Contents
General Information
Word of Welcome . . . . . . .
Rational . . . . . . . . . . . .
Prerequisite Study . . . . . .
Requirements . . . . . . . . .
Study Material . . . . . . . .
How the Subject is Presented
How to Use the Study Guide .
How to Use the Textbooks . .
How to Study . . . . . . . . .
Mathematical Symbols . . . .
Action Words . . . . . . . . .
Module Plan and Timetable .
Evaluation . . . . . . . . . . .
Marks . . . . . . . . . . . . .
Examination . . . . . . . . .
Passing Requirements . . . .
Module Outcomes . . . . . . .
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1 Functions
1.1 Solving of Inequalities . . . . . . . . . .
1.2 Absolute Value Function . . . . . . . . .
1.3 Functions and Operations with Functions
1.4 Inverse Functions . . . . . . . . . . . . .
1.5 Exponential Function . . . . . . . . . . .
1.6 Logarithmic Function . . . . . . . . . . .
1.7 Radian Measure . . . . . . . . . . . . . .
1.8 Trigonometric Functions . . . . . . . . .
1.9 Inverse Trigonometric Functions . . . . .
iii
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v
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1
6
9
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21
iv
2 Limits of Functions
2.1 Informal Definition of the Limit of a Function . . . . . . . .
2.2 Limit Identities . . . . . . . . . . . . . . . . . . . . . . . . .
2.3 Asymptotes of Functions . . . . . . . . . . . . . . . . . . . .
2.4 Limits of Exponential and Logarithmic Functions . . . . . .
2.5 Limits of Trigonometric and Inverse Trigonometric Functions
2.6 Formal Definition of the Limit of a Function . . . . . . . . .
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25
30
34
35
36
36
36
3 Continuity
39
3.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.2 Limits of the form f (x)g(x) . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
4 Differentiation
4.1 Definition of Derivative . . . . . . . . . . . . . . . . .
4.2 Rules for Differentiation . . . . . . . . . . . . . . . .
4.3 Chain Rule . . . . . . . . . . . . . . . . . . . . . . .
4.4 Implicit Differentiation . . . . . . . . . . . . . . . . .
4.5 Derivatives of Inverse Functions . . . . . . . . . . . .
4.6 Derivatives of Exponential and Logarithmic Functions
4.7 Logarithmic Differentiation . . . . . . . . . . . . . . .
4.8 Derivatives of Trigonometric Functions . . . . . . . .
4.9 Derivatives of Inverse Trigonometric Functions . . . .
4.10 L’Hôspital’s Rule . . . . . . . . . . . . . . . . . . . .
4.11 Applications of Differentiation . . . . . . . . . . . . .
5 Integration
5.1 Integration . . . . . . . . . . . . . . . . . . . . . . . .
5.2 Integrals of Exponentials and Logarithmic Functions
5.3 Integrals of Trigonometric Functions . . . . . . . . .
5.4 Integrals of Inverse Trigonometric Functions . . . . .
5.5 Substitution Rule . . . . . . . . . . . . . . . . . . . .
6 Numbers systems
6.1 Real Number System . . . . .
6.1.1 Real Numbers . . . . .
6.1.2 Natural Numbers . . .
6.1.3 Integer Numbers . . .
6.1.4 Rational and Irrational
6.2 Complex Number System . .
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Numbers
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49
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105
v
6.2.1
6.2.2
6.2.3
6.2.4
6.2.5
Complex Numbers as a Number System; Basic Operations; Standard Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Complex Conjugate and Modulus or Absolute value . . . . . . . .
Geometric Representation of Complex Numbers. . . . . . . . . . .
The Polar and Exponential Form of Complex Numbers . . . . . .
Powers and Exponents of Complex Numbers . . . . . . . . . . . .
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105
108
111
111
112
A List of Unacceptable Errors
115
B Infinity and the Calculation of Limits
117
C List of Formulas
119
D List of Words
123
vi
General Information
Word of Welcome
The Euclidian geometry was perhaps the greatest mathematical achievement of the antique times. It has developed over the centuries by practical observation and applications
to the level that we are nowadays getting it in schools. The theory is constructed by
basic definitions and axioms which were formulated after thorough observation and
experimentation through the ages, followed by further theorems that were proven and
by conclusions that were derived from them. The theory has therefore developed over
many years and has been formalised to the product that we have today.
During the 17th century a discovery was made that is probably the greatest mathematical
discovery of all times, namely the differential and integral calculus. Independently
from each other, Newton in England and Leibniz in Germany established the foundation
for the development of this theory during the last quarter of the seventeenth century.
They did this while working on the limits of functions. The ideas of the derivative of
a function and of the integral of a function followed naturally. Although these discoveries led to the solution of problems from the earlier centuries, the main motivation for the
development of the differential and integral calculus was to solve the major problems of
the seventeenth century. Eventually, after many years and after much contributing work
by many mathematicians, especially French mathematicians, a new discipline evolved
which grew from practical applications to something which resembled geometry - in the
sense that the theory is also constructed from definitions, theorems and conclusions. This
discipline is what is today called differential and integral calculus. In our study we shall
therefore put much emphasis on the correct representation of definitions, wording of theorems, proof of some theorems and finally, of course, the application of all this.
Welcome and enjoy the study.
Rationale
Calculus forms part of a wider subject field, namely MATHEMATICS. At university level,
Calculus is basically a combination of Algebra and Trigonometry as you’ve come to know
it at school level... we shall just slightly expand on your knowledge of these subjects.
Calculus can be divided into two parts (with which you will still become familiar), i.e.
Differential Calculus and Integral Calculus.
vii
viii
General Information
In MTHS111 and MTHS121 the basic techniques are introduced upon which the mathematics of later study years as well as adjacent disciplines depend. You will have the
privilege to learn and apply some of the mathematical techniques which have been developed through many centuries. You will also acquire the theoretical foundation of the
techniques so that you will know which technique to choose to solve a certain mathematical problem. In this module you will be acquainted with the available mathematical
techniques and afforded the opportunity to refine your skills in the use of these techniques.
You must see yourself as a mathematical apprentice, this includes all learners who take
mathematical subjects, and the mathematical techniques as your tools in your toolbox
with which you are going to walk through life to solve any mathematical problem that
crosses your path. A good craftsman knows how to use his tools in order to solve problems. Therefore, use your time as efficiently as possible to acquire the necessary skills
needed for success in the implementation of your tools.
In particular, you will learn to differentiate and to integrate in MTHS111. These are
two very important mathematical techniques with which infinitely many problems can be
solved. In mathematics physical phenomena are represented by functions. Differentiation
is used mainly to study the change of physical phenomena. When functions are differentiated (the derivatives are calculated), expressions are obtained that give an indication
of the change of the physical phenomena. If the displacement of an object is known,
the velocity and acceleration of it can be determined by differentiating the displacement.
For us as modern people it is very important to know how fast something may and will
happen or change. Integration is the opposite process of differentiation. It is mainly used
to calculate areas and in solving mathematical equations in which derivatives of functions
appear. In most mathematical models phenomena are described in terms of their rate of
change, for example, Newton’s second law gives the acceleration of an object as the sum
of all the external forces acting on the object. To determine the velocity and the displacement of the object, the acceleration is integrated. You will also learn the theoretical
foundation of differentiation and integration so that you will use the techniques correctly.
MTHS111 forms an essential base for the following modules which are presented in the
Faculty of Natural Sciences: Electricity, Magnetism, Optics, Atom and Nuclear Physics
(NPHY121), Statics and Mathematical Modelling (APPM121), Mathematical Modelling
and Vector Algebra (APPM122) and Introductory Algebra and Analysis II (MTHS121).
These modules again form an essential base for even more modules. It is therefore important that you pass this module this semester. If you do not pass, you not only have lost
a lot of important mathematical techniques, but you close many doors for yourself and
prolong your studies by at least one year. If you have not passed MTHS111, you will not
be allowed to proceed with the mentioned modules.
Prerequisite Study
You should have obtained a mark of not less than 60% (level 5) in the Grade 12 examination, or a mark of not less than 70 % (level 6) in any other examination judged by Senate
to be equivalent to the above.
General Information
ix
Requirements
You will need the following for this module:
1. Work Schedule (on eFundi)
2. Textbook (see Study Material)
3. Study guide
4. Access to the internet for using eFundi
5. Workbook for class work
6. Writing pad (loose sheets) for submission of assignments
Study Material
Prescribed textbooks:
Title:
Calculus
Edition:
Seventh Edition (2011) or Eighth Edition (2016)
Author:
James Stewart
Editors:
Cencage Learning
AND
Title:
Introductory Algebra,
Edition:
Second Edition (1990)
Author:
De la Rosa et al.
Editors:
Lexicon Publishers
Recommended Book for Additional Use:
Mathematics Review Manual, Miroloav Lovric, http://ms.mcmaster.ca/lovric/rm.html
Unimaths Intro Workbook, Christopher Mills, Paarlmedia, 2013
Beginning Calculus, Elliott Mendelson, 2008
Understanding Calculus Concepts, Eli Passow, 1996
How to solve Word Problems in Calculus: a Solved Problem Approuch, Eugene Don &
Benay Don, 2001
How the Subject is Presented
1. Preparation: Review the work that will be discussed during the contact session
beforehand so that you will know what to expect. This is your first exposure to
the work. A schedule indicating which work will be discussed during each contact
session, is available on eFundi. It is important that first exposure to the work takes
place before and not during the contact session.
2. Contact session: During the contact session recognition and understanding have to
take place. To support these processes, the contact sessions are presented interactively. During the contact session
x
General Information
(a) the lecturer explains
i. concepts (definitions),
ii. how techniques (rules or theorems) follow from concepts (definitions),
iii. how concepts and techniques (rules or theorems) are used to solve problems, and
iv. the outline of the proofs of theorems.
(b) you get the opportunity to use concepts and techniques to solve problems.
3. Inculcation: To inculcate the work discussed during the contact session, you have
to
(a) go through and learn the formal writing (wording) of concepts (definitions),
(b) go through and learn the formal writing (wording) of techniques (rules or theorems),
(c) step by step work through and learn the proofs of theorems, and
(d) exercise the use of concepts and techniques to solve problems.
How to Use the Study Guide
Each study unit clearly outlines the work it covers, under the following headings: Time
Allocation, Prerequisite Study, Study Outcomes, References, Preparation, Exercising of
Skills and Test Yourself.
1. Time Allocation. The time allocation specified is the approximate time you will
require to master the work of the particular study unit. It includes the time you will
need for preparation for the contact sessions on the study unit and do the homework
One contact session is 50 minutes and 1 tutorial session is 90 minutes.
2. Prerequisite Study. The prerequisite study is the knowledge and skills you will
require to be able to do the work in this particular study unit. It will be assumed
that you are conversant with the theory and skills listed here. We shall not treat
this material in class and should you not possess over the prerequisite knowledge
and skills for a particular study unit, it will be your responsibility to acquire it
before we embark on that study unit.
3. Study Outcomes. The study outcomes provide you with a summary of what you
should master in this particular study unit.
4. References. Study the prescribed study material which is referred to in the study
guide.
5. Preparation. All activities listed under Preparation should be done before attending the relevant contact session. Make notes of any difficulties you may come
across.
General Information
xi
6. Exercising of Skills. Practise the skills of the study unit before the work of the
next study unit will be treated in the next contact session. If you are capable of
doing all the assigned problems independently, you have achieved that part of the
outcomes of that study unit which bears on problem solving (i.e. the application
of the theorems and definitions). All that remains is for you to memorise the given
definitions, theorems and the proofs of the theorems. After this you will have
achieved all the outcomes of the particular study unit.
7. Test Yourself. The questions listed under Test Yourself can be used to test whether
you have achieved the outcomes of the study unit. It will help you to test whether
you really understand the content of the specific study unit. Learning without
understanding, has no long term value. There is no memorandum available for
the questions listed under Test Yourself. The answers to these questions are given
during the contact sessions and/or can be found in the study guide and/or textbook.
At the end of the study guide a number of addenda have been added to compliment the
studying of the module.
1. In Addendum A a list of errors is given which you should avoid. Study this list well
and make sure you will not lose unnecessary marks during tests and exams.
2. In Addendum B a summary of calculations of infinite limits are given.
3. In Addendum C you are given a list of formulas that you should know and others
which will be supplied to you in tests and in exams. You are expected to be able to
derive all the formulas given in the list. Make sure you are aware which formulas
you should know from memory and study them well.
4. Addendum D contains a list of general mathematical terms to be used in this module,
in both English and Afrikaans.
You should spend about 120 quality hours to the study outlined in this module and to
the completion of all the activities prescribed in it.
How to Use the Textbooks
The textbooks are used to illustrate the theory and for the study of examples. The
textbooks are complementary to the knowledge given in the study guide. They should be
used to get a better understanding of the theory and to work through examples in order
to better understand the problem solving techniques. Both textbooks are also to be used
for the exercising of skills. The lectures are offered both from the textbook and from the
study guide. Be sure to have both these books (and yourself) in class every period.
How to Study
During contact sessions both theory and applications will be treated. In order to benefit
at all from these sessions, you will have to read the prescribed material at least once
xii
General Information
before the contact session and have performed the preparatory activities. You should also
have completed the homework on the previous contact session and have ensured that you
have mastered the work of the previous session. Normally new work will be done during
every contact session. Following every contact session, you will have new definitions,
theorems and techniques to work through and to master. The contact sessions will be
offered making use of both the textbook and the study guide. Make sure you attend and
bring both to every contact session. During the contact sessions the proof of theorems,
discussions of theory and examples will be presented. You should keep a good, solid
notebook for taking notes during the contact sessions. The reason is that aspects of the
work that cannot be found in the textbook will often be discussed.
You will be expected to work through examples. In the process of working through an
example, you should be guided by the following procedure:
1. Read the problem outline and establish precisely what is asked and what you should
calculate.
2. Read the first step of the solution and make use of your preknowledge (everything
you have previously learned about mathematics) as well as your new knowledge (that
which you have just covered in the section immediately preceding the example) to
determine why the writer wrote this specific step in the calculation. You should try
to see the relationship between this step and the preceding theory. Should you not
understand it or not be able to grasp the reason for the step, you should not leave
it at that. Once again revise the work preceding the example. Should you still not
understand it, consult a class-mate, or your facilitator or the lecturer.
3. Read the second step of the solution and make use of your preknowledge and of
your new knowledge to determine how the writer arrives at this next step. You
should always try to see a relationship between the step and the theory preceding
the example. Should you not understand it or not be able to see a relationship,
do not resign yourself to it and proceed. First then revise the theory preceding the
example. If you still do not understand, consult a fellow student, your facilitator or
your lecturer.
4. Repeat these steps until you have worked through the entire example.
You must be willing to put in an effort! Memorise the definitions and wording of theorems
- understanding follows on knowledge. Work through the examples. When you have done
that, you should not experience difficulties with the homework problems. If you find that
you spend too much time on a particular problem, skip it for the time being and try
to find someone to assist you. But, for heavens sake, resist the temptation to copy the
homework from someone else! You lose all the way if you should do that! The homework
is supposed to be a preparation for your weekly tutorial, and eventually for your semester
test and the examination.
The renowned mathematician George Polya gave much attention to the doing and learning
of mathematics. His book on this topic is well known: How to solve it. The writers
Daepp and Gorkin write about the solution of problems in their book Reading, Writing
and Proving - a closer look at mathematics and they give the following summary of the
hints of Polya for the solution of mathematical problems:
xiii
General Information
1. Understand the problem: Firstly you should understand all the words and terminology in the problem - look it up if necessary. Secondly you should determine
what is given. Make a sketch if applicable. Finally you should establish what is
asked: Must something be calculated? Must a statement be proven wrongly? Must
an example be presented?
2. Design a plan: How should I approach the problem? Is the problem known from
work done previously? Check your earlier notes, problems, theorems. Revise the
proofs of preceding theorems. Try to simplify the problem.
3. Execute your plan: Solve it and study your answer. Is every deduction or operation correct? Leave the problem for a while and return to it later.
4. Revise: Check everything. Even try a different technique or proof and see if you
get the same result. Discuss your solution with your fellow students - even if the
person does not know any mathematics. Check and re-check.
Mathematical Symbols
The following is a list of all the symbols that will be used in this module. Study them
well and make sure you know them.
Symbol
∼ or ¬
→ or ⇒
↔ or ⇔
Σ
∈
∀
∃
∄
∴
∋
T
S
Meaning
not
implies
if and only if
sigma-notation
element
for all
there exists
there doesn’t exist
that is to say
so that
intercept
union
Action Words
Example
∼ p, ¬p, means not p
p → q (p implies q). If p is valid, then q is valid
p ↔ q (p valid if and only if q valid)
Σ4i=1 xi means x1 + x2 + x3 + x4
x ∈ X, means x is an element of X
∀ x ∈ X, means for all x in X
∃ p, means there exists a p
∄ p, means there doesn’t exist a p
∴x=1
∋ 0 < |x
T− a| < δ ⇒ |f (x) − L| < ǫ
(−2, 3] S[0, 4) = [0, 3]
(−2, 3] [0, 4) = (−2, 4)
1. Understand: Grasp it with your mind. See what someone means.
2. Master: Conquer, overpower. Master some study material.
3. Determine: Find a value. Determine the distance, time, date.
4. Calculate: Arrive at an answer using arithmetic or algebra. We have to calculate
the cost.
5. Describe: Explain in words, present a word portrait. Describe the nature of roots.
xiv
General Information
6. Statement: Something claimed, or an opinion, often without proof. This statement
still needs to be proven.
7. Proof: Evidence that something is true, arguments confirming a theorem. Proof of
a statement.
8. Define: Carefully describe, determine the precise meaning(s). Define a concept.
9. Formulate: Express in the form of a formula; or clearly, briefly and precisely
express in words A well formulated theorem.
10. Illustrate: Explain, clarify, elucidate. Illustrate an idea or thought or statement.
11. Check: Revise, verify. Please verify all the numbers, or all the facts.
12. Exercise: Practise to perform, an assignment. Receive two exercises for homework.
13. Develop: Work it out, construct a theory, form a structure through study, add
knowledge. Develop your knowledge.
14. Solve: Find a solution, get an answer. Solve a riddle.
15. Sketch: A rough drawing giving only the main lines, or main characteristics. Sketch
the graph of the polynomial.
16. State: Express it in the correct words. State and prove the mean value theorem.
17. Test: Investigate in general. A test is performed, or to pass a test.
18. Verify: Investigate the validity, determine the correctness, compare to check, test.
Verify the statement.
19. Understand: Know well, be well familiar with. He understands mathematics well.
20. Find: Determine, calculate, get an answer, look for. Find a point on the line.
21. Example: Something to clarify, elucidate. The question is clarified using an example.
Module Plan and Timetable
You will need approximately 120 quality study hours (12 credits) in order to master this
module. Each week consists of 6 contact sessions of 50 minutes each and one tutorial
session of 100 minutes. This will occupy 80 study hours. Usually one contact session at
the beginning of the semester is allocated for an introductory session to the module, 4
contact sessions for the writing of tests and 2 contact sessions at the end of the semester
for revision. About 120 minutes per week are required for the preparation, homework and
preparation of tests. This will occupy 24 study hours. In addition a further 16 hours will
have to be spent on studying for the exams and writing the exams.
xv
General Information
Theme
Functions
Limits
Continuity
Differentiation
Integration
Number systems
Time allocation
2 weeks
2 weeks
2
weeks
3
3 weeks
1 31 weeks
3 weeks
Evaluation
In the evaluation of your knowledge and skills in this subject, the following aspects will
be assessed:
1. Knowledge: Whether you can reproduce definitions, theorems and proof of theorems. Included here is whether you possess over the knowledge to answer a question
or to solve a problem and whether you have the ready knowledge of derivatives and
integrals of standard functions.
2. Insight: To know which technique should be used, how to apply a particular definition or theorem, what line of arguments should be followed or to be aware that a
certain function should be manipulated in a certain fashion to solve the problem.
3. Application: Any question or example not drilled in, falls in this category.
4. Analysis: Consider the problem well. You should realise which components make
up the problem. This is the first thing you should do when solving questions consisting of problems. Although no direct marks are normally awarded for this analysis,
you won’t be able to do the rest of the solution (which lies at a lower level) if you
have not managed this correctly. A sketch is a good indication that an analysis had
been done and bonus points may be awarded if it is correct.
5. Synthesis: Any problem requiring of a learner to venture beyond his/her present
knowledge; demanding that the learner should recognise a pattern and should be
able to write a general formula representing it.
6. Evaluation: Ask yourself what does the answer imply. Explain the meaning of the
answer. E.g. state that a person is driving at 20 km/h in an easterly direction, in
stead of simply writing −20 km/h. Do an evaluation of the answer which you have
calculated. Test the validity of your answer by doing a test, even though it may not
have been asked. You will usually be awarded bonus points for this.
In most cases more than one of these categories or levels occur in the same question. Even
the reproduction of a definition, theorem or proof of a theorem is not limited to the level
of knowledge only, because a certain logical order is required in the writing of it, for which
some insight is required.
The following prescriptions are used when marking tutorials, tests and examination papers:
xvi
General Information
1. The answers in the memorandum are only model answers. If an answer does not
correspond exactly with the answer on the memorandum, but is mathematically
correct and within the prescriptions of the questions, all the marks are awarded for
the answer.
2. If certain steps have been omitted for which there are marks allocated on the memorandum, without discrediting the mathematical logic layout of the answer, all the
marks are awarded for the answer, for example when calculating integrals where
substitution (number of steps) or standard form (one step) can be used.
3. If some sign, calculation or writing errors have been made, without changing the
essence of the problem, and the procedure that has been followed, is correct, half
a mark is deducted for each error and the rest of the marks are awarded for the
answer.
4. If sign, calculation or writing errors have been made, and in the process the problem
is made easier, at most half the marks are awarded for the answer.
5. If sign, calculation or writing errors have been made, and the essence of the problem
has changed, no further marks are awarded for the answer.
6. If the method that has to be followed answering a question, is specified, and another,
but correct method, has been followed, no marks are awarded for the answer.
Marks
The assessment plan and the way in which the participation mark will be calculated, will
be presented to you at the beginning of the semester. The participation mark will count
50% and the examination mark 50% of the module mark. The pass mark is 50% with
a minimum of 40% in the examination. A participation mark of not less than 40% is
required to get admission to the examination.
Examination
For admission to the examination, a participation mark of not less than 40% is required. A
further prerequisite for entry to the MTHS121 module is that you have passed MTHS111.
This participation mark is compiled from marks awarded during the course of the semester
by your participation in contact sessions, work assignments, tutorial tests and semester
tests. The examination consists of two papers of one and half hours each. The final mark
for the module is calculated from the examination mark and the participation mark in
a ratio of 1 : 1. To pass the module an examination mark of not less than 40% and a
module mark of not less than 50% must be obtained.
A learner obtaining an adequate participation mark for a module, has two opportunities
to write the examination. The examination consists of two papers. Both papers have to be
written per examination opportunity. If only one paper is written per opportunity, zero is
assigned to the other paper. The examination mark is the average of the two papers. The
General Information
xvii
learner may use any one or both of these opportunities, provided that should the learner
use both opportunities, the mark obtained in the second examination will determine the
module mark. Furthermore a learner who, after both examination opportunities have gone
bye, irrespective of whether one or both of those opportunities have been utilised, have
not passed the module, will have to repeat the module in its entirety. We wish to point
out that it will be extremely risky not to utilise the first examination opportunity. If such
a learner then misses the second examination opportunity because of sickness for instance,
there will be no further special (i.e. a third) opportunity to write the examination. Such
a learner will have to register again for that module, pay the class fees and attend lectures
to accumulate a new participation mark in order to get admitted to the next scheduled
examination.
Passing Requirements
It is the responsibility of the lecturer who presents the subject to decide whether the
students have mastered the module satisfactorily or not. There are two ways of doing
this: formative (continuous) and summative (reviewing) assessment. If a student obtains
a participation mark of at least 50% for the formative assessment, an examination mark
of at least 50% in the summative assessment and a module mark (the average of the two
assessments) of at least 60%, the lecturer can declare with certainty that the student
has satisfactorily mastered the module. The university requires a minimum participation
mark of 40%, a minimum examination mark of 40% and a minimum module mark of 50%
to ensure that students are not adversely affected by circumstances during the semester
or when writing examinations.
A module mark of 80% clearly indicates that a student has mastered the module well
enough to begin with the following module. A module mark of 60% is acceptable and
indicates that a student is ready to continue with the next module if a greater effort is
made. The university requires 50% as a pass mark. As explained before, this is to ensure
that students are not adversely affected by circumstances during the semester or when
writing examinations. Any mark below 50% indicates that a student has not mastered
the work sufficiently.
Module Outcomes
After completing this module, you should:
1. know about functions and operations involving functions and the concepts of limits
and continuity;
2. understand differential and integral calculus and introductory algebra (number systems);
3. be able to demonstrate that you have mastered mathematical proof techniques, by
being able to prove selected theorems from Calculus and/or Algebra;
4. have developed the ability to solve problems as a result of frequent exercises;
xviii
General Information
5. have experienced the boundaries of this discipline and be able to realise which
propositions are considered true or false and why.
More general outcomes which you should also achieve.
1. You should be able to demonstrate that you are capable to take responsibility for
your own learning and study within a structured, supervised environment.
2. You should be able to make decisions and take responsibility for your actions and
for completing your own assignments.
3. You should be able to evaluate your own performance against certain given criteria.
The presentation and evaluation of this module is constructed in such a way that, should
you not achieve the above outcomes, you will not pass this module.
Study Unit 1
Functions
Time Allocation
17 hours 20 minutes (9 hours 30 minutes contact time and 2 tutorial sessions).
Prerequisite Study
1. Number sets: natural numbers, integers and real numbers.
2. Set theory: union and intersection of sets.
3. Solving of inequalities.
Study Outcomes
When you have completed this section, you should be able to do the following, using
standard mathematical notation.
1. Determine which values of x will satisfy any given inequality.
2. Define and use the concept of absolute value (Definition 1) to sketch graphs of absolute value functions, know the properties (Theorem 1) there of and solve equations
and inequalities.
3. Formulate, prove and use Theorems 2 and 3 to solve equations and inequalities.
4. Formulate and use Theorems 4 and 5 to solve equations and inequalities.
5. Define and use the concepts of function (Definition 2), value of a function (Definition
4), domain (Definition 3), range (Definition 5), even function (Definition 6), and odd
function (Definition 7) to determine the domain and range of functions, algebraic
combinations of functions and composite functions (Definition 8).
6. Define and use the concepts of a biunique function (Definition 9) and the inverse of
a function (Definitions 10 and 11) to determine inverse functions.
1
2
Functions
7. Define and use exponential and logarithmic functions (Definitions 12 and 13) to
sketch graphs of exponential and logarithmic functions, know the properties (Theorems 9, 10 and 11) there of and solve equations and inequalities.
8. Define and use the concepts of radian measure (Definition 14) and one radian (Definition 15) to convert from degrees to radians and inversely from radians to degrees.
9. Define and use trigonometric functions (Definitions 16 and 17) to sketch graphs of
trigonometric functions and solve equations and inequalities.
10. Define and use inverse trigonometric functions (Definition 18) to sketch graphs of
inverse trigonometric functions (arcsin x, arccos x, arctan x) and solve equations.
References
Stewart: Sections 1.1, 1.3, 6.1, 6.2, 6.3, 6.6, Appendix A, D.
Study guide: Sections 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9.
Mathematics Review Manual: Chapters 1, 3, 5, 6, 7.
Preparation
Read the Sections referred to in the textbook and in the study guide.
Work through the following examples from Stewart:
Appendix A: 1, 2, 3, 4;
Section 1.1: 1, 2, 6;
Section 1.3: 1, 2, 3, 4, 6;
Section 6.1: 1, 2, 3;
Section 6.3: 1, 2;
Appendix D: 1, 2, 3;
Section 6.6: 1.
Answer the following questions and submit your answers at the beginning of the first
contact session of this study unit:
1. Solve the following equations.
(a) −2 < 2x + 3 ≤ 5
(b) sin x cos x = 14 , x ∈ [0, 2π] (sin 2x = 2 sin x cos x)
(c) 2 · 3x = 54
(d) log2 x = 5
2. Draw the following graphs.
(a) y + 2 = 2(x − 1)
(b) y − 3 = (x + 2)2
Functions
3
Exercising of Skills
Work through the following examples from Stewart:
Appendix A: 5, 6, 7, 8, 9;
Section 1.1: 7, 8, 9;
Section 1.3: 5, 7, 8, 9;
Section 6.1: 4, 5;
Section 6.2: 1;
Section 6.3: 4, 5, 6, 7, 8;
Appendix D: 4, 5, 6;
Section 6.6: 3.
Do the following exercises:
Stewart (8th edition):
Appendix A: 28, 32, 45, 46, 47, 48, 52, 53, 55;
Appendix D: 37, 38, 65, 67, 68, 73, 74;
Ex. 1.1: 3, 4, 7 - 11, 31 - 40, 43, 55, 56;
Ex. 1.3: 31 - 33, 35, 38, 43 - 48, 52, 53, 59(a), 63, 64;
Ex. 6.1: 1, 2, 5 - 20, 23, 25;
Ex. 6.2: 1, 2, 4, 8, 9, 10;
Ex. 6.3: 1, 2(a), (c), 4(b), 5(a), 8(b), 10 - 12, 14 - 18, 27 - 36, 55(a), 59, 62;
Ex. 6.6: 1 - 6.
Stewart (7th edition):
Appendix A: 28, 32, 45, 46, 47, 48, 52, 53, 55;
Appendix D: 37, 38, 65, 67, 68, 73, 74;
Ex. 1.1: 3, 4, 7 - 11, 31 - 46, 49, 55, 56;
Ex. 1.3: 29 - 31, 34, 36, 41 - 46, 50, 51, 57(a), 61, 62;
Ex. 6.1: 1, 2, 5 - 20, 23, 25;
Ex. 6.2: 1, 2, 4, 7, 9, 10;
Ex. 6.3: 1, 2(a), (c), 4(a), 5(b), 8(b), 10 - 12, 14 - 18, 27 - 36, 55(a), 59, 60;
Ex. 6.6: 1 - 6.
Test Yourself
1. Solve equations and inequalities in which polynomials and rational functions occur,
and write the answer in interval and set notation. (Appendix A, Stewart)
2. Define the concept absolute value function and draw a sketch to elucidate the definition.
3. Draw graphs of absolute value functions of the form y = m|x − a| + b. (Appendix
A, Stewart)
4. Explain the meaning of |x − a| = b, |x − a| < b and |x − a| > b in terms of distance.
5. Prove the following.
√
√
(a) x2 = |x| and − x2 = −|x|
(b) |x| ≤ a ⇔ −a ≤ x ≤ a, a ≥ 0
4
Functions
6. Complete the following.
(a) |a2 | =
(b) |ab| =
(c) |a/b| =
(d) | − a| =
7. Is it always true that |a + b| = |a| + |b|? Motivate your answer.
8. Solve equations and inequalities in which the absolute value function occurs, and
write the answer in interval and set notation. (Appendix A, Stewart)
9. Define the concept function and draw a sketch to elucidate the definition.
10. Draw a graph that does not satisfy the definition in question 9.
11. Define the concepts domain and range.
12. Define the concepts even and odd function and draw sketches to elucidate the definitions.
13. Name the five elementary functions from which all other functions can be composed.
14. Determine the domain and range of functions. (Ex 1.1, Stewart)
15. Name six ways how other functions can be formed out of the elementary functions.
16. Define each as named in question 15 and give their domains.
17. Given two functions and their domains, determine the sum, difference, product,
quotient and composition as well as the domains thereof. (Ex 1.3, Stewart)
18. Define the concept one-to-one function and draw a sketch to elucidate the definition.
19. Define the concept inverse function in two different ways and draw a sketch to
elucidate the definitions.
20. Which of the definitions in question 19 will be used to
(a) determine the inverse of a function?
(b) test whether two given functions are inverses of each other?
21. Determine the inverse functions of given functions. (Ex 6.1, Stewart)
22. Draw sketches of the form ax for 0 < a < 1 and a > 0.
23. Complete the following.
(a) ax ay =
(b) ax /ay =
(c) (ax )y =
(d) ax bx =
24. What is special about the function ex ?
25. Simplify expressions in which exponential functions occur. (Ex 6.2, Stewart)
26. Solve equations in which exponential functions occur. (Ex 6.2, Stewart)
Functions
5
27. Define the logarithmic function (domain included).
28. Which two properties of exponential and logarithmic functions follow from question
27?
29. Draw sketches of the form loga x for 0 < a < 1 and a > 0 and indicate on the
sketches that it is the inverse function of ax .
30. Complete the following.
(a) loga x + loga y =
(b) loga x − loga y =
(c) loga xy =
(d) loga a =
(e) loga 1 =
(f) loga b =
31. What is special about the function ln x?
32. Simplify expressions in which logarithmic functions occur. (Ex 6.3, Stewart)
33. Solve equations in which logarithmic functions occur. (Ex 6.3, Stewart)
34. Define the concepts radial measure and one radian and draw sketches to elucidate
the definitions.
35. Do conversions between radians and degrees. (Appendix D, Stewart)
36. Why is the concept radial measure developed?
37. Define the functions sin x and cos x and draw a sketch to elucidate the definitions.
38. Define the functions tan x, csc x, sec x and cot x (domain included).
39. Draw sketches for the functions sin x, cos x, tan x, csc x, sec x and cot x on the
interval [−2π, 2π].
40. Solve equations and inequalities in which trigonometric functions occur. (Appendix
D, Stewart)
41. Define the functions sin−1 x, cos−1 x, tan−1 x, csc−1 x, sec−1 x and cot−1 x (domain
included).
42. Draw sketches of the functions sin−1 x, cos−1 x and tan−1 x indicate on the sketches
that it is the inverse function of sin x, cos x and tan x respectively.
43. Determine values of inverse trigonometric functions in fixed points. (Ex 6.6, Stewart)
44. Solve equations in which inverse trigonometric functions occur. (Ex 6.6, Stewart)
6
1.1
Functions
Solving of Inequalities
(45 minutes contact time)
You should study Appendix A (Stewart) and Chapter 3 (Mathematics Review Manual)
together with this section.
The following are examples of the solutions of some inequalities.
Example 1: Which values of x satisfy the inequality given below?
1
>2
x−1
To solve this inequality, the two cases x − 1 > 0 and x − 1 < 0 should be examined.
1
>2
x−1
x − 1 > 0 and 1 > 2(x − 1)
3
⇒ x > 1 and x <
2
3
⇒ 1<x<
2
or
1
x − 1 < 0 and
>2
x−1
x − 1 < 0 and 1 < 2(x − 1)
3
⇒ x < 1 and x >
2
⇒ x∈∅
x − 1 > 0 and
The values of x satisfying the given inequality, are therefore
1<x<
3
or x ∈ (1, 3/2).
2
This inequality can also be solved as follows.
1
>2
x−1
1
⇒
−2>0
x−1
3 − 2x
⇒
>0
x−1
This inequality will be satisfied if both (3 − 2x) and (x − 1) is positive and negative,
therefore for
3
1 < x < or x ∈ (1, 3/2).
2
This inequality can further also be solved as follows.
1
>2
x−1
7
Functions
1
· (x − 1)2 > 2(x − 1)2
x−1
⇒ x − 1 > 2x2 − 4x + 2
⇒ 2x2 − 5x + 3 < 0
⇒ (2x − 3)(x − 1) < 0
⇒
This inequality will be satisfied if (2x − 3) and (x − 1) have different signs, therefore for
1<x<
3
or x ∈ (1, 3/2).
2
Example 2: Which values of x satisfy the inequality given below?
1≤
1
<5
x−4
This inequality is a combination of the two inequalities
1≤
1
1
and
<5
x−4
x−4
where the two cases x − 4 > 0 and x − 4 < 0 have to be considered.
1
First consider the inequality 1 ≤ x−4
.
1
x−4
x − 4 > 0 and x − 4 ≤ 1
⇒ x > 4 and x ≤ 5
⇒ 4<x≤5
or
1
x − 4 < 0 and 1 ≤
x−4
x − 4 < 0 and x − 4 ≥ 1
⇒ x < 4 and x ≥ 5
⇒ x∈∅
x − 4 > 0 and 1 ≤
The values of x satisfying the inequality 1 ≤
1
x−4
4 < x ≤ 5.
Now consider the inequality
1
x−4
are therefore
(1)
< 5.
1
<5
x−4
21
⇒ x > 4 and x >
5
21
⇒ x>
5
x − 4 > 0 and
8
Functions
or
1
<5
x−4
21
⇒ x < 4 and x <
5
⇒ x<4
x − 4 < 0 and
The values of x satisfying the inequality
1
x−4
x < 4 or x >
< 5 are therefore
21
.
5
(2)
Inequalities (1) and (2) have to be satisfied simultaneously. The values of x satisfying the
inequality are therefore
21
< x ≤ 5 or x ∈ (21/5, 5].
5
This inequality can also be solved as follows.
1
<5
x−4
1
1
and
<5
1≤
x−4
x−4
1
1
1−
≤ 0 and
−5 <0
x−4
x−4
21 − 5x
x−5
≤ 0 and
<0
x−4
x
−
4
21
4 < x ≤ 5and x < 4 or x >
5
21
<x≤5
5
1≤
⇒
⇒
⇒
⇒
⇒
This inequality can further also be solved as follows.
1
<5
x−4
1
1
1≤
and
<5
x−4
x−4
(x − 4)2 ≤ x − 4 and x − 4 < 5(x − 4)2
x2 − 9x + 20 ≤ 0 and 0 < 5x2 − 41x + 84
(x − 4)(x − 5) ≤ 0 and 0 < (5x − 21)(x − 4)
21
4 < x ≤ 5 and x < 4 or x >
5
21
<x≤5
5
1≤
⇒
⇒
⇒
⇒
⇒
⇒
You have to give the solutions of inequalities in terms of interval and set notation and
represent it on a number line as summarised in Table 1 (Appendix A, Stewart). Study
Examples 1, 2, 3 and 4 (Appendix A, Stewart).
9
Functions
1.2
Absolute Value Function
(1 hour 30 minutes contact time)
You should study Appendix A (Stewart) and Chapter 3 (Mathematics Review Manual)
together with this section.
Definition 1
The absolute value function is defined as follows:
x,
x≥0
|x| =
−x, x < 0
y = |x|
Study Examples 5 and 6 (Appendix A, Stewart).
Properties of absolute value functions that follows from the definition are:
Theorem 1
|a|2
|a||b|
|a|
|b|
| − a|
= a2
= |ab|
a
=
b
= |a|
From the above it also follows that
Theorem 2
Proof:
Similarly
√
x2 = |x| and −
√
x2 =
√
x2 = −|x|.
x,
x≥0
= |x|.
−x, x < 0
√
− x2 = −|x|.
✷
Below are some theorems on inequalities and absolute values.
10
Functions
Theorem 3
|x| ≤ a ⇔ −a ≤ x ≤ a, a ≥ 0
Proof:
Suppose |x| ≤ a. Then
(x ≤ a if x ≥ 0) or (−x ≤ a if x < 0)
⇒ (x ≤ a if x ≥ 0) or (x ≥ −a if x < 0)
⇒ (0 ≤ x ≤ a) or (−a ≤ x < 0)
⇒ −a ≤ x ≤ a.
Suppose −a ≤ x ≤ a. Then
−a ≤ x and x ≤ a
⇒ (−a ≤ x < 0) or (0 ≤ x ≤ a)
⇒ (−x ≤ a if x < 0) or (x ≤ a if x ≥ 0)
⇒ |x| ≤ a.
✷
Theorem 4
|x| ≥ a ⇔ x ≤ −a or x ≥ a
Study Example 7 and 8 (Appendix A, Stewart).
Theorem 5
|x + y| ≤ |x| + |y|
Proof: For x and y we have that
−|x| ≤ x ≤ |x| and − |y| ≤ y ≤ |y|
⇒ −(|x| + |y|) ≤ x + y ≤ |x| + |y|
⇒ |x + y| ≤ |x| + |y|
according to Theorem 3.
Study Example 9 (Appendix A, Stewart).
The following theorems are given for completeness.
✷
Theorem 6
|x − y| ≤ |x| + |y|
Proof:
|x − y| = |x + (−y)| ≤ |x| + | − y| = |x| + |y|
according to Theorem 5.
✷
11
Functions
Theorem 7
||x| − |y|| ≤ |x − y|
Proof:
|x| = |x − y + y| ≤ |x − y| + |y|
according to Theorem 5. From this it follows that
|x| − |y| ≤ |x − y|.
(1)
Similarly it follows that
|y| = |y − x + x| ≤ |y − x| + |x| ⇒ |y| − |x| ≤ |y − x|
⇒ −(|x| − |y|) ≤ |x − y|
⇒ −|x − y| ≤ |x| − |y|.
(2)
From (1) and (2) it follows that
−|x − y| ≤ |x| − |y| ≤ |x − y| ⇒ ||x| − |y|| ≤ |x − y|
according to Theorem 3.
✷
Theorem 8
||x| − |y|| ≤ |x + y|
Proof:
||x| − |y|| = ||x| − | − y|| ≤ |x − (−y)| = |x + y|
according to Theorem 7.
Theorems 5, 6, 7 and 8 may be summarised as
✷
||x| − |y|| ≤ |x ± y| ≤ |x| + |y|.
1.3
Functions and Operations with Functions
(1 hour 30 minutes contact time)
You should study Sections 1.1 and 1.3 (Stewart) and Chapter 5 (Mathematics Review
Manual) together with this section.
Definition 2
A function f : A → B is a ruling whereby for each element in set A exactly one element
is assigned in set B.
Notation: f : A → B is the function f with domain A and range B.
12
Functions
Elementary functions:
c ∈ R,
1. Constant function: f (x) = c,
2. Power function: f (x) = xn ,
n ∈ N,
3. Sine function: f (x) = sin x,
x∈R
4. Cosine function: f (x) = cos x,
x∈R
x∈R
x∈R
5. Exponential function: f (x) = ax ,
a > 0,
x∈R
Definition 3
The domain of a function refers to all points where the function is defined.
• Natural domain: refers to all possible points where the function is defined, for instance
if
x+6
f (x) =
(x − 1)(x + 2)
then
Df = {x ∈ R : x 6= 1, x 6= −2}
= (−∞, −2) ∪ (−2, 1) ∪ (1, ∞).
• Restricted (limiting) domain: the domain is restricted (limited) to only part, or
parts, of the natural domain, for instance if
g(x) =
x+6
,
(x − 1)(x + 2)
x≥3
then
Dg = {x ∈ R : x ≥ 3}
= [3 , ∞) .
NOTE: Although f and g are identical expressions, they are different functions because
their domains differ.
Definition 4
The number f (x) is the value of the function f at the point x.
Definition 5
The range of a function is all possible values of the function for all points in the domain.
13
Functions
Example: Consider a balloon with radius r = 10 cm. The volume of the balloon is given
by
V
4
π(r)3
3
4
π(10)3
=
3
= 4188.8 cm3 .
=
If the balloon should slowly deflate, the volume V decreases proportional to the radius r.
Note that 0 ≤ r ≤ 10 and V (10) = 4188.8, V (0) = 0 and V (5) = 523.6 etc.
1. DV = [0, 10] is therefore the domain of V by the physical restriction.
2. The value of V at 5 is given by V (5) = 34 π53 = 523.6.
3. The set {V (r) : r ∈ DV } is called the range of V .
Also note that at any given instant t there is only one value for the radius of the balloon.
Therefore r is a function of time. The formula for this function is not known and we write
r = r(t). Because of this the volume is naturally also a function of time,
4
V (t) = π[r(t)]3 .
3
Study Example 1, 2 and 6 (Section 1.1, Stewart) and Examples 1, 2, 3, 4 and 5 (Section
1.3, Stewart).
Definition 6
The function f is an even function if f (−x) = f (x) for all x ∈ Df .
Definition 7
The function f is an odd function if f (−x) = −f (x) for all x ∈ Df .
Definition 8
Let f and g be two functions. Define
(a) the sum f + g as (f + g)(x) = f (x) + g(x) where Df +g = Df ∩ Dg .
(b) the difference f − g as (f − g)(x) = f (x) − g(x) where Df −g = Df ∩ Dg .
(c) the product f · g as (f · g)(x) = f (x)g(x) where Df ·g = Df ∩ Dg .
(d) the quotient f /g as (f /g)(x) =
f (x)
g(x)
where Df /g = {x ∈ Df ∩ Dg : g(x) 6= 0}.
(e) the composite f ◦g as (f ◦g)(x) = f (g(x)) where Df ◦g = {x ∈ R : x ∈ Dg and g(x) ∈
Df }.
Study Examples 6, 8 and 9 (Section 1.3, Stewart).
Example: Consider the functions f (x) and g(x), where
f (x) = 2x + 1,
2 ≤ x < 5,
14
Functions
g(x) =
Determine f ◦ g and Df ◦g .
Solution:
√
x − 3.
√
(f ◦ g)(x) = f (g(x)) = 2 x − 3 + 1
Df ◦g = {x ∈ R : x ∈ Dg and g(x) ∈ Df }
√
⇒ x ≥ 3 and 2 ≤ x − 3 < 5
4 ≤ x − 3 < 25
7 ≤ x < 28
From this it follows that
Df ◦g = {x ∈ R : x ≥ 3 and 7 ≤ x < 28} = {x ∈ R : 7 ≤ x < 28} = [7 , 28) .
Study Example 7 (Section 1.3, Stewart).
Note that f n (x) = [f (x)]n = f (x)f (x) . . . f (x) (n factors) if n is a positive integer, but
1
f −1 (x) 6= f (x)
- The notation f −1 is used for the inverse of f . Use the notation
(1/f )(x) =
1
f (x)
for the reciprocal of the function. Inverse functions are discussed in Section 1.4.
A piecewise defined function is a function defined in ”pieces”, for example
2x − 5,
x ≤ 4,
f (x) =
2
3x − 7x, x > 4.
Study Example 7, 8 and 9 (Section 1.1, Stewart).
1.4
Inverse Functions
(45 minutes contact time)
You should study Section 6.1 (Stewart) together with this section.
A function is given in the form y = f (x), for instance y = tan x, which gives the ratio of
two sides of a rectangular triangle for a given angle. The function f (x) gives the value y
for a value of x, for instance tan(π/4) = 1. The inverse is sometimes required: for which
x will the function y assume a particular value? For instance, what should the angle be
for the ratio between the two sides to be 1? For this case the inverse process should be
performed and x should be treated as a function of y.
Definition 9
A Function f is a one-to-one function if it never assumes the same value more than
once. Therefore f (x1 ) 6= f (x2 ) for x1 6= x2 .
Study Examples 1 and 2 (Section 6.1, Stewart).
15
Functions
Definition 10
Let f : A → B be a one-to-one function. The inverse function f −1 : B → A is defined
by
f −1 (y) = x ⇔ f (x) = y
for any y ∈ B.
Study Example 3 (Section 6.1, Stewart).
The inverse function gives the mirror image about the line y = x. The function f has to
be a one-to-one function so that f −1 can also be a function. An inverse function can also
be defined as follows.
Definition 11
If f and g are two functions such that
f (g(x)) = x for all x ∈ Dg
and
g(f (x)) = x for all x ∈ Df ,
then f is called the inverse of g and g is called the inverse of f . The notation g = f −1
and f = g −1 is normally used.
Study Examples 4 and 5 (Section 6.1, Stewart).
Definition 10 is used to determine the inverse of a function and Definition 11 is used to
test whether two functions are the inverse of each other.
1.5
Exponential Function
(1 hour 15 minutes contact time)
You should study Section 6.2 (Stewart) and Chapter 7 (Mathematics Review Manual)
together with this section.
Exponential functions describe processes of growth and decay, such as bacterial growth,
the discharge of capacitors and radioactive decay. Logarithmic functions give the opposite.
Definition 12
The exponential function is
f (x) = ax ,
a > 0,
x ∈ R.
In the exponential function, the exponent of the function is the variable. This is opposite
to the power function f (x) = xn where the base is the variable.
16
Functions
The graph of y = ax , a > 1 is given below.
y
y = ax
1
x
1
The graph of y = ax , 0 < a < 1 is given below.
y
y = ax
1
x
1
From the graphs above it is evident that
f (0) = a0 = 1 for x = 0,
f (x) = ax = 1 for a = 1.
In Figure 3 (Section 6.2, Stewart) a range of exponential functions are illustrated. Study
Example 1 (Section 6.2, Stewart).
Theorem 9
Properties of exponential functions:
ax ay
ax
ay
x y
(a )
(ab)x
= ax+y ,
= ax−y ,
= axy ,
= ax bx .
17
Functions
The natural exponential function is
f (x) = ex
where e ≈ 2.718281828 is the number so that the gradient of the curve in the point x = 0
is one.
1.6
Logarithmic Function
(1 hour 15 minutes contact time)
You should study Section 6.3 (Stewart) and Chapter 7 (Mathematics Review Manual)
together with this section.
Definition 13
The logarithmic function f −1 (x) = loga x, a 6= 1, x > 0 is the inverse function of the
exponential function f (x) = ax , a 6= 1, x ∈ R.
From the definition of the inverse function it follows that
Theorem 10
loga x = y ⇐⇒ ay = x, x > 0,
loga (ax ) = x, x ∈ R,
aloga x = x, x > 0.
Study Example 1 (Section 6.3, Stewart).
The graph of y = loga x, a > 1 is given below.
y
y = ax
1
x
1
y = loga x
18
Functions
The graph of y = loga x, 0 < a < 1 is given below.
y
y = ax
1
x
1
y = loga x
From the graphs it is evident that
f (1) = loga 1 = 0 for x = 1.
In Figure 2 (Section 6.3, Stewart) a range of logarithmic functions are illustrated. Study
Example 8 (Section 6.3, Stewart).
Theorem 11
The properties of logarithmic functions:
loga x + loga y = loga xy,
x
loga x − loga y = loga
,
y
loga (xr ) = r loga x,
loga a = 1,
loga 1 = 0,
logb x
loga x =
.
logb a
Study Examples 2, 4, 5, 6 and 7 (Section 6.3, Stewart).
The natural logarithmic function
ln x = loge x
is the inverse of the natural exponential function ex , and consequently
ln x = y ⇐⇒ ey = x, x > 0,
ln(ex ) = x, x ∈ R,
eln x = x, x > 0,
ln e = 1,
ln 1 = 0,
ln x
loga x =
.
ln a
This is the logarithmic function of which the gradient at the point x = 1 is equal to 1.
19
Functions
1.7
Radian Measure
(30 minutes contact time)
You should study Appendix D (Stewart) and Chapter 6 (Mathematics Review Manual)
together with this section.
There are two standard measuring systems for the measurement of angular quantities,
namely measuring in degrees and measuring in radians. You are well familiar with
measuring angles in degrees. In radian measure the angle size is given by the length of
an arc on a circle with unit radius, which subtends the angle at the centre of the circle.
The reason why radian measure is usually used in calculus when dealing with angles, is
because many of the formulas are simplified by the use of radians rather than degrees.
For convenience, therefore, in calculus, the concept of radian measure is introduced and
the trigonometric functions are defined on subsets of the set of real numbers.
|a|
r
θ
r
Definition 14
Suppose an angle θ is subtended at the centre of a circle with radius r by an arc of length
|a| (see the sketch). The real number ar is called the radian measure of the angle θ. The
radius r is always taken as positive. The number a is taken as positive (so that ar > 0)
if the angle is measured anti-clockwise and negative ( ar < 0) if the angle is measured
clockwise.
Definition 15
An angle of one radian is that angle which is subtended at the centre of the circle by an
arc of length (measured anti-clockwise) equal to the radius of the circle.
If θ = 360◦ , then the arc length is a = 2πr and consequently θ = a/r = 2πr/r = 2π.
π
π
Therefore 2π rad = 360◦ , π rad = 180◦ , rad = 90◦ = ,
rad = 1◦ , etc.
2
180
Remark: Please note that if we should choose r = 1 in the above definition, the radian
measure of the angle would be exactly equal to a. Usually it is convenient to use unity
circles. The length measurements are of course expressed in terms of some fixed scale.
Study Examples 1 and 2 (Appendix D, Stewart).
20
1.8
Functions
Trigonometric Functions
(1 hour 15 minutes contact time)
You should study Appendix D (Stewart) and Chapter 6 (Mathematics Review Manual)
together with this section.
Pt (xt , yt )
|t|
1
t
O
1
A
Definition 16
To define the trigonometric functions sin and cos, consider a circle with centre on
the origin of a set of axes and with unit radius. (See the sketch given above). Let t be
a real number. Now measure the distance |t| from the point A in the sketch along the
circumference of the circle (anti-clockwise if t > 0 and clockwise if t < 0) and mark the
point Pt on the circumference. Suppose Pt is the point (xt ; yt). The functions sin and cos
are then defined as sin t = yt and cos t = xt for all real numbers t.
If now, for instance, 2nπ < t < 2(n + 1)π, then the distance t − 2nπ on the arc, from
A to Pt = (xt ; yt ), is measured anti-clockwise. One can see from the definition (and
from the sketch) that if the angle θ has radian measure t, then sin t = yt = sin θ and
cos t = xt = cos θ. These two functions are defined on R with range [−1, 1] in both cases.
The other trigonometric functions are then defined as follows on subsets of R:
Definition 17
tan x =
sin x
π
, x 6= (2n + 1) , n ∈ Z
cos x
2
cot x =
cos x
, x 6= nπ, n ∈ Z
sin x
sec x =
1
π
, x 6= (2n + 1) , n ∈ Z
cos x
2
csc x =
1
, x 6= nπ, n ∈ Z
sin x
Study Examples 3, 4 and 5 (Appendix D, Stewart).
All identities that were taught in school are still valid in this “new” context of radian
measure. Some of them can be derived directly from the sketch of the unity circle, by
their definitions and the rest are proved as was done in secondary school. You are assumed
to be familiar with all the identities. Nevertheless, a revision of the identities as well as
their proofs are given in Appendix D (Stewart). Study Example 6 (Appendix D, Stewart).
In Appendix D (Stewart) the graphs of these functions are also presented and discussed.
You are expected to be able to sketch the graphs of all the trigonometric functions.
21
Functions
1.9
Inverse Trigonometric Functions
(45 minutes contact time)
You should study Section 6.6 (Stewart) together with this section.
As functions have to be one-to-one for inverse functions to be determined, the trigonometric functions have in certain cases to be restricted to certain intervals when inverse
trigonometric functions are determined.
1
Notation: For the inverse sin use sin−1 x or bgsinx or arcsinx. For
use (sin x)−1 .
sin x
Definition 18
(a) The function f −1 (x) = arcsin x, x ∈ [−1, 1] is the inverse function of f (x) = sin x,
x ∈ [−π/2, π/2].
y
f −1 (x) = arcsin x
π
2
f (x) = sin x
1
x
− π2 −1
π
2
1
−1
− π2
(b) The function g −1(x) = arccos x, x ∈ [−1, 1] is the inverse function of g(x) = cos x,
x ∈ [0, π].
y
π
g −1 (x) = arccos x
π
2
1
g(x) = cos x
x
−1
1
−1
π
2
π
22
Functions
(c) The function h−1 (x) = arctan x, x ∈ (−∞, ∞) is the inverse function of h(x) = tan x,
x ∈ (− π2 , π2 ).
y
π
2
1
h−1 (x) = arctan x
x
− π2 −1
π
2
1
−1
− π2
h(x) = tan x
(d) The function j −1 (x) = arccsc x, x ∈ (−∞, −1] ∪ [1, ∞) is the inverse function of
].
j(x) = csc x, x ∈ (0, π2 ] ∪ (π, 3π
2
y
3π
2
j(x) = csc x
j −1 (x) = arccsc x
π
π
2
1
x
− 3π
2
−π
− π2 −1
1
−1
− π2
−π
− 3π
2
π
2
π
3π
2
23
Functions
(e) The function k −1 (x) = arcsec x, x ∈ (−∞, −1] ∪ [1, ∞) is the inverse function of
).
k(x) = sec x, x ∈ [0, π2 ) ∪ [π, 3π
2
y 3π
2
k(x) = sec x
π
k −1 (x) = arcsec x
π
2
1
x
− 3π
2
−π
− π2 −1
π
π
2
1
−1
− π2
3π
2
−π
− 3π
2
(f ) The function m−1 (x) = arccot x, x ∈ (−∞, ∞) is the inverse function of m(x) =
cot x, x ∈ (0, π).
y
π
m−1 (x) = arccot x
m(x) = cot x
π
2
1
x
−π
−1
1
−1
π
2
π
−π
You are expected to be able to draw the graphs of arcsin x, arccos x and arctan x.
Study Examples 1 and 3 (Section 6.6, Stewart).
24
Functions
Study Unit 2
Limits of Functions
Time Allocation
16 hours 20 minutes (9 hours contact time and 2 tutorial sessions).
Prerequisite Study
1. Knowledge of polynomial, rational, root, piecewise defined, absolute value, exponential, logarithmic, trigonometric and inverse trigonometric functions.
2. Factorisation of polynomials and simplification of rational functions.
3. Solution of inequalities.
4. Study unit 1, MTHS111.
Study Outcomes
When you have completed this section, you should be able to do the following, using
standard mathematical notation.
1. Use the basic techniques for the calculation of limits (Theorems 12 and 13) of
polynomial functions, rational functions, root functions, piecewise defined functions,
absolute value functions, exponential and logarithmic functions, and trigonometric
and inverse trigonometric functions to calculate limits of functions.
2. Formulate and use the squeeze theorem (Theorem 14) to calculate limits of functions.
3. Formulate, prove and use the identities lim sinx x = 1 and lim cos xx−1 = 0 (Theorem
x→0
x→0
15) to calculate limits.
4. Define and use the concepts of vertical asymptote (Definition 19) and horizontal
asymptote (Definition 20) to find the vertical and horizontal asymptotes of functions.
5. Define and use limits of the general form lim f (x) = L (Definition 21), lim− f (x) =
x→a
x→a
L (Definition 22) and lim+ f (x) = L (Definition 23) to prove the existence of limits
x→a
of linear functions and root functions and the sum rule for limits (Theorem 16).
25
26
Limits of functions
References
Stewart: Sections 1.4, 1.5, 1.6, 1.7, 2.4, 3.4, 6.2, 6.3, 6.6.
Study guide: Sections 2.1, 2.2, 2.3, 2.4, 2.5, 3.2, 2.6.
Mathematics Review Manual: Chapter 8.
Unimaths Intro Workbook: Section 4.
Preparation
Read the Sections referred to in the textbook and in the study guide.
Work through the following examples from Stewart:
Section 1.4: 1, 2, 3;
Section 1.5: 1, 2, 3, 4, 5, 6;
Section 1.6: 1, 2;
Section 3.4: 1, 2, 3;
Section 6.2: 1;
Section 6.3: 8;
Section 6.6: 4;
Section 1.7: 1.
Answer the following questions and submit your answers at the beginning of the first
contact session of this study unit:
1. Draw the following graphs and say what happens with the value of the graph when
the value of x approaches the following values.
(i)
(ii)
(iii)
(iv)
(v)
2
0 from the right hand side
0 from the left hand side
∞
−∞
(a)
(b)
(c)
(d)
(e)
y
y
y
y
y
= 2x − 1
= 3x3
1
= x−1
= 3x
= log2 x
Exercising of Skills
Work through the following examples from Stewart:
Section 1.5: 7, 8, 9, 10;
Section 1.6: 3, 4, 5, 6, 7, 8, 9, 11;
Section 2.4: 5, 6;
Section 3.4: 4, 5, 6, 7, 8, 9, 10;
Section 6.2: 6;
Section 1.7: 2, 3.
Limits of functions
27
Do the following exercises:
Stewart (8th edition):
Ex. 1.5: 1 - 9, 11 - 18, 29 - 39;
Ex. 1.6: 1, 2, 10, 11 - 32, 35, 37 - 39, 41 - 47, 49, 50, 52;
Ex. 2.4: 39 - 50;
Ex. 3.4: 4, 7, 17, 21, 35, 36, 40;
Ex. 6.2: 23 - 30;
Ex. 6.3: 47 - 51;
Ex. 6.6: 43 - 46;
Ex. 1.7: 1 - 4, 15, 17, 19, 22, 23, 24, 28.
Stewart (7th edition):
Ex. 1.5: 1 - 9, 11 - 18, 29 - 37;
Ex. 1.6: 1, 2, 10, 11 - 32, 35, 37 - 39, 41 - 50;
Ex. 2.4: 39 - 48;
Ex. 3.4: 4, 7, 17, 19, 33, 34, 38;
Ex. 6.2: 23 - 30;
Ex. 6.3: 47 - 51;
Ex. 6.6: 43 - 46;
Ex. 1.7: 1 - 4, 15, 17, 19, 21, 23, 24, 28.
Test Yourself
1. Describe in your own words what the following symbols mean and draw sketches to
elucidate the descriptions.
(a) lim f (x)
x→a
(b) lim+ f (x)
x→a
(c) lim− f (x)
x→a
2. What do you know when the value obtained at question 1b is equal to the value
obtained at question 1c?
3. Name a few practical problems that can be solved using limits.
4. What is the difference between the determination of a limit in a point and the
calculation of a function value in a point?
5. In what does the ”power” or value of limits lie?
6. Determine limits of functions graphically. (Ex 1.5, Stewart)
7. Limits can be calculated numerically by substituting values closer and closer to the
point in which the limit must be determined. Give an example of a limit for which
it does not work.
8. When is a limit said to exist?
9. Give the seven limit rules (theorems) which can be used to determine limits with
finite values.
28
Limits of functions
10. Consider two limits with infinite values (∞). Which of the limit rules (theorems)
in question 9 can not be applied here?
11. How do we interpret limits of which the answer is ±∞?
12. Consider two limits of which the value is zero. Which of the limit rules (theorems)
in question 9 can not be applied here?
13. How must limits of polynomials for x → ±∞ be handled?
14. What has to be taken in consideration when limits of polynomials in root functions
√
(for example square root) for x → ±∞ are determined, for example lim
x6 /x3 ?
x→±∞
15. How must limits of rational functions be handled when the nominator as well as the
2 +x−2
denominator approach zero, for example lim x x−1
?
x→1
16. How must limits of the quotient of two functions of which the nominator as well
as the denominator approach
zero and the nominator contains one or more square
√
3−x−1
roots, for example lim x−2 , be handled?
x→2
17. How must the limit of a piecewise defined function at the point where the function
changes expression, be handled?
18. Determine the limits of polynomial, rational, root and piecewise defined functions,
if it exists. If it does not exist, explain why. Use the symbols ±∞ where necessary.
(Ex 1.4, 1.5 and 1.6, Stewart)
19. Consider the squeeze theorem and answer the following questions.
(a) What are the conditions of the theorem?
(b) What is the outcome of the theorem?
(c) Which kind of limits will this theorem be used for?
(d) Give an example of such a limit.
20. Determine limits of functions using the squeeze theorem. (Ex 1.6, Stewart)
21. Write down two limit identities that contain trigonometric functions.
22. Extend the limits named in question 21 to more general cases.
23. Determine limits using the identities in question 21. (Ex 2.4, Stewart)
24. Define the concepts horizontal and vertical asymptotes and draw sketched to elucidate the definitions.
25. Determine the horizontal and vertical asymptotes of functions and show the limit
calculations so that the behaviour of the function at the asymptotes is clear.
26. Determine the limits in which exponential and logarithmic functions occur, if it
exists. If it does not exist, explain why. Use the symbols ±∞ where necessary. (Ex
6.2 and 6.3, Stewart)
27. Determine the limits in which trigonometric and inverse trigonometric functions
occur, if it exists. If it does not exist, explain why. Use the symbols ±∞ where
necessary. (Ex 6.6, Stewart)
29
Limits of functions
28. Define the concepts limit, left limit and right limit and draw sketches to elucidate
the definitions.
29. Use the definitions named √
in question 28 and prove
√ the existence of limits of the
form lim (bx+c) = L, lim+ bx + c = L and lim− bx + c = L. (Ex 1.7, 7th edition
x→a
x→a
x→a
or Ex 2.4, 6th edition of Stewart)
30. Use the definition of a limit and prove the sum rule for limits.
30
2.1
Limits of functions
Informal Definition of the Limit of a Function
(2 hours 30 minutes contact time)
You should study Sections 1.4, 1.5 and 1.6 (Stewart), Chapter 8 (Mathematics Review
Manual) and Section 4 (Unimaths Intro Workbook) together with this section.
The limit limx→a f (x) is a number which the function value approaches when the value
of the variable x approaches the value a. However, this number is independent of the
value f (a) of the function at the point a. In order to facilitate the investigation about the
existence and value of the limit, we introduce the concepts limit from the left (left
limit) and limit from the right (right limit). The following is often used:
Theorem 12
lim f (x) = L ⇐⇒ lim+ f (x) = lim− f (x) = L.
x→a
x→a
x→a
This theorem states that a limit only exists if it has a finite value (L ∈ R and not ±∞)
and if it has the same value from both sides. Left limits and right limits exist if they have
finite values (L ∈ R and not ±∞).
Suppose that the equation of a curve is given by y = f (x). The tangent to the curve at
the point P = (a, f (a)) on the curve (if it exists) may be described as follows: Consider
another point Q = (x, f (x)) on the curve. The straight line through the points (a, f (a))
and (x, f (x)) is called a line of intersection. If x approaches a (so that Q on the curve
approaches P ), the line of intersection will rotate about the point P , up to a limit. The
line corresponding to this limit is called a tangent to the curve at the point P . Study
Examples 1, 2 and 3 (Section 1.4, Stewart).
Let y = f (x) be the equation of a curve above the x axis, such that a ≤ x ≤ b. Suppose
that we need to determine the area under the curve. This surface area can be approximated by choosing a finite number of inscribed rectangles, all of the same width, and
adding their surface areas. The more such inscribed rectangles are used, the better the
approximation, because they will tend to fill the area under the curve ever closer. In the
limit, when the number of rectangles tend to infinity, the real value of the surface area is
obtained.
sin x
In Example 3 (Section 1.5, Stewart) the limit lim
is investigated numerically, by
x→0 x
alternatively choosing positive and negative values of x ever closer to 0. From the table
it is clear that
sin x
sin x
lim−
= 1 = lim+
.
x→0
x→0
x
x
Please note that this numerical treatment still does not prove that the limit is truly 1.
This limit is proved in Section 2.4 (Stewart).
Examples 4 and 5 (Section 1.5, Stewart) serve as a warning not to rely too easily on
numerical calculations. Two of the most common causes why limits may not exist, are
oscillations and unboundedness of functions.
Study Examples 6 and 7 (Section 1.5, Stewart).
The rules for the calculations with limits of functions, (or their properties), are summarised in:
31
Limits of functions
Theorem 13
Let lim represent one of the limits lim , lim− , lim+ , lim or lim . If both lim f (x) = L1
x→a x→a
x→a
x→∞
x→−∞
and lim g(x) = L2 exist, with L1 , L2 ∈ R (finite values) and if k is a constant, then we
have that
(a) lim k = k.
(b) lim[kf (x)] = k lim f (x) = kL1 .
(c) lim[f (x) + g(x)] = lim f (x) + lim g(x) = L1 + L2 .
(d) lim[f (x) − g(x)] = lim f (x) − lim g(x) = L1 − L2 .
(e) lim[f (x)g(x)] = lim f (x) lim g(x) = L1 L2 .
lim f (x)
L1
f (x)
=
=
, provided that L2 6= 0.
(f ) lim
g(x)
lim g(x)
L2
1
1
1
(g) lim f (x) n = [lim f (x)] n = L1n , where L1 ≥ 0 for even n.
Study Examples 1, 2 and 4 (Section 1.6, Stewart).
Example: The absolute value function is defined by
|x| =
x, x ≥ 0
−x, x < 0
such that
lim+
x→0
|x|
=
x
=
x
x→0 x
lim+ 1
lim+
x→0
= 1
and
lim−
x→0
|x|
=
x
=
−x
x→0
x
lim− −1
lim−
x→0
= −1.
|x|
|x|
|x|
6= lim−
, the limit lim
does not exist in the point 0. Study
x→0
x→0
x→0
x
x
x
Examples 7 and 8 (Section 1.6, Stewart).
Seeing that lim+
32
Limits of functions
Consider the limits of
1. Polynomials
lim (y 6 − 12y + 1) = 14
y→−1
lim (c0 + c1 x + · · · + cn−1 xn−1 + cn xn )
c
c1
cn−1
0
n
= lim
+
+
·
·
·
+
+
c
n x
x→∞ xn
xn−1
x
∞,
cn > 0
= lim cn xn =
−∞, cn < 0
x→∞
where n is the degree (i.e. highest power) of the polynomial.
x→∞
lim (c0 + c1 x + · · · + cn−1 xn−1 + cn xn )
c
c1
cn−1
0
n
+
+
·
·
·
+
+
c
= lim
n x
x→−∞ xn
xn−1
x
∞,
cn > 0, n even



−∞, cn > 0, n odd
= lim cn xn =
x→−∞
−∞,
cn < 0, n even



∞,
cn < 0, n odd
where n is the degree (i.e. highest power) of the polynomial.
x→−∞
lim (x−2 + 3 + x3 ) = lim x3 = −∞
x→−∞
x→−∞
Study Examples 8 and 9 (Section 3.4, Stewart).
2. Rational functions
y 2 − 2y
3
lim
=
y→3 y + 1
4
x2 − 16
(x − 4)(x + 4)
= lim
= lim (x + 4) = 8
x→4 x − 4
x→4
x→4
x−4
lim
c0 + c1 x + · · · + cn xn
cn xn
=
lim
x→∞ d0 + d1 x + · · · + dm xm
x→∞ dm xm
where n is the degree of the polynomial in the numerator position and
m the degree of the polynomial in the denominator position.
lim
c0 + c1 x + · · · + cn xn
cn xn
=
lim
x→−∞ d0 + d1 x + · · · + dm xm
x→−∞ dm xm
where n is the degree of the polynomial in the numerator position and
m the degree of the polynomial in the denominator position.
lim
x
x3
x−2 + 3 + x3
=∞
= lim
= lim −
lim
x→−∞ −3x2
x→−∞
x→−∞ x−1 + 2x − 3x2
3
33
Limits of functions
NB It is important to note that
1
1
= 0 and lim
= 0.
x→∞ x
x→−∞ x
lim
Study Examples 3 and 5 (Section 1.6, Stewart). Also study Examples 3 and 10
(Section 3.4, Stewart).
3. Root functions
√
√
lim x3 − 3x − 1 = 109
x→5
√
√ 2
√ 2
√
√
3x4 + x
3x4
3|x |
3x
3
=
3
=
lim
=
lim
=
lim
=
lim
lim
x→∞ x2
x→∞
x→∞ x2
x→∞
x→∞ x2 − 8
x2
√
√
√
√
√
3x2 + x
3x2
3|x|
= lim
= lim
= lim (− 3) = − 3
lim
x→−∞
x→−∞
x→−∞
x→−∞
x−8
x
x
√
Study Example 6 (Section 1.6, Stewart). Also study Example 5 (Section 3.4, Stewart).
4. Piecewise defined functions
lim+ f (x) = 0 and lim− f (x) = −2 for f (x) =
x→0
x→0
x2 ,
x ≥ 0,
x − 2, x < 0.
Study Example 9 (Section 1.6).
In some cases it may be difficult to calculate the limit according to the rules of calculation
specified in Theorem 13. In such cases use can be made of the squeeze theorem.
Theorem 14
Squeeze theorem. Let f , g and h be functions and lim represents one of the limits lim ,
x→a
lim− , lim+ , lim or lim . Suppose an open interval I exists, such that f (x) ≤ g(x) ≤
x→a
x→a
x→∞
x→−∞
h(x) for all x ∈ I. If
lim f (x) = lim h(x) = L,
then
lim g(x) = L.
1
oscillates between -1 and 1. If x → 0, the function oscillates
x
1
ever more rapidly, so that lim sin does not exist. It is however possible to calculate
x→0
x
1
lim x sin by making use of the squeeze theorem.
x→0
x
Example: The function sin
−1 ≤ sin
1
≤1
x
34
Limits of functions
⇒
but
⇒
and
but
⇒
⇒
1
≤ x, x > 0
x
lim+ (−x) = 0 = lim+ x
−x ≤ x sin
x→0
x→0
1
lim+ x sin = 0
x→0
x
1
−x ≥ x sin ≥ x, x < 0
x
lim− (−x) = 0 = lim− x
x→0
x→0
lim− x sin
x→0
lim x sin
x→0
1
=0
x
1
=0
x
Study Example 11 (Section 1.6, Stewart).
2.2
Limit Identities
(50 minutes contact time)
You should study Section 2.4 (Stewart) together with this section.
For the trigonometric functions sin and cos we have lim sin θ = 0 and lim cos θ = 1.
θ→0
θ→0
Theorem 15
Limit identities
cos x − 1
= 0;
x→0
x
sin x
= 1;
x→0 x
lim
lim
ex − 1
= 1;
x→0
x
lim
lim (1 + x)1/x = e
x→0
sin x
= 1 is proved in Section 2.4 (Stewart). You do not have to know
x
sin x
this proof for examination purposes. The identity lim
= 1 is used to prove the
x→0 x
cos x − 1
= 0. This is proved in Section 2.4 (Stewart). You should be able
identity lim
x→0
x
ex − 1
to prove this. The identities lim
= 1 and lim (1 + x)1/x = e are proved in Section
x→0
x→0
x
4.6 of the Study guide.
These four identities are used in the calculation of other limits. Let lim represent any of
the limits lim , lim± or lim . If lim f (x) = 0, it follows that
The identity lim
x→0
x→±∞
x→a x→a
lim
sin f (x)
= 1;
f (x)
lim
cos f (x) − 1
= 0;
f (x)
lim
ef (x) − 1
= 1;
f (x)
It further follows that
−1
sin[f (x)]
f (x)
= lim
= 1−1 = 1
lim
sin[f (x)]
f (x)
lim[1+f (x)]1/f (x) = e
35
Limits of functions
and
lim
and
sin[f (x)]
1
tan[f (x)]
= lim
× lim
=1×1=1
f (x)
f (x)
cos[f (x)]
−1
f (x)
tan[f (x)]
lim
= lim
= 1−1 = 1.
tan[f (x)]
f (x)
Example:
sin(x2 − 4)
sin(x2 − 4)
= lim (x + 2)
x→2
x→2
x−2
(x − 2)(x + 2)
sin(x2 − 4)
= lim (x + 2) lim
x→2
x→2
x2 − 4
= 4.
lim
Study Examples 5 and 6 (Section 2.4, Stewart).
2.3
Asymptotes of Functions
(50 minutes contact time)
You should study Section 3.4 (Stewart) together with this section.
Vertical asymptotes of functions are vertical lines which a function approaches, but never
touches or intersects. Typical a function will approach ±∞ as is illustrated in Figure 14
(Section 1.5, Stewart).
Definition 19
The line x = x0 is a vertical asymptote of the graph of f if
lim f (x) = ∞, or lim− f (x) = ∞, or lim f (x) = ∞,
x→x+
0
or
x→x0
x→x0
lim f (x) = −∞, or lim− f (x) = −∞, or lim f (x) = −∞.
x→x+
0
x→x0
x→x0
The left- and right limits of a function at a vertical asymptote should be considered
separately, because they need not necessarily have the same values. Consider Examples
8, 9 and 10 (Section 1.5, Stewart).
A function has a horizontal asymptote if the function approaches a constant value when
x approaches ±∞. In this case it is possible that the function may intersect the line for
x values close to zero, as is shown in Figure 4 (Section 3.4, Stewart).
Definition 20
The line y = y0 is a horizontal asymptote of the graph of f if
lim f (x) = y0 or
x→∞
lim f (x) = y0 .
x→−∞
Study Examples 1, 2 and 4 (Section 3.4, Stewart).
36
2.4
Limits of functions
Limits of Exponential and Logarithmic Functions
(50 minutes contact time)
You should study Sections 6.2 and 6.3 (Stewart) together with this section.
Have a look at the graphs of the exponential and logarithmic functions in Study unit 1
again. From these graphs it follows that
lim ax = 0 and lim ax = ∞ for a > 1,
x→−∞
x→∞
x
lim a = ∞ and lim ax = 0 for 0 < a < 1,
x→−∞
x→∞
lim+ loga x = −∞ and lim loga x = ∞. for a > 1,
x→∞
x→0
lim loga x = ∞ and lim loga x = −∞. for 0 < a < 1,
x→0+
x
x→∞
x
lim e = 0 and lim e = ∞,
x→−∞
x→∞
lim+ ln x = −∞ and lim ln x = ∞.
x→0
x→∞
Study Examples 1 and 6 (Section 6.2, Stewart). Also study Example 8 (Section 6.3,
Stewart).
2.5
Limits of Trigonometric and Inverse Trigonometric Functions
(50 minutes contact time)
You should study Section 6.6 (Stewart) together with this section.
You have to know the graphs of trigonometric and inverse trigonometric very well to
be able to calculate limits containing trigonometric and inverse trigonometric functions.
Have a look at the graphs of the trigonometric and inverse trigonometric functions in
Study unit 1 again. Study Examples 6 and 7 (Section 3.4, Stewart). Also study Example
4 (Section 6.6, Stewart).
2.6
Formal Definition of the Limit of a Function
(3 hours 30 minutes contact time)
You should study Section 1.7 (Stewart) together with this section.
Before starting to prove any of the rules of Theorem 13, it is necessary that we should
formulate mathematically correct definitions of limits. When we calculate the value
of a limit, the estimated limit is still not sufficient proof that the limit does exist and
that the calculated value is indeed the true value of the limit.
However, before you start with the study of the formal definitions of limits and their
applications, it is important that you should first possess a good understanding of the
left- and right limit.
37
Limits of functions
Definition 21
The limit lim f (x) = L if for every ǫ > 0 a number δ > 0 exists, such that if 0 < |x − a| <
x→a
δ implies that |f (x) − L| < ǫ, and where f is defined on an open interval including
(a − δ, a) ∪ (a, a + δ). The number L is called the limit of the function f at the point a.
Definition 22
The limit lim− f (x) = L if for every ǫ > 0 a number δ > 0 exists, such that if a−δ < x < a
x→a
implies that |f (x) − L| < ǫ, and where f is defined on an open interval which includes
(a − δ, a). The number L is called the left limit of the function f at the point a.
Definition 23
The limit lim+ f (x) = L if for every ǫ > 0 a number δ > 0 exists, such that if a < x < a+δ
x→a
implies that |f (x) − L| < ǫ, and where f is defined on an open interval which includes
(a, a + δ). The number L is called the right limit of the function f at the point a.
Differential and integral calculus is based on these definitions. Other than the theoretical
use of the definitions in the proofs of limit theorems, they are also used to validate that an
estimated limit is truly the limit of a function at a particular point. A simple illustration
is presented below.
Example: Prove that
lim 5x = 15.
x→3
Solution: (Find a δ > 0 such that if 0 < |x − 3| < δ then |5x − 15| < ǫ.)
Now, take any ǫ > 0.
⇔
⇔
|5x − 15| < ǫ
5|x − 3| < ǫ
|x − 3| < ǫ/5
Proof: ∀ ǫ > 0, ∃ δ = ǫ/5 > 0 so that if
0 < |x − 3| < δ
⇒ 5|x − 3| < 5δ
⇒ |5x − 15| < ǫ
⇒ lim 5x = 15.
x→3
Study Examples 1 and 2 (Section 1.7, Stewart).
Example: Prove that
√
lim− 3 − x = 0.
x→3
√
Solution: (Find a δ > 0 such that if 3 − δ < x < 3 then | 3 − x − 0| < ǫ.)
Now, take any ǫ > 0.
√
| 3 − x − 0| < ǫ
38
Limits of functions
⇔
⇔
⇔
⇔
Proof: ∀ ǫ > 0, ∃ δ = ǫ2 > 0 so that if
⇒
⇒
⇒
⇒
⇒
⇒
√
0< 3−x<ǫ
0 < (3 − x) < ǫ2
−3 < −x < ǫ2 − 3
3 − ǫ2 < x < 3
3−δ <x<3
−δ < x − 3 < 0
0<3−x<δ
√
√
3−x< δ
√
√
3 − x < ǫ2
√
| 3 − x − 0| < ǫ
√
lim− 3 − x = 0.
x→3
1
1
= 0 and lim
= 0.
x→∞ x
x→−∞ x
Study Example 3 (Section 1.7, Stewart).
The sum rule for limits is proved by making use of Definition 21.
NB: It is important to note that lim
Theorem 16
If lim f (x) = L1 and lim g(x) = L2 both exist, then
x→a
x→a
lim [f (x) + g(x)] = lim f (x) + lim g(x) = L1 + L2 .
x→a
x→a
x→a
Proof:
Let ǫ > 0 be given.
(Find a δ > 0 such that if 0 < |x − a| < δ then |f (x) + g(x) − (L1 + L2 )| < ǫ.)
If lim f (x) = L1 exists, there exists for every ǫ > 0 a δ1 > 0 such that if
x→a
0 < |x − a| < δ1 then
ǫ
|f (x) − L1 | < .
(A)
2
If lim g(x) = L2 exists, there exists for every ǫ > 0 a δ2 > 0 such that if
x→a
0 < |x − a| < δ2 then
ǫ
|g(x) − L2 | < .
(B)
2
Take δ = min{δ1 , δ2 }, then both (A) and (B) are true for 0 < |x − a| < δ and
|f (x) + g(x) − (L1 + L2 )| = |f (x) − L1 + g(x) − L2 |
≤ |f (x) − L1 | + |g(x) − L2 |
ǫ
+ ǫ2
<
2
= ǫ
(triangular inequality)
Therefore
lim [f (x) + g(x)] = lim f (x) + lim g(x).
x→a
x→a
x→a
✷
Study Unit 3
Continuity
Time Allocation
6 hours 45 minutes (4 hours contact time and 1 tutorial session).
Prerequisite Study
Study units 1 and 2, MTHS111.
Study Outcomes
When you have completed this section, you should be able to do the following, using
standard mathematical notation.
1. Define and use the concepts continuity (Definitions 24, 25, 27, 28, 29 and 30) and
discontinuity (Definition 26) to determine the continuity and discontinuity of functions at points and on intervals.
2. Formulate, prove and use the continuity theorems (Theorem 18) to determine the
continuity and discontinuity of functions at points and on intervals.
3. Formulate and use the intermediate value function (Theorem 19) in the proofs of
theorems in the second semester and in problem solving.
4. Formulate and use Theorem 17 to calculate limits of functions of the form f (x)g(x) .
References
Stewart: Section 1.8.
Study guide: Sections 3.1.
Unimaths Intro Workbook: Section 5.
39
40
Continuity
Preparation
Read the Sections referred to in the textbook and in the study guide.
Work through the following examples from Stewart:
Section 1.8: 1, 2, 4.
Answer the following questions and submit your answers at the beginning of the first
contact session of this study unit:
1. Describe continuity and discontinuity in your own words.
2. Draw a function which is continuous.
3. Draw a functions which is discontinuous.
Exercising of Skills
Work through the following examples from Stewart:
Section 1.8: 5, 6, 7, 8, 9.
Do the following exercises:
Stewart (8th edition):
Ex. 1.8: 2 - 5, 11 - 47, 53.
Stewart (7th edition):
Ex. 1.8: 2 - 5, 11 - 46, 51.
Test Yourself
1. Define the following concepts.
(a) Continuous in a point (in two different ways).
(b) Discontinuous in a point.
(c) Left continuous in a point.
(d) Right continuous in a point.
(e) Continuous on an open interval.
(f) Continuous on a closed interval.
2. Draw graphs of functions that is discontinuous because of the following.
(a) The limit of the function does not exist in the point and the function does not
exist in the point.
(b) The limit of the function does not exist in the point, but the function exists in
the point.
(c) The limit of the function exists in the point, but the function does not exist in
the point.
(d) The limit of the function exists in the point and the function exists in the
point, but the limit is not equal to the function in the point.
Continuity
41
3. Name and draw three types of discontinuities and explain at each why it is a discontinuity.
4. Use the definitions in question 1 to prove the continuity and discontinuity of functions. (Ex 1.8, Stewart)
5. Write down the limit rule (theorem) to determine limits of composite functions.
6. Give an example of a limit of a composite function for which the rule in question 5
can not be used and explain why.
7. State the ten continuity rules (theorems).
8. Prove the first six rules of Theorem 18.
9. Use the continuity rules (theorems) named in question 7 to prove the continuity and
discontinuity of functions. (Ex 1.8, Stewart)
10. Consider the intermediate value theorem.
(a) What are the conditions of the theorem?
(b) What is the outcome of the theorem?
(c) Draw a sketch to illustrate the theorem.
(d) Draw a sketch to illustrate a case when there are more than one point on the
interval for which the theorem holds.
11. Use the intermediate value theorem to prove that functions have roots on certain
intervals. (Ex 1.8, Stewart)
12. Use the definition of continuity to determine expressions of piecewise defined functions so that the functions are continuous everywhere. (Ex 1.8, Stewart)
13. How can limits of the form f (x)g(x) be determined and why?
14. Determine limits of the form f (x)g(x) , if it exists. If it does not exist, explain why.
Use the symbols ±∞ where necessary.
42
3.1
Continuity
Continuity
(3 hours contact time)
You should study Section 1.8 (Stewart) and Section 5 (Unimaths Intro Workbook) together with this section.
The dynamics of physical objects usually have an unbroken line for their trajectory. Such
curves are called continuous. Mathematical modelling of physical phenomena (such as
the trajectory of an object) is an important aspect in science. It is therefore incumbent
that there should be a mathematically correct definition for the concept of a continuous
function.
Definition 24
A function f is called continuous at the point a if lim f (x) = f (a).
x→a
This definition implies that
(a) f (a) exists (f is defined at the point a),
(b) lim f (x) exists and
x→a
(c) lim f (x) = f (a).
x→a
Continuity can also be defined as follows:
Definition 25
A function f is called continuous at the point a if lim f (a + h) = f (a).
h→0
This definition is illustrated in Figure 1 (Section 1.8, Stewart).
Now that we have defined the concept of continuity, we can state and apply the limit
theorem for composite functions.
Theorem 17
If f is continuous at the point b and lim g(x) = b, then
x→a
lim f (g(x)) = f lim g(x) = f (b).
x→a
x→a
Example: Calculate
lim
x→1
√
2x − 1.
√
Solution: Let f (x) = x and g(x) = 2x−1. Then lim (2x−1) = 1 and f (x) is continuous
x→1
at the point 1. According to the limit theorem for composite functions we have that
q
√
lim 2x − 1 = lim (2x − 1) = 1.
x→1
x→1
Counter example: Calculate
lim
x→0
√
2x − 1.
43
Continuity
Solution: Let f (x) =
√
x and g(x) = 2x − 1. Then
lim (2x − 1) = −1.
x→0
But f (x) is not continuous at the point -1, because f (x) does not exist at the point -1.
In this case we cannot use the limit theorem for composite functions, else we get
q
√
√
lim 2x − 1 = lim (2x − 1) = −1
x→0
x→0
which is not defined.
Definition 26
If one or more of the conditions for continuity at a point a is not satisfied, then f is
discontinuous at the point a.
Example: Establish whether the function
( 2
f (x) =
x−1
2,
, if x 6= 1
if x = 1
is continuous at the point x = 1.
Solution:
•
f (1) = 2 and therefore it exists.
2
2
= +∞ and lim− f (x) = lim−
= −∞
•
lim+ f (x) = lim+
x→1 x − 1
x→1
x→1 x − 1
x→1
Therefore lim f (x) does not exist (left limit 6= right limit) and therefore the function is
x→1
discontinuous at the point x = 1.
Example: Determine whether the function

 x2 − 1
, if x 6= 1
g(x) =
x
−
1
 2,
if x = 1
is continuous or discontinuous at the point x = 1.
Solution:
•
g(1) = 2 and therefore it exists.
x2 − 1
•
lim+ g(x) = lim+
= lim+ (x + 1) = 2 and
x→1 x − 1
x→1
x→1
lim− (x + 1) = 2
lim− g(x) = lim−
x→1
x→1
x2 − 1
=
x−1
x→1
Therefore lim g(x) = 2 and it exists.
x→1
•
lim g(x) = 2 = g(1)
x→1
The function g is therefore continuous at the point x = 1.
Study Examples 1 and 2 (Section 1.8, Stewart).
One can test these conclusions by making use of the definition of continuity, but it is
much easier to make use of the following general formulations.
44
Continuity
Theorem 18
(a) If the function f is continuous at the point a and k is a constant, then kf is also
continuous at the point a.
(b) If the functions f and g are both continuous at the point a, then f +g is also continuous
at the point a.
(c) If the functions f and g are both continuous at the point a, then f −g is also continuous
at the point a.
(d) If the functions f and g are both continuous at the point a, then f ·g is also continuous
at the point a.
(e) If the functions f and g are both continuous at the point a, and we are given that
g(a) 6= 0, then f /g is also continuous at the point a.
(f ) If g is continuous at the point a and f is continuous at the point g(a), then f ◦ g is
continuous at the point a.
(g) Polynomials are continuous at every point x ∈ R.
(h) Rational functions are continuous at every point in their domain.
(i) Root functions are continuous at every point in their domain.
(j) Trigonometric functions are continuous at every point in their domain.
This theorem is proved by means of the limit theorems (Theorem 13 and Theorem 17).
You should be able to prove items (a) to (f) above.
Example: Prove Theorem 18(b).
Proof:
lim (f + g)(x) = lim [f (x) + g(x)] Definition 8(a)
x→a
x→a
= lim f (x) + lim g(x) Theorem 13(c)
x→a
x→a
= f (a) + g(a) = (f + g)(a)
because f and g are both continuous at the point a. Therefore f + g is continuous at the
point a.
✷
Example: Prove Theorem 18(f).
Proof:
lim (f ◦ g)(x) = lim f (g(x)) Definition 8(e)
x→a
x→a
= f (lim g(x)) Theorem 17
x→a
= f (g(a)) = (f ◦ g)(a)
because g is continuous at the point a.Therefore f ◦ g is continuous at the point a.
How do we use Theorem 18?
✷
45
Continuity
Example: Consider the function h(x) = |x|.
Then
x,
if x ≥ 0,
h(x) =
.
−x, if x < 0
Therefore h(x) = x for all x > 0 is a polynomial on the interval (0, ∞) and is therefore
continuous at every x > 0.
Similarly h(x) = −x for all x < 0 is a polynomial and is continuous at every x < 0.
Furthermore, lim h(x) = 0 = h(0), and therefore h is also continuous at 0.
x→0
Therefore h is continuous for every x ∈ R.
Example: If f (x) = | 7 − x3 |, let g(x) = 7 − x3 and h(x) = | x| so that f (x) = h(g(x)).
Then g is a polynomial and therefore continuous at every point x ∈ R and h is continuous
everywhere, also in every point g(x) = 7 − x3 .
Therefore f = h ◦ g is continuous at every x ∈ R.
Study Example 5 (Section 1.8, Stewart).
Definition 27
A function f is called left continuous at the point a if lim− f (x) = f (a).
x→a
Definition 28
A function f is called right continuous at the point a if lim+ f (x) = f (a).
x→a
Definition 29
A function f is called continuous on the open interval (a, b) if f is continuous at
every point x ∈ (a, b).
Definition 30
A function f is called continuous on the closed interval [a, b] if f is continuous on
(a, b), left continuous at the point b and right continuous at the point a.
Study Example 4 (Section 1.8, Stewart).
The continuity of a function on an interval can be proved from the definition of continuity,
or by making use of the continuity theorems.
Example : Make use of the continuity theorems (Theorem 18) to prove that
√
f (x) = 2x + 25 − x2
is continuous on its domain.
Solution: The domain of f is Df = [−5, 5].
Let
√
g(x) = 2x, h(x) = x, j(x) = 25 − x2
and
k(x) = (h ◦ j)(x) = h(j(x)).
The function j is a polynomial and is continuous at every point x ∈ R.
46
Continuity
The function h is a root function and is continuous at every point x ≥ 0.
From the above it follows that the function k is continuous at every point x ∈ [−5, 5].
The function g is a polynomial and is continuous at every point x ∈ R.
As f (x) = g(x) + k(x) is the sum of two functions which are both continuous at every
point x ∈ [−5, 5], the function f will also be continuous at every point x ∈ [−5, 5].
Example: Make use of the definition of continuity and the properties of limits (Theorems
13 and 17) and prove that
√
f (x) = 2x + 25 − x2
is continuous on its domain.
Solution: The domain of f is Df = [−5, 5].
Consider any a ∈ (−5, 5).
√
Then f (a) = 2a + 25 − a2 exists and
lim f (x) = lim (2x +
x→a
x→a
√
25 − x2 )
r
2
= 2 lim x + 25 − lim x
x→a
x→a √
= 2a + 25 − a2 = f (a)
and consequently the function f is continuous on the open interval (−5, 5).
Similar calculations illustrate that
√
lim− f (x) = lim− (2x + 25 − x2 ) = 10 = f (5)
x→5
x→5
and
lim + f (x) = lim + (2x +
x→−5
x→−5
√
25 − x2 ) = −10 = f (−5)
so that the function f is left continuous at the point 5 and right continuous at the point
−5.
From this it follows that the function f is continuous on the closed interval [−5, 5].
Study Examples 6, 7 and 8 (Section 1.8, Stewart).
Theorem 19
Intermediate value theorem. If f is continuous on the closed interval [a, b] and N is
any number between f (a) and f (b) with f (a) 6= f (b), then a number c ∈ (a, b) exists such
that f (c) = N.
Example: Show that the equation cos x = x has at least one root (solution) on the
interval (0, π).
Solution: Let f (x) = cos x − x. We have to show that f (x) has at least one root on
the interval (0, π). Put in another way, we have to show that f (x) = 0 has at least one
solution on the interval (0, π).
f (0) = cos 0 − 0 = 1
47
Continuity
f (π) = cos π − π = −1 − π
Because f (x) is continuous, there is at least one number c ∈ (0, π) so that f (c) = 0
because 0 ∈ (−1 − π, 1).
Study Example 9 (Section 1.8, Stewart).
Other applications of continuity:
Example: Determine the value of k for which the following piecewise defined function
will be continuous on R.
x − 3, x < 5
f (x) =
3x + k, x ≥ 5
Solution:
lim f (x) = f (5) = lim+ f (x)
x→5−
x→5
lim (x − 3) = 3(5) + k
x→5−
2 = 15 + k
k = −13
3.2
Limits of the form f (x)g(x)
(1 hour contact time)
A logarithm has the property to simplify calculations. Multiplication and division are
simplified to addition and subtraction:
a
log ab = log a + log b and log = log a − log b.
b
Taking a power of a number to multiplication:
log ab = b log a.
Since exponential and logarithmic functions are inverse functions of each other, it holds
that
g(x)
f (x)g(x) = eln f (x)
= eg(x) ln f (x) .
According to Theorem 17
lim f (x)g(x) = lim eg(x) ln f (x) = elim g(x) ln f (x)
if lim g(x) ln f (x) exists since exponential functions are continuous at all real values.
Example: Calculate
x
lim (sin x)e
x→π/2
(sin x)
ex
x
ln(sin x)e
=e
= ee
x
ln(sin x)
en lim ex ln(sin x) = 0
x→π/2
x
The function e is continuous at 0, therefore
x
lim (sin x)e = lim ee
x→π/2
x→π/2
x
ln(sin x)
lim ex ln(sin x)
= ex→π/2
= e0 = 1
48
Continuity
Study Unit 4
Differentiation
Time Allocation
26 hours (14 hours contact time and 3 tutorial sessions).
Prerequisite Study
Study units 1, 2 and 3, MTHS111.
Study Outcomes
When you have completed this section, you should be able to do the following, using
standard mathematical notation.
1. Define and use the concepts average rate of change (Definition 31), instantaneous
rate of change (Definition 32) and tangent to a curve at a point (Definition 33) in
problem solving.
2. Define and use the concept derivative of a function at a point (Definition 34) to
prove the differentiation rules (Theorem 23), determine the derivatives of functions
and in problem solving.
3. Define and use the concept of differentiability of a function (Definitions 35, 36, 37,
38, 39 and 40) to determine the differentiability of functions.
4. Formulate and use the theorems on differentiation (Theorems 20 and 22) to determine the differentiability of functions.
5. Formulate, prove and use the differentiation theorem (Theorem 21) to determine
the continuity of functions and prove Theorem 22, using standard mathematical
notation
6. Formulate, prove and use the differentiation rules (Theorem 23) for sums, products
and quotients of differentiable functions to determine the derivatives of functions
and in problem solving.
49
50
Differentiation
7. Formulate, prove and use the chain rule for composite, differentiable functions (Theorem 24) to determine the derivatives of functions and in problem solving.
8. Use the technique of implicit differentiation to determine the derivatives of implicit
functions and in problem solving.
9. Formulate, prove and use Theorem 25 to determine the derivatives inverse functions.
10. Formulate, prove and use the derivatives of exponential functions (Theorems 26 and
27) and logarithmic functions (Theorems 28 and 29) to determine the derivatives of
functions and in problem solving.
11. Use the technique of logarithmic differentiation to prove the power rule (Theorem
23(b)), determine the derivatives of functions and in problem solving.
12. Formulate, prove and use the derivatives of trigonometric functions (Theorem 30)
to determine the derivatives of functions and in problem solving.
13. Formulate, prove and use the derivatives of inverse trigonometric functions (Theorem 31) to determine the derivatives of functions and in problem solving.
14. Formulate and use L’Hôspital’s rule (Theorem 32) to calculate limits of functions.
References
Stewart: Sections 2.1, 2.2, 2.3, 2.4, 2.5, 2.6, 2.7, 6.1, 6.2, 6.4, 6.6, 6.8.
Study guide: Sections 4.1, 4.2, 4.3, 4.4, 4.5, 4.6, 4.7, 4.8, 4.9, 4.10, 4.11.
Mathematics Review Manual: Chapter 8.
Unimaths Intro Workbook: Section 7.
Preparation
Read the Sections referred to in the textbook and in the study guide.
Work through the following examples from Stewart:
Section 2.1: 1, 2, 3;
Section 2.2: 1, 2;
Section 2.3: 1, 2, 3, 6, 8;
Section 2.4: 1, 2;
Section 2.5: 1, 2;
Section 2.6: 1, 2;
Section 2.7: 1, 3, 5, 7;
Section 6.1: 6;
Section 6.2: 2, 3;
Section 6.4: 1, 2, 3, 4;
Section 6.6: 5;
Section 6.8: 1, 2, 3, 4, 5.
Answer the following questions and submit your answers at the beginning of the first
contact session of this study unit:
Differentiation
51
1. Determine the average change of f (x) = 2x − 1 over the interval [1, 3].
2. Determine the instantaneous change of f (x) = 2x − 1 in the point 4.
3. Determine the equation of the tangent to the function y = 2x2 − 3 in the point
(1, −1).
Exercising of Skills
Work through the following examples from Stewart:
Section 2.1: 4, 5, 6, 7;
Section 2.2: 3, 4, 5;
Section 2.3: 4, 5, 7, 9, 10, 11, 12, 13;
Section 2.4: 3, 4;
Section 2.5: 3, 4, 5, 6, 7, 8;
Section 2.6: 3, 4;
Section 2.7: 2, 4, 6, 8;
Section 6.1: 7;
Section 6.2: 4, 5;
Section 6.4: 5, 12, 13, 15, 16;
Section 6.6: 6;
Section 6.8: 6, 8, 9, 10.
Do the following exercises:
Stewart (8th edition):
Ex. 2.1: 1, 3, 5 - 8, 15, 16, 20, 21, 23, 27;
Ex. 2.2: 19, 21, 39 - 42, 57, 59, 62;
Ex. 2.3: 1 - 45, 63, 69 - 72, 75, 76, 79, 87(a), 97;
Ex. 2.4: 1 - 21, 33;
Ex. 2.5: 1 - 46, 61, 63, 68, 70, 71;
Ex. 2.6: 1, 5 - 24, 26;
Ex. 2.7: 1, 3, 29 - 31;
Ex. 6.1: 35, 37, 41, 43;
Ex. 6.2: 31 - 53;
Ex. 6.4: 2 - 27, 30, 37, 43 - 54;
Ex. 6.6: 19, 21 - 30;
Ex. 6.8: 15, 16, 47, 59, 60, 62, 67.
Stewart (7th edition):
Ex. 2.1: 1, 3, 5 - 8, 15, 16, 18, 19, 21, 23;
Ex. 2.2: 19, 20, 21, 35 - 38, 51, 53, 54;
Ex. 2.3: 1 - 45, 63, 67 - 70, 73, 74, 77, 85(a), 93;
Ex. 2.4: 1 - 21, 33;
Ex. 2.5: 1 - 46, 61, 63, 68, 70, 71;
Ex. 2.6: 1, 5 - 24, 27;
Ex. 2.7: 1, 3, 29 - 31;
Ex. 6.1: 35, 37, 41, 43;
52
Differentiation
Ex.
Ex.
Ex.
Ex.
6.2:
6.4:
6.6:
6.8:
31 - 53;
2 - 27, 30, 37, 43 - 54;
19, 21 - 30;
11, 13, 45, 57, 58, 60, 62.
Test Yourself
1. Define the following concepts and draw a sketch to elucidate the definitions.
(a) Average rate of change
(b) Gradient of a secant
(c) Instantaneous rate of change
(d) Gradient of a tangent
2. Which of the concepts in question 1 are determined by the same calculation?
3. With what does x in the expression of Definition 32 has to be replaced to get the
expression in Definition 33?
4. Use the definitions in question 1 to calculate the average and instantaneous change
of a physical phenomenon relative to a variable on which it is dependent. (Ex 2.1,
Stewart)
5. Define the following concepts and use it to determine the differentiability of functions. (Ex 2.2, Stewart)
(a) Derivative of a function
(b) Left derivative of a function
(c) Right derivative of a function
(d) Differentiability of a function in a point
(e) Differentiability of a function on an open interval
(f) Differentiability of a function on a closed interval
6. What is the difference between Definition 33 and 34?
7. Which of the concepts in question 1 can also be determined using the derivative of
a function?
8. Why is the gradient of a tangent to a function f (x) in a point a defined as
lim [f (x) − f (a)]/(x − a)? Draw a sketch to elucidate your answer.
x→a
9. Why is the derivative of a function defined as lim [f (x+h)−f (x)]/h? Draw a sketch
to elucidate your answer.
h→0
10. Name and draw three ways for a function not to be differentiable and explain at
each why it is not differentiable.
11. Determine the derivative of continuous functions using the definition of a derivative.
(Ex 2.2, Stewart)
Differentiation
53
12. Determine the derivatives of piece-wise defined functions. (Ex 2.2, Stewart)
13. What is the relation between the continuity and the differentiability of functions?
Draw sketches to elucidate your answer.
14. Prove that if a function is differentiable, it will be continuous.
15. State (write down) the eight differentiation rules.
16. Prove the following using the definition of a derivative. Suppose that the two functions f and g are both differentiable and that both g and g ′ are not zero.
(a) Dx k = 0 where k is constant.
(b) Dx kf (x) = kf ′ (x) where k is constant.
(c) (f + g)′ (x) = f ′ (x) + g ′(x)
(d) (f − g)′ (x) = f ′ (x) − g ′ (x)
(e) (f · g)′ (x) = f ′ (x) · g(x) + f (x) · g ′ (x)
(f) (f /g)′(x) = [f ′ (x)g(x) − f (x)g ′(x)]/[g(x)]2
(g) (f ◦ g)′ (x) = f ′ (g(x))g ′(x)
17. Draw a sketch to illustrate the concepts change along a curve f (x), change along a
secant of a curve f (x) and change along a tangent to a curve, so that the differences
and similarities are clear.
18. Use differentiation rules to determine the derivatives and higher derivatives of functions. (Ex 2.3 and 2.5, Stewart)
19. Use differentiation rules to calculate the instantaneous change of a physical phenomenon relative to a variable on which it is dependent. (Ex 2.3 and 2.5, Stewart)
20. Use differentiation rules to determine the equations of tangents to curves. (Ex 2.3
and 2.5, Stewart)
21. Determine the derivatives of and equations of tangents to curves that are defined
implicitly. (Ex 2.6, Stewart)
22. If a function f (x) is one-to-one and differentiable, derive an expression for the
derivative of the inverse function f −1 (x) using implicit differentiation.
23. Determine the derivatives of inverse functions in general. (Ex 6.1, Stewart)
24. Determine the derivative of the function ex using the definition of a derivative.
25. Derive expressions for the derivatives of the functions ax , ln x and loga x where
a > 0.
26. Determine the derivatives of functions in which exponential and logarithmic functions occur. (Ex 6.2 and 6.4, Stewart)
27. Determine the derivatives of functions of the form f (x)g(x) . (Ex 6.4, Stewart)
28. Prove the power rule for differentiation using logarithmic differentiation.
29. What is the gradient of the function ln x in the point 1?
54
Differentiation
30. Derive an expression with which the number e can be calculated by comparing the
expression that is obtained when the derivative of the function ln x in the point 1
is determined by using the definition of a derivative, is compared to the answer in
question 29.
31. Derive expressions for the derivatives of the functions sin x and cos x using the
definition of a derivative.
32. Derive expressions for the derivatives of the functions tan x, csc x, sec x and cot x
using differentiation rules.
33. Derive expressions for the derivatives of the functions sin−1 x, cos−1 x and tan−1 x.
34. Determine the derivatives of functions in which trigonometric and inverse trigonometric functions occur. (Ex 2.4 and 2.6, Stewart)
35. State (write down) l’Hôspital’s rule which is used to calculated limits.
36. When may l’Hôspital’s rule be used to calculate limits?
37. Use l’Hôspital’s rule to prove the following limit identities,
(a) lim
sin x
x
(b) lim
cos x−1
x
x→0
x→0
ex −1
x→0 x
(c) lim
=1
=0
=1
(d) lim (1 + h)1/h = e
h→0
38. Use the definition of a derivative and the identity in question 37c to determine the
gradient of the function ex in the point 1.
→0 →±∞
39. Determine limits of the form →0
, →±∞ , (→ 0)·(→ ±∞), (→ ∞)−(→ ∞), (→ 0)(→0) ,
(→ ±∞)(→0) and (→ 1)(→±∞) . (Ex 6.8, Stewart)
40. Solve various real world problems using differentiation. (Ex 2.7, Stewart)
55
Differentiation
4.1
Definition of Derivative
(2 hours 20 minutes contact time)
You should study Sections 2.1 and 2.2 (Stewart), Chapter 8 (Mathematics Review Manual)
and Section 7 (Unimaths Intro Workbook) together with this section.
In this paragraph we shall study physical phenomena such as average velocity, instantaneous velocity and more generally rate of change. This serves as an introduction to,
and motivation for, the introduction of the concept derivative of a function. We shall
also investigate the geometric presentation of it as the slope of a tangent to a relevant
curve.
The following geometric interpretations are generally attached to the concepts average
rate of change and instantaneous rate of change.
Definition 31
Let y = f (x). The average change of y relative to x over the interval [a, b] is
the slope msec of the secant through the points (a, f (a)) and (b, f (b)) on the graph of f ;
therefore
f (b) − f (a)
msec =
.
b−a
Definition 32
Let y = f (x). The instantaneous change of y relative to x at the point a is the
slope mtang of the tangent to the curve at the point (a, f (a)); therefore
mtang = lim
x→a
f (x) − f (a)
.
x−a
This definition is illustrated in Figure 1 (Section 2.1, Stewart). Study Example 1 (Section
2.1, Stewart).
A tangent can also be defined in the following way.
Definition 33
Let P (a, f (a)) be a point on the graph of f . The tangent to the graph of f at the point
P is the straight line through P with slope
f (a + h) − f (a)
,
h→0
h
m = lim
provided that this limit exists.
This definition is illustrated in Figure 3 (Section 2.1, Stewart). Study Examples 2 and 3
(Section 2.1, Stewart).
f (x + h) − f (x)
exists, a new function is
In the set of all x ∈ Df for which the limit lim
h→0
h
now defined as follows.
56
Differentiation
Definition 34
The function f ′ defined by
f (x + h) − f (x)
,
h→0
h
f ′ (x) = lim
is called the derivative of f w.r.t. x. The domain of f ′ is the set of all x ∈ Df for which
the limit exists.
Study Examples 4, 5, 6 and 7 (Section 2.1, Stewart). Also study Examples 1, 2, 3 and 4
(Section 2.2, Stewart).
Definition 35
f is differentiable at the point a if a ∈ Df ′ .
or
Definition 36
f is differentiable at the point a if f ′ (a) exists.
The derivative of a function f may be interpreted geometrically as the function f ′ of
which the value f ′ (x) at x is equal to the slope of the tangent to the curve y = f (x) at
the point (x, f (x)).
√
Example: Let f (x) = x, x ≥ 0. Then f is differentiable at each x > 0 and
√
√
x+h− x
′
f (x) = lim
h→0
h
√
√
√
√
x+h− x
x+h+ x
×√
= lim
√
h→0
h
x+h+ x
√
√ √
√
( x + h − x)( x + h + x)
√
= lim
√
h→0
h( x + h + x)
h
= lim √
√
h→0 h( x + h +
x)
1
= lim √
√
h→0
x+h+ x
1
= √ .
2 x
The derivative of a function also represents the rate at which the function changes relative
to the variable on which the function depends.
Definition 37
The function f is right differentiable at the point x if the right derivative
f+′ (x) = lim+
h→0
exists.
f (x + h) − f (x)
h
57
Differentiation
Definition 38
The function f is left differentiable at the point x if the left derivative
f−′ (x) = lim−
h→0
f (x + h) − f (x)
h
exists.
Right and left derivatives in the point a can also be calculated by using the expressions
f+′ (a) = lim+
x→a
f (x) − f (a)
f (x) − f (a)
and f−′ (a) = lim−
.
x→a
x−a
x−a
Theorem 20
The function f is differentiable at the point x if and only if both f−′ (x) and f+′ (x) exist
and have the same value.
Definition 39
A function is called differentiable on an open interval (a, b) if the function is differentiable at every point in (a, b).
Definition 40
A function f is called differentiable on a closed interval [a, b] if the following conditions are satisfied:
(a) f is differentiable on the open interval (a, b),
(b) f is right differentiable at the point a and
(c) f is left differentiable at the point b.
Study Example 5 (Section 2.2, Stewart).
Theorem 21
If the function f is differentiable at the point a, then f is continuous at the point a.
f (a+h)−f (a)
h
h→0
Thus have to prove that if f ′ (a) = lim
then lim f (a + h) = f (a).
h→0
Proof:
The function f is differentiable at the point a and consequently the limit
f ′ (a) = lim
h→0
f (a + h) − f (a)
h
exists according to Definition 34.
It then follows that
lim f (a + h) = lim [f (a + h) − f (a) + f (a)] ·
h→0
h→0
h
h
58
Differentiation
f (a + h) − f (a)
f (a)h
= lim
·h+
h→0
h
h
f (a + h) − f (a)
= lim
· lim h + lim f (a)
h→0
h→0
h→0
h
= f ′ (a) · 0 + f (a) = f (a)
and consequently f is continuous at the point a, by Definition 25.
A direct consequence of the theorem is (the contrapositive of negation):
✷
Theorem 22
If f is discontinuous at the point a, then f is not differentiable at the point a.
Example: The function f (x) = | x| is continuous at the point 0 because
lim | x| = 0 = f (0).
x→0
However, it is not differentiable at the point 0 because the right derivative
f+′ (0) = lim+
h→0
|0 + h| − |0|
|h|
f (0 + h) − f (0)
= lim+
= lim+
=1
h→0
h→0
h
h
h
and the left derivative
f−′ (0) = lim−
h→0
f (0 + h) − f (0)
|0 + h| − |0|
|h|
= lim−
= lim−
= −1
h→0
h→0
h
h
h
do not have the same value. We write
x,
x≥0
f (x) = |x| =
−x, x < 0
1,
x>0
f ′ (x) =
−1, x < 0
f+′ (0) = 1
f−′ (0) = −1
It is therefore possible that although a function is continuous at a point, it is not differentiable at that point because the left- and right derivatives may not be equal at that
point. This happens at a point where the function forms an angle (has a kink).
59
Differentiation
y = |x|
dy
dx
1◦
=
d|x|
dx
◦ −1
Example:
f (x) =
3x − 2, x < 1
x2 ,
x≥1
For this function the derivative is


 lim f (x+h)−f (x) , x < 1
 lim 3(x+h)−2−(3x−2)
, x<1
h
h
3, x < 1
′
h→0
h→0
f (x) =
=
=
(x)
(x+h)2 −x2
2x,
x>1
 lim f (x+h)−f

, x>1
lim
,
x>1
h
h
h→0
h→0
f (1 + h) − f (1)
3(1 + h) − 2 − 12
f−′ (1) = lim−
= lim−
=3
h→0
h→0
h
h
(1 + h)2 − 12
f (1 + h) − f (1)
= lim+
=2
f+′ (1) = lim+
h→0
h→0
h
h
using Definitions 34, 37 and 38.
Example:
g(x) =
3x, x < 1
x2 , x ≥ 1
For this function the derivative is


 lim g(x+h)−g(x) , x < 1
 lim 3(x+h)−3x
, x<1
h
h
3, x < 1
′
h→0
h→0
g (x) =
=
=
g(x+h)−g(x)
(x+h)2 −x2
2x,
x>1
 lim
 lim
, x>1
, x>1
h
h
h→0
h→0
3(1 + h) − 12
g(1 + h) − g(1)
= lim−
= −∞
= lim−
h→0
h→0
h
h
g(1 + h) − g(1)
(1 + h)2 − 12
′
g+
(1) = lim+
= lim+
=2
h→0
h→0
h
h
′
g−
(1)
using Definitions 34, 37 and 38.
Example:
k(x) =
3x, x 6= 1
2, x = 1
60
Differentiation
For this function the derivative is
3(x + h) − 3x
k(x + h) − k(x)
= lim
= 3, x 6= 1
h→0
h→0
h
h
3(1 + h) − 2
k(1 + h) − k(1)
′
= lim−
= −∞
k−
(1) = lim−
h→0
h→0
h
h
k(1 + h) − k(1)
3(1 + h) − 2
′
k+
(1) = lim+
= lim+
=∞
h→0
h→0
h
h
k ′ (x) = lim
using Definitions 34, 37 and 38.
4.2
Rules for Differentiation
(1 hour 30 minutes contact time)
You should study Section 2.3 (Stewart) and Chapter 8 (Mathematics Review Manual)
together with this section.
It is often difficult and tedious to determine the derivative of a function by means of the
definition. It is therefore useful and essential to know the rules for the differentiation of
sums, products, quotients and composite functions and be able to apply them.
The techniques are summarized in the following theorem.
Theorem 23
Rules for differentiation.
(a) If f (x) = k where k is a constant, then f is differentiable at every point x ∈ Df and
d
[k] = 0
dx
for all x ∈ Df .
(b) If f (x) = xn , then
d n
[x ] = nxn−1
dx
for all x, n ∈ R.
(c) If f is differentiable at the point x and k is a constant, then
d
d
[kf (x)] = k [f (x)] = kf ′ (x).
dx
dx
(d) If f and g are differentiable at the point x, then f + g is differentiable at the point x
and
(f + g)′ (x) =
d
d
d
d
[(f + g)(x)] =
[f (x) + g(x)] =
[f (x)] + [g(x)] = f ′ (x) + g ′ (x).
dx
dx
dx
dx
61
Differentiation
(e) If f and g are differentiable at the point x, then f − g is also differentiable at the point
x and
(f − g)′(x) =
d
d
d
d
[(f − g)(x)] =
[f (x) − g(x)] =
[f (x)] − [g(x)] = f ′ (x) − g ′ (x).
dx
dx
dx
dx
(f ) If f and g are differentiable at the point x, then f · g is differentiable at the point x and
d
d
[(f · g)(x)] =
[f (x)g(x)]
dx
dx
d
d
[f (x)]g(x) + f (x) [g(x)] = f ′ (x)g(x) + f (x)g ′(x).
=
dx
dx
(f · g)′(x) =
(g) If f and g are differentiable at the point x and g(x) 6= 0, then f /g is differentiable at
the point x and
′
f
d f (x)
d
f
(x) =
(x) =
g
dx
g
dx g(x)
d
d
[f (x)]g(x) − f (x) dx
[g(x)]
f ′ (x)g(x) − f (x)g ′(x)
dx
=
=
.
[g(x)]2
[g(x)]2
All the rules are proven by making use of the definition of the derivative of a function,
Definition 34.
You should be able to do all the proofs, except the proof of the power rule.
The proofs of the constant multiplier rule, the sum rule and the product rule are given in
Section 2.3 (Stewart).
You should write out the proofs for the constant function rule and the difference rule.
Proof of the quotient rule:
From the Definition 34 it follows that
d
f
1
f
f
(x) = lim
(x + h) −
(x)
h→0 h
dx
g
g
g
1 f (x + h) f (x)
= lim
−
h→0 h g(x + h)
g(x)
1 f (x + h)g(x) − f (x)g(x + h)
= lim
h→0 h
g(x)g(x + h)
1 f (x + h)g(x) − f (x)g(x) + f (x)g(x) − f (x)g(x + h)
= lim
h→0 h
g(x)g(x + h)
1
[f (x + h) − f (x)]g(x) − f (x) h1 [g(x + h) − g(x)]
= lim h
h→0
g(x)g(x + h)
1
lim [f (x + h) − f (x)]g(x) − f (x) h1 [g(x + h) − g(x)]
h→0 h
=
lim g(x)g(x + h)
h→0
=
lim h1 [f (x + h) − f (x)] lim g(x) − lim f (x) lim h1 [g(x + h) − g(x)]
h→0
h→0
h→0
lim g(x)g(x + h)
h→0
h→0
62
Differentiation
=
f ′ (x)g(x) − f (x)g ′(x)
.
[g(x)]2
✷
Because a derivative is a function, it can be differentiated again to get the second order
derivative and so on.
d
f (x) = f ′ (x)
dx
d
d
d2
f (x)
=
f (x) = f ′′ (x) = f (2) (x)
dx dx
dx2
2
d
d3
d
f
(x)
=
f (x) = f ′′′ (x) = f (3) (x)
2
3
dx dx
dx
Study Examples 1 to 13 (Section 2.3, Stewart).
4.3
Chain Rule
(1 hour 30 minutes contact time)
You should study Section 2.5 (Stewart) and Chapter 8 (Mathematics Review Manual)
together with this section.
Functions one meets in practice are seldomly in an elementary form. They are usually
d
(f (g(x))) if f and
composite. It is therefore essential that you should know what is dx
g are two differentiable functions for which the derivatives are known. For example,
the derivatives of f (x) = sin x (f ′ (x) = cos x) and g(x) = 4x3 + x (g ′ (x) = 12x2 + 1)
are known. How can we make use of this information to determine the derivative of
h(x) = sin(4x3 + x)? Note that h = f ◦ g.
In this case the important result is: The chain rule
If g is differentiable at x and f is differentiable at g(x), then the composite function f ◦ g
is differentiable at x and
d
[f (g(x))] = f ′ (g(x))g ′(x).
dx
If we put y = f (u) and u = g(x), the expression reduces to
dy du
dy
=
· .
dx
du dx
Study Examples 1 to 8 (Section 2.5, Stewart).
As composite functions are often regarded as “a function within a function”, you might
find it easier to remember the following wording for the chain rule: The value of the
derivative of f ◦ g at the point x is equal to the value of the derivative of the outer
function (namely f ) at the point g(x), multiplied by the value of the derivative of the
inner function (namely g) at the point x.
63
Differentiation
Theorem 24
If g is differentiable at x and f is differentiable at g(x), then the composite function f ◦ g
is differentiable at x and
d
[f (g(x))] = f ′ (g(x))g ′(x).
dx
Proof:
Suppose g is differentiable in c and f is differentiable in g(c).
Because g(c) is differentiable in the point c, it is continuous in the point c according to
Theorem 21 and lim g(x) = g(c) according to Definition 24.
x→c
According to 32 is
(f ◦ g)(x) − (f ◦ g)(c)
x−c
f (g(x)) − f (g(c))
lim
x→c
x−c
f (g(x)) − f (g(c)) g(x) − g(c)
·
lim
x→c
g(x) − g(c)
x−c
f (g(x)) − f (g(c))
g(x) − g(c)
lim
· lim
x→c
x→c
g(x) − g(c)
x−c
f (g(x)) − f (g(c))
g(x) − g(c)
· lim
lim
x→c
g(x)→g(c)
g(x) − g(c)
x−c
′
′
f (g(c))g (c)
(f ◦ g)′ (c) = lim
x→c
=
=
=
=
=
✷
4.4
Implicit Differentiation
(1 hour contact time)
You should study Section 2.6 (Stewart) together with this section.
Sometimes a curve is not defined explicitly by an expression of the form y = f (x), but
rather by an equation in which the variables x and y appear.
Example: Suppose that a differentiable function y of x is defined by the equation
x3 y + xy 3 = 10.
The point (2, 1) falls on this curve (test it by substituting x = 2 and y = 1 into the
equation). To find the tangent to the curve at the point (2, 1), the slope dy/dx has
to be calculated at the point (2, 1). This may be done by using the so-called implicit
differentiation: Note that
d 3
d
[x y + xy 3 ] =
[10].
dx
dx
Therefore
d
d 3
[x y] + [xy 3 ] = 0,
dx
dx
64
Differentiation
and therefore, according to the product rule and the chain rule, and because of the fact
that y is a function of x, it follows that
3x2 y + x3
dy
dy
+ y 3 + 3xy 2
= 0.
dx
dx
Consequently
dy
3x2 y + y 3
=−
dx
3xy 2 + x3
and therefore if x = 2 and y = 1 then
dy
dx
(2,1)
=−
3 × 22 × 1 + 13
13
=− .
2
3
3×2×1 +2
14
The equation of the tangent is therefore given by y − 1 = − 13
(x − 2).
14
Study Examples 1, 2, 3 and 4 (Section 2.6, Stewart).
4.5
Derivatives of Inverse Functions
(30 minutes contact time)
You should study Section 6.1 (Stewart) together with this section.
Differentiability of f on Df means in broad terms that the curve y = f (x) does not have
any acute angles. Seeing that the graph of f −1 is the mirror image of the graph of f ,
reflected about the line y = x, f −1 will also be differentiable for all x where f ′ (f −1 (x)) 6= 0.
It is in fact possible to prove the following important result.
Theorem 25
Suppose f has an inverse and f ′ (f −1 (x)) 6= 0 for all x ∈ I. Then f −1 is differentiable on
I and
1
.
(f −1 )′ (x) = ′ −1
f (f (x))
Proof:
Let y = f −1 (x). Then
x = f (y).
Differentiate to x.
d f (y)
dx
=
dx
dx
dy
⇒ 1 = f ′ (y)
dx
1
1
dy
= ′
= ′ −1
⇒
dx
f (y)
f (f (x))
✷
The determination of the derivatives of all inverse functions follow in the same way.
Study Examples 6 and 7 (Section 6.1, Stewart).
65
Differentiation
4.6
Derivatives of Exponential and Logarithmic Functions
(1 hour 30 minutes contact time)
You should study Sections 6.2 and 6.4 (Stewart) together with this section.
The natural exponential function f (x) = ex is the exponential function of which the
gradient (derivative) in the point x = 0 is one, therefore
f (0 + h) − f (0)
e0+h − e0
eh − 1
= lim
= lim
=1
h→0
h→0
h→0
h
h
h
f ′ (0) = lim
which is the limit identity given in Theorem 15.
The number e can be presented as a limit by calculating the derivative of lnx at the point
x = 1. According to Theorem 25 taking f (x) = ex and f −1 (x) = ln x
d ln x
dx
= (f −1 )′ (1) =
x=1
1
f ′ (f −1 (1))
=
1
e0
since f (0) = 1 implies f −1 (1) = 0. From this it follows that
d ln x
dx
x=1
1
= lim [f (1 + h) − f (1)]
h→0 h
1
= lim [ln(1 + h) − ln 1]
h→0 h
1
= lim ln(1 + h)
h→0 h
= lim ln(1 + h)1/h
h→0
i
h
= ln lim (1 + h)1/h = 1
h→0
and
i
h
ln lim (1 + h)1/h = 1
h→0
consequently
lim (1 + h)1/h = e
h→0
which is the limit identity given in Theorem 15.
These identities are used to determine the derivatives of ex and ln x.
Theorem 26
dex
deu(x)
= ex and
= eu(x) u′(x)
dx
dx
Proof:
dex
ex+h − ex
= lim
h→0
dx
h
66
Differentiation
ex eh − ex
h→0
h
h
e −1
= ex lim
h→0
h
= ex · 1 = ex
= lim
✷
Derivatives in general:
Theorem 27
d x
d u(x)
(a ) = ax ln a and
(a ) = au(x) u′(x) ln a
dx
dx
Proof:
d x
d ln a x
(a ) =
[(e ) ]
dx
dx
d x ln a
(e
)
=
dx
d
= ex ln a · (x ln a)
dx
x ln a
= e
· ln a
x
= a ln a
✷
Study Examples 2, 3, 4 and 5 (Section 6.2, Stewart).
Theorem 28
1
d
1
u′ (x)
d ln x
=
and
[ln u(x)] =
· u′ (x) =
dx
x
dx
u(x)
u(x)
Proof:
Let y = ln x so that ey = x. Then we have that
d y
d
(e ) =
(x)
dx
dx
dy
=1
⇒ ey
dx
1
1
dy
= y = .
⇒
dx
e
x
✷
Derivatives in general:
Theorem 29
d
1
d
1
u′(x)
(loga x) =
and
[loga u(x)] =
· u′ (x) =
dx
x ln a
dx
u(x) ln a
u(x) ln a
67
Differentiation
Proof:
d
d ln x
(loga x) =
dx
dx ln a
1 d
=
(ln x)
ln a dx
1 1
·
=
ln a x
1
=
x ln a
✷
This theorem can also be proved using the definition of a derivative.
ln(x + h) − ln x
d ln x
= lim
h→0
dx
h 1
x+h
= lim ln
h→0 h
x
1
= lim ln (1 + h/x)
h→0 h
1
ln(1 + u)
= lim
u→0 xu
1
lim ln(1 + u)(1/u)
=
u→0
x
i
1 h
(1/u)
ln lim (1 + u)
=
u→0
x
1
1
=
ln e =
x
x
✷
Study Examples 1, 2, 3, 4 and 12 (Section 6.4, Stewart).
4.7
Logarithmic Differentiation
(50 minutes contact time)
You should study Section 6.4 (Stewart) together with this section.
The function f (x) = xx is neither a power function, nor an exponential function. We
can therefore not make use of the rules for the differentiation of power functions or exponential functions. The derivative of the function f (x) = xx is determined by means of
logarithmic differentiation.
y = xx ⇒ ln y = x ln x
d
d
⇒
(ln y) =
(x ln x)
dx
dx
1
1 dy
= ln x + x ·
⇒
y dx
x
68
Differentiation
dy
= y(ln x + 1)
dx
dy
= xx (ln x + 1)
⇒
dx
⇒
In the case where y = [f (x)]g(x) logarithmic differentiation yields the following.
⇒
⇒
⇒
⇒
ln y = g(x) ln f (x)
d
d
(ln y) =
[g(x) ln[f (x)]
dx
dx
f ′ (x)
1 dy
= g ′(x) ln[f (x)] + g(x) ·
y dx
f (x)
dy
f ′ (x)
′
= y g (x) ln[f (x)] + g(x) ·
dx
f (x)
f ′ (x)
dy
′
g(x)
g (x) ln[f (x)] + g(x) ·
= [f (x)]
dx
f (x)
Logarithmic differentiation may also be used for the calculation of the derivatives of
complicated functions containing products, quotients and powers.
Example:
√
(x2 − 8)1/3 x3 + 1
y=
,
x6 − 7x + 5
Then
ln |y| = (1/3) ln |x2 − 8| + (1/2) ln |x3 + 1| − ln |x6 − 7x + 5|.
Therefore
dy
1
=
(1/y)
dx
3
This yields
2x
2
x −8
1
+
2
3x2
x3 + 1
−
6x5 − 7
.
x6 − 7x + 5
√
2
dy
(x − 8)1/3 x3 + 1
2x
3x2
6x5 − 7
=
+
−
.
dx
3(x2 − 8) 2(x3 + 1) x6 − 7x + 5
x6 − 7x + 5
Study Examples 15 and 16 (Section 6.4, Stewart).
The power rule for differentiation is proved by making use of logarithmic differentiation.
d n
(x ) = nxn−1
dx
Proof of the power rule:
⇒
⇒
y = xn
ln |y| = n ln |x|
n
1 dy
=
y dx
x
69
Differentiation
⇒
dy
n
= xn · = nxn−1
dx
x
✷
Differentiation is one of the most important techniques that you will meet in Calculus.
You can only master this technique successfully by getting lots of practice. You should
therefore do as many problems as possible!
4.8
Derivatives of Trigonometric Functions
(50 minutes contact time)
You should study Section 2.4 (Stewart) together with this section.
From the basic definition of derivatives, Definition 34, and the results about the limits of
trigonometric functions, we can derive that:
Theorem 30
d
d
(sin x) = cos x and
[sin u(x)] = [cos u(x)]u′(x)
dx
dx
d
d
(cos x) = − sin x and
[cos u(x)] = [− sin u(x)]u′ (x)
dx
dx
d
d
2
(tan x) = sec x and
[tan u(x)] = [sec2 u(x)]u′ (x)
dx
dx
d
d
(cot x) = − csc2 x and
[cot u(x)] = [− csc2 u(x)]u′(x)
dx
dx
d
d
(sec x) = sec x tan x and
[sec u(x)] = [sec u(x) tan u(x)]u′ (x)
dx
dx
d
d
(csc x) = − csc x cot x and
[csc u(x)] = [− csc u(x) cot u(x)]u′(x)
dx
dx
Proof:
d
cos(x + h) − cos x
[cos x] = lim
h→0
dx
h
cos x cos h − sin x sin h − cos x
= lim
h→0
h cos h − 1
sin h
= lim cos x
− sin x
h→0
h
h
cos h − 1
sin h
= cos x lim
− sin x lim
h→0
h→0
h
h
= − sin x.
✷
The derivation of the derivative of sin x follows in a similar fashion. By making use of the
70
Differentiation
rules for differentiation we can now determine the derivatives of the other trigonometric
functions.
Example: Prove that f ′ (x) = sec x tan x if f (x) = sec x.
Solution:
d
1
d
′
f (x) =
[sec x] =
dx
dx cos x
0 − (− sin x)
=
cos2 x
= sec x tan x.
You are expected to be able to do the proofs of all the derivatives in Theorem 30.
By means of the chain rule we have that
d
[sin(4x3 + x)] = (12x2 + 1) cos(4x3 + x).
dx
Example: You might prefer to use the second version of the theorem.
d √ 3
Suppose that
[ 4x + sin x] has to be calculated.
dx
√
Then let u = 4x3 + sin x and y = u = u1/2 , so that
dy du
1
d √ 3
[ 4x + sin x] =
·
= u−1/2 (12x2 + cos x)
dx
du dx
2
1
(4x3 + sin x)−1/2 (12x2 + cos x).
=
2
Example:
d
[sin5 3x] = 5(sin4 3x)(cos 3x)(3) = 15 sin4 3x cos 3x.
dx
Study Examples 1, 2, 3 and 4 (Section 2.4, Stewart).
4.9
Derivatives of Inverse Trigonometric Functions
(50 minutes contact time)
You should study Section 6.6 (Stewart) together with this section.
Theorem 31
The derivatives of inverse trigonometric functions:
d
u′ (x)
d
1
1
′
√
p
p
· u (x) =
and
(arcsin x) =
(arcsin u(x)) =
dx
dx
1 − x2
1 − [u(x)]2
1 − [u(x)]2
71
Differentiation
−u′ (x)
−1
−1
d
d
· u′ (x) = p
(arccos x) = √
(arccos u(x)) = p
and
dx
dx
1 − x2
1 − [u(x)]2
1 − [u(x)]2
1
d
1
u′ (x)
d
′
(arctan x) =
and
(arctan
u(x))
=
·
u
(x)
=
dx
1 + x2
dx
1 + [u(x)]2
1 + [u(x)]2
d
−1
d
−1
−u′ (x)
′
( arccot x) =
and
(
arccot
u(x))
=
·
u
(x)
=
dx
1 + x2
dx
1 + [u(x)]2
1 + [u(x)]2
1
1
d
d
p
( arcsec x) = √
( arcsec u(x)) =
· u′ (x)
and
2
2
dx
dx
x x −1
u(x) [u(x)] − 1
u′ (x)
p
=
u(x) [u(x)]2 − 1
−1
d
−1
d
p
· u′ (x)
( arccsc x) = √
( arccsc u(x)) =
and
2
2
dx
dx
x x −1
u(x) [u(x)] − 1
−u′ (x)
p
=
u(x) [u(x)]2 − 1
Proof for arccos x:
Let y = arccos x, y ∈ [0, π] so that x = cos y. Then
⇒
⇒
d
d
(x) =
(cos y)
dx
dx
dy
1 = − sin y
dx
dy
1
1
1
=−
= −p
= −√
.
2
dx
sin y
1 − x2
1 − cos y
because sin y ≥ 0 for y ∈ [0, π].
You should be able to reproduce the proofs for arccos x, arcsin x and arctan x.
Study Examples 5 and 6 (Section 6.6, Stewart).
4.10
✷
L’Hôspital’s Rule
(2 hours 20 minutes contact time)
You should study Section 6.8 (Stewart) together with this section.
Appendix B should be studied together with this section.
In this section ”lim” is used for any of the following limits:
lim , lim+ , lim− ,
x→a
x→a
x→a
lim ,
x→+∞
lim .
x→−∞
In Calculus I you have learned that there are certain limits that cannot be calculated
directly by the first rules for finding limits. These are limits of the form
lim
f (x)
g(x)
72
Differentiation
where lim f (x) = lim g(x) = 0 or lim f (x) = lim g(x) = ±∞. These limits of the form 00
±∞
and ±∞
are known as undetermined forms. The rule of L’Hôspital provides a method for
calculating these limits.
Theorem 32
L’Hôspital’s rule. Suppose lim f (x) = lim g(x) = 0 or lim f (x) = lim g(x) = ±∞. If
′ (x)
′ (x)
= L with L ∈ R or lim fg′ (x)
= ±∞, then we have that
lim fg′ (x)
lim
Proof of Theorem 15:
f (x)
f ′ (x)
= lim ′ .
g(x)
g (x)
cos x
sin x
= lim
=1
x→0
x→0 x
1
cos x − 1
− sin x
lim
= lim
=0
x→0
x→0
x
1
lim
✷
Example: Consider the limit
1 − cos(x − 5)
.
x→5
(x − 5)2
lim
The limit is of the form 00 . It appears as if the limit cannot be calculated, but making use
of your knowledge of Calculus I the limit can be found:
1 − cos(x − 5)
1 − cos(x − 5) 1 + cos(x − 5)
=
lim
×
x→5
x→5
(x − 5)2
(x − 5)2
1 + cos(x − 5)
2
1 − cos (x − 5)
= lim
x→5 (x − 5)2 [1 + cos(x − 5)]
sin2 (x − 5)
= lim
x→5 (x − 5)2 [1 + cos(x − 5)]
2
sin(x − 5)
1
= lim
· lim
x→5 (x − 5)
x→5 1 + cos(x − 5)
1
.
=
2
lim
Making use of l’Hôspital’s rule, the limit can be calculated as follows:
1
1 − cos(x − 5)
sin(x − 5)
= lim
= .
2
x→5
x→5 2(x − 5)
(x − 5)
2
lim
Evidently, l’Hôspital’s rule simplifies the calculation of limits considerably. Consider
Examples 1, 2, 3, 4 and 5 (Section 6.8, Stewart).
According to Appendix B, limits of the form 0 × (±∞) are also undetermined. Suppose
lim f (x) = 0 and lim g(x) = ±∞, then
lim f (x)g(x) = lim
g(x)
1/f (x)
73
Differentiation
is of the form
±∞
.
±∞
Or
lim f (x)g(x) = lim
f (x)
1/g(x)
which is of the form 00 . In both these cases we can use l’Hôspital’s rule to find the limit.
In Example 6 (Section 6.8, Stewart), lim+ x ln x is calculated by means of this method.
x→0
Undetermined limits of the form ∞ − ∞ (see Appendix B) may also be calculated by
means of l’Hôspital’s rule. The function is factorised to get it in the form ±∞ × (±∞)
or 0 × (±∞). In the first case the value of the limit is ±∞ and in the second case we use
the method that was discussed in the previous paragraph. Study Example 8 (Section 6.8,
Stewart).
L’Hôspital’s rule is also used for calculating undetermined limits of the form
00 , ±∞0 , 1±∞ .
These limits are all of an undetermined form. When taking the logarithm of these limits,
they are all of the undetermined form 0 · ∞. Study Examples 9 and 10 (Section 6.8,
Stewart).
4.11
Applications of Differentiation
(50 minutes contact time)
Study the different applications of differentiation as discussed in Section 2.7 (Stewart).
74
Differentiation
Study Unit 5
Integration
Time Allocation
11 hours 30 minutes (7 hours contact time and 1 tutorial session).
Prerequisite Study
Study units 1, and 4, MTHS111.
Study Outcomes
When you have completed this section, you should be able to do the following, using
standard mathematical notation.
1. Define and use the concept of an indefinite integral (Definition 41) to calculate
indefinite integrals of functions.
2. Formulate and use the integration rules (Theorems 33, 34, 35 and 41) to calculate
indefinite integrals of functions.
3. Formulate and use the integrals of exponential functions (Theorems 36 and 37) and
logarithmic functions (Theorem 38) to calculate indefinite integrals of functions.
R
R
R
R
4. Formulate, prove (only tan x dx, cot x dx, sec x dx and csc x dx) and use
the integrals of trigonometric functions (Theorem 39) and inverse trigonometric
functions (Theorem 40), to calculate indefinite integrals of functions.
References
Stewart: Sections 3.9, 4.1, 4.4, 4.5, 6.2, 6.4, 6.6.
Study guide: Sections 5.1, 5.2, 5.3 5.4, 5.5.
75
76
Integration
Preparation
Read the Sections referred to in the textbook and in the study guide.
Work through the following examples from Stewart:
Section 3.9: 1, 2;
Section 4.1: 1;
Section 4.4: 1;
Section 4.5: 1, 2.
Answer the following questions and submit your answers at the beginning of the first
contact session of this study unit:
1. The following are all derivatives of functions. How did these functions look like
before they were differentiated?
(a) f (x) = 2x − 1
(b) g(x) = ex
(c) h(x) =
1
x
(d) j(x) = sin x
Exercising of Skills
Work through the following examples from Stewart:
Section 3.9: 3, 4;
Section 4.4: 2;
Section 4.5: 3, 4, 5;
Section 6.2: 8;
Section 6.4: 9, 11;
Section 6.6: 8, 9.
Do the following exercises:
Stewart (8th edition):
Ex. 3.9: 2, 4, 5, 23, 26, 27;
Ex. 4.4: 1 - 18;
Ex. 4.5: 1 - 34;
Ex. 6.2: 83 - 94, 96;
Ex. 6.4: 76 - 82;
Ex. 6.6: 63, 65 - 70.
Stewart (7th edition):
Ex. 3.9: 2, 4, 5, 21, 23, 25;
Ex. 4.4: 1 - 18;
Ex. 4.5: 1 - 34;
Ex. 6.2: 79 - 90, 92;
Ex. 6.4: 76 - 82;
Ex. 6.6: 63, 65 - 70.
Integration
77
Test Yourself
1. What is the two main motivations for the development of the infinitesimal calculus?
2. Of which mathematical manipulation is integration the inverse?
R
3. What is the meaning of the expression f (x) dx?
4. If the displacement as function of time is given, which mathematical manipulation
can be used to determine the velocity and acceleration?
5. If the acceleration as function of time is given, which mathematical manipulation
can be used to determine the velocity and displacement?
6. If the work done as function of time is given, which mathematical manipulation can
be used to determine the power?
7. If the power as function of time is given, which mathematical manipulation can be
used to determine the work done?
8. If the profit as function of number of products sold, is given, which mathematical
manipulation can be used to determine the marginal profit?
9. If the marginal profit as function of number of products sold, is given, which mathematical manipulation can be used to determine the profit?
10. Define the concept indefinite integral.
11. If the function g(x) is an indefinite integral of the function f (x), which other functions are also indefinite integrals of f (x)?
12. Why is the use of the integration constant when determining indefinite integrals,
important? Give an example to elucidate your answer.
13. Write down five integration rules (theorems).
14. Write down the integration standard forms for the following functions.
(a) Power function
(b) Exponential function
(c) Logarithmic function
(d) Six trigonometric functions
(e) Six inverse trigonometric functions
15. Determine the integrals of a variety of functions using the rules in question 13 and
standard forms in question 14. (Ex 3.9, 4.4, 4.5, 6.2, 6.4 and 6.6, Stewart)
78
5.1
Integration
Integration
(1 hour 40 minutes contact time)
You should study Sections 3.9 and 4.1 (Stewart) together with this section.
Up to now we have given attention only to the first important problem in calculus: determine the slope of the tangent to a curve y = f (x) at a given point (a, f (a)). - It is this
basic problem that led to the development of differential calculus.
The second important problem in calculus is the problem of area: If f is a continuous,
non-negative function on the interval [a, b], determine the area of the region bounded by
the graph of f and the lines x = a, x = b and y = 0. This problem is addressed mainly
in MTHS121.
If we chose an arbitrary x0 ∈ [a, b], then the area bounded by the curve y = f (x) and the
lines x = a, x = x0 and y = 0, may be presented by A(x0 ) (where A(a) = 0) and if x0 is
varied over the interval [a, b], we get a function A(x) which is such that A′ (x) = f (x).
Example: If f (x) = x4 , a = 0 and b = 2, then a possible choice for the function is
A(x) = 51 x5 , and consequently A′ (x) = x4 . But, of course, we realise that for A(x) =
1 5
x + C we also get A′ (x) = x4 for C being any real number. If we require that A(0) = 0,
5
then it means that C = 0 is the right choice and A(x) = 51 x5 will be the value of the area
under the curve y = f (x) and above [0, x]. The area under the curve and above [0, 2] is
therefore given by
1
32
A(2) = 25 =
units squared.
5
5
Study Example 1 (Section 4.1, Stewart).
From the brief discussion above, one can see that to determine the area under a curve
y = f (x) (for example) it will be important that we should find a function g which is such
that g ′ (x) = f (x). This leads to the following definition.
Definition 41
A function g is called an indefinite integral (or also anti-derivative) of a function f
on an interval I if g ′(x) = f (x) for all x ∈ I.
It is written as
g(x) =
From this it follows that
g(x) =
Z
Z
f (x) dx.
f (x) dx =
and
Z
g ′ (x) dx
Z
d
f (x) dx
f (x) = g (x) =
dx
which shows clearly that differentiation and integration are inverse processes.
Note that an indefinite integral is not unique. The following theorem will clarify everything.
′
79
Integration
Theorem 33
If g is an indefinite integral of f on I, then g(x) + C is also an indefinite integral of f
on I for each value of the constant C. In fact, every indefinite integral of f on I may be
written in the form g(x) + C for an arbitrary constant C.
The phrase “g(x) + C is an indefinite integral of f ” is written in mathematical notation
as
Z
f (x) dx = g(x) + C.
Study Examples 1, 2, 3 and 4 (Section 3.9, Stewart).
Example:
Z
1
x4 dx = x5 + C.
5
This is an example of the power rule of integration.
Because
dxn
d[u(x)]n
= nxn−1 and
= n[u(x)]n−1 u′ (x)
dx
dx
it follows that
Theorem 34
Power rule.
Z
xn dx =
for n 6= −1, n ∈ R.
1
xn+1 + C and
n+1
Z
[u(x)]n u′ (x) dx =
1
[u(x)]n+1 + C
n+1
From this it is also clear that differentiation and integration are inverse processes: do
the inverse actions in the inverse order. According to the power rule for differentiation
d n
x = nxn−1 which can be stated in words as: ”Multiply with the exponent andR subtract
dx
one from the exponent.” In contrast to this the power rule for integration is xn dx =
1
xn+1 + C which can be stated in words as: ”Add one to the exponent and divide by
n+1
the exponent plus one.”
Examples:
Z
Z
1
1
1
dx = x−6 dx =
x−6+1 + C = − 5 + C.
6
x
−6 + 1
5x
Z
Z
1
(2x + 1)9 dx =
(2x + 1)9 · 2 · dx
2
Z
1
(2x + 1)9 · 2 dx
=
2
1 1
=
(2x + 1)10 + C
2 10
1
=
(2x + 1)10 + C
20
To determine an indefinite integral of a function f on a set D, means to search for a
function g which is so that g ′ (x) = f (x) for all x ∈ D.
Some basic properties of indefinite integrals are summarised in the following theorem.
80
Integration
Theorem 35
R
Integration
rules. Let f and g be functions with indefinite integrals f (x) dx and
R
g(x) dx and let k be a constant. Then we have that
(a)
(b)
(c)
R
R
R
kf (x) dx = k
R
f (x) dx
[f (x) + g(x)] dx =
[f (x) − g(x)] dx =
R
R
f (x) dx +
f (x) dx −
R
R
g(x) dx
g(x) dx
Example:
Z
√
(x + x) dx =
3
Z
3
x dx +
Z
√
2
1
x dx = x4 + x3/2 + C.
4
3
Consider in general the composite function h(g(x)). The derivative of h(g(x)) relative to
x is
d h(g(x))
= h′ (g(x))g ′(x)
dx
according to the chain rule (Theorem 24). From this it follows that
Z
h′ (g(x))g ′(x) dx = h(g(x)) + C.
As with the power rule (Theorem 34) it is here also clear that differentiation and integration are inverse processes: do the inverse actions in the inverse order. According to the
d
[h(g(x))] = h′ (g(x))g ′(x) which can be stated in words as:
chain rule for differentiation dx
”Differentiate the outer function, with the inner functions still inside and multiply then
with the
R derivative of the inner function.” In contrast to this the chain rule for integration is h′ (g(x))g ′(x) dx = h(g(x)) + C which can be stated in words as: ”Divide by the
derivative of the inner function and integrate the outer function with the inner function
still inside.”
This is known as the standard form of integrals. If an expression is in this form, it is only
the outer function h′ (x) that integrates back to h(x). In the formula list (Appendix C)
there are more examples.
Integration and differentiation are the two most important mathematical techniques you
will meet in Calculus. As is the case for differentiation, much exercise is required to
master integration. You should therefore do as many problems as possible.
81
Integration
5.2
Integrals of Exponentials and Logarithmic Functions
(1 hour 40 minutes contact time)
You should study Sections 6.2 and 6.4 (Stewart) together with this section.
Seeing that
dex
deu(x)
= ex and
= eu(x) u′ (x)
dx
dx
it follows that
Theorem 36
Z
x
x
e dx = e + C and
Z
eu(x) u′(x) dx = eu(x) + C.
As
d x
d u(x)
(a ) = ax ln a and
(a ) = au(x) u′ (x) ln a
dx
dx
it follows that
Theorem 37
Z
ax
+ C and
a dx =
ln a
x
Z
au(x) u′(x) dx =
au(x)
+ C.
ln a
Study Example 8 (Section 6.2, Stewart).
As
1
d
u′ (x)
d ln x
=
and
[ln u(x)] =
dx
x
dx
u(x)
it follows that
Theorem 38
Z
1
dx = ln |x| + C and
x
Z
u′(x)
dx = ln |u(x)| + C.
u(x)
Study Examples 9 and 11 (Section 6.4, Stewart).
82
Integration
5.3
Integrals of Trigonometric Functions
(1 hour contact time)
You should study Section 4.4 (Stewart) together with this section.
The integrals of the trigonometric functions look as follows.
Theorem 39
Z
Z
Z
Z
Z
Z
Z
Z
[cos u(x)]u′ (x) dx = sin u(x) + C
Z
sin xdx = − cos x + C and
[sin u(x)]u′ (x) dx = − cos u(x) + C
Z
tan xdx = − ln | cos x| + C and
[tan u(x)]u′ (x) dx = − ln | cos u(x)| + C
Z
cot xdx = ln | sin x| + C and
[cot u(x)]u′(x) dx = ln | sin u(x)| + C
Z
sec xdx = ln | sec x + tan x| + C and
[sec u(x)]u′ (x) dx
cos xdx = sin x + C and
csc xdx = ln | csc x − cot x| + C and
sec2 xdx = tan x + C and
Z
Z
= ln | sec u(x) + tan u(x)| + C
[csc u(x)]u′ (x) dx
= ln | csc u(x) − cot u(x)| + C
[sec2 u(x)]u′ (x) dx = tan u(x) + C
Z
Z
2
csc xdx = − cot x + C and
[csc2 u(x)]u′ (x) dx = − cot u(x) + C
Z
Z
sec x tan xdx = sec x + C and
[sec u(x) tan u(x)]u′ (x) dx = sec u(x) + C
Z
Z
csc x cot xdx = − csc x + C and
[csc u(x) cot u(x)]u′ (x) dx = − csc u(x) + C
Examples:
Z
Z
(sec x tan x + cos x) dx =
Z
sec2 x dx = tan x + C
sec x tan x dx +
Z
cos x dx = sec x + sin x + C.
R
R
Note that tan xdx = ln | sec x| + C and csc xdx = − ln | csc x + cot x| + C.
You have to be able to proof the integrals for tan x, csc x, sec x and cot x.
Study Examples 1 and 2 (Section 4.4, Stewart).
83
Integration
5.4
Integrals of Inverse Trigonometric Functions
(1 hour 40 minutes contact time)
You should study Section 6.6 (Stewart) together with this section.
From the derivatives of the inverse trigonometric functions we may derive the integrals.
Theorem 40
Z
Z
u′ (x)
1
√
p
dx = arcsin u(x) + C
dx = arcsin x + C and
1 − x2
1 − [u(x)]2
Z
Z
1
u′ (x)
√
p
dx = − arccos u(x) + C
dx = − arccos x + C and
1 − x2
1 − [u(x)]2
Z
Z
1
u′ (x)
dx
=
arctan
x
+
C
and
dx = arctan u(x) + C
1 + x2
1 + [u(x)]2
Z
Z
u′ (x)
1
dx
=
−
arccot
x
+
C
and
dx = − arccot u(x) + C
1 + x2
1 + [u(x)]2
Z
Z
u′ (x)
1
p
√
dx = arcsec u(x) + C
dx = arcsec x + C and
x x2 − 1
u(x) [u(x)]2 − 1
Z
Z
1
u′(x)
√
p
dx = − arccsc u(x) + C
dx = − arccsc x + C and
x x2 − 1
u(x) [u(x)]2 − 1
x
dx we may do the following:
4 − 3x4
Z
Z
x
1
x
√
p
dx
dx = √
4
4 − 3x
4
1 − 3x4 /4
Z
1
x
q
=
√
2 dx
2
2
1−
3x /2
√
Z
3x
1
q
(∗)
= √
√
2 dx
2 3
2
1−
3x /2
√ 2 1
3x /2 + C.
= √ arcsin
2 3
Example: In the calculation of
Z
√
In stead of step (∗) above, we could also have used substitution. Let
√ 2
3x
u=
2
in the previous step. Then we have that
√
du √
= 3x and du = 3xdx
dx
84
Integration
so that
1
2
Z
q
1−
x
√
3x2 /2
1
dx
=
2
2
=
Z
√
1
√
Z
1
1
√ du
1 − u2 3
√
1
du
1 − u2
2 3
1
= √ arcsin u + C
2 3
√
1
3x2 /2 + C
= √ arcsin
2 3
Study Examples 8 and 9 (Section 6.6, Stewart).
5.5
Substitution Rule
(1 hour contact time)
You should study Section 4.5 (Stewart) together with this section.
To integrate composite functions, we use the substitution rule.
Theorem 41
Substitution rule. If u = g(x) is a differentiable function with an interval I as range,
and f is continuous on I, and h is an indefinite integral of f , then we have that
Z
Z
′
f (g(x))g (x) dx = f (u) du = h(u) + C = h(g(x)) + C.
The substitution rule for integration is equivalent to the chain rule for differentiation.
We can also explain the substitution rule in the following way. Consider the composite
d
[h(g(x))] = h′ (g(x))g ′(x).
function h(g(x)). By making use of the chain rule we see that
dx
It then follows that
d h(g(x)) = h′ (g(x))g ′(x) dx
Z
Z
⇒
d h(g(x)) = h′ (g(x))g ′(x) dx
Z
⇒ h(g(x)) + C = h′ (g(x))g ′(x) dx
From this you clearly see that the substitution rule for integration is the inverse of the
chain rule for differentiation. If the integrand (the function to be integrated) is in standard
form, then only the outer function is integrated. Standard form means a function with a
function enclosed, multiplied by the derivative of the enclosed function. The substitution
rule can be expressed in words as follows: If an integrand consists of a function with a
function enclosed, multiplied by the derivative of the enclosed function, then the integral
of the integrand is determined by integrating the outer function only.
85
Integration
Example:
Z
2x sin(x2 )dx = − cos(x2 ) + C.
Sometimes the integrand differs from a standard form by only a constant. In such a case
that particular integrand may be multiplied by the constant to convert it to the standard
form. But in order to not change the original integral that is to be calculated, one then
has to multiply the integral by the reciprocal of the constant involved.
Example:
Z
Z
1
1
2
x sin(x )dx =
2x sin(x2 )dx = − cos(x2 ) + C.
2
2
It may also happen that the integrand is not in the standard form and that it cannot be
converted to the standard form by multiplication of a constant. In this case the integral
may be calculated by making a suitable substitution - i.e. by changing the integration
variable.
R √
u−1
Example: If x2 1 + 2x dx has to be calculated, let u = 1+2x and note that x =
2
du
= 2, so that du = 2dx. Therefore
and
dx
Z
Z
√
(u − 1)2 √ 1
2
u du
x 1 + 2x dx =
4
2
Z
1
u5/2 − 2u3/2 + u1/2 du
=
8
1 2 7/2 4 5/2 2 3/2
+C
u − u + u
=
8 7
5
3
1
1
1
(1 + 2x)7/2 − (1 + 2x)5/2 + (1 + 2x)3/2 + C.
=
28
10
12
Study Examples 1, 2, 3, 4 and 5 (Section 4.5, Stewart).
86
Integration
Study Unit 6
Numbers systems
Time Allocation
26 hours (14 hours contact time and 3 tutorial sessions).
Prerequisite Study
Natural, integer, rational, irrational and real number system
Study Outcomes
When you have completed this section, you should be able to do the following, using
standard mathematical notation.
1. Give brief summaries of the characteristic properties and uses of the real number
systems (Section 1.2).
2. Formulate the basic definitions and assumptions on which the algebraic structure
as well as the ordering structure of the real number system are founded and derive
lists of other well known properties from them and present these lists without proof
(Section 1.3).
3. Formulate the basic properties of the natural numbers (Section 1.4.1) and deduct
the proofing method of mathematical induction from it.
4. Describe the structure of proofs by means of mathematical induction and prove
mathematical statements in this way (Section 1.4.2).
5. Describe which properties are added and which are lost if a transition is made from
natural numbers to integers (Definition 1.5.1).
6. Formulate and implement the algorithm for division (Theorem 1.5.2) for positive as
well as negative integers.
7. Define the concept GCD of two integers (Definition 1.5.3) and calculate it by means
of the Euclidian algorithm (Section 1.5.4) of integers.
8. Formulate, prove and use Theorem 1.5.5 to determine whether a diophantic equation
has a solution, and, if it has a solution, determine the solution thereof.
87
88
Numbers systems
9. Isolate rational numbers as quotients of integers from R (Definition 1.6.1) and point
out which properties are added and which are lost when a transition is made from
integers to the rational numbers.
10. Prove that R contains elements which are not rational (Theorem 1.6.2).
11. Describe the difference between rational and irrational numbers by referring to their
decimal representations.
12. Explain why special symbols that often occur, are used for irrational numbers.
13. Define and use the concepts complex number system (Definition 3.2.1), standard
form of complex numbers (Definition 3.3.3), real and imaginary parts of complex
numbers (Definition 3.3.4) and powers of complex numbers (Definition 3.3.5) to
calculate sums, products and powers of complex numbers.
14. Formulate and use the algebraic properties, with special reference to the zero element, opposites, identity element and inverses (Section 3.2.3) to calculate differences
and sums of complex numbers.
15. Show that real numbers is a subset of complex numbers (Section 3.3.1).
16. Solve equations which are unsolvable in R, in C.
17. Define and use the concepts complex conjugate (Definition 3.4.1) and modulus or
absolute value (Definition 3.4.2) to calculate complex conjugates, modulus or absolute values and quotients of complex numbers.
18. Formulate, prove and use the properties of the conjugate and modulus of a complex
number as given in Theorems 3.4.3, 3.4.4 and 3.4.5, in proofs of other properties.
19. Formulate, prove and use Theorem 3.4.6 to determine complex roots of real polynomials.
20. Represent complex numbers geometrically as points or as vectors in a rectangular
set of axes.
21. Define the polar form of complex numbers (Definition 3.6.2) and formulate, prove
and use Theorem 3.6.1 to do inversions between standard form and polar form of
complex numbers.
22. Formulate, prove and use multiplication (Theorem 3.7.1) and division (Corollary
3.7.3) of complex numbers in polar form to calculate the product and quotient of
complex numbers.
23. Formulate, prove and use the theorem of de Moivre (Theorem 3.7.2) for integer
exponents to calculate powers of complex numbers.
24. Formulate and use Theorem 3.7.4. to calculate n-th power roots (n ∈ N) of complex
numbers presenting it geometrically.
25. Give the exponential representation r expiθ of the modulus argument form and use
it in calculations of complex numbers.
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Numbers systems
References
Introductory Algebra: Chapters 1 and 3.
Study guide: Sections 6.1, 6.2.
Unimaths Intro Workbook: Section 1.
Preparation
Read the Sections referred to in the textbook and in the study guide.
Work through the following examples from Introductory Algebra:
Section 1.4: 1;
Section 1.5: 1;
Section 3.2: 1, 2;
Section 3.3: 1;
Section 3.4: 1, 2;
Section 3.5: 1;
Section 3.6: 1;
Section 3.7: 1, 3.
Answer the following questions and submit your answers at the beginning of the first
contact session of this study unit:
1. If ab = 0, what do you know of a and b?
2. If a + c = b + c, what do you know of a and b?
3. If ac = bc, what do you know of a and b?
4. Complete
(a) a(−b) =
(b) −(−a) =
(c) (−a)(−b) =
(d) −(a + b) =
(e) −(a − b) =
(f) a(b − c) =
(g) (a + b)c =
(h) (a + b)(c + d) =
a
b
+
c
b
=
(j) b, d 6= 0,
a
b
a
b
+
c
d
(i) b 6= 0,
(k) b, d 6= 0,
(l) b, c, d 6= 0,
·
a
b
c
d
=
=
÷
c
d
=
5. If a > 0 and b > 0, what do you know of ab?
6. If a > 0 and b < 0, what do you know of ab?
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Numbers systems
7. If a > b, what do you know of a + c and b + c?
8. If a > b, what do you know of a − c and b − c?
9. If a > b, what do you know of 1/a and 1/b?
10. If a > b and c > 0, what do you know of ac and bc?
11. If a > b and c > 0, what do you know of a/c and b/c?
12. If a > b and c < 0, what do you know of ac and bc?
13. If a > b and c < 0, what do you know of a/c and b/c?
14. Calculate the following.
(a) (1 − i3) + (−2 + i4)
(b) (1 − i3) − (−2 + i4)
(c) (1 − i3) · (−2 + i4)
(d) (1 − i3) ÷ (−2 + i4)
15. Write in polar form.
(a) 2 + 3i
(b) −5 − i
16. Write in standard form.
(a) 3 cis π3
(b) 2 cis 3π
4
Exercising of Skills
Work through the following examples from Introductory Algebra:
Section 1.4: 2;
Section 3.2: 3;
Section 3.3: 2, 3, 4, 5, 6;
Section 3.4: 3, 4;
Section 3.5: 2;
Section 3.6: 2, 3;
Section 3.7: 2, 4, 5, 6.
Do the following exercises:
Introductory Algebra:
Ex. 1.4: 1, 3, 6, 13, 15, 17, 19, 20, 21, 22;
Ex. 1.5: 1(4), 2(2), 2(4), 3, 5;
Ex. 3.3: 1(6), 1(7), 2, 3(2), 3(3), 3(4), 3(5), 4;
Ex. 3.4: 1, 2(2), 2(5), 4, 7, 8, 9 , 10, 12;
Ex. 3.5: 1, 2, 3;
Ex. 3.6: 1, 2;
Ex. 3.7: 2, 5, 10, 11.
Numbers systems
91
Test Yourself
1. Define the real number system.
2. Make a list of the algebraic properties of real numbers.
3. What is the zero element of real numbers?
4. What is the identity element of real numbers?
5. Write down the opposite of the real number a.
6. Write down the reciprocal of the real number a.
7. Define the set R+ .
8. Use the set R+ to explain the following concepts for real numbers:
(a) Positive
(b) Negative
(c) Greater than
(d) Less than
9. What does it mean that R is unbounded.
10. What occurs between each two real numbers?
11. Define the natural number system.
12. Make a list of the algebraic properties of natural numbers.
13. What is the zero element of natural numbers?
14. What is the identity element of natural numbers?
15. Write down the opposite of the natural number a.
16. Write down the reciprocal of the natural number a.
17. Which property of natural numbers makes mathematical induction as proofing technique possible?
18. Describe mathematical induction as proofing technique in words.
19. Use mathematical induction to proof a variety of mathematical truths. (Ex 1-4, De
la Rosa et-al.)
20. Define the integer number system.
21. Make a list of the algebraic properties of integers.
22. What is the zero element of integers?
23. What is the identity element of integers?
24. Write down the opposite of the integer a.
25. Write down the reciprocal of the integer a.
26. Which shortcoming of natural numbers is fulfilled by integers?
27. Write down the division algorithm for integers.
28. Show by way of an example, how the division algorithm works.
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Numbers systems
29. Use the Euclidian algorithm to determine the GCD between two integers and write
the GCD as linear combination of the two numbers. (Ex 1-5, De la Rosa et-al.)
30. Write down the Bezout identity.
31. What is an diophantine equation?
32. What is guaranteed by the Bezout identity?
33. Write down the theorem stating when a diophantine equation will have a solution.
34. When will a diophantine equation have a solution?
35. Prove the theorem stated in question 33.
36. Determine solutions of diophantine equations. (Ex 1-5, De la Rosa et-al.)
37. Define the rational number system.
38. Make a list of the algebraic properties of rational numbers.
39. What is the zero element of rational numbers?
40. What is the identity element of rational numbers?
41. Write down the opposite of the rational number a.
42. Write down the reciprocal of the rational number a.
43. Which shortcoming of integers is fulfilled by rational numbers?
44. What notations are used for rational numbers?
45. Which decimal numbers are rational numbers?
46. Which shortcoming of rational numbers is fulfilled by irrational numbers?
47. Which decimal numbers are irrational numbers?
48. What is the intercept between the rational and irrational numbers?
49. What is the union of the rational and irrational numbers?
50. What does it mean that natural numbers are properly included in integers, integers
properly included in rational numbers, rational numbers properly included in real
numbers and irrational numbers properly included in real numbers?
51. How can that stated in question 50 be represented symbolically?
52. What can be done with rational numbers which can not be done with irrational
numbers? Two things.
53. Give examples of special symbols used for irrational numbers.
54. Why are special symbols used for irrational numbers?
√
55. Prove that 2 is not a rational number.
56. Define the complex number system.
57. Why has the complex number system been developed?
58. What is the meaning (value) of the number i?
59. Give four different notations of complex numbers.
Numbers systems
93
60. When will two complex numbers be equal?
61. Indicate that the real numbers are a subset of complex numbers.
62. Make a list of the algebraic properties of complex numbers.
63. What is the zero element of complex numbers?
64. What is the identity element of complex numbers?
65. Derive an expression for the opposite of the complex number z = a + ib.
66. Derive an expression for the reciprocal of the complex number z = a + ib.
67. Write down the following for the complex number z = a + ib:
(a) Opposite
(b) Reciprocal
(c) Absolute value
(d) Complex conjugate
68. Determine the opposite, reciprocal, sum, difference, product, quotient, absolute
value, complex conjugate and combinations thereof for complex numbers. (Ex 3-3
and 3-4, De la Rosa et-al.)
69. For the equation ax2 + bx + c = 0, describe the nature of the solutions when
(a) b2 − 4ac = 0.
(b) b2 − 4ac > 0.
(c) b2 − 4ax < 0.
70. Solve equations in C. (Ex 3-3 and 3-4, De la Rosa et-al.)
71. Prove the following properties for complex numbers:
(a) z + z̄ = 2Re(z)
(b) z − z̄ = i2Im(z)
(c) |z| = 0 ⇔ z = 0
(d) |z̄| = |z|
(e) zz̄ = |z|2
(f) z̄¯ = z
(g) z ± w = z̄ ± w̄
(h) zw = z̄ w̄
(i) z/w = z̄/w̄, w 6= 0
(j) z w̄ + z̄w = 2Re(z w̄) = 2Re(z̄w)
(k) z w̄ − z̄w = i2Im(z w̄) = −i2Im(z̄w)
(l) |zw| = |z||w|
(m) |z/w| = |z|/|w|
(n) Re z ≤ |z|
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Numbers systems
(o) Im z ≤ |z|
(p) |z + w| ≤ |z| + |w|
72. Use the properties listed in question 71 to prove a variety other properties. (Ex 3-4,
De la Rosa et-al.)
73. Prove that the complex roots of polynomials with real coefficients, occur in conjugate
pairs.
74. Use the theorem stated in question 73 to determine the roots of a polynomial with
real coefficients, if one of the complex roots is given. (Ex 3-4, De la Rosa et-al.)
75. Represent complex numbers and sets of complex numbers geometrically. (Ex 3-5,
De la Rosa et-al.)
76. Convert complex numbers given in standard form, to polar form. (Ex 3-6, De la
Rosa et-al.)
77. Convert complex numbers given in polar form, to standard form. (Ex 3-6, De la
Rosa et-al.)
78. Prove that if z1 = r1 cis θ1 and z2 = r2 cis θ2 then z1 z2 = r1 r2 cis (θ1 + θ2 ).
79. Prove that if z = r cis θ 6= 0 then z n = r n cis (nθ), n ∈ Z.
80. Prove that if z1 = r1 cis θ1 and z2 = r2 cis θ2 6= 0 then
z1
z2
=
r1
r2
cis (θ1 − θ2 ).
81. Determine products, powers, quotients and roots for complex numbers using polar
form. (Ex 3-7, De la Rosa et-al.)
82. Solve equations for z ∈ C. (Ex 3-7, De la Rosa et-al.)
83. Write complex numbers in the form reiθ . (Ex 3-7, De la Rosa et-al.)
84. Determine products, powers, quotients and roots for complex numbers using the
form reiθ . (Ex 3-7, De la Rosa et-al.)
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Numbers systems
6.1
Real Number System
The primary aim in this section is to give you an overview on distinguishing properties
of two of the number systems that are often used in mathematics. Also to explain why
natural numbers, (the number system which may be considered to be a subsystem of
the real numbers), continue to be maintained independently as a number system. While
discussing these topics, we shall familiarise you with mathematical induction as a method
of proof and stress the importance of it for the entire discipline.
The second aim of this section is to demarcate integers as a subset of R and to provide
you with the ability to use one of the special calculation skills which is based on the
typical properties of integers. The algorithm for division and the Euclidian algorithm are
powerful arithmetic tools which come into play in several of the fields of mathematics and
anybody possessing a specialised knowledge of mathematics should be able to use them.
The third aim of this section is to establish an appreciation with you, the learner, for the
role of rational numbers and irrational numbers within the context of R, by illustrating
how rational numbers can be written by means of integers. In doing so we shall stress the
point that the rational number system is the most important mathematical tool in any
mathematical environment.
6.1.1
Real Numbers
(50 minutes contact time)
You should study Chapter 1 (Introductory Algebra), Sections 1.1, 1.2 and 1.3 together
with this section.
The real number system is going to be studied as an algebraic structure. To define
the elements of the structure, there first has to be stated how the elements look like and
secondly how the calculations addition and multiplication work. Real numbers are defined
as
1. set with elements real numbers with
2. addition a + b and
3. multiplication ab.
The basic properties (axioms) of real numbers are listed below. All other properties are
derived from this. For addition the following holds.
1. Closed: If a, b ∈ R then a + b ∈ R.
2. Associative: (a + b) + c = a + (b + c),
3. Commutative: a + b = b + a,
∀a, b, c ∈ R.
∀a, b ∈ R.
4. Zero element: The number 0 so that a + 0 = a,
∀a ∈ R.
5. Opposite: ∀a ∈ R there is a number −a so that a + (−a) = 0.
For multiplication the following holds.
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Numbers systems
1. Closed: If a, b ∈ R then ab ∈ R.
2. Associative: (ab)c = a(bc),
3. Commutative: ab = ba,
∀a, b, c ∈ R.
∀a, b ∈ R.
4. Identity element: The number 1 so that a · 1 = a,
5. Reciprocal: ∀a ∈ R, a 6= 0 there is a number
1
a
∀a ∈ R.
= a−1 so that a ·
1
a
= 1.
Addition is distributive over multiplication.
a(b + c) = ab + ac, ∀a, b, c ∈ R.
The difference between two real numbers is defined as the sum between one element and
the opposite of the other.
a − b = a + (−b)
The quotient between two real numbers is defined as the product between one element
and the reciprocal of the other.
a
1
= a · , b 6= 0
b
b
From the axioms the following properties can be deducted and proved. Study the proofs of
the properties illustrated in the textbook. It will improve your understanding of algebraic
structures and the properties of numbers. The proofs of these properties are not for
examination purposes.
a·0=0
a(−b) = −ab
−(−a) = a
(−a)(−b) = ab
−(a + b) = −a − b
−(a − b) = b − a
ad + bc
a c
+ =
b d
bd
a c
ac
· =
b d
bd
To determine the ordering (greater than, smaller than and equal) of real numbers, the
concepts positive and negative are needed. For this purpose the set of positive real
numbers R+ is defined.
1. x > 0 ⇔ x ∈ R+
2. x, y ∈ R+ ⇒ x + y ∈ R+ , xy ∈ R+
3. ∀x ∈ R ⇒ x ∈ R+ of/or x = 0 of/or − x ∈ R+
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Numbers systems
R+ is a subset of R. A real number is positive if x ∈ R+ and negative if −x ∈ R+ . From
this is follows that for g, k ∈ R
g > k, g − k ∈ R+ ,
g < k, k − g ∈ R+ ,
g = k, g − k = 0.
From this it follows that
1 ∈ R+
a > b or a = b or a < b
1
a>0⇒ >0
a
a > 0, b > 0 ⇒ ab > 0
a > 0, b < 0 ⇒ ab < 0
a>b⇒ a+c>b+c
a>b⇒ a−c> b−c
1
1
a>b⇒ <
a
b
a > b, c > 0 ⇒ ac > bc
b
a
a > b, c > 0 ⇒ >
c
c
a > b, c < 0 ⇒ ac < bc
b
a
a > b, c < 0 ⇒ <
c
c
∀ a, b ∈ R, a < b, ∃ c ∈ R ∋ a < c < b
R is unbounded
Study the proofs of the properties illustrated in the textbook. It will improve your understanding of algebraic structures and the properties of numbers. The proofs of these
properties are not for examination purposes.
In real numbers the following set notation is used.
(a, b) = {x ∈ R|a < x < b},
[a, b) = {x ∈ R|a ≤ x < b},
(a, b] = {x ∈ R|a < x ≤ b},
[a, b] = {x ∈ R|a ≤ x ≤ b},
6.1.2
open interval
semi closed interval
semi closed interval
closed interval
Natural Numbers
(1 hour 30 minutes contact time)
You should study Chapter 1 (Introductory Algebra), Sections 1.4.1 and 1.4.2 together
with this section.
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Numbers systems
We have a very natural feeling for natural numbers since these are the numbers we learned
to count.
1, 2, 3, 4, 5, · · · , N ⊂ R
The natural numbers have characteristic mathematical properties that leads to handy
mathematical techniques. For natural numbers the basic properties (axioms) are very
similar to that of real numbers since it is a subset of real numbers.
1. Addition and multiplication as R
2. No opposite
3. Subtraction not closed m > n, m, n ∈ N, n − m ∈/N
4. No reciprocal
5. Division not closed
6. Identity element 1 ∈ N
7. n ≥ 1, ∀n ∈ N
8. x ∈ N ⇒ x + 1 ∈ N
9. x ∈ N ∃/ y ∈ N ∋ x < y < x + 1
10. M ⊂ N ∋ 1 ∈ M, x ∈ M, x + 1 ∈ M ⇒ M = N
The last property is known as the principle of mathematical induction. It is used to prove
that statements in which natural numbers occur, holds for all natural numbers. This
proofing method can summarised as follows.
1. Show statement holds for n = 1.
2. Accept statement holds for n > 1.
3. Prove statement holds for n + 1.
Mathematical Induction
Adding to the textbook, some more examples of mathematical induction are given.
Mathematical induction can be used to confirm an expectation about the outcome of a
process which follows a regular pattern.
Example 1:
Use mathematical induction to prove that for all n ∈ N:
1+2+···+n =
n(n + 1)
2
Proof:
1. The statement is true for n = 1, because:
RS =
2
1(1 + 1)
= = 1 = LS.
2
2
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Numbers systems
2. Suppose the statement is true for n = k, then:
1+2+···+k =
k(k + 1)
, ∀ k ∈ N.
2
3. We shall now prove that the statement is true for n = k + 1 :
{We therefore have to prove that:
1 + 2 + · · · + (k + 1) = (k+1)((k+1)+1)
= (k+1)(k+2)
, ∀ k ∈ N.}
2
2
k(k + 1)
+ (k + 1)
2
k(k + 1) (k + 1)2
=
+
2
2
(k + 1)(k + 2)
= RS.
=
2
LS = 1 + 2 + · · · + k + (k + 1) =
∴ The statement is true.
Use mathematical induction to prove a divisibility property.
Example 2:
For all n ∈ N , 22n − 1 is divisible by 3.
Proof:
1. The proposition is true for n = 1, because:
22(1) − 1 = 4 − 1 = 3 and we know that 3 is divisible by 3.
2. Suppose the proposition is true for n = k, then:
22k − 1 is divisible by 3 i.e. an m ∈ N exists, such that 22k − 1 = 3m
3. We shall now prove that the proposition is true for n = k + 1 :
{We therefore have to prove that: 22(k+1) − 1 is divisible by 3.}
Consider:
22(k+1) − 1 =
=
=
=
=
=
=
22k+2 − 1
22k .22 − 1
22k .4 − 1
22k .(3 + 1) − 1
22k .3 + (22k − 1)
22k .3 + 3m
3 22k + m
From (2) it follows that 22k − 1 is divisible by 3, but 22k .3 is a multiple of 3 and
therefore it is also divisible by 3, i.e. 22(k+1) − 1 is the sum of 2 quantities which are
both divisible by 3 and therefore it is also divisible by 3.
∴ The proposition is true.
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Numbers systems
Use of mathematical induction to prove inequalities.
Example 3:
For all natural numbers n ≥ 3 it is true that 2n + 1 < 2n .
Proof:
1. The inequality is true for n = 3, because:
LS: 2(3) + 1 = 7
RS: 23 = 8 i.e. LS= 7 < 8 =RS
2. Suppose the inequality is true for n = k, then:
2k + 1 < 2k , for all natural numbers k ≥ 3.
3. We shall now prove that the inequality is true for n = k + 1 :
{I.e. we aim to prove that 2(k + 1) + 1 < 2k+1 }
2(k + 1) + 1
=
=
2k + 2 + 1
2k + 1 + 2
hypothesis
<
<
=
=
2k + 2
2k + 2k because 2 < 2k for all natural numbers k ≥ 2
2.2k
2k+1
∴ the inequality is true.
Use mathematical induction to prove that a sequence possesses a certain property.
Example 4:
Define a sequence a1 , a2 , a3 , · · · as follows: a1 = 2 and ak = 5ak−1 for all natural numbers
k ≥ 2.
Use mathematical induction to prove that the terms of the sequence satisfy the formula
an = 2.5n−1 for all natural numbers n ≥ 1.
Proof:
1. The formula is true for n = 1, because:
LS: a1 = 2
RS: 2.51−1 = 2.50 = 2.1 = 2.
LS = RS
2. Suppose the formula is true for n = k, then:
ak = 2.5k−1 for all natural numbers k ≥ 1.
3. We shall now prove that the inequality is true for n = k + 1 :
{This means that we have to prove that: ak+1 = 2.5(k+1)−1 = 2.5k }
ak+1
def. of sequence
=
=
induction hypothesis
=
=
=
∴ The formula is true.
5a(k+1)−1
5ak
5.(2.5k−1)
2(5.5k−1)
2.51+k−1 = 2.5k .
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Numbers systems
6.1.3
Integer Numbers
(1 hour 30 minutes contact time)
You should study Chapter 1 (Introductory Algebra), Section 1.5 together with this section.
A shortcoming of N is that it is not closed with respect to subtraction.
m > n, m, n ∈ N, n − m ∈/N
To address this shortcoming integers Z are defined as (Definition 1.5.1)
Z = {−n, n ∈ N} ∪ {0} ∪ N.
Integer numbers are closed for addition, subtraction and multiplication, but not for division.
Division algorithm for integers states that for any
a, b ∈ Z, b > 0 ∃ q, r ∈ Z, ∋ a = qb + r, 0 ≤ r < b.
(Theorem 1.5.2, Introductory Algebra)
An algorithm is a set of certain steps that is repeated until a satisfactory answer is
achieved. It is usually programmable. The division algorithm can be programmed as
follows. For a > 0
r1 = a − 1b, 0 ≤ r1 < b ×
r2 = a − 2b, 0 ≤ r2 < b ×
..
.
√
rn = a − nb, 0 ≤ rn < b
Example: 45 ÷ 7
a = nb + rn , 0 ≤ rn < b
45 − 1 · 7 = 38,
45 − 2 · 7 = 31,
45 − 3 · 7 = 24,
45 − 4 · 7 = 17,
45 − 5 · 7 = 10,
45 − 6 · 7 = 3,
0 ≤ 38 < 7 ×
0 ≤ 31 < 7 ×
0 ≤ 24 < 7 ×
0 ≤ 17 < 7 ×
0 ≤ 10 < 7 ×
√
0≤3<7
45 = 6 · 7 + 3
For a < 0
r1 = a + 1b, 0 ≤ r1 < b ×
r2 = a + 2b, 0 ≤ r2 < b ×
..
.
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Numbers systems
rn = a + nb, 0 ≤ rn < b
√
a = −nb + rn , 0 ≤ rn < b
Example: −45 ÷ 7
−45 + 1 · 7 = −38,
−45 + 2 · 7 = −31,
−45 + 3 · 7 = −24,
−45 + 4 · 7 = −17,
−45 + 5 · 7 = −10,
−45 + 6 · 7 = −3,
−45 + 7 · 7 = 4,
0 ≤ −38 < 7 ×
0 ≤ −31 < 7 ×
0 ≤ −24 < 7 ×
0 ≤ −17 < 7 ×
0 ≤ −10 < 7 ×
0 ≤ −3 < 7 ×
√
0≤4<7
−45 = −7 · 7 + 4
The algorithm can be adopted for b < 0.
Definition: The greatest common divider (GCD) (Definition 1.5.3, Introductory
Algebra) is
d = gcd(a, b), ∀c|a, b ⇒ c|d, a, b, c, d ∈ Z.
The GCD can be determined using prime factors. For large numbers it can get very
tedious. Think about the following statements for
a = bq + r, a, b, q, r ∈ Z.
1. Every common factor of b and r is a factor of a and therefore a common factor of a
and b.
2. Every common factor of a and b is a factor of r and therefore a common factor of r
and b.
3. Therefore (a and b) and (r and b) have these same common factors.
4. Therefore gcd(a, b)=gcd(b, r).
Study the example in Section 1.5.3. (Introductory Algebra). In general we have the
Euclidian algorithm that is based on the division algorithm.
a>0
a = qb + r0 , 0 ≤ r0 < b
b = q0 r0 + r1 , 0 ≤ r1 < r0
r0 = q1 r1 + r2 , 0 ≤ r2 < r1
..
.
rn−2 = qn−1 rn−1 + rn , 0 ≤ rn < rn−1
rn−1 = qn rn + 0
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Numbers systems
The last non-zero remainder is the GCD so that the gcd(a, b) = rn . It is possible to write
the GCD as a linear combination of the numbers a and b
gcd(a, b) = ua + vb, v, u ∈ Z.
This is called the Bezout identiy. The Bezout identiy guarantees solutions for diophantine
equations (Theorem 1.5.5, Introductory Algebra).
Theorem:
ax + by = c, a, b > 0, a, b, c ∈ Z have a solution x, y ∈ Z ⇔ gcd(a, b)|c
Proof:
Let d = gcd(a, b)
⇒ a = a′ d, b = b′ d, a′ , b′ ∈ Z.
Suppose ax + by = c has a solution in Z. Therefore
∃x, y ∈ Z, ∋ c = ax + by
= a′ dx + b′ dy
= d(a′ x + b′ y)
Therefore d|c.
Suppose d|c.
Therefore c = c′ d, c′ ∈ Z.
d = ua + vb
⇒ c = c′ (ua + vb)
= a(c′ u) + b(c′ v)
The equation therefore have the solution
x = c′ u, y = c′ v, x, y ∈ Z.
✷
Study Example 1 (Section 1.5.4, Introductory Algebra).
6.1.4
Rational and Irrational Numbers
(50 minutes contact time)
You should study Chapter 1 (Introductory Algebra), Section 1.6 together with this section.
Rational Numbers
A shortcoming of N is that it is not closed with respect to subtraction. To address this
shortcoming rational numbers Q are defined (Definition 1.6.1) as
na
o
Q=
, a, b ∈ Z, b 6= 0 .
b
Rational numbers Q are now also closed for division. Rational numbers can be written
as a fraction or a finite/repeating decimal number. All finite/repeating decimal numbers
can be written as the ratio of two integers. The division algorithm makes it possible.
104
Numbers systems
Irrational numbers
A shortcoming of Q is that it is not closed with respect to extracting roots. To address this
shortcoming partly irrational numbers are defined as all infinite, non-repeating decimal
numbers. Irrational numbers can therefore not be written as the ratio of two integer
numbers. It can also not be represented fully as a decimal number. Therefore special
symbols are used to present irrational numbers for example
√ √
3
π, e, 2, 4.
Note that π ≈ 22/7.
For rational and irrational numbers it hold that
I ⊂ R,
Q ∪ I = R,
Q ∩ I = ∅.
Study the following theorem and its proof. This theorem and its proof is not for examination purposes, but is needed for the followup theorem.
Theorem:
a|b ⇔ a|b2 , a, b ∈ Z, a prime
Proof:
1.
a|b ⇒ a|b2 :
Suppose a|b
⇒ b = ac, c ∈ Z
⇒ b2 = (ac)2 = a(ac2 )
⇒ a|b2
2.
a|b2 ⇒ a|b :
Suppose a|b2
Let b = aq + r, q, r ∈ Z
⇒ b2 = a2 q 2 + 2aqr + r 2 = a(aq 2 + 2qr) + r 2
But a|b2
⇒ r 2 = 0 ⇒ r = 0 ∋ b2 = a(aq 2 ) = (aq)2
⇒ b = aq
⇒ a|b
✷
√
√
The following theorem is illustrated for 2, but can be done for any a where a is prime
(Definition 1.4.4, Introductory Algebra).
Theorem:
√
2 ∈ I ( ∈/Q)
Proof:
Suppose
√
2∈Q
105
Numbers systems
√
m
, m, n ∈ Z, gcd(m, n) = 1
n
⇒ 2n2 = m2
⇒ 2|m2 ⇒ 2|m
⇒ m = 2p, p ∈ Z
⇒ m2 = 4p2 = 2n2 ⇒ n2 = 2p2
⇒ 2|n2 ⇒ 2|n
⇒ cd(m, n) = 2
⇒
2=
This is in conflict with
gcd(m, n) = 1.
Therefore
√
√
2 ∈/Q ⇒ 2 ∈ I
✷
6.2
6.2.1
Complex Number System
Complex Numbers as a Number System; Basic Operations; Standard Notation
(1 hour 30 minutes contact time)
You should study Sections 3.1, 3.2 and 3.3 (Introductory Algebra) and Section 1 (Unimaths Intro Workbook) together with this section.
In this section we aim to illustrate that there is a final expansion of the number system.
The equation x2 + 1 = 0 does not have a solution in R, but a root can always be found in
the complex number system, written as C, which we are now going to study. However, the
complex numbers lose the properties of ordering which we have in R. In C two operations
are defined, that of addition and multiplication and with respect to these operations the
complex numbers possess all the algebraic properties of R.
What is the solution of the equation x2 + 1 = 0?
Define i =
but ii = 1.
In general
√
x2 + 1 = 0 ⇒ x2 = −1
√
⇒ x = ± −1
⇒ x = ±i
−1 so that i2 = −1 and the solution of x2 + 1 = 0 is x = ±i. Note: i · i = −1
√
−a =
√
√
i2 a = i a,
a ≥ 0.
Definition: Complex numbers are defined as
z = a + ib, a, b ∈ R
106
Numbers systems
with real part
Re (z) = a
and imaginary part
Im (z) = b.
Complex numbers are indicated with C and are the set of ordered numbers a + ib of real
numbers with addition
(a + ib) + (c + id) = (a + c) + i(b + d)
and multiplication
(a + ib)(c + id) = ac + i2 bd + iad + ibc = (ac − bd) + i(ad + bc).
(Definition 3.3.3 and 3.3.4, De la Rosa, et al.)
In the addition and multiplication of complex numbers, the addition and multiplication
of real numbers are used. Real numbers are the subset of complex numbers.
a = a + i0
Two complex numbers are equal when their corresponding components are equal.
a + ib = c + id ⇔ a = c and b = d
Study Examples 1, 2 and 3 (Section 3.3, Introductory Algebra).
Sometimes the notation a + ib = (a, b) (Definition 3.2.1, Introductory Algebra) is used.
It is convenient when presenting complex numbers graphically. Study Examples 1 and 2
(Section 3.2, Introductory Algebra).
The following algebraic properties follows from the properties of real numbers.
For addition the following hold.
1. Closed: If z, w ∈ C then z + w ∈ C.
2. Associative: (z + w) + v = z + (w + v),
∀z, w, v ∈ C.
∀z, w ∈ C.
3. Commutative: z + w = w + z,
4. Zero element: The number 0 = 0 + i0 so that z + 0 = z,
∀z ∈ C.
5. Opposite: ∀z = a + ib ∈ C there is a number −z = −a − ib so that z + (−z) = 0.
The opposite is determined by solving the complex number x + iy in the equation (a +
ib) + (x + iy) = 0 + i0. Difference is defined as the sum of the one element with the
opposite of the other element.
z − w = z + (−w)
For multiplication the following hold.
1. Closed: If z, w ∈ C then zw ∈ C.
2. Associative: (zw)v = z(wv),
∀z, w, v ∈ C.
107
Numbers systems
3. Commutative: zw = wz,
∀z, w ∈ C.
4. Identity element: The number 1 = 1 + i0 so that z · 1 = z,
∀z ∈ C.
5. Reciprocal: ∀z = a + ib ∈ C there is a number
a
−b
1
= z −1 = 2
+i 2
2
z
a +b
a + b2
so that z ·
1
z
= 1.
Addition is distributive over multiplication:
z(w + v) = zw + zv,
∀z, w, v ∈ C.
The reciprocal is determined by solving the complex number x + iy in the equation
(a + ib)(x + iy) = 1 + i0. Quotient is defined as the product of the one element with the
reciprocal of the other element.
z
1
=z·
w
w
Study Example 3 (Section 3.2, Introductory Algebra). All these properties of complex
numbers can be proved using the properties of real numbers.
Example:
⇒
zw = (a + ib)(c + id)
= (ac − bd) + i(ad + bc)
wz = (c + id)(a + ib)
= (ca − db) + i(da + cb)
= (ac − bd) + i(ad + bc)
zw = wz
Definition: Further it also holds that ∀z ∈ C
z 0 = 1, z 6= 0
z n = z · z · z . . . z, n faktore/factors, n ∈ N
1
z −m = m , m ∈ N, z 6= 0
z
(Definition 3.3.5, Introductory Algebra)
Study Examples 4, 5 and 6 (Section 3.3, Introductory Algebra). You have to know the
derivation in Example 5.
108
6.2.2
Numbers systems
Complex Conjugate and Modulus or Absolute value
(1 hour 30 minutes contact time)
You should study Section 3.4 (Introductory Algebra) together with this section.
The aim of this section is to investigate the two concepts of complex conjugate and
modulus of a complex number. Division by a complex number cannot be performed
easily. To do this one has to make use of the complex conjugate.
Definition: The complex conjugate is defined as
z̄ = a − ib for z = a + ib.
(Definition 3.4.1, Introductory Algebra)
Definition: The modulus or absolute value is defined as
√
|z| = a2 + b2 for z = a + ib.
(Definition 3.4.2, Introductory Algebra)
Absolute value of real numbers is a special case of this. For z = a + i0
√
a,
a≥0
|z| = a2 =
−a, a < 0
Compare this to Theorem 2.
The complex conjugate can be used to determine quotients. If z = a + ib and w = c + id
it follows that
z
z w̄
(a + ib)(c − id)
ac + bd
bc − ad
1
=
=
= 2
+i 2
=z· .
2
2
w
w w̄
(c + id)(c − id)
c +d
c +d
w
Study Examples 1, 2, and 3 (Section 3.3, Introductory Algebra).
The following properties (Theorems 3.4.3, 3.4.4 and 3.4.5, Introductory Algebra) holds
for complex conjugate and modulus. You have to know all the proofs.
1. z + z̄ = 2Re(z)
Proof:
Let z = a + ib.
z + z̄ = a + ib + a − ib = 2a = 2Re(z)
✷
2. z − z̄ = i2Im(z)
Proof:
Let z = a + ib.
z − z̄ = a + ib − (a − ib) = i2b = i2Im(z)
✷
109
Numbers systems
3. |z| = 0 ⇔ z = 0
Proof:
Let z = a + ib.
√
a2 + b2 = 0 ⇒ a = b = 0 ⇒ z = 0.
√
z = 0 ⇒ a + ib = 0 + i0 ⇒ a = b = 0 ⇒ |z| = a2 + b2 = 0.
|z| = 0 ⇒
✷
4. |z̄| = |z|
Proof:
Let z = a + ib.
|z̄| = |a − ib| =
√
a2 + b2 = |z|
✷
5. zz̄ = |z|2
Proof:
Let z = a + ib.
zz̄ = (a + ib)(a − ib) = a2 + b2 = |z|2
✷
6. z̄¯ = z
Proof:
Let z = a + ib.
z̄¯ = a − ib = a + ib = z
✷
7. z ± w = z̄ ± w̄
Proof:
Let z = a + ib, w = c + id
z±w =
=
=
z̄ ± w̄ =
=
=
(a + ib) ± (c + id)
a ± c + i(b ± d)
a ± c − i(b ± d)
a + ib ± c + id
a − ib ± (c − id)
a ± c − i(b ± d)
✷
110
Numbers systems
8. zw = z̄ w̄
See proof in textbook.
9. z/w = z̄/w̄, w 6= 0
Proof:
z̄ = (z/w)w = (z/w)w̄ ⇒ z̄/w̄ = (z/w)
✷
10. |zw| = |z||w|
Proof:
|zw|2 =
=
=
=
⇒
(zw)(zw)
(zw)(z̄ w̄)
(zz̄)(w w̄)
|z|2 |w|2
|zw| = |z||w|
✷
11. |z/w| = |z|/|w|, w 6= 0
Proof:
|z| = |(z/w)w| = |(z/w)||w|
⇒ |z|/|w| = |z/w|
✷
12. Re z ≤ |z| and Im z ≤ |z|
Proof:
Let z = a + ib
√
Re z = a ≤ a2 + b2 = |z|
√
Im z = b ≤ a2 + b2 = |z|
✷
111
Numbers systems
Proof:
13. |z + w| ≤ |z| + |w|
|z + w|2 =
=
=
=
=
≤
=
=
=
⇒ |z + w| ≤
(z + w)(z + w)
(z + w)(z̄ + w̄)
zz̄ + z w̄ + wz̄ + w w̄
|z|2 + z w̄ + z w̄ + |w|2
|z|2 + 2Re(z w̄) + |w|2
|z|2 + 2|z w̄| + |w|2
|z|2 + 2|z||w̄| + |w|2
|z|2 + 2|z||w| + |w|2
(|z| + |w|)2
|z| + |w|
✷
Study Example 4 (Section 3.4, Introductory Algebra).
According to Theorem 3.4.6 (Introductory Algebra) the roots of polynomials with real
coefficients occurs in conjugate pairs. Compare examples 4, 5 and 6 (Section 3.3, Introductory Algebra). Study this theorem and proof. You should know this theorem and
proof for examination purposes.
6.2.3
Geometric Representation of Complex Numbers.
(50 minutes contact time)
You should study Sections 3.5 (Introductory Algebra) together with this section.
One can represent complex numbers geometrically as points, or as vectors in a rectangular
set of axes. The distance between the point and the origin then represents the modulus
of that number. To represent complex numbers geometrically, the real part comes on
the horizontal axis and the imaginary part comes on the vertical axis. Study Example 1
(Section 3.5, Introductory Algebra).
Absolute value indicates distance. |z| is the distance from the origin to z. |z − w| is the
distance between z and w. Study Example 2 (Section 3.5, Introductory Algebra).
6.2.4
The Polar and Exponential Form of Complex Numbers
(1 hour 30 minutes contact time)
You should study Section 3.6 (Introductory Algebra) together with this section.
In this section the aim is to illustrate that every point in the complex plane is also uniquely
determined by the distance r from the point to the origin and the angle θ ∈ (−π, π]
between the positive x-axis and the line from the origin to that point. This representation
in polar coordinates is especially useful in exponentiation and evolution.
112
Numbers systems
Then we also want to familiarise you with the performing of the operations of multiplication, division, exponentiation and evolution in C when these numbers are presented in
the polar form.
Use polar form also for the representation of complex numbers.
z = a + ib = r(cos θ + i sin θ)
with
r = |z|, a = r cos θ, b = r sin θ, θ ∈ (−π, π]
where θ is called the argument of z, indicated by arg z. (Definition 3.6.2, Introductory
Algebra). The choice of θ ∈ (−π, π] is based on the even and uneven properties of the
cosine and sine functions respectively. Study Theorem 3.6.1 (Introductory Algebra) with
proof. You should know this theorem with proof for examination purposes.
Short hand writing
z = r(cos θ + i sin θ) = r cis θ.
This notation is useful when powers and roots are calculated. In this notation two complex
numbers are equal when
r cis θ = t cis α ⇔ r = t and α = θ + 2kπ, k ∈ Z.
Study Examples 1, 2 and 3 (Section 3.6, Introductory Algebra).
Complex numbers can also be represented by
z = reiθ
where r and θ have the same meaning and are determined in the same way as for z = r cis θ.
6.2.5
Powers and Exponents of Complex Numbers
(4 hours contact time)
You should study Section 3.7 (Introductory Algebra) together with this section.
In this section the aim is to familiarise you with the performing of the operations of multiplication, division, exponentiation and evolution in C when these numbers are presented
in the polar form. The product of two complex numbers written in polar form, is given
by Theorem 3.7.1 (Introductory Algebra).
Theorem:
If z1 = r1 cis θ1 , z2 = r2 cis θ2 then z1 z2 = r1 r2 cis (θ1 + θ2 ).
You have to know the proof of this theorem. De Moivre’s theorem describes the exponentiation of complex numbers for integer powers.
113
Numbers systems
De Moivre’s Theorem:
If z = r cis θ 6= 0 then z n = r n cis (nθ), n ∈ Z.
(Theorem 3.7.2, Introductory Algebra)
The proof is divided into three parts, for n = 0, n > 0 and n < 0. For n > 0 mathematical
induction and Theorem 3.7.1 (Introductory Algebra) are used. For n < 0 z n is taken as
1/z m with n = −m, m > 0. The reciprocal 1/z m is determined by multiplying by the
complex conjugate cis (−mθ). You have to know this proof.
From De Moivre’s theorem (Theorem 3.7.2, Introductory Algebra) it follows that:
Theorem:
If z1 = r1 cis θ1 , z2 = r2 cis θ2 6= 0 then
z1
r1
= cis (θ1 − θ2 )
z2
r2
(Theorem 3.7.3, Introductory Algebra).
Study Examples 1 and 2 (Section 3.7, Introductory Algebra).
In R the equation
x2 = −4
is not solvable, but it can be solved in C with x = ±i2. Theorem 3.7.4 (Introductory
Algebra) describes how complex roots can be determined (extracting roots).
Theorem: The equation
z n = a, n ∈ N, a ∈ C
has n different roots
√
n
z=
√
n
r cis
θ + 2kπ
, k = 0, 1, 2, . . . , n − 1.
n
Study the proof of this theorem. It will improve your understanding of complex numbers.
This proof is not for examination purposes. Study Examples 3, 4, 5 and 6 (Section 3.7,
Introductory Algebra).
In the exponential form z = r cis θ = reiθ these theorems are written as
z1 z2 = r1 eiθ1 r2 eiθ2 = r1 r2 ei(θ1 +θ2 ) ,
z1
r1 eiθ1
r1
=
= ei(θ1 −θ2 ) ,
iθ
2
z2
r2 e
r2
z n = r n einθ ,
√
n
z=
√
n
rei(θ+2kπ)/n , k = 0, 1, 2, . . . , n − 1
which follow directly from exponential properties.
114
Numbers systems
Closed concerning calculations for all number systems can be summarised as follows.
+
N
Z
Q
R
C
×
N
Z
Q
R
C
−
Z
Q
R
C
÷ xy
N x, y ∈ N
x, y ∈ Z
Q
x, y ∈ Q
R
x, y ∈ R
C C x, y ∈ C
These sets are properly included - there are elements that are in the one that are not in
the other.
N⊂Z⊂Q⊂R⊂C
Appendix A
List of Unacceptable Errors
Should you make any of the following errors in a test or in the examination, you will be
penalised by at least one mark.
1.
2.
3.
4.
5.
6.
7.
8.
9.
1
= 0 =⇒ x = ∓a.
x±a
1 1
1
= ± .
x±a
x a
√
2
2
y = a ± x2 =⇒ y = a ± x or a2 ± x2 = a ± x.
log(x + a) = log x + log a or ln(x + a) = ln x + ln a.
log(x + a) = x + a or ln(x + a) = x + a.
ln x
log x
= 1 or
= 1.
x
x
log x
ln x
= 0 or
= 0.
x
x
The = sign omitted.
The lim part omitted, eg.
lim x(x − 1) = x2 − x = . . . in stead of lim x(x − 1) = lim (x2 − x) = lim . . .
x→a
10.
11.
12.
x→a
x→a
x→a
Setting a function equal to its derivative, eg.
−1
1
−1
1
=
in stead of f (x) =
=⇒ f ′ (x) =
.
f (x) =
2
x+2
(x + 2)
x+2
(x + 2)2
R
Omitting the sign, eg.
Z
Z
Z
2
x(x + 1) dx = (x + x) dx in stead of
x(x + 1) dx = (x2 + x) dx.
The dx omitted, eg.
Z
Z
Z
Z
2
x(x + 1) dx = (x + x) in stead of
x(x + 1) dx = (x2 + x) dx.
115
116
Unacceptable Errors
Appendix B
Infinity and the Calculation of Limits
1. Suppose lim f (x) = L > 0 and lim g(x) = ∞ where lim represents each of the
following
lim , lim+ , lim− , lim , lim .
x→a
x→a
x→a
x→+∞
x→−∞
Then we have that:
lim[f (x) + g(x)] = L + ∞ = ∞
lim[f (x) − g(x)] = L − ∞ = −∞
lim[±g(x) · f (x)] = ±∞ · L = ±∞
±∞
±g(x)
=
= ±∞
lim
f (x)
L
f (x)
L
lim
=
= 0.
±g(x)
±∞
2. Suppose lim f (x) = L < 0 and lim g(x) = ∞ where lim represents any of the
following
lim , lim+ , lim− , lim , lim .
x→a
x→a
x→a
x→+∞
x→−∞
Then we have that:
lim[f (x) + g(x)] = L + ∞ = ∞
lim[f (x) − g(x)] = L − ∞ = −∞
lim[±g(x) · f (x)] = ±∞ · L = ∓∞
±∞
±g(x)
=
= ∓∞
lim
f (x)
L
f (x)
L
lim
=
= 0.
±g(x)
±∞
3. Suppose lim f (x) = 0 and lim g(x) = ∞ where lim represents any of the following
lim , lim+ , lim− ,
x→a
x→a
x→a
117
lim ,
x→+∞
lim .
x→−∞
118
Infinity and the Calculation of Limits
Then we have that:
lim[f (x) + g(x)] = 0 + ∞ = ∞
lim[f (x) − g(x)] = 0 − ∞ = −∞
0
f (x)
=
= 0.
lim
±g(x)
±∞
The following is an indefinite form which cannot be calculated directly:
lim[±g(x) · f (x)] = ±∞ · 0.
4. Suppose lim f (x) = ∞ and lim g(x) = ∞ where lim represents any of the following
lim , lim+ , lim− ,
x→a
x→a
x→a
lim ,
x→+∞
lim .
x→−∞
The we have that:
lim[f (x) + g(x)] = ∞ + ∞ = ∞
lim[±f (x) · g(x)] = ±∞ · ∞ = ±∞.
The following are indefinite forms which cannot be calculated directly:
lim[f (x) − g(x)] = ∞ − ∞
±f (x)
±∞
lim
=
.
g(x)
∞
Appendix C
List of Formulas
You should be able to derive all the formulas.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
sin(x ± y) = sin x cos y ± cos x sin y
cos(x ± y) = cos x cos y ∓ sin x sin y
tan x ± tan y
tan(x ± y) =
1 ∓ tan x tan y
1
cos2 x = (1 + cos 2x)
2
1
2
sin x = (1 − cos 2x)
2
d
{[u(x)]n } = n[u(x)]n−1 u′(x)
dx
d
[sin u(x)] = [cos u(x)]u′ (x)
dx
d
[cos u(x)] = [− sin u(x)]u′ (x)
dx
d
[tan u(x)] = [sec2 u(x)]u′ (x)
dx
d
[csc u(x)] = [− csc u(x) cot u(x)]u′ (x)
dx
d
[sec u(x)] = [sec u(x) tan u(x)]u′ (x)
dx
d
[cot u(x)] = [− csc2 u(x)]u′(x)
dx
u′ (x)
d
[arcsin u(x)] = p
dx
1 − [u(x)]2
u′ (x)
d
[arccos u(x)] = − p
dx
1 − [u(x)]2
u′ (x)
d
[arctan u(x)] =
dx
1 + [u(x)]2
119
120
Formulas
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
d
u′ (x)
p
[ arccsc u(x)] = −
dx
u(x) [u(x)]2 − 1
u′ (x)
d
p
[ arcsec u(x)] =
dx
u(x) [u(x)]2 − 1
u′(x)
d
[ arccot u(x)] = −
dx
1 + [u(x)]2
d u(x)
[a ] = au(x) u′ (x) ln a
dx
d u(x)
[e ] = eu(x) u′ (x)
dx
d u′ (x)
′
v(x)
v(x)
v (x) ln[u(x)] + v(x) ·
[u(x)]
= [u(x)]
dx
u(x)
′
d
u (x)
[loga u(x)] =
dx
u(x) ln a
′
u (x)
d
[ln u(x)] =
dx
u(x)
Z
1
[u(x)]n+1 + C
[u(x)]n u′(x) dx =
n+1
Z
[cos u(x)]u′ (x) dx = sin u(x) + C
Z
[sin u(x)]u′ (x) dx = − cos u(x) + C
Z
[sec2 u(x)]u′ (x) dx = tan u(x) + C
Z
[csc u(x) cot u(x)]u′(x) dx = − csc u(x) + C
Z
[sec u(x) tan u(x)]u′ (x) dx = sec u(x) + C
Z
[csc2 u(x)]u′ (x) dx = − cot u(x) + C
Z
[tan u(x)]u′ (x) dx = ln | sec u(x)| + C
Z
[csc u(x)]u′(x) dx = ln | csc u(x) − cot u(x)| + C
Z
[sec u(x)]u′(x) dx = ln | sec u(x) + tan u(x)| + C
Z
[cot u(x)]u′ (x) dx = ln | sin u(x)| + C
Z
u′ (x)
p
dx = arcsin u(x) + C
1 − [u(x)]2
Z
u′ (x)
p
dx = − arccos u(x) + C
1 − [u(x)]2
121
Formulas
37.
38.
39.
40.
41.
42.
43.
Z
Z
Z
Z
Z
Z
Z
u′ (x)
dx = arctan u(x) + C
1 + [u(x)]2
u′ (x)
p
dx = − arccsc u(x) + C
u(x) [u(x)]2 − 1
u′ (x)
p
dx = arcsec u(x) + C
u(x) [u(x)]2 − 1
u′ (x)
dx = − arccot u(x) + C
1 + [u(x)]2
au(x)
au(x) u′ (x) dx =
+C
ln a
eu(x) u′ (x) dx = eu(x) + C
u′ (x)
dx = ln |u(x)| + C
u(x)
122
Formulas
Appendix D
List of Words
absolute extreme value − absolute ekstreemwaarde
absolute maximum value − absolute maksimumwaarde
absolute minimum value − absolute minimumwaarde
absolute value function − absolute waarde-funksie
acceleration − versnelling
analysis − analise
arc length − booglengte
area − oppervlakte
area under a curve − oppervlakte onder ’n kromme
asymptote − asimptoot
average cost function − gemiddelde kostefunksie
average rate of change − gemiddelde tempo van verandering
average value − gemiddelde waarde
calculate − bereken
calculus − differensiaal- en integraalrekening
chain rule − kettingreël
circle − sirkel
closed interval − geslote interval
composition − samestelling
concave downward − konkaaf af
concave upward − konkaaf op
concavity test − konkawiteitstoets
constant − konstant
continuity − kontinuı̈teit
continuous − kontinu
cost function − kostefunksie
critical point − kritieke punt
cube − kubus
curve sketching − krommesketsing
decreasing − dalend
definite integral − bepaalde integraal
definition − definisie
demand function − aanvraagfunksie
derive − aflei
123
124
List of Words
derivative − afgeleide
determine − bepaal
difference − verskil
differentiable − differensieerbaar
differential − differensiaal
differentiate − differensieer
differentiation − differensiasie
discontinuous − diskontinu
displacement − verplasing
domain − definisieversameling
exponential function − eksponensiale funksie
extreme value theorem − ekstreemwaardestelling
finite − eindig
function − funksie
global extreme value − globale ekstreemwaarde
global maximum value − globale maksimumwaarde
global minimum value − globale minimumwaarde
gradient − gradiënt, helling
higher order derivative − hoër orde afgeleide
hyperbolic function − hiperboliese funksie
implicit − implisiet
increasing − stygend
increment − inkrement
indefinite integral − onbepaalde integraal
infinite − oneindig
inflection point − buigpunt
instantaneous rate of change − oombliklike tempo van verandering
integrable − integreerbaar
integrand − integrand
integration − integrasie
integration by parts − faktorintegrasie
intermediate value theorem − tussenwaardestelling
limit − limiet
linear approximation − liniêre benadering
logarithmic differentiation − logaritmiese differensiasie
logarithmic function − logaritmiese funksie
local extreme value − lokale ekstreemwaarde
local maximum value − lokale maksimumwaarde
local minimum value − lokale minimumwaarde
marginal cost − marginale koste
marginal profit function − marginale winsfunksie
marginal revenue function − marginale inkomstefunksie
mean value theorem − middelwaardestelling
number of units − aantal eenhede
obtuse cone − stompkeël
one-to-one function − een-eenduidige funksie
open interval − oop interval
operations − bewerkings
List of Words
partial fraction decomposition − parsiële breukontbinding
piecewise defined function − stuksgewys gedefinieerde funksie
polynomial − polinoom
power function − magsfunksie
power rule − magreël
product − produk
production level − produksievlak
profit function − winsfunksie
quotient − kwosiënt
radian − radiaal
radian measure − radiaalmaat
rational function − rasionale funksie
range − waardeversameling
radical function− wortelfunksies
ratio test − verhoudingstoets
rectangle − reghoek
reduction formula − herleidingsformule
related rates − verwante tempo’s
relative extreme value − relatiewe ekstreemwaarde
relative maximum value − relatiewe maksimumwaarde
relative minimum value − relatiewe minimumwaarde
remainder term − resterm
revenue function − inkomstefunkse
root test − worteltoets
secant − snylyn
series − reeks
set − versameling
slope − helling, gradiënt
smooth − glad
solid of revolution − omwentelingsliggaam
square − vierkant
squeeze theorem − knyptangstelling
stationary point − stasionêre punt
straight line − reguitlyn
substitution rule − substitusie reël
sum − som
surface − oppervlak
tangent − raaklyn
Taylor series − Taylorreeks
theorem − stelling
triangle − driehoek
trigonometric function − trigonometriese funksie
velocity − snelheid
125
126
List of Words
aantal eenhede − number of units
aanvraagfunksie − demand function
absolute ekstreemwaarde − absolute extreme value
absolute maksimumwaarde − absolute maximum value
absolute minimumwaarde − absolute minimum value
absolute waarde-funksie − absolute value function
aflei − derive
afgeleide − derivative
analise − analysis
asimptoot − asymptote
bepaal − determine
bepaalde integraal − definite integral
bereken − calculate
bewerkings − operations
booglengte − arc length
buigpunt − inflection point
dalend − decreasing
differensiasie − differentiation
differensieer − differentiate
differensieerbaar − differentiable
definisie − definition
definisieversameling − domain
differensiaal − differential
differensiaal- en integraalrekening − calculus
diskontinu − discontinuous
driehoek − triangle
een-eenduidige funksie − one-to-one function
eindig − finite
eksponensiale funksie − exponential function
ekstreemwaardestelling − extreme value theorem
faktorintegrasie − integration by parts
funksie − function
gemiddelde kostefunksie − average cost function
gemiddelde tempo van verandering − average rate of change
gemiddelde waarde − average value
geslote interval − closed interval
glad − smooth
globale ekstreemwaarde − global extreme value
globale maksimumwaarde − global maximum value
globale minimumwaarde − global minimum value
gradiënt − gradient, slope
helling − slope, gradient
herleidingsformule − reduction formula
hiperboliese funksie − hyperbolic function
hoër orde afgeleide − higher order derivative
implisiet − implicit
inkomstefunksie − revenue function
inkrement − increment
List of Words
integrand − integrand
integrasie − integration
integreerbaar − integrable
keël − cone
kettingreël − chain rule
konkaaf af − concave downward
konkaaf op − concave upward
konkawiteitstoets − concavity test
konstant − constant
kontinu − continuous
kontinuı̈teit − continuity
kostefunksie − cost function
knyptangstelling − squeeze theorem
kritieke punte − critical point
krommesketsing − curve sketching
kubus − cube
kwosiënt − quotient
limiet − limit
liniêre benadering − linear approximation
logaritmiese differensiasie − logarithmic differentiation
logaritmiese funksie − logarithmic function
lokale ekstreemwaarde − local extreme value
lokale maksimumwaarde − local maximum value
lokale minimumwaarde − local minimum value
magreël − power rule
magsfunksie − power function
marginale inkomstefunksie − marginal revenue function
marginale koste − marginal cost
marginale winsfunksie − marginal profit function
middelwaardestelling − mean value theorem
omwentelingsliggaam − solid of revolution
onbepaalde integraal − indefinite integral
oneindig − infinite
oombliklike tempo van verandering − instantaneous rate of change
oop interval − open interval
oppervlak − surface
oppervlakte − area
oppervlakte onder ’n kromme − area under a curve
parsiële breukontbinding − partial fraction decomposition
polinoom − polynomial
produk − product
produksievlak − production level
raaklyn − tangent
radiaal − radian
radiaalmaat − radian measure
rasionale funksie − rational function
reeks − series
reghoek − rectangle
127
128
List of Words
reguitlyn − straight line
relatiewe ekstreemwaarde − relative extreme value
relatiewe maksimumwaarde − relative maximum value
relatiewe minimumwaarde − relative minimum value
resterm − remainder term
samestelling − composition
sirkel − circle
snelheid − velocity
snylyn − secant
som − sum
stasionêre punt − stationary point
stelling − theorem
stompkeël − obtuse cone
stuksgewys gedefinieerde funksie − piecewise defined function
stygend − increasing
substitusiereël − substitution rule
Taylorreeks − Taylor series
trigonometriese funksie − trigonometric function
tussenwaardestelling − intermediate value theorem
verhoudingtoets − ration test
verplasing − displacement
versameling − set
verskil − difference
versnelling − acceleration
verwante tempo’s − related rates
vierkant − square
waardeversameling − range
winsfunksie − profit function
wortelfunksie − radical function
worteltoets − root test