[PDF] Exercises and Problems in Calculus - Portland State University

PDF document for free

1. PDF document for free

40312_2CALCULUS_pdf.pdf

Exercises and Problems in Calculus

John M. Erdman

Portland State University

Version August 1, 2013

c

2010 John M. Erdman

E-mail address:erdman@pdx.edu

Contents

Prefaceix

Part 1. PRELIMINARY MATERIAL1

Chapter 1. INEQUALITIES AND ABSOLUTE VALUES

3

1.1. Background3

1.2. Exercises4

1.3. Problems5

1.4. Answers to Odd-Numbered Exercises

6

Chapter 2. LINES IN THE PLANE

7

2.1. Background7

2.2. Exercises8

2.3. Problems9

2.4. Answers to Odd-Numbered Exercises

10

Chapter 3. FUNCTIONS

11

3.1. Background11

3.2. Exercises12

3.3. Problems15

3.4. Answers to Odd-Numbered Exercises

17

Part 2. LIMITS AND CONTINUITY19

Chapter 4. LIMITS

21

4.1. Background21

4.2. Exercises22

4.3. Problems24

4.4. Answers to Odd-Numbered Exercises

25

Chapter 5. CONTINUITY

27

5.1. Background27

5.2. Exercises28

5.3. Problems29

5.4. Answers to Odd-Numbered Exercises

30
Part 3. DIFFERENTIATION OF FUNCTIONS OF A SINGLE VARIABLE31

Chapter 6. DEFINITION OF THE DERIVATIVE

33

6.1. Background33

6.2. Exercises34

6.3. Problems36

6.4. Answers to Odd-Numbered Exercises

37

Chapter 7. TECHNIQUES OF DIFFERENTIATION

39
iii iv CONTENTS

7.1. Background39

7.2. Exercises40

7.3. Problems45

7.4. Answers to Odd-Numbered Exercises

47

Chapter 8. THE MEAN VALUE THEOREM

49

8.1. Background49

8.2. Exercises50

8.3. Problems51

8.4. Answers to Odd-Numbered Exercises

52

Chapter 9. L'H

^OPITAL'S RULE53

9.1. Background53

9.2. Exercises54

9.3. Problems56

9.4. Answers to Odd-Numbered Exercises

57

Chapter 10. MONOTONICITY AND CONCAVITY

59

10.1. Background

59

10.2. Exercises60

10.3. Problems65

10.4. Answers to Odd-Numbered Exercises

66

Chapter 11. INVERSE FUNCTIONS

69

11.1. Background

69

11.2. Exercises70

11.3. Problems72

11.4. Answers to Odd-Numbered Exercises

74

Chapter 12. APPLICATIONS OF THE DERIVATIVE

75

12.1. Background

75

12.2. Exercises76

12.3. Problems82

12.4. Answers to Odd-Numbered Exercises

84
Part 4. INTEGRATION OF FUNCTIONS OF A SINGLE VARIABLE87

Chapter 13. THE RIEMANN INTEGRAL

89

13.1. Background

89

13.2. Exercises90

13.3. Problems93

13.4. Answers to Odd-Numbered Exercises

95

Chapter 14. THE FUNDAMENTAL THEOREM OF CALCULUS

97

14.1. Background

97

14.2. Exercises98

14.3. Problems102

14.4. Answers to Odd-Numbered Exercises

105

Chapter 15. TECHNIQUES OF INTEGRATION

107

15.1. Background

107

15.2. Exercises108

15.3. Problems115

15.4. Answers to Odd-Numbered Exercises

118

CONTENTS v

Chapter 16. APPLICATIONS OF THE INTEGRAL

121

16.1. Background

121

16.2. Exercises122

16.3. Problems127

16.4. Answers to Odd-Numbered Exercises

130

Part 5. SEQUENCES AND SERIES131

Chapter 17. APPROXIMATION BY POLYNOMIALS

133

17.1. Background

133

17.2. Exercises134

17.3. Problems136

17.4. Answers to Odd-Numbered Exercises

137

Chapter 18. SEQUENCES OF REAL NUMBERS

139

18.1. Background

139

18.2. Exercises140

18.3. Problems143

18.4. Answers to Odd-Numbered Exercises

144

Chapter 19. INFINITE SERIES

145

19.1. Background

145

19.2. Exercises146

19.3. Problems148

19.4. Answers to Odd-Numbered Exercises

149

Chapter 20. CONVERGENCE TESTS FOR SERIES

151

20.1. Background

151

20.2. Exercises152

20.3. Problems155

20.4. Answers to Odd-Numbered Exercises

156

Chapter 21. POWER SERIES

157

21.1. Background

157

21.2. Exercises158

21.3. Problems164

21.4. Answers to Odd-Numbered Exercises

166

Part 6. SCALAR FIELDS AND VECTOR FIELDS169

Chapter 22. VECTOR AND METRIC PROPERTIES ofRn171

22.1. Background

171

22.2. Exercises174

22.3. Problems177

22.4. Answers to Odd-Numbered Exercises

179

Chapter 23. LIMITS OF SCALAR FIELDS

181

23.1. Background

181

23.2. Exercises182

23.3. Problems184

23.4. Answers to Odd-Numbered Exercises

185
Part 7. DIFFERENTIATION OF FUNCTIONS OF SEVERAL VARIABLES187 vi CONTENTS

Chapter 24. PARTIAL DERIVATIVES

189

24.1. Background

189

24.2. Exercises190

24.3. Problems192

24.4. Answers to Odd-Numbered Exercises

193
Chapter 25. GRADIENTS OF SCALAR FIELDS AND TANGENT PLANES 195

25.1. Background

195

25.2. Exercises196

25.3. Problems199

25.4. Answers to Odd-Numbered Exercises

201

Chapter 26. MATRICES AND DETERMINANTS

203

26.1. Background

203

26.2. Exercises207

26.3. Problems210

26.4. Answers to Odd-Numbered Exercises

213

Chapter 27. LINEAR MAPS

215

27.1. Background

215

27.2. Exercises217

27.3. Problems219

27.4. Answers to Odd-Numbered Exercises

221

Chapter 28. DEFINITION OF DERIVATIVE

223

28.1. Background

223

28.2. Exercises224

28.3. Problems226

28.4. Answers to Odd-Numbered Exercises

227
Chapter 29. DIFFERENTIATION OF FUNCTIONS OF SEVERAL VARIABLLES 229

29.1. Background

229

29.2. Exercises232

29.3. Problems234

29.4. Answers to Odd-Numbered Exercises

237

Chapter 30. MORE APPLICATIONS OF THE DERIVATIVE

239

30.1. Background

239

30.2. Exercises241

30.3. Problems243

30.4. Answers to Odd-Numbered Exercises

244

Part 8. PARAMETRIZED CURVES245

Chapter 31. PARAMETRIZED CURVES

247

31.1. Background

247

31.2. Exercises248

31.3. Problems255

31.4. Answers to Odd-Numbered Exercises

256

Chapter 32. ACCELERATION AND CURVATURE

259

32.1. Background

259

32.2. Exercises260

32.3. Problems263

CONTENTS vii

32.4. Answers to Odd-Numbered Exercises

265

Part 9. MULTIPLE INTEGRALS267

Chapter 33. DOUBLE INTEGRALS

269

33.1. Background

269

33.2. Exercises270

33.3. Problems274

33.4. Answers to Odd-Numbered Exercises

275

Chapter 34. SURFACES

277

34.1. Background

277

34.2. Exercises278

34.3. Problems280

34.4. Answers to Odd-Numbered Exercises

281

Chapter 35. SURFACE AREA

283

35.1. Background

283

35.2. Exercises284

35.3. Problems286

35.4. Answers to Odd-Numbered Exercises

287

Chapter 36. TRIPLE INTEGRALS

289

36.1. Background

289

36.2. Exercises290

36.3. Answers to Odd-Numbered Exercises

293

Chapter 37. CHANGE OF VARIABLES IN AN INTEGRAL

295

37.1. Background

295

37.2. Exercises296

37.3. Problems298

37.4. Answers to Odd-Numbered Exercises

299

Chapter 38. VECTOR FIELDS

301

38.1. Background

301

38.2. Exercises302

38.3. Answers to Odd-Numbered Exercises

304

Part 10. THE CALCULUS OF DIFFERENTIAL FORMS305

Chapter 39. DIFFERENTIAL FORMS

307

39.1. Background

307

39.2. Exercises309

39.3. Problems310

39.4. Answers to Odd-Numbered Exercises

311

Chapter 40. THE EXTERIOR DIFFERENTIAL OPERATOR

313

40.1. Background

313

40.2. Exercises315

40.3. Problems316

40.4. Answers to Odd-Numbered Exercises

317

Chapter 41. THE HODGE STAR OPERATOR

319

41.1. Background

319

41.2. Exercises320

viii CONTENTS

41.3. Problems321

41.4. Answers to Odd-Numbered Exercises

322

Chapter 42. CLOSED AND EXACT DIFFERENTIAL FORMS

323

42.1. Background

323

42.2. Exercises324

42.3. Problems325

42.4. Answers to Odd-Numbered Exercises

326

Part 11. THE FUNDAMENTAL THEOREM OF CALCULUS327

Chapter 43. MANIFOLDS AND ORIENTATION

329

43.1. Background|The Language of Manifolds

329

Oriented points330

Oriented curves330

Oriented surfaces

330

Oriented solids331

43.2. Exercises332

43.3. Problems334

43.4. Answers to Odd-Numbered Exercises

335

Chapter 44. LINE INTEGRALS

337

44.1. Background

337

44.2. Exercises338

44.3. Problems342

44.4. Answers to Odd-Numbered Exercises

343

Chapter 45. SURFACE INTEGRALS

345

45.1. Background

345

45.2. Exercises346

45.3. Problems348

45.4. Answers to Odd-Numbered Exercises

349

Chapter 46. STOKES' THEOREM

351

46.1. Background

351

46.2. Exercises352

46.3. Problems356

46.4. Answers to Odd-Numbered Exercises

358

Bibliography359

Index361

Preface

This is a set ofexercisesandproblemsfor a (more or less) standard beginning calculus sequence. While a fair number of theexercisesinvolve only routine computations, many of theexercisesand most of theproblemsare meant to illuminate points that in my experience students have found confusing. Virtually all of theexerciseshave ll-in-the-blank type answers. Often anexercisewill end with something like, \ ...so the answer isap3 + b wherea=andb=." One advantage of this type of answer is that it makes it possible to provide students with feedback on a substantial number of homework exercises without a huge investment of time. More importantly, it gives students a way of checking their work without giving them the answers. When a student works through theexerciseand comes up with an answer that doesn't look anything likeap3+ b , he/she has been given an obvious invitation to check his/her work. The major drawback of this type of answer is that it does nothing to promote good communi- cation skills, a matter which in my opinion is of great importance even in beginning courses. That is what theproblemsare for. They require logically thought through, clearly organized, and clearly written up reports. In my own classes I usually assignproblemsfor group work outside of class. This serves the dual purposes of reducing the burden of grading and getting students involved in the material through discussion and collaborative work. This collection is divided into parts and chapters roughly by topic. Many chapters begin with a \background" section. This is most emphaticallynotintended to serve as an exposition of the relevant material. It is designed only to x notation, denitions, and conventions (which vary widely from text to text) and to clarify what topics one should have studied before tackling the exercisesandproblemsthat follow. The ood of elementary calculus texts published in the past half century shows, if nothing else, that the topics discussed in a beginning calculus course can be covered in virtually any order. The divisions into chapters in these notes, the order of the chapters, and the order of items within a chapter is in no way intended to re ect opinions I have about the way in which (or even if) calculus shouldbe taught. For the convenience of those who might wish to make use of these notes I have simply chosen what seems to me one fairly common ordering of topics. Neither theexercisesnor the problemsare ordered by diculty. Utterly trivial problems sit alongside ones requiring substantial thought. Each chapter ends with a list of the solutions to all the odd-numberedexercises. The great majority of the \applications" that appear here, as in most calculus texts, are best regarded as jests whose purpose is to demonstrate in the very simplest ways some connections between physical quantities (area of a eld, volume of a silo, speed of a train,etc.) and the mathematics one is learning. It doesnotmake these \real world" problems. No one seriously imagines that some Farmer Jones is really interested in maximizing the area of his necessarily rectangular stream-side pasture with a xed amount of fencing, or that your friend Sally just happens to notice that the train passing her is moving at 54:6 mph. To my mind genuinely interesting \real world" problems require, in general, way too much background to t comfortably into an already overstued calculus course. You will nd in this collection just a very few serious applications, problem 15 in Chapt er 29
, for example, where the background is either minimal or largely irrelevant to the solution of the problem. ix x PREFACE I make no claims of originality. While I have dreamed up many of the items included here, there are many others which are standard calculus exercises that can be traced back, in one form or another, through generations of calculus texts, making any serious attempt at proper attribution quite futile. If anyone feels slighted, please contact me. There will surely be errors. I will be delighted to receive corrections, suggestions, or criticism at erdman@pdx.edu

I have placed the the L

ATEX source les on my web page so that anyone who wishes can download the material, edit it, add to it, and use it for any noncommercial purpose.

Part 1

PRELIMINARY MATERIAL

CHAPTER 1

INEQUALITIES AND ABSOLUTE VALUES

1.1. Background

Topics:inequalities, absolute values.

1.1.1. Denition.Ifxandaare two real numbers thedistancebetweenxandaisjxaj. For

most purposes in calculus it is better to think of an inequality likejx5j<2 geometrically rather then algebraically. That is, think \The numberxis within 2 units of 5," rather than \The absolute value ofxminus 5 is strictly less than 2." The rst formulation makes it clear thatxis in the open interval (3;7).

1.1.2. Denition.Letabe a real number. Aneighborhoodofais an open interval (c;d)

inRwhich containsa. An open interval (a;a+) which is centered atais asymmetric neighborhood(or a-neighborhood) ofa.

1.1.3. Denition.Adeleted(orpunctured)neighborhoodof a pointa2Ris an open

interval aroundafrom whichahas been deleted. Thus, for example, the deleted-neighborhood about 3 would be (3;3 +)n f3gor, using dierent notation, (3;3)[(3;3 +).

1.1.4. Denition.A pointais anaccumulation pointof a setBRif every deleted neigh-

borhood ofacontains at least one point ofB.

1.1.5. Notation(For Set Operations).LetAandBbe subsets of a setS. Then

(1)x2A[Bifx2Aorx2B(union); (2)x2A\Bifx2Aandx2B(intersection); (3)x2AnBifx2Aandx =2B(set dierence); and (4)x2Acifx2SnA(complement). If the setSis not specied, it is usually understood to be the setRof real numbers or, starting in

Part 6, the setRn, Euclideann-dimensional space.

3

4 1. INEQUALITIES AND ABSOLUTE VALUES

1.2. Exercises

(1) The inequalit yjx2j<6 can be expressed in the forma < x < bwherea=and b=. (2) The inequalit y15x7 can be expressed in the formjxaj bwhere a=andb=. (3)

Solv ethe e quationj4x+ 23j=j4x9j. Answer:x=.

(4)

Find all n umbersxwhich satisfyjx2+ 2j=jx211j.

Answer:x=andx=.

(5)

Solv ethe inequalit y

3xx

2+ 21x1. Express your answer in interval notation.

Answer: [,)[[2,) .

(6)

Solv ethe e quationjx2j2+ 3jx2j 4 = 0.

Answer:x=andx=.

(7) The inequalit y4x10 can be expressed in the formjxaj bwherea=and b=. (8)

Sk etchthe graph of the equat ionx2 =jy3j.

(9) The inequalit yjx+ 4j<7 can be expressed in the forma < x < bwherea=and b=. (10) Solv ethe inequalit yj3x+ 7j<5. Express your answer in interval notation.

Answer: (,).

(11)

Fin dall n umbersxwhich satisfyjx29j=jx25j.

Answer:x=andx=.

(12)

Solv ethe inequalit y2x2314

 12 . Express your answer in interval notation.

Answer: [,].

(13) Solv ethe inequalit yjx3j 6. Express your answer in interval notation.

Answer: (,][[,) .

(14)

Solv ethe inequalit y

xx+ 2x+ 3x4. Express your answer in interval notation.

Answer: (,)[[,).

(15) In in tervalnotation the solution set for the in equality x+ 1x2x+ 2x+ 3is (1;)[[;2). (16)

Solv ethe inequalit y

4x2x+ 19x

3+x2+ 4x+ 41. Express your answer in interval notation.

Answer: (,].

(17)

Solv ethe equation 2 jx+ 3j215jx+ 3j+ 7 = 0.

Answer:x=,x=,x=, andx=.

(18)

Solv ethe inequalit yx1 +2x

. Express your answer in interval notation.

Answer: [,0)[[,).

1.3. PROBLEMS 5

1.3. Problems

(1)

Let a,b2R. Show thatjjaj jbjj  jabj.

(2)

Let a,b2R. Show thatjabj 12

(a2+b2).

6 1. INEQUALITIES AND ABSOLUTE VALUES

1.4. Answers to Odd-Numbered Exercises

(1)4, 8 (3)74 (5) [ 12 ;1)[[2;1) (7) 3, 7 (9)11, 3 (11)p7, p7 (13) ( 1;3][[9;1) (15) ( 1;3)[[74 ;2) (17)10,72 ,52 , 4

CHAPTER 2

LINES IN THE PLANE

2.1. Background

Topics:equations of lines in the plane, slope,x- andy-intercepts, parallel and perpendicular lines.

2.1.1. Denition.Let (x1;y1) and (x2;y2) be points in the plane such thatx16=x2. Theslope

of the (nonvertical straight) lineLwhich passes through these points is m

L:=y2y1x

2x1:

TheequationforLis

yy0=mL(xx0) where (x0;y0) is any point lying onL. (If the lineLis vertical (that is, parallel to they-axis) it is common to say that it hasinnite slopeand writemL=1. The equation for a vertical line is x=x0where (x0;y0) is any point lying onL.) Two nonvertical linesLandL0areparallelif their respective slopesmLandmL0are equal. (Any two vertical lines are parallel.) They areperpendicularif their respective slopes are negative reciprocals; that is, ifmL0=1m L. (Vertical lines are always perpendicular to horizontal lines.) 7

8 2. LINES IN THE PLANE

2.2. Exercises

(1) The equation of the line passing through the p oints( 7;3) and (8;2) isay=x+bwhere a=andb=(2)The equation of the p erpendicularbisector of the li nesegmen tjoining the p oints( 2;5) and (4;3) isax+by+ 1 = 0 wherea=andb=. (3) Let Lbe the line passing through the point (4;9) with slope34 . Thex-intercept ofLisand itsy-intercept is. (4) The equation of the line whic hpasses through the p oint(4 ;2) and is perpendicular to the linex+ 2y= 1 isax+by+ 1 = 0 wherea=andb=. (5) The equation of the line whic his parallel to the line x+32 y=52 and passes through the point (1;3) is 2x+ay+b= 0 wherea=andb=.

2.3. PROBLEMS 9

2.3. Problems

(1) The to wnof Plaineld is 4 miles e astand 6 miles north of Burlington. Allen townis 8 miles west and 1 mile north of Plaineld. A straight road passes through Plaineld and Burlington. A second straight road passes through Allentown and intersects the rst road at a point somewhere south and west of Burlington. The angle at which the roads intersect is=4 radians. Explain how to nd the location of the point of intersection and carry out the computation you describe. (2) Pro vethat the line segmen tjoining the midp ointsof t wosides of a triangl eis half t he length of the third side and is parallel to it.Hint.Try not to make things any more complicated than they need to be. A thoughtful choice of a coordinate system may be helpful. One possibility: orient the triangle so that one side runs along thex-axis and one vertex is at the origin.

10 2. LINES IN THE PLANE

2.4. Answers to Odd-Numbered Exercises

(1) 3, 2 (3)8, 6 (1)

3, 11

CHAPTER 3

FUNCTIONS

3.1. Background

Topics:functions, domain, codomain, range, bounded above, bounded below, composition of functions.

3.1.1. Denition.IfSandTare sets we say thatfis afunctionfromStoTif for everyx

inSthere corresponds one and only one elementf(x) inT. The setSis called thedomainoff and is denoted by domf. The setTis called thecodomainoff. Therangeoffis the set of allf(x) such thatxbelongs toS. It is denoted by ranf. The wordsfunction, map, mapping, and transformationare synonymous. A functionf:A!Bis said to bereal valuedifBRand is called afunction of a real variable ifAR. The notationf:S!T:x7!f(x) indicates thatfis a function whose domain isS, whose codomain isT, and whose value atxisf(x). Thus, for example,f:R!R:x7!x2denes the real valued function whose value at each real numberxis given byf(x) =x2. We use domfto denote the domain offand ranfto denote its range.

3.1.2. Denition.A functionf:S!Risbounded aboveby a numberMisf(x)Mfor

everyx2S, It isbounded belowby a numberKifKf(x) for everyx2S. And it is boundedif it is bounded both above and below; that is, if there existsN >0 such thatjf(x)j N for everyx2S.

3.1.3. Denition.Letfandgbe real valued functions of a real variable. Dene thecomposite

ofgandf, denoted bygf, by (gf)(x) :=g(f(x)) for allx2domfsuch thatf(x)2domg. The operationis calledcomposition.

For problem

2 , the following fact may be useful.

3.1.4. Theorem.Every nonempty open interval inRcontains both rational and irrational numbers.

11

12 3. FUNCTIONS

3.2. Exercises

(1)

Let f(x) =11 +

11 + 1x . Then: (a)f(12 ) =. (b) The domain of fis the set of all real numbers except,, and. (2)

Let f(x) =7px

29p25x2. Then domf= (,][[,).

(3) Find the domain and range of the fu nctionf(x) = 2p4x23.

Answer: domf= [,] and ranf= [,].

(4) Let f(x) =x34x211x190. The set of all numbersxsuch that jf(x)40j<260 is (,)[(,). (5) Let f(x) =x+ 5,g(x) =px, andh(x) =x2. Then (g(h(gf)))(4) =. (6)

Let f(x) =1121 +

11x. (a)

Find f(1=2). Answer..

(b) Find the domain of f. Answer. The domain offis the set of all real numbersexcept,, and. (7)

Let f(x) =px

245p36x2. Then, in interval notation, that part of the domain offwhich

is to the right of the origin is [2;a)[(a;b] wherea=andb=. (8)

Let f(x) = (x27x10)1=2.

(a)

Then f(3) =.

(b)

The domain of fis (,) .

(9) Let f(x) =x34 for all real numbersx. Then for allx6= 0 dene a new functiongby g(x) = (2x)1(f(1 +x)f(1x)). Theng(x) can be written in the formax2+bx+c wherea=,b=, andc=. (10) The cost of making a widget is 75 cen ts.If they are sold for $1.95 eac h,3000 widgets can be sold. For every cent the price is lowered, 60 more widgets can be sold. (a) If xis the price of a widget in cents, then the net prot isp(x) =ax2+bx+cwhere a=,b=, andc=. (b) The \b est"price (that is, the price that maximizes prot) is x= $.. (c)

A tthis b estprice the prot is $ .

(11) Let f(x) = 3p25x2+ 2. Then domf= [,] and ranf= [,]. (12) Fin da form ulaexhibiting the ar eaAof an equilateral triangle as a function of the length sof one of its sides.

Answer:A(s) =.

(13) Let f(x) = 4x318x24x+33. Find the largest setSon which the functionfis bounded above by 15 and below by15.

Answer:S= [,][[,][[,] .

3.2. EXERCISES 13

(14) Let f(x) =px,g(x) =45x, andh(x) =x2. Find (h((hgf)f))(4).

Answer:.

(15) Let f(x) =x+ 7,g(x) =px+ 2, andh(x) =x2. Find (h((fg)(gf)))(7).

Answer:.

(16) Let f(x) =p5x,g(x) =px+ 11,h(x) = 2(x1)1, andj(x) = 4x1.

Then (f(g+ (hg)(hj)))(5) =.

(17) Let f(x) =x2,g(x) =p9 +x, andh(x) =1x2. Then (h(fggf))(4) =. (18)

Let f(x) =x2,g(x) =p9 +x, andh(x) = (x1)1=3.

Then (h((fg)(gf)))(4) =.

(19)

Let f(x) =5x

,g(x) =px, andh(x) =x+ 1. Then (g(fg) + (gfh))(4) =. (20)

Let g(x) = 5x2,h(x) =px+ 13, andj(x) =1x

. Then (jhg)(3) =. (21)

Let h(x) =1px+ 6,j(x) =1x

, andg(x) = 5x2. Then (gjh)(3) =. (22)

Let f(x) =x2+2x

,g(x) =22x+ 3, andh(x) =p2x. Then (hgf)(4) =. (23)

Let f(x) = 3(x+ 1)3,g(x) =x5+x4x+ 1, andh(x) =px.

Then (h(g+ (hf)))(2) =.

(24) Let f(x) =x2,g(x) =px+ 11,h(x) = 2(x1)1, andj(x) = 4x1.

Then (f((hg) + (hj)))(5) =.

(25) Let f(x) =x35x2+x7. Find a functiongsuch that (fg)(x) = 27x3+90x2+78x2.

Answer:g(x) =.

(26) Let f(x) = cosxandg(x) =x2for allx. Write each of the following functions in terms of fandg.Example.Ifh(x) = cos2x2, thenh=gfg. (a)

If h(x) = cosx2, thenh=.

(b)

If h(x) = cosx4, thenh=.

(c)

If h(x) = cos4x2, thenh=.

(d)

If h(x) = cos(cos2x), thenh=.

(e)

If h(x) = cos2(x4+x2), thenh=.

(27) Let f(x) =x3,g(x) =x2, andh(x) = sinxfor allx. Write each of the following functions in terms off,g, andh.Example.Ifk(x) = sin3(x2)3, thenk=fhfg. (a)

If k(x) = sin3x, thenk=.

(b)

If k(x) = sinx3, thenk=.

(c)

If k(x) = sin(x32), thenk=.

(d)

If k(x) = sin(sinx2), thenk=.

(e)

If k(x) = sin3(sin3(x2), thenk=.

(f)

If k(x) = sin9(x2), thenk=.

(g)

If k(x) = sin(x38), thenk=.

(h)

If k(x) = sin(x36x2+ 12x8), thenk=.

14 3. FUNCTIONS

(28) Let g(x) = 3x2. Find a functionfsuch that (fg)(x) = 18x236x+ 19.

Answer:f(x) =.

(29) Let h(x) = arctanxforx0,g(x) = cosx, andf(x) = (1x2)1. Find a numberpsuch that (fgh)(x) = 1 +xp. Answer:p=. (30) Let f(x) = 3x2+5x+1. Find a functiongsuch that (fg)(x) = 3x4+6x34x27x+3.

Answer:g(x) =.

(31) Let g(x) = 2x1. Find a functionfsuch that (fg)(x) = 8x328x2+ 28x14.

Answer:f(x) =.

(32)

Fin dt wosolution sto the equ ation

8cos

3((x2+83

x+ 2)) + 16cos2((x2+83 x+ 2)) + 16cos((x2+83 x+ 2)) = 13:

Answer:and.

(33) Let f(x) = (x+ 4)1=2,g(x) =x2+ 1,h(x) = (x3)1=2, andj(x) =x1.

Then (j((gh)(gf)))(5) =.

(34) Let g(x) = 3x2. Find a functionfsuch that (fg)(x) = 18x236x+ 19.

Answer:f(x) =.

(35) Let f(x) =x35x2+x7. Find a functiongsuch that (fg)(x) = 27x3+90x2+78x2.

Answer:g(x) =.

(36) Let f(x) =x2+ 1. Find a functiongsuch that (fg)(x) = 2 +2x +1x 2.

Answer:g(x) =.

(37) Let f(x) =x2+ 3x+ 4. Find two functionsgsuch that (fg)(x) = 4x26x+ 4.

Answer:g(x) =andg(x) =.

(38) Let h(x) =x1andg(x) =px+ 1. Find a functionfsuch that (fgh)(x) = x 3=2+ 4x1+ 2x1=26.

Answer:f(x) =.

(39) Let g(x) =x2+x1. Find a functionfsuch that (fg)(x) =x4+ 2x33x24x+ 6.

Answer:f(x) =.

(40)

Let S(x) =x2andP(x) = 2x.

Then (SSSSPP)(1) =.

3.3. PROBLEMS 15

3.3. Problems

(1)

Do there exist functions fandgdened onRsuch that

f(x) +g(y) =xy for all real numbersxandy? Explain. (2) Y ourfriend Susan has b ecomein terestedin functions f:R!Rwhich preserve both the operation of addition and the operation of multiplication; that is, functionsfwhich satisfy f(x+y) =f(x) +f(y) (3.1) and f(xy) =f(x)f(y) (3.2) for allx;y2R:Naturally she started her investigation by looking at some examples. The trouble is that she was able to nd only two very simple examples:f(x) = 0 for allxand f(x) =xfor allx. After expending considerable eort she was unable to nd additional examples. She now conjectures that there are no other functions satisfying ( 3.1 ) and ( 3.2 ). Write Susan a letter explaining why she is correct. Hint.You may choose to pursue the following line of argument. Assume thatfis a function (not identically zero) which satises ( 3.1 ) and ( 3.2 ) above. (a)

Sho wthat f(0) = 0. [In (3.1) lety= 0.]

(b)

Sh owthat if a6= 0 anda=ab, thenb= 1.

(c) Sho wthat f(1) = 1. [How do we know that there exists a numbercsuch that f(c)6= 0? Letx=candy= 1 in (3.2).] (d)

Sh owthat f(n) =nfor every natural numbern.

(e) Sho wthat f(n) =nfor every natural numbern. [Letx=nandy=nin (3.1).

Use (d).]

(f) Sho wthat f(1=n) = 1=nfor every natural numbern. [Letx=nandy= 1=n in ( 3.2 ).] (g) Sho wthat f(r) =rfor every rational numberr. [Ifr0 writer=m=nwhere mandnare natural numbers; then use (3.2), (d), and (e). Next consider the case r <0.] (h)

Sh owthat if x0, thenf(x)0. [Writexaspx

pxand use (3.2).] (i) Sh owthat if xy, thenf(x)f(y). [Show thatf(x) =f(x) holds for all real numbersx. Use (h).] (j) No wpro vethat fmust be the identity function onR. [Argue by contradiction: Assumef(x)6=xfor some numberx. Then there are two possibilities: eitherf(x)> x orf(x)< x. Show that both of these lead to a contradiction. Apply theorem3.1.4 to the two casesf(x)> xandf(x)< xto obtain the contradictionf(x)< f(x).] (3) Let f(x) = 1xandg(x) = 1=x. Taking composites of these two functions in all possible ways (ff,gf,fgfff,ggfgff, etc.), how many distinct functions can be produced? Write each of the resulting functions in terms offandg. How do you know there are no more? Show that each function on your list has an inverse which is also on your list. What is the common domain for these functions? That is, what is the largest set of real numbers for which all these functions are dened? (4) Pro veor dispro ve:comp ositionof functions is comm utative;that is gf=fgwhen both sides are dened. (5) Let f,g,h:R!R. Prove or disprove:f(g+h) =fg+fh. (6) Let f,g,h:R!R. Prove or disprove: (f+g)h= (fh) + (gh).

16 3. FUNCTIONS

(7) Let a2Rbe a constant and letf(x) =axfor allx2R. Show thatff=I(whereI is the identity function onR:I(x) =xfor allx).

3.4. ANSWERS TO ODD-NUMBERED EXERCISES 17

3.4. Answers to Odd-Numbered Exercises

(1) (a) 34
(b)1,12 , 0 (3) [ 2;2], [3;1] (5)p13 (7)p11, 6 (9)

1, 0, 3

(11) [ 5;5], [2;17] (13) [ 32 ;1][[1;2][[4;92 ] (15) 36
(17) 16 (19) 6 (21)4 (23) 5 (25)

3 x+ 5

(27) (a) fh (b)hf (c)hgf (d)hgh (e)fhfhg (f)ffhg (g)hggggf (h)hfg (29)2 (31)x34x2+ 3x6 (33) 917
(35)

3 x+ 5

(37)2x, 2x3 (39)x22x+ 3

Part 2

LIMITS AND CONTINUITY

CHAPTER 4

LIMITS

4.1. Background

Topics:limit off(x) asxapproachesa, limit off(x) asxapproaches innity, left- and right-hand limits.

4.1.1. Denition.Suppose thatfis a real valued function of a real variable,ais an accumulation

point of the domain off, and`2R. We say that`is thelimit off(x)asxapproachesaif for every neighborhoodVof`there exists a corresponding deleted neighborhoodUofawhich satises the following condition: for every pointxin the domain offwhich lies inUthe pointf(x) lies inV. Once we have convinced ourselves that in this denition it doesn't matter if we work only with symmetric neighborhoods of points, we can rephrase the denition in a more conventional algebraic fashion:`is thelimit off(x)asxapproachesaprovided that for every >0 there exists >0 such that if 04.1.2. Notation.To indicate that a number`is the limit off(x) asxapproachesa, we may write either limx!af(x) =lorf(x)!`asx!a: (See problem 2 .) 21

22 4. LIMITS

4.2. Exercises

(1) lim x!3x

313x2+ 51x63x

34x23x+ 18=a5

wherea=. (2) lim x!0px

2+ 9x+ 93x

=a2 wherea=. (3) lim x!1x

3x2+ 2x2x

3+ 3x24x=3a

wherea=. (4) li m t!0tp4t2=. (5) lim x!0px+ 93x =1a wherea=. (6) lim x!2x

33x2+x+ 2x

3x6=1a

wherea=. (7) lim x!2x

3x28x+ 12x

310x2+ 28x24=a4

wherea=. (8) lim x!0px

2x+ 42x

2+ 3x=1a

wherea=. (9) lim x!1x

3+x25x+ 3x

34x2+ 5x2=.

(10) lim x!3x

34x23x+ 18x

38x2+ 21x18=.

(11) lim x!1x

3x25x3x

3+ 6x2+ 9x+ 4=4a

wherea=. (12) lim x!02xsinx1cosx=. (13) lim x!01cosx3xsinx=1a wherea=. (14) lim x!0tan3xsin3xx 3=a2 wherea=. (15) lim h!0sin2h5h2+ 7h=. (16) lim h!0cot7hcot5h=. (17) lim x!0secxcosx3x2=1a wherea=. (18) lim x!1(9x86x5+ 4)1=2(64x12+ 14x77)1=3=a4 wherea=. (19) lim x!1px(px+ 3px2) =a2 wherea=. (20) lim x!17x+ 2x23x35x44 + 3xx2+x3+ 2x4=a2 wherea=. (21) lim x!1(2x4137)5(x2+ 429)10=.

4.2. EXERCISES 23

(22) lim x!1(5x10+ 32)3(12x6)5=a32 wherea=. (23) lim x!1 px

2+xx

=1a wherea=. (24) lim x!1x(256x4+ 81x2+ 49)1=4=1a wherea=. (25) lim x!1xp3x2+ 22p3x2+ 4 =apawherea=. (26) lim x!1x23 (x+ 1)13 x13
=1a wherea=. (27) lim x!1 qx+pxqxpx  =. (28)

Let f(x) =8

< :2x1;ifx <2; x

2+ 1;ifx >2.Then limx!2f(x) =and lim

x!2+f(x) =. (29)

Let f(x) =jx1jx1. Then limx!1f(x) =and lim

x!1+f(x) =. (30)

Let f(x) =8

< :5x3;ifx <1; x

2;ifx1.Then limx!1f(x) =and lim

x!1+f(x) =. (31)

Let f(x) =8

< :3x+ 2;ifx <2; x

2+ 3x1;ifx 2.Then limx!2f(x) =and lim

x!2+f(x) =. (32) Su pposey=f(x) is the equation of a curve which always lies between the parabola x

2=y1 and the hyperbolayx+y1 = 0. Then limx!0f(x) =.

24 4. LIMITS

4.3. Problems

(1)

Find lim

x!0+ e1=xsin(1=x)(x+ 2)3 (if it exists) and give a careful argument showing that your answer is correct. (2)

The notation lim

x!af(x) =`that we use for limits is somewhat optimistic. It assumes the uniqueness of limits. Prove that limits, if they exist, are indeed unique. That is, suppose thatfis a real valued function of a real variable,ais an accumulation point of the domain off, and`,m2R. Prove that iff(x)!`asx!aandf(x)!mas x!a, thenl=m. (Explain carefully why it was important that we requireato be an accumulation point of the domain off.) (3) Let f(x) =sinxx+ 1for allx6=1. The following information is known about a functiong dened for all real numbersx6= 1: (i)g=pq wherep(x) =ax2+bx+candq(x) =dx+efor some constantsa;b;c;d;e; (ii) the only x-intercept of the curvey=g(x) occurs at the origin; (iii)g(x)0 on the interval [0;1) and is negative elsewhere on its domain; (iv)ghas a vertical asymptote atx= 1; and (v)g(1=2) = 3.

Either nd lim

x!1g(x)f(x) or else show that this limit does not exist. Hints.Write an explicit formula forgby determining the constantsa:::e. Use (ii) to ndc; use (ii) and (iii) to nda; use (iv) to nd a relationship betweendande; then use (v) to obtain an explicit form forg. Finally look atf(x)g(x); replace sinx by sin((x1) +) and use the formula for the sine of the sum of two numbers. (4)

Ev aluatelim

x!0pjxjcos(1=x2)2 + px

2+ 3(if it exists). Give a careful proof that your conclusion is

correct.

4.4. ANSWERS TO ODD-NUMBERED EXERCISES 25

4.4. Answers to Odd-Numbered Exercises

(1)4 (3) 5 (5) 6 (7) 5 (9)4 (11) 3 (13) 6 (15) 27
(17) 3 (19) 5 (21) 32
(23) 2 (25) 3 (27) 1 (29)1, 1 (31)4,3

CHAPTER 5

CONTINUITY

5.1. Background

Topics:continuous functions,intermediate value theorem.extreme value theorem. There are many ways of stating theintermediate value theorem. The simplest says thatcon- tinuous functions take intervals to intervals.

5.1.1. Denition.A subsetJof the real lineRis anintervalifz2Jwhenevera,b2Jand

a < z < b.

5.1.2. Theorem(Intermediate Value Theorem).LetJbe an interval inRandf:J!Rbe

continuous. Then the range offis an interval.

5.1.3. Denition.A real-valued functionfon a setAis said to have amaximumat a pointain

Aiff(a)f(x) for everyxinA; the numberf(a) is themaximum valueoff. The function has aminimumataiff(a)f(x) for everyxinA; and in this casef(a) is theminimum valueof f. A number is anextreme valueoffif it is either a maximum or a minimum value. It is clear that a function may fail to have maximum or minimum values. For example, on the open interval (0;1) the functionf:x7!xassumes neither a maximum nor a minimum. The concepts we have just dened are frequently calledglobal(orabsolute)maximumand global(orabsolute)minimum.

5.1.4. Denition.Letf:A!RwhereAR. The functionfhas alocal(orrelative)

maximumat a pointa2Aif there exists a neighborhoodJofasuch thatf(a)f(x) whenever x2Jandx2domf. It has alocal(orrelative)minimumat a pointa2Aif there exists a neighborhoodJofasuch thatf(a)f(x) wheneverx2Jandx2domf.

5.1.5. Theorem(Extreme Value Theorem).Every continuous real valued function on a closed

and bounded interval inRachieves its (global) maximum and minimum value at some points in the interval.

5.1.6. Denition.A numberpis afixed pointof a functionf:R!Riff(p) =p.

5.1.7. Example.Iff(x) =x26 for allx2R, then 3 is a xed point off.

27

28 5. CONTINUITY

5.2. Exercises

(1)

Let f(x) =x32x22x3x

34x2+ 4x3forx6= 3. How shouldfbe dened atx= 3 so that it

becomes a continuous function on all ofR?

Answer:f(3) =a7

wherea=. (2)

Let f(x) =8

>>>< > >>:1 ifx <0 xif 0< x <1

2xif 1< x <3

x4 ifx >3. (a) Is it p ossibleto dene fatx= 0 in such a way thatfbecomes continuous atx= 0?

Answer:. If so, then we should setf(0) =.

(b) Is it p ossibleto dene fatx= 1 in such a way thatfbecomes continuous atx= 1?

Answer:. If so, then we should setf(1) =.

(c) Is it p ossibleto dene fatx= 3 in such a way thatfbecomes continuous atx= 3?

Answer:. If so, then we should setf(3) =.

(3)

Let f(x) =8

>>>< > >>:x+ 4 ifx <2 xif2< x <1 x

22x+ 1 if 1< x <3

102xifx >3.

(a) Is it p ossibleto dene fatx=2 in such a way thatfbecomes continuous at x=2? Answer:. If so, then we should setf(2) =. (b) Is it p ossibleto dene fatx= 1 in such a way thatfbecomes continuous atx= 1?

Answer:. If so, then we should setf(1) =.

(c) Is it p ossibleto dene fatx= 3 in such a way thatfbecomes continuous atx= 3?

Answer:. If so, then we should setf(3) =.

(4) The e quationx5+x3+ 2x= 2x4+ 3x2+ 4 has a solution in the open interval (n;n+ 1) wherenis the positive integer. (5) The equation x46x253 = 22x2x3has a solution in the open interval (n;n+1) where nis the positive integer. (6) The equation x4+x+ 1 = 3x3+x2has solutions in the open intervals (m;m+ 1) and (n;n+ 1) wheremandnare the distinct positive integersand. (7) The equation x5+ 8x= 2x4+ 6x2has solutions in the open intervals (m;m+ 1) and (n;n+ 1) wheremandnare the distinct positive integersand.

5.3. PROBLEMS 29

5.3. Problems

(1)

Pro vethat the equation

x

180+841 +x2+ cos2x= 119

has at least two solutions. (2) (a) Fin dall the xed p ointsof the f unctionfdened in example5.1.7 . Theorem:Every continuous functionf: [0;1]![0;1] has a xed point. (b) Pro veth epreceding theorem. Hint.Letg(x) =xf(x) for 0x1. Apply the intermediate value theorem5.1.2to g. (c) Let g(x) = 0:1x3+0:2 for allx2R, andhbe the restriction ofgto [0;1]. Show that hsatises the hypotheses of the theorem. (d) F orth efunction hdened in (c) nd an approximate value for at least one xed point with an error of less than 10 6. Give a careful justication of your answer. (e) Let gbe as in (c). Are there other xed points (that is, points not in the unit square where the curvey=g(x) crosses the liney=x)? If so, nd an approximation to each such point with an error of less than 10 6. Again provide careful justication. (3) Dene fon [0;4] byf(x) =x+ 1 for 0x <2 andf(x) = 1 for 2x4. Use the extreme value theorem5.1.5to sho wthat fis not continuous. (4) Giv ean example of a function dened on [0 ;1] which has no maximum and no minimum on the interval. Explain why the existence of such a function does not contradict the extreme value theorem5.1.5. (5) Giv ean example of a con tinuousfunction d enedon the in terval(1 ;2] which does not achieve a maximum value on the interval. Explain why the existence of such a function does not contradict theextreme value theorem5.1.5. (6) Giv ean example of a con tinuousfunction on the closed in terval[3 ;1) which does not achieve a minimum value on the interval. Explain why the existence of such a function does not contradict theextreme value theorem5.1.5. (7) Dene fon [2;0] byf(x) =1(x+ 1)2for2x <1 and1< x0, andf(1) =3. Use theextreme value theorem5.1.5to sho wthat fis not continuous. (8)

Let f(x) =1x

for 0< x1 andf(0) = 0. Use theextreme value theorem5.1.5to sho w thatfis not continuous on [0;1].

30 5. CONTINUITY

5.4. Answers to Odd-Numbered Exercises

(1) 13 (3) (a) y es,2 (b) no, | (c) y es,4 (5) 3 (7) 1, 2

Part 3

DIFFERENTIATION OF FUNCTIONS OF

A SINGLE VARIABLE

CHAPTER 6

DEFINITION OF THE DERIVATIVE

6.1. Background

Topics:denition of the derivative of a real valued function of a real variable at a point

6.1.1. Notation.Letfbe a real valued function of a real variable which is dierentiable at a

pointain its domain. When thinking of a function in terms of its graph, we often writey=f(x), callxtheindependent variable, and callythedependent variable. There are many notations for the derivative offata. Among the most common are

Df(a); f0(a);dfdx

 a; y0(a);_y(a);anddydx  a: 33

34 6. DEFINITION OF THE DERIVATIVE

6.2. Exercises

(1)

Sup posey oukno wthat the deriv ativeof

pxis12 px for everyx >0. Then lim x!9px3x9=1a wherea=: (2)

Sup posey oukno wthat the deriv ativesof x13

is13 x23 for everyx6= 0. Then lim x!8 x8  13 1x8=1a wherea=: (3) Sup posey oukno wthat the deriv ativeof exisexfor everyx. Then lim x!2e xe2x2=: (4)

Sup posey oukno wthat the deriv ativeof ln xis1x

for everyx >0. Then lim x!elnx33xe=: (5) Sup posey oukno wthat the deriv ativeof tan xis sec2xfor everyx. Then lim x!4 tanx14x=: (6) Sup posey oukno wthat the deriv ativeof arctan xis11 +x2for everyx. Then lim x!p3

3arctanxxp3

=: (7) Sup posey oukno wthat the deriv ativeof c osxissinxfor everyx. Then lim x!3

2cosx13x=1a

wherea=: (8) Sup posey oukno wthat the deriv ativeof c osxissinxfor everyx. Then lim t!0cos( 6 +t)p3 2 t =1a wherea=. (9) Sup posey oukno wthat the deriv ativeof s inxis cosxfor everyx. Then lim x!=4p2sinx+ 14x+=1a wherea=. (10) Su pposey oukno wthat the deriv ativeof sin xis cosxfor everyx. Then lim x!712

2p2sinxp3112x7=1pa

b wherea=andb=. (11)

Let f(x) =8

>< > :x

2;forx1

1;for 1< x3

52x;forx >3. Thenf0(0) =,f0(2) =, andf0(6) =.

6.2. EXERCISES 35

(12) Su pposethat the tangen tline to the graph of a function fatx= 1 passes through the point (4;9) and thatf(1) = 3. Thenf0(1) =. (13) Su pposethat gis a dierentiable function and thatf(x) =g(x)+5 for allx. Ifg0(1) = 3, thenf0(1) =. (14) Su pposethat gis a dierentiable function and thatf(x) =g(x+5) for allx. Ifg0(1) = 3, thenf0(a) = 3 wherea=. (15) Su pposethat fis a dierentiable function, thatf0(x) =2 for allx, and thatf(3) = 11. Find an algebraic expression forf(x). Answer:f(x) =. (16) Su pposethat fis a dierentiable function, thatf0(x) = 3 for allx, and thatf(3) = 3. Find an algebraic expression forf(x). Answer:f(x) =.

36 6. DEFINITION OF THE DERIVATIVE

6.3. Problems

(1)

Let f(x) =1x

21anda=3. Show how to use the denition ofderivativeto nd

Df(a).

(2) Let f(x) =1px+ 7. Show how to use the denition ofderivativeto ndf0(2). (3) Let f(x) =1px+ 3. Show how to use the denition ofderivativeto ndf0(1). (4)

Let f(x) =px

25. Show how to use the denition ofderivativeto ndf0(3).

(5) Let f(x) =p8x. Show how to use the denition ofderivativeto ndf0(1). (6) Let f(x) =px2. Show how to use the denition ofderivativeto ndf0(6). (7)

Let f(x) =xx

2+ 2. Show how to use the denition ofderivativeto ndDf(2).

(8) Let f(x) = (2x23)1. Show how to use the denition ofderivativeto ndDf(2). (9)

Let f(x) =x+ 2x2sin1x

forx6= 0 andf(0) = 0. What is the derivative offat 0 (if it exists)? Is the functionf0continuous at 0?

6.4. ANSWERS TO ODD-NUMBERED EXERCISES 37

6.4. Answers to Odd-Numbered Exercises

(1) 6 (3)e2 (5) 12 (7)p3 (9) 4 (11)

0, 0, 2

(13) 3 (15)2x+ 5

CHAPTER 7

TECHNIQUES OF DIFFERENTIATION

7.1. Background

Topics:rule for dierentiating products,rule for dierentiating quotients,chain rule, tangent lines, implicit dierentiation.

7.1.1. Notation.We usef(n)(a) to denote the nthderivative offata.

7.1.2. Denition.A pointain the domain of a functionfis astationary pointoffis

f

0(a) = 0. It is acritical pointoffif it is either a stationary point offor if it is a point where

the derivative offdoes not exist. Some authors use the termsstationary pointandcritical pointinterchangeably|especially in higher dimensions. 39

40 7. TECHNIQUES OF DIFFERENTIATION

7.2. Exercises

(1) If f(x) = 5x1=2+ 6x3=2, thenf0(x) =axp+bxqwherea=,p=, b=, andq=. (2)

If f(x) = 105px

3+126 px

5, thenf0(x) =axp+bxqwherea=,p=,

b=, andq=. (3) If f(x) = 9x4=3+ 25x2=5, thenf00(x) =axp+bxqwherea=,p=, b=, andq=. (4)

If f(x) = 186px+84

px

3, thenf00(x) =axp+bxqwherea=,p=,

b=, andq=. (5) Find a p ointasuch that the tangent line to the graph of the curvey=pxatx=ahas y-intercept 3. Answer:a=. (6) Let f(x) =ax2+bx+cfor allx. We know thatf(2) = 26,f0(2) = 23, andf00(2) = 14.

Thenf(1) =.

(7) Find a n umberksuch that the liney= 6x+ 4 is tangent to the parabolay=x2+k.

Answer:k=.

(8) The equation for the t angentline to the curv ey=x3which passes through the point (0;2) isy=mx+bwherem=andb=. (9)

Let f(x) =14

x4+13 x33x2+74 . Find all pointsx0such that the tangent line to the curve y=f(x) at the point (x0;f(x0)) is horizontal. Answer:x0=,, and. (10) In the land of Oz there is an enormou sstatue of the Go odW itchGlind a.Its b aseis 20 feet high and, on a surveyor's chart, covers the region determined by the inequalities 1y24x2: (The chart coordinates are measured in feet.) Dorothy is looking for her little dog Toto. She walks along the curved side of the base of the statue in the direction of increasingx and Toto is, for a change, sitting quietly. He is at the point on the positivex-axis 7 feet from the origin. How far from Toto is Dorothy when she is rst able to see him?

Answer: 5paft. wherea=.

(11)

Let f(x) =x32

x

2+ 2andg(x) =x2+ 1x

2+ 2. At what values ofxdo the curvesy=f(x) and

y=g(x) have parallel tangent lines? Answer: atx=andx=. (12) The tangen tline to th egraph of a function fat the pointx= 2 hasx-intercept103 and y-intercept10. Thenf(2) =andf0(2) =. (13) The tangen tline to the graph of a function fatx= 2 passes through the points (0;20) and (5;40). Thenf(2) =andf0(2) =. (14) Su pposethat the tangen tline to the graph of a function fatx= 2 passes through the point (5;19) and thatf(2) =2. Thenf0(2) =. (15)

Let f(x) =8

>< > :x

2;forx1

1;for 1< x3

52x;forx >3. Thenf0(0) =,f0(2) =, andf0(6) =.

(16) Su pposethat gis a dierentiable function and thatf(x) =g(x+5) for allx. Ifg0(1) = 3, thenf0(a) = 3 wherea=.

7.2. EXERCISES 41

(17) Let f(x) =j2 jx1jj 1 for every real numberx. Then f

0(2) =,f0(0) =,f0(2) =, andf0(4) =.

(18)

Let f(x) = tan3x. ThenDf(=3) =.

(19)

Let f(x) =1x

csc21x . ThenDf(6=) =2a  pb 1 wherea=andb=. (20) Let f(x) = sin2(3x5+ 7). Thenf0(x) =ax4sin(3x5+ 7)f(x) wherea=and f(x) =. (21) Let f(x) = (x4+ 7x25)sin(x2+ 3). Thenf0(x) =f(x)cos(x2+ 3) +g(x)sin(x2+ 3) wheref(x) =andg(x) =. (22) Let j(x) = sin5(tan(x2+6x5)1=2). ThenDj(x) =p(x)sinn(g(a(x)))f(g(a(x)))h(a(x))a(x)
r where f(x) =, g(x) =, h(x) =, a(x) =, p(x) =, n=, and r=. (23) Let j(x) = sin4(tan(x33x2+6x11)2=3). Thenj0(x) = 8p(x)f(g(a(x)))cos(g(a(x)))h(a(x))a(x)
r where f(x) =, g(x) =, h(x) =, a(x) =, p(x) =, r=. (24)

Let j(x) = sin11(sin6(x37x+ 9)3). Then

Dj(x) = 198(3x2+b)sinp(g(a(x)))h(g(a(x)))sinq(a(x))h(a(x))a(x)
r where g(x) =, h(x) =, a(x) =, p=, q=, r=, and b=, (25)

Let f(x) = (x2+ sinx)100. Thenf0(1) =.

(26) Let f(x) = (x215)9(x217)10. Then the equation of the tangent line to the curve y=f(x) at the point on the curve whosex-coordinate is 4 isy=ax+bwherea=andb=.

42 7. TECHNIQUES OF DIFFERENTIATION

(27) Let f(x) = (x23)10(x3+9)20. Then the equation of the tangent line to the curvey=f(x) at the point on the curve whosex-coordinate is2 isy=ax+bwherea=andb=. (28) Let f(x) = (x39)8(x37)10. Then the equation of the tangent line to the curvey=f(x) at the point on the curve whosex-coordinate is 2 isy=ax+bwherea=and b=. (29) Let f(x) = (x210)10(x28)12. Then the equation of the tangent line to the curve y=f(x) at the point on the curve whosex-coordinate is 3 isy=ax+bwherea=andb=. (30) Let h=gfandj=fgwherefandgare dierentiable functions onR. Fill in the missing entries in the table below.xf(x)f

0(x)g(x)g

0(x)h(x)h

0(x)j(x)j

0(x)0311

32
103
201
2 (31) Let f=ghandj=g
hwheregandhare dierentiable functions onR. Fill in the missing entries in the table below.xg(x)g

0(x)h(x)h

0(x)f(x)f

0(x)j(x)j

0(x)02463

12442

2441324419

Also,g(4) =andg0(4) =.

(32) Let h=gf,j=g
f, andk=g+fwherefandgare dierentiable functions onR. Fill in the missing entries in the table below.xf(x)f

0(x)g(x)g

0(x)h(x)h

0(x)j(x)j

0(x)k(x)k

0(x)1244200011

12206
(33)

Let y= log3(x2+ 1)1=3. Thendydx

=2xa(x2+ 1)wherea=. (34) Let f(x) = ln(6 + sin2x)10(7 + sinx)3. ThenDf(=6) =a5 wherea=. (35) Let f(x) = ln(lnx). What is the domain off? Answer: (,). What is the equation of the tangent line to the curvey=f(x) at the point on the curve whose x-coordinate ise2? Answer:ya=1b (xe2) wherea=andb=. (36)

Fin dwhen y= (tanx)sinxfor 0< x < =2. Thendydx

= (tanx)sinx(f(x) + cosxlntanx) wheref(x) =.

7.2. EXERCISES 43

(37)

Fin dwhen y= (sinx)tanxfor 0< x < =2. Thendydx

= (sinx)tanx(a+f(x)sec2x) where a=andf(x) =. (38) ddx pxlnx=xp(1 +g(x)) wherep=andg(x) =. (39) If f(x) =x3ex, thenf000(x) = (ax3+bx2+cx+d)exwherea=,b=, c=, andd=. (40) Let f(x) =x2cosx. Then (ax2+bx+c)sinx+ (Ax2+Bx+C)cosxis an antiderivative off(x) ifa=,b=,c=,A=,B=, andC=. (41) Let f(x) = (x4x3+x2x+ 1)(3x32x2+x1). Use the rule for dierentiating products to ndf0(1). Answer:. (42)

Let f(x) =x3=2x3xx1=2. Thenf0(4) =9a

wherea=. (43)

Fin da p ointon the curv ey=x2x

32where the tangent line is parallel to the line 4x+

6y5 = 0. Answer: (,).

(44) Let f(x) = 5xcosxx2sinx. Then (ax2+bx+c)sinx+ (Ax2+Bx+C)cosxis an antiderivative off(x) ifa=,b=,c=,A=,B=, and C=. (45) Let f(x) = (2x3)cscx+(2+3xx2)cotxcscx. Thenax2+bx+csinxis an antiderivative off(x) ifa=,b=, andc=. (46)

Let f(x) =x210x

28. Find the equation of the tangent line to the curvey=f(x) at

the point on the curve whosex-coordinate is 3. Answer:y=ax+bwherea=andb=. (47) Let f(x) = (x4+x3+x2+x+ 1)(x5+x3+x+ 2). Find the equation of the tangent line to the curvey=f(x) at the point on the curve whosex-coordinate is1. Answer: y=ax+bwherea=andb=. (48)

Let y=x22x+ 1x

3+ 1. Thendydx

 x=2=2a wherea=. (49)

Let f(x) =x32

x

2+ 2andg(x) =x2+ 1x

2+ 2. At what values ofxdo the curvesy=f(x) and

y=g(x) have parallel tangent lines? Answer: atx=and. (50) Let f(x) =xsinx. Find constantsa,b,A, andBso that (ax+b)cosx+(Ax+B)sinxis an antiderivative off(x). Answer:a=,b=,A=, andB=. (51) Fin d ddx  1x d 2dx 2

11 +x

=abx+ 1x

2(1 +x)bwherea=andb=.

(52) ddx  1x

2
d2dx

2 1x 2 =axpwherea=andp=. (53) Let f(x) =x+ 34x. Findf(15)(x). Answer:7n!(4x)pwheren=andp=. (54) Let f(x) =xx+ 1. Thenf(4)(x) =a(x+ 1)pwherea=andp=.

44 7. TECHNIQUES OF DIFFERENTIATION

(55) Let f(x) =x+ 12x. Thenf(4)(x) =a(2x)pwherea=andp=. (56) Fin dthe equation of the tangen tline to the curv e2 x6+y4= 9xyat the point (1;2).

Answer: 23y=ax+bwherea=andb=.

(57)

F orth ecurv ex3+ 2xy+13

y3=113 , nddydx andd2ydx

2at the point (2;1).

Answer:y0(2) =andy00(2) =a5

wherea=. (58) Fin d dydx andd2ydx

2for the devil's curvey4+ 5y2=x45x2at the point (3;2).

Answer:y0(3) =andy00(3) =.

(59) Fin d dydx ,d2ydx

2, andd3ydx

3at the point (1;8) on the astroidx2=3+y2=3= 5.

Answer:y0(1) =:y00(1) =a6

wherea=; andy000(1) =b24 where b=. (60) Fin dthe p ointof in tersectionof th etangen tlines to the curv ex2+y33x+3yxy= 18 at the points where the curve crosses thex-axis. Answer: (,). (61) Fin dthe equation of the t angentline t othe curv exsiny+x3= arctaney+x4 at the point (1;0). Answer:y=ax+bwherea=andb=. (62) The equation of the tangen tline to th elemniscate 3( x2+y2)2= 25(x2y2) at the point (2;1) isy1 =m(x2) wherem=. (63) The p ointson the o valsof Cass ini( x2+y2)24(x2y2)+3 = 0 where there is a horizontal tangent line are pa b pb ;1b pb  wherea=andb=. (64) The p ointson the o valsof Cas sini( x2+y2)24(x2y2)+3 = 0 where there is a vertical tangent line are (pa;b) and (c;b) wherea=,b=, andc=. (65) A tthe p oint(1 ;2) on the curve 4x2+ 2xy+y2= 12,dydx =and d2ydx 2=. (66) Let fandgbe dierentiable real valued functions onR. We know that the points (4;1) and (3;4) lie on the graph of the curvey=f(x) and the points (4;3) and (3;2) lie on the graph ofy=g(x). We know also thatf0(4) = 3,f0(3) =4,g0(4) =2, and g

0(3) = 6.

(a)

If h=f
g, thenh0(4) =.

(b)

If j= (2f+ 3g)4, thenj0(3) =.

(c)

If k=fg, thenk0(4) =.

(d)

If `=fg

, then`0(3) =. (67)

Let f(x) = 5sinx+ 3cosx. Thenf(117)() =.

(68)

Let f(x) = 4cosx7sinx. Thenf(87)(0) =.

7.3. PROBLEMS 45

7.3. Problems

(1) Let ( x0;y0) be a point inR2. How many tangent lines to the curvey=x2pass through the point (x0;y0)? What are the equations of these lines?Hint.Consider the three cases: y

0> x02,y0=x02, andy0< x02.

(2) F orthe purp osesof this pr oblemy ouma yassume that the dieren tialequation y

00+y= 0 ()

has at least one nontrivial solution on the real line. (That is, there exists at least one twice dierentiable functiony, not identically zero, such thaty00(x) +y(x) = 0 for allx2R.) (a) Sho wthat if uandvare solutions of () anda,b2R, thenw=au+bvandu0are also solutions of (). (b) Sh owthat if yis a solution of () theny2+ (y0)2is constant. (c) Sho wthat if yis a nontrivial solution of (), then eithery(0)6= 0 ory0(0)6= 0.Hint. Argue by contradiction. Show that ifyis a solution of () such that bothy(0) = 0 andy0(0) = 0, theny(x) = 0 for allx. (d) Sh owthat there exists a solution sof () such thats(0) = 0 ands0(0) = 1.Hint. Letybe a nontrivial solution of (). Look for a solutionsof the formay+ by

0(witha,b2R) satisfying the desired conditions.

(e) Sho wthat if yis a solution of () such thaty(0) =aandy0(0) =b, theny=bs+as0. Hint.Letu(x) =y(x)bs(x)as0(x) and show thatuis a solution of () such that u(0) =u0(0) = 0. Use (c). (f) Dene c(x) =s0(x) for allx. Show that (s(x))2+ (c(x))2= 1 for allx. (g) Sho wthat sis an odd function and thatcis even.Hint.To see thatsis odd let u(x) =s(x) for allx. Show thatuis a solution of (). Use (e). Once you know thatsis odd, dierentiate to see thatcis even. (h) Sh owthat s(a+b) =s(a)c(b) +c(a)s(b) for all real numbersaandb.Hint.Let y(x) =s(x+b) for allx. Show thatyis a solution of (). Use (e). (i) Sho wthat c(a+b) =c(a)c(b)s(a)s(b) for all real numbersaandb.Hint.Dierentiate the formula fors(x+b) that you derived in (h). (j) Dene t(x) =s(x)c(x)and(x) =1c(x)for allxsuch thatc(x)6= 0. Show that t

0(x) = ((x))2and0(x) =t(x)(x) whereverc(x)6= 0.

(k)

Sho wthat 1 + ( t(x))2= ((x))2whereverc(x)6= 0.

(l) E xplaincarefully what the (mathematical) p ointof this problem is. (3) Sup posethat fis a dierentiable function such thatf0(x)32 for allxand thatf(1) = 2.

Prove thatf(5)8.

(4) Sup poseth atfis a dierentiable function such thatf0(x)3 for allxand thatf(0) =4.

Prove thatf(3)5.

(5) Sup posethat fis a dierentiable function such thatf0(x) 2 for allx2[0;4] and that f(1) = 6. (a)

Pro vethat f(4)0.

(b)

Pr ovethat f(0)8.

(6)

Giv ea carefu lpro ofthat sin xxfor allx0.

(7)

Giv ea carefu lpro ofthat 1 cosxxfor allx0.

(8)

Pro vethat if x2=1y21 +y2, thendxdy

 2 =1x41y4at points wherey6=1.

46 7. TECHNIQUES OF DIFFERENTIATION

(9) F orthe circle x2+y21 = 0 use implicit dierentiation to show thaty00=1y 3and y

000=3xy

5. (10)

Exp lainho wto calculate

d2ydx

2at the point on the folium of Descartes

x

3+y3= 9xy

where the tangent line is parallel to the asymptote of the folium. (11) Exp laincarefully ho wto nd the curv epassing through the p oint(2 ;3) which has the following property: the segment of any tangent line to the curve contained between the (positive) coordinate axes is bisected at the point of tangency. Carry out the computation you have described.

7.4. ANSWERS TO ODD-NUMBERED EXERCISES 47

7.4. Answers to Odd-Numbered Exercises

(1)52 ,32 , 9,12 (3)

4, 23

,6,85 (5) 36
(7) 13 (9)3, 0, 2 (11)1, 2 (13)

4, 12

(15)

0, 0, 2

(17)1, 1,1, 1 (19) 9, 3 (21)

2 x5+ 14x310x, 4x3+ 14x

(23) sin

3x, tanx, sec2x,x33x2+ 6x11

23
,x22x+ 2,12 (25)

100(2 )

(27)

200, 401

(29)

12, 35

(31)3, 0,1, 1 (rst row)

2, 2, 1, 1 (second row)

1, 3 (third row)

13, 8 (33) 3 ln3 (35)

1, 1, ln2, 2e2

(37)

1, ln sinx

(39)

1, 9, 18, 6

(41) 8 (43) 2, 23
(45)

1, 3,2

(47)

11, 10

(49)1, 2 (51)2, 4 (53)

15, 16

(55)

72, 5

(57)2, 4 (59)2, 5,25 (61)4, 4 (63)

15, 2

48 7. TECHNIQUES OF DIFFERENTIATION

(65)2,43 (67)5

CHAPTER 8

THE MEAN VALUE THEOREM

8.1. Background

Topics:Rolle's theorem, themean value theorem, theintermediate value theorem.

8.1.1. Denition.A real valued functionfdened on an intervalJisincreasingonJiff(a)

f(b) whenevera,b2Jandab. It isstrictly increasingonJiff(a)< f(b) whenevera, b2Janda < b. The functionfisdecreasingonJiff(a)f(b) whenevera,b2Jandab. It isstrictly decreasingonJiff(a)> f(b) whenevera,b2Janda < b. NOTE:In many texts the word \nondecreasing" is used where \increasing" in these notes; and \increasing" is used for \strictly increasing". 49

50 8. THE MEAN VALUE THEOREM

8.2. Exercises

(1) Let M >0 andf(x) =x3for 0xM. Find a value ofcwhich satises the conclusion of themean value theoremfor the functionfover the interval [0;M]. Answer:c=Ma wherea=. (2) Let f(x) =x4+x+ 3 for 0x2. Find a pointcwhose existence is guaranteed by the mean value theorem. Answer:c= 2pwherep=. (3) Let f(x) =pxfor 4x16. Find a pointcwhose existence is guaranteed by themean value theorem. Answer:c=. (4)

Let f(x) =xx+ 1for12

x12 . Find a pointcwhose existence is guaranteed by the mean value theorem. Answer:c=a2 1 wherea=.

8.3. PROBLEMS 51

8.3. Problems

(1) Use Rolle's theoremto derive themean value theorem. (2) Use the mean value theoremto deriveRolle's theorem. (3) Use th emean value theoremto prove that if a functionfhas a positive derivative at every point in an interval, then it is increasing on that interval. (4) Let a2R. Prove that iffandgare dierentiable functions withf0(x)g0(x) for every xin some interval containingaand iff(a) =g(a), thenf(x)g(x) for everyxin the interval such thatxa. (5) Sup posethat fis a dierentiable function such thatf0(x) 2 for allx2[0;4] and that f(1) = 6. (a)

Pro vethat f(4)0.

(b)

Pro vethat f(0)8.

(6) Y ourfriend F redis confused. The function f:x7!x23 takes on the same values atx=1 and atx= 1. So, he concludes, according toRolle's theoremthere should be a pointcin the open interval (1;1) wheref0(c) = 0. But he cannot nd such a point. Help your friend out. (7)

Consider the equation c osx= 2x.

(a) Use the intermediate value theoremto show that the equation has at least one solution. (b) Use the mean value theoremto show that the equation has at most one solution. (8) Let m2R. UseRolle's theoremto show that the functionfdened byf(x) =x33x+m can not have two zeros in the interval [1;1]. (9) Use the mean value theoremto show that if 0< x=3, then12 xsinxx. (10) Use the mean value theoremto show that on the interval [0;=4] the graph of the curve y= tanxlies between the linesy=xandy= 2x. (11)

Let x >0. Use themean value theoremto show thatxx

2+ 1 (12)

Use the mean value theoremto show that

x+ 1< ex<2x+ 1 whenever 0< xln2. (13) Sh owthat the equ ationex+x= 0 has exactly one solution. Locate this solution between consecutive integers. (14) Pro vethat the equation sin x= 12xhas exactly one solution. Explain how theinter- mediate value theoremcan be used to produce an approximation to the solution which is correct to two decimal places. (15) Giv ea careful p roofthat at one time y ourheigh t(in inc hes)w asexactly equal to y our weight (in pounds). Be explicit about any physical assumptions you make.

52 8. THE MEAN VALUE THEOREM

8.4. Answers to Odd-Numbered Exercises

(1)p3 (3) 9

CHAPTER 9

L'H ^OPITAL'S RULE

9.1. Background

Topics:l'H^opital's rule.

53

54 9. L'H

^OPITAL'S RULE

9.2. Exercises

(1) lim x!013 x3+ 2x2sinx4x3=1a wherea=. (2) lim x!0

1sinx1x

 =. (3) lim t!1nt n+1(n+ 1)tn+ 1(t1)2=whenn2. (4) lim x!0tanxxxsinx=. (5) lim x!2x

44x3+ 5x24x+ 4x

44x3+ 6x28x+ 8=a6

wherea=. (6) Let nbe a xed integer. Then the functionfgiven byf(x) =sin(n+12 )xsin 12 xis not dened at pointsx= 2mwheremis an integer. The functionfcan be extended to a function continuous on all ofRby dening f(2m) =for every integerm. (7) lim x!1x

516x54x3+x3=5a

wherea=. (8) Sup posethat ghas derivatives of all orders, thatg(0) =g0(0) =g00(0) = 0, thatg000(0) = 27, and that there is a deleted neighborhoodUof 0 such thatg(n)(x)6= 0 wheneverx2U andn0. Denef(x) =x4g(x)(1cosx) forx6= 0 andf(0) = 0. Thenf0(0) =a4 wherea=. (9) Sup posethat ghas derivatives of all orders, thatg(0) =g0(0) = 0, thatg00(0) = 10, thatg000(0) = 12, and that there is a deleted neighborhood of 0 in whichg(x),g0(x), xg

0(x)g(x)5x2, andg00(x)10 are never zero. Letf(x) =g(x)x

forx6= 0 and f(0) = 0. Thenf00(0) =. (10) Su pposethat ghas derivatives of all orders, thatg(0) =g0(0) =g00(0) =g000(0) = 0, and thatg(4)(0) = 5. Denef(x) =xg(x)2cosx+x22forx6= 0 andf(0) = 0. Thenf0(0) =a2 wherea=. (11) lim x!0x

2+ 2cosx2x

4=1a wherea=. (12) lim x!0cosx+12 x215x4=1a wherea=. (13) lim x!0x

2+ 2ln(cosx)x

4=1a wherea=. (14) lim x!1

152x

4x =. (15) lim x!1 lnxx  1=lnx =. (16) lim x!0 sinxx 31x 2 =1a wherea=.

9.2. EXERCISES 55

(17) lim x!1

1x1lnx(x1)2

=1a wherea=. (18) lim x!1

12(x1)1(x1)2+lnx(x1)3

=1a wherea=. (19) lim x!0+ r1 x 2+1x r1 x 21x ! =. (20) lim x!0xsinxxarctanx=. (21) Let f(x) =x2e1=xfor allx6= 0. Then limx!0f(x) =and lim x!0+f(x) =. (22) lim x!1(xln(5x))3=lnx=. (23) lim x!0 1x 2+2x

4lncosx

=. (24) lim x!0+(sinx)tanx=. (25) lim x!2 (secxtanx) =. (26) lim x!2 (sec2xtan2x) =. (27) lim x!2 (sec3xtan3x) =. (28) lim x!1(lnx)25x =. (29) lim x!1

157x

2x =ea=7wherea=. (30) lim x!1 3xe

2x+ 7x2

1=x =eawherea=. (31) lim x!1 lnxx  1=lnx =eawherea=.

56 9. L'H

^OPITAL'S RULE

9.3. Problems

(1) Is the follo winga correct application of l'H^ opital'srule? Exp lain. lim x!12x33x+ 1x

41= limx!16x234x3= limx!112x12x2= limx!11x

= 1: (2) Let tbe the measure of a central angle\AOBof a circle. The segmentsACandBC are tangent to the circle at pointsAandB, respectively. The triangular region4ABCis divided into the region outside the circle whose area isg(t) and the region inside the circle with areaf(t). Find limt!0f(t)g(t).x ab cd egxh jgxh(3)Let f(x) =x(x1)1forx >0,x6= 1. How shouldf(1) be dened so thatfis continuous on (0;1)? Explain your reasoning carefully. (4) Sho wthat the curv ey=x(lnx)2does not have a vertical asymptote atx= 0. (5) Dene f(x

Derivatives Documents PDF, PPT , Doc

[PDF] about derivatives in finance

1. Business

2. Finance

3. Derivatives

[PDF] about derivatives market

[PDF] about derivatives market in india

[PDF] about derivatives pdf

[PDF] about derivatives with examples

[PDF] afterpay derivatives

[PDF] all about derivatives

[PDF] amide derivatives amino acids

[PDF] amide derivatives as prodrug

[PDF] amide derivatives drugs

12345 Next 200000 acticles

Politique de confidentialité -Privacy policy