very basic calculus

8/6/2019 Very Basic Calculus

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Integrated PrecalculusCalculus

David A. [email protected]

January 2, 2010 Version
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Contents

Preface v

1 Preliminaries 1

1.1 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.5 Infinitesimals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Infinitely Large Quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141.7 Distance on the Plane and Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 An Informal View of Graphs of Functions 19

2.1 Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.1.1 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.1.2 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.1.3 Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.1.4 Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.1.5 Poles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.1.6 Monotonicity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.1.7 Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2.2 Transformations of Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.2.1 Translations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.2.2 Reflexions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262.2.3 Distortions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

3 Introduction to the Infinitesimal Calculus 29

3.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.2 Graphical Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.3 The Strong Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.4 Derivatives and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

3.5 Graphical Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

3.6 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

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4 Polynomial Functions 54

4.1 Differentiating Power Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.2 Graphs of Power Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

4.3 Integrals of Power Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4.4 Differentiating Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

4.5 Integrating Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

4.6 Affine Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.7 Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

4.8 Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

4.9 Splitting Set of a Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

4.10 Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

4.11 Ruffinis Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

4.12 Graphs of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

4.13 Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

5 Rational Functions 92

5.1 Reciprocal Power Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

5.2 The Quotient Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

5.3 Graphing Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

5.4 Taylor Expansions and the Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

5.5 Partial Fraction Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

5.6 Integrating Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

6 Inversion of Functions and Algebraic Functions 106

6.1 Invertibility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

6.2 Differentiating and Graphing Algebraic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

6.3 Integrating Algebraic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

6.4 Asymptotic Expansions of Algebraic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
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7 Exponential and Logarithmic Functions 1157.1 The Natural Logarithm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1187.2 Integrals and Asymptotic Expansions Involving Logarithms . . . . . . . . . . . . . . . . . . . . . . . 118

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1197.3 The Natural Exponential Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122

7.4 Integrals and Asymptotic Expansions Involving Exponentials . . . . . . . . . . . . . . . . . . . . . . 122Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1247.5 Other Logarithmic and Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1277.6 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1327.7 Integration by Hyperbolic Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

8 Goniometric Functions 1348.1 Arcsine and Arccosine . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1398.2 The Sine and Cosine Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1428.3 Addition Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

8.4 Equations in Sines and Cosines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

8.5 Other Goniometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

8.6 Eulers Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

8.7 Integration by Trigonometric Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165Homework . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

A Answers to Selected Problems 167

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Preface

These notes started the fall of 2004, when I taught Maths 165, Differential Calculus, at Community College ofPhiladelphia.

The students at that course were Andrea BATEMAN, Kelly BLOCKER, Alexandra LOUIS, Cindy LY, ThorayaSABER, Stephanilee MAHONEY, Brian McCLINTON, Jessica MENDEZ, Labaron PALMER, Leonela TROKA, and

Samneak SAK. I would like to thank them for making me a better teacher with their continuous input and ques-tions.I have profitted from conversations with Jose Mason and Alain Schremmer regarding approaches to teaching

this course.

David A. Santos

Please send comments to [email protected]
mailto:[email protected]:[email protected]


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Copyright c 2007 David Anthony SANTOS. Permission is granted to copy, distribute and/or modifythis document under the terms of the GNU Free Documentation License, Version 1.2 or any later ver-sion published by the Free Software Foundation; with no Invariant Sections, no Front-Cover Texts, andno Back-Cover Texts. A copy of the license is included in the section entitled GNU Free Documenta-tion License.


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1 Preliminaries

This chapter introduces essential notation and terminology that will be used throughout these notes. We will often

use the symbol for if and only if, and the symbol , implies. The symbol means approximately.

1.1 Real Numbers

In these notes we consider the following sets of numbers, assigning to them special notation.

1 Definition The setN = {0 1 2 3 4 }

is the set ofnatural numbers.

Natural numbers allow us to count objects. The sum and product of two natural numbers is also a natural number,

and so we say that natural numbers are closed under addition and subtraction. So, for example, 1 + 1 is a naturalnumber, which we write as 1 + 1 N, read one times one belongs to the natural numbers. Similarly, 1 1 N.The natural numbers are not closed under subtraction or division, since, for example, 1 2 N, which we readone minus two is not in the natural numbers. In order to have a set closed under subtraction, we must adjointhe opposite of the natural numbers, creating thus the following set.

2 Definition The setZ = { 4 32 1 0 1 2 3 4 }

is the set ofintegers.

The integers are not closed under division, since for example, 1 2 Z. Starting from the integers, in order tohave a set closed under division, we must adjoin all the quotients of integers, creating thus the following set.

3 Definition The set

Q =

: Z Z = 0

is the set ofrational numbers. This is read as Q is the set of all fractions over such that is an integer, is aninteger, and is different from 0.

In other words, fractions, that is, rational numbers, are divisions that we are too lazy to perform.

Notice that we do not allow division by 0. What would happen if we were not that lazy, and actually performedthe implicit divisions in the rational numbers? We would get objects like

2

1

= 2

0

1

2

= 0

5

1

3

= 0

33333

= 0

3

where in this last division, the division goes on forever, but we see that we repeatedly obtain 3 in the quotient,that is, we get a periodic decimal. It can be provedbut we will not do so in these notesthat the set of rationalnumbers Q is precisely the set of numbers whose decimal expansion is either finite, or a periodic decimal.

In Q we have a very elegant system of numbers that is closed under addition, subtraction, multiplication, anddivision (except division by 0). Do we need more numbers? What happens with numbers like the Champernowne-

Mahler constant0123456789101112131415161718192021


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Intervals

which is the decimal number obtained by consecutively writing all the natural number? This number is clearlynot a periodic decimal, and hence it is not rational.

To accommodate infinite non-periodic decimals, we must create the following set.

4 Definition The set R is the set of real numbers, that is, the set of all numbers with either

1. a finite decimal expansion, or

2. an infinite periodic decimal expansion, or

3. an infinite non-periodic decimal.

A real number which is not rational is called irrational.

We must remark that looking into the decimal expansion of a number is not enough to prove that a number

is irrational. For example, it was known since the times of Pythagoras that the number

2 is irrational. Thisguarantees that its decimal expansion

2 = 141421356 does not repeat. If we started, however, with the number 141421356 we would not know whether it is rationalor irrational, for, it may have a very long decimal period, so long that our calculators and computers could not

store it. Again, although Archimedes suspected that

= 314159265

was irrational, a proof of this was not obtained until the eighteenth century by Lambert.

Homework

1.1.1 Problem Give an example of a rational number between1

10= 01 and

1

9= 01. Give an example of an

irrational number between1

10= 01 and

1

9= 01.

1.2 Intervals5 Definition An interval I is a subset of the real numbers with the following property: if I and I, and if < < , then I. In other words, intervals are those subsets of real numbers with the property that everynumber between two elements is also contained in the set. Since there are infinitely many decimals between twodifferent real numbers, intervals with distinct endpoints contain infinitely many members. Table 1.1 shews thevarious types of intervals.

Observe that we indicate that the endpoints are included by means of shading the dots at the endpoints and thatthe endpoints are excluded by not shading the dots at the endpoints. 1

1.3 Inequalities

!Vocabulary Alert! We will call a number positive if 0 and strictly positive if > 0.Similarly, we will call a number negative if 0 and strictly negative if < 0. This usagediffers from most Anglo-American books, who prefer such newspeak terms as non-negative and non-positive.

1It may seem like a silly analogy, but think that in [; ] the brackets are arms hugging and , but in ]; [ the arms are repulsed.Hugging is thus equivalent to including the endpoint, and repulsing is equivalent to excluding the endpoint.
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Interval Notation Set Notation Graphical Representation

[ ; ] { R : }2

] ; [ { R : < < }

[ ; [

{

R :

<

} ] ; ] { R : < }

] ; +[ { R : > }

+[ ; +[ { R : }

+] ; [ { R : < }

] ; ] { R : }

] ; +[ R

+

Table 1.1: Intervals.

The set of real numbers R is endowed with a relation > which satisfies the following axioms.

6 Axiom (Trichotomy Law) For all real numbers exactly one of the following holds:

> = or >

7 Axiom (Transitivity of Order) For all real numbers ,

if > and > then >

8 Axiom (Preservation of Inequalities by Addition) For all real numbers ,

if > then + > +

9 Axiom (Preservation of Inequalities by Positive Factors) For all real numbers ,

if > and > 0 then >

10 Axiom (Inversion of Inequalities by Negative Factors) For all real numbers ,

if > and < 0 then <

! < means that > . means that either > or = , etc.

The above axioms allow us to solve several inequality problems.

11 Example Solve the inequality2 3 < 13
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Inequalities

Solution: We have

2 3 < 13 2 < 1 3 + 3 2 < 10

The next step would be to divide both sides by 2. Since 2 > 0, the sense of the inequality is preserved, whence

2 < 10 < 102

< 5

The solution set is thus the interval ] ; 5[.

12 Example Solve the inequality

2 3 13

Solution: We have

2 3 13 2 1 3 + 3 2 10

The next step would be to divide both sides by 2. Since 2 < 0, the sense of the inequality is inverted, and so

2 10 102 5

The solution set is therefore [5 ; +[.

The method above can be generalised for the case of a product of linear factors. To investigate the set on theline where the inequality

(1 + 1) ( + ) > 0 (1.1)holds, we examine each individual factor. By trichotomy, for every , the real line will be split into the threedistinct zones

{ R : + > 0} { R : + = 0} { R : + < 0}

Here the sign , read union means that elements of all the sets involved are considered. We will call the real linewith punctures at =

and indicating where each factor changes sign the sign diagram corresponding to the

inequality (1.1).

13 Example Consider the inequality

2 + 2 35 < 0

1. Form a sign diagram for this inequality.

2. Write the set { R : 2 + 2 35 < 0} as an interval or as a union of intervals.

3. Write the set R : 2 + 2 35 0 as an interval or as a union of intervals.4. Write the set

R : + 7

5 0

as an interval or as a union of intervals.

5. Write the set

R : + 7

5 2

as an interval or as a union of intervals.

Solution:
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Chapter 1

1. Observe that 2 + 2 35 = ( 5)( + 7), which vanishes when = 7 or when = 5. Inneighbourhoods of = 7 and of = 5, we find:

] ; 7[ ]7 ; 5 [ ]5 ; +[

+ 7 + +

5 +( + 7)( 5) + +

On the last row, the sign of the product ( +7)( 5) is determined by the sign of each of the factors + 7and 5.

2. From the sign diagram above we see that

{ R : 2 + 2 35 < 0} = ]7 ;5[

3. From the sign diagram above we see that

R : 2 + 2 35 0 = ] ; 7] [5 ; +[ Notice that we include both = 7 and = 5 in the set, as ( + 7)( 5) vanishes there.

4. From the sign diagram above we see that R : + 7

5 0

= ] ; 7] ]5 ; +[

Notice that we include = 7 since + 7 5 vanishes there, but we do not include = 5 since there the

fraction + 7

5 would be undefined.5. We must add fractions:

+ 7 5 2 + 7 5 + 2 0 + 7 5 + 2 10 5 0 3 3 5 0

We must now construct a sign diagram puncturing the line at = 1 and = 5:

] ; 1[ ]1 ; 5[ ]5 ; +[

3 3 + +

5 +3 3 5 + +

We deduce that R : + 7

5 2

= [1 ;5[

Notice that we include = 1 since3 3 5 vanishes there, but we exclude = 5 since there the fraction

3 3 5 is undefined.


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Functions

Homework

1.3.1 Problem

1. Determine a sign diagram for the set

{ R : ( + 1)( 1) < 0}

2. Using the sign diagram obtained, write the set

{

R : ( + 1)(

1) < 0

}as a union of intervals.

3. Write the set

R : ( 1)

+ 1 0

as a union of intervals.

4. Write the set

R : ( 1)

+ 1 1

as an interval.

1.4 Functions

We will take a very narrow approach here to the definition of a function, one that will be useful to our purposes.

14 Definition A function : Dom ()

R,

() from the set Dom () to the real numbers is the collection of

the following ingredients:

1. A collection of inputs Dom () called the domain of definition of the function, being the largest set of realnumbers for which the output of the function is a real number.

2. A name for the function, typically , etc.

3. A name for a typical input, normally .

4. An assignment rule or formula, that associates to every input a unique output ().

We will give here some examples of functions.

15 Example Consider the function

:R R

2

Observe that the domain of definition is R since the expression 2 is a real number for every R. We have,for instance,

(0) = 002 = 0 (2) = 222 = 2 (1+

2) = 1+

2(1+

2)2 = 1 +

2(1+2

2+2) = 2

2


:R\ {0 1} R

1 2

Since the denominator vanishes when 2 = 0 (1 ) = 0 {0 1}, the domain of definitionis all real numbers excluding those two numbers, which we write as R \ {0 1}, pronounced R without the setcontaining 0 and 1. We have, for instance,

1

2

=

1

1

2 1

4

= 4


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Chapter 1


:] ; 7] [5 ; +[ R

2 + 2 35

By example 13, the quantity under the square root is strictly negative for

]

7 ;5[, and hence, the formula will

not give a real number output. We must consider only those numbers for which the quantity under the squareroot is positive and hence the domain of definition of the function is ] ; 7] [5 ; +[.

18 Definition Let and be two functions and let the point be in both of their domains. Then + is theirsum, defined at each point by

( + )() = () + ()

The difference is defined by( )() = () ()

and their product is defined by()() = () ()

Furthermore, if()

= 0, then their quotient is defined as

() =

()

()

The composition ( composed with ) is defined at the point by

( )() = (())

19 Example Let

:R\ {0} R

1

:[5 ;5] R

25 2

Find

1. ( + )(2)

2. ()(2)

3. ( )(2)4. ()(2)

Solution: We have

1. (+ )(2) = (2) + (2) =1

2+

21

2. ()(2) =1

2

21 =21

2

3. ( )(2) = ((2)) = (

21) =121

4. ()(2) = ((2)) =

1

2

=

25 1

4=

3

11

2


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Functions

We conclude this section with some special names that will be used throughout these notes.

20 Definition A function : R R whose formula is of the form

() = + 11 + + 1 + 0

where the R are constants, = 0, is called a polynomial function of degree . If = 0 then

() =

for some real constant numbers , this is called a constant function. If = 1 then

() = +

for some real numbers , with = 0, this is called an affine function. If = 2 then

() = + + 2

for some real numbers , with = 0, this is called a quadratic function.

21 Definition Let () and () be two polynomials with real coefficients, and let Z = { R : () = 0}. Thefunction : R\ Z R with formula () =

()

()is called a rational function.

22 Definition The function with assignment rule () is said to be algebraic if = () is a solution of anequation of the form

() + + 1() + 0() = 0

where the 0() 1() () are polynomials in . A function that satisfies no such equation is said to betranscendental.

23 Example The function

:R R

3 + 1

is a polynomial function of degree 3.

The function

:R R

2 + 2 + 2

is a rational function.

The function

:[0 ; +[ R

+

+

is a algebraic function.


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Chapter 1

Homework

1.4.1 Problem Determine the domain of definition of the assignment rule + .

1.4.2 Problem Determine the domain of definition of the assignment rule 1 +

.

1.4.3 Problem Determine the domain of definition of the assignment rule + 1( 1) .

1.4.4 Problem Consider the functions

:R R

2 + 1

:] ; 1] [1 ; +[ R

2 1

1. Determine (+ )(2).

2. Determine ()(2).

3. Determine ( )(2).4. Determine ()(2).

1.5 Infinitesimals

We now want to informally introduce the concepts ofnearness and smallness.What does it mean for one point to be near another point? We could argue that 1 is near to 0, but, for some

purposes, this distance could be too far. We could certainly say that 05 is closer to 0 than 1 is, but then again, forsome purposes, even this distance could be far. Mentioning a specific number near 0, like 1 or 05 fails in whatwe desire for nearness because mentioning a specific point immediately gives a static quality to nearness:once you mention a specific point, you could mention infinitely many more points which are closer than the pointyou mentioned. The points in the sequence

01 001 0001 00001

get closer and closer to 0 with an arbitrary precision. Notice that this sequence approaches 0 through values > 0.This arbitrary precision is what will be the gist of our concept of nearness. Nearness is dynamic: it involvesthe ability of getting closer to a point with any desired degree of accuracy. It is not static.

Again, the points in the sequence

12

14

18

116

are arbitrarily close to 0, but they approach 0 from the left. Once again, the sequence

+ 12

13

+ 14

15

approaches 0 from both above and below. After this long preamble, we may formulate our first definition.

24 Definition The notation , read tends to , means that is very close, with an arbitrary degree ofprecision, to . Here can approach through values smaller or larger than . We write + (read tendsto from the right) to mean that approaches through values larger than and we write (read tends to from the left) we mean that approaches through values smaller than .


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Infinitesimals

25 Definition Given a function and a point R, we write(+) for the value that() attains when +.Similarly, we write () for the value that() attains when .

26 Example Given : R R,

() =

4 if < 4

1 if = 4

3 if > 4

Then(4) =

4 4 = 0 (4) = 1 (4+) = 3

Observe that also, for example,(0) = 2 (0) = 2 (0+) = 2

and(5) = 3 (5) = 3 (5+) = 3

+

+

|

Figure 1.1: A neighbourhood of.

27 Definition A neighbourhood of a point is an interval containing .

Notice that the definition of neighbourhood does not rule out the possibility that may be an endpoint of the theinterval. Our interests will be mostly on arbitrarily small neighbourhoods of a point. Schematically we have adiagram like figure 1.1.

28 Definition (Infinitesimal) We say that a given formula () in the variable is infinitesimal as 0, written() = o (1) if() 0 as 0. We also say that is small oh of1.

We make the following observations, which we will take as axioms.

29 Axiom If 0 and if 1, N, then = o (1).

30 Axiom (The Sum of Infinitesimals is an Infinitesimal) If 0 and if() = o (1) and () = o (1), then() + () = o (1).

31 Axiom (The Product of an Infinitesimal and a Constant is an Infinitesimal) If 0 and if() = o (1) and R, = 0, then () = o (1).

32 Axiom (The Product of Infinitesimals is an Infinitesimal) If 0 and if() = o (1) and () = o (1), then()() = o (1).
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33 Example Justify the assertion that() = 3 = o (1) as 0.

Solution: We have 3 = o (1) and = o (1) as 0 by Axiom 29. Also, = o (1) by Axiom 31.Finally, 3 = 3 + () = o (1) by Axiom 30.

We would like to compare now how fast two infinitesimals formulas tend toward zero. In order to motivatethe following definitions, let us consider a numerical example. Consider a very small number, say,

00000000100023 =1

108+

2

1012+

3

1013

In comparison to1

108, the sum

2

1012+

3

1013is very small, practically negligible. In other words, the number

1

108

overwhelms or dominates the sum2

1012+

3

1013.

34 Definition Let and be non-zero natural numbers with < . We say that dominates as 0.We also say that goes faster to 0 than , or that goes slower to 0 than . We write this as = o (). Ingeneral, if

()

() 0, we write () = o

()

, and if() () = o

()

, we write () = () + o

()

.

The term o () is called the error term.35 Axiom Let R, = 0, and () = o (1) and () = o (1). The o () symbol has the following properties: as 0,

o () = o () (1.2)o () = o () (1.3)

o () o () = o () (1.4)oo () = o () (1.5)()() = o () (1.6)()() =

o () (1.7)o () + o () = o () (1.8)

o () = o () 2 N = 1 2 1 (1.9)o () = o (()) 1 N (1.10)

(())o () = o (())+1 N (1.11)o (())

()= o

(())1

2 N (1.12)

36 Example As 0 we have o3

+ o4

= o3

, since the 3 term dominates over the 4 for small .

37 Definition Let 1 < 2 < < be natural numbers. The dominant term as 0 of the polynomial() = 1

1 + 2 2 + +

is 1 1 . We write this as () 1 1 , read of is asymptotic to sub one times to the power sub 1.

38 Example Given that () = 22 + 3 + o3

and () = 44 + o4

, estimate the product ()()as 0.


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Infinitely Large Quantities

Solution: We have

()() =

22 + 3 + o3

44 + o4

= 23 86 + o

26

+ 4 47 + o7

+ o4

+ o47

+ o

7

= 23 + 4 + o 4

since all other error terms are faster than o 4.

39 Example As 0, 32 + 2 + .

40 Axiom (Product of Asymptotic Estimates) If () and () are polynomials, with () and () as 0, then ()() + as 0.

41 Example As 0, (2 + 32) ( 3 + 4) ( 5 + 6) 2 3 5 = 202.

Homework

1.5.1 Problem Given () = 2 + 2 + o 2 and () = + o (), as 0, estimate ()() as 0.1.6 Infinitely Large Quantities

42 Definition We use the symbol + to denote a quantity that is larger than any positive real number.

Thus ifP R, no matter how large P > 0 might be, we always have0 < P < +

43 Definition We use the symbol to denote a quantity that is smaller than any negative real number.

Thus ifN R, N < 0, no matter how large |N| might be, we always have < N < 0

0 1 2 3 4 5 6 701234567 +

Figure 1.2: The Real Line.

Geometrically, each real number can be viewed as a point on a straight line. We make the convention that weorient the real line with 0 as the origin, the positive numbers increasing towards the right from 0 and the negativenumbers decreasing towards the left of0, as in figure 1.2. We append the object +, which is larger than anyreal number, and the object , which is smaller than any real number. Letting R, we make the followingconventions.

(+) + ( +) = + (1.13)

() + () = (1.14)
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Chapter 1

+ (+) = + (1.15)

+ () = (1.16)

(+) = + if > 0 (1.17)

(+) = if < 0 (1.18)

() = if > 0 (1.19)

() = + if < 0 (1.20)

= 0 (1.21)

Observe that we leave the following undefined:

(+) + () 0()

44 Definition The notation +, read tends to plus infinity means that increases without bound, even-tually surpassing any preassigned large positive constant. The notation ,read tends to minus infinitymeans that decreases without bound, eventually surpassing any preassigned large in magnitude negative con-stant.

45 Definition Let and be non-zero natural numbers with < . We say that dominates as .We also say that goes faster to than , or that goes slower to than . We write this as =o (). In general, if ()

() 0, we write () = o

()

as and if() () = o

()

, we write

() = () + o (). The term o () is called the error term.The properties expounded in the preceding section for little oh as 0 also hold for little oh as ,

except for 29, which must be replaced by the following.

46 Axiom If and if 1, N, then 1 = o (1).

We now change the definition of dominant term.

47 Definition Let 1 < 2 < < be natural numbers. The dominant term as of the polynomial

() = 1 1

+ 2 2

+ +

is . We write this as () , read of is asymptotic to sub times to the power sub .

48 Example As + we have o3

+ o4

= o4

, since the 4 term dominates over the 3 for large .

49 Example Given that () = 23 + 2 + o2

and () = 4 4 + o (), estimate the product ()() as +.


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Distance on the Plane and Lines

Solution: We have

()() =

23 + 2 + o2

4 4 + o ()

= 27 84 + o

24

+ 6 43 + o3

+ o6

+ o43

+ o

3

= 2

7 +

6 +

o 6 as + since all other error terms are weaker than o

6

.

50 Example As +, 32 + 2 + 32.

51 Axiom (Product of Asymptotic Estimates) If () and () are polynomials, with () and () as 0, then ()() + as .

52 Example As +, (2 + 32) ( 3 + 4) ( 5 + 6) 32 4 6 = 725.

Homework1.6.1 Problem Given () = 3 + 22 + o

2

and () = + o (1) as +, estimate ()() as +.

1.7 Distance on the Plane and Lines

We now introduce some concepts from analytic geometry that will be used throughout these notes.

53 Theorem (Distance Between Two Points on the Plane) The distance between the points A = (1 1) B =(2 2) in R2 is given by

AB = d

(1 1) (2 2)

:= (1 2)

2 + (1

2)2

Proof: Consider two points on the plane, as in figure 1.3. Constructing the segments CA and BC withC = (2 1), we may find the length of the segment AB, that is, the distance from A to B, by utilising thePythagorean Theorem:

AB2 = AC2 + BC2 AB =

(2 1)2 + (2 1)2

u

54 Example The length of the line segment joining the points (1 2) and (3 4) on the plane is

(1 (3))2 + (2 4)2 = 16 + 4 = 2555 Definition Let and be real number constants. A vertical line on the plane is a set of the form

{( ) R2 : = }

Similarly, a horizontal line on the plane is a set of the form

{( ) R2 : = }
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Definition 55 is striking. It provides a link between Algebra and Geometry, and viceversa, between Geometry andAlgebra. For example, if we are given the equation of a horizontal line (Algebra), then we can graph it (Geometry).Conversely, if we are given the picture of a horizontal line on the coordinate plane (Geometry), we may find itsequation (Algebra).

B(2 2)

A(1 1) C(2 1)|2 1|

|2 1|

Figure 1.3: Distance between two points.

Figure 1.4: A vertical line.

Figure 1.5: A horizontal line.

(1 1)

( )

(2 2)

2

1

1

12 1

Figure 1.6: Theorem 56.

56 Theorem The equation of any non-vertical line on the plane can be written in the form = + , where and are real number constants. Conversely, any equation of the form = + , where are fixed realnumbers has as a line as a graph.

Proof: If the line is parallel to the -axis, that is, if it is horizontal, then it is of the form = , where is a

constant and so we may take = 0 and = . Consider now a line non-parallel to any of the axes, as in figure1.6, and let ( ), (1 1), (2 2) be three given points on the line. By similar triangles we have

2 12 1 =

1 1

which, upon rearrangement, gives

=

2 12 1

1

2 12 1

+ 1
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and so we may take

=2 12 1 = 1

2 12 1

+ 1

Conversely, consider real numbers 1 < 2 < 3, and let P = (1 1 + ), Q = (2 2 + ), andR = (3 3 + ) be on the graph of the equation = + . We will shew that

dP Q + dQ R = dP RSince the points P Q R are arbitrary, this means that any three points on the graph of the equation = + are collinear, and so this graph is a line. Then

dP Q =

(2 1)2 + (2 1)2 = |2 1|

1 + 2 = (2 1)

1 + 2

dQ R =

(3 2)2 + (3 2)2 = |3 2|

1 + 2 = (3 2)

1 + 2

dP Q =

(3 1)2 + (3 1)2 = |3 1|

1 + 2 = (3 1)

1 + 2

from wheredP Q + dQ R = dP R

follows. This means that the points PQ and R lie on a straight line, which finishes the proof of the theorem.u

Figure 1.7: > 0 Figure 1.8: < 0 Figure 1.9: = 0 Figure 1.10: =

57 Definition The quantity =2 12 1 in Theorem 56 is the slope or gradient of the line passing through (1 1)

and (2 2). Since = (0) + , the point (0 ) is the -intercept of the line joining (1 1) and (2 2). Figures1.7 through 1.10 shew how the various inclinations change with the sign of.

58 Example By Theorem 56, the equation = represents a line with slope 1 and passing through the origin.Since = , the line makes a 45 angle with the -axis, and bisects quadrants I and III. See figure 1.11

59 Example A line passes through (3 10) and (6 5). Find its equation and draw it.

Solution: The equation is of the form = + . We must find the slope and the -intercept. To find we compute the ratio

=10 (5)3 6 =

5

3

Thus the equation is of the form = 53

+ and we must now determine . To do so, we substitute either

point, say the first, into = 53

+ obtaining 10 = 53

(3) + , whence = 5. The equation sought
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Chapter 1

is thus = 53

+ 5. To draw the graph, first locate the -intercept (at (0 5)). Since the slope is 53

, move

five units down (to (0 0)) and three to the right (to (3 0)). Connect now the points (0 5) and (3 0). The graphappears in figure 1.12.

123

12345

1 2 312345

Figure 1.11: Example 58.

12345

12

1 2 312


123

12345

1 2 312345


60 Example Three points (4 ) (1

1) and (

3

2) lie on the same line. Find .

Solution: Since the points lie on the same line, any choice of pairs of points used to compute the gradientmust yield the same quantity. Therefore

(1)4 1 =

1 (2)1 (3)

which simplifies to the equation + 1

3=

1

4

Solving for we obtain = 14

. See figure 1.13.

B(2 2)

A(1 1) C(2 1)

MA

MB

( )

Figure 1.14: Midpoint of a line segment.

61 Theorem (Midpoint of a Line Segment) The point

1 + 2

2

1 + 22

lies on the line joining A(1 1) and

B(2 2), and it is equidistant from both points.

Proof: First observe that it is easy to find the midpoint of a vertical or horizontal line segment. The interval

[ ; ] has length . Hence, its midpoint is at + 2

= +

2.


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Let ( ) be the midpoint of the line segment joining A(1 1) and B(2 2). With C(2 1), form thetriangle ABC, right-angled at C. From ( ), consider the projections of this point onto the line segmentsAC and BC. Notice that these projections are parallel to the legs of the triangle and so these projections passthrough the midpoints of the legs. Since AC is a horizontal segment, its midpoint is at MB = ( 1+22 1). AsBC is a horizontal segment, its midpoint is MA = (2 1+22 ). The result is obtained on noting that ( )must have the same abscissa as MB and the same ordinate as MA. u

Homework1.7.1 Problem Find the distance of the line segments having (1 2) and (3 4) as endpoints.

1.7.2 Problem Find the equation of the line passing through (1 2) and (3 4).

1.7.3 Problem What is the slope of the line with equation

+

= 1?

1.7.4 Problem Let ( ) R2. Find the equation of the straight line joining ( ) and ( ).

1.7.5 Problem Which points on the line with equation = 6 2 are equidistant from the axes?

1.7.6 Problem In figure 1.15, point Mhas coordinates (2 2), points A Sare on the -axis, point B is on the -axisSM A is isosceles at M, and the line segment SM has slope 2. Find the coordinates of points ABS.

M

AS

B

Figure 1.15: Problem 1.7.6.
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2 An Informal View of Graphs of Functions

2.1 Graphs of Functions62 Definition The graph of a function is the set on the plane

= {( ) R2 : Dom () = ()}


Id :R R

called the identity function. By example 58, we know that its graph is the straight line that appears in figure 1.11.


:R R

53

+ 5

By example 59, we know that its graph is the straight line that appears in figure 1.12.

1

0

1

1 0 1

Figure 2.1: Example 65. = () Figure 2.2: Fails the vertical line test.Not a function.

Figure 2.3: Fails the vertical line test.Not a function.

65 Example Demonstrate that the graph of the function

:[1 ;1] R

1

2

is the upper semicircle in figure 2.1.

Solution: If = () =

1 2, then

2 + 2 = 1

( 0)2 + ( 0)2 = 1which means that any point ( ) on the graph of the function is at distance 1 from the origin. Since 0, the

points satisfying these requirement form the aforementioned semicircle.
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Graphs of Functions

!By the definition of the graph of a function, the -axis contains the set of inputs and -axis has the set of outputs.Since in the definition of a function every input goes to exactly one output, wee see that if a vertical line crossestwo or more points of a graph, the graph does not represent a function . We will call this the vertical linetest for a function. See figures 2.2 and 2.3.

Given a function , it is generally difficult to know a priori what its graph looks like. Most of the material onthis chapter will be dedicated to the investigation of tools for sketching graphs of functions. In the subsections

below we will informally discuss some features of graphs of (algebraic) functions, which will be elaborated lateron.

2.1.1 Symmetry

66 Definition A function is even if for all in its domain, is also in its domain, and it is verified that() = (), that is, if the portion of the graph for < 0 is a mirror reflexion of the part of the graph for > 0. This means that the graph of is symmetric about the -axis. A function is odd if for all , is alsoin its domain, and it is verified that () = (), in other words, is odd if it is symmetric about the origin.This implies that the portion of the graph appearing in quadrant I is a 180 rotation of the portion of the graph

appearing in quadrant III, and the portion of the graph appearing in quadrant II is a 180 rotation of the portionof the graph appearing in quadrant IV.

67 Example The function whose graph appears in figure 2.4 is even. The function whose graph appears in figure2.5 is odd.

Figure 2.4: The graph of an even function. Figure 2.5: The graph of an odd function.

2.1.2 Continuity

Heuristically speaking, a continuous function at the point = is one whose graph has no breaks at the point = .

What could possibly go wrong and make a function discontinuos? Let us consider the following three typesof discontinuities. Let be a function and R. Assume that is defined in a neighbourhood of , but notprecisely at = . Which value can we reasonably assign to ()? Consider the situations depicted in figures2.6 through 2.8. In figure 2.6 it seems reasonably to assign (0) = 0. What value can we reasonably assign

in figure 2.7? (0) =1 + 1

2= 0? In figure 2.8, what value would it be reasonable to assign? (0) = 0?,

(0) = +?, (0) = ? The situations presented here are typical, but not necessarily exhaustive. Thus of thethree discontinuous functions in figures 2.6 through 2.8, only that of figure 2.6 has a chance of being continuous,and that happens only if one defines (0) as 0.
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2.1.3 Intercepts

68 Definition Given a function, the set

{( ()) R2 : Dom () () = 0}

is called the set of-intercepts (zeroes or roots) of. If is defined at 0, then the point (0 (0)) is called the -interceptof. In other words, the intercepts are the places where the graph of a function touches the axes.

69 Example In figure 2.9, the points ABC are the -intercepts of the function and the point L is the -intercept.

Figure 2.6: = ().

Figure 2.7: = ().

Figure 2.8: = ().

A

B

C

L

Figure 2.9: Intercepts

2.1.4 Asymptotes

70 Definition Given a function we write () for the value that may eventually approach for large (inabsolute value) and negative inputs and(+) for the value that may eventually approach for large (in absolutevalue) and positive input. The line = is a (horizontal) asymptote for the function if either

() = or (+) =

Figure 2.10: () as +.

Figure 2.11: () as +.

Figure 2.12: () as .


Heuristically, the line = is an asymptote of the function, if eventually, the graph of the function gets closerand closer to it without actually touching it. Figures 2.10 through 2.13 shew different ways in which the line = can be the asymptote of a function.
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71 Example Find the asymptotes, if any, of the function

:R\ {1 1} R

(2 + 1)2(1 3)2

4 1

Solution: To do this, we find the dominant terms of the numerator and denominator as : =

(2 + 1)2(1 3)24 1

(2)2(3)24

= 36

which means that = 36 is its horizontal asymptote.


:R\ {1 1} R

(2 + 1)3(1 3)2

4 1

Solution: To do this, we find the dominant terms of the numerator and denominator as :

=(2 + 1)3(1 3)2

4 1 (2)3(3)2

4= 72

which tends to as , and hence does not have a horizontal asymptote.1


:R\ {1 1} R

(2 + 1)(1 3)4

1

Solution: To do this, we find the dominant terms of the numerator and denominator as :

=(2 + 1)(1 3)

4 1 (2)(3)

4= 6

2 0

as , and hence = 0 is its horizontal asymptote.

2.1.5 Poles

74 Definition Let > 0be an integer. A function has apole of order at the point = if()1() as , but ( )() as is finite. Some authors prefer to use the term vertical asymptote, rather thanpole.

75 Example Find the poles, if any, of the function

:R\ {1 1} R

(2 + 1)(1 3)4 1

1The expert will note that this curve has a slanted asymptote, but we will discuss these later in the text.


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Chapter 2

Solution: To do this, we find the real zeroes of the denominator:

4 1 = 0 ( 1)( + 1)(2 + 1) {1 1}from where the lines = 1 and = 1 are poles.

76 Example Find the poles, if any, of the function

: R\ {1 1} R (2 + 1)(1 3)

4 + 1

Solution: To do this, we find the real zeroes of the denominator. But since 1 + 4 = 1 + (2)2 is 1 plus areal square, and real squares are always positive, the denominator is never 0, as it is, in fact, always 1. Thismeans that does not have any poles.

=

Figure 2.14: () as +.

=

Figure 2.15: () +as +.

=


=

Figure 2.17: () +as .

Figures 2.14 through 2.17 exhibit the various ways a function may have a pole.

2.1.6 Monotonicity

77 Definition A function is said to be increasing (respectively, strictly increasing) if < () ()(respectively, < () < ()). A function is said to be decreasing (respectively, strictly decreasing) if < () () (respectively, < () < ()). A function is monotonic if it is either (strictly)increasing or decreasing. By the intervals of monotonicity of a function we mean the intervals where the functionmight be (strictly) increasing or decreasing.

! If the function is (strictly) increasing, its opposite is (strictly) decreasing, and viceversa.The following theorem is immediate.

78 Theorem A function is (strictly) increasing if for all < for which it is defined

() () 0 (respectively

() () > 0)

Similarly, a function is (strictly) decreasing if for all < for which it is defined

() () 0 (respectively

() () < 0)
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Chapter 2

2. has zeroes at = 3, = 1, and = 2

3. () = (+) =

4. (1) = 2

5. (5) = 16

2.1.2 Problem Draw the graph of a function meeting the following conditions:

1. is everywhere continuous;

2. (2) = (0) = (2) = 0;

3. () = 0,(+) = +

4. (4) = 1,(1) = 1 ,(1) = 1,(4) = 1

2.1.3 Problem Draw the graph of a function meeting the following conditions:

1. is continuous in R\ {0};

2. has a pole at = 0, with(0) = and(0+) = +;

3. () = ,(+) = +

12345

123456

1 2 3 4 5123456

Figure 2.20: for Problem 2.1.4.

12345

123456

1 2 3 4 5123456

Figure 2.21: for Problem 2.1.4.

2.1.4 Problem Consider the functions and in the figure below. You may assume that they are composed solelyof straight line segments.

1. Draw + .

2. Draw .

3. Draw 2 + .


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Transformations of Graphs of Functions

2.2 Transformations of Graphs of Functions

2.2.1 Translations

81 Theorem Let be a function and let and be real numbers. If(0 0) is on the graph of, then (0 0 + )is on the graph of, where () = () + , and if(1 1) is on the graph of , then (1 1) is on the graphof, where() = ( + ).

Proof: Let denote the graphs of respectively.

(0 0) 0 = (0) 0 + = (0) + 0 + = (0) (0 0 + ) Similarly,

(1 1) 1 = (1) 1 = (1 + ) 1 = (1 ) (1 1) u

82 Definition Let be a function and let and be real numbers. We say that the curve = () + is a verticaltranslation of the curve = (). If > 0 the translation is up, and if < 0, it is units down. Similarly, we saythat the curve = ( + ) is a horizontal translation of the curve = (). If > 0, the translation is units left,

and if < 0, then the translation is units right.

83 Example Figures 2.23 through 2.25 shew various translations of the graph of a function .

543210

12345

5

4

3

2

101 23 45

Figure 2.22: = ().

543210

12345

5

4

3

2

101 23 45

Figure 2.23: = () + 1.

543210

12345

5

4

3

2

101 23 45

Figure 2.24: = ( + 1).

543210

12345

5

4

3

2

101 23 45

Figure 2.25: = ( + 1 ) + 1.

2.2.2 Reflexions

84 Theorem Let be a function If(0 0) is on the graph of, then (0 0) is on the graph of, where () =(), and if(1 1) is on the graph of, then (1 1) is on the graph of, where() = ().

Proof: Let denote the graphs of respectively.

(0 0)

0 = (0)

0 =

(0)

0 = (0)

(0

0)

Similarly,

(1 1) 1 = (1) 1 = (1) 1 = (1) (1 1)

u

85 Definition Let be a function. The curve = () is said to be the reflexion of about the -axis and the curve = () is said to be the reflexion of about the -axis.
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Chapter 2

86 Example Figure 2.26 shews the graph of the function . Figure 2.27 shews the graph of its reflexion aboutthe -axis, and figure 2.28 shews the graph of its reflexion about the -axis.

5432

1012345

5432101 23 45

Figure 2.26: = ().

5432

1012345

5432101 23 45

Figure 2.27: = ().

5432

1012345

5432101 23 45

Figure 2.28: = ().

5432

1012345

5432101 23 45

Figure 2.29: = ().

2.2.3 Distortions

87 Theorem Let be a function and let V = 0 and H = 0 be real numbers. If(0 0) is on the graph of, then(0 V0) is on the graph of, where () = V(), and if(1 1) is on the graph of, then

1H

1 is on thegraph of, where() = (H ).Proof: Let denote the graphs of respectively.

(0 0) 0 = (0) V 0 = V(0) V 0 = (0) (0 V 0)

Similarly,

(1 1) 1 = (1) 1 =

1H

H

1 =

1H

1H

1

u

88 Definition Let V > 0, H > 0, and let be a function. The curve = V() is called a vertical distortion of thecurve = (). The graph of = V () is a vertical dilatation of the graph of = () ifV > 1 and a verticalcontraction if0 < V < 1 The curve = (H ) is called a horizontal distortion of the curve = () The graph of = (H ) is a horizontal dilatation of the graph of = () if0 < H < 1 and a horizontal contraction ifH > 1

76543

210

1234567

76543210 1 2 3 4 5 6 7

Figure 2.30: = ()

76543

210

1234567

76543210 1 2 3 4 5 6 7

Figure 2.31: =()

2

76543

210

1234567

76543210 1 2 3 4 5 6 7

Figure 2.32: = (2)

76543

210

1234567

76543210 1 2 3 4 5 6 7

Figure 2.33: =(2)

2

89 Example Consider the function whose graph appears in figure 2.30.
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3 Introduction to the Infinitesimal Calculus

3.1 Continuity

90 Definition A function is continuous at = if as 0,

( + ) = () + o (1)

This means, essentially, that if , then() ().

91 Example Let K R. Demonstrate that the constant function

:R R

K

is continuous.

Solution: Let R be fixed. Then

( + ) = K = K + 0 = K + o (1) = () + o (1)

proving that a constant function is continuous.

92 Example Demonstrate that the identity function

Id :

R

R

is continuous.

Solution: Let R be fixed. ThenId( + ) = +

Since 0 means the same as = o (1), we have proved that

Id( + ) = + = Id() + o (1)

and so the identity function is continuous.

93 Example Demonstrate that the square function

Sq :R R

2

is continuous.


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Chapter 3

Proof: We are given that( + ) = () + o (1)

The quantity o (1) will eventually (as 0) be smaller in absolute value than, say |()|2

. But then this means

that

(+) ()+ |()|2

() ()2

< (+) < ()+()

2 ()

2< (+) 0 1 + > > 0

and so

0 < 2( 1 + 2) 1 < 1 + 2 + 23

1 < 1 2 + 22 + 23

1 < (1 + )(1 ) + ( 1 + )22

11 +

< 1 + 22

as was to be shewn. Hence, for 12

< < 0, we have

1 < 11 +

< 1 + 22

which gives the lemma in the remaining case. u

98 Theorem If is continuous at = and () = 0, then 1

is also continuous at = .


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Proof: By Theorem 96, for sufficiently close 0, ( + ) = 0. Hence1

( + ) =

1

( + )

=1

() + o (1)=

1

()

1

1 + o 1()

=

1

() 1

1 + o (1)=

1

() 1 + o (1)

=1

()+ o

1

()

=

1

()+ o (1)

by Lemma 97. This proves that1

is continuous at = .

u

99 Corollary If the functions and are continuous at = and if() = 0, then

is continuous at = .

Proof: Immediate from theorems 95 and 98. u

100 Corollary (Polynomials are continuous functions) Let : R R be a polynomial with real coefficients.Then is continuous at every R.

Proof: The proof is by induction on the degree on the degree of . Ifdeg = 0, then is a constant function,which is continuous by example 91.

If deg = 1 then () = + , for some constants . Since the identity function is continuous byexample 92, the function is continuous being the product of two continuous functions and . We then see that is the sum of the continuous functions and .

Assume now that any polynomial of degree 1 is continuous at = . Then if() = + 11 + + 1 + 0 = (1 + 12 + + 1) + 0 := () + 0where deg () = 1. By the induction hypothesis, () is continuous at = . By example 92, theidentity functions is continuous, and so () is the product of two continuous functions, and hencecontinuous. Since the constant function 0 is continuous, is then the sum of two continuous functionsand hence continuous.

u

101 Corollary (Rational Functions are Continuous) Let () and ()be two polynomials with real coefficients.Let Z = { R : () = 0}. Then the rational function

:R \ Z R

()

()


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Chapter 3

is continuous.

Proof: This is immediate from Corollary 100 and Theorem 98. u

()

()

Figure 3.1: Intermediate Value Theorem.

()

()

Figure 3.2: Intermediate Value Theorem.

102 Theorem Let be continuous at = and continuous at = (). Then is continuous at = .

Proof: We are given that as 0, then

(() + ) = (()) + o (1) ( + ) = () + o (1)

We must shew that( )( + ) = ( )() + o (1)

We have

( )( + ) = (( + )) = (() + o (1)) = (()) + o (1) = ( )() + o (1)

as was to be shewn. u

103 Definition Let : Dom () R and assume ] ; [ Dom (). We say that is continuous in ] ; [ if iscontinuous at every point ] ; [.

104 Definition Let : Dom () R and assume [ ; ] Dom (). We say that is continuous in [ ; ] if iscontinuous at every point ] ; [ and if(+) = () and() = ().

105 Theorem (Darbouxs Intermediate Value Theorem) Let : [ ; ] R be continuous in [ ; ]. Let =

min(() ()) and M = max(() ()). Then, for every [ ; M], there exists [ ; ] such that() = .

Proof: We shall not give a rigorous proof of this assertion, but refer the reader to figures 3.1 and 3.2. Sincethere are no breaks in the graph, the graph will not jump over the horizontal line = . u

106 Corollary A continuous function defined on an interval maps that interval into an interval.

Proof: This follows at once from the Intermediate Value Theorem and the definition of an interval. u
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Continuity

107 Theorem (Bolzanos Theorem) If : [ ; ] R is continuous and ()() < 0, then there is a ] ; [such that() = 0.

Proof: This follows at once from the Intermediate Value Theorem by putting = min(() ()) < 0 and = max(() ()) > 0 . u

108 Corollary Every polynomial of odd degree and real coefficients has at least one real root.

Proof: Since () and (+) have opposite sign, this follows by the Intermediate Value Theorem. u

109 Example Demonstrate that the polynomial () = 5 1 has a root in [1 ;2]. Further, find an interval

of length1

10or smaller that contains this root.

Solution: Observe that (1) = 1 < 0 and (2) = 29 > 0, and so, since the polynomial changes sign in[1 ;2], by Bolzanos Theorem, the polynomial has a root [1 ;2]. In fact, we find

(11) 05 (12) 03

and so [11 ; 12], which is an interval of length1

10 containing the zero. Still

(115) 14 (117) 002

meaning that [115 ; 117]. This interval has length 117 115 = 002 = 150

, so we have done better

than the required1

10.

110 Theorem (Weierstrass Theorem) A continuous function : [ ; ] R attains a maximum and a minimumon [ ; ].

Proof: By Darbouxs Theorem and by Bolzanos Theorem, ([ ; ]) is a closed finite interval. The minimumsought is the left endpoint of this interval and the maximum sought is the right endpoint of this interval. u

Homework

3.1.1 Problem Given that the equation 7 6 24 + 3 + 1 = 0 has exactly three different real roots, findintervals, of length 1 or shorter, containing each root.

3.1.2 Problem Let () () be polynomials with real coefficients such that

(2 + + 1) = ()()

Prove that must have even degree.

3.1.3 Problem A function defined over all real numbers is continuous and for all real satisfies()

()() = 1Given that(1000) = 999, find(500).

3.1.4 Problem Suppose that : [0 ;1] [0 ;1] is continuous. Prove that there is a number in [0 ;1] such that() = 1 .


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Chapter 3

3.1.5 Problem (Universal Chord Theorem) Suppose that is a continuous function of[0 ;1] and that(0) = (1).Let be a strictly positive integer. Prove that there is some number [0 ;1] such that() = ( + 1/)

3.2 Graphical Differentiation

In this section we will take a very informal approach to differentiation. A more formal approach will be given in

the next section.

111 Definition Let be a continuous function and let A( ()) be a point on the graph of the function. We saythat is smooth at A if upon imagining a particle travelling at some steady speed along the curve, then the particledoes not experience an abrupt change of direction.

A

Figure 3.3: Smooth curveat A.

A

Figure 3.4: Corner at A.

A

Figure 3.5: Cusp at A.

A

Figure 3.6: Tangent line at = .

112 Example Figure 3.3 gives an example of a smooth curve. The curve in figure 3.4 has a sharp corner at Aand it is not smooth. The curve in figure 3.5 is a cusp.

5

4321

0

1

2

3

4

5

5 4 3 2 1 0 1 2 3 4 5


5

4321

0

1

2

3

4

5

5 4 3 2 1 0 1 2 3 4 5


113 Definition If is an affine function, we define the tangent line at any point A( ()) to be the line that is thegraph of. Otherwise, if is a smooth function at the point A( ()), then tangent line at A is the unique line withthe following properties:


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Graphical Differentiation

1. for a sufficiently small neighbourhood ofA, the line just touches the curve at A.

2. for this sufficiently small neighbourhood ofA, the portion of the curve inside the neighbourhood is on onlyone side of the line.

The derivative of at = , denoted by(), is the slope of the tangent line at = .

114 Example Figure 3.6 gives an example of the tangent line to a curve.

115 Example For the curve in figure 3.8, find, approximately, the value of(1).

Solution: We draw a line through the point (1 (1)) that just grazes the curve, as in figure 3.8. We computethe slope of this line:

3 (1)2 0 = 2

Hence we find(1) = 2.

116 Definition If a function is differentiable at every point of its domain, then the derivative function of ,

denoted by, is the function with assignment rule ().

Given the graph of a smooth curve, we can approximately obtain the graph of its derivative by taking thefollowing steps.

1. We divide up the domain of into intervals of the same length.

2. For each endpoint of an interval above, we look at the point ( ()) on the graph of.

3. We place a ruler so that it is tangent to the curve at ( ()).

4. We find the slope of the ruler. Recall that any two points on the tangent line (the ruler) can be used to findthe slope.

5. We tabulate the slopes obtained and we plot these values, obtaining thereby an approximate graph of.

Figure 3.9: Increasing and convex. Figure 3.10: Decreasing and convex.

To further aid our graphing of the derivative, we make the following observations.

1. If the function increases, then the slope of the tangent is positive.

2. If the function decreases, then the slope of the tangent is negative.
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Chapter 3

3. If the curve is convex, then the slope of the tangent increases.

4. If the curve is concave, then the slope of the tangent decreases.

See figures 3.9 through 3.12 for several examples.

Figure 3.11: Increasing and concave. Figure 3.12: Decreasing and concave.

2

1

0

1

2

2 1 0 1 2

Figure 3.13: Example 117. = ()

2

1

0

1

2

2 1 0 1 2


117 Example Find an approximate graph for the derivative of given in figure 3.13.

Solution: Observe that from the remarks following figure 3.10, we expect to be positive in [14; 06],since increases there. We expect to be 0 at = 06, since appears to have a (local) maximum there. Weexpect to be negative in [06; 06] since decreases there. We expect to be 0 at = 06, since appears to

have a (local) minimum there. Finally we expect to be positive for[0

6; 1

4]

since is increasing there.In our case we obtain the following (approximate) values for().

14 12 1 08 06 04 02 0 02 04 06 08 1 12 14

() 488 332 2 092 008 052 088 1 088 052 008 092 2 332 488


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Graphical Differentiation

An approximate graph of the strong derivative appears in figure 3.14.

118 Example Figure 3.16 gives an approximate graph of the derivative of the graph appearing in figure 3.15. Wesee that the curve is increasing and so the derivative is positive. As the curve travels the interval ]1 ;0[, the slopeof the tangent is going from almost horizontal (slope 0) to almost vertical (infinite slope). At = 0 the tangent isvertical, and so the derivative has a pole there. As the curve travels the interval ]0 ; +1[, the slope of the tangentgoing from almost vertical to almost 0.

To obtain the graph of the derivative above, we have served ourselves of the table below.

10 08 06 04 02 00 02 04 06 08 10

() 03 04 05 06 10 + 10 06 05 04 03

1

0

1

1 0 1


1

0

1

1 0 1


Homework

3.2.1 Problem Use the graph in figure 3.17 below to complete the following table for a function . Also, make arough sketch of.

1 04 02 0 02 04 1

()

()

1

0

1

1 0 1


3.2.2 Problem Draw an example of a curve that is everywhere continuous, but fails to be differentiable for {1 0 1}.
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Chapter 3

3.2.3 Problem Draw an example of a curve that is everywhere continuous, differentiable inR\{2 2}, and suchthat() = 1 for < 2,() = 0 for 2 < < 2 and() = 1 for > 2.

3.2.4 Problem Draw an example of a curve that is continuous and differentiable inR\{0}, and such that(0+) =+,(0+) = +,(0) = ,(0) = ,() = 0,(+) = 0,(1) = 1 and(1) = 1.

3.2.5 Problem Draw an example of an everywhere differentiable curve = () with

(2) = (0) = (2) = 0 (1) = 4 (1) = 2 (1) = (0) = (1) = (2) = 0

(+) = 0 () = 2 () > 0 for ] ; 2[ () < 0 for

+5

2; +

or explain why such a curve does not exist.

3.2.6 Problem Use the graph in figure 3.18 below to complete the following table for a function . Also, make arough sketch of.

1 04 02 0 02 04 1

()

()

1

0

1

1 0 1


3.3 The Strong Derivative

A

C

B

Figure 3.19: A few points do not a graph determine.

Given a finite number of points, we can find infinitely many curves passing through them. See for examplefigure 3.19, where we see three very different curves each simultaneously passing through the points A, B,C. Thus plotting a few points of the graph of a function can give a misleading picture.

By the same token, given a formula, the plotting of a few points does not give the salient features of a graph.For example, let us say that we wanted to graph = 4 3. In figures 3.20 through 3.23 we have chosen a few


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The Strong Derivative

selected points on the curve and interpolated between them through lines. But relying on this method does notgive proof that the graph will not have more turns or bends, say, or that it will grow indefinitely for values of oflarge magnitude.

Figure 3.20: Fourplot points.

Figure 3.21: Sevenplot points.

Figure 3.22: Ten plotpoints.

Figure 3.23: Onethousand plot points.

But for all its faults, the progression of shapes in figures 3.20 through 3.23 suggests that a reasonable graphcan be approximated by a series of straight lines. By a reasonable graph we mean one that does not havemany sharp turns, does not oscillate wildly, does not have many jumps or many asymptotes, and that it is mostlycontinuous and smooth. Admittedly, these concepts are vague, but we will gain more insight into them as weprogress.

( ())

Figure 3.24: Left secant through

( ()).

( ())

Figure 3.25: Right secant through

( ()).

( ())

Figure 3.26: Line grazing ( ()).

How do we choose the lines to approximate a given reasonable curve? Given a function consider the point( ()) on the graph of the function. What happens around this point? If we approached through values < and joined the line with endpoints ( ()) and ( ()), we would obtain a secant line like that of figure 3.24.If we approached through values > and joined the line with endpoints ( ()) and ( ()), we wouldobtain a secant line like that of figure 3.25. Eventually, on getting closer to ( ()) we obtain a line just barelygrazing the curvethat is, tangent to the curveat the point ( ()), as in figure 3.26.

In the simplest of cases, if our curve is the line L : = + , then in a neighbourhood of the point = the tangent line to L should be itself! Given now the formula for a function and a point ( ()) on the graph of

, how do we determine the tangent line to at ( ())? Recall that if

0, then + is in a neighbourhood of. The slope of the secant line joining ( + ( + )) and ( ()) is

( + ) () + =

( + ) ()

(3.1)

We denote the valueif there is oneof3.1 as by(). Hence for fixed but small we have

( + ) ()

() ( + ) () +()
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Derivatives and Graphs

3.4 Derivatives and Graphs

In this section we prove the remarks preceding figures 3.9 through 3.12, which will help us with the graphing insubsequent chapters.

123 Theorem If is strongly differentiable at then is continuous at .

Proof: We have( + ) = () +()+ o (). If > 0 and 0 then( + ) = (+) and similarlyif < 0 and 0( + ) = (). Hence we have(+) = () = (), and is continuous at . u

124 Theorem Let be strongly differentiable at . If () > 0 then is increasing in a neighbourhood of , if() < 0 then is decreasing in a neighbourhood of.

Proof: We have( + ) () = ()+ o (). For very small, this means that

( + ) () ()

that is, the sign of ( + ) () is the same as the sign of (). Thus if > 0 and () > 0, then( + ) > (), that is, is increasing. If > 0 and() < 0, then( + ) < (), that is, is decreasing.

Similar conclusions are reached when considering < 0

and the theorem is proved.u

125 Definition If is strongly differentiable at and() = 0, then we say that is a stationary point of.

126 Definition If there is a point for which () () for all in a neighbourhood centred at = thenwe say that has a local maximum at = . Similarly, if there is a point for which () () for all in aneighbourhood centred at = then we say that has a local minimum at = .

127 Theorem If is strongly differentiable at = ,() = 0, and changes fr

very basic calculus

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