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Basics of pre-universitymathematics

J. A. Angelova, D. A. Kolev

Dept. of MathematicsUniversity of Chemical Technology and Metallurgy

8 Kliment Ohridsky, blvd.,Sofia 1756, BULGARIA

2011

Sofia

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Preface iv

Notations and abbreviations v

1. Sets and basic operations 11.1. Subsets 21.2. Unions 31.3. Intersections 41.4. Complements 41.5. Cartesian products 51.6. Basic properties of set operations 6Exercises 6

2. Numbers 82.1. Real numbers 82.2. Integers 102.3. Rational numbers 112.4. Irrational numbers 12Exercises 13

3. Functions, graph representations, classification 143.1. Functions and mappings 143.2. Representation of functions 163.3. Functions classification 173.4. Composition of functions 183.5. Inverse functions 18Exercises 19

4. Linear and quadratic functions. Inequalities 204.1. Linear functions 204.2. Quadratic functions 21Exercises 24

5. Exponentiation and power functions 265.1. Powers 265.2. Exponentiation 265.3. Identities and properties 27

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CONTENTS ii

5.4. Power functions 285.5. Basic irrational equations and inequalities 29Exercises 31

6. Systems of equations and inequalities 326.1. Systems of two equations in two variables 326.2. Polynomial functions and their signs 366.3. Systems of algebraic inequalities 37Exercises 42

7. Exponential and logarithmic functions 447.1. Exponentiation, identities and properties 447.2. Exponential functions 457.3. Exponential equations and inequalities in one variable 457.4. Logarithms, identities and properties 477.5. Logarithmic functions 487.6. Logarithmic equations and inequalities in one variable 49Exercises 51

8. Trigonometry 528.1. Definition of trigonometric functions 528.2. Classification of trigonometric functions 558.3. Identities 578.4. Inverse trigonometric functions 598.5. Trigonometric equations and inequalities 61Exercises 65

9. Sequences and progressions 669.1. Number sequences 669.2. Arithmetic progression 699.3. Geometric progressions 71Exercises 72

10. Limits and continuity of functions. 7410.1. Limits of functions 7410.2. Continuity of functions 76Exercises 79

11. Derivatives. Extrema. Inflection points 8111.1. Definition of derivatives 8111.2. Interpretations of derivatives 8211.3. Differentiation rules and table of derivatives 8311.4. Extreme values 8511.5. Inflection points 87Exercises 88

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CONTENTS iii

Appendix A. Notes on the pronunciation of some basic formulae 90

Appendix B. Plane geometry 92Vectors 92Angles 92Triangles 93Quadrilaterals 96Regular polygons 97Circles 98

Appendix C. Solid geometry 100Dihedral angles 100Polyhedrons 100Cylinders, cones and spheres 103

Bibliography 107

Index 113

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This textbook contains basics of school mathematics, essential forstudents of technical universities. The main purposes of the proposedmaterials are: to acquire some basic knowledge in the field of mathemat-ics, to calibrate school knowledge of algebra and analysis, to popularizemathematical language, so the students can meet the universitys re-quirements.

Lessons contain fundamentals of: sets, numbers, functions, trigonom-etry, number sequences, limits and derivatives of functions. They includealso important topics as algebraic equations and inequalities along withtheir methods of solving. All elementary functions are given with theirgraphs, properties, limits and derivatives. The topic for sequences in-cludes arithmetic and geometric progressions. The essentials of elemen-tary plane and space geometry one can find in appendicies as vademe-cum.

All topics covered in this book are illustrated by appropriate exam-ples, and some of them are supplied by graphical solutions. The basicdefinitions and concepts of the classical algebra and analysis are pre-sented strictly and by sufficient precision.

The materials are formatted in full compliance with the UCTM-Sofia programs of mathematics. The authors would be delighted if thisbook would become a favorite reading for anyone who will choose theengineering as a future profession.

iv

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empty setN set of natural numbers, N = {1, 2, . . . }Z set of integers, Z = {0,1,2, . . . }Q set of rational numbers, Q = {pq : p Z, q N}I set of irrational numbers, such as 5

7, 2, e, R set of reals, R = (,+)[a, b] closed interval, {x R : a x b}; a, b R, a < b(a, b) open interval, {x R : a < x < b}; a, b R, a < b[a, b) left closed right open interval,

{x R : a x < b}; a, b R, a < b(a, b] left open right closed interval,

{x R : a < x b}; a, b R, a < b(x0, ) -neighborhood of a point x0 - (x0 , x0 + )AB Cartesian product of sets A and BA B union of sets A and BA B intersection of sets A and BA B a set A is a subset of BB \A the complement of A in Bf : X Y a function f with domain X and range Yf1 : Y X inverse function of the function fF u(x) composite function F[u(x)]f (x) =

df(x)

dxfirst order derivative

universal quantifier - for all existential quantifier - there exists= follows equivalent tends approximately equalgcd greatest common divisorlcm least common multipleTIT Test Interval TechniqueAP Arithmetic ProgressionGP Geometric Progression

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Sets and basic operations

The base of modern set theory is set by Georg Cantor and RichardDedekind in the 1870s.

Definition 1.1. A set is a collection of distinct objects, that arecalled elements or members.

A set that has no members is called the empty set and is denotedby the symbol .

The elements of a set can be anything: numbers, people, cars, lettersof an alphabet, other sets, and so on. A set cannot contain multiplecopies of an element, i.e. {a, a, a, b}={a, b}. Sets are conventionallydenoted with capital letters, A, B, ..., and their members by small letters.If a is an element of A we write a A.

Definition 1.2. Two sets A and B are said to be equal, A = B, ifthey have the same elements.

All sets can be precisely described in words, lists or mathematicalnotations.

Example 1.1. Let consider the following sets:A is the set whose members are the first four positive integers,B is the set whose elements are the colors of the Bulgarian flag,C = {4, 2, 1, 3},D = {white, green, red},E = {n2 10 : n is a prime number and n 9}, the colon (:) meanssuch that.The sets A, B, D are defined in words and the sets C and E - bymathematical notations.

One and the same set can be defined in different ways. The sets Aand C defined above, are identical, since they have the same members.Similarly, B = D, and E = {6, 1, 5, 39}, as n = 2, 3, 5, 7. Georg Ferdinand Ludwig Philipp Cantor (1845 - 1918) was a German mathemati-cian, best known as the inventor of a set theory. He defined infinite and well-orderedsets, the cardinal and ordinal numbers and their arithmetic. Cantors work is of greatphilosophical interest.

Julius Wilhelm Richard Dedekind (1831 - 1916) was a German mathematicianwho did important work in abstract algebra, algebraic number theory and the foun-dations of the real numbers. The notion Dedekind cut, named after him, now is astandard definition of the real numbers. 1

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1.1. SUBSETS 2

Note 1.1. A prime number is a positive integer greater than 1 andcan be divided only by 1 and itself.

Set identity does not depend on the order in which the elements arelisted, nor on their repetitions in the list.

Special sets. There are some special sets, such as:N denotes the set of all natural numbers, i.e. N = {1, 2, 3, ...}.Z denotes the set of all integers positive, negative or zero, so

Z = {0,1,2, . . . }.Q denotes the set of all rational numbers - proper and improper

fractions, i.e. Q =

{p

q: p Z , q N

}. For example, 34 Q, 1117 Q,

and integers are in this set, too.I is the set of all irrational numbers, numbers that are not rational,

i.e. non-repeating (non-periodic) or non-terminating decimals. Exam-ples of irrational numbers are roots of numbers that are not perfect roots( 416), such as

2, 3

5 and transcendental numbers like and e.R is the set of all real numbers, R = (,). This set includes all

rational numbers Q, together with set of all irrational numbers I.C is the set of all complex numbers.

0 1

Figure 1.1. The set of real numbers R

The real set R can be compared with an axis, known as the realaxis, i.e. a straight line so that every point on the line corresponds tothe unique real number (see Figure 1.1).

The sets may have finitely or infinitely many elements and may becountable or uncountable.

The next sets contain finite number of members, thus they are finite:the empty set: , the set of natural numbers less than 5: {1, 2, 3, 4},{1, 1.5, , e}, {2;3; 23 ;5, 1}.

Infinite sets are: the natural numbers N, the set of even positivenumbers B = {2, 4, 6, 8, ...}, the primes: {2, 3, 5, 7, 11, . . . }, the rationalnumbers: Q, an interval of the reals (0, 1), the set of all real numbers R.

The first four sets are countable, but the last two are uncountable.

1.1. Subsets

Definition 1.3. The set A is said to be a subset of B, A B, ifevery member of the set A is a member of the set B (A is containedin B). Equivalently, we can write B A (B includes or contains A).

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1.2. UNIONS 3

The relationship between sets established by is called inclusion orcontainment.

Definition 1.4. If A is a subset of B but not equal to B, then A iscalled a proper subset of B, A B or B A (B is proper superset ofA), see Figure 1.2.

Figure 1.2. A is a subset of B

Example 1.2. The set of all men is a proper subset of the set of allpeople.

The following inclusion is valid {1, 4, 5} {1, 2, 3, 4, 5}.The empty set is a subset of every set and every set is a subset of

itself: A, A A.Intervals of reals. Intervals of reals are subsets of R.Given two real numbers a and b, a < b, the closed interval [a, b] is

the set of all real numbers x, such that a x b, and the open interval(a, b) = {x R : a < x < b}, see Figures 1.3 and 1.4.

Figure 1.3. Closed interval [a, b] Figure 1.4. Open interval (a, b)

For all type of intervals - [a, b]; half-open - [a, b), (a, b]; (a, b); thenumbers a and b are called respectively, lower and upper endpoint. In-tervals (, a] and [a,) are also half-open. When x [a, b], we say xis between a and b, if x (a, b) - x is strictly between a and b.

1.2. Unions

Definition 1.5. Let A and B are two sets. The union of A and B,A B, is the set of all elements which are members of either A or B,i.e. A B = {x : x A or x B}, see Figure 1.5.

Some examples on unions.

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1.4. COMPLEMENTS 4

Figure 1.5. The union of A and B, A B

Example 1.3. The following identities are valid:{1, 2} {red, white} = {1, 2, red, white},{1, 2, green} {red, white, green} = {1, 2, red, white, green},{1, 2} {1, 2} = {1, 2}.

1.3. Intersections

Definition 1.6. Let A and B be two sets. The intersection, AB,is the set of members of both A and B, i.e. AB = {x : x A and x B}, see Figure 1.6

If A B = , then A and B are said to be disjoint.

A B

Figure 1.6. The intersection of A and B, A B

Some examples on intersections.

Example 1.4. Find the next sets:{1, 2} {red, white} = ,{1, 2, green} {red, white, green} = {green},{1, 2} {1, 2} = {1, 2}.

1.4. Complements

Definition 1.7. Let A and B be two sets. The set of all elementswhich are elements of B, but not members of A is said to be the relativecomplement of A in B, B \A = {x : x / A and x B}, see Figure 1.7.

Note that it is valid to subtract members of the set A from theelements of B.

Some examples on complements:

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1.5. CARTESIAN PRODUCTS 5

Figure 1.7. The relative complement of A in B, B \A

Example 1.5. The next equalities are fulfilled:{1, 2} \ {red, white} = {1, 2},{1, 2, green} \ {red,white, green} = {1, 2},{1, 2} \ {1, 2} = .Obviously, if A B, then B \ (B \A) = A, see Figure 1.2.

1.5. Cartesian products

Definition 1.8. The Cartesian product of two sets A and B (prod-uct set, or cross product) is the set of all ordered pairs whose first com-ponent is an element of A and whose second component is a member ofB, i.e.

AB = {(a, b) : a A , b B}.Example 1.6. Let A = {,,} and B = {a, b} be given. Their

Cartesian product A B consists of all ordered pairs, as it is shownbelow:

AB = {(, a), (, b), (, a), (, b), (, a), (, b)}.Note 1.2. We use the same symbol (a, b) for an open interval and

ordered pair, but from the context it would be clear whether it is aninterval or ordered pair.

Similarly as reals correspond to points on a line, the ordered pairsof reals correspond to points on a plane and ordered triples of realscorrespond to points on a space. By Definition 1.8 follows, that the realplane and 3-dimensional space are, respectively:

R2 = R R = {(x, y) : x, y R} , andR3 = R R R = {(x, y, z) : x, y, z R} .

In plane: the x-axes is the set of all points of the form (x, 0), the y-axesis the set of all points of the form (0, y), and the origin is the point (0, 0).

The Cartesian product is named after Rene Descartes (1596 - 1650) latinized formof Renatus Cartesius. He was a French philosopher (Father of Modern Philosophy)and writer who contributed greatly to analytic geometry. The Cartesian coordinatesystem was named after him.

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EXERCISES 6

1.6. Basic properties of set operations

Let A, B, C and D be arbitrary sets. Then the following relationsare valid:

A B = B A and A B = B A - commutative law; (A B) C = A (B C) and (A B) C = A (B C) -associative law;

(AB)C = (AC) (B C) and (AB)C = (AC)(B C)- distributive law;

A A = A and A = A; A A = A and A = ; A \B = B \A and A \A = ; A = and AB = B A; A(BC) = (AB)(AC), (AB)C = (AC)(BC)and A (B C) = (A B) (A C), (A B) C = (A C) (B C);

(AB)(CD) = (AC)(BD) and (AB)(CD) =(A C) (B D);

(A\B)C = (AC)\(BC) and A\(BC) = (AB)\(AC); C\(AB) = (C\A)(C\B) and C\(AB) = (C\A)(C\B)- De Morgans laws, see Figure 1.8.

A B

C

A B

C

Figure 1.8. De Morgans laws

On this lesson the following books and sites were used: [9] - [11],[14, 15], [27], [35], [44, 45], [67], [79], [107] and [115].

Exercises

1.1. Let consider the sets: A = {1, 3, 5, a,m, p, q}, B = {5,m, q},C = {a, p, q, 1, 2, 4}, D = {1, 3, 4, 5, p, q}. Find the following sets: M =A B; N = C D; R = B C D; S = (A B) (C D).

Augustus De Morgan (1806 - 1871) was a British mathematician and logician. Heformulated De Morgans laws and introduced the term mathematical induction. Thecrater De Morgan on the Moon is named after him.

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EXERCISES 7

Answer: M = {1, 3, 5, a,m, p, q}; N = {1, 4, p, q};R = {a, p, q,m, 1, 2, 3, 4, 5};

S = {5,m, q} {1, 4, p, q} = {1, 4, 5, p, q,m}.1.2. Given a segment AB. A pair of points M and N lies on the

segment between the origin A and the end point B so thatM stays beforeN . Find the following sets: P = {AM} {AN}, Q = {AN} {MB}and R = {MN} {MB}.

Answer: P = {MN}; Q = {AB}; R = {MB}.1.3. In a box there are 5 red balls enumerated as {ri}5i=1 = R, 8

white balls enumerated as {wi}8i=1 = W and 10 blue balls enumeratedas {bi}10i=1 = B. Find the following sets: A = R W , B = R W , andC = {r1, r3, r5, w2, w4, w8, b3, b7, b8} {r2, r4, r5, w4, w5, w6, b1, b2, b3}.

Answer: A = {r1, . . . , r5, w1, . . . , w8}; B = {}; C = {r5, w4, b3}.1.4. Let roll once two distinct dice. Consider the sets: A = {the

sum of points on the upper faces is 5}; B = {the sum of points onthe upper faces is even}; C = {the sum of points on the upper faces is 10}. Find the following sets: A B; B C; (A B) C.

Answer: A B = {(1, 1) , (1, 3) , (3, 1)};B C = {(4, 6) , (6, 4) , (5, 5) , (6, 6)}; (A B) C = B C.

1.5. Consider following intervals: A = (1, 2), B = [0, 3), C =[6, 2], D = [2, /2], E = [1,). Find: C D; (A B) C;C \B; E \A.

Answer: C D = [6, /2]; (A B) C = (1, 2];C \B = (6, 0) E \A = {1} [2,) .

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Numbers

2.1. Real numbers

The set of reals, R, includes rational numbers Q, irrational numbersI, either algebraic (real roots) and transcendental numbers (such as and e).

Axiomatic definition of reals. The set of reals R is a set of num-bers, which satisfies the following axioms.Extend axiom states that R has at least two distinct members.Field axioms concern operations addition + and multiplication .

Axioms 2.1. Let a, b, c R then:(1) a+ b R (closure law of addition) and a+ b = b+ a (commu-

tative law of addition).

(2) a+(b+c) = (a+b)+c = a+b+c (associative law of addition).

(3) There exists a real number 0 R (zero), such that a + 0 =0 + a = a (existence of additive identity).

(4) For each real number a there exists a unique opposite numbera R, that possesses the property a + (a) = a a = 0(existence of additive inverse).

(5) a b R (closure law of multiplication) and a b = b a (com-mutative law of multiplication).

(6) a (b c) = (a b) c = a b c (associative law of multiplication).(7) There exists a real number 1 R (unit), such that a 1 = a

(existence of multiplicative identity).

(8) For each real number a = 0 there exists a unique inverse numbera1 1/a R, such that aa1 = 1 (existence of multiplicativeinverse).

(9) a(b+ c) = ab+ ac (distributive law - combined axioms).

Relations (1)-(4) are the axioms for operation addition (+), and (5)-(8) are axioms for multiplication. The next statements are implicationsfrom the field axioms.

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2.1. REAL NUMBERS 9

(1) From a+ c = b+ c it follows that a = b.(2) a 0 = 0.(3) From c a = c b and c = 0 it follows that a = b.

Order axioms concern order relation less than (

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2.2. Integers

The set Z, of all integers positive, negative and 0, includes the setof all natural numbers N, N Z.

Similarly the set of natural numbers, Z is closed under the operationsof addition and multiplication, i.e. the sum and product of any twointegers are integers. Including negative natural numbers and zero, Z(unlike N) is also closed under subtraction. The set of integers is notclosed under the operation of division, since the quotient of two integers(e.g., 1/2), need not be an integer.

Let a, b, c Z. Then, the basic properties of addition a+ b, a+ b+ cand multiplication ab, abc are analogous to relations (1)-(7) and (9) fromAxioms 2.1. It is clear, that if ab = 0, then a = 0 or/and b = 0.

In number theory, the fundamental theorem of arithmetic (or theunique prime factorization theorem) states the following.

Theorem 2.1. Any integer n N, greater than 1 can be representedas a unique product of prime numbers, as

n = p11 p22 . . . p

kk ,

where p1, p2, . . . , pk are primes and 1, 2, . . . , k N are their frequen-cies.

Remark 2.1. The fundamental theorem of arithmetic is a corollaryof the first principle of Euclid that states if a prime p divides a productmn, m,n N, then p divides m or p divides n.

For example, next two numbers can be written as the product ofprime numbers

6936 = 23 3 172 and 3600 = 24 32 52 .Finding the prime factorization of an integer allows derivation of all itsdivisors, both prime and non-prime. By the prime factorizations of twointegers their greatest common divisor (gcd) and least common multiple(lcm) can be found quickly.

Example 2.1. Find lcm and gcd of 48 and 180.Solution. Factorizing these numbers we have:

48 = 24 3 and 180 = 22 32 5 .Hence, lcm(48, 180) = 24 32 5 = 720 and gcd(48, 180) = 22 3 = 12.

Euclid of Alexandria (about 325 - about 265 BC) was an ancient Greek, knownas the Father of Geometry. His Elements, is divided into 13 books, that coverEuclidean geometry and the ancient Greek version of elementary number theory.

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2.3. Rational numbers

Definition 2.1. A real number r is said to be rational if and onlyif it can be represented by a fraction (ratio) r = pq , where p and q are

integers and q = 0. The number p is called numerator, and q denomi-nator.

A fraction pq is said to be proper (improper) fraction if p < q (p q).The set of all rational numbers, Q, is an ordered field. Obviously,

N Z Q R.Decimal representations of rationals. A decimal representation

written with a repeating final 0 is said to terminate before these zeros,e.g. 123/40 = 3.07500 = 3.075. The following theorem is valid.

Theorem 2.2. Any rational number pq Q with a denominatorwhose only prime factors are 2 and/or 5, q = 2m5n; m,n N; hasa finite decimal expansion (that is called terminating decimal) and viseversa.

Definition 2.2. Two numbers are called relatively prime, or co-prime if their greatest common divisor equals 1.

Definition 2.3. A decimal representation of a real number is calleda repeating decimal (or recurring decimal) if there is some finite sequenceof digits that is repeated indefinitely.

By example, 19 = 0.111 = 0.(1).Note 2.1. A number has a terminating or repeating expansion if and

only if it is rational and this does not depend on the base. A numberthat terminates in one numeral base may repeat in another, by example0.310 = 0.0100110011001...2 = 0.0(1001)2.

Theorem 2.3. Any rational number pq Q, where p and q are co-primes, and q has at least one multiplier different from 2 and 5 can beuniquely rewritten as infinite periodic decimal fraction.

Example 2.2. The numbers 1,2, 0.5, 53 , 38 = 0.375 Q, as wellas 111 = 0.(09) Q, 217 = 0.(1176470588235294) Q. But the numbers3, / Q.Converting decimals to fractions. Terminating decimals are eas-

ily converted to fraction by multiplying and dividing by 10 to the powerof number of digits after decimal point and simplifying the fraction tolowest common term.

Example 2.3. Convert 1.375 to a fraction. We have:

1.375 =1375

1000=

53 11

53 8=

11

8.

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2.4. IRRATIONAL NUMBERS 12

Repeating decimals are converted to fractions by the following rule.The decimal is equal to a fraction, which nominator is a difference be-tween number with cutting digits after the first period and number withcutting digits before the first period; and denominator is equal to anumber, which begins with digits 9 as many times, as numbers are inthe period, and finishing with 0s as many times as digits are betweendecimal point and the first period.

Example 2.4. Convert 0.096(21) to a fraction. We have:

0.096(21) =9621 9699000

=9525

99000=

3 52 127

3 52 1320=

127

1320.

2.4. Irrational numbers

In Mathematics, it is not quite true that what is not rational isirrational. Irrationality is a term reserved for a very special kind ofnumbers. There are numbers which are neither rational or irrational(for example, infinitesimal numbers are neither rational nor irrational).

Definition 2.4. An irrational number is a real number, that cannot be expressed as a fraction p/q of two integers p and q, q = 0.

For instance, the numbers2 (Pythagorass constant - length of

the hypotenuse of an isosceles right triangle with legs of length one),53, are irrational numbers. The irrational numbers consists of reals:algebraic numbers, e.g. n-th root of non-perfect powers, and transcen-dental numbers, that are not the root of any integer polynomial, seechapter 6, section 6.3.

By using the mathematical analysis, it can be proved that everyirrational number s can be approximated by two rational numbers aand b (a < b), such that a < s < b.

The following statements are valid:

Between any two different rational numbers a and b there is atleast one other rational number.

Between any two different irrational numbers a and b there isat least one other rational number.

Between any two different rational numbers a and b there is atleast one other irrational number.

Between any two different irrational numbers a and b there isat least one other irrational number.

Pythagoras of Samos (570 - 495 BC) was an ancient Greek, philosopher, mathe-matician, and founder of the religious movement called Pythagoreanism. He is bestknown for the Pythagorean theorem and Pythagorean triples, integers m,n, p satis-fying m2 + n2 = p2, e.g. 3, 4, 5.

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EXERCISES 13

Remark 2.2. In all four statements the existence of a single num-ber can be replaced with the assertion of existence of infinitely manynumbers.

Example 2.5. The irrational number2 satisfies 1.4140 2, respectively.

Tabular. By tablex . . . xk . . .

y = f(x) . . . yk . . ..

3.3. Functions classification

Classification of functions are performed according to their prop-erties. There are different strategies of classification besides the onesproposed here, e.g. according their: analytical representation - elemen-tary and non-elementary, differentiability and others.Bounded functions

Definition 3.6. A function f : X Y is called bounded in X, ifthe set of its values Y is bounded, i.e

the function f is bounded above, if there exists constantM R,such that f(x) M , x X,

the function f is bounded below if there exists constant m R,such that m f(x), x X,

the function f is bounded if it has upper and lower bounds.By example sine and cosine functions are bounded in R, as | sinx| 1

and | cosx| 1, x R.Even and odd functions

Definition 3.7. Let f : X Y ; X,Y R. A function f is saidto be:

even in X, if and only if x X follows that the oppositex X and f(x) = f(x). Geometrically, an even functionis symmetric with respect to the y-axis, i.e. its graph remainsunchanged after reflection about the y-axis.

odd in X, if and only if x X follows that the oppositex X and f(x) = f(x). Geometrically, an odd functionis symmetric with respect to the origin of the coordinate system,i.e. its graph remains unchanged after rotation of (180) aboutthe origin.

Monotone functions

Definition 3.8. Let f : X Y ; = X,Y R. Then f is monotone increasing (decreasing) over X, if for all x1, x2 X and x1 < x2, we have f(x1) f(x2) (f(x1) f(x2)).

f is strictly monotone increasing (strictly monotone decreasing)over X, if for all x1 < x2, x1, x2 X, we have f(x1) < f(x2)(f(x1) > f(x2)), i.e. if the graph of the function is going upfrom left to right (if the graph is going down from left to right).

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3.5. INVERSE FUNCTIONS 18

f is monotone over X, if f is either increasing or decreasingover X.

f is strictly monotone over X, if f is either strictly increasingor decreasing over X.

Periodic functions

Definition 3.9. Let f : X Y , = X,Y R. A function f(x)is said to be periodic in X with least period T > 0 if x X followsthat x+ T X and f(x) = f(x+ T ).

Remark 3.2. It is clear that kT , k = 2, 3 . . . , are periods too. Forexample, the sine and cosine functions are periodic with period 2, andalso sin(x+ 2k) = sinx and cos(x+ 2k) = cosx, x R and k N.

The constant function is periodic with any period for all nonzeroreal numbers, so there is no concept analogous to the least period forconstant functions.

3.4. Composition of functions

Consider two functions : T X and f : X Y and ( =T,X, Y R), where the function is defined on the interval T andf is defined on the interval X. This define a function F : T Y(composite function - F = f ) as a composition of functions andf by equality y = F (t) = f [(t)], where t T , x = (t) X thereforey = f(x) Y and y = f [(t)] Y .

3.5. Inverse functions

Definition 3.10. Let f : X Y be a one-to one map (bijection)from set X to set Y , = X,Y R, where x X, y = f(x) Y . Wecall that mapping f1 : Y X, x = f1(y) is the inverse function off if and only if the following equalities are valid:

x = f1(f(x)) and y = f(f1(y)) .

We note that the existence of the inverse function is not alwayspossible, but injective functions can be reversed.

Example 3.5. Let consider the linear function

y = ax+ b f(x), then x = y ba= f1(y) for a = 0.

On this topics the following textbooks and sites were used: [1]-[9],[14], [24], [31], [34], [46], [58], [76], [82], [102], [104], [118] and [120].

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EXERCISES 19

Exercises

3.1. Determined the domains Dfi of the following functions: f1(x) =x 1; f2(x) = x

x+ 2; f3(x) =

1x.

Answer: Df1 = [1,); Df2 = R \ {2}; Df3 = (0,).3.2. Consider the next functions: f1(x) = x

3 5x, f2(x) = x12 5x6 + x2, f3(x) = 7x

5 x3 + 1. Prove, that f1 is odd, f2 is even, f3 isneither even nor odd.

3.3. Prove that the function f(x) =1

xwith a domain Df = [0.1,)

is a monotone decreasing in Df , and the function g(x) =x+ 1 is

monotone increasing in the domain Dg = [0,).3.4. Let consider the functions: f1(x) = x

2 + 1; f2(x) = tanx

and f3(x) = cotx; f4(x) = ex; f5(x) =

a

x, a R; f6(x) = loga x,

a (0,) \ {1}.Which of them are: bounded, odd or even, monotone, periodic and

in what sets? Find answers in the next chapters.

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Linear and quadratic functions. Inequalities

4.1. Linear functions

In mathematics the linear function is of the form f(x) = ax + b,where a, b R. Linear functions always have as domain and codomainthe set of all reals. The function is strongly increasing for a > 0 andstrongly decreasing for a < 0. The graph of the linear function is astraight line in the coordinate plane, see Figure 4.1, 4.2 and 4.3.

x

y

0

y=b, b>0

y=b, b 0, ( a < 0, ) the straight line is upwards (downwards) sloping.- If a = 0, the function is stationary, f(x) = b =const.

Signs of linear functions. The signs of linear functions are shownfor a > 0 and a < 0 on Figures 4.2 and 4.3, respectively.

Therefore, by example, a linear inequality ax + b > 0 has solution:x > ba for a > 0, x < ba for a < 0, x for a = 0 and b 0, anyx for a = 0 and b > 0.

Similarly any linear inequality can be solved.

Example 4.1. Given the functions f(x) = x1 and g(x) = x2.What are the signs of the functions values for both functions.

20

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4.2. QUADRATIC FUNCTIONS 21

Figure 4.2. Graphs of linear functions for a > 0

Answer. We have: f(1) = 0, f(x) < 0 for x (, 1), and f(x) > 0,x (1,), see Figure 4.2. For g it follows that g(2) = 0; g(x) > 0,x (,2); and g(x) < 0, x (2,), see Figure 4.3.

Figure 4.3. Graphs of linear functions for a < 0

4.2. Quadratic functions

A quadratic function, is a function of the form f(x) = ax2 + bx+ c,where a, b, c R, a = 0.Parabola

The graph of a quadratic function f(x) = ax2 + bx + c is calleda parabola. The form of parabola depends mainly on the sign of thecoefficient a, see Figures 4.4 and 4.5: if a > 0, the parabola opens upward(convex function), if a < 0, the parabola opens downward (concavefunction).Quadratic formula

If the quadratic function is set to be equal to zero, then we have aquadratic equation. The roots of a quadratic equation can be computedusing the quadratic formula. The quadratic formula is derived by themethod of completing the square.

0 = f(x) = ax2 + bx+ c a[x2 +

b

ax+

b2

4a2 b

2

4a2+

c

a

]= 0 ,

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4.2. QUADRATIC FUNCTIONS 22

which is equivalent to

f(x) = a

[(x+

b

2a

)2 b

2 4ac4a2

]= a(x x1)(x x2) = 0 , (4.1)

where

x1,2 =bb2 4ac

2a. (4.2)

The expression D = b2 4ac is called the discriminant of a quadraticequation.Vietes formulas

Vietes formulas give a simple relation between the roots of a poly-nomial and its coefficients. In the case of quadratic function, they canderive from (4.1) and take the following form:

x1 + x2 = ba,

x1 x2 = ca.

(4.3)

This yields a convenient expression when graphing a quadratic function,as the function is symmetric vertically about the vertex.

Example 4.2. Given the quadratic equation

ax2 4x+ c = 0,that possesses two different roots x1 = 1 and x2 = 3. Find the un-known coefficients a and c.Solution. By Vietes formulas (4.3) follows x1 + x2 = ba = 2, andhence a = 2; by x1x2 = ca = c2 = 3 we derive c = 6.

Having in mind the above stated example we conclude, that thequantities a, b, c, x1, x2 are connected each other by two equations, thatis, the pair of Vietes formulae. Hence if we know three of the abovestated five variables, then by (4.3) one can find the rest two unknowns.Vertex

The place where the parabola turns is called the turning point or thevertex of the parabola (the extreme point). From (4.1) we have that the

x-coordinate of the vertex is xv = b2a

. If there are two real roots the

vertexs x-coordinate is located at the arithmetic mean of the roots, i.e.

xv =x1 + x2

2. The y-coordinate can be obtained by substituting the

Francois Viete (1540 - 1603), a Frenchman known by his Latinized name FranciscusVieta. He practiced as a lawyer and was a member of the kings council, serving kingHenry III and Henry IV. Vieta made significant contributions to mathematics inthe areas of arithmetic, algebra, trigonometry, and geometry. In arithmetic Vietarecommended to use decimal fractions, instead of sexagesimal. His main contributionwas in algebra and relationships between coefficients and roots of a polynomial werenamed after him.

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4.2. QUADRATIC FUNCTIONS 23

above result into the function. Hence, yv = D4a

and the vertex point

is V = (xv, yv) =

( b2a

,D4a

).

The vertex is the maximum if a < 0, or the minimum if a > 0.

Signs of quadratic functions. Signs of quadratic functions areestablished by (4.1):

- If D < 0, then af(x) > 0 for x R,- If D = 0, then af(x) = a2(x x1)2 > 0 for x R \ {x1},- If D > 0, then af(x) = a2(x x1)(x x2) > 0, x1 < x2, for a > 0

follows x (, x1) (x2,) and for a < 0 we have x (x1, x2).The graphics of quadratic function and their corresponding signs are

presented on Figures 4.4 and 4.5, respectively for D > 0, D = 0 andD < 0.

Figure 4.4. Signs of quadratic functions for a > 0

By using signs of quadratic functions we can solve any quadraticinequality.

Example 4.3. Given the inequality

x2 100 > 0 .Find out the solutions of this inequality.Solution. As x2 100 = (x + 10)(x 10) > 0, then x (,10) (10,+), see Figure 4.4.

Example 4.4. Find the solutions of the following inequality:

x2 + x 2 > 0 .Solution. By using (4.1) it follows x2 + x 2 = [(x 12)2 + 74 ] < 0for any reals, therefore x , see Figure 4.5.

Example 4.5. Find the solutions of the following inequality:

2x2 + 5x+ 3 0 .

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EXERCISES 24

Solution. We have 2x2 + 5x + 3 = 2(x + 12)(x 3) 0, hencex [12 , 3], see Figure 4.5.

Figure 4.5. Signs of quadratic functions for a < 0

Biquadratic equations. These equations are used for a quarticequation (fourth-order algebraic equation) having no odd powers, i.e.

ax4 + bx2 + c = 0 .

Such equations are easy to solve, since they reduce by substitution x2 =t 0 to a quadratic equation in the variable t,

at2 + bt+ c = 0 ,

and then can be solved using the quadratic formula (4.2) and, finally, interms of the original variable x by taking the square roots.

Example 4.6. Find solutions of the equation

x4 5x+ 4 = 0.Solution. By substituting x2 = t 0 we obtain the equation t25t+4 =(t1)(t4) = 0 with two real roots, and therefore t2 = 1 = x1,2 = 1and t2 = 4 = x3,4 = 2.

On this topic the following textbooks and sites were used: [1]-[9],[14], [34], [46], [58], [76], [88], [102], [104] and [114].

Exercises

4.1. Prove that the solutions of the inequality x 3 0 belong tothe interval [3,).

4.2. Solve the inequality 3x 2 < x+ 5.Answer. x (, 3.5).

4.3. Solve the inequality |x| > 0.Answer. x R \ {0}.

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EXERCISES 25

4.4. What has to be the parameter k in the quadratic equation

2x2 (k 1)x+ 2 = 0,so that one of the roots to be x1 = 3. Find also the second root of thisequation.

Answer. k = 23/3, x2 = 1/3.

4.5. Prove that the solutions of the inequality x2 4 0 belong tothe set (;2] [2;).

4.6. Solve the inequality x2 x 2 < 0.Answer. x (1; 2).

4.7. Solve the inequality x2 + x+ 1 > 0.

Answer. x R.4.8. Solve the inequality x2 + x+ 1 < 0.

Answer. It has no real solutions, x .4.9. Solve the inequality x2 4x+ 4 0.

Answer. x R.4.10. Solve the inequality 2x2 x 1 0.

Answer. x [1/2, 1].4.11. Find the solutions of the following inequality:

x2 + 2x 3 0.Answer. x [3, 1].

4.12. Find the solutions of the following inequality:

x2 + x 5 0.Answer. It has no real solution, x .

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Exponentiation and power functions

5.1. Powers

In mathematics, exponentiation - operation of raising a number a toa power n, is a process of iterated multiplication, i.e.

an = a a a n

.

Exponentiation involves two numbers, the base a and the exponent n.The expression an is therefore known as a to the n-th power. The inverseof exponentiation is the logarithm; exponentiation is sometimes calledthe antilogarithm - the inverse function of logarithm.

The power may be an integer, rational, real or complex number.However, the power of a real number to a non-integer exponent is not

to be a real number. For example x12 , is real only for x 0.

5.2. Exponentiation

Exponentiation with integer exponents. The exponentiationoperation with integer exponents defined as a repeated multiplication

Example 5.1. We have 35 = 3 3 3 3 3 = 243 and(1

2

)4=

1

16.

Here, 3 and1

2are bases, 5 and 4 are the exponents, respectively.

Traditionally a2 = a a is called the square and a3 = a a a is calledthe cube.Positive integer exponents. The powers with positive integer expo-nents are defined by the initial condition

a1 = a and the recurrence relation an+1 = a an , n N.Remark 5.1. Any number to the power 1 is itself, a1 = a. A number

other than 0 taken to the power 0 is defined to be 1, a0 = 1.

Negative integer exponents. Raising a nonzero number to the 1power produces its reciprocal, i.e.

a1 =1

a, thus: an = (an)1 =

1

an.

26

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5.3. IDENTITIES AND PROPERTIES 27

A negative integer exponent can also be seen as repeated division by thebase, but raising 0 to a negative power would imply division by 0, andso is undefined.

Fractional (rational) exponents.

Definition 5.1. Let introduce the even and odd root.

Even root. Let a 0 and n be a positive even integer. Theunique non-negative solution of the equation

xn = a (5.1)

is called n-th root of the number a and is denoted by x = na.

Odd root. Let a be a real number and n be a positive oddinteger. The unique solution of the equation (5.1) is called n-throot of the number a.

The number n is said to be the index of the root.Exponentiation with a simple fractional exponent 1n , with integer

n > 2, can be defined as taking n-th roots

a1n = n

a

Exponentiation with a rational exponent mn , m Z, n N, can nowbe defined as

amn =

(na)m

.

Example 5.2. There are valid: 813 = 2 and 27

23 = 9.

5.3. Identities and properties

The rules for combining quantities containing powers are called theexponent laws. Let a, b > 0, and x and y be rational numbers or realnumbers. The most important identities satisfied by exponentiation are:

Example 5.3. Simplify the expression

0.000000025

500000 0.00000005 .Solution. By decimal representation of reals and properties of exponen-tiation it follows

0.00000000025

500 0.00000005 =25 109

5 102 5 108 = 103 = 0.001 .

Example 5.4. Compare numbers 234 and 3

23 .

Solution. As 2 < 3 but 34 >23 , we suppose that 2

34 < 3

23 . By Table

5.1 we estimate 234 from above and 3

23 from below, as follows: 2

34 =

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5.4. POWER FUNCTIONS 28

ax > 0 ax+y = ax ay

a0 = 1(ab

)x=

ax

bx

ax =1

axaxy =

ax

ay

(ab)x = axbx (ax)y = (ay)x = axy

ax < ay for a > 1, x < y ax > ay for 0 < a < 1, x < y.

Table 5.1. Identities and properties for exponentiation

20.75 < 20.8 and 323 = 30.(6) > 30.6. Finally we have 2

34 < 20.8 = (28)

110 =

256110 < 729

110 = (36)

110 = 30.6 < 3

23 . Hence, 2

34 < 3

23 . Really, 2

34

1.6818 < 2.0801 3 23 .

5.4. Power functions

In this section some examples of the function y = xn, n Q, arepresented. Let sketch the function y = x3, thus we shall use the Ta-ble 5.2. On Figure 5.1 graphics of power functions y = xn with power

x 2 1.5 1 0.5 0 0.5 1 1.5 2y = x3 8 3.375 1 0.125 0 0.125 1 3.375 8

Table 5.2. Values of the function y = x3

n = 1, 2, 3 are shown. The functions y = x and y = x3 are increasingand odd for x R.The function y = x2 is even for x R, and decreasing for x (, 0]and increasing for x [0,).

On Figure 5.2 are presented the graphs of the functions y =x and

y = 3x. The function y =

x is increasing for x [0,) and y = 3x

is odd, increasing for x R and 3x 0 for x 0. Figure 5.3 illus-trates behavior of functions y =

1x

. The function y =1

xis decreasing

and odd for x R\{0} and y = 1x

is increasing and odd for x R\{0}.

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Figure 5.1. Graphics of function xn, n = 1, 2, 3

5.5. Basic irrational equations and inequalities

Let consider functions f, g : R R, if the functions are not de-fined everywhere we have final solution intersects with domains of f andg. According the definition of n-th root, and monotone increasment ofthe corresponding function we consider the next algorithms.

Algorithms for solving basic irrational equations.

2n+1f(x) = g(x) , n N , f(x) = g2n+1(x),

Figure 5.2. Graphics of functionsx and 3

x

Figure 5.3. Graphics of functions1x and

1x

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5.5. BASIC IRRATIONAL EQUATIONS AND INEQUALITIES 30

2nf(x) = g(x) , n N , f(x) 0g(x) 0f(x) = g2n(x)

.

Algorithms for solving basic irrational inequalities.

2n+1f(x) g(x) , n N , f(x) g2n+1(x), 2nf(x) < g(x) , n N ,

f(x) 0g(x) > 0f(x) < g2n(x)

;

2nf(x) > g(x) , n N , g(x) < 0f(x) 0f(x) < g2n(x)

; or, i.e.

union () with the solution of g(x) < 0f(x) 0 .

Example 5.5. Solvex2 4 + 2x = 5.

Solution. Rewriting this equation asx2 4 = 52x and applying the

corresponding algorithm we obtain:x2 4 05 2x 0x2 4 = (5 2x)2

x (,2) (2,)x (, 52 ]x2 4 = (5 2x)2

x (,2) (2, 52 ] = D3x2 20x+ 29 = 0 .

The quadratic equation 3x220x+29 = 0 has two roots 1013

3 . Finally

we obtain the solution x = 1013

3 2.1315 D, because the other rootof quadratic equation 10+

13

3 4.5352 / D.Remark 5.2. It is possible to solve equations and inequalities with-

out finding its domains, but raising expressions to powers can lead toadditional solutions. Therefore the obtained result has to be checkedwhether it satisfies the equality or inequality and leave values that aretrue solutions.

Example 5.6. Solve the inequality 3x x.

Solution. To remove the radical we raise both sides on cube, and receivex3 x 0. The signs of the function f(x) = (x + 1)x(x 1) in inter-vals (,1), (1, 0), (0, 1) and (1,) are respectively , +, , +.Therefore x (,1] [0, 1]. For more particularities see Chapter 6,Example 6.8.

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Example 5.7. Solve the inequalityx < x+ 1.

Solution. First we determine the domain of the inequality, that is, x 0,therefore x [0,). The second step is to square both sides of the aboveinequality {

x}2

0.Solving the latter obtain that x R and conclude that the solution ofthe first inequality is x [0,) (,+).

Examples 5.5, 5.6 and 5.7 can be solved by graphing or using signcharts and test interval technique, see next chapter.

On this topic the following textbooks and sites were used: [2], [7],[9], [14], [49] and [118].

Exercises

5.1. Simplify an expression 62+

3+

5.

Answer.3 + (3/2)

2 (1/2)30.

5.2. Solve the inequalityx 1 0.

Answer. x = 1.

5.3. Solve the inequalityx 1 x.

Answer. x [1,).5.4. Find solutions of equations: a) (2x2 + 1)1/5 = 2, b) x +

5x+ 19 = 1, c)2x+ 1 =

3x 5.

Answer. a) x1,2 =

312 ; b) x1 = 3, x2 = 6; c) x = 3.

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Systems of equations and inequalities

A system of linear equations is a collection of two and more thantwo linear equations in the same variables.

6.1. Systems of two equations in two variables

Systems of two linear equations in two unknowns. Thesesystems are systems of type a1x+ b1y = c1a2x+ b2y = c2 , ai , bi , ci R ; a2i + b2i > 0 ; i = 1 , 2 . (6.1)This is inhomogeneous system when c21 + c

22 > 0, i.e. c1 and c2 are not

vanishing simultaneously, otherwise - homogeneous.

Definition 6.1. A solution to a linear system is an assignment ofnumbers to the variables, such that each of the equations is satisfied.The set of all possible solutions is called the solution set.

Definition 6.2. Two linear systems in the same variables are equiv-alent if they have the one and the same solution set, i.e. if each of theequations in the second system can be derived (by addition of some equa-tions multiplying with non-zero constant) from the equations in the firstsystem, and vice-versa.

The systems in two variables of type (6.1) can be solved using graph-ing, substitution method or addition method.

GraphingSketch the graphs of linear functions a1x+ b1y = c1 and a2x+ b2y = c2in the same coordinate plane. The lines may be: parallel - no solution,intersecting - unique solution is the crossing point, and one and the sameline - infinitely many solutions

Substitution method

Let a1 = 0, from the first equation of (6.1) we obtain x, so x = c1 b1ya1

and substituting it in the second equation we derive, if a1b2 a2b1 = 0,32

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6.1. SYSTEMS OF TWO EQUATIONS IN TWO VARIABLES 33

the solution of system (x0, y0):

x0 =b2c1 b1c2a1b2 a2b1 , y0 =

a1c2 a2c1a1b2 a2b1 . (6.2)

Addition method

We are going to eliminate y and x from the first and second equation,respectively, in system (6.1). Multiplying first equation by b2 and secondby b1 and after that summarizing equations we eliminate y by analogywe eliminate x, i.e.

+

a1x+ b1y = c1a2x+ b2y = c2 b2(b1)(a1b2 a2b1)x = b2c1 b1c2

+

a1x+ b1y = c1a2x+ b2y = c2 (a2)a1(a1b2 a2b1)y = a1c2 a2c1

If a1b2 a2b1 = 0 the solution (x0, y0) is derived via formulas (6.2).

If a1b2 a2b1 = 0, we have:- inconsistent system (the system does not have a solution) for b2c1

b1c2 = 0 or a1c2 a2c1 = 0,- undefined system (the system posses infinitely many solutions) for

b2c1 b1c2 = 0 or a1c2 a2c1 = 0.Example 6.1. Solve the linear system x y = 12x+ y = 2 .

Solution. First combine both equations so that after summing obtain

3x = 3 = x = 1, y = 0 , or the solution is (1, 0) .Example 6.2. Solve the linear system x 3y = 22x 6y = 4 .

Solution. We note that the second equation is obtained from the firstone by multiplying both sides by the number 2. Hence there is only oneindependent linear equation x 3y = 2 depending on two unknownsx and y. Then we substitute either x or y as a parameter p R,for instance y = p, and get x = 2 + 3p. Therefore, the system underconsideration has infinitely many solutions depending on one parameterp R.

Example 6.3. Solve the linear system x+ y = 13x+ 3y = 5 .

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6.1. SYSTEMS OF TWO EQUATIONS IN TWO VARIABLES 34

Solution. Multiply the first equation by 3. Combine both equationsso that after summing obtain

0 = 5,that is impossible. Therefore the system is inconsistent, i.e. has nosolution.

Remark 6.1. Systems of three linear equations in three variables ofthe typea1x+ b1y + c1z = p1a2x+ b2y + c2z = p2a3x+ b3y + c3z = p3

, ai , bi , ci , pi R ; a2i+b2i+c2i > 0 ; i = 1, 2, 3 ,

are not a subject for consideration here. Such systems can be solvedusing graphing, substitution or addition method, as mentioned above,or apply Gaussian elimination algorithm. This is an object of futurestudding at an university and strongly recommended for using.

Special systems of two equations in two unknowns. Thesesystems are systems with one linear equation and other equation of aspecial type, or with two bivariate quadratic equations. One variable isexpressed from the linear equation and substituted in the other equationof the system. If there are two bivariate quadratic equations of specialtype we can apply Vietes formulas.

Systems of two equations with one linear equation, that aresystems with one linear equation and other equation of special type, asbivariate of second order, or exponential, or trigonometric or etc. Letconsider following examples.

Example 6.4. Solve the system: xy = 6x+ y = 1 .Solution. This system can be solved in two ways.

Substituting y = 1 x in the first equation of the system, we getx2x 6 = 0 with roots x1 = 2 and x2 = 3. Therefore we obtain twopairs of solutions (x1, y1) = (2, 3) and (x2, y2) = (3,2).

The same result follows supposing that x and y are roots of a qua-dratic equation. Applying Vietes formulas we find this equation, namely

Johann Carl Friedrich Gauss (1777 - 1855) was a German mathematician and sci-entist who contributed significantly to many fields, such as number theory, statistics,analysis, differential geometry, geodesy, geophysics, electrostatics, astronomy and op-tics. Sometimes referred to him as the Prince of Mathematicians or greatestmathematician since antiquity. Gauss referred to mathematics as the queen of sci-ences. Many things are named in his honor, such as theorems, functions, notions andothers, e.g. the unit of magnetic flux density, 1 gauss in the centimeter-gram-secondsystem, is equal to 1 maxwell per square centimeter, or 104 tesla.

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6.1. SYSTEMS OF TWO EQUATIONS IN TWO VARIABLES 35

z2 z 6 = 0, with roots z1 = 2 and z2 = 3, and hence solutions are(z1, z2) and (z2, z1).

Example 6.5. Solve the system: 3 2x + 2y = 22x y = 1 .Solution. Expressing y from the second equation and substituting in thefirst one, we obtain: y = 2x+ 12(2x)2 + 3 2x 2 = 0 .Introducing a new variable z = 2x > 0 the second equation is reduced to2z2+3z2 = 0 with solutions z1 = 21 = 2x > 0 and z2 = 2 = 2x < 0,that is impossible in real analysis. Therefore the solution is x = y = 1.

Example 6.6. Solve the system: sinx+ sin y = 1x+ y = 2 .Solution. By second equation we have y = 2 x and substituting infirst yields the trigonometric equation

sinx+ sin(

2 x) = 1 sinx+ cosx = 1 .

For the solutions of this trigonometric equation see Chapter 8, Example8.7: x1 = 2k, x2 =

2+2k; k Z. Therefore y1 = 22k, y2 = 2k.

Systems with two bivariate quadratic equations are systems withequations in two variable x and y of second order, see next example.

Example 6.7. Solve the system: 9x2 + y2 = 45xy = 6 .Solution. Multiplying second equation of the system by 6 and sum-marizing with first gives: (3x y)2 = 9xy = 6

3x y = 3xy = 6 y = 3x+ 3x2 + x 2 = 0 and y = 3x 3x2 x 2 = 0 .

The solutions of first system are x1 = 2, y1 = 3; x2 = 1, y2 = 6, andof the second - x3 = 1, y3 = 6; x4 = 2, y4 = 3.

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6.2. Polynomial functions and their signs

Any system of linear, quadratic, rational, irrational equations andinequalities by some mathematical operation as multiplying, raising andetc, can be reduced to equality or inequality of a type

p(x) 0 , p(x) 0 , p(x) 0 ,where p(x) is a polynomial. In case of equality, the real roots of theequation p(x) = 0 has to be derived, if there is an inequality - a testsigns of a polynomial function p is realized.

Definition 6.3. A function of one variable p : R R of the form

p(x) = anxn + an1xn1 + + a1x+ a0 =

ni=0

anixni , (6.3)

where: ai R , an = 0; i = 0, 1, 2, ..., n; n = 0, 1, 2 . . . , is said to be apolynomial function of degree n (degp = n) with coefficients an, an1, . . . ,a1, a0.

Any non-zero constant is a polynomial of degree 0, a linear functionax+ b, a = 0, is a polynomial of degree 1, and etc.

Signs of a polynomial function are determined by test interval tech-nique (TIT), after finding functions zeros and factorizing it. Its zeros- branch points separate reals in test intervals. In each test interval atest point is chosen, after that the sign of each binomial is determined.Finally we assign a sign to test interval on the line chart, that is the signof p(x) on this interval.

On graphical representations of inequalities, we use the followingnotations: if the inequality is strict the end points of testing interval aremarked by circle (), otherwise - by solid-circle (). The graph of p issketched by dash-line, if the inequality is strict, otherwise by solid line.

Remark 6.2. The roots of integer polynomial (coefficients are inte-gers) equation are found by using the Gauss (see footnote on the page34) theorem and the table of Horner scheme.

The next example is an illustration of TIT.

Example 6.8. Solve the inequality f(x) = (x+ 1)x(x 1) 0.Solution. Test points and signs of binomials are shown in the Table 6.1.Using the sign of binomials we derive the sign chart, which is equivalentto the graph of f with corresponding functions signs. Therefore f(x) 0 for x (,1] [0, 1]. William George Horner (1786 - 1837) was a British mathematician and schoolmas-ter. The invention of the zoetrope, in 1834 and under a different name (Daedaleum),has been attributed to him. Horner published a method of solving numerical equa-tions of any degree, now known as Horners scheme.

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6.3. SYSTEMS OF ALGEBRAIC INEQUALITIES 37

Test points Sign of binomialsx+ 1 x x 1

2 (,1) - - -1/2 (1, 0) + - -1/2 (0, 1) + + -2 (1,) + + +

Test intervals and points Graph and signs of the function

Table 6.1. Application of TIT for f(x) = (x+ 1)x(x 1)

6.3. Systems of algebraic inequalities

The well studied objects at school mathematics are so called alge-braic equations and inequalities, we introduce the next definition of analgebraic function.

Algebraic functions in one variable.

Definition 6.4. A real valued function of one variable y = f(x) issaid to be algebraic if it satisfies a polynomial equation, whose coeffi-cients p0(x), p1(x), . . . , pn(x) are polynomials in x with rational coeffi-cients, i.e. y is a solution of an equation of the form

pn(x)yn + pn1(x)yn1 + + p0(x) = 0 , n N.

A function that satisfies no such equation is said to be transcendental.

By Definition 6.4 follows that:- Any polynomial p(x) is an algebraic function, since polynomials

are simply the solutions for y of the equation y p(x) = 0.- Any rational function p(x)q(x) , where p, q are polynomials, is algebraic,

being the solution of the equation q(x)y p(x) = 0.- The n-th root n

p(x) of any polynomial is an algebraic function, as

it is a solution of the equation yn p(x) = 0. If n = 2, we have modularequation |y(x)| = p(x).

Algebraic inequalities in one variable. Let f(x) be an algebraicfunction with domain D. The algebraic inequalities can take any of theforms:

f(x) < a , f(x) > a , f(x) a, f(x) a , a R .Definition 6.5. A solution of an inequality is any number which

substituting as the variable makes the inequality a true statement.

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6.3. SYSTEMS OF ALGEBRAIC INEQUALITIES 38

Two inequalities are said to be equivalent, if they have the same solu-tions.

By applying some algebraic operations, the algebraic inequality isreduced to equivalent polynomial ones (see (6.3)) in D. To solve suchinequalities you have to determine the sign of obtained polynomial inD.

Remark 6.3. The elementary operations, that yield the equivalentinequalities are:

- addition any real number to both sides of an inequality,- multiplication or division both sides of an inequality by any positive

number,- multiplication or division both sides of an inequality by a negative

number, reverse the direction of the inequality sign.

Remark 6.4. The irrational equation of the form

a n

f(x) + b mg(x) = 0,

where f(x) and g(x) are polynomials; a, b R; m,n Q, can be re-duced to another polynomial equation with accompanying domain of theequality, more general any irrational inequality by appropriate transfor-mations is reduced to polynomials ones.

Systems of linear inequalities in one variableThe systems linear inequalities are simplest systems of algebraic onesand often appear as solutions of polynomial inequalities.

To solve a system of two inequalities in one variable one have only tointersect the domains of first and second inequality. The next examplesdemonstrate this fact.

Example 6.9. Solve the system x 1 > 0x+ 3 > 0 . (6.4)Solution. From the first equation obtain x > 1, and from the second onex > 3, whence x (1,) (3,) = (1,).

Example 6.10. Solve the system x 2 0x 7 0 . (6.5)Solution. From the first equation obtain x 2, and from the second onex 7, whence x [2,) (, 7] = [2, 7].

Example 6.11. Solve the system x 5 > 0x+ 1 < 0 . (6.6)

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6.3. SYSTEMS OF ALGEBRAIC INEQUALITIES 39

Solution. From the first equation obtain x > 5, and from the second onex < 1, whence x (5,) (,1) = . Thus conclude that thesystem is inconsistent.

Systems of quadratic inequalities in one variableTo solve a system of two quadratic inequalities in one variable by us-ing the signs of quadratic functions determine the domains of first andsecond inequality and after that intersect them. The next examplesillustrate an algorithm.

Example 6.12. Solve the system of inequalities: x2 6x+ 5 < 0x2 + 3x 0 .Solution. The graphs and signs of quadratic functions p1(x) = x

26x+5and p2(x) = x2 + 3x are sketched on Figure 6.1.

Figure 6.1. Graphs of functionsx2 6x+ 5 and x2 + 3x

Figure 6.2. Graph of functionf(x) = (x2 6x+ 5) (x2 + 3x)

We have p1(x) < 0 for x (1, 5) and p2(x) 0 for x [0, 3],thus the solution is x (1, 3]. The same result yields by TIT andsigns of the function f(x) = p1(x)p2(x), see Figure 6.2, where we haveto choose some of the following intervals: (, 0], (1, 3], (5,), thatis accomplished by checking signs of p1 or p2 in mentioned intervals.By example p1(1) > 0, p1(2) < 0, p1(6) > 0, hence the solution isx (1, 3].Systems of rational inequalities in one variable

Consider a rational inequality r(x) = p1(x)p2(x) 0, where p1 and p2 arepolynomials. The domain D of r is all reals except zeros of the de-nominator. All rational inequalities, are reduced to equivalent ones

in D by multiplying and dividing r by p2. As r(x) =p1(x)p2(x)

p22(x)we

have to solve the inequality about numerator in the same direction,

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6.3. SYSTEMS OF ALGEBRAIC INEQUALITIES 40

f(x) = p1(x)p2(x) 0, because p22(x) > 0, x D, see the next exam-ple for illustration.

Example 6.13. Find solutions of the inequalityx+ 1

x2 4x 5 1.As the denominator is vanishing for x = 1, 5, it follows that the domainof inequality is R \ {1, 5}. The inequality is equivalent to

x+ 1

x2 4x 5 1 0 x2 5x+ 4x2 4x 5 0.

Let p1(x) = x2 5x+4 and p2(x) = x2 4x 5. Consider the function

f(x) = p1(x)p2(x) = (x 1)(x 4)(x + 1)(x 5). The graphs withfunctions signs of p1 and p2 are presented on Figure 6.3. The signs of

Figure 6.3. Graphs of functionsx2 5x+ 4 and x2 4x 5

Figure 6.4. Graph of functionf(x) = (x2 5x+4) (x2 4x 5)

function f is shown on Figure 6.4. Hence the inequality is fulfilled forx (,1) [1, 4] (5,).Modular inequalities of a type

|f(x)| g(x) , |f(x)| g(x) , |f(x)| g(x) .are solved according the following schemes:

|f(x)| > g(x) f(x) > g(x) or f(x) < g(x);

f(x) = g(x) g(x) 0f(x) = g(x) or

g(x) 0f(x) = g(x) ; f(x) < g(x)

g(x) > 0f(x) < g(x) and g(x) > 0f(x) > g(x) .

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6.3. SYSTEMS OF ALGEBRAIC INEQUALITIES 41

Figure 6.5. Graphics of functions |x2 2x| and x

Example 6.14. Solve the inequality |x2 2x| x.Solution. A graphical solution is presented on Figure 6.5. Applying the

algorithm for solving modular inequalities we have:

x 0x2 2x xand

x 0x2 2x x . These systems of inequalities are equivalent,respectively, to: x 0x2 3x 0 and

x 0x2 x 0 .For x 0, we have: x(x3) 0 = x 3, and x(x1) 0 = x 1.Therefore, x [1, 3].

Systems of linear inequalities in two variables. Let considersystems of two and three linear inequalities. These systems have theform: a1x+ b1y p1a2x+ b2y p2 ,

a1x+ b1y p1a2x+ b2y p2a3x+ b3y p3

, (6.7)

where ai, bi, pi R and a2i + b2i > 0 (i = 1, 2, 3). Some of the symbols or each one of them in the systems (6.7) could be replaced by strict >.The solutions can be obtained by sketching the graphs of the systems inthe plane.

Example 6.15. Solve x y 1.Solution. First graph the line x y = 1. This line separate planeinto two half-planes: above the line which contains positive y-axesand below - the opposite one. Next test a point (0, 1), whether itcoordinates satisfy the inequality x y 1. As the inequality 1 1 istrue we choose the upper half-plane, see Figure 6.6.

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EXERCISES 42

Figure 6.6. Domain defined by xy 1

Figure 6.7. Domain defined by x+y > 0, x y 1, y 2

Example 6.16. Sketch the system

0 < x+ yx y 1y 2

.

Solution.- Sketch all the three inequalities in the same coordinate plane.- The graph of x+y > 0 is the half-plane above the dashed line x+y = 0.- The graph of x y 1 is the half-plane on and above the solid linex y = 1.- The inequality y 2 defines the half-plane on and below the solid liney = 2.- The solution set of the system is the intersection of the three half planesas shown in the Figure 6.7.

On this topic the following textbooks and sites were used: [1], [7],[9], [21], [25], [31], [55], [61], [83], [91] and [103].

Exercises

6.1. Solve the systems: a)

1x2 + 1y+3 = 13x 2y = 2 ;b)

x2 xy = 10y2 xy = 6 ; c) x2 + y2 + 2x 6y + 5 = 0x2 + y2 2y 9 = 0 .

Answer. a) (7145

6 ,3145

4 ); b) (5,3); c) (3, 2), (1, 4).

6.2. Solve the systems of inequalities: a)

2x2 + 9x 5 > 0x2 + 5x 6 < 0 ;b)

x+ 1 0x+ 3 0

x2 x 6 < 0; c)

x2 4 0

x2 + 3x < 02 x x2 0

.

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EXERCISES 43

Answer. a) x (6,5) (1/2, 1); b) x (2,1]; c) x [2, 0).

6.3. Solve the rational inequalities: a)x2 2x 15

2x 5 < 0;

b)x2 x 6

x2 4x+ 5 > 0; c)x 4x 2 3x; d)

5x2 + x 3x2 3x 4 1.

Answer. a) x (,3) (5/2, 5); b) x (5,2) (1, 3);c) x [1, 4/3] (2,); d) x (,1) {1/2} (4,).

6.4. Find domains Df , Dg, Dfg, Dfg, Df/g, where:

f(x) =

(x 3)(x 2)

x 1 , g(x) =x2 16x 7

x2 9 .

Answer: Df = (1, 2] [3,); Dg = (,3] [3, 7) (7,);Dfg = Dfg = Df Dg; Df/g = (3, 4) (4,).

6.5. Solve the systems of inequalities:

a)

x+ 4

2x 3 > 0x+ 4

2x 3 1

2

; b)

x+ 8

x+ 2> 2

lg(x 1) 1.

Answer. a) x (,4); b) x (1, 4).6.6. Solve the next modular inequalities:

a)

x 2x+ 1 2; b)

3x+ 1x 3 < 3; c) |x2 5x| < 6;

d)|x 3|

x2 5x+ 6 2; e)x2 5x+ 4x2 4

1.Answer. a) x [4,1) (1, 0]; b) x (, 4/3);

c) x (1, 2) (3, 6); d) x [3/2, 2); e) x [0, 8/5] [5/2,).6.7. Sketch the solutions to systems of inequalities in two variables:

a)

x 2y 43x y < 3 ; b)

x 2y 24x 3y > 6

x+ y < 5; c)

2x+ y 24x+ 3y 12

0 x 2y > 0

.

Answer.

a) b) c)

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Exponential and logarithmic functions

In Chapter 5 exponentiation with rational exponent has been con-sidered. The next section extended this notion.

7.1. Exponentiation, identities and properties

Exponentiation with real exponents. Since any real numbercan be approximated by rational numbers, exponentiation to an arbi-trary real exponent can be defined by continuity.

Let x be an irrational number, using mathematical analysis it canbe proved:

For a > 0 there exists a unique real number A, such that ar < A A > as, where r and s are rational numbers and r < x < s.

Example 7.1. Since3 1.732, we can assume 5

3 51.732.

Identities and properties. The rules for combining quantitiescontaining powers are called the exponent laws. Let a, b > 0, and xand y be rational numbers or real numbers. The most important iden-tities satisfied by exponentiation are given in Table 5.1 chapter 5.

Remark 7.1. The number e, mathematical constant of John Napier,is defined as the limit of a series of integer powers

e = limn

(1 +

1

n

)n,

where e 2.718281828459.... An integer power x of e is

ex =

(lim

m

(1 +

1

m

)m)x= lim

m

(1 +

x

mx

)mx= lim

n

(1 +

x

n

)n.

The right hand side generalizes the meaning of ex so that x does nothave to be an integer but can be a fraction, a real number, a complexnumber, or a square matrix.

John Napier (1550 - 1617) also signed as Neper, Nepair named Marvellous Mer-chiston, was a Scottish mathematician, physicist, astronomer and astrologer, and alsothe 8th Laird of Merchistoun. He was the inventor of logarithms, introduced constante, denoted sometimes by exp, and natural logarithm.

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7.3. EXPONENTIAL EQUATIONS AND INEQUALITIES IN ONE VARIABLE 45

Example 7.2. Compare reals ( 110)e and 102

2.

Solution. As e 2.7183 ans 2 1.4142, we have(1

10

)e 102.7183 > 102.8284 102

2 .

This inequality follows by the fact, that the function 10x is monotoneincreasing and 2.7183 > 2.8284.

7.2. Exponential functions

The graphs of some exponential functions y = ax are given on Figure7.1. As a > 0, then ranges of exponential functions is (0,).

For 0 < a < 1 the exponential functions ax are monotone decreasingfor all reals. So, ax1 > ax2 if and only if x1 < x2.

For a > 1 the exponential functions ax are monotone increasing inR. So, ax1 < ax2 if and only if x1 < x2.

Figure 7.1. Exponential functions

7.3. Exponential equations and inequalities in one variable

Exponential equations in one unknown. Let: a, b > 0; a, b = 1and c, p, q be real constants and f and g are real valued functions withcommon domain = D R. Then the following schemes for solvingequalities in one variable are valid.

af(x) = ag(x) f(x) = g(x), x D. af(x) = bg(x) (after logarithmic transformation with abase of a or b) f(x) = g(x) loga b, x D, or f(x) logb a = g(x),x D.

af(x) = cbf(x) (ab

)f(x)= c, x D.

a2f(x)+ paf(x)+ g = 0 for x D, substituting y = af(x) > 0 weobtain

af(x) = y > 0y2 + py + q = 0 .

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7.3. EXPONENTIAL EQUATIONS AND INEQUALITIES IN ONE VARIABLE 46

paf(x) + qbf(x) = c for x D, if ab = 1, the substitutionaf(x) = y > 0 yields

af(x) = y > 0py2 cy + q = 0 .Example 7.3. Solve the equation 8x 4x+0.5 2x + 2 = 0.

Solution. Rewriting equation as

(2x)3 2(2x)2 2x + 2 = 2x[(2x)2 1] 2[(2x)2 1] =[(2x)2 1](2x 2) = 0 ,

leads to 2x = 1, 2x = 1 and 2x = 2. The first equality is impossible inreal analysis, the last ones yield solutions x1 = 0 and x2 = 1.

Exponential inequalities in one variable. Let: f be a real val-ued function with non-empty domain D R, a > 0 and b R. Considerthe next inequalities:

af(x) > b , (7.1)

af(x) < b . (7.2)

Solutions depend on a and b.Solutions of (7.1) are derived according the following schemes:- For b 0 the inequality is valid x R.- For b > 0 and a (0, 1) we have

loga af(x) < loga b f(x) < loga b , x D .

- For b > 0 and a > 1 we have

loga af(x) > loga b f(x) > loga b , x D .

The inequality (7.2) is solved in the following way:- For b 0 the inequality is not valid, x .- For b > 0 and a (0, 1) we have

loga af(x) > loga b f(x) > loga b , x D .

- For b > 0 and a > 1 we have

loga af(x) < loga b f(x) < loga b , x D .

Remark 7.2. If we have a2f(x) + paf(x) + q < 0; p, q R; a substi-tution y = af(x) > 0 yields y2 + py + q > 0.

Example 7.4. Find solutions of the inequality 0.53x1 0.5x+1.We represent the inequality as

(0.5x)3 0.25 0.5x 0.5x[(0.5x)2 0.25] 0 0.5x[0.5x + 0.5](0.5x 0.5) 0 .

As the ax > 0 for any x, then the sign of inequality is determined by0.5x 0.5 0. Therefore we have x 1, because the function 0.5x ismonotone decreasing x R.

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7.4. LOGARITHMS, IDENTITIES AND PROPERTIES 47

7.4. Logarithms, identities and properties

Logarithms. The logarithm is the mathematical operation that isthe inverse of exponentiation, or raising a number (the base) to a power.

Definition 7.1. The logarithm of a number b, b > 0, in base a,a > 0, a = 1, is the unique solution of the equations ax = b, see Figure7.2.

Figure 7.2. Definition of a logarithm

The logarithm is the number x R such that ax = b. It is usuallywritten as x = loga b.

Example 7.5. We have: log2 64 = 6 since 26 = 64; log3 81 = 4 since

34 = 81 and log0.1 0.01 = 2 since 0.12 = 0.01.

If x is a positive integer n, an means multiplying a by itself n times.The most widely used bases for logarithms are 10, the mathematical

constant e 2.71828... and 2, respectively their notations are: common logarithm (lg = log10) in engineering and when loga-

rithm tables are used to simplify hand calculations; natural logarithm (ln = loge) in mathematical analysis; binary logarithm (ld = log2) in information theory and musical

intervals.

Identities and properties of logarithms. Logarithms were in-troduced to make numerical calculations easier. For instance, loga xy =loga x + loga y because a

x ay = ax+y. The following trivial identitiesare valid loga 1 = 0 because a

0 = 1, also loga a = 1 because a1 = a. The

logarithm is defined only for positive reals because loga 0 is undefined,there is no number x such that ax = 0.

Let: a, b R \ {1} , m R and x, y > 0, the most importantidentities satisfied by logarithms are (see Table 7.1):

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7.5. LOGARITHMIC FUNCTIONS 48

aloga x = x loga ax = x

loga xy = loga x+ loga y logaxy = loga x loga y

loga xm = m loga x logam x =

1m loga x, m = 0

loga x =logb xlogb a

loga b =1

logb a

loga x < loga y, a > 1 loga x > loga y, 0 < a < 1

Table 7.1. Properties of logarithms

7.5. Logarithmic functions

Change of base is used to find a logarithm with base b > 0, b = 1,using any other base a > 0, a = 1 (common or natural), according to theequality logb x =

loga xloga b

. This result implies that all logarithm functions

are similar to each other, they are scaled by loga b, see Figure 7.3.

Example 7.6. To draw a graph of the function y = ln x let calcu-late some of its values, see next table:

x 1 e e2 e3 e1

y = ln x 0 1 2 3 1

Figure 7.3. Graphs of logarithmic functions

Let 0 < a < 1, then the logarithmic function loga x is monotonedecreasing for x > 0. So, loga x1 > loga x2 0 < x1 < x2.

Let a > 1, then the logarithmic function loga x is monotone increas-ing for x > 0. So, loga x1 < loga x2 0 < x1 < x2.

The antilogarithm function is another name for the inverse of thelogarithmic function. It is written antiloga(x) and means the same asax.

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7.6. Logarithmic equations and inequalities in one variable

Logarithmic equations in one unknown. Let the following con-ditions hold: a, b, x > 0; a, b = 1; c, p, q be real constants; f(x), g(x),f1(x), f2(x), f3(x) be positive valued functions with common domainD R.

loga f(x) = loga g(x) f(x) = g(x), x D.

loga f(x) = c aloga f(x) = axf(x) > 0 f(x) = axf(x) > 0 , x D.

loga f1(x) + loga f2(x) = loga f3(x) loga f1(x) f2(x) = loga f3(x) f1(x) f2(x) = f3(x) , x D ,

or

loga f1(x) loga f2(x) = loga f3(x)

logaf1(x)

f2(x)= loga f3(x)

f1(x)

f2(x)= f3(x) , x D .

loga f(x) = logb g(x)

loga f(x) =loga g(x)

loga b loga b loga f(x) = loga g(x)

loga f(x)loga b = loga g(x) f(x)loga b = g(x) , x D, .

log2a f(x) + p loga f(x) + q = 0 loga f(x) = yy2 + py + q = 0 , x D .

Example 7.7. Solve the equation log1x 3 + log3(1 x) = 2.5.Solution. The domain of this equality is 1 x > 0 and 1 x = 1, hencex D = (, 0) (0, 1). Using properties of the logarithmic functionwe have:

log1x 3 + log3(1 x) = 2.5 1

log3(1 x)+ log3(1 x) = 2.5 .

The last one is a quadratic equation

log23(1 x) 2.5 log3(1 x) + 1 = 0with roots log3(1 x1) = 12 and log3(1 x1) = 2, and applying antilog-arithm function we derive: x1 = 1

3 D and x2 = 8 D.

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Logarithmic inequalities in one variable. Let: f be a positivevalued function with domain = D R; a > 0, a = 1, and b R.Consider the next inequalities:

loga f(x) > b (7.3)

loga f(x) < b . (7.4)

. Solutions depend on a and b.Solutions of (7.3) are derived according the following schemes:- For 0 < a < 1 taking the antilogarithm of both sides we obtain

aloga af(x)

< ab f(x) < abf(x) > 0 , x D .- For a > 1 taking the antilogarithm of both sides we obtain

aloga af(x)

> ab f(x) > abf(x) > 0 , x D .

The inequality (7.4) is solved in the following way:- For 0 < a < 1 taking the antilogarithm of both sides we obtain

aloga af(x)

> ab f(x) > abf(x) > 0 , x D .

- For a > 1 taking the antilogarithm of both sides we obtain

aloga af(x)

< ab f(x) < abf(x) > 0 , x D .

Remark 7.3. If we have log2a f(x) + p loga f(x) + q < 0; p, q R; asubstitution y = loga f(x) yields a quadratic equation y

2 + py + q < 0.

Example 7.8. Solve the inequality log 12(x+ 5)2 log 1

2(3x 1)2.

Solution. As (x+5)2 and (3x1)2 are positive for all reals except points5 and 13 , the domain of this inequality is D = (,5) (5, 13) (13 ,). By using that logarithmic function with base < 1 is decreasing,we obtain (x + 5)2 (3x 1)2 (x + 1)(x 3) 0. The signs ofthis quadratic function is nonnegative in (,1] [3,). The finalsolution is the intersection (,1][3,)D = (,5)(5,1][3,).

On this theme the following textbooks and sites were used: [2], [9],[14], [49], [66, 68] and [118].

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Exercises

7.1. Solve the following inequalities:

a) 33x+1 > 1; b) 2x21 > 1; c) 91/x 3x; d) 2x < 321/x;

e) (1/3)|x|4 < 9.

Answer. a) x (1/3,); b) x (,1) (1,);c) x [2; 0) [2,); d) x (,10) (0,10);

e) x (,2) (2,).7.2. Solve the following equations:

a) xx = x (x > 0); b) 3x2+1 + 3x

21 = 270; c) 4x 2x+2 32 = 0;d) 4x+1,5 2x = 1, 5; e) 32x+5 3x+2 = 2.

Answer. a) x = 1; b) x = 2; c) x = 3; d) x = 1; e) x = 2.7.3. Solve the following inequalities:

a) logx32 +

3

7 log2(x+ 3);

c) log2 x > 1; d) log1/3 x > 2.

Answer. a) x (0; 1); b) x (4,); c) x (2,); d) x (0, 1/9).7.4. Solve the following equations:

a) log1/5 x2 = 3; b) lg(3x 8) + lg(2 x) = 5; c) log3

(x 6)2 = 4;

d) logx(2x2 5x+ 6) = 2; e) log(x1)(x2 5x+ 10) = 2.

Answer. a) x =5/25; b) it has no solution; c) x1 = 87, x2 = 75;

d) x1 = 2, x2 = 3; e) x = 3.

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Trigonometry

The trigonometric functions, that are known in mathematics as cir-cular functions, may introduced by different ways. We starting acquain-tance with them as functions of an angle, and later they extended asfunctions of reals.

8.1. Definition of trigonometric functions

Right-angled triangle definition. Consider the right-angled tri-angle ABC with sides of length: BC = a > 0, AC = b > 0, ACBCand hypotenuse AB = c > 0 as shown in Figure 8.1. For the angle

Figure 8.1. A rectangular triangle

= BAC, we have:sin =

b

c=

opposite

hypotenuse, cos =

a

c=

adjacent

hypotenuse,

tan =b

a=

opposite

adjacent=

sin

cos , sec =

c

a=

hypotenuse

adjacent=

1

cos ,

cot =a

b=

adjacent

opposite=

cos

sin , csc =

c

b=

hypotenuse

opposite=

1

sin .

There are few angles for which all trigonometric functions may be foundusing the triangles shown in the Figure 8.1. Their values are given inthe Table 8.1 and on the Figure 8.9.

52

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8.1. DEFINITION OF TRIGONOMETRIC FUNCTIONS 53

Angle Trigonometric functions(radians) sin cos tan cot csc sec

0 0 1 0 not defined not defined 1

6

1

2

3

2

3

3

3 2

23

3

4

2

2

2

21 1

2

2

3

3

2

1

2

3

3

3

23

32

21 0 not defined 0 1 not defined

Table 8.1. Special values in trigonometric functions

Measure of an angle. Any real number can be interpreted asthe radian measure of the angle constructed as follows: wrap a piece ofstring of length s units around the unit circle (counterclockwise if 0,clockwise if < 0) with initial point P (1, 0) and terminal point Q(x, y),see Figure 8.2. This gives rise to the central angle with vertex O(0, 0)and sides through the points P and Q. If instead of wrapping a length

Figure 8.2. Measure of an angle

s of string around the unit circle, we decide to wrap it around a circleof radius r, the angle (in radians) will satisfy the following relations = r.

Note 8.1. Note that the length s of string gives the measure of theangle only when r = 1.

Usually we measure angles in degrees, which are defined by parti-tioning one whole revolution into 360 equal parts, each of which is thencalled one degree. So, one whole revolution around the unit circle mea-sures 2 radians and also 360 degrees (or 360), i.e.:

360 = 2 radians, or 180 = radians .

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8.1. DEFINITION OF TRIGONOMETRIC FUNCTIONS 54

Each degree may be further subdivided into 60 parts, called minutes,and in turn each minute may be subdivided into another 60 parts, calledseconds: 1 = 60 minutes = 60 and 1 = 60 seconds = 60.

As it wellknown the number is irrational, 3.14 < < 227 , and aremarkably accurate approximation of was given by Archimedes.

Figure 8.3. Sine and tangent axes Figure 8.4. Cosine and cotangentaxes

Unit circle definition. All six trigonometric functions are definedin terms of the coordinates of the point Q(x, y), as follows:

cos = x ; tan =y

x, x = 0 ; sec = 1

x, x = 0 ;

sin = y ; cot =x

y, y = 0 ; csc = 1

y, y = 0 .

Since Q(x, y) is a point on the unit circle, we know that x2 + y2 =1. This fact and the definitions of the trigonometric functions yieldPythagorean, see Footnote on the page 12, and reciprocal identities,see third section.

This modern notation for trigonometric functions is due to Leonard

Archimedes of Syracuse (287 - 212 BC) was an ancient Greek, mathematician,physicist, engineer, inventor, and astronomer. Among his advances in physics arethe foundations of hydrostatics, statics and an explanation of the principle of thelever. He used the method of exhaustion to calculate the area under the arc of aparabola with the summation of an infinite series. He also defined the spiral bearinghis name, formulae for the volumes of surfaces of revolution and an ingenious systemfor expressing very large numbers. Archimedes had proven that the sphere has twothirds of the volume and surface area of the cylinder, and regarded this as the greatestof his mathematical achievements.

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8.2. CLASSIFICATION OF TRIGONOMETRIC FUNCTIONS 55

Euler (1748).Periodic functions. Let an angle corresponds to a point Q(x, y) onthe unit circle. It is not hard to see that the angle + 2 correspondsto the same point Q(x, y), and hence

cos( + 2) = cos , sin( + 2) = sin . (8.1)

Moreover, 2 is the smallest positive angle for which equations (8.1) aretrue for any angle . In general, we have for all angles :

cos( + 2n) = cos , sin( + 2n) = sin , n = 0,1,2, . . . .(8.2)

We call the number 2 the period of the trigonometric functions sineand cosine, and refer to these functions as being periodic, see Figures8.5 and 8.6. Both secant and cosecant are periodic functions with period2, while tangent and cotangent are periodic with period , see Figures8.7 and 8.8.

8.2. Classification of trigonometric functions

The main properties of trigonometric functions - sine, cosine, tangentand cotangent are accomplished according classification given in Chapter3, Section 3.3.

Sine function. The sine function is a surjective map of reals onto[1, 1], sin : R [1, 1], with following properties:

Bounded function, | sinx| 1, on R. Odd function, sin(x) = sinx, on R (odd identity). Periodic function with period 2, sin(x+ 2) = sinx, x R. Strictly increasing on [2 , 2 ], strictly decreasing on [2 , 32 ], andon rotated intervals over 2k, k Z.

Figure 8.5. Graph of sine function

Leonhard Euler (1707 - 1783) was a Swiss mathematician and physicist. He madeimportant discoveries in infinitesimal calculus and graph theory. He also introducedmuch of the modern mathematical terminology and notation, such as the notion of amathematical function. He is also renowned for his work in mechanics, fluid dynamics,optics, and astronomy. Euler spent most of his adult life in St. Petersburg, Russia,and in Berlin, Prussia. He is considered to be one of the greatest mathematician ofall time. He is also one of the most prolific mathematicians ever - his collected worksfill 6080 quarto volumes.

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8.2. CLASSIFICATION OF TRIGONOMETRIC FUNCTIONS 56

The graph of sinx is shown on Figure 8.5.

Cosine function. The cosine function is a surjective map of realsonto [1, 1], cos : R [1, 1], with following properties:

Bounded function, | cosx| 1, on R. Even function, cos(x) = cosx, on R (even identity). Periodic function with period 2, cos(x+ 2) = cosx, x R. Strictly increasing on [, 0], strictly decreasing on [0, ], andon wrapped intervals over 2k, k Z.

The graph of cosx is shown on Figure 8.6.

Figure 8.6. Graph of sine function

Tangent function. . The tangent function is a surjective map ofD = R \ {2 + k , k Z} onto R, tan : D R, with followingproperties:

Unbounded function, x D. Odd function, tan(x) = tanx, on D (odd identity). Periodic function with period , tan(x+ ) = tanx, x D. Strictly increasing on (2 , 2 ), and on wrapped intervals overk, k Z.

Figure 8.7. Graph of tangent Figure 8.8. Graph of cotangent

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8.3. IDENTITIES 57

The graph of tangent function is shown on Figure 8.7.

Cotangent function. The cotangent function is a surjective mapofD = R\{k , k Z} onto R, cot : D R, with following properties:

Unbounded function, x D. Odd function, cot(x) = cotx, on D (odd identity). Periodic function with period , cot(x+ ) = cotx, x D. Strictly decreasing on (0, ), and on wrapped intervals over k,k Z.

The graph of cotangent function is presented on Figure 8.8.

8.3. Identities

Fundamental identities

Reciprocal identities: sinx =1

cscx, cosx =

1

secx, tanx =

1

cotx.

Pythagorean identities: sin2 x+ cos2 x = 1, 1 + tan2 x = sec2 x, 1 +cot2 x = csc2 x.

Addition formulas

Sum-difference formulas:

sin(x y) = sinx cos y cosx sin y , tan(x y) = tanx tan y1 tanx tan y ,

cos(x y) = cosx cos y sinx sin y , cot(x y) = cotx cot y 1cotx cot y .

Using the addition formulas, we obtain the co-function identities:

sin(x+

2

)= cosx , cos

(x+ 2

)= sinx , cos

(2 x

)= sinx ,

sin(2 x

)= cosx , cot

(2 x

)= tanx , tan

(2+ x

)= cotx ,

sin( x) = sinx , cos( x) = cosx , cot( x) = cotx .Applying co-function and even-odd identities, we can extent the spe-

cial values in sine and cosine functions, see Figure 8.9.Double-angle and half-angle formulasDouble-angle formulas

sin 2x = 2 sinx cosx , sin 2x =2 tanx

1 + tan2 x,

cos 2x = cos2 x sin2 x , cos 2x = 2 cos2 x 1 , cos 2x = 1 2 sin2 x ,

cos 2x =1 tan2 x1 + tan2 x

, tan 2x =2 tanx

1 tan2 x , cot 2x =cot2 x 12 cotx

.

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8.3. IDENTITIES 58

Figure 8.9. Special values in sine and cosine functions

Example 8.1. Simplify the expression 2 sin sin 2+ cos 3.Solution. Consequently applying sum and difference formulas for cosinewe have:

2 sin sin 2+ cos 3 = 2 sin sin 2+ cos(2+ ) =

2 sin sin 2+ cos 2 cos sin 2 sin =cos 2 cos+ sin sin 2 = cos .

Half-angle formulas

sin2x

2=

1 cosx2

, cos2x

2=

1 + cosx

2,

tan2x

2=

1 cosx1 + cosx

=1 cosxsinx