxω0f.x/x2D 2 thenlimxω0f.x/Dlimxω0x2f.x/x2D0.2/D0,and similarly,limxω0f.x/xDlimxω0xf.x/x2D0.2/D0.69.y0.40.20.20.40.60.81.01.2x3 2 11 2yDsinxxFig. 1.2.69limxω0sinxxD170.y0.40.20.20.40.60.8x0.08 0.040.04 0.08yDsin.2x/sin.3x/Fig. 1.2.70limxω0sin.2x/=sin.3x/D2=371.y0.10.10.20.30.40.50.60.70.8x0.2 0.4 0.6 0.8 1.0yDsinp1xp1x2Fig. 1.2.71limxω1sinp1xp1x2π0:707127www.konkur.in

SECTION 1.2 (PAGE 71)ADAMS and ESSEX: CALCULUS 872.y1.21.00.80.60.40.2x0.2 0.4 0.6 0.8yDxpxpsinxFig. 1.2.72limxω0CxpxpsinxD 173.y0.20.10.1x0.2 0.10.1yD xyDxsin.1=x/yDxFig. 1.2.73f.x/Dxsin.1=x/oscillates innitely often asxapproaches 0, but the amplitude of the oscillations decreasesand, in fact, limxω0f.x/D0. This is predictable becausejxsin.1=x/j jxj.(See Exercise 95 below.)74.Sincep52x2f.x/p5x2for1x1, andlimxω0p52x2Dlimxω0p5x2Dp5, we havelimxω0f.x/Dp5 by the squeeze theorem.75.Since 2x2g.x/2cosxfor allx, and sincelimxω0.2x2/Dlimxω02cosxD2, we havelimxω0g.x/D2 by the squeeze theorem.76.a)y123x2 11yDx2yDx4.1;1/.1;1/Fig. 1.2.76b) Since the graph offlies between those ofx2andx4, and since these latter graphs come together at.1;1/and at.0;0/, we have limxω1f.x/D1and limxω0f.x/D0 by the squeeze theorem.77.x1=3x3on.1;1/and.0;1/. The graphs ofx1=3andx3intersect at.1;1/,.0;0/, and.1;1/. If the graph of?.x/lies between those ofx1=3andx3, then we can determinelimxωa?.x/foraD 1,aD0, andaD1 by thesqueeze theorem. In factlimxω1?.x/D 1;limxω0?.x/D0;limxω1?.x/D1:78.f.x/Dssin1xis dened for allx6D0; its domain is.1;0/[.0;1/. Sincejsintj 1 for allt, we havejf.x/j jxjandjxj f.x/ jxjfor allx6D0.Since limxω0D.jxj/D0Dlimxω0jxj, we havelimxω0f.x/D0 by the squeeze theorem.79.jf.x/j g.x/)g.x/f.x/g.x/Since limxωag.x/D0, therefore 0limxωaf.x/0.Hence, limxωaf.x/D0.If limxωag.x/D3, then either3limxωaf.x/3 orlimxωaf.x/does not exist.Section 1.3 Limits at Innity and InniteLimits (page 78)1.limxω1x2x3Dlimxω112.3=x/D122.limxω1xx24Dlimxω11=x1.4=x2/D01D028www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION 1.3 (PAGE 78)3.l?mxω13x35x2C78C2x5x3Dl?mxω135xC7x38x3C2x25D 354.l?mxω1x22xx2Dl?mxω112x21x1D11D 15.l?mxω1x2C3x3C2Dl?mxω11xC3x31C2x3D06.l?mxω1x2Cs??xx2Cc?sxDl?mxω11Cs??xx21Cc?sxx2D11D1We ?ave used t?e fact t?at l?mxω1s??xx2D0 (a?d s?m?la?ly f?? c?s??e) because t?e ?ume?at?? ?s b?u?ded w??let?e de??m??at?? g??ws la?ge.7.l?mxω13xC2px1xDl?mxω13C2px1x1D 38.l?mxω12x1p3x2CxC1Dl?mxω1x21xjxjr3C1xC1x2(butjxj Dxasxω 1/Dl?mxω121xr3C1xC1x2D2p39.l?mxω12x1p3x2CxC1Dl?mxω121xr3C1xC1x2D 2p3,becausexω 1?m?l?es t?atx<0 a?d s?px2D x.10.l?mxω12x5j3xC2jDl?mxω12x5.3xC2/D2311.l?mxω313xd?es ??t ex?st.12.l?mxω31.3x/2D113.l?mxω313xD 114.l?mxω3C13xD 115.l?mxω5=22xC55xC2D0252C2D016.l?mxω2=52xC55xC2d?es ??t ex?st.17.l?mxω.2=5/2xC55xC2D 118.l?mxω2=5C2xC55xC2D 119.l?mxω2Cx.2x/3D120.l?mxω1xp1x2D 121.l?mxω1C1jx1jD 122.l?mxω11jx1jD 123.l?mxω2x3x24xC4Dl?mxω2x3.x2/2D 124.l?mxω1Cpx2xxx2Dl?mxω1C1px2xD 125.l?mxω1xCx3Cx51Cx2Cx3Dl?mxω11x2C1Cx21x3C1xC1D 126.l?mxω1x3C3x2C2Dl?mxω1xC3x21C2x2D 127.l?mxω1xpxC1-1p2xC3·76xC4x2Dl?mxω1x2 r1C1x→ 1pxr2C3x→x27x26xC4D1.p2/4D 14p228.l?mxω1x2xC1x2x1Dl?mxω12x2x21D 229www.konkur.in

SECTION 1.3 (PAGE 78)ADAMS and ESSEX: CALCULUS 829.l?mxω1⎷x2C2x⎷x22xDl?mxω1.x2C2x/.x22x/px2C2xCpx22xDl?mxω14x.x/ r1C2xCr12x→D 41C1D 230.l?mxω1⎷x2C2x⎷x22xDl?mxω1x2C2xx2C2xpx2C2xCpx22xDl?mxω14xxr1C2xCxr12xDl?mxω14r1C2xCr12xD42D231.l?mxω11px22xxDl?mxω1px22xCx.px22xCx/.px22xx/Dl?mxω1px22xCxx22xx2Dl?mxω1x.p1.2=x/C1/2xD22D 132.l?mxω11px2C2xxDl?mxω11jxj.p1C.2=x/C1D033.By Exe?c?se 35,yD 1 ?s a ????z??tal asym?t?te (at t?e??g?t) ?fyD1px22xx. S??cel?mxω11px22xxDl?mxω11jxj.p1.2=x/C1D0;yD0?s als? a ????z??tal asym?t?te (at t?e left).N?wpx22xxD0 ?f a?d ??ly ?fx22xDx2, t?at?s, ?f a?d ??ly ?fxD0. T?e g?ve? fu?ct??? ?s u?de?edatxD0, a?d w?e?ex22x<0, t?at ?s, ?? t?e ??te?val[0;2]. Its ??ly ve?t?cal asym?t?te ?s atxD0, w?e?el?mxω01px22xxD 1.34.S??ce l?mxω12x5j3xC2jD23a?d l?mxω12x5j3xC2jD 23,yD .2=3/a?e ????z??tal asym?t?tes ?fyD.2x5/=j3xC2j.T?e ??ly ve?t?cal asym?t?te?sxD 2=3, w??c? makes t?e de??m??at?? ze??.35.l?mxω0Cf.x/D136.l?mxω1f.x/D 137.y1123x1 2 3 456yDf.x/F?g. 1.3.37l?mxω2Cf.x/D138.l?mxω2f.x/D239.l?mxω3f.x/D 140.l?mxω3Cf.x/D 141.l?mxω4Cf.x/D242.l?mxω4f.x/D043.l?mxω5f.x/D 144.l?mxω5Cf.x/D045.l?mxω1f.x/D146.????z??tal:yD1; ve?t?cal:xD1,xD3.47.l?mxω3Cbxc D348.l?mxω3bxc D249.l?mxω3bxcd?es ??t ex?st50.l?mxω2:5bxc D251.l?mxω0Cb2xc Dl?mxω2bxc D152.l?mxω3bxc D 453.l?mtωt0C.t/DC.t0/exce?t at ??tege?st0l?mtωt0C.t/DC.t0/eve?yw?e?el?mtωt0CC.t/DC.t0/?ft06Da? ??tege?l?mtωt0CC.t/DC.t0/C1:5 ?ft0?s a? ??tege?30www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION 1.4 (PAGE 87)yx6:004:503:001:501 2 3 4yDC.t/F?g. 1.3.5354.l?mxω0Cf.x/DL(a) Iff?s eve?, t?e?f.x/Df.x/.He?ce, l?mxω0f.x/DL.(b) Iff?s ?dd, t?e?f.x/Df.x/.T?e?ef??e, l?mxω0f.x/D L.55.l?mxω0Cf.x/DA;l?mxω0f.x/DBa) l?mxω0Cf.x3x/DB(s??cex3x<0 ?f 00 ?f10 f??01?s c??t??u?us eve?yw?e?eexce?t atxD1, w?e?e ?t ?s left c??t??u?us but ??t ??g?tc??t??u?us because 0:9876D1. Cl?se, as t?ey say, but ??c?ga?.3∞www.konkur.in

SECTION 1.4 (PAGE 87)ADAMS and ESSEX: CALCULUS 811.T?e least ??tege? fu?ct???dxe?s c??t??u?us eve?yw?e?e??Rexce?t at t?e ??tege?s, w?e?e ?t ?s left c??t??u?usbut ??t ??g?t c??t??u?us.12.C.t/?s d?sc??t??u?us ??ly at t?e ??tege?s. It ?s c??t??u?us ?? t?e left at t?e ??tege?s, but ??t ?? t?e ??g?t.13.S??cex24x2DxC2 f??x6D2, we ca? de?e t?efu?ct??? t? be 2C2D4 atxD2 t? make ?t c??t??u?ust?e?e. T?e c??t??u?us exte?s??? ?sxC2.14.S??ce1Ct31t2D.1Ct/.1tCt2/.1Ct/.1t/D1tCt21tf??t6D 1, we ca? de?e t?e fu?ct??? t? be 3=2 attD 1t? make ?t c??t??u?us t?e?e. T?e c??t??u?us exte?s??? ?s1tCt21t.15.S??cet25tC6t2t6D.t2/.t3/.tC2/.t3/Dt2tC2f??t6D3,we ca? de?e t?e fu?ct??? t? be 1=5 attD3 t? make ?tc??t??u?us t?e?e. T?e c??t??u?us exte?s??? ?st2tC2.16.S??cex22x44D.xp2/.xCp2/.xp2/.xCp2/.x2C2/DxCp2.xCp2/.x2C2/f??x6Dp2, we ca? de?e t?e fu?ct??? t? be 1=4 atxDp2 t? make ?t c??t??u?us t?e?e. T?e c??t??u?usexte?s??? ?sxCp2.xCp2/.x2C2/. (N?te: ca?cell??g t?exCp2 fact??s ???v?des a fu?t?e? c??t??u?us exte?s??? t?xD p2.17.l?mx!2Cf.x/Dk4 a?d l?mx!2f.x/D4Df.2/.T?usfw?ll be c??t??u?us atxD2 ?fk4D4, t?at ?s,?fkD8.18.l?mx!3g.x/D3ma?dl?mx!3Cg.x/D13mDg.3/. T?usgw?ll be c??t??u?us atxD3 ?f 3mD13m, t?at ?s, ?fmD 1.19.x2?as ?? max?mum value ??10 ??.1;0/a?d.1;1/.f.x/<0 ??.1;1/a?d.0;1/.26.f.x/Dx2C4xC3D.xC1/.xC3/f.x/ >0 ??.1;3/a?d.1;1/f.x/<0 ??.3;1/.27.f.x/Dx21x24D.x1/.xC1/.x2/.xC2/fD0 atxD 1.f?s ??t de?ed atxD 2.f.x/ >0 ??.1;2/,.1;1/, a?d.2;1/.f.x/<0 ??.2;1/a?d.1;2/.28.f.x/Dx2Cx2x3D.xC2/.x1/x3f.x/ >0 ??.2;0/a?d.1;1/f.x/<0 ??.1;2/a?d.0;1/.29.f.x/Dx3Cx1,f.0/D1,f.1/D1.S??cef?s c??t??u?us a?d c?a?ges s?g? betwee? 0 a?d 1,?t must be ze?? at s?me ????t betwee? 0 a?d 1 by IVT.30.f.x/Dx315xC1 ?s c??t??u?us eve?yw?e?e.f.4/D3;f.3/D19;f.1/D13;f.4/D5:Because ?f t?e s?g? c?a?gesf?as a ze?? betwee?4a?d3, a??t?e? ze?? betwee?3 a?d 1, a?d a??t?e?betwee? 1 a?d 4.32www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALSECTION 1.5 (PAGE 92)31.F.x/D.xa/2.xb/2Cx. Without loss o{ }enerality?we can assume thata? andg.∞/?. Bein} even?fmust there{ore be de×nedon [?;?]. I{xD y? thenlimxω?f.x/Dlimyω?Cf.y/Dlimyω?Cf.y/Df.?/:Thus?fis continuous on the le{t atxD?. Bein} continuous on both sides? it is there{ore continuous.34.fodd,f.x/Df.x/fcontinuous on the ri}ht,limxω?Cf.x/Df.?/There{ore? lettin}tD x? we obtainlimxω?f.x/Dlimtω?Cf.t/Dlimtω?Cf.t/Df.?/Df.?/Df.?/:There{orefis continuous at ? andf.?/D?.35.max ∞:593 at?:83∞? min?:75? at ?:?2936.max ?:∞33 atxD∞:437; min?:232 atxD∞:8?537.max ∞?:333 atxD3; min 4:7?2 atxD∞:2??38.max ∞:5∞? atxD?:4?5; min ? atxD?andxD∞39.rootxD?:?8240.rootxD?:73941.rootsxD ?:?37 andxD∞:4∞?42.rootsxD ?:72449∞959? andxD∞:22?744?8543.{solve }ives an a⎷⎷roximation to the sin}le real root to∞? si}ni×cant ×}ures; solve }ives the three roots (includin} a com⎷lex con|u}ate ⎷air? in exact {orm involvin} thequantity∞?8C∞2p?9∞=3; eval{(solve? }ives a⎷⎷roximations to the three roots usin} ∞? si}ni×cant ×}ures {orthe real and ima}inary ⎷arts.Section 1.5 The Formal Denition of Limit(page 92)1.We require 39:9L4?:∞. Thus39:939:?C?:?25T4?:∞?:3?:?25T?:5∞2T2?:The tem⎷erature should be ke⎷t between ∞2δ+ and 2?δ+.2.Since ∞.2% o{ 8???? is 9?? we require the ed}e len}thxo{ the cube to satis{y 79?4x38?9?. It is su{×cientthat ∞9:92?x2?:?79. The ed}e o{ the cube must bewithin ?.?79 cm o{ 2? cm.3.3?:?22x∞3C?:?23:982x4:?2∞:99x2:?∞4.4?:∞x24C?:∞∞:9749x2:??245.∞?:∞px∞:∞?:8∞x∞:2∞6.2?:?∞∞x 2C?:?∞∞2:?∞x∞∞:99?:5?25x?:49757.We need?:?3.3xC∞/7?:?3? which is equivalentto?:?∞x2?:?∞ ThusD?:?∞ will do.8.We need?:?∞p2xC33?:?∞. Thus2:99p2xC33:?∞8:94?∞2xC39:???∞2:97??5x3:?3??53?:?2995x3?:?3??5:HereD?:?2995 will do.9.We need 8?:2x38:2? or ∞:9832x2:?∞?5.Thus? we need?:?∞?8x2?:?∞?5. HereD?:?∞?5 will do.33www.konkur.in

SECTION 1.5 (PAGE 92)ADAMS and ESSEX: CALCULUS 810.We ?eed 10.051/(xC1)1C0.05,?? 1.0526xC10.9524. T??s w?ll ?ccu? ?f0.0476x0.0526. I? t??s case we ca? takeδD0.0476.11.T? be ???ved: l?mx!1(3xC1)D4.P???f: Let? >0 be g?ve?. T?e?j(3xC1)4j0 be g?ve?. T?e?j(52x)1j0 be g?ve?. T?e?jx20j< ???lds ?fjx0j D jxj< δDp?.14.T? be ???ved: l?mx!2x21Cx2D0.P???f: Let? >0 be g?ve?. T?e?x21Cx20Djx2j1Cx2 jx2j< ????v?dedjx2j< δD?.15.T? be ???ved: l?mx!1/214x212xD2.P???f: Let? >0 be g?ve?. T?e? ?fx6D1/2 we ?ave14x212x2D j(1C2x)2jD j2x1j D2x12< ????v?dedjx12j< δD?/2.16.T? be ???ved: l?mx!2x2C2xxC2D 2.P???f: Let? >0 be g?ve?. F??x6D 2 we ?avex2C2xxC2(2)DjxC2j< ????v?dedjxC2j< δD?. T??s c?m?letes t?e ????f.17.T? be ???ved: l?mx!11xC1D12.P???f: Let? >0 be g?ve?. We ?ave1xC112D1x2(xC1)Djx1j2jxC1j.Ifjx1j<1, t?e? 01. LetδDm??(1,2?). Ifjx1j< δ, t?e?1xC112Djx1j2jxC1j<2?2D?.T??s establ?s?es t?e ?equ??ed l?m?t.18.T? be ???ved: l?mx!1xC1x21D 12.P???f: Let? >0 be g?ve?. Ifx6D 1, we ?avexC1x2112D1x112DjxC1j2jx1j.IfjxC1j<1, t?e?21. Le?δDm??(1,2?). If 01 a?djxC1j<2?. T?usxC1x2112DjxC1j2jx1j<2?2D?.T??s c?m?letes t?e ?equ??ed ????f.19.T? be ???ved: l?mx!1pxD1.P???f: Let? >0 be g?ve?. We ?avejpx1j Dx1pxC1 jx1j< ????v?dedjx1j< δD?. T??s c?m?letes t?e ????f.20.T? be ???ved: l?mx!2x3D8.P???f: Let? >0 be g?ve?. We ?avejx38j D jx2jjx2C2xC4j. Ifjx2j<1,t?e? 10 t?e?e ex?sts a ?umbe?δ >0, de?e?d??g ???, suc? t?ataδ 0 t?e?e ex?sts a ?umbe?R>0, de?e?d??g ???, suc? t?atx0 t?e?e ex?sts a?umbe?δ >0, de?e?d??g ??B, suc? t?at0

INSTRUCTOR'S SOLUTIONS MANUALSECTION 1.5 (PAGE 92)24.We say t?at l?mxω1f.x/D 1?f t?e f?ll?w??g c??d?t?????lds: f?? eve?y ?umbe?B>0 t?e?e ex?sts a ?umbe?R>0, de?e?d??g ??B, suc? t?atx>R?m?l?esf.x/ >B:25.We say t?at l?mxωaCf.x/D 1?f t?e f?ll?w??g c??d?t??? ??lds: f?? eve?y ?umbe?B>0 t?e?e ex?sts a?umbe?◦ >0, de?e?d??g ??R, suc? t?ata0 t?e?e ex?sts a?umbe?◦ >0, de?e?d??g ??B, suc? t?ata◦ B:27.T? be ???ved: l?mxω1C1x1D 1. P???f: LetB>0be g?ve?. We ?ave1x1>B?f 00be g?ve?. We ?ave1x1x1>1=B,t?at ?s, ?f 1◦ 0be g?ve?. We ?ave1px2C1D1px2C1<1x< ???v?dedx>R, w?e?eRD1=. T??s c?m?letes t?e????f.30.T? be ???ved: l?mxω1pxD 1. P???f: LetB>0 beg?ve?. We ?avepx>B?fx>Rw?e?eRDB2. T??sc?m?letes t?e ????f.31.T? be ???ved: ?f l?mxωaf.x/DLa?d l?mxωaf.x/DM, t?e?LDM.P???f: Su???seL6DM. LetD jLMj=3. T?e? >0. S??ce l?mxωaf.x/DL, t?e?e ex?sts◦1>0 suc? t?atjf.x/Lj< ?fjxaj< ◦1. S??ce l?mxωaf.x/DM, t?e?eex?sts◦2>0 suc? t?atjf.x/Mj< ?fjxaj< ◦2.Let◦Dm??.◦1;◦2/. Ifjxaj<◦, t?e?3D jLMj D j.f.x/M/C.Lf.x/jjf.x/Mj C jf.x/Lj< CD2:T??s ?m?l?es t?at 3<2, a c??t?ad?ct???. T?us t?e ???g??al assum?t??? t?atL6DMmust be ??c???ect. T?e?ef??eLDM.32.T? be ???ved: ?f l?mxωag.x/DM, t?e? t?e?e ex?sts◦ >0suc? t?at ?f 00 suc? t?at ?f 00 be g?ve?. S??ce l?mxωaf.x/DL, t?e?eex?sts◦1>0 suc? t?atjf.x/Lj< =.2.1CjMj//?f 00 suc? t?atjg.x/Mj< =.2.1CjLj//?f00suc? t?atjg.x/j<1CjMj?f 00 suc? t?at ?f 0jMj=2.P???f: By t?e de??t??? ?f l?m?t, t?e?e ex?sts◦ >0 suc?t?at ?f 0jMj.jMj=2/DjMj=2, as?equ??ed.35www.konkur.in

SECTION 1.5 (PAGE 92)ADAMS and ESSEX: CALCULUS 835.T? be ???ved: ?f l?mxωag.x/DMw?e?eM6D0, t?e?l?mxωa1g.x/D1M.P???f: Let >0 be g?ve?. S??ce l?mxωag.x/DM6D0,t?e?e ex?sts◦1>0 suc? t?atjg.x/Mj< jMj2=2 ?f00suc? t?atjg.x/j>jMj=2?f 00 be g?ve?. S??cef?s c??t??u?us atL,t?e?e ex?sts a ?umbe?

, t?e?jf.y/f.L/j<. S??ce l?mxωcg.x/DL, t?e?e ex?sts◦ >0 suc? t?at ?f 0

.Tak??gyDg.x/, ?t f?ll?ws t?at ?f 00), a?d ?f l?mxωaf.x/Dl?mxωah.x/DL, t?e?als? l?mxωag.x/DL.P???f: Let >0 be g?ve?. S??ce l?mxωaf.x/DL,t?e?e ex?sts◦2>0 suc? t?at ?f 00 suc? t?at ?f 0

INSTRUCTOR'S SOLUTIONS MANUALCHALLENGING PROBLEMS 1 (PAGE 94)10.l?mxω2x24x24xC4Dl?mxω2xC2x2D 111.l?mxω2Cx24x2C4xC4Dl?mxω2Cx2xC2D 112.l?mxω42pxx4Dl?mxω44x.2Cpx/.x4/D 1413.l?mxω3x29pxp3Dl?mxω3.x3/.xC3/.pxCp3/x3Dl?mxω3.xC3/.pxCp3/D12p314.l?mhω0hpxC3hpxDl?mhω0h.pxC3hCpx/.xC3h/xDl?mhω0pxC3hCpx3D2px315.l?mxω0Cpxx2D016.l?mxω0pxx2d?es ??t ex?st becausepxx2?s ??t de?ed f??x<0.17.l?mxω1pxx2d?es ??t ex?st becausepxx2?s ??t de?ed f??x>1.18.l?mxω1pxx2D019.l?mxω11x23x2x1Dl?mxω1.1=x2/13.1=x/.1=x2/D1320.l?mxω12xC100x2C3Dl?mxω1.2=x/C.100=x2/1C.3=x2/D021.l?mxω1x31x2C4Dl?mxω1x.1=x2/1C.4=x2/D122.l?mxω1x4x24Dl?mxω1x21.4=x2/D123.l?mxω0C1pxx2D 124.l?mxω1=21pxx2D1p1=4D225.l?mxω1s??xd?es ??t ex?st; s??xtakes t?e values1 a?d 1?? a?y ??te?val.R;1/, a?d l?m?ts, ?f t?ey ex?st, must beu??que.26.l?mxω1c?sxxD0 by t?e squeeze t?e??em, s??ce1xc?sxx1xf?? allx>0a?d l?mxω1.1=x/Dl?mxω1.1=x/D0.27.l?mxω0xs??1xD0 by t?e squeeze t?e??em, s??cejxj xs??1x jxjf?? allx6D0a?d l?mxω0.jxj/Dl?mxω0jxj D0.28.l?mxω0s??1x2d?es ??t ex?st; s??.1=x2/takes t?e values1a?d 1 ?? a?y ??te?val.◦;◦/, w?e?e◦ >0, a?d l?m?ts, ?ft?ey ex?st, must be u??que.29.l?mxω1[xCpx24xC1]Dl?mxω1x2.x24xC1/xpx24xC1Dl?mxω14x1x jxjp1.4=x/C.1=x2/Dl?mxω1x[4.1=x/]xCxp1.4=x/C.1=x2/Dl?mxω14.1=x/1Cp1.4=x/C.1=x2/D2:N?te ??w we ?ave usedjxj D x(?? t?e sec??d lastl??e), becausexω 1.30.l?mxω1[xCpx24xC1]D 1 C 1 D 131.f.x/Dx34x2C1 ?s c??t??u?us ?? t?e w??le ?eal l??ea?d s? ?s d?sc??t??u?us ??w?e?e.32.f.x/DxxC1?s c??t??u?us eve?yw?e?e ?? ?ts d?ma??,w??c? c??s?sts ?f all ?eal ?umbe?s exce?txD 1. It ?sd?sc??t??u?us ??w?e?e.33.f.x/Dx2?fx>2x?fx2?s de?ed eve?yw?e?e a?d d?sc??t??u?us atxD2, w?e?e ?t ?s, ??weve?, left c??t??u?uss??ce l?mxω2f.x/D2Df.2/.34.f.x/Dx2?fx>1x?fx1?s de?ed a?d c??t??u?us eve?yw?e?e, a?d s? d?sc??t??u?us ??w?e?e. Obse?ve t?atl?mxω1f.x/D1Dl?mxω1Cf.x/.35.f.x/DH.x1/Dn1 ?fx10 ?fx<1?s de?ed eve?yw?e?ea?d d?sc??t??u?us atxD1 w?e?e ?t ?s, ??weve?, ??g?tc??t??u?us.36.f.x/DH.9x2/Dn1?f3x30 ?fx<3 ??x>3?s de?edeve?yw?e?e a?d d?sc??t??u?us atxD 3. It ?s ??g?tc??t??u?us at3 a?d left c??t??u?us at 3.37.f.x/D jxjCjxC1j?s de?ed a?d c??t??u?us eve?yw?e?e.It ?s d?sc??t??u?us ??w?e?e.38.f.x/Dnjxj=jxC1j?fx6D 11 ?fxD 1?s de?ed eve?yw?e?ea?d d?sc??t??u?us atxD 1 w?e?e ?t ?s ?e?t?e? left ?????g?t c??t??u?us s??ce l?mxω1f.x/D 1, w??lef.1/D1.37www.konkur.in

CHALLENGING PROBLEMS 1 (PAGE 94)ADAMS and ESSEX: CALCULUS 8Challenging Problems 1 (page 94)1.Let 0 (aCb)/2.Proof: Sincea22abCb2D(ab)2>0, we have4a2C4abC4b2>3a2C6abC3b2a2CabCb23>a2C2abCb24DaCb22sa2CabCb23>aCb2.2.Forxnear 0 we havejx1j D1xandjxC1j DxC1.Thuslimx!0xjx1j jxC1jDlimx!0x(1x)(xC1)D12.3.Forxnear 3 we havej52xj D2x5,jx2j Dx2,jx5j D5x, andj3x7j D3x7. Thuslimx!3j52xj jx2jjx5j j3x7jDlimx!32x5(x2)5x(3x7)Dlimx!3x34(3x)D14.4.LetyDx1/6. Then we havelimx!64x1/34x1/28Dlimy!2y24y38Dlimy!2(y2)(yC2)(y2)(y2C2yC4)Dlimy!2yC2y2C2yC4D412D13.5.UseabDa3b3a2CabCb2to handle the denominator.We havelimx!1p3Cx23p7Cx2Dlimx!13Cx4p3CxC2(7Cx)2/3C2(7Cx)1/3C4(7Cx)8Dlimx!1(7Cx)2/3C2(7Cx)1/3C4p3CxC2D4C4C42C2D3.6.rC(a)D1Cp1Caa,r(a)D1p1Caa.a) lima!0r(a)does not exist. Observe that the rightlimit is1and the left limit is1.b) From the following table it appears thatlima!0rC(a)D1/2, the solution of the linear equation 2x1D0 which results from settingaD0 inthe quadratic equationax2C2x1D0.a rC(a)1 0.414210.1 0.488100.1 0.513170.01 0.498760.01 0.501260.001 0.499880.001 0.50013c) lima!0rC(a)Dlima!0p1Ca1aDlima!0(1Ca)1a(p1CaC1)Dlima!01p1CaC1D12.7.TRUE or FALSEa) If limx!a{(x)exists and limx!a}(x)does notexist, then limx!a{(x)C}(x)does not exist.TRUE, because if limx!a{(x)C}(x)were toexist thenlimx!a}(x)Dlimx!a{(x)C}(x){(x)Dlimx!a{(x)C}(x)limx!a{(x)would also exist.b) If neither limx!a{(x)nor limx!a}(x)exists, thenlimx!a{(x)C}(x)does not exist.FALSE. Neither limx!01/xnor limx!0(1/x)exist, but limx!0(1/x)C(1/x)Dlimx!00D0exists.38www.konkur.in

INSTRUCTOR'S SOLUTIONS MANUALCHALLENGING PROBLEMS 1 (PAGE 94)c) Iffis continuous ata, then so isjfj.TRUE. For any two real numbersuandvwe havejuj jvj juvj.This follows fromjuj D juvCvj juvj C jvj,andjvj D jvuCuj jvuj C juj D juvj C juj.Now we havejf(x)j jf(a)jjf(x)f(a)jso the left side approaches zero whenever the rightside does. This happens whenx!aby the conti-nuity offata.d) Ifjfjis continuous ata, then so isf.FALSE. The functionf(x)D1 ifx<01 ifx0isdiscontinuous atxD0, butjf(x)j D1everywhere,and so is continuous atxD0.e) Iff(x) 1.fis continuous wherever it is defined, that is at allpoints exceptxD 1.fhas left and right limits1and 1, respectively, atxD1, and has left and right limits1 and1, respectively, atxD 1. It is not, however,discontinuous at any point, since1 and 1 are not in itsdomain.10.f(x)D1xx2D114-14xCx2·D114-x12·2.Observe thatf(x)f(1/2)D4for allxin(0,1).11.Supposefis continuous on [0,1] andf(0)Df(1).a) To be proved:f(a)Df(aC12)for someain [0,12].Proof: Iff(1/2)Df(0)we can takeaD0 and bedone. If not, letg(x)Df(xC12)f(x).Theng(0)6D0andg(1/2)Df(1)f(1/2)Df(0)f(1/2)Dg(0).Sincegis continuous and has opposite signs atxD0 andxD1/2, the Intermediate-Value The-orem assures us that there existsabetween 0 and1/2 such thatg(a)D0, that is,f(a)Df(aC12).b) To be proved: if?>2 is an integer, thenf(a)Df(aC1?)for someain [0,11?].Proof: Letg(x)Df(xC1?)f(x). Considerthe numbersxD0,xD1/?,xD2/?,...,xD(?1)/?.Ifg(x)D0 for any of these num-bers, then we can letabe that number. Otherwise,g(x)6D0 at any of these numbers. Suppose that thevalues ofgat all these numbers has the same sign(say positive). Then we havef(1)>f(?1?) >>f(2?) >1?>f(0),which is a contradiction, sincef(0)Df(1). There-fore there existsjin the setf0,1,2,...,?1gsuchthatg(j/?)andg((jC1)/?)have opposite sign. Bythe Intermediate-Value Theorem,g(a)D0for someabetweenj/?and(jC1)/?, which is what we hadto prove.39www.konkur.in

SECTION 2.1 (PAGE 100)ADAMS and ESSEX: CALCULUS 8CHAPTER 2. DIFFERENTIATIONSection 2.1 Tangent Lines and Their Slopes(page 100)1.Slope ofy=3x-1 at.1;2/ism=limh→03.1+h/-1-.3×1-1/h=limh→03hh=3:The tangent line isy-2=3.x-1/, ory=3x-1. (Thetangent to a straight line at any point on it is the samestraight line.)2.Sincey=x=2 is a straight line, its tangent at any point.a;a=2/on it is the same liney=x=2.3.Slope ofy=2x2-5 at.2;3/ism=limh→02.2+h/2-5-.2.22/-5/h=limh→08+8h+2h2-8h=limh→0.8+2h/=8Tangent line isy-3=8.x-2/ory=8x-13.4.The slope ofy=6-x-x2atx= -2 ism=limh→06-.-2+h/-.-2+h/2-4h=limh→quotesdbs_dbs50.pdfusesText_50