McGILL UNIVERSITY FACULTY OF SCIENCE … Optional Reference Book . . . . 5 ... where the tutor works problems at the ... # Lecture Day Begins Ends Room Tutor (tentative) or Lab # Weeks

McGILL UNIVERSITY

FACULTY OF SCIENCE

DEPARTMENT OFMATHEMATICS AND STATISTICS

MATHEMATICS 189–140A

CALCULUS I

Notes Distributed to Students(Fall Term, 1999)

W. G. Brown and D. Sussman

November 21, 1999

Notes Distributed to Students in Mathematics 189-140A (1999) mmmmmi

(Items marked ‡ not distributed in hardcopy)

Contents

1 General Information 11.1 Instructors and Times . . . . . 11.2 Calendar Description . . . . . . 11.3 Self-Supervision . . . . . . . . . 21.4 Final Examination . . . . . . . 21.5 Quizzes . . . . . . . . . . . . . 21.6 Tutorials . . . . . . . . . . . . . 21.7 Homework . . . . . . . . . . . . 31.8 Calculators . . . . . . . . . . . 41.9 Final Grade . . . . . . . . . . . 41.10 Text-Book . . . . . . . . . . . . 41.11 Optional Reference Book . . . . 51.12 Supplemental Examination . . 51.13 Additional Work Option . . . . 51.14 Machine Scoring . . . . . . . . 51.15 Syllabus . . . . . . . . . . . . . 51.16 Tutors’ Office Hours . . . . . . 61.17 Website . . . . . . . . . . . . . 6

2 First Problem Assignment 8

3 Solutions, First Problem Assign-ment 11

4 Second Problem Assignment 21

5 Notes and reminders 235.1 Laboratory-type tutorials and of-

fice hours . . . . . . . . . . . . 235.2 “Infinite limits” . . . . . . . . . 235.3 Composition of functions . . . 245.4 Differentiability implies continu-

ity . . . . . . . . . . . . . . . . 255.5 Critical points and extrema . . 25

6 Third Problem Assignment 27

7 Fourth Problem Assignment 30

8 Fifth Problem Assignment 32

9 Solutions, Third Problem Assign-ment‡ 34

A Notes Specifically for Students inSection 1 1001A.1 Timetable for Section 1 of Math-

ematics 189-140A . . . . . . . . 1001A.2 Inverse functions; exponentials

and logarithms . . . . . . . . . 1003A.2.1 Invertible and inverse func-

tions . . . . . . . . . . . 1003A.2.2 The derivative of the in-

verse function . . . . . . 1004A.2.3 The exponential function

ex . . . . . . . . . . . . 1004A.2.4 The natural logarithm . 1005A.2.5 The exponential function

ax, for any positive base a1005A.2.6 The logarithm loga x, to

any positive base a . . . 1006

B Notes Specifically for Students inSection 2 2001B.1 Timetable for Section 2 of Math-

ematics 189-140A . . . . . . . . 2001

C Assignments, Tests and Examina-tions from Previous Years 3001C.1 1998 Problem Assignments . . 3001C.2 1998 Class Quiz, with Solutions 3001C.3 1996 Final Examination in 189-

122A . . . . . . . . . . . . . . . 3004C.4 1997 Final Examination in 189-

140A . . . . . . . . . . . . . . . 3006C.5 1998 Final Examination in 189-

140A . . . . . . . . . . . . . . . 3009C.6 1998 Supplemental Examination

in 189-140A . . . . . . . . . . . 3011

Notes Distributed to Students in Mathematics 189-140A (1999) 1

1 General Information

Distribution Date: Wednesday, September 1st, 1999(all information is subject to change )

1.1 Instructors and Times

INSTRUCTOR: Prof. W. G. Brown Prof. D. SussmanOFFICE: BURN1224 BURN 1110OFFICE HRS. W 14:30→15:20 h.; by appointment(subject to F 10→11 h.change or by appointment

TELEPHONE: 398–3836 398–3817E-MAIL: BROWN@ SUSSMAN@

MATH.MCGILL.CA MATH.MCGILL.CACLASSROOM: ADAMS AUD ADAMS AUDCLASS HOURS: MWF 8:30–9:30 h. MW 16:30–18:00 h.

1.2 Calendar Description

189-140A CALCULUS I. (3 credits) (3 hours lecture; tutorial sessions) (Prerequisite:High School Calculus.) (Not open to students who have taken 189-120, 189-122, 189-139 or CEGEP course 201–103.) Review of functions and graphs. Limits, continuity,derivative. Differentiation of elementary functions. Antidifferentiation. Applications.

The topics covered in the course may deviate slightly from the preceding list: An-tidifferentiation may possibly be deferred to course 189-141. Also, the topic of InverseFunctions, which was intended when the above description was prepared, but not explic-itly listed, will be covered at least to the extent that it is discussed (briefly, in severallocations) in the text-book.

In addition to 189-140, the Department of Mathematics and Statistics offers two otherbeginning calculus courses:

• 189-150 Calculus A. This course, together with its sequel, 189-151 Calculus B, coversapproximately the material of courses 189-140/139, 189-141 Calculus II, 189-222 Calcu-lus III, in only two semesters. A course in Vector Geometry (e.g. 189-133) is pre- orcorequisite.

• 189-139 Calculus. This course is intended for students who have never had a coursein calculus. In order to apply for a password to register in this course students mustbring their complete transcripts to Professor Brown before the end of the Course ChangePeriod. This course covers approximately the same material as 189-140.

UPDATED TO November 21, 1999


1.3 Self-Supervision

1. The monitoring of your progress before the final examination is largely your ownresponsibility. While the tutors and instructors are available to help you, theycannot do so unless and until you identify the need for help. Assignments andquizzes are designed to assist you in doing this.

2. Be sure to budget enough time to attend lectures and tutorials, for private study,and for the preparation of assignments; do not short-change your mathematicsstudies because of assignments and tests in other courses. This course requiresa consistent investment of time throughout the semester, rather than time forcramming just before the final examination.

3. The prerequisite shown in §1.2 above is “high school calculus”. Implicit in this is thenormal prerequisite for high shool calculus — a course in functions, which shouldinclude the elements of trigonometry. If you have not had such a “precalculus”course, you should probably be in Mathematics 189-112: see your academic advisorimmediately. If you have had a precalculus course, but have forgotten some of thematerial, you must take steps to restore your competence, as such skills will beassumed in the present course.

The first assignment in this course is intended to help you identify gaps in yourbackground, so that you may undertake a review.

1.4 Final Examination

There will be one 3-hour-long final examination, at a time to be announced, during theregular examination period for the fall term (December 7th, 1999 through December21st, 1999). You are advised not to make any travel arrangements that would preventyou from being present on campus at any time during this period.

1.5 Quizzes

One or two short quizzes may be administered during regular class hours. Quizzes, ifany, are intended to help students to pace their work, and will carry no weight in thecomputation of the final grade.

1.6 Tutorials

Tutorials in 189-140A will be offered in two forms: as two-hour-long “lecture-type”sessions, where the tutor works problems at the chalkboard; and as two-hour-long“laboratory-type” sessions, where students work alone or in groups, with a tutor available


to answer questions. One of each type of tutorials will be offered on each of Monday,Tuesday, Wednesday afternoons, and one “lecture-type” will be offered on Thursdayafternoons, beginning the week of September 8th, 1999, and ending the week of Novem-ber 29th, 1999 (inclusive). Attendance at the tutorials is highly recommended, butcompletely voluntary; space permitting, interested students may attend more than onetutorial. The “laboratory-type” tutorials may be preceded and/or followed by tutors’office-hours in the same room: it is expected that, eventually, there will be tutors onduty in one or other capacity in BURN911 on Mondays through Thursdays between thehours of 12:30 and 5:30.

Schedule and Locations of Tutorials# Lecture Day Begins Ends Room Tutor (tentative)

or Lab # Weeks 1-6 Weeks 7-12T01 Lecture M 13:30 15:30 BURN 1B39 Ms. L. Ma Mr. M. Al-RefaiT02 Lecture T 12:00 14:00 BURN 1214 Mr. I. Stewart Mr. I. StewartT03 Lecture W 13:30 15:30 BURN 1B39 Mr. C.E. M’lan Mr. S. SirisupT04 Lecture Th 12:00 14:00 ARTS W-125 Mr. J. Morton Mr. J. MortonT05 Lab M 14:30 16:30 BURN 911 Mr. M. Al-Refai Ms. L. MaT06 Lab T 15:30 17:30 BURN 911 Mr. M. Kharenko Mr. M. KharenkoT07 Lab W 13:30 15:30 BURN 911 Mr. S. SirisupT07 Lab W 14:30 16:30 BURN 911 Mr. C. E. M’lan

1

1.7 Homework

There will be 5 or 6 homework assignments. The numerical grade recorded for theassignments is relatively insignificant; but students should be sure that they understandthe problems and their solutions, and to successfully complete other exercises for masteryof the subject matter. Students are advised that tutors may be instructed to grade onlycertain problems on each assignment.

With the exception of Assignment #1, which will not be graded2, homework is to behanded in AT THE LECTURES. While students are not discouraged from discussingassignment problems with their colleagues, the written solutions that are handed inshould be each student’s own work.3

1Subject to change; corrected as of November 21, 19992This is a diagnostic assignment; solutions will be made available.3From the Handbook on Student Rights and Responsibilities:

“No student shall, with intent to deceive, represent the work of another person as his orher own in any...assignment submitted in a course or program of study or represent ashis or her own an entire essay or work of another, whether the material so represented



In preparing your assignments you should:

1. Use paper approximately 812

inches × 11 inches.

2. STAPLE the pages securely.

3. Include a COVER PAGE that contains

• your NAME

• your STUDENT NUMBER

• the COURSE NUMBER, 189-140A, and

• the ASSIGNMENT NUMBER.

4. PRINT your name and student number on every page.

You can minimize the possibility that your assignment is lost or fragmented.

1.8 Calculators

The use of calculators, computers, notes, or other aids will be permitted neither at theexamination, nor at quizzes.

1.9 Final Grade

The final grade will be a letter grade, computed from the maximum of

Final Examination Mark/100

and

(.9×(Final Examination Mark/100)) + (Homework Mark/10)

1.10 Text-Book

[3] C. H. Edwards, Jr., and D. E. Penney, CALCULUS WITH ANALYTIC GEOM-

ETRY, SINGLE VARIABLE: early transcendentals, Fifth Edition, PrenticeHall, Englewood Cliffs, NJ (1997). ISBN 0-13-793092-5.4

constitutes a part or the entirety of the work submitted.”

4Some students may prefer to purchase the full version of this text-book: [1] C. H. Edwards, Jr., andD. E. Penney, CALCULUS WITH ANALYTIC GEOMETRY: early transcendentals, Fifth

Edition, Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-793076-3. This book also covers thematerial for Calculus III; whether this is the text-book used in such courses at McGill depends on theircurrent instructors’ decisions.



1.11 Optional Reference Book

[4] C. H. Edwards, Jr., and D. E. Penney, STUDENT SOLUTION MANUAL FOR

SINGLE VARIABLE CALCULUS WITH ANALYTIC GEOMETRY, Fifth

Edition, Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-095147-1.5

1.12 Supplemental Examination

There will be a supplemental examination in this course. (Often the examination in189–140B serves a supplemental examination for students who have received permissionto write it.)

1.13 Additional Work Option

“Will students with marks of D, F, or J have the option of doing additional work toupgrade their mark?” No.

1.14 Machine Scoring

“Will the final examination be machine scored?” It is expected that parts of the FinalExamination will be Multiple Choice, and may be machine scored.

1.15 Syllabus

In the following list section numbers refer to the text-book.

Chapter 1: Functions and Graphs. This is review material, and will not be dis-cussed in the lectures.

Chapter 2: Prelude to Calculus. §§2.1 – 2.4.

Chapter 3: The Derivative. §§3.1 – 3.9. (It is suggested that students read §3.10, butit is not examination material, and will probably not be discussed in the lectures.)

Chapter 7: Exponential and Logarithmic Functions. §§7.1 – 7.4.

Chapter 8: Further Calculus of Transcendental Functions. §§8.1 – 8.4. (Omit§5, which is material to be studied in 189-141.)

5For the “full” version of the textbook the solution manual is [2] C. H. Edwards, Jr., and D. E. Pen-ney, STUDENT SOLUTION MANUAL FOR CALCULUS WITH ANALYTIC GEOMETRY,

Fifth Edition, Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-079875-4.


Chapter 4: Additional Applications of the Derivative. §§4.1 – 4.7, (omitting pp.273 – 275).

1.16 Tutors’ Office Hours

Tutors will hold their office hours in BURN911. Following is the tentative timetable forthe presence of tutors from 189-140A in BURN911, as of November 21, 1999; the two-hour periods shown in boxes on Mondays, Tuesdays, and Wednesays are laboratory-typetutorials.

6 Sept. to 15 Oct. 18 Oct. to 3 Dec.Starts M T W Th F M T W Th F12:30 Al-Refai Sirisup M’lan Stewart Morton Al-Refai Sirisup M’lan Stewart Morton

13:30 Al-Refai SirisupT07

SirisupKharenko

Stewart Morton Ma SirisupM’lan

KharenkoStewart Morton

14:30T05

Al-RefaiSirisup

T07Sirisup

KharenkoStewart

MortonMa

T05Ma

SirisupT07M’lan

KharenkoStewart

MortonMa

15:30T05

Al-RefaiT06

KharenkoM’lan

KharenkoMa

T05Ma

T06Kharenko

T07M’lan

KharenkoMa

16:30 Al-RefaiT06

KharenkoM’lan Ma Al-Refai

T06Kharenko

M’lan

1.17 Website

These notes, and other materials distributed to students in this course, will be accessibleat the following URL:

http://www.math.mcgill.ca/brown/math140a.html

The notes for the current year will be “pdf” (.pdf) form, which can be read using theAdobe Acrobat reader, which many users have on their computers. Notes for previousyears, where available (including old examinations) will be in PostScript (.ps) form,which requires a less accessible type or reader (Ghostview, or similar software). Thisweb page is being offered as a convenience — the materials are not essential for thecourse. It is expected that most computers in campus labs should have the necessarysoftware to read the posted materials.

Where revisions are made to distributed printed materials — for example these in-formation sheets — we expect that the last version will be posted on the web page.



(This page intentionally left blank.)



2 First Problem Assignment

Distribution Date: Wednesday, September 1st, 1999

Instructions

• This assignment is intended to help you determine gaps in your background. Itwill not be graded, but the solutions will be made available on the Web. If youhave difficulty with the problems or the solutions which will be posted, you shoulddiscuss these with one of the tutors.

• Do not use a calculator when solving these problems, even for simple arithmetic.(You may, however, wish to use a calculator afterwards to verify whether youranswers are “reasonable”.)

• Do not attempt to approximate square roots, or π. Where possible, however,formulæ should always be simplified .

• Unless specifically stated in degrees, all angles should be assumed to be expressedin radians . (Remember that the straight angle, 180◦ is equal to π radians, so anangle of 1◦ is equal to π

180radians, and d◦ is equal to πd

180radians.)

1. (a) Give a right-angled triangle in which one angle is α = π3, and use the triangle

to determine the values of sinα, cosα, tanα.

(b) Give a right-angled triangle in which one angle is β = π4, and use the triangle

to determine the values of sin β, cos β, cot β, sec β.

It is not sufficient to state the values: you should explain how you determine themin terms of the lengths of the sides of your triangle.

2. State, without proof,

(a) a formula which expresses sin(x + y) in terms of the sines and cosines of xand y.

(b) a relationship between sinx and sin(−x), and another between cosx andcos(−x).

(c) By applying the results of parts 2b, 2a above, determine a formula that ex-presses sin(x− y) in terms of the sines and cosines of x and y.

The formulæ you state should be valid for all values of x and y.

3. (a) By substituting x = α = π3

and y = β = π4

in the formula in 2c, determinethe value of sin 15◦.


(b) Using the result of 3a, determine the value of cos π12

.

4. State, without proof, formulæ which express cos(x+ y) and cos(x− y) in terms ofthe sines and cosines of x and y.

5. (a) Specialize the formula in Problem 4 (by making a suitable choice for y interms of x) to express cos 2x in terms of sinx and cos x.

(b) Use the identity sin2 x + cos2 x = 1 to express cos 2x in terms of cos x alone;and in terms of sinx alone.

(c) Apply your formulæ in 5b to determine the values of cos π6

and cos π2

from thesines of π

12and π

4.

6. Find an equation for each of the following lines in the plane:

(a) The line through the points (−1, 4) and (−2, 1).

(b) The line through the point (4,−1) which is perpendicular to the line

2x+ y = −7 . (1)

7. (a) It is claimed that the equation (2x + y + 7)2 = 0 represents the same pointsas the equation 2x+ y + 7 = 0. Determine whether the claim is correct.

(b) Determine what is represented by the equation (2x+ y)2 = (−7)2, i.e. by theequation obtained by squaring both sides of equation (1).

8. Determine the centre and radius of the circle x2 + y2 + 4x− 6y + 3 = 0.

9. Simplify completely the following formula, leaving your answer free of negative

exponents or radicals:6√b ·

(3√a3b2

√a6b3

)−1

.

10. Solve the equation1

x+

1

x− 2=

4

3. (2)

11. Determine the “natural” (i.e. largest possible) domain of definition of each of thefollowing functions. Explain your results.

(a) f(x) =

√x

2− sinx.

(b)1

log3(x+ 1).


12. Factorize the following polynomial completely:

(x2 + 4xy + 4y2)− (3x+ 6y)

13. Solve the system of equations:

3x− 7y = 1

4x+ 3y = 5

14. Find all solutions to the equation

log2(3x+ 2) + log2(x+ 1) = 2 (3)

15. Determine all real numbers x such that

(a) 5 tan2 x− sec2 x = 11

(b) cos x+ secx = 52

16. Show that1

3≤ x2 − x+ 1

x2 + x+ 1≤ 3 (4)

for any real number x.


3 Solutions, First Problem Assignment

Distribution Date: Monday, September 13th, 1999(Problems were assigned on Wednesday, September 1st, 1999.)

Caveat lector! 6As in any duplicated materials, there could be misprints or errors.

Original Instructions

• “This assignment is intended to help you determine gaps in your background. Itwill not be graded, but the solutions will be made available on the Web. If youhave difficulty with the problems or the solutions which will be posted, you shoulddiscuss these with one of the tutors.

• “Do not use a calculator when solving these problems, even for simple arithmetic.(You may, however, wish to use a calculator afterwards to verify whether youranswers are ‘reasonable’.)

• “Do not attempt to approximate square roots, or π. Where possible, however,formulæ should always be simplified .

• “Unless specifically stated in degrees, all angles should be assumed to be expressedin radians . (Remember that the straight angle, 180◦ is equal to π radians, so anangle of 1◦ is equal to π

180radians, and d◦ is equal to πd

180radians.)”

1. (a) Give a right-angled triangle in which one angle is α = π3, and use the triangle

to determine the values of sinα, cosα, tanα.

(b) Give a right-angled triangle in which one angle is β = π4, and use the triangle

to determine the values of sin β, cos β, cot β, sec β.

It is not sufficient to state the values: you should explain how you determine themin terms of the lengths of the sides of your triangle.

Solution:

(a) We take the right half of an equilateral triangle, each of whose sides has length2. In ∆DBC \C = π

3= α. DB bisects \ADC and meets AC in its midpoint,

B. Then

sinα =|BD||CD|

=

√3

2

cosα =|BC||CD|

=1

2

tanα =|BD||BC|

=

√3

1=√

3

6Let the reader beware!


��

TTTTTTTTTTTTTTTTT

A B C

D

1 1

2 2√

3

60◦90◦

@@@@@@@@@@

E G

F

1

1√

2

45◦90◦

(b) Here we take an isosceles ∆EFG, in which EF = EG = 1, and \F = 45◦ = β.

sinβ =|EG||FG|

=1√2

cos β =|EF ||FG|

=1√2

cot β =|EF ||EG|

= 1

sec β =|FG||EF |

=

√2

1=√

2

(Note: Some students may have been taught never to leave a surd in thedenominator, as in the fraction 1√

2above. This convention derives from the

difficulties, in the days before calculators, of working with numbers like√

2.While it still is useful, for hand calculations, to confine surds to the numerator,you may work with fractions like 1√

2if you are happy with them.)

2. State, without proof,

(a) a formula which expresses sin(x + y) in terms of the sines and cosines of xand y.

(b) a relationship between sinx and sin(−x), and another between cosx andcos(−x).


(c) By applying the results of parts 2b, 2a above, determine a formula that ex-presses sin(x− y) in terms of the sines and cosines of x and y.

The formulæ you state should be valid for all values of x and y.

Solution:

(a)sin(x+ y) = sinx cos y + cosx sin y (5)

[1, A-15 (6)]

(b) sin(−x) = − sinx; cos(−x) = − cos x [1, A-15 (4)]

(c) Replacing y by −y throughout (5), one obtains

sin(x− y) = sinx cos(−y) + cos x sin(−y)

= sinx cos y + cos x(− sin y)

= sinx cos y − cosx sin y (6)

3. (a) By substituting x = α = π3

and y = β = π4

in the formula in 2c, determinethe value of sin 15◦.

(b) Using the result of 3a, determine the value of cos π12

.

Solution:

(a)

sinπ

12= sin

(π3− π

4

)= sin

π

3cos

π

4− cos

π

3sin

π

4

=

(√3

2

)·(

1√2

)−(

1

2

)·(

1√2

)=

√3− 1

2√

2=

√6−√

2

4.

(b) cos2 π12

= 1− (√

3−1)2

8= 2+

√3

4; hence cos π

12is one of the square roots of 2+

√3

4.

Since the angle is in the first quadrant, the sign is positive. It follows that

cos π12

=√

2+√

34

. This can be seen — you were not expected to see this —

that the square root is√

3+12√

2.


4. State, without proof, formulæ which express cos(x+ y) and cos(x− y) in terms ofthe sines and cosines of x and y.

Solution:

cos(x+ y) = cos x cos y − sinx sin y (7)

cos(x− y) = cos x cos y + sinx sin y (8)

[1, A-15 (7)]

5. (a) Specialize the formula in Problem 4 (by making a suitable choice for y interms of x) to express cos 2x in terms of sinx and cosx.

(b) Use the identity sin2 x + cos2 x = 1 to express cos 2x in terms of cos x alone;and in terms of sinx alone.

(c) Apply your formulæ in 5b to determine the values of cos π6

and cos π2

from thesines of π

12and π

4.

Solution:

(a) Taking y := x in (7) yields

cos 2x = cos2 x− sin2 x (9)

(b)

cos 2x = cos2 x− (1− cos2 x) = 2 cos2 x− 1 (10)

= (1− sin2 x)− sin2 x = 1− 2 sin2 x (11)

(c)

cosπ

6= cos 2

( π12

)= 1− 2 sin2

( π12

)= 1− 2 · (

√3− 1)2

8= 1− 2 · 1 + 3− 2

√3

8=

√3

2

cosπ

2= 1− 2 sin2 π

4= 1− 2

(1√2

)2

= 0

6. Find an equation for each of the following lines in the plane:

(a) The line through the points (−1, 4) and (−2, 1).

(b) The line through the point (4,−1) which is perpendicular to the line

2x+ y = −7 . (12)


Solution: (This problem is from the 1998 Final examination in 189-112A.)

(a) The slope of the line will be 1−4(−2)−(−1)

= 3, so an equation is y−4 = 3(x−(−1))or y = 3x+ 7.

(b) The slope of the line we seek will be the negative reciprocal of the line y =−2x − 7, which has slope −2; thus the line will have slope 1

2. One equation

will be y − (−1) = 12(x− 4), or x− 2y = 6.

7. (a) It is claimed that the equation (2x + y + 7)2 = 0 represents the same pointsas the equation 2x+ y + 7 = 0. Determine whether the claim is correct.

(b) Determine what is represented by the equation (2x+ y)2 = (−7)2, i.e. by theequation obtained by squaring both sides of equation (12).

Solution:

(a) The equation of a line is a constraint satisfied by the coordinates of its points,and satisfied by the coordinates of no other points. As (2x + y + 7)2 maybe zero if and only if 2x + y + 7 = 0, this is another equation for the sameline. (Some authors would, however, say that this is “the equation of twocoincident lines”.)

(b) (The symbol ⇔ means that the statements which it connects are logicallyequivalent.)

(2x+ y)2 = (−7)2 ⇔ (2x+ y)2 − (−7)2 = 0

⇔ (2x+ y − 7)(2x− y + 7) = 0

⇔ 2x+ y − 7 = 0 or 2x+ y + 7 = 0

Thus the given equation is satisfied by the coordinates of the points on eitherof the parallel lines 2x+ y−7 = 0 and 2x+ y+ 7 = 0; the equation representsthe union of the sets of points on the two lines.

8. Determine the centre and radius of the circle x2 + y2 + 4x− 6y + 3 = 0.

Solution: (This problem is from the 1998 Final examination in 189-112A.) Wecomplete the squares separately.

x2 + y2 + 4x− 6y + 3 = 0

⇔ (x2 + 4x) + (y2 − 6y) + 3 = 0

⇔

(x2 + 4x+

(4

2

)2)−(

4

2

)2

+

(y2 − 6y +

(−6

2

)2)−(−6

2

)2

+ 3 = 0

⇔ (x+ 2)2 + (y − 3)2 = 4 + 9− 3 = 10

⇔ (x− (−2))2 + (y − 3)2 = 4 + 9− 3 = (√

10)2


Hence the centre of the circle is (−2, 3), and its radius is√

10.

9. Simplify completely the following formula, leaving your answer free of negative

exponents or radicals:6√b ·

(3√a3b2

√a6b3

)−1

.


6√b ·

(3√a3b2

√a6b3

)−1

=6√b ·√a6b3

3√a3b2

=b

16a3b

32

a1b23

=a3b

16

+ 32

ab23

=a3b

53

ab23

= a3−1b53− 2

3 = a2b

10. Solve the equation1

x+

1

x− 2=

4

3. (13)

Solution: (This problem is from the 1998 Final examination in 189-112A.) If wemultiply both sides of the equation7 by 3x(x−2), we obtain the polynomial equation3(x−2)+3x = 4x(x−2), which reduces to 4x2−14x+6 = 0, then to 2x2−7x+3 = 0,which is equivalent to (2x − 1)(x − 3) = 0. The product on the left can be zeroonly with at least one of the factors is zero, i.e. when either x = 1

2or x = 3. These

are the only solutions to (13).

11. Determine the “natural” (i.e. largest possible) domain of definition of each of thefollowing functions. Explain your results.

(a) f(x) =

√x

2− sinx.

(b)1

log3(x+ 1).

7We should verify, when we have found the alleged solutions to the equation, that this operation ofmultiplying both sides by the same quantity did not entail the introduction of any values which werenot solutions. For example, if we had multiplied by 3x2(x−2), the subsequent computations would haveproduced a 3rd degree equation whose solutions would be 0, 1

2 and 3 — but x = 0 is evidently not asolution, since the left side is not even defined there. Similarly, if we had multiplied by 3x(x−2)(x+11),we would have obtained x = 11 as one of the solutions of the polynomial equation, but it is certainlynot a solution of the original equation. While the procedure could be made more rigorous, the safestpolicy is to verify that all claimed solutions are indeed solutions of the original equation — particularlywhen, in the course of your solution, you have multiplied both sides of an equation by an expressionthat could be 0.



(a) The denominator is defined for all x. The numerator is defined for all non-negative x. The ratio function is defined only for non-zero denominators,so we must ensure that 2 − sinx is not zero; but this can never happen, as| sinx| ≤ 1. Thus the largest possible domain for this function is the set of allnon-negative real numbers.

(b) Logarithms are defined only for positive numbers, so we must require that x+1 > 0, i.e. that x > −1. But the logarithm here appears in the denominator,which can assume any value except 0. log3(x+1) = 0 precisely when x+1 = 1,i.e. x = 0. Thus the largest possible domain for this function is the union ofthe set −1 < x < 0 with the set x > 0 of positive real numbers.

12. Factorize the following polynomial completely:

(x2 + 4xy + 4y2)− (3x+ 6y)

Solution: (This problem is from the 1998 Final examination in 189-112A.) Studentswere expected to observe that the first summand is a perfect square.

(x2 + 4xy + 4y2)− (3x+ 6y) = (x+ 2y)2 − 3(x+ 2y)

= (x+ 2y)(x+ 2y − 3)

is the desired factorization.

13. Solve the system of equations:

3x− 7y = 1

4x+ 3y = 5

Solution: (This problem is from the 1998 Final examination in 189-112A.) Studentswill study the systematic solution of systems of linear equations in courses onlinear algebra and matrices. This problem is simply to detect whether you areable to solve a routine small system, even if your methods are not systematic.Subtracting 3 times the second equation from 4 times the first equation yields−37y = −11, implying that, if there is a solution, y = 11

37; this, substituted into

the first equation, yields x = 3837

. (This solution can be verified by substitutingin the second equation. While that is hardly necessary in the present problem,substitution should be undertaken in large linear systems, unless care has beentaken to avoid the possibility that some constraint has been lost. This is beyondthe present course, but will be studied in courses in linear algebra. What is atissue is the possibility that a system of equations may have no solutions at all, ormay have infinitely many solutions.)


14. Find all solutions to the equation

log2(3x+ 2) + log2(x+ 1) = 2 (14)


log2(3x+ 2) + log2(x+ 1) = 2

⇒ log2((3x+ 2)(x+ 1)) = 2 = log2 22

⇔ 3x2 + 5x+ 2 = 4

⇔ 3x2 + 5x− 2 = 0

⇔ (3x− 1)(x+ 2) = 0

⇔ x =1

3or x = −2

But the value x = −2 is extraneous, as the logarithm is not defined at −2 + 1 =−1 < 0, or at 3 · (−2) + 2 = −4 < 0. We conclude that the only possiblesolution to (14) is x = 1

3, and verify that it does indeed satisfy the equation, as

log2 3 + log243

= log2 4 = log2 22 = 2.

15. Determine all real numbers x such that

(a) 5 tan2 x− sec2 x = 11

(b) cos x+ secx = 52

Solution:

(a)

5 tan2 x− sec2 x = 11 ⇔ 5 tan2 x− (tan2 x+ 1) = 11

⇔ 4 tan2 x = 12

⇔ tanx = ±√

3

The smallest solutions in absolute value are x = ±π3. From the period-

icity of the tangent function we conclude that the set of all solutions is{(n± 1

3)π : n ∈ Z

}where Z is the set of all integers.

(b)

cos x+ secx =5

2⇔ cosx+

1

cosx=

5

2⇔ 2 cos2 x− 5 cos x+ 2 = 0

⇔ (2 cos x− 1)(cos x− 2) = 0


⇔ cosx =1

2or cos x = 2

⇒ cosx =1

2as | cosx| ≤ 1

⇔ x =

(2n± 1

3

)π

16. Show that1

3≤ x2 − x+ 1

x2 + x+ 1≤ 3 (15)

for any real number x.

Solution: Let us first prove the inequalities

x2 + x+ 1

3≤ x2 − x+ 1 ≤ 3(x2 + x+ 1)

i.e. equivalently, the two inequalities

x2 + x+ 1 ≤ 3(x2 − x+ 1)

x2 − x+ 1 ≤ 3(x2 + x+ 1)

The first of these is equivalent to 2(x− 1)2 ≥ 0, which is true because the left sideis a square, hence non-negative; the second is equivalent to 2(x+ 1)2 ≥ 0, which isnon-negative for the same reason. In order to pass from these two inequalities to(15) we need to divide by x2 + x + 1. This is a quadratic polynomial, having noreal roots; its sign is that of the leading coefficient x2, i.e. it is positive; this canbe shown by observing that x2 + x + 1 = (x + 1

2)2 + 3

4≥ 0 + 3

4> 0. So dividing

the members of an inequality by this positive quantity preserves the inequalities,thereby yielding the desired pair of inequalities.

A more elegant solution can be found if we define z = x2−x+1x2+x+1

, and transform thisequation to yield a quadratic equation for x in terms of z:

x2 − x+ 1− z(x2 + x+ 1) = 0

⇔ (1− z)x2 − (1 + z)x+ (1− z) = 0

For this equation to admit a solution for every x, the discriminant must be non-negative, i.e.

(1 + z)2 − 4(1− z)2 ≥ 0

This is equivalent to

[(1 + z)− 2(1− z)][(1 + z) + 2(1− z)] ≥ 0


which is equivalent to −(z − 1

3

)(z − 3) ≥ 0. For the product

(z − 1

3

)(z − 3) to

be non-positive, z must lie between the roots, i.e. z must be such as to make oneof the factors negative and the other positive. This can be achieved only with13≤ z ≤ 3, as required.


4 Second Problem Assignment

Distribution Date: Monday, September 13th, 1999Solutions are to be submitted on Monday, September 27th, 1999

1. Determine if limx→1+

f(x), limx→1−

f(x), and/or limx→1

f(x) exist for each of the following

functions, and evaluate those that do exist.

(a) f(x) =√x2 + x− 2

(b) f(x) =

{x2 − 3 for x ≤ 12− x for x > 1

(c) f(x) =

{ √2− x for x < −3

x3 − x for 2 > x ≥ −3

2. In each of the following cases, evaluate the limit, or show that it does not exist.

(a) limx→3

(1

x2 − x− 6− 1

x− 3

)(b) lim

x→3

(5

x2 − x− 6− 1

x− 3

)3. 8In each of the following cases, evaluate the limit, or show that it does not exist.

(a) limx→∞

3x+ 2√x

1− x

(b) limx→−∞

|2x− 3|x+ 2

4. Evaluate limx→0

x+1

x2

2

x2− 3x2

5. 9Find a value for a so that

limx→∞

(3x+ 1− ax

2 + 1

x+ 1

)exists as a finite limit, and evaluate that limit.

8Added 21.09.99: This problem should be omitted. It is based upon [1, §§4.5,4.7], which materialmay not have been discussed in the lectures before the due date of the assignment.

9See footnote 8.


6. Sketch the graph of f(x) =

2 when x ≤ −2

x+ 4 when −2 < x < 420− x2 when 4 ≤ x

.

Find the values of x for which f(x) is not continuous.

7. Find values of a and b which will make the following function continuous:

f(x) =

x2 + 2a when 0 < x < 1bx+ a when 1 ≤ x < 22b

x− 3 when 2 ≤ x

.

8. Let f(x) =x2 − 2x− 3

4− x2. Determine where f(x) is not continuous, and where

f(x) = 0. Then solve the inequalityx2 − 2x− 3

4− x2≥ 0. [Hint: One method to solve

this problem uses the Intermediate Value Theorem.]

9. Solve the inequality3

x+ 1≤ 1− 4x

x− 1. [Hint: Bring everything to one side of the

equation and simplify first.]


5 Notes and reminders

Distribution Date: Monday, October 4th, 1999

5.1 Laboratory-type tutorials and office hours

Students with questions are advised that there are tutors on duty in BURN 911 duringthe hours of 12:30 to 17:30 MTWF and 12:30 to 15:30 Thursdays until 15th October;and 12:30 to 17:30 MTW, 12:30 to 15:30 Thursdays, 12:30 to 16:30 Friday from 18thOctober to 3rd December. On MTW there will be a 2-hour “laboratory-type tutorial”during these hours, where the tutors will assist students who wish to work examples;during the other hours the tutors will be holding “office hours” in the room, and arethere to assist you.

5.2 “Infinite limits”

The rigorous definition of limx→a

f(x) = L [1, §2.1] — a definition we do not expect students

in this course to be completely comfortable with — applies only for a real number L.That is, we do not permit L to be infinite in this original definition. Where there existsa real number L such that lim

x→af(x) = L, we say that lim

x→af(x) exists ; if there is no such

L, we say that limx→a

f(x) does not exist .

Statements like

limx→a

f(x) = ∞ (16)

or limx→a

f(x) = −∞ (17)

[1, pp. 83–84] are instances were we take an established definition and extend it to abroader situation; we give definitions below. Students should recognize that, even whenone or other of these statements is true, we still say that the limit does not exist . Theprecise definitions in these cases are easier to understand than the definition of the finitelimit. We say that (16) holds for a function f(x) if, no matter what positive number Nwe take, it is possible to find an interval (a − h, a + h) which surrounds the point a onthe x-axis so that

1. f is defined at all points of (a− h, a+ h), except possibly at a; and

2. the function values for points distinct from a in the interval are all greater thanN ;


similarly, we say that (17) holds for a function f(x) if, no matter what negative numberN we take, it is possible to find an interval (a − h, a + h) which surrounds the point aon the x-axis so that

1. f is defined at all points of (a− h, a+ h), except possibly at a; and

3. the function values for points distinct from a in the interval are all less than N .

Thus, limx→0

1

|x|= ∞; but we cannot make either statement about the function 1

xas

x → 0, since this fraction becomes large positively only on the positive side of 0, andlarge negatively only on the negative side of 0: for no N is there an interval (−h, h)surrounding 0 with the desired property. We could, however, speak of one-sided infinitelimits, and say that

limx→0−

1

x= −∞ (18)

or limx→0+

1

x= ∞ (19)

where one-sided infinite limits are defined in the obvious ways.

5.3 Composition of functions

Suppose that we follow the action of a function g by a function f . We could think ofthe original “independent” variable as being named x, of this being first mapped on tou = g(x), and of u then being mapped on to y = f(u). Thus y = f(g(x)); we oftendefine the composition f ◦ g of the functions in this way: (f ◦ g)(x) = f(g(x)). We mayrepresent this situation with an arrow notation as

xg7→ u

f7→ y

in which the function f appears to the right of g, even though it is g that is appliedfirst.10

10It is perhaps this confusion that caused the authors of the textbook to interchange the names of thefunctions in [1, (10) p. 128]. The correct form of that statement is

dy

dx= lim

∆x→0

∆y∆x

= lim∆x→0

f(g(x+ ∆x))− f(g(x))∆x

.


5.4 Differentiability implies continuity

Suppose that f(x) is differentiable at x = x0. Then the following limit exists:

lim∆x→0

f(x0 + ∆x)− f(x0)

∆x. (20)

It is the value of this limit which is defined to be f ′(x0). For the limit of this quotientto exist, the quotient must be defined “near” x = x0. Hence, f(x0) must be defined —i.e. x0 must be in the domain of f ; and f(x0 + ∆x) must be defined for ∆x “near” 0 —or, equivalently, f(x) must be defined for x near x0. But then

f(x0 + ∆x) = (f(x0 + ∆x)− f(x0)) + f(x0)

=

(f(x0 + ∆x)− f(x0)

∆x

)∆x+ f(x0)

In this last sum lim∆x→0

∆x = 0, and the quotient is known to have a limit, so we may apply

the product law to the product; the second summand, f(x0), is constant. It follows bythe Product and Sum Laws for limits that

lim∆x→0

f(x0 + ∆x) = f ′(x0) · 0 + f(x0) = 0 + f(x0) = f(x0) .

We have proved that f is continuous at x0.Note that the existence of limit (20) also implies that f(x) is defined “near” x0.11

5.5 Critical points and extrema

The term critical point is used in somewhat different ways by authors of different calculusbooks. As the definition in our textbook [1, p. 144] is ambiguous, we will state thedefinition we propose to use in this course.

Definition 5.1 Let f(x) be a function which is continuous over an interval; here wepermit all types of intervals — finite or infinite, and with or without end-points. A pointc is said to be a critical point if c is not an end-point of the interval, and if either

1. f is not differentiable at c; or

2. f is differentiable at c, and f ′(c) = 0.

11Thus the definition in the textbook [1, Theorem 2, p. 143] that begins “If f is differentiable at cand is defined on an open interval containing c...” is redundant. It would have been sufficient to say “Iff is differentiable at c...”.


We call these points critical because we can determine the maximum and minimumvalues of a function f that is continuous over a closed interval just by studying itsvalues at these points and at the end-points of the interval. But N.B., this statementapplies only for finite, closed intervals: not for intervals that may be missing one orother end-point, nor for infinite intervals; note also that the function must be known tobe continuous throughout the interval.

An algorithm for finding the extrema of a function f continuous12 over a closedinterval [a, b], is given in your textbook [1, p. 144]. The extrema can occur only atcritical points or at the end-points a, b. Thus it suffices to determine the values of f atall of these candidate points, and to compare them: the largest value may be attainedat more than one point — these are all maximum points , and the value is the maximum(value) of the function; an analogous statement may be made about minimum pointsand the minimum (value). (Where the function is being considered over a finite intervalthat may not be closed, or over an interval which is infinite, the behavior of the functionas the end-points or ±∞ are approached must also be considered: these problems aremore difficult than the problem for a finite closed interval, and are not even addressedin some calculus books.)

In all of this discussion we are considering only functions continuous throughout aninterval.

12Recall that f is continuous at the end-point a of interval [a, b] if (1) f is defined at a, (2) limx→a+

f(x)

exists, and (3) limx→a+

f(x) = f(a). For continuity at b replace conditions (2) and (3) by (2′) limx→b−

f(x)

exists, and (3′) limx→b−

f(x) = f(b).


6 Third Problem Assignment

Distribution Date: Wednesday, October 13th, 1999Solutions are to be submitted on Monday, October 25th, 1999.

1. (a) Apply the definition of the derivative as a limit to determine the derivative of

f(x) =

√7x− 1

3for any x in the domain.

(b) Determine the value of f ′(7) from the result of 1a.

2. You will find below the definitions for a number of functions. If the domain ofdefinition is not stated, then you are to assume the domain to be as large aspossible.

• Determine the derivative of each function wherever it is defined in the domain.

• List any points in the domain where the function does not possess a derivative,and explain precisely why there fails to be a derivative.

• If possible, determine, for those functions where requested, the equation ofthe tangent or the normal to the graph of the function at the given point.

(a) f1(x) =x

5− 8x; tangent at (1, f1(1)).

(b) f2(x) = (5 + 6x2 + x4)− (1 + 2x+ x2)2 + 4x(1 + x); normal at (5, f2(5)).

(c) f3(x) = (1− x)2.

(d) f4(x) = −√

10x− x2; tangent at (1,−3).

(e) f5(x) = |3− 2x|; tangent at x =3

2.

(f) f6(x) =

x3

|x|when x 6= 0

0 when x = 0; tangent and normal at (0, 0).

(g) f7(x) =

{x2 − 6x when x < 1−1− 4x when x ≥ 1

; tangent at (1,−5).

(h) f8(x) =

x2 − 6x when x < 1

0 when x = 1−1− 4x when x > 1

; normal at (1,−5).

3. (The functions referred to below were defined in Problem 2 above.)


(a) Determine the value of the function f7 ◦f7 at any point x in its domain. [Thisproblem is technically difficult. Try to understand what makes the problemdifficult. Your solution to this problem will not be graded.]

(b) Determine the value of the functions f1 ◦ f4 and f4 ◦ f1 at all points in theirdomains. Find the derivatives of these functions in two ways:

i. by differentiation of the formulæ you have determined

ii. by applying the chain rule

and show that the results are the same.

(c) Suppose that y =1 + s

1− s, s = t − 1

t, t =

√x. Determine the value of

dy

dxat

x = 2.

4. For each of the following functions, and for the given closed interval,

• find all critical points;

• find all local maxima and local minima;13

• find all global maxima and global minima, or explain why none of either exists.

All claims should be supported by careful reasoning; show all your work.

(a) g1(x) = (x− 3)5; interval [2, 4]

(b) g2(x) = 6− 36x+ 15x2 − 2x3; inverval

[5

2, 5

]

(c) g3(x) =

0 when x = 01

xwhen −1 ≤ x < 0

1

xwhen 0 < x ≤ 1

; interval [−1, 1]

(d) g4(x) =

0 when x = 01

|x|when −1 ≤ x < 0

1

|x|when 0 < x ≤ 1

; interval [−1, 1]

(e) g5(x) =√|x|; interval [−4, 9]

13NOT TO BE GRADED: It was announced in the lectures that this topic will not be discussed until[1, Chapter 4]; accordingly this part of the question is to be omitted.



5. [5, Examples XLVI.17] (Cambridge Math. Tripos 1930) The graph of the function

h(x) =ax+ b

(x− 1)(x− 4)has (2,−1) as a critical point. Determine a and b, and show

that the critical point is a local maximum. (Note: It is intended that this problembe solved without using concepts from Chapter 4 of your textbook (involving higherderivatives.))

6. Determine values of the constants a, b, c which will cause the curve y = ax3 +bx2 +cx + d to pass through the points (2, 6) and (−1, 6) and to be tangent to the liney = 3x+ 1 at the point (1, 4).

7. Find all lines with slope −3 which are normal to the curve 64y = x3.

8. Find the volume of the uncovered box of greatest volume that can be made bycutting equal squares out of the corners of a piece of sheet metal which is 21 cm.× 5 cm., and turning up the sides.

9. Show that, among all right-angled triangles whose hypotenuse is 10 units long, thetriangle whose area is maximum is isosceles.


7 Fourth Problem Assignment

Distribution Date: Wednesday, October 27th, 1999Solutions are to be submitted on Monday, November 8th, 1999

1. Finddy

dxif

(a) y = ln(1 + x2

)(b) y = e−x

2

(c) y = x√x

2. Find an equation for the tangent line at x = 0 to the graph of y = ex−e−x .

3. Find the greatest value of f(x) =1

x2lnx .

4. If f(x) = e−x sinx , find the values of x where

(a) f ′(x) = 0 ;

(b) f ′′(x) = 0 .

5. (a) Finddy

dxif y =

sin−1 2x

sin−1 x.

(b) Show that if z = tan−1

(x+ 1

x− 1

)+ tan−1 x , then

dz

dx= 0 .

6. A picture a metres high is placed on a wall with its base b metres abovean observer’s eye. If the observer stands x metres away from the wall, find

(a) the angle α subtended by the picture at the observer’s eye; and

(b) the distance x which will give the maximum value for α .

7. Use the tangent line approximation to find an approximate value for

(a) ln(0.94) ;

(b) cos−1(0.47) .

8. Find an equation for the line tangent to the curve x4 + x2y2 + y4 = 21 at thepoint (1, 2) .

9. If x3 + y3 + 6xy = −5 , findd2y

dx2at the point (−1, 2) .


10. Evaluate limx→1

lnx

x− 1. [Hint: Write down the limit definition of f ′(1) , where

f(x) = lnx .]


8 Fifth Problem Assignment

Distribution Date: Wednesday, November 3rd, 1999Solutions are to be submitted on Monday, November 22nd, 1999

1. For each of the functions,

f(x) =(x2 − 1)2

4x2g(x) =

1− sinx

cos x

h(x) = xe−x2

k(x) =√

lnx

determine all of the information requested below, and sketch a graph of the func-tion. Show all your work.

(a) For any discontinuities, determine whether they are removable.

(b) Where is the function increasing? Where is it decreasing?

(c) Where is the graph concave upwards? Where is it concave downwards?

(d) Where, if any, are the local extrema? In each case you should indicate whetherthe extremum is a local maximum or a local minimum, and provide justifica-tion for your choice.

(e) Where, if any, are the global extrema? Again, justification is expected.

(f) Where, if any, are the intercepts (with the coordinate axes)?

(g) Where, if any, are the inflection points?

(h) Is the graph symmetric about any vertical line x = a? (The graph of F issymmetric about x = a if the equation remains unchanged under the trans-formation x − a → a − x; i.e. if F (x) = F (2a − x). In particular, the graphis symmetric about the y-axis if F (x) = F (−x). Such a function is said to beeven.)

(i) Is the graph symmetric under rotation about the origin? (The graph of F issymmetric under rotation about the origin if the equation remains unchangedunder the transformation (x, y) → (−x,−y); i.e. if F (−x) = −F (x). Such afunction is said to be odd .)

(j) Determine all horizontal and all vertical asymptotes to the graph.

2. (a) Show that at least one of the hypotheses of the Mean Value Theorem fails tohold for the following functions; show also that the conclusion of the theoremfails to hold for each function, where the interval is [a, b] = [−1, 1].

i. f1(x) =

{1x

when x 6= 00 when x = 0

.


ii. f2(x) = x23 .

(b) [1, Problem 4.3.48] [3, Problem 4.3.48] Show that the function f2(x) defined inProblem 2(a)ii above does satisfy the conclusion of the Mean Value Theoremon the interval [−1, 27].

(c) Using the Mean Value Theorem, show carefully that the equation x7 + x− 12has exactly one real solution. Then show — without using a calculator —that this solution lies between x = 1 and x = 2.

3. Consider the function f3(x) = x cosx.

(a) Determine the intercepts of the graph of f3 on the coordinate axes.

(b) Show that the critical points of f3 occur at points of intersection of y = tanx

with the curve y =1

x.

(c) Show that, wherever it is defined, the function tanx − 1

xis increasing. Use

this fact to show that there is exactly one critical point of f3 in each interval((2n− 1)π

2,(2n+ 1)π

2

), where n is any integer. [Hint: Use a Corollary to

the Mean Value Theorem.]

(d) Sketch the graph of f3.

4. Use the Second Derivative Test in finding all points on the curve y =1

2x2which

are closest to the origin.

5. 14In each of the following cases, evaluate the limit, or show that it does not exist.

(a) limx→∞

3x+ 2√x

1− x

(b) limx→−∞

|2x− 3|x+ 2

6. Find a value for a so that

limx→∞

(3x+ 1− ax

2 + 1

x+ 1

)exists as a finite limit, and evaluate that limit.

14This problem and the next were originally included in Assignment 2, but were subsequently omitted,because they are based upon [1, §§4.5,4.7], which material had not been discussed in the lectures beforethe due date of Assignment 2.


9 Solutions, Third Problem Assignment

Solutions were to be submitted on Monday, October 25th, 1999.Caveat lector! There could be misprints and other errors in these draft solutions.

1. (a) Apply the definition of the derivative as a limit to determine the derivative of

f(x) =

√7x− 1

3for any x in the domain.

Solution:

f ′(x) = limh→0

√7(x+h)−1

3−√

7x−13

h

= limh→0

√7(x+h)−1

3−√

7x−13

h·

√7(x+h)−1

3+√

7x−13√

7(x+h)−13

+√

7x−13

= limh→0

7(x+h)−13

− 7x−13

h· 1√

7(x+h)−13

+√

7x−13

= limh→0

7h3

h· 1√

7(x+h)−13

+√

7x−13

= limh→0

7

3· 1√

7(x+h)−13

+√

7x−13

=7

3limh→0

1√7(x+h)−1

3+√

7x−13

=7

6

1√7x−1

3

(b) Determine the value of f ′(7) from the result of 1a.

Solution: 7

6√

49−13

= 724

2. You will find below the definitions for a number of functions. If the domain ofdefinition is not stated, then you are to assume the domain to be as large aspossible.

• Determine the derivative of each function wherever it is defined in the domain.


• List any points in the domain where the function does not possess a derivative,and explain precisely why there fails to be a derivative.

• If possible, determine, for those functions where requested, the equation ofthe tangent or the normal to the graph of the function at the given point.

(a) f1(x) =x

5− 8x; tangent at (1, f1(1)).

Solution: The domain of f1 is R−{58}, the set of all non-zero real numbers x

for which the denominator is non-zero; the function is continuous and differ-entiable at every point of the domain. f ′1(x) = 1(5−8x)−x(−8)

(5−8x)2 = 5(5−8x)2 , at every

point x in the domain, i.e. at all real numbers except x = 58. As f ′1(1) = 5

9,

the equation of the tangent at (1,−13) is y− (−1

3) = 5

9(x−1), i.e. 5x−9y = 8.

(b) f2(x) = (5 + 6x2 + x4)− (1 + 2x+ x2)2 + 4x(1 + x); normal at (5, f2(5)).

Solution: As f2 is a polynomial, its domain is the entire real line; it is dif-ferentiable (hence continuous) at all points of R also. f ′2(x) = (12x + 4x3)−2(1 + 2x+ x2)(2 + 2x) + (4 + 8x) = −12x2 + 8x; hence f ′2(5) = −260, and theslope of the normal at the point (x, y) = (5, f2(5)) is 1

260. The equation of the

normal is y − (−396) = 1260

(x− 5).

(c) f3(x) = (1− x)2.

Solution: f3 is a polynomial, and is both defined and differentiable at all realnumbers. f ′3(x) = −2(1− x).

(d) f4(x) = −√

10x− x2; tangent at (1,−3).

Solution: The domain of f4 consists of all real numbers x such that 10x−x2 ≥0, i.e. where the product x(x−10) is non-positive: that is, all x in the interval0 ≤ x ≤ 10. The derivative will be defined at every point in the domain exceptat the end-points of the interval, at one of which the tangent approaches thevertical; f ′4(x) = −1

2(10x − x2)−

12 (10 − 2x) = x−5√

10x−x2 . At the point x = 1,

f ′ = −12· 1

3·8 = −4

3. The equation of the tangent there is y−(−3) = −4

3(x−1),

i.e. 4x+ 3y + 5 = 0.

(e) f5(x) = |3− 2x|; tangent at x =3

2.

Solution: The function is defined for all real x. As

f5(x) =

{3− 2x 3− 2x ≥ 02x− 3 3− 2x < 0

,

f5(x) =

{−2 x < 3

2

2 x > 32

.


The derivative does not exist at the point x = 32, since the difference quotient

approaches +2 on one side of this point, and −2 on the other, so there is no2-sided limit to the difference quotient.

(f) f6(x) =

x3

|x|when x 6= 0

0 when x = 0; tangent and normal at (0, 0).

Solution: The function has all real numbers in its domain; the only point thatmight have caused difficulty would have been 0, as the given quotient is notdefined there; however, the value at x = 0 is given separately. Differentiatingf within each of the defining half-lines gives

f ′6(x) =

{2x when x > 0−2x when x < 0

To determine the value of the derivative at x = 0 we have to appeal to first

principles. f ′6(0) = limh→0

h3

|h|−0

h. Since lim

h→0+

h3

|h|−0

h= lim

h→0+

h2−0h

= limh→0+

h = 0,

and the limit from the left is also equal to 0, we can assert that the 2-sidedlimit exists, so f ′6(0) = 0. The line through the origin with slope 0 is thex-axis, which is tangent to the curve at that point.

(g) f7(x) =

{x2 − 6x when x < 1−1− 4x when x ≥ 1

; tangent at (1,−5).

Solution: f7 is defined for all x. The function was created by piecing togethera branch of a parabola for x < 1, and a ray of a straight line for x ≥ 1.Strictly within each of the constituent half-lines the function is obviouslydifferentiable: for x < 1, f ′7(x) = 2x − 6; while, for x > 1, f ′7(x) = −4; notethat we were unable to make any assertion about x = 1. Indeed, at x = 1the limit from the right of the difference quotient is again seen to be −4. Thelimit from the left is the value 2(1) − 6 = −4, which can be seen from firstprinciples. As here again the limits from left and right are equal, the limitexists, and the function is differentiable at x = 1. The line through (1,−5)with slope 4 has equation y + 5 = 4(x− 1), or 4x− y = 9.

(h) f8(x) =

x2 − 6x when x < 1

0 when x = 1−1− 4x when x > 1

; normal at (1,−5).

Solution: f8 is similar to f7, except for the definition of the function at x = 1.This time, when we attempt to determine the derivative at x = 1 by firstprinciples, we find that the quotient f(1+h)−f(1)

h= f(1+h)

hhas infinite limits —

i.e. does not have a (real) limit from the two sides. As the limit does notexist, the function is not differentiable at x = 1, and there can be no normal.


3. (The functions referred to below were defined in Problem 2 above.)

(a) Determine the value of the function f7 ◦f7 at any point x in its domain. [Thisproblem is technically difficult. Try to understand what makes the problemdifficult. Your solution to this problem will not be graded.]

Solution:

f7◦f7(x) =

(x2 − 6x)2 − (x2 − 6x) if x2 − 6x < 1 and x < 1

(−1− 4x)2 − (−1− 4x) if x ≥ 1 and −1− 4x < 1−1− 4(x2 − 6x) if x < 1 and x2 − 6x ≥ 1−1− 4(−1− 4x) if x ≥ 1 and −1− 4x ≥ 1

To study the four combinations of inequalities we observe that x2 − 6x =(x− 3)2 − 9. The inequalities simplify to

f7 ◦ f7(x) =

(x2 − 6x)2 − (x2 − 6x) if 3−

√10 < x < 1

(−1− 4x)2 − (−1− 4x) if x ≥ 1

−1− 4(x2 − 6x) if x ≤ 3−√

10−1− 4(−1− 4x) never.

i.e.

f7 ◦ f7(x) =

(x2 − 6x)(x2 − 6x− 1) if 3−

√10 < x < 1

2(8x2 + 6x+ 1) if x ≥ 1

−4x2 + 24x− 1 if x ≤ 3−√

10

(b) Determine the value of the functions f1 ◦ f4 and f4 ◦ f1 at all points in theirdomains. Find the derivatives of these functions in two ways:

i. by differentiation of the formulæ you have determined

ii. by applying the chain rule

and show that the results are the same.

Solution:

(f1 ◦ f4)(x) =−√

10x− x2

5− 8(−√

10x− x2)

= − 1

8 + 5(10x− x2)−12

(f4 ◦ f1)(x) = −

√10

(x

5− 8x

)−(

x

5− 8x

)2

= −

√x(50− 81x)

(5− 8x)2


Both methods of differentiation should lead to the same results:

(f1 ◦ f4)′(x) =5(x− 5)

5 + 8√

10x− x2

(f4 ◦ f1)′(x) =205x− 125

(5− 8x)2√x(50− 81x)

Don’t panic! These computations are much to difficult for an examination!The purpose was to test your perseverence, under conditions where two meth-ods had to lead to the same answer, so you could verify your work.

(c) Suppose that y =1 + s

1− s, s = t − 1

t, t =

√x. Determine the value of

dy

dxat

x = 2.

Solution:

dy

dx=

dy

ds· dsdt· dtdx

=2

(1− s)2·(

1− 1

t2

)· 1

2√x

When x = 2, t =√

2, s = 1√2. Hence

dy

dx=

2(1− 1√

2

)2 ·(

1− 1

2

)· 1

2√

2=

3√

2 + 4

2.

4. For each of the following functions, and for the given closed interval,

• find all critical points;

• find all local maxima and local minima;15

• find all global maxima and global minima, or explain why none of either exists.

All claims should be supported by careful reasoning; show all your work.

(a) g1(x) = (x− 3)5; interval [2, 4]

Solution: g′1(x) = 5(x − 3)4 for 2 < x < 4. The derivative is thus definedat all points in the interior of the interval; we cannot speak of a derivativeat the end points, since the behavior of the function must be known on both

15THIS PROBLEM WAS NOT TO BE GRADED: It was announced in the lectures that this topicwould not be discussed until [1, Chapter 4]; accordingly this part of the question was to be omitted.


sides of a point where a derivative is to be determined; (we could speak of1-sided derivatives, but have not done so in this course). The critical pointswill be those points where the derivative is zero, i.e. only the point x = 3. Todetermine local and global extrema we must consider the value of the functionat the critical point and also at the end-points of the interval. Since g1(3) = 0,g1(2) = −1, and g1(4) = 1, we see that the global maximum, of value 1, occursat x = 4, and that the global minimum, of value −1, occurs at x = 2. Thepoint x = 3 is neither a local minimum nor a local maximum: for x < 3 thefunction value is less than g1(3), while, for x > 3, the function value is greaterthan g1(3).

(b) g2(x) = 6− 36x+ 15x2 − 2x3; inverval

[5

2, 5

]Solution: The function g2, being a polynomial, is defined at all points. As thedefinition of g2 has been specified as

[52, 5], the function will have a derivative

at all points in the interval(

52, 5); we have no information about g2 for x < 5

2,

nor for x > 5, so we do not have a derivative at either of those end-points.Where it is defined, g′2(x) = −36 + 30x − 6x2 = −6(x − 2)(x − 3), whichvanishes at x = 2 and x = 3. But x = 2 is not in the domain of definitionof the function, so this point is irrelevant. We thus have just one criticalpoint, x = 3; there g2(3) = 6 − 108 + 135 − 54 = −21. At the end-points,g2(5

2) = −33

2, g2(5) = −49. As g2(5

2) > g2(3) > g2(5), the global maximum is

at 52, and the global minimum at 5.

(c) Solution: g′3(x) =

− 1

x2when −1 ≤ x < 0

− 1

x2when 0 < x ≤ 1

. For h 6= 0, the value of

the difference quotient g3(h)−0h

is 1h2 , which has no finite limit as h→ 0; hence

the function is not differentiable at x = 0. The derivative is not zero anywherein the domain; thus the only critical point is x = 0. Evaluating the functionat this critical point, and at the two end-points, we obtain g3(−1) = −1,g3(0) = 0, g(1) = 1. We may not apply [3, Theorem 3, p. 144] here, becausethis function is not continuous throughout the domain of definition. In fact,it is discontinuous at x = 0. As x → 0−, g3(x) → −∞; and, as x → 0+,g3(x) → ∞; these one-sided limits are not equal real numbers, so no limitexists; for continuity they must be equal, and must equal the function value,here 0. Because the function becomes positively and negatively infinite insidethe domain of definition, it can have neither a global maximum nor a globalminimum. There is no need to evaluate it at the critical point, nor at theend-points of the interval of definition.


(d) Solution: This problem differs from the preceding problem in that the limitas x → 0, while still not existing as a real number, is ∞. In this case therecan be no global maximum. The cited theorem still does not apply, as there isstill a discontinuity at x = 0. The function values at x = ±1 are both 1, and,from examination of the behavior of the function near these points, might besuspected of being global minima. However, the value at x = 0 is lower. Theglobal minimum is thus 0, assumed only at the point x = 0; everywhere elsethe function value is strictly larger.

(e) Solution: The domain of definition of g5 is the whole real line. The derivativeof√x for x > 0 is 1

2√x; similarly, the derivative of

√−x for x < 0 is − 1

2√−x .

Investigation of the difference quotient g5(h)−0h

as h→ 0, shows that the limitdoes not exist as a finite number; in fact is it ∞, and the function may besaid to have a vertical tangent. Thus x = 0 is a critical point, because thefunction lacks a derivative there. The function is, however, continuous, so wemay determine the global extrema by studying its behavior at the end-pointsand the critical point: g5(−4) = 2, g5(0) = 0, g5(9) = 3. The global maximumis thus at x = 9, and the global minimum is at x = 0.

5. [5, Examples XLVI.17] (Cambridge Math. Tripos 1930) The graph of the function

h(x) =ax+ b

(x− 1)(x− 4)has (2,−1) as a critical point. Determine a and b, and show

that the critical point is a local maximum. (Note: It is intended that this problembe solved without using concepts from Chapter 4 of your textbook (involving higherderivatives.))

Solution: For x different from 1 and 4, h′(x) = a(x2−5x+4)−(ax+b)(2x−5)(x−1)2(x−4)2 . As the

derivative exists at x = 2, it can be a critical point only because the derivative iszero; this implies that a(22−5·2+4)−(a·2+b)(2·2−5) = 0, i.e. b = 0. We are alsotold that h(2) = −1, hence 2a+0

(2−1)(2−4)= −1, so a = 1. Thus h(x) = x

(x−1)(x−4). As

h(2)−h(x) = − (x−2)2

(x−1)(x−4), this difference is non-negative in an interval surrounding

x = 2 (since the numerator is non-negative, and the denominator is negative). Thush(x) ≤ h(2), so 2 is a local maximum.

6. Determine values of the constants a, b, c which will cause the curve y = ax3 +bx2 +cx + d to pass through the points (2, 6) and (−1, 6) and to be tangent to the liney = 3x+ 1 at the point (1, 4).

Solution: We are given four kinds of information: three points through which thecurve passes, and the equation of the tangent at one of these points. Imposing thecondition that the curve pass through the points (2, 6), (−1, 6), and (1, 4) yields


three linear equations:

8a+ 4b+ 2c+ d = 6

−a+ b− c+ d = 6

a+ b+ c+ d = 4

Finally, as the tangent at x = 1 has slope 3, we know that 3ax2 + 2bx+ c|x=1 = 3,i.e. 3a+2b+c = 3. Solving the 4 linear equations simultaneously yields (a, b, c, d) =(−1, 3, 0, 2).

7. Find all lines with slope −3 which are normal to the curve 64y = x3.

Solution: At the point (x, 164x3) on the curve the slope of the tangent is 3

64x2, so

the slope of the normal is − 643x2 . Imposing the condition that this equal −3, we

obtain that x = ±83. Through the point

(±8

3,± 8

27

)the equation of the line with

slope 3 is y ∓ 827

= −3(x∓ 8

3

), i.e. 3x+ y = ±8·28

27.

8. Find the volume of the uncovered box of greatest volume that can be made bycutting equal squares out of the corners of a piece of sheet metal which is 21 cm.× 5 cm., and turning up the sides.

Solution: If the side of the square cut from each corner is of lengh x, where 0 ≤ x ≤52, then the volume obtained after the sides are folded up is x · (21− 2x)(5− 2x) =

4x3−52x2 +105x. Setting the derivative, 12x2−104x+105 (i.e. (2x−15)(6x−7))equal to zero, we find that x = 15

2or x = 7

6. The second of these is the only critical

point; the first is not in the interval of definition of the function. To determine themaximum we compare the value of the volume at x = 7

6with the volume at the

end-points, both of which give volume 0. At x = 76

the volume is 76· 56

3· 8

3= 1568

27

cm.3, which exceeds the value of 0 at the end-points. This is the largest volume.

9. Show that, among all right-angled triangles whose hypotenuse is 10 units long, thetriangle whose area is maximum is isosceles.

Solution: Denote the lengths of the non-hypotenuse sides by a and b. As these areconstrained by the equation x2 + b2 = 102 = 100, we know that b =

√100− a2.

The area is therefore 12a√

100− a2. We may take the domain to be 0 ≤ a ≤ 10.

The derivative of the area function is, after reduction, 50−a2√

100−a2 , which is zero when

a = ±5√

2. Of these two values, only the positive one is in the domain of definitionof the function. At the end-points of the domain the function is zero; while, atthe critical point we have found, the function value is positive; this, then, is themaximum point. For this value of a, b = a, which was to be proved.

Notes Distributed to Students in Section 1 of Mathematics 189-140A (1999) 1001

A Notes Specifically for Students in Section 1

A.1 Timetable for Section 1 of Mathematics 189-140A

Distribution Date: (Original version) Wednesday, September 1st, 1999(1st revision) Monday, November 1st, 1999

(All information is subject to change.)16

MONDAY WEDNESDAY FRIDAY

SEPTEMBER1 §2.1 1� 3 §2.2

6 LABOUR DAY 8 §2.2, §2.3 10 §2.3Course changes must be completed on MARS by September 14

13 §2.4; Misc. Prob-lems pp. 96-97 2�

15 §3.1 17 §3.2

Deadline for withdrawal with fee refund = September 2120 NO LECTURES OR

INSTRUCTORS’OFFICE HOURS INTHIS COURSE

22 §3.3 24 §3.4

27 §3.5 2© 29 §3.6

OCTOBER1 §3.6

4 §3.7 6 §3.8 8 §3.9Verification Period: October 12–15

11 THANKSGIVING 13 X 3� 15 §7.1, §7.2Deadline for withdrawal (with W) from course via MARS = Oct. 17

18 §7.3, §7.4; Misc.Problems pp. 474–476.

20 §8.1, §8.2 22 §8.2

25 §4.1, §4.2 3© 27 §4.3 4� 29 §4.4

16Notation: #� = distribution of assignment #

n© = assignment #n due

R© = Read Only

X = reserved for eXpansion or review

Section numbers refer to the text-book.


MONDAY WEDNESDAY FRIDAYNOVEMBER

1 §4.4, §4.5 3 §4.5 5� 5 §4.68 §4.6 4© 10 §4.7 12 §4.715 Misc. Problems

pp. 278-28017 §8.3 19 §8.4

22 Assignment #3 5© 24 Misc. Problemspp. 510-511.

26 §5.1, §5.2

29 X

DECEMBER1 X 3 X


A.2 Inverse functions; exponentials and logarithms

Distribution Date: Friday, October 15th, 1999

A.2.1 Invertible and inverse functions

If f is a function whose domain is a set A, and takes its values in the set B, we oftenspeak of the function f : A → B. The set of values that f takes may not be all of B;where it is, we say that f is onto or surjective. We say that f is one-to-one or injective[1, p. 478] if distinct points of A are always mapped on to distinct points of B; i.e. iff(x1) = f(x2) implies that x1 = x2. We say that a function g : B → A is an inverse off : A→ B if, for all x in the domain of f ,

(g ◦ f)(x) = x (21)

(i.e. g(f(x)) = x) and, for every y in the domain of B (which is the image17 of f),

(f ◦ g)(y) = y , (22)

(i.e. f(g(y)) = y). When an inverse for f exists we call f invertible. Following is a listof facts about inverses that are often proved in algebra courses:

Theorem A.1 Let a function f : A→ B be given.

1. If f is invertible, then f is surjective (onto).

2. If f is invertible, then f is injective (one-to-one).

3. If f is invertible, then it has a unique inverse.

4. If f is injective and surjective, then f is invertible.

Since the inverse, if it exists, is unique, we may then speak of the inverse, and will oftendenote it by a symbol like f−1. But not every function f has an inverse. For real-valuedfunctions f whose domain is in the real numbers, the condition of being one-to-one isequivalent to the fact that the graph of y = f(x) crosses horizontal lines at most once.(Remember that the graph of any function crosses vertical lines at most once.) Not everyfunction has this property. For example, the function f(x) =

√1− x2, whose domain is

the interval [−1, 1], and whose graph is a semicircle, crosses lines y = c (0 ≤ c < 1) twice,and the function is therefore not invertible. Another function which is not invertible issinx, whose graph crosses all lines y = c (−1 ≤ c ≤ 1) infinitely often. (In this case,however, we will first define a new function which is invertible by restricting ourselvesto a narrow interval of the domain of sin, rather than to the whole real line.18)

17also called the range [1, p. 2]18The function that we will eventually call the inverse sine and denote by sin−1 is really not the

inverse of the sine function — the notation and name are misleading!



Exercise A.1 Suppose that f(x) = 5x− 9 for all x. By solving for x in terms of f(x),determine a formula for the inverse of f . [Answer: f−1(x) = x+9

5.]

Exercise A.2 Explain why the function |x| is not invertible.

A.2.2 The derivative of the inverse function

Suppose that y = f(x), and that x = g(y). Then g(f(x)) = x for all x; differentiatingthis equation by the chain rule, with respect to x, we obtain

g′(f(x)).f ′(x) = 1

or simply

g′(y) =1

f ′(x). (23)

which may be written more compactly as

dx

dy=

1

dy

dx

(24)

which appears to be a relation between fractions.19

A.2.3 The exponential function ex

To define the exponential functions rigorously we require concepts from Calculus II. Weare therefore relying on students’ intuition in introducing these functions at this time. Inthe lectures we saw an argument that there should exist a real number which we denoteby e, with the property that

d

dxex = ex . (25)

If we also make the “reasonable” assumption that

e0 = 1 , (26)

we can actually develop the properties of the function from these two equations. We willnot fully prove these properties at this time. For any real numbers x1, x2,

(ex1)x2 = ex1x2 (27)

ex1 · ex2 = ex1+x2 (28)

e−x1 =1

ex1(29)

19But dydx and dx

dy are not fractions; the suggestive notation, however, makes it easy to rememberproperties related to the chain rule.


A.2.4 The natural logarithm

By virtue of (25), the graph of y = ex crosses all horizontal lines y = c, where c > 0,exactly once. Hence it has an inverse, whose domain is 0 < x < ∞; we call the inversefunction the (natural) logarithm, and denote it by lnx or loge x, and even just log x.Equations (21) and (22) become

ln ex = x for all real x (30)

elnx = x for all positive x (31)

We can show that (27), (28), (29), (30), (31) together imply the familiar properties

ln(x1x2) = lnx1 + lnx2 (32)

ln (xx21 ) = x2 lnx1 (33)

ln

(x1

x2

)= ln x1 − lnx2 (34)

where x1 and x2 are any positive real numbers. Differentiating (31) yields elnx· ddx

lnx = 1,hence

d

dxlnx =

1

xfor all positive x (35)

A.2.5 The exponential function ax, for any positive base a

For any positive real number a we define the function ax, whose domain is the set R ofall real numbers, and whose image (=range) is the set of all positive real numbers, byax = e(ln a)x.(We often write the exponent in the order x ln a, so that no parentheses arethen needed.)20 Analogous to (25) we have

d

dxax = ax ln a . (36)

20We need parentheses in (ln a)x to avoid confusion with ln(ax), which usually has a different value.

Exercise A.3 [Difficult] Suppose that a is a fixed real number. Investigate whether there can be realnumbers x such that (ln a)x = ln(ax).Solution to Exercise A.3:

(ln a)x = ln(ax)⇔ ln (ax) = ln(ax) logarithm of a power⇔ ln (ax)− ln(ax) = 0

⇔ ln(ax

ax

)= 0 logarithm of a quotient

⇔ eln( axax ) = e0

⇔ ax

ax= 1


A.2.6 The logarithm loga x, to any positive base a

The logarithm loga x may be defined to be the inverse of the function ax, and the prop-erties developed analogously. Thus

loga ax1 = x1 (37)

aloga x1 = x1 (38)

loga(x1x2) = loga x1 + loga x2 (39)

loga (xx21 ) = x2 loga x1 (40)

loga

(x1

x2

)= loga x1 − loga x2 (41)

where x1 and x2 are any positive real numbers. Taking logarithms of both sides of (38)to base b yields

loga x1 · logb a = logb x1 . (42)

Hence, in particular, when x1 = b, loga b · logb a = logb b = 1 so

loga b =1

logb a(43)

We can also show that

d

dxloga x =

1

x ln a(44)

= (loga e)1

x(45)

There will be exactly one number x with this property: it is the intersection of the line y = x with thetranslated exponential curve y = ax−1.


B Notes Specifically for Students in Section 2

B.1 Timetable for Section 2 of Mathematics 189-140A

Distribution Date: Wednesday, September 1st, 1999(All information is subject to change.)21

MONDAY WEDNESDAY FRIDAY

SEPTEMBER1 1�

6 LABOUR DAY 8Course changes must be completed on MARS by September 14

13 2� 15Deadline for withdrawal with fee refund = September 21

20 NO LECTURES ORINSTRUCTORS’OFFICE HOURS INTHIS COURSE

22

27 2© 29

OCTOBER

4 6Verification Period: October 12–15

11 THANKSGIVING 13 3�

Deadline for withdrawal (with W) from course via MARS = Oct. 1718 2025 3© 27 4�

NOVEMBER1 3 5�

8 4© 1015 1722 5© 2429

DECEMBER1

21Notation: #� = distribution of assignment #

n© = assignment #n due


C Assignments, Tests and Examinations from Pre-

vious Years

C.1 1998 Problem Assignments

(For most of these problems the answer is in the back of the text-book, but the fullsolution is not in the Student Solution Manual.)

Assignment Due Exercise NumbersNumber Date

1 25 Sept./98 Chapter 2 Miscellaneous Problems:2, 5, 9, 12, 17, 21, 26, 28, 32, 35, 38,40, 50, 56, 57, 59, 61

2 9 Oct./98 §3.1: 11, 17, 20, 36, 41§3.2: 2, 5, 9, 15, 21, 33, 45, 48, 53§3.3: 14, 15, 17, 44, 50, 53

3 23 Oct./98 §3.4: 45, 50§3.5: 2, 6, 9, 21, 27, 45§3.6: 21, 26, 32§3.7: 62, 68, 74

4 6 Nov./98 §3.8: 17, 21, 26, 29, 35, 47(deferred to §3.9: 3, 8, 11, 23, 27, 50, 51, 59

9 Nov.) Chapter 7 Miscellaneous Problems:3, 11, 17, 21Chapter 8 Miscellaneous Problems:5, 9

5 23 Nov./98 §4.2: 3, 8, 13, 20, 26, 40, 41, 42§4.3: 15, 23, 29, 33§4.4: 5, 11, 15, 30§4.5: 21, 32§4.6: 31, 49§4.7: 2, 6, 9, 15, 29, 47

C.2 1998 Class Quiz, with Solutions

(This quiz was given to both sections of 189-140A in the middle of October, 1998. Nogrades were recorded. The solutions were posted on the Web.)

1. Determine the global maxima and global minima of the function


f(x) = 5x2/3 − x5/3 on the interval −2 ≤ x ≤ 4. (You may take 1.6 as an ap-

proximation to 413 . You may find it helpful to make a rough sketch, but your

solutions must not depend on that sketch.)

Solution:

f ′(x) = 5 · 2

3· x−

13 − 5

3· x

23

=5(2− x)

3x13

,

provided x 6= 0. When x = 0, the function lacks a derivative. The derivativevanishes (i.e. is equal to zero) if and only if x = 2. Thus the list of critical points22

for this function, whose domain is the closed interval [−2, 4], is

Points where there is no derivative:

(0, f(0)) = (0, 0)

Points where the derivative vanishes:

(2, f(2)) = (2, 3 · 223 )

End-points of the domain of definition:

(−2, f(−2)) = (−2, 7 · 413 )

(4, f(4)) = (4, 1 · 423 )

Both the global maximum and the global minimum must be attained at pointsamong these 4. Comparing the function values — if necessary students could haveused the approximation of 1.6 for 4

13 — shows that the maximum occurs at the

point (−2, 7 · 4 13 ), and the minimum at the point (0, 0).

(A graph of this function may be found in [1, Example 6 of §3.5].)

2. A rectangle, with its base on the x-axis, is constructed so that its upper two verticesare on the semi-circle x2 + y2 = 4, y ≥ 0. Determine the maximum possible areafor such a rectangle, and determine where the maximum value is attained.

Solution: There are several ways of approaching this problem; we present two.

22The definition [1, p. 144] of critical point in the text-book is unclear as to whether or not the end-points of the interval of definition are to be considered critical . If you choose to consider end-pointsas not being included in this definition, then you must include them with the (other) critical points inyour list of candidates for extrema.


(a) A rectangle inscribed in the semicircle must have its upper side parallel to thex-axis, so the coordinates of the upper vertices may be taken to be (±x, y);as x2 + y2 = 4, and y is non-negative, y =

√4− x2. Thus the vertices of the

rectangle will be (±x,√

4− x2) and (±x, 0). The area, which we may denoteby A(x) is then given by

A(x) = |(2x) ·√

4− x2| .

(The absolute signs are needed since the intention of the problem is thatthe area should be non-negative.) We can now approach the problem in twoequivalent ways.

i. We may take the domain of A(x) to be −2 ≤ x ≤ 2, since the circle hasradius

√4 = 2. Differentiating yields

A′(x) = 2√

4− x2 + 2x · 1

2

1√4− x2

· (−2x)

=4− 2x2

√4− x2

However, this is valid only for x > 0, because of the absolute signs; forx < 0

A′(x) =2x2 − 4√

4− x2.

Note that there is no derivative at x = 0.Thus the critical points are

(0, A(0)) = (0, 0)

(−√

2, A(−√

2)) = (−√

2, 4)

(√

2, A(√

2)) = (√

2, 4)

and the end points,

(−2, f(−2)) = (−2, 0)

(2, f(2)) = (2, 0)

Comparing the values, we find the maximum of A = 2 to be attained atboth of the points x = ±

√2. The minimum value of 0 is attained at 3

points: x = ±2 and x = 0.

ii. The preceding approach can be modified, in that we can restrict thedomain to be 0 ≤ x ≤ 2, by the symmetry of the function. In this


approach the only critical points are

(√

2, A(√

2)) = (√

2, 4)

and the end points,

(0, A(0)) = (0, 0)

(2, A(2)) = (2, 0)

and we obtain the same extreme values.

(b) A more elegant approach would observe that all points on the semi-circle havecoordinates of the form (2 cos t, 2 sin t), where 0 ≤ t ≤ π. Here again we couldtreat two cases, according as we permit the domain to be the full intervalstated, or only 0 ≤ t ≤ π

2. In either case the area is α(t) = |4 cos t · 2 sin t| =

|4 sin 2t|. For convenience we will take the domain to be 0 ≤ t ≤ π2. Then,

observing that α(t) = 4 sin 2t, we have α′(t) = 4(cos 2t) · 2, which vanishesonly where 2t = π

2, so t = π

4. Other than the end-points of the interval, this

is the only critical point, and the area there is α(π4) = 4 sin π

2= 4. At the

end-points we have α(0) = sin 0 = 0, and α(π2) = 0. Thus the maximum value

of 4 occurs only at t = π4, and the minimum value of 0 occurs at t = 0 and

t = π2.

C.3 1996 Final Examination in 189-122A

1. [4 MARKS] Evaluate limx→2

1x− 1

2

x− 2.


x− 9√x− 3

.


x cot 3x .

4. [5 MARKS] Find an equation for the straight line which is normal to the graph off(x) = x2 at x = −3 .

5. [8 MARKS] The surface area of a sphere is increasing at a rate of 10 square metresper hour. At what rate is the volume increasing, when the radius is 2 metres?(Note: The volume of a sphere of radius r is 4

3πr3 ; the surface area is 4πr2 .)

6. [8 MARKS] Show that the equation 6x4 − 7x + 1 = 0 does not have more thantwo distinct real roots.


7. [10 MARKS] A right circular cylinder is inscribed in a right circular cone of heightH and radius R. Determine the dimensions of the cylinder with the largest possiblevolume. What is that largest volume?

8. [8 MARKS] Show that the function f(x) = 2 + (1− x3)15 has an inverse. Deter-

mine f−1(x) .

9. The function f(x) = x− cosx (−π ≤ x ≤ π) is differentiable.

(a) [2 MARKS] Show that this function has an inverse.

(b) [4 MARKS] Calculate the derivative of the inverse function.

(c) [4 MARKS] Evaluate the derivative of f−1(x) at all points x such that f(x) =−1 .

10. [8 MARKS] Use the mean value theorem to show that, whenx > 1 ,

x− 1

x< lnx < x− 1 .

11. Let f(x) = ln

(x4

x− 1

).

(a) [3 MARKS] Specify the domain of f .

(b) [3 MARKS] Determine the interval(s) where f increases, and the interval(s)where f decreases.

(c) [3 MARKS] Determine the concavity of the graph of f , and find the pointsof inflection.

(d) [3 MARKS] Sketch the graph, using the information determined above.

12. Let f(x) = xe−x .

(a) [3 MARKS] Specify the domain of f .

(b) [3 MARKS] Determine the interval(s) where f increases, and the interval(s)where it decreases.

(c) [3 MARKS] Determine the concavity of the graph of f , and find the pointsof inflection.

(d) [3 MARKS] Sketch the graph, using the information determined above.

13. [7 MARKS] Given the curve x3 + y3 = 1 + 3xy2 , verify that the point (x, y)

= (0, 1) is on the curve. Finddy

dxand

d2y

dx2at (x, y) = (0, 1) .



1. [8 MARKS] For the function f(x) =x2 − 4

|x− 2|

(a) Find the left-hand and right-hand limits at x = 2 .

(b) Determine whether the two-sided limit exists at x = 2 .

(c) Sketch the graph of y = f(x) .

2. [2 MARKS] Multiple Choice: Circle the correct answer (A, B, C, D, or E.)

limx→0

√9 + x−

√9− 3x

x

A B C D E

= 0 =2

3= 1 does not exist (or =

∞, or = −∞).is a real number r dif-

ferent from 0,2

3, 1.

3. [8 MARKS] Apply the intermediate value property of continuous functions to provethat the equation x3 − 4x2 + 1 = 0 has at least three different solutions. (Hint:Denoting f(x) = x3 − 4x2 + 1 , we have f(+1) = −2.)

4. [8 MARKS] At time t, the radius of a leaking spherical balloon is r =60− t

12centimetres. Determine the rate (in cm.3/second) at which the volume is decreasingwhen t = 30 . (Hint: You may assume that the volume of a sphere of radius r is43πr3.)


5. [8 MARKS] Determine an equation for the straight line that passes through thepoint (1, 5) and is tangent to the curve y = x3 .


limx→0

x− 4 tanx

sinx

A B C D E

= 0 =2

3= 1 does not exist (or =

∞, or = −∞).is a real number r dif-

ferent from 0,2

3, 1.

7. [8 MARKS] Showing all your work, determine the maximum area of a rectanglewith a base that lies on the x-axis, and with two upper vertices that lie on thegraph of the equation y = 4− x2.

8. [8 MARKS] A tank is in the shape of an inverted right circular cone of height 800cm., whose top is a disk of radius 160 cm. Water is running out of a small holeat the vertex (apex) of the cone, which is at the bottom. Showing all your work,determine the rate of change of volume V with respect to height h, at a time whenthe height is 600 cm. (Hint: You may assume that the volume of a right circularcone of height h, whose base has radius r, is 1

3πr2h.)


If y = sin 2x cos 3x , the value ofdy

dxwhen x =

π

2is

A B C D E

02

31 −1 different from 0,

2

3, 1, −1.

10. [8 MARKS] A covered rectangular box is to be constructed with volume 576 cubiccentimetres, with its bottom twice as long as it is wide. Determine the dimensionsof the box that will minimize its total surface area (including the cover).

11. [5 MARKS] Showing all your work, determine the maximum and minimum valuesof f(x) = 3− |x− 2| on the interval [1, 4] .



The function f(x) =(2x2 − 3x

)e−x has a global maximum on the half-line x ≥ 0 .

The maximum value is

A B C D E

−√e − 1√

e

9

e2

9

8e−

34 none of the preceding

4 values.

13. [8 MARKS] Showing all your work, sketch the graph of the function

f(x) =e2x

e2x + 3,

identifying asymptotes, critical points, and inflection points. Show clearly wherethe graph is concave upward and where it is concave downward.

14. [8 MARKS]

(a) State the Mean Value Theorem.

(b) Given that f(x) = |x| for −1 ≤ x ≤ 1 , does it follow from the Mean ValueTheorem that there exists a point c such that −1 ≤ c ≤ 1 and f ′(c) = 0 ?Explain your answer.


If x5 − y5 = 2x2y2 , the value ofdy

dxwhen (x, y) = (1,−1) is

A B C D E

−1 1 −5

3

5

3none of −1, 1, −5

3,

5

3.

16. [10 MARKS] Showing all your work, sketch the graph of the function

f(x) =x2 − x− 2

(1− x)2− 1 ,

identifying asymptotes, critical points, and inflection points. Show clearly wherethe graph is concave upward and where it is concave downward.



1. [4 MARKS] Showing your work, find limx→3

x3 − 27

x− 3.

2. [5 MARKS] Given that a function f has the property that

|f(x)− 2| ≤ (x− 1)2 ,

determine limx→1

f(x).

3. (a) [4 MARKS] Define v(x) = x3 − 4x2 + x + 3 . Use the Intermediate ValueTheorem to show that the equation v(x) = 0 has a solution between x = 1and x = 2.

(b) [4 MARKS] By examining the behavior of v(x) as x → ∞ and as x → −∞,or otherwise, discuss the existence of other solutions to v(x) = 0.

4. [8 MARKS] Find equations for all straight lines which are both normal to the curvey =√x− 3 and parallel to the straight line y = −2x+ 11.

5. For the function g(x) = |1− x2|,

(a) [3 MARKS] Sketch the graph of g.

(b) [3 MARKS] Show that g is continuous at x = 1.

(c) [2 MARKS] Determine whether g is differentiable at x = 1.

6. (a) [3 MARKS] Determine h′(x) , if h(x) =sinx

1− 2 cos x.

(b) [3 MARKS] If u(x) = ln ((2x− 1)3), determine u′(x).

7. [12 MARKS] Determine the slope of the curve

(x+ y)2 − (x− y)2 = x4 + y4

at all point(s), other than the origin, where the curve meets the line x = y.

8. [6 MARKS] A child is building a snowman by rolling a snowball on the ground; itsvolume is increasing at the rate of 8 cubic centimetres per minute. Find the rateat which the radius is increasing when the snowball is 75 centimetres in diameter.(The volume of a sphere of radius r is 4

3πr3.)


9. [8 MARKS] The domain of the function F is to be taken to be the interval[−2, 1

2

].

On that interval,F (x) = x3 + x2 − x+ 1 .

Determine the global maximum ( = absolute maximum), global minimum ( = abso-lute minimum), local maxima ( = relative maxima), and local minima ( = relativeminima).

10. [10 MARKS] A closed box with a square base is to have a volume of 2, 000 cubiccentimetres. The material for the top and bottom of the box costs $3 per squarecentimetre, while the material for the sides costs $1.50 per square centimetre.Determine the dimensions of the least expensive box.

11. For the function f(x) = x2

x2 − 4,

(a) [2 MARKS] Determine the (largest possible) domain.

(b) [4 MARKS] Determine all local extrema, and all points of inflection.

(c) [2 MARKS] Determine precisely where the function is increasing, decreasing,concave upward, concave downward.

(d) [2 MARKS] Determine all horizontal and all vertical asymptotes, if any.

(e) [1 MARK] Sketch the graph.

12. It is given that the function f , defined by f(x) = x + x3 has an inverse, denotedby f−1.

(a) [3 MARKS] Determine the value of f−1(2).

(b) [3 MARKS] Determine the value of(ddx

(f−1))

(2).

13. [4 MARKS] Determine the derivative of the function

m(x) = xx .

14. [4 MARKS] Determine the value of the following limit, if it exists:

limx→0

ex3 − 1

x− sinx


C.6 1998 Supplemental Examination in 189-140A

1. (a) Evaluate limx→2

x2 − 4

x− 2.

(b) Let f be the function with domain ∞ < x <∞ given by

f(x) =

{x2 − 2x if x ≤ 1x− 2 if x > 1

i. Is f continuous at x = 1? Explain (justify) your answer.

ii. Is f differentiable at x = 1? Explain (justify) your answer.

2. Find the derivative of each of the following functions. (You need not simplify youranswers.)

(a) F (x) =sin(2x2 − 1)

(x2 + 1)3

(b) G(x) = tan−1

(1√

x3 + 1

)(c) H(x) = x2e−2x lnx

3. It is given that the function f , defined by f(x) = x + x3 has an inverse, denotedby f−1.

(a) Determine the value of f−1(2).

(b) Determine the value of(ddx

(f−1))

(2).

4. (a) Find an equation for the line tangent at the point (1, 1) to the curve y2 =x3(2− x).

(b) A woman who is 1.75 metres tall walks at a rate of 1.5 metres per secondaway from a lamp that is at the top of a 4-metre high lamp post. At whatrate is her shadow lengthening when she is 30 metres from the lamp post.

5. Find the area of the largest rectangle which can be inscribed in a semicircle ofradius 1, where one side of the rectangle lies on the diameter of the semicircle, andthe other two vertices lie on the semicircle.

6. Consider the function h(x) = 4x3 − 15x2 + 12x+ 7, with domain −∞ < x <∞.

(a) Find all points at which h has a local maximum, a local minimum, or a pointof inflection. Justify all of your answers.


(b) Find the global (absolute) maximum and the global (absolute) minimum of hon the interval [0, 3].

7. Sketch the curve y =2x2

x2 − 1, indicating any horizontal or vertical asymptotes.

8. Let the function u be defined by u(x) = ln(2x+ 1).

(a) What is the (maximum possible) domain of u. (Explain.)

(b) Sketch the graph of u(x) = ln(2x+ 1).


D References

[1] C. H. Edwards, Jr., and D. E. Penney, Calculus with Analytic Geometry, EarlyTranscendentals Version, Fifth Edition. Prentice Hall, Englewood Cliffs, NJ (1997).ISBN 0-13-793076-3.

[2] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Calculus withAnalytic Geometry, Early Transcendentals Version, Fifth Edition. Prentice Hall,Englewood Cliffs, NJ (1997). ISBN 0-13-079875-4.

[3] C. H. Edwards, Jr., and D. E. Penney, Single Variable Calculus with AnalyticGeometry, Early Transcendentals Version, Fifth Edition. Prentice Hall, EnglewoodCliffs, NJ (1997). ISBN 0-13-793092-5.

[4] C. H. Edwards, Jr., and D. E. Penney, Student Solutions Manual for Single VariableCalculus with Analytic Geometry, Early Transcendentals Version, Fifth Edition.Prentice Hall, Englewood Cliffs, NJ (1997). ISBN 0-13-095247-1.

[5] G. H. Hardy, A Course of Pure Mathematics, 10th edition. Cambridge UniversityPress (1967).

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