Idea Transcript
Russian Ministry of Education and Science GUBKIN RUSSIAN STATE UNIVERSITY OF OIL AND GAS Department of High Mathematics
Yu. O. KOROLEVA A. V. KOROLEV
BRIDGING COURSE IN MATH Educational Handbook
Moscow 2015
ББК 51 (075)
R e vi e w by: Prof. V. V. Kalinin Prof. S. V. Shaposhnikov
Koroleva Yu. O., Korolev A. V. Bridging course in Math: Educational Handbook. – М.: Publishing Center Gubkin Russian State University of Oil and Gas, 2015. – 65 p. The present book deals with some background knowledge in mathematics that covers school level. Many exercises are solved with detailed explanations. The book is oriented for students of any specialization and those who listens the preparatory course. This publication is the owned by Gubkin Russian State University of Oil and Gas and reproduction without the agreement of the university is prohibited.
© Yu. O. Koroleva, A.V. Korolev, 2015 © Gubkin Russian State University of Oil and Gas, 2015
INTRODUCTION Independently what profession you chose for your future education, any university program always includes a math course. In order to have a good start for studying Math at a High school it is very useful to refresh the school math program to “build the bridge” from school to University. The present educational manuscript contains all necessary background which must know every student to successfully study and pass the university math course. The book collects theoretical material that generalizes the school math topics as well a great number of practical exercises with detailed solutions. There are lists of exercises at the end of each section to practice more yourself. We hope the presented educational material will be useful to a certain audience. In particular, the book could be interesting for those who are preparing for graduated school exams. We wish good luck to everyone in the studying of mathematics!
Yu. O. Koroleva, A.V. Korolev The work was partially supported by the grant of President of Russian Federation supporting young Russian scientists (project MK-4615.2015.1).
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Content I. NUMERICAL EXPRESSIONS……………………………………………………....5 How to compare numbers ............................................................................................ 8 Properties of integer powers ........................................................................................ 9 Short multiplication formulas ..................................................................................... 10 Exercises ........................................................................................................................ 11 II. FUNCTIONS ................................................................................................................ 12 Domain and codomain ................................................................................................. 12 Some important functions ............................................................................................ 13 1. Linear function .................................................................................................. 14 2. Quadratic function ............................................................................................ 15 3. Inversely proportional function ....................................................................... 18 4. Exponential function ......................................................................................... 19 5. Logarithmic function......................................................................................... 20 6. The root function √𝒙𝒙 ......................................................................................... 21 7. The absolute value function |х| ........................................................................ 21 8. The piece-vice function...................................................................................... 22 Shifts and stretches of functions .................................................................................. 23 Drawing of graph for fractional- linear function ...................................................... 25 Composition of functions ............................................................................................. 27 Exercises ........................................................................................................................ 29 III. POLYNOMIALS ........................................................................................................ 30 How to divide polynomials ........................................................................................... 31 Factorization of polynomials ....................................................................................... 33 Equations ....................................................................................................................... 36 n = 2 or about the quadratic polynomial .................................................................... 37 Method of intervals....................................................................................................... 38 Exercises ........................................................................................................................ 41 IV. EXPONENTIAL AND LOGARITHMIC EXPRESSIONS ................................... 42 Exponential equations and inequalities…………………………………………….. 43 Logarithms .................................................................................................................... 44 The simplest logarithmic equations and inequalities ................................................ 45 Exercises ........................................................................................................................ 48 V. TRIGONOMETRY ...................................................................................................... 49 The main formulas of trigonometry............................................................................ 49 Trigonometric equations .............................................................................................. 54 Inverse trigonometric functions .................................................................................. 54 The simplest trigonometric equations ........................................................................ 55 Homogeneous trigonometric equations………………………………………….......57 Graphs for trigonometric functions ............................................................................ 58 1. y(𝒙𝒙)=sin 𝒙𝒙, 𝒚𝒚(𝒙𝒙)=cos 𝒙𝒙 ................................................................................... 59 2. y(𝒙𝒙)=𝒕𝒕𝒈𝒈 𝒙𝒙, 𝒚𝒚(𝒙𝒙)=𝒄𝒄𝒕𝒕𝒈𝒈 𝒙𝒙 ................................................................................... 60 Exercises ........................................................................................................................ 61 Bibliography
................................................................................................................ 63
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I. NUMERICAL EXPRESSIONS Numbers 1,2,3,..., are called natural. The set of all natural numbers is denoted by ℕ. The set {0, ±1, ±2, ±3, … . } is the set of all integer numbers. Obviously, natural numbers are subset of integer, i.e. any natural number is integer one as well. Rational numbers ℚ are numbers that can be expressed as irreducible frac𝑚𝑚
tion, , here m is an integer number, and n is natural. For example, 𝑛𝑛
2
5
,
3
11
1
,− . 2
By the way, integer numbers are rational as well, because any integer can be 𝑚𝑚 represented as , 𝑚𝑚 ∈ ℤ. The set of rational numbers ℚ is bigger than ℤ: there 1
exist many fractions which are not integers, for example:
21 5
3
, . 8
The fraction is just a symbolic notation of a number. The same number can be represented by different ways: by common fractions as well as decimals. The decimal is a positional form of the number which looks as follows: ±𝑎𝑎1 𝑎𝑎2 … 𝑎𝑎𝑛𝑛 , 𝑏𝑏1 𝑏𝑏2 …, for example, 1,263. A part that comes before the comma is the integer part of the number, the part coming after comma is the fractional one. In order to convert the rational fraction to the decimal one needs to divide 1
the nominator by the denominator, for example, =
5
2
10
= 0,5 𝑜𝑜𝑜𝑜
= 23
145
.
1 3
= 0,3333. ..
In particular, the obtained decimal can have infinite number of decimals after the comma, then it is called the infinite decimal. To convert the decimal into the rational fraction, one needs to rewrite its fractional part as natural number divided by the corresponding power of 10. Then one shall write the integer part in front of the result. For example, 23, 145 = 23 +
145
1000
1000
One shall to learn how to convert the decimal numbers to the rational fractions since it is more convenient to make operations with the last ones. Indeed, it is not clear at all how to add two decimals with infinite number of digits after the comma. With such approach one could to add the averages of the original decimals, however, this leads to not precise answer. The main property of the fraction is as follows. If one multiply the numerator and denominator of the fraction by the same number, then the fraction keeps the same value, while the fractions can be different, for example: 5
3
=
4
9
=
12
12 16
.
And vice verse, if both the nominator and denominator of the fraction have the common factor, then both parts can be divided by that. This operation is called reducing by a factor. 12 12:4 3 For example: 16 = 16:4 = 4 – here both nominator and denominator is re-
duced by common factor 4. The sum, multiplication and division of fractions are hold by the rules: 𝑚𝑚 𝑛𝑛
𝑝𝑝
± = 𝑞𝑞
𝑚𝑚𝑚𝑚 ±𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛
𝑚𝑚
,
𝑛𝑛
∙
𝑝𝑝 𝑞𝑞
=
𝑚𝑚𝑚𝑚 𝑛𝑛𝑛𝑛
,
𝑚𝑚 𝑛𝑛
:
𝑝𝑝 𝑞𝑞
=
𝑚𝑚 𝑞𝑞 𝑛𝑛𝑛𝑛
.
Fractions with identical denominators are very simple to add, it is enough to add nominators, while denominator keeps the same. For example, it is evident that 1 8
5
6
3
+ = = [𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑏𝑏𝑏𝑏 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑓𝑓𝑓𝑓𝑓𝑓𝑡𝑡𝑡𝑡𝑡𝑡 𝑓𝑓𝑓𝑓𝑓𝑓 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 6 𝑎𝑎𝑎𝑎𝑎𝑎 8] = . 8
8
It is much more difficult to add, for example,
1 8
4
1
+ , since denominators are dif4
ferent. Adding fractions, it is extremely important to be able to find the common denominator that can be evaluated as LCM (the least common multiple) of fractions. LCM of numbers a and b (denoted in the sequel by LCM(a ,b)) is the lowest number that can be divided by both a and b. In our example НОК(8 ,4)=8. Thus, one should represent the factor
1 4
as
1 4
2
= , to have the 8
common denominator. After that it remains only to add the nominators: 1 8
1
1
2
3
+ = + = . 4
8
8
8
Example. Evaluate: 7 1 + , 8 2
Solution. 7
1. tio
11 8
8
1
7 1 ∙ , 8 2
7 1 : . 8 2
+ = ( 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 8 ) = 2
7+1∙4 8
=
11 8
. The ra-
is the sum of fractions. Besides this, the answer can be written in the form
“integer part plus the remain”: 7
1
2. ∙ = 8 2 7 1
3. : = 8 2
7
.
16 7∙2 8∙1
7 8
1
+ = 2
11 8
3
=1 . 8
7
= [𝑐𝑐𝑐𝑐𝑐𝑐 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 2𝑎𝑎𝑎𝑎𝑎𝑎 8 𝑏𝑏𝑏𝑏 𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 2] = .
Example. To evaluate
2 3
4
+ 0,01.
Solution. One can see that the task is to add fractions of different types: the rational fraction and decimal. The first step one needs to do is convert the terms to the unique type. It is always more convenient to work with rational 6
fractions, because this way allows to find the common denominators more easy. Thus, represent the number 0,01 𝑎𝑎𝑎𝑎 1
1 1 2 −� − � 2 4 3
. Hence,
100 2∙100+3
|𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡𝑡𝑡 𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑| = Example. Evaluate the value
1
.
3∙100
=
203 300
.
2 3
2
+ 0,01 = + 3
1
100
=
Solution. Let us calculate it by steps. Firstly, one needs to compute the value in parenthesis:
Then
1 2 1 ∙ 3 − 2 ∙ 4 −5 − = = . 4 3 12 12
1 1 1 12 1 = = [𝐿𝐿𝐿𝐿𝐿𝐿 (12 ,2 ) = 12] = = =1 . 1 1 2 1 −5 6 + 5 11 11 2 − �4 − 3� 2 − 12 12
Example. Simplify the expression
1 1 𝑎𝑎 𝑏𝑏 1 1 �𝑎𝑎 −𝑏𝑏 �(𝑎𝑎+𝑏𝑏)
� + �(𝑎𝑎−𝑏𝑏)
.
Solution. Let us simplify the expression by using the fraction properties.
𝑚𝑚 𝑛𝑛
1 1 𝑎𝑎 + 𝑏𝑏 �𝑎𝑎 + � (𝑎𝑎 − 𝑏𝑏) (𝑎𝑎 − 𝑏𝑏) (𝑎𝑎 + 𝑏𝑏)(𝑎𝑎 − 𝑏𝑏)𝑎𝑎𝑎𝑎 𝑎𝑎 − 𝑏𝑏 𝑏𝑏 = 𝑎𝑎𝑎𝑎 = = = −1. 1 1 𝑏𝑏 − 𝑎𝑎 𝑎𝑎𝑎𝑎(𝑏𝑏 − 𝑎𝑎)(𝑎𝑎 + 𝑏𝑏) 𝑏𝑏 − 𝑎𝑎 �𝑎𝑎 − � (𝑎𝑎 + 𝑏𝑏) (𝑎𝑎 + 𝑏𝑏) 𝑏𝑏 𝑎𝑎𝑎𝑎
Irrational numbers are real number that can not be represented as fraction , (where m∈ ℤ, 𝑛𝑛 ∈ ℕ), i.e. any irrational number is not the rational ones. It
can be written as infinity nonperiodic decimal fraction. For example, the numbers √2, 𝑒𝑒 and 𝜋𝜋 are irrational ones. The union of all rational numbers together with irrational ones is the set of all real numbers which is denoted by ℝ. The real line ℝ is illustrated on the figure. It includes both natural, integer, rational and irrational numbers.
The distance between the number 𝑎𝑎 and origin is called the absolute value of the number 𝒂𝒂 and is denoted by |𝑎𝑎|. According to such definition, −𝑎𝑎, 𝑖𝑖𝑖𝑖 𝑎𝑎 < 0 |𝑎𝑎| = � . 𝑎𝑎, 𝑖𝑖𝑖𝑖 𝑎𝑎 ≥ 0 7
For example, |𝜋𝜋 − 3| = 𝜋𝜋 − 3, since 𝜋𝜋 − 3 > 0. Example. Find a number 𝑎𝑎, such that |𝑎𝑎 − 1| = 2. Solution. One needs to find such point 𝑎𝑎 on the real line that the distance from it to 1 (the point a=1) equals 2. There are two points satisfying such properties: 𝑎𝑎 = −1 or 𝑎𝑎 = 3.
How to compare numbers Sometimes it is quite important to point the numbers on the real line in the correct order with respect to each other. This is a simple task if the numbers are integer or rational. In the case the numbers are irrational, this question is not trivial: one needs to know how to compare numbers with each other without a calculator! Example. To compare numbers 1 + 2√2 and √14. Solution. It is clear that «approximately» both numbers are greater than 3, but less than 4. We need to prove that one number is greater than the other one. During the compare procedure we will write the symbol «v» between the numbers that substitute the unknown sign “>”, “» since 32 > 25. Therefore, 1 + 2√2 > √14.
1 + 2√2 v √14
Example. Compare numbers
−1−√5 2
and −√2.
Solution. Denote by «^» the symbol opposite to «v». Then −1−√5 2
v −√2 ⇔ −1 − √5 v
−2√2
⇔ 1 + √5 ^
2√2 ⇔
⇔ 2√5 ^ 2 ⇔ 20 ^ 4. 1 + 2√5 + 5 ^ 8 The symbol «^» is «>» since 20 > 4. Hence, the opposite unknown symbol
«v» is « 0 lowing system of conditions to determine the domain: � ⇔ cos 2𝑥𝑥 ≠ 0 𝑥𝑥 > 0 𝜋𝜋 𝜋𝜋 𝑛𝑛 �𝑥𝑥 ≠ 𝜋𝜋 + 𝜋𝜋 𝑛𝑛 , 𝑛𝑛 ∈ ℤ. Thus, 𝐷𝐷 = ℝ+ ⧵ � + � , 𝑛𝑛 ∈ ℤ. 4 2 4
2
𝑐𝑐) The function arcsin is given in the formula, therefore one get the restriction −1 ≤ 𝑥𝑥 2 − 1 ≤ 1. Observe that even the formula involves the even order root. One do not get any extra conditions because of that. Indeed, the positive constant value 2 is the argument of the square root, not the function of x. There-
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fore, the domain can be given by the only restriction −1 ≤ 𝑥𝑥 2 − 1 ≤ 1 ⇔ 0 ≤ 𝑥𝑥 2 ≤ 2 ⇔ |𝑥𝑥| ≤ √2. Hence, 𝐷𝐷 = �−√2, √2�. 𝑑𝑑) The given function has no any restrictions to domain, because log 3 5 is
the constant value, not a function, and the denominator of the fraction always differ from zero (it does not depend on x as well). Thus, 𝐷𝐷 = ℝ.
2𝑥𝑥+5 3
is
Even and odd functions
One says that the function 𝑓𝑓(𝑥𝑥) is even if 𝑓𝑓(−𝑥𝑥) = 𝑓𝑓(𝑥𝑥). If 𝑓𝑓(−𝑥𝑥) = −𝑓𝑓(𝑥𝑥), then the function is odd. The graph of even function is symmetric with respect to OY, while the odd function is symmetric with respect to origin. Neither an even nor odd function is called the general type function.
Some important functions Let us consider main types of functions which are studying in the school math course. 1. Linear function Let the function 𝑦𝑦(𝑥𝑥) is given by formula 𝑦𝑦 = 𝑘𝑘𝑘𝑘 + 𝑏𝑏. The domain and codomain are the whole real line: 𝐷𝐷 = 𝐸𝐸 = (−∞, +∞). The graph is straight. Here 𝑘𝑘 = tg 𝛼𝛼 is a tangent of the angle between the straight and positive direc𝜋𝜋 tion of OX. If 𝑘𝑘 > 0, then 0 < 𝛼𝛼 < , i.e. the angle is acute, if 𝑘𝑘 < 0, then vice 2
versa, the angle is obtuse. The value 𝑏𝑏 shows the shift up (if 𝑏𝑏 > 0) or down (if 𝑏𝑏 < 0) along OY – axis.
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There are two degenerate cases: • 𝑦𝑦 = 𝑏𝑏 - is a straight parallel to OX. In this case the argument 𝑥𝑥 varies on the whole real line (−∞, +∞), while 𝑦𝑦 is always equal 𝑏𝑏, therefore 𝐸𝐸 = {𝑏𝑏}. • 𝑥𝑥 = 𝑎𝑎 – is a straight parallel to OY. In this case the variable 𝑥𝑥 keeps the same value 𝑎𝑎, therefore 𝐷𝐷 = {𝑎𝑎}, while 𝑦𝑦 may varies on (−∞, +∞).
One can easily see that two given straights 𝑦𝑦 = 𝑘𝑘𝑥𝑥 + 𝑏𝑏1 and 𝑦𝑦 = 𝑘𝑘𝑘𝑘 + 𝑏𝑏2 are parallel to each other since their tangential coefficients are identical. It is known also that two points are sufficient to draw the straight, i.e. two different points corresponds the unique straight passing through them. 2. Quadratic function The function 𝑦𝑦 = 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐, where 𝑎𝑎 ≠ 0 is called the quadratic one whose graph is quadratic parabola. The branches of the parabola have updirections for 𝑎𝑎 > 0 and down- directions for 𝑎𝑎 < 0 (see the figure). The domain of the function is the whole real line. The vertex is the point where the graph has its extremum (maximum or minimum). Thus, the vertex is the point with coordinates �−
𝑏𝑏
2𝑎𝑎
, 𝑦𝑦 �−
𝑏𝑏
2𝑎𝑎
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��. The important value is the number
𝐷𝐷 = 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 called discriminant of the quadratic function 𝑦𝑦 = 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐. The sign of D let to understand how the parabola is located with respect to OX-axis.
Let 𝑎𝑎 > 0, i.e. the branches have up-directions. If 𝐷𝐷 < 0, then the equation has no real solutions 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0 what means the graph does not cross OX. Since 𝑎𝑎 > 0 this implies the graph is fully located above OX. If 𝐷𝐷 = 0, then the graph intersect the OX-axis exactly in one point which are vertex of the parabola. If 𝐷𝐷 > 0, then there exist two intersections with OX. The abscises of these points can be find as solutions of the equation 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0, i.e. can be calculated by formula 𝑥𝑥1,2 =
−𝑏𝑏±√𝐷𝐷 2𝑎𝑎
.
Assume now 𝑎𝑎 < 0, then parabola’s branches are directed down. If 𝐷𝐷 < 0, then equation 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0 has no solutions what implies that the graph does not intersect OX-axis, i.e. graph is located below OX. Analogously to the case 𝑎𝑎 > 0, if 𝐷𝐷 = 0, then the graph has the only intersection with OX at vertex. 16
If 𝐷𝐷 > 0, then we have two intersections with OX, those can be evaluated by 𝑥𝑥1,2 =
−𝑏𝑏±√𝐷𝐷 2𝑎𝑎
.
The useful approach is to complete the square in the quadratic polynomials 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐. One shall do it as follows: 2
2
𝑏𝑏 𝑐𝑐 −𝑏𝑏 −𝑏𝑏 2 𝑐𝑐 𝑎𝑎𝑥𝑥 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 𝑎𝑎 �𝑥𝑥 + 2 𝑥𝑥 + � = 𝑎𝑎 �𝑥𝑥 − � �� − 𝑎𝑎 � � + 𝑎𝑎 = 2𝑎𝑎 𝑎𝑎 2𝑎𝑎 2𝑎𝑎 𝑎𝑎 2
2
2
−𝑏𝑏 𝑏𝑏 2 − 4𝑎𝑎𝑎𝑎 = 𝑎𝑎 �𝑥𝑥 − � �� − � �. 2𝑎𝑎 4𝑎𝑎
This implies directly that the value of the function in the vertex point 𝑏𝑏 2 −4𝑎𝑎𝑎𝑎
−𝑏𝑏
𝑥𝑥 = � � equals − � 2𝑎𝑎 function is
𝑏𝑏 2 −4𝑎𝑎𝑎𝑎
𝐸𝐸 = �− �
4𝑎𝑎
4𝑎𝑎
�=
4𝑎𝑎𝑎𝑎 −𝑏𝑏 2 4𝑎𝑎
. Then the co-domain of the quadratic 𝑏𝑏 2 −4𝑎𝑎𝑎𝑎
� , +∞) as 𝑎𝑎 > 0 and 𝐸𝐸 = (−∞, − �
4𝑎𝑎
�� as 𝑎𝑎 < 0.
To drawing the graph for the parabola it is sufficient to take a few symmetric abscises, since the quadratic function is symmetric with respect to the line 𝑥𝑥 = −
𝑏𝑏
2𝑎𝑎
. The detailed examples of drawing are presented below.
Remark. The quadratic function is a partial case of the polynomial function 𝑦𝑦 = 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑛𝑛−1 + ⋯ + 𝑎𝑎0 . Namely, this is the second order polynomial.
The graph of the quadratic function can be drawn by using the table with some “easy” values of x and corresponding values of y. Example. Draw the function 𝑦𝑦 = 2𝑥𝑥 2 . 17
Solution. The given quadratic function, whose graph is parabola with updirected branches, since 𝑎𝑎 = 2 > 0. The parabola vertex has coordinates 𝑏𝑏
�− 2𝑎𝑎 ,
4𝑎𝑎𝑎𝑎 −𝑏𝑏 2 4𝑎𝑎
� = (0,0) . Fill in the tables of auxiliary x- values. Let us chose
the “easy” values of 𝑥𝑥: 0,1,2,3 and also symmetric ones: -1,-2,-3. Substituting them into the formula one gets the corresponding values of 𝑦𝑦. Fill in the y-row into the table. x y
0 0
±1 2
±2 8
±3 18
3. Inversely proportional function Consider the inversely proportional function that is given by equation 𝑘𝑘
𝑦𝑦 = , where 𝑘𝑘 is some number belonging to ℝ. One can see that the formula 𝑥𝑥
involves the denominator which can equals to zero as 𝑥𝑥 = 0. Therefore the domain D of the function is ℝ ⧵ {0}. The graph is hyperbola which consists of two branches (see the figures for positive and negative values of 𝑘𝑘). If 𝑘𝑘 > 0, then the function is decreasing on the domain D, i.e. the bigger the value of 𝑥𝑥, the less the value of 𝑦𝑦. The branches of hyperbola are located in coordinate domains 𝐼𝐼 = {𝑥𝑥 > 0, 𝑦𝑦 > 0} and 𝐼𝐼𝐼𝐼𝐼𝐼 = {𝑥𝑥 < 0, 𝑦𝑦 < 0}. The case 𝑘𝑘 < 0, is the opposite one: the bigger the 𝑥𝑥 - value, the bigger is 𝑦𝑦 – value what means the function is increasing. In this case the hyperbola is located in coordinate domains 𝐼𝐼𝐼𝐼 = {𝑥𝑥 < 0, 𝑦𝑦 > 0} and 𝐼𝐼𝐼𝐼 = {𝑥𝑥 < 0, 𝑦𝑦 < 0}. The straights 𝑦𝑦 = 0 (axis OX) and 𝑥𝑥 = 0 (axis OY) are asymptotes, i.e. the lines such that the distance between the line and the graph tends to zero for large (infinite) values of |𝑥𝑥| and |𝑦𝑦|. However the asymptote never crosses the graph even if they are very close to each other!
18
Remark. Both linear function and inversely proportional functions can be generalized to the fractional liner function of the type 𝑦𝑦(𝑥𝑥) =
𝑎𝑎𝑎𝑎 +𝑏𝑏 𝑐𝑐𝑐𝑐 +𝑑𝑑
.
Indeed, if 𝑐𝑐 = 0, then 𝑦𝑦(𝑥𝑥) becomes linear function given by formula 𝑎𝑎
𝑏𝑏
𝑎𝑎
𝑦𝑦 = 𝑥𝑥 + , here the tangential coefficient equals . 𝑑𝑑
𝑑𝑑
If 𝑎𝑎 = 0, then 𝑦𝑦(𝑥𝑥) takes the form 𝑦𝑦(𝑥𝑥) = 𝑏𝑏
𝑏𝑏
𝑐𝑐𝑐𝑐 +𝑑𝑑
𝑑𝑑
=
𝑏𝑏
𝑑𝑑 𝑐𝑐
𝑐𝑐�𝑥𝑥+ �
, that is the in-
versely proportional function where 𝑘𝑘 = , while the argument а is shifted 𝑐𝑐
units left.
𝑑𝑑 𝑐𝑐
4. Exponential function Given the number 𝑎𝑎 > 0, the function 𝑦𝑦(𝑥𝑥) = 𝑎𝑎 𝑥𝑥 is called exponential with the base 𝑎𝑎. One should assume 𝑎𝑎 ≠ 1, otherwise the function is generated to linear 𝑦𝑦(𝑥𝑥) = 1. Due to such definition there is no any restrictions for the domain, what means 𝐷𝐷 = ℝ. Moreover, 𝑎𝑎 𝑥𝑥 > 0, therefore 𝐸𝐸 = (0, +∞). Depending on value of 𝑎𝑎 the function can be increasing (as 𝑎𝑎 > 1) or decreasing (as 0 1, it increases to the whole real line ℝ. 5. Logarithmic function Let 𝑎𝑎 > 0, 𝑎𝑎 ≠ 1. For any value of 𝑥𝑥 > 0 we define the logarithmic function: 𝑦𝑦(𝑥𝑥) = log 𝑎𝑎 𝑥𝑥. The condition 𝑎𝑎 ≠ 1 is stated to exclude the trivial case: log1 𝑥𝑥 =1 for any 𝑥𝑥. The domain 𝐷𝐷 = ℝ+ is the set of all positive numbers. It is clear that in case 𝑎𝑎 > 1 the bigger the argument of logarithm the bigger the value of the function itself, i.e. the function is increasing. When 0 < 𝑎𝑎 < 1 the situation is the opposite one. The graphs are illustrated below.
20
Since always log a 1 = 0, the graph passes through the point (1,0). One can define more complicated logarithmic function whose base a can also varies depending on x: 𝑦𝑦 = log a(x) 𝑓𝑓(𝑥𝑥). The domain of such a domain must satisfy several conditions: both functions 𝑎𝑎(𝑥𝑥) and 𝑓𝑓(𝑥𝑥) must be strictly positive, moreover, 𝑎𝑎(𝑥𝑥) ≠ 1. Mathematically this can be written as: 𝑎𝑎(𝑥𝑥) > 0, 𝑎𝑎(𝑥𝑥) ≠ 1 𝐷𝐷 = �𝑥𝑥 ∈ ℝ � � �. 𝑓𝑓(𝑥𝑥) > 0 6. The root function √𝒙𝒙
The function 𝑓𝑓(𝑥𝑥) = √𝑥𝑥 is defined for nonnegative 𝑥𝑥, therefore the domain 𝐷𝐷 = [0, +∞). The square root (and in general the even power root) is always nonnegative and increasing when 𝑥𝑥 increases, thus the co-domain is 𝐸𝐸 = [0, +∞). The graph is illustrated below.
7. The absolute value function |𝒙𝒙|
By the definition, the absolute value is the function 𝑥𝑥, 𝑖𝑖𝑖𝑖 𝑥𝑥 ≥ 0 𝑓𝑓(𝑥𝑥) = |𝑥𝑥| = � . The given definition implies that the graph of −𝑥𝑥, 𝑖𝑖𝑖𝑖 𝑥𝑥 < 0 |𝑥𝑥| consists of two parts: the lines 𝑦𝑦 = ±𝑥𝑥, that coincides at point 𝑥𝑥 = 0. For the figure see below.
21
The function is even: |−𝑥𝑥| = |𝑥𝑥|, therefore the graph is symmetric with respect to OY-axis. 8. The piece-vice function Let the domain for 𝑦𝑦(𝑥𝑥) is the union as follows: 𝑓𝑓(𝑥𝑥) 𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐷𝐷1 𝐷𝐷 = 𝐷𝐷1 ∪ 𝐷𝐷2 . Define 𝑦𝑦(𝑥𝑥)=� . Then the graph of the 𝑔𝑔(𝑥𝑥) 𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐷𝐷2 function consists of the graph for 𝑓𝑓(𝑥𝑥), restricted to the set 𝐷𝐷1 and the graph for 𝑔𝑔(𝑥𝑥) on the set 𝐷𝐷2 .
1 𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ (−∞, −1) Example. Draw the graph for 𝑦𝑦(𝑥𝑥) = � 2 . 2𝑥𝑥 𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ [−1, +∞) Solution. We shall draw the function 𝑦𝑦 = 1 in coordinates XY, then one shall remove (erase) the part of graph which comes out of the interval (−∞, −1). Thus, the remain part is the function 𝑦𝑦(𝑥𝑥) = 1 𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ (−∞, −1). Analogously, one draw the parabola 𝑦𝑦(𝑥𝑥) = 2𝑥𝑥 2 and erase the part outside the set [−1, +∞). The remain part of graph is the function 𝑦𝑦(𝑥𝑥) = 2𝑥𝑥 2 as 𝑥𝑥 ∈ [−1, +∞). Summing, the graph consists of two parts, see the figure below. One can observe that the values of function to the left of point 𝑥𝑥 = −1 and to the right do not coincide. This is an example of discontinuous function. 22
Shift and stretch of functions Here we summarize the operations on graphs. Translations = y f (x + k)
y = f (x) - k y = f (x + h) y = f (x - h)
up k units down k units left h units right h units
Stretches/Shrinks y = m·f (x) stretch vertically by a factor of m y= y=f(
·f (x) x)
shrink vertically by a factor of m (stretch by
)
stretch horizontally by a factor of n
y = f (nx)
shrink horizontally by a factor of n (stretch by
Reflections y = - f (x) y = f (- x) x = f (y)
reflect over x -axis (over line y = 0 ) reflect over y -axis (over line x = 0 ) reflect over line y = x
)
We can combine operations, as long as we pay attention to the order in which we alter inputs and outputs. Operations on outputs follow the order of operations, and operations on inputs follow the reverse order of operations (since we have to "undo" them). For a given function 𝑦𝑦 = 𝑓𝑓(𝑥𝑥), the equation 𝑦𝑦 = 𝑚𝑚𝑚𝑚�𝑘𝑘(𝑥𝑥 − 𝑎𝑎)� + 𝑏𝑏 define the new function with the graph that can be obtained from graph 𝑦𝑦 = 𝑓𝑓(𝑥𝑥) by its translation a units right along OX, then shrinking it horizontally by a factor k, stretching vertically by a factor of m and finally by translation b units up along OY. For example, the equation of a function stretched vertically by a factor of 2 and then shifted 3 units up is y = 2f (x) + 3, and the equation of a function stretched horizontally by a factor of 2 and then shifted 3 units right is 1
1
3
𝑦𝑦 = 𝑓𝑓 � (𝑥𝑥 − 3)� = 𝑓𝑓 � 𝑥𝑥 − �. 2 2 2
23
Example: Make composition of transformations for the function f (x) = 2x 2 as follows: •
Stretch f vertically by a factor of 2, and then shift f up 3 units: 2f (x) + 3 = 2(2x 2) + 3 = 4x 2 + 3. • Shrink f horizontally by a factor of 5, and then shift f right 2 units: f (5(x - 2)) = 2(5(x - 2))2 = 2(25)(x - 2)2 = 50(x - 2)2 . • Stretch f vertically by a factor of 3, stretch f horizontally by a factor of 6, and shift f down 2 units: 1
1
1
1
3f � 𝑥𝑥� - 2 = 3(2( 𝑥𝑥)2) = 6 x 2 = x 2 . •
6
6
36
6
Shrink f vertically by a factor of 4, shrink f horizontally by a factor of 2, and shift f left 6 units: 1 4
f (2(x + 6)) =
1 4
∙ (2(x + 6))2 =
1 4
∙4(x + 6)2 = (x + 6)2 .
Example. To draw the graph of function 𝑦𝑦 = 3(𝑥𝑥 + 2)2 + 1.
Solution. Consider the function 𝑦𝑦 = 𝑥𝑥 2 which is quadratic, the graph is parabola whose branches are positively oriented, the vertex is the point (0,0). Then the graph of function 𝑦𝑦 = 3(𝑥𝑥 + 2)2 + 1 = 3(𝑥𝑥 − (−2))2 + 1 can be obtained from by the following transformations 𝑦𝑦 = 𝑥𝑥 2 : • translation 2 units left along OX • the graph stretches along OY by a factor of 3 • translation 1 unit up along OY. Let us consider the figure to see the steps of drawing.
24
Example. To draw the graph of function 𝑦𝑦 = Solution. Draw the graph of 𝑦𝑦 =
2
2
𝑥𝑥−1
− 3.
and make the following transformations:
𝑥𝑥
• shift 1 unit right along OX • translate 3 units down along OY.
The result is presented on the figure.
Drawing of graph for fractional- linear function. Let us consider now how to draw the graph of fractional-linear function 𝑦𝑦(𝑥𝑥) =
𝑎𝑎𝑎𝑎 +𝑏𝑏 𝑐𝑐𝑐𝑐 +𝑑𝑑
by using such operations as translations and stretches. We shall
assume that 𝑐𝑐 ≠ 0, otherwise the function becomes linear, and its graph we know how to draw. In general case one needs to do the following steps: • Simplify the expression as follows: 𝑎𝑎𝑎𝑎 +𝑏𝑏 𝑐𝑐𝑐𝑐 +𝑑𝑑
=
𝑎𝑎 𝑎𝑎𝑎𝑎 (𝑐𝑐𝑐𝑐 +𝑑𝑑)+𝑏𝑏− 𝑐𝑐 𝑐𝑐
𝑐𝑐𝑐𝑐 +𝑑𝑑
𝑎𝑎
= + 𝑐𝑐
𝑏𝑏𝑏𝑏 −𝑎𝑎𝑎𝑎
𝑐𝑐(𝑐𝑐𝑐𝑐 +𝑑𝑑)
𝑎𝑎
= + 𝑐𝑐
𝑏𝑏𝑏𝑏 −𝑎𝑎𝑎𝑎 𝑐𝑐 2
1
𝑑𝑑 𝑐𝑐
𝑥𝑥−�− �
.
Now one can observe that if 𝑏𝑏𝑏𝑏 − 𝑎𝑎𝑎𝑎 = 0, then both nominator and denomi𝑎𝑎 nator are proportional, and the function simply equals to (naturally we assume 𝑐𝑐
𝑑𝑑
that 𝑥𝑥 ≠ − , otherwise this point does not belong to the domain of 𝑦𝑦(𝑥𝑥)). 𝑐𝑐
𝑑𝑑
• Let us draw the vertical line 𝑥𝑥 = − which is asymptote for the graph. 𝑐𝑐
𝑎𝑎
• Next we draw the horizontal line 𝑦𝑦 = , which is horizontal asymptote, 𝑐𝑐
since the graph approaches to that line when 𝑥𝑥 and − 𝑥𝑥 tend to infinity. • Taking these asymptotes as new coordinate axis one shall draw the graph
𝑘𝑘 𝑥𝑥
, where 𝑘𝑘 =
𝑏𝑏𝑏𝑏 −𝑎𝑎𝑎𝑎 𝑐𝑐 2
.
25
The function is decreasing if 𝑏𝑏𝑏𝑏 − 𝑎𝑎𝑎𝑎 > 0 and increasing in the opposite case (when 𝑏𝑏𝑏𝑏 − 𝑎𝑎𝑎𝑎 < 0).
2𝑥𝑥+1
Example. Draw the graph of function y=
5−𝑥𝑥
.
Solution. Following our instruction, we shall begin with simplifying of the given algebraic expression. Let us separate the integer and fractional part: y=
2𝑥𝑥+1 5−𝑥𝑥
=−
2𝑥𝑥+1 𝑥𝑥−5
=−
2(𝑥𝑥−5)+11 𝑥𝑥−5
= −2 −
11
𝑥𝑥−5
. This implies that the vertical
asymptote is the straight 𝑥𝑥 = 5, while the horizontal line is the line 𝑦𝑦 = −2. Now, taking the asymptotes as new coordinate lines, one needs to draw the 11
graph 𝑦𝑦 = − . The function is increasing since 𝑘𝑘 = −11 < 0 (see the figure). 𝑥𝑥
26
Composition of functions. Consider the functions 𝑓𝑓(𝑥𝑥) and 𝑔𝑔(𝑥𝑥), while the domain of 𝑔𝑔(𝑥𝑥) coincides with the co-domain of 𝑓𝑓(𝑥𝑥). Then if one take 𝑓𝑓(𝑥𝑥) as the argument of 𝑔𝑔, one gets a new function ℎ1 (𝑥𝑥) = 𝑔𝑔�𝑓𝑓(𝑥𝑥)� called composition of functions. Analogously, one can consider the other composition, defined as ℎ2 (𝑥𝑥) = 𝑓𝑓�𝑔𝑔(𝑥𝑥)�. Making the compositions, one needs to control that the domain of 𝑓𝑓 be subset of the co-domain of 𝑔𝑔. Let us consider the examples. Examples. What are the compositions 𝑓𝑓(𝑔𝑔(𝑥𝑥)) and 𝑔𝑔�𝑓𝑓(𝑥𝑥)�, where 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 2 , and 𝑔𝑔(𝑥𝑥) = ln(1 + 𝑥𝑥)?
Solution. To write down the first composition, one needs to substitute the argument x of 𝑓𝑓 with formula for 𝑔𝑔 .As the result one gets 𝑓𝑓�𝑔𝑔(𝑥𝑥)� = 2(ln(1 + 𝑥𝑥))2 . Analogously, in the second case we shall write the formula for 𝑓𝑓(𝑥𝑥) instead of argument in 𝑔𝑔: 𝑔𝑔�𝑓𝑓(𝑥𝑥)� = ln(1 + 2𝑥𝑥 2 ). Making such formal substitutions, one needs always to control that the domain for the outer function coincide with the co-domain of the inner function. In the considered case everything works well. Example. What are the compositions 𝑓𝑓(𝑔𝑔(𝑥𝑥)) and 𝑔𝑔�𝑓𝑓(𝑥𝑥)�, if 𝑓𝑓(𝑥𝑥) = −2𝑥𝑥 2 , and 𝑔𝑔(𝑥𝑥) = ln 𝑥𝑥 ? Solution. Following the formal procedure (as in the last example), we get that 𝑓𝑓�𝑔𝑔(𝑥𝑥)� = −2(ln 𝑥𝑥)2 , while the second composition is not trivial at all. The formal substitution gives the formula 𝑔𝑔�𝑓𝑓(𝑥𝑥)� = ln(−2𝑥𝑥 2 ), which has no any sense: the logarithmic function is defined only for positive arguments while the value of −2𝑥𝑥 2 is never positive. Such “troubles” appears because the domain for 𝑔𝑔 and co-domain of 𝑓𝑓 has no any intersections. The conclusion is that the function 𝑔𝑔�𝑓𝑓(𝑥𝑥)� is not defined. It is more clear now that in general 𝑓𝑓�𝑔𝑔(𝑥𝑥)� and 𝑔𝑔�𝑓𝑓(𝑥𝑥)� is not the same.
However, there exist some cases when 𝑓𝑓�𝑔𝑔(𝑥𝑥)� = 𝑔𝑔�𝑓𝑓(𝑥𝑥)�. Such situation is presented in the next example. 27
Example. Write down the composition 𝑓𝑓(𝑔𝑔(𝑥𝑥)) and 𝑔𝑔�𝑓𝑓(𝑥𝑥)�, if 𝑓𝑓(𝑥𝑥) = log a 𝑥𝑥, and 𝑔𝑔(𝑥𝑥) = 𝑎𝑎 𝑥𝑥 .
Solution. Taking into account the properties of the logarithmic function, it is not difficult to see that 𝑓𝑓�𝑔𝑔(𝑥𝑥)� = log a 𝑎𝑎 𝑥𝑥 = 𝑥𝑥 and 𝑔𝑔�𝑓𝑓(𝑥𝑥)� = 𝑎𝑎log a 𝑥𝑥 = 𝑥𝑥. 𝑡𝑡
Example. Write the formula for 𝐶𝐶 if 𝑈𝑈 = 𝐸𝐸0 �1 − 𝑒𝑒 −𝑅𝑅𝑅𝑅 �.
Solution. Here the constant 𝐶𝐶 is “hidden” into a few compositions. In order 𝑡𝑡
to express 𝐶𝐶, one needs first to get the formula for 𝑒𝑒 −𝑅𝑅𝑅𝑅 , then applying the loga-
rithmic function to exponent, we can obtain the formula for power −
𝑡𝑡
𝑅𝑅𝑅𝑅
. Having
that formula, one can easily express 𝐶𝐶. Thus, let us follow all these steps: 𝑡𝑡
𝑈𝑈 = 𝐸𝐸0 �1 − 𝑒𝑒 −𝑅𝑅𝑅𝑅 � ⇔ 𝑡𝑡
ln 𝑒𝑒 −𝑅𝑅𝑅𝑅 = ln �1 − 𝑅𝑅𝑅𝑅 = −
𝑡𝑡
𝑡𝑡 𝑈𝑈 = 1 − 𝑒𝑒 −𝑅𝑅𝑅𝑅 𝐸𝐸0
𝑡𝑡
⇔ 𝑒𝑒 −𝑅𝑅𝑅𝑅 = 1 −
𝑈𝑈 ⇔ 𝐸𝐸0
𝑡𝑡 𝑈𝑈 𝑈𝑈 = ln �1 − � ⇔ � ⇔ − 𝑅𝑅𝑅𝑅 𝐸𝐸0 𝐸𝐸0
𝑈𝑈 ln �1 − � 𝐸𝐸0
⇔ 𝐶𝐶 = −
28
𝑡𝑡
𝑈𝑈 R ln �1 − � 𝐸𝐸0
.
Exercises. 1.
Find
the
domains
for
functions:
𝑏𝑏) 𝑦𝑦 = 2𝑥𝑥 5 +cos 2𝑥𝑥 + ln(𝑥𝑥 − 2), 𝑐𝑐) 𝑦𝑦 =
√𝑥𝑥+2 . ln (3𝑥𝑥+1)
2. Draw the graphs of functions: 𝑎𝑎) 𝑦𝑦(𝑥𝑥) = 2𝑥𝑥 − 1, 𝑏𝑏) 𝑦𝑦(𝑥𝑥) = −2𝑥𝑥 2 + 1, 𝑑𝑑)𝑦𝑦(𝑥𝑥) = 𝑒𝑒 𝑥𝑥 ,
|𝑥𝑥|, 𝑖𝑖𝑖𝑖 𝑥𝑥 ≤ 1 𝑒𝑒) 𝑦𝑦(𝑥𝑥) = � , 1, 𝑖𝑖𝑖𝑖 𝑥𝑥 > 1.
𝑎𝑎)
1
𝑥𝑥
√𝑥𝑥 2 −9 𝑥𝑥+5
𝑐𝑐) 𝑦𝑦(𝑥𝑥) = log 1 𝑥𝑥, 𝑓𝑓) 𝑦𝑦(𝑥𝑥) =
3. Write down the compositions 𝑓𝑓(𝑔𝑔(𝑥𝑥)) and 𝑔𝑔�𝑓𝑓(𝑥𝑥)�, if 𝑎𝑎)𝑓𝑓(𝑥𝑥) = , 𝑔𝑔(𝑥𝑥) = ln|𝑥𝑥| ,
𝑦𝑦 =
3
2𝑥𝑥 . 𝑥𝑥 − 1
𝑏𝑏) 𝑓𝑓(𝑥𝑥) = 2𝑥𝑥 4 + 3𝑥𝑥, 𝑔𝑔(𝑥𝑥) = √𝑒𝑒 𝑥𝑥 + 1 ?
4. Write down the expressions for 𝑎𝑎) 𝐸𝐸0 , 𝑏𝑏) 𝑈𝑈, 𝑐𝑐) 𝑅𝑅 if 𝐶𝐶𝐶𝐶ln �1 −
𝑈𝑈 � + 𝑡𝑡 = 0. 𝐸𝐸0
29
+ 1,
III. POLYNOMIALS Consider a function given by 𝑃𝑃𝑛𝑛 (𝑥𝑥)= 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑛𝑛−1 + ⋯ + 𝑎𝑎0 , where 𝑥𝑥 ∈ ℝ is variable and coefficients 𝑎𝑎𝑖𝑖 are some fixed values. 𝑃𝑃𝑛𝑛 (𝑥𝑥) is called the polynomial of power 𝑛𝑛. For example, the expression 2𝑥𝑥 3 + 𝑥𝑥 2 + 1 is the polynomial of power 3, here 𝑎𝑎0 = 2, 𝑎𝑎1 = 1, 𝑎𝑎2 = 0 ( since the term with 𝑥𝑥 1 absent), 𝑎𝑎3 = 1. We remark that all real numbers are polynomials of power 0. Indeed, 𝑃𝑃0 (𝑥𝑥) = 𝑎𝑎0 is a umber belonging to ℝ independently on value of x. Thus we see that the notion of polynomials generalizes the notion of number. Analogously with operation on the real numbers, one can define operations on the polynomials: addition, subtraction, multiplication and division. In order to add two polynomials one need to sum the coefficients in front of the same powers of 𝑥𝑥. Example. To find the sum of polynomials 2𝑥𝑥 4 + 6𝑥𝑥 3 − 3𝑥𝑥 2 − 𝑥𝑥 + 1 and 𝑥𝑥 2 − 2𝑥𝑥 + 3. Solution. (2𝑥𝑥 4 + 6𝑥𝑥 3 − 3𝑥𝑥 2 − 𝑥𝑥 + 1) + (𝑥𝑥 2 − 2𝑥𝑥 + 3)= |sum the coefficients in front of the same power of 𝑥𝑥| = 2𝑥𝑥 4 + 6𝑥𝑥 3 + (−3 + 1)𝑥𝑥 2 + (−1 − 2)𝑥𝑥 + (1 + 3) = 2𝑥𝑥 4 + 6𝑥𝑥 3 − 2 𝑥𝑥 2 − 3𝑥𝑥 + 4. The obtained polynomial is the resulting sum. Let us discuss now how to get the product of polynomials. Example. To multiple the polynomials (𝑥𝑥 + 1) and (𝑥𝑥 2 − 2𝑥𝑥 + 3).
Solution. One needs to write it in the form (𝑥𝑥 + 1)(𝑥𝑥 2 − 2𝑥𝑥 + 3), then to simplify the expression as follows: multiply consistently each term of the first parenthesis with each term in the second one. We shall apply the rule: 𝑎𝑎𝑥𝑥 𝑚𝑚 ∙ 𝑏𝑏𝑥𝑥 𝑛𝑛 = 𝑎𝑎𝑎𝑎𝑥𝑥 𝑚𝑚 +𝑛𝑛 .
Thus, (𝑥𝑥 + 1)(𝑥𝑥 2 − 2𝑥𝑥 + 3) = 𝑥𝑥 ∙ 𝑥𝑥 2 + 𝑥𝑥 ∙ (−2𝑥𝑥) + 𝑥𝑥 ∙ 3 + 1 ∙ 𝑥𝑥 2 + 1 ∙ (−2𝑥𝑥) + +1 ∙ 3 = 𝑥𝑥 3 − 2𝑥𝑥 2 + 3𝑥𝑥 − 2𝑥𝑥 + 3 = 𝑥𝑥 3 − 2𝑥𝑥 2 + 𝑥𝑥 + 3. 30
The obtained polynomial is the product of two given. One can observe that the product has power 3. In general case, when one multiply the polynomials of power 𝑚𝑚 and 𝑛𝑛, the resulting polynomial has the power 𝑚𝑚 + 𝑛𝑛 (by the rule for multiplication of powers). How to divide polynomials.
Now we discuss the division of polynomials. It is known that the real numbers can be divided Остановимся поподробнее на операции деления. Как известно, целые числа можно делить друг на друга «уголком» с остатком. Let 𝑚𝑚, 𝑛𝑛 be integer or natural numbers. Then to divide 𝑚𝑚 by 𝑛𝑛 with remainder means to find such integer 𝑝𝑝 (the quotient) and q called reminder, satisfying 0 ≤ 𝑞𝑞 < 𝑛𝑛, such that 𝑚𝑚 = 𝑛𝑛𝑛𝑛 + 𝑞𝑞. For example, the result of division 16 by 3 with the reminder is as follows: 16 = 3 ∙ 5 + 1 what is equivalent to
16 3
1
=5+ . 3
Analogously the operation of division with remainder is defined for polynomials. To divide polynomial 𝑃𝑃(𝑥𝑥) by 𝑄𝑄(𝑥𝑥) means to find polynomials 𝑝𝑝(𝑥𝑥) (the quotient) and q(𝑥𝑥) (the remainder) such that 𝑃𝑃(𝑥𝑥) = 𝑄𝑄(𝑥𝑥)𝑝𝑝(𝑥𝑥) + 𝑞𝑞(𝑥𝑥), and the power of 𝑞𝑞(𝑥𝑥) is strictly less than power of 𝑄𝑄(𝑥𝑥). Let us show how it works. Example. Divide 2𝑥𝑥 4 + 6𝑥𝑥 3 − 3𝑥𝑥 2 − 𝑥𝑥 + 1 by 𝑥𝑥 2 − 2𝑥𝑥 + 3.
Solution. The polynomial 2𝑥𝑥 4 + 6𝑥𝑥 3 − 3𝑥𝑥 2 − 𝑥𝑥 + 1 is called dividend while 𝑥𝑥 2 − 2𝑥𝑥 + 3 is divider. One needs to divide the highest power term of the dividend by the highest power term of the divider: 2𝑥𝑥 4 divide by 𝑥𝑥 2 . The quotient is 2𝑥𝑥 2 . Write it down in the line for the result, multiply by divider and write the result under the dividend.
Analogously to the division of the natural numbers, one needs to put minus sign, draw the horizontal line and do the subtraction. Further one shall write down the next term of the dividend: (− 𝑥𝑥). We get the polynomial 31
10𝑥𝑥 3 − 9𝑥𝑥 2 − 𝑥𝑥, which is a new dividend. Make analogous operations with the new dividend, namely we divide the term of the highest order 10𝑥𝑥 3 by 𝑥𝑥 2 , the result is 10𝑥𝑥.
Thus, we obtain the next term in the quotient. We add this with sign «+» and multiply it by divider. In this way one get the polynomial 10𝑥𝑥 3 − 20𝑥𝑥 2 + 30𝑥𝑥, write it under the dividend 10𝑥𝑥 3 − 9𝑥𝑥 2 − 𝑥𝑥 and subtract. After that the polynomial 11𝑥𝑥 2 − 31𝑥𝑥 is obtained, now write down the last term of the dividend.
Now it remains to divide 11𝑥𝑥 2 − 31𝑥𝑥 + 1 by 𝑥𝑥 2 − 2𝑥𝑥 + 3. Dividing the terms of the highest order, we get 11. Multiply this number by divider and subtract. The polynomial of the first order −9𝑥𝑥 − 32 is obtained.
−9𝑥𝑥 − 32 is the reminder since the degree of this polynomial is strictly less than power of 𝑥𝑥 2 − 2𝑥𝑥 + 3. Hence, 2𝑥𝑥 4 + 6𝑥𝑥 3 − 3𝑥𝑥 2 − 𝑥𝑥 + 1 = (𝑥𝑥 2 − 2𝑥𝑥 + 3)(2𝑥𝑥 2 + 10𝑥𝑥 + 11) − 9𝑥𝑥 − 32. 32
Consider now one nontrivial example. Example. To divide polynomial 𝑥𝑥 − 2𝑥𝑥 5 + 1 by 𝑥𝑥 − 1.
Solution. First one shall write the polynomial 𝑥𝑥 − 2𝑥𝑥 5 + 1 in the standard form, i.e. to place his term according to the decreasing order of powers. The absents of terms with 𝑥𝑥 2 , 𝑥𝑥 3 , 𝑥𝑥 4 means that corresponding coefficients are zero. Hence, 𝑥𝑥 − 2𝑥𝑥 5 + 1 ≡ −2𝑥𝑥 5 + 0 ∙ 𝑥𝑥 4 + 0 ∙ 𝑥𝑥 3 + 0 ∙ 𝑥𝑥 2 + 𝑥𝑥 + 1. It remains to divide by usual technique:
The final result is as follows: 𝑥𝑥 − 2𝑥𝑥 5 + 1 = (𝑥𝑥 − 1)(−2𝑥𝑥 4 − 2𝑥𝑥 3 − 2𝑥𝑥 2 − 2𝑥𝑥 − 1). Factorization of polynomials. Assume one needs to factorize a polynomial 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑛𝑛−1 + ⋯ + 𝑎𝑎0 , i.e. to represent it as product of “the simplest” factors: 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑛𝑛 + 𝑎𝑎1 𝑥𝑥 𝑛𝑛−1 + ⋯ + 𝑎𝑎0 = 𝑎𝑎𝑛𝑛 (𝑥𝑥 − 𝑥𝑥1 )𝑛𝑛 1 ⋯ (𝑥𝑥 − 𝑥𝑥𝑙𝑙 )𝑛𝑛 𝑙𝑙 (𝑥𝑥 2 + 𝑏𝑏1 𝑥𝑥 + 𝑐𝑐1 ) ⋯ (𝑥𝑥 2 + 𝑏𝑏𝑘𝑘 𝑥𝑥 + 𝑐𝑐𝑘𝑘 ), where each quadratic polynomial 𝑥𝑥 2 + 𝑏𝑏𝑖𝑖 𝑥𝑥 + 𝑐𝑐𝑖𝑖 , 𝑖𝑖 = 1, ⋯ , 𝑘𝑘 is not factorized (such situation can be only if 𝐷𝐷 < 0). Here the numbers 𝑥𝑥1 , 𝑥𝑥2 , ⋯ , 𝑥𝑥𝑙𝑙 , are called roots of polynomial. One can use the following approaches:
33
1. Short multiplication formulas (see Section I) together with “clever” grouping terms. As the result one can find the common factor (parenthesis). As soon as we find one factor, on should divide the original polynomial by this factor. 2. Try to guess a root of the polynomial (let us call it 𝑥𝑥1 ), and then we divide the polynomial by factor 𝑥𝑥 − 𝑥𝑥1 . The result is a new polynomial of power 𝑛𝑛 − 1, and one needs to factorize it similar. 3. To apply advises 1 and 2, discussed above, to the obtained polynomial of the less power. Let us discuss the examples. Example. Factorize polynomials: a) 𝑥𝑥 3 − 𝑥𝑥 2 + 𝑥𝑥 − 1. Solution.
b) 𝑥𝑥 4 + 2𝑥𝑥 3 + 3𝑥𝑥 2 − 2𝑥𝑥 − 4
a) The given polynomial allows two different ways of grouping terms: 𝑥𝑥 3 − 𝑥𝑥 2 + 𝑥𝑥 − 1 = (𝑥𝑥 3 − 𝑥𝑥 2 ) + (𝑥𝑥 − 1) = 𝑥𝑥 2 (𝑥𝑥 − 1) + (𝑥𝑥 − 1) = |𝑡𝑡ℎ𝑒𝑒 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝ℎ𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝 𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑎𝑎𝑎𝑎 𝑡𝑡ℎ𝑒𝑒 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑡𝑡ℎ𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠ℎ 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑡𝑡ℎ𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 , 𝑖𝑖. 𝑒𝑒. 𝑒𝑒𝑒𝑒𝑒𝑒ℎ 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠ℎ𝑎𝑎𝑎𝑎𝑎𝑎 𝑏𝑏𝑏𝑏 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑏𝑏𝑏𝑏 𝑡𝑡ℎ𝑖𝑖𝑖𝑖 𝑓𝑓𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎| = (𝑥𝑥 − 1)(𝑥𝑥 2 + 1).
This is the final factorization since the polynomial 𝑥𝑥 2 + 1 can not be longer factorized. The second way of grouping is the following: 3 𝑥𝑥 − 𝑥𝑥 2 + 𝑥𝑥 − 1 = (𝑥𝑥 3 − 1) + (−𝑥𝑥 2 + 𝑥𝑥) = (𝑥𝑥 − 1)(𝑥𝑥 2 + 𝑥𝑥 + 1) − 𝑥𝑥(𝑥𝑥 − 1) = (𝑥𝑥 − 1)(𝑥𝑥 2 + 𝑥𝑥 + 1 − 𝑥𝑥) = (𝑥𝑥 − 1)(𝑥𝑥 2 + 1). One can see that two different ways gives the same result (what is naturally expected). One can easily guess the root 𝑥𝑥 = 1, indeed if 𝑥𝑥 = 1 then 𝑥𝑥 4 + 2𝑥𝑥 3 + 3𝑥𝑥 2 − 2𝑥𝑥 − 4 = 14 + 2 ∙ 13 + 3 ∙ 12 − 2 ∙ 1 − 4 = 0. Denote 𝑥𝑥1 = 1. Now we shall divide the original polynomial by 𝑥𝑥 − 1:
b)
34
As the result one gets 𝑥𝑥 4 + 2𝑥𝑥 3 + 3𝑥𝑥 2 − 2𝑥𝑥 − 4 = (𝑥𝑥 − 1)(𝑥𝑥 3 + 3𝑥𝑥 2 + 6𝑥𝑥 + 4). Further, we need to factorize 𝑥𝑥 3 + 3𝑥𝑥 2 + 6𝑥𝑥 + 4. One can guess again the root: 𝑥𝑥2 = −1 satisfies the equation 𝑥𝑥 3 + 3𝑥𝑥 2 + 6𝑥𝑥 + 4 = 0. Divide the polynomial 𝑥𝑥 3 + 3𝑥𝑥 2 + 6𝑥𝑥 + 4 by 𝑥𝑥 − (−1) = 𝑥𝑥 + 1:
Thus, 𝑥𝑥 3 + 3𝑥𝑥 2 + 6𝑥𝑥 + 4 = (𝑥𝑥 + 1)(𝑥𝑥 2 + 2𝑥𝑥 + 4). The next step is try to factorize the obtained polynomial 𝑥𝑥 2 + 2𝑥𝑥 + 4. However, the discriminant equals -12, therefore 𝑥𝑥 2 + 2𝑥𝑥 + 4 > 0 for all 𝑥𝑥, hence, this polynomial can not be longer factorized. The final factorization looks like 𝑥𝑥 4 + 2𝑥𝑥 3 + 3𝑥𝑥 2 − 2𝑥𝑥 − 4 = (𝑥𝑥 − 1)(𝑥𝑥 + 1)(𝑥𝑥 2 + 2𝑥𝑥 + 4).
There are many cases when it is not possible at all to guess the roots of polynomials, for example, the roots can be rational or irrational number. One fails to find the roots “by hand” without computer help. Nevertheless in such cases one can estimate the roots, i.e. to find intervals where they are located. Such question are out of the aim of the present book, but we study one example for those who are interested.
35
Example. Factorize the polynomial 𝑥𝑥 4 − 8𝑥𝑥 3 + 17𝑥𝑥 2 − 8𝑥𝑥 + 1.
Solution. One needs to solve the equation 𝑥𝑥 4 − 8𝑥𝑥 3 + 17𝑥𝑥 2 − 8𝑥𝑥 + 1 = 0 to factorize. Substituting integer numbers instead of x one can soon understand that it is impossible to guess the roots. However, if 𝑥𝑥 = 1 then 𝑓𝑓(𝑥𝑥) = 𝑥𝑥 4 − 8𝑥𝑥 3 + 17𝑥𝑥 2 − 8𝑥𝑥 + 1 > 0, while 𝑓𝑓(𝑥𝑥) < 0 when 𝑥𝑥 = 2. Consequently, one of the root belongs to the interval (1,2). It is clear now that to guess it will be impossible task. Evidently, 𝑥𝑥 can not be equal zero. Thus, the equation may be divided by 𝑥𝑥 2 . As the result one gets the equation 8 1 𝑥𝑥 2 − 8𝑥𝑥 + 17 − + 2 = 0, 𝑥𝑥 𝑥𝑥 Which is equivalent to the original (i.e. all roots of both original and obtained equations are coincide). Grupping the terms in a “clever” way, rewrite the equation as follows 𝑥𝑥 2 +
1
1 2
1
1
+ 2 − 8 �𝑥𝑥 + � + 15 = 0 ⇔ �𝑥𝑥 + � − 8 �𝑥𝑥 + � + 15 = 0. 𝑥𝑥 2 𝑥𝑥 𝑥𝑥 𝑥𝑥 1
Substituting 𝑦𝑦 = 𝑥𝑥 + , one can reduce it to the equation 𝑥𝑥
1
1
𝑦𝑦 2 − 8𝑦𝑦 + 15 = 0, and derive that 𝑥𝑥 + = 3 or 𝑥𝑥 + = 5. It is left now to 𝑥𝑥
𝑥𝑥
solve the equations with respect to 𝑥𝑥 and evaluate the roots: 𝑥𝑥1,2 =
3±√5 2
, 𝑥𝑥3,4 =
5±√21 2
.
Finally, the factorization takes the form: 𝑥𝑥 4 − 8𝑥𝑥 3 + 17𝑥𝑥 2 − 8𝑥𝑥 + 1 = �𝑥𝑥 −
3+√5 � �𝑥𝑥 2
−
3−√5 � �𝑥𝑥 2
−
5+√21 � �𝑥𝑥 2
−
5−√21 �. 2
Equations. Let the function 𝑓𝑓(𝑥𝑥) is given. Then the solutions (roots) of the equation 𝑓𝑓(𝑥𝑥) = 0 are the set of 𝑥𝑥 belonging to the domain 𝑓𝑓 such that 𝑓𝑓(𝑥𝑥) = 0.
In particular, if 𝑓𝑓(𝑥𝑥) = 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑛𝑛 + 𝑎𝑎𝑛𝑛−1 𝑥𝑥 𝑛𝑛−1 + ⋯ + 𝑎𝑎0 is some polynomial of power 𝑛𝑛, then the formula 𝑎𝑎𝑛𝑛 𝑥𝑥 𝑛𝑛 + 𝑎𝑎𝑛𝑛 −1 𝑥𝑥 𝑛𝑛−1 + ⋯ + 𝑎𝑎0 = 0 is called the equation of power 𝑛𝑛, and it is not always solvable in real numbers. For example, there is no 𝑥𝑥 ∈ ℝ which would solve the equation 𝑥𝑥 2 + 1 = 0. In general situation, to solve the equation of power 𝑛𝑛 one needs to factorize the given polynomial and then equal to zero each factor. The concrete formulas of roots are known for cases 𝑛𝑛 = 1 and 𝑛𝑛 = 2. If 𝑛𝑛 = 1, then the corresponding equation 36
reads as 𝑎𝑎𝑎𝑎 + 𝑏𝑏 = 0 and is called linear. The solution is unique and is given by 𝑏𝑏
formula 𝑥𝑥 = − , 𝑎𝑎 ≠ 0 . The case 𝑛𝑛 = 2 should be discussed more detailed. 𝑎𝑎
𝒏𝒏 = 𝟐𝟐 𝐨𝐨𝐨𝐨 about the quadratic polynomial
1. The equation 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0 has real solutions iff 𝐷𝐷 = 𝑏𝑏2 − 4𝑎𝑎𝑎𝑎 ≥ 0.
2. The equation 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0 has no more than 2 solutions as 𝑎𝑎 ≠ 0;
the solutions can be evaluated by 𝑥𝑥1,2 =
−𝑏𝑏±√𝐷𝐷 2𝑎𝑎
. If 𝐷𝐷 = 0, then 𝑥𝑥1 = 𝑥𝑥2 .
3. The roots of the equation 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0 are called by roots of the corresponding quadratic polynomial 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐. 4. The Viet theorem: Assume the equation 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0, is given, 𝑏𝑏
𝑥𝑥1 + 𝑥𝑥2 = − 𝑎𝑎 . 𝑎𝑎 ≠ 0. If 𝑥𝑥1 , 𝑥𝑥2 are the solutions, then � 𝑐𝑐 𝑥𝑥1 𝑥𝑥2 = 𝑎𝑎
5. If the equation 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 0 has different solutions 𝑥𝑥1 , 𝑥𝑥2 , then the following identity holds: 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 𝑎𝑎(𝑥𝑥 − 𝑥𝑥1 )(𝑥𝑥 − 𝑥𝑥2 ). In case the root is unique (i.e. 𝑥𝑥1 = 𝑥𝑥2 ), the previous identity reads as 𝑎𝑎𝑥𝑥 2 + 𝑏𝑏𝑏𝑏 + 𝑐𝑐 = 𝑎𝑎(𝑥𝑥 − 𝑥𝑥1 )2 .
Example. Find the roots of quadratic polynomial 2𝑥𝑥 2 − 2𝑥𝑥 + 1. Solution. To find the roots of given polynomial, one needs to solve the equation 2𝑥𝑥 2 − 4𝑥𝑥 + 1 = 0. Here 𝑎𝑎 = 2, 𝑏𝑏 = −4, 𝑐𝑐 = 1. Evaluate the discriminant 𝐷𝐷 = (−4)2 − 4 ∙ 2 ∙ 1 = 8 > 0 , it is positive, hence there exist 2 solutions: 𝑥𝑥1 =
−(−4) + √8 −(−4) − √8 , 𝑥𝑥2 = . 2∙2 2∙2
After some natural simplifications one gets: 𝑥𝑥1,2 =
4 ± 2√2 2 ± √2 = . 2∙2 2
Example. Solve the equation 𝑥𝑥 2 − 𝑥𝑥 − 2 = 0.
37
Solution. One can use either the general formula via discriminant or apply the Viet theorem, what gives the following: the sum of roots is and the product is
𝑐𝑐
𝑎𝑎
=
−2 1
−𝑏𝑏 𝑎𝑎
=
−(−1) 1
=1
= −2. It is easily seen that the roots -1 и 2 satisfy,
since 2 + (−1) = 1, 2 ∙ (−1) = −2. Thus, 𝑥𝑥1 = −2, 𝑥𝑥2 =1.
After the solutions are obtained one may check yourself: substitute these roots into the original equation and see if the equation gives zero or not. The equation becomes zero means our solutions are correct. Example. How many solutions does the equation 2𝑥𝑥 2 − 𝜋𝜋𝑥𝑥 + √𝑒𝑒 = 0 have? Solution. Evaluate the discriminant: 𝐷𝐷 = (−𝜋𝜋)2 − 4 ∙ 2 ∙ √𝑒𝑒 = 𝜋𝜋 2 − 8√𝑒𝑒.
The number of roots depends on the sign of 𝐷𝐷. Therefore one needs to compare it with zero. Here we will widely apply our knowledge in comparison of numbers. 𝜋𝜋 2 − 8√𝑒𝑒 v 0 ⇔
𝜋𝜋 2 v 8√𝑒𝑒
𝜋𝜋 4 v 64𝑒𝑒 .
⇔
Since 3 < 𝜋𝜋 < 4, then 𝜋𝜋 4 < 81. At the same moment, 2 < 𝑒𝑒 < 3, therefore 64𝑒𝑒 > 64 ∙ 2 = 128. Consequently, 𝜋𝜋 4 < 64𝑒𝑒, i.e. the symbol «v» shall be substituted by «0,
𝑓𝑓(𝑥𝑥)
𝑔𝑔(𝑥𝑥)
≤ 0,
𝑓𝑓(𝑥𝑥)
𝑔𝑔(𝑥𝑥)
< 0,
(1)
where the functions 𝑓𝑓(𝑥𝑥) and 𝑔𝑔(𝑥𝑥) are factorized already. Discuss now the method of solution (interval method) for a given example. Example. Solve the inequality
(𝑥𝑥−3)2 𝑥𝑥 3 (𝑥𝑥−1) (𝑥𝑥+2)4 (𝑥𝑥+1)
≥ 0.
Solution. First, one needs to check that all parenthesizes contains variable 𝑥𝑥 with positive coefficient, i.e. it should be (𝑥𝑥 − 3), but not (3 − 𝑥𝑥), it should be 38
(𝑥𝑥 − 1), and not (1 − 𝑥𝑥). One can always make such by multiplying the inequality with -1 and changing its sign as much as it is required.
Second, on real line we should mark by black points zeros of the nominator and by whites -zeroes of denominator. In other words, we mark the point which gives zero in each parenthesis (see figure. 1). If the parenthesis has even power, then we call the corresponding point by zero of even power. Analogously we introduce zeroes of odd power. Would the inequality be strong (i.e. would the sign be < or >), then all points should be marked as whites.
After that we draw the “wave” starting from the right up to the left (Fig. 2) according to the rule: the wave starts from the right up, change the sign at any zero of odd power and keeps the same sign at zero of even order. It remains now to write down the answer taking into account only the intervals with sign «+», and also zeroes of nominator: 𝑥𝑥 ∈ (−1,0] ∪ {1} ∪ [3, +∞). Observe that the round parenthesis corresponds to white points while the quadratic one is reserved for black points. The solution is complete. Let us prove now why the answer is exactly as we obtained. We start from the most right interval. If 𝑥𝑥 > 3, then all factors both of nominator and denominator are strictly positive. Hence, the whole fraction is positive what corresponds to the upper position of wave. The next interval we shall look in is (1,3). Passing through point 3 all factors of the fraction keep the same sign except parenthesis (𝑥𝑥 − 3). The parenthesis (𝑥𝑥 − 3) change the sign to the opposite, therefore, due to the odd power, the factor (𝑥𝑥 − 3)3 change the sign itself, hence the wave should comes down.
The next interval is (0,1). Compare with the previous case, the only factor (𝑥𝑥 − 1) change the sigh, while the rest factors keeps remains the same. The even power is applied to the parenthesis (𝑥𝑥 − 1), consequently, the sign will not be changed and the waves keeps the same negative direction. The further steps are similar. Independently on the fraction sign inside a given interval (between two points), it is always the factor (𝑥𝑥 − 𝑥𝑥0 ) that changes its sign passing over 𝑥𝑥0 . Moreover, if it is an even power factor, then the sign does not change and the 39
wave keeps the same direction. In the opposite, if the parenthesis (𝑥𝑥 − 𝑥𝑥0 ) has odd power, then the sign changes and the waves changes its direction.
The sign of factor on each interval can also be checked “by hands”, taking the concrete values of 𝑥𝑥 from the considered interval. However, this approach is not always convenient, especially when coefficients of factors or ends of the intervals are rational or irrational numbers. Example. Solve the inequality
(1−2𝑥𝑥)(𝑥𝑥 2 +6𝑥𝑥+9) 𝑥𝑥
< 0.
Solution. First, let us simplify the nominator. We shall reduce the parenthe1
sis (1 − 2𝑥𝑥) to the form (1 − 2𝑥𝑥) = −(2𝑥𝑥 − 1) = −2 �𝑥𝑥 − �. Next, the pa2
renthesis (𝑥𝑥 2 + 6𝑥𝑥 + 9) can also be factorized: Thus, we have obtained:
(𝑥𝑥 2 + 6𝑥𝑥 + 9) = (𝑥𝑥 + 3)2 .
1 1 −2 �𝑥𝑥 − 2� (𝑥𝑥 + 3)2 �𝑥𝑥 − 2� (𝑥𝑥 + 3)2 (1 − 2𝑥𝑥)(𝑥𝑥 2 + 6𝑥𝑥 + 9) 0.
IV. EXPONENTIAL AND LOGARITHMIC EXPRESSIONS. Exponential equations and inequalities We have already mentioned the exponential functions 𝑦𝑦(𝑥𝑥) = 𝑎𝑎 𝑥𝑥 in Section II. Let us list the main properties of exponential functions that are very useful to simplify different algebraic expressions. 9. 𝑎𝑎 𝑥𝑥 > 0 for all 𝑎𝑎 > 0 and 𝑥𝑥 ∈ ℝ. 10. 𝑎𝑎0 = 1, 𝑎𝑎1 = 𝑎𝑎 𝑚𝑚
11. 𝑎𝑎 𝑛𝑛 = 𝑛𝑛√𝑎𝑎𝑚𝑚 , where 𝑚𝑚 and 𝑛𝑛 are natural numbers. 1 12. 𝑎𝑎−𝑥𝑥 = 𝑥𝑥 𝑎𝑎 𝑥𝑥
𝑎𝑎 𝑦𝑦
𝑦𝑦
𝑎𝑎
13. 𝑎𝑎 ∙ 𝑎𝑎 = 𝑎𝑎 𝑥𝑥+𝑦𝑦 14.
𝑥𝑥
= 𝑎𝑎 𝑥𝑥−𝑦𝑦
15. (𝑎𝑎 𝑥𝑥 )𝑦𝑦 = (𝑎𝑎 𝑦𝑦 )𝑥𝑥 = 𝑎𝑎 𝑥𝑥𝑥𝑥 16. (𝑎𝑎𝑎𝑎)𝑥𝑥 = 𝑎𝑎 𝑥𝑥 𝑏𝑏 𝑥𝑥 𝑎𝑎 𝑥𝑥
𝑎𝑎 𝑥𝑥
17. � � = 𝑥𝑥 . 𝑏𝑏 𝑏𝑏
The definition of the exponential function implies directly the property if 𝑎𝑎 > 0, 𝑏𝑏 > 0, 𝑎𝑎 ≠ 1 log 𝑎𝑎 𝑥𝑥, 𝑥𝑥 if 𝑎𝑎 = 1, 𝑏𝑏 ≠ 1 . 18. 𝑎𝑎 = 𝑏𝑏 ⇔ 𝑥𝑥 = � ∅, any number, if 𝑎𝑎 = 𝑏𝑏 = 1 Here and further the symbol ⇔ has the meaning "𝑖𝑖𝑖𝑖 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒; 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ". The simplest exponential equations and inequalities can be resolved by the following formulas: 19. 𝑎𝑎 𝑓𝑓(𝑥𝑥) = 𝑎𝑎 𝑔𝑔(𝑥𝑥) , 𝑎𝑎 > 0, 𝑎𝑎 ≠ 1 ⇔ 𝑓𝑓(𝑥𝑥) = 𝑔𝑔(𝑥𝑥) 𝑓𝑓(𝑥𝑥) ≤ 𝑔𝑔(𝑥𝑥) as 0 < 𝑎𝑎 < 1 20. 𝑎𝑎 𝑓𝑓(𝑥𝑥) ≥ 𝑎𝑎 𝑔𝑔(𝑥𝑥) ⇔ � 𝑓𝑓(𝑥𝑥) ≥ 𝑔𝑔(𝑥𝑥) 𝑎𝑎𝑎𝑎 𝑎𝑎 > 1 𝑓𝑓(𝑥𝑥) ≥ 𝑔𝑔(𝑥𝑥) as 0 < 𝑎𝑎 < 1 . 21. 𝑎𝑎 𝑓𝑓(𝑥𝑥) ≤ 𝑎𝑎 𝑔𝑔(𝑥𝑥) ⇔ � 𝑓𝑓(𝑥𝑥) ≤ 𝑔𝑔(𝑥𝑥) as 𝑎𝑎 > 1 Example. Solve the equation 2(𝑥𝑥+1) = 4−𝑥𝑥 .
Solution. Transform the given equation to satisfy formula 11, i.e. the left and right hand sides of the equation should be the functions with the same base. 2(𝑥𝑥+1) = 4−𝑥𝑥 ⇔ 2(𝑥𝑥+1) = (22 )−𝑥𝑥 42
⇔ 2(𝑥𝑥+1) = 2−2𝑥𝑥 ⇔
𝑥𝑥 + 1 = −2𝑥𝑥 ⇔ 3𝑥𝑥 = −1 2 1 𝑥𝑥 −𝑥𝑥
Example. Solve the inequality � � 3
1 ⇔ 𝑥𝑥 = − . 3
≤ 9𝑥𝑥 .
Solution. Analogously to the previous example, one shall first reduce the inequality to the type with exponents of the same base.
Since
1 𝑥𝑥 � � 3 1 3
2 −𝑥𝑥
≤9
𝑥𝑥
1 𝑥𝑥 ⇔ � � 3
2 −𝑥𝑥
1 −2 ≤ �� � � 3
𝑥𝑥
⇔
1 𝑥𝑥 � � 3
2 −𝑥𝑥
1 −2𝑥𝑥 ≤ � � 3
< 1, the exponential function is decreasing, therefore
𝑥𝑥 2 − 𝑥𝑥 ≥ −2𝑥𝑥 ⇔ 𝑥𝑥 2 + 𝑥𝑥 ≥ 0
⇔ 𝑥𝑥(𝑥𝑥 + 1) ≥ 0. Applying the interval method to the obtained inequality, one gets the answer: 𝑥𝑥 ∈ (−∞, −1] ∪ [0, +∞). Logarithms. By the definition, log 𝑎𝑎 𝑏𝑏 (reads as «logarithm with the base 𝑎𝑎 of the number») is the number 𝑐𝑐 such that 𝑎𝑎𝑐𝑐 = 𝑏𝑏. Moreover, the definition of the exponential function 𝑎𝑎𝑐𝑐 the following conditions must be satisfied: 𝑎𝑎 > 0, 𝑏𝑏 > 0, 𝑎𝑎 ≠ 1. In other words, the logarithm with base 𝑎𝑎 of number b is such a power that must be applied to a in order to obtain b. Mathematically the same can be written as 𝑎𝑎 > 0, 𝑏𝑏 > 0, 𝑎𝑎 ≠ 1. 𝑎𝑎log 𝑎𝑎 𝑏𝑏 = 𝑏𝑏, The following notations will be widely used: ln 𝑏𝑏 ≡ log 𝑒𝑒 𝑏𝑏, lg 𝑏𝑏 ≡ log10 𝑏𝑏.
Example. What are numbers: a) log 2 8, b) log 3 81?
Solution. One needs to think according to logarithm definition: log 2 8 is such a power that should be applied to exponent with base 2 in order to obtain 8. Since 23 = 8, this means log 2 8 = 3. Analogously, to compute log 3 81 it is enough to remember that 34 = 81, hence log 3 81 = 4. 43
The main properties of the logarithmic function. 1. log a 1 = 0, log a 𝑎𝑎 = 1, 𝑎𝑎 > 0, 𝑎𝑎 ≠ 1. 2. log 𝑎𝑎 (𝑏𝑏c)= log 𝑎𝑎 𝑏𝑏 + log 𝑎𝑎 𝑐𝑐, 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 > 0, 𝑎𝑎 ≠ 1. 𝑏𝑏
3. log 𝑎𝑎 � �=log 𝑎𝑎 𝑏𝑏 − log 𝑎𝑎 𝑐𝑐, 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 > 0, 𝑎𝑎 ≠ 1. 𝑐𝑐 4. log 𝑎𝑎 𝑏𝑏 𝑥𝑥 = 𝑥𝑥log 𝑎𝑎 𝑏𝑏, 𝑎𝑎, 𝑏𝑏 > 0, 𝑎𝑎 ≠ 1. 1
5. log 𝑎𝑎 𝑥𝑥 𝑏𝑏= log 𝑎𝑎 𝑏𝑏, 𝑎𝑎, 𝑏𝑏 > 0, 𝑎𝑎 ≠ 1, 𝑥𝑥 ≠ 0. 𝑥𝑥
6. log 𝑎𝑎 𝑏𝑏 2𝑛𝑛 = 2𝑛𝑛log 𝑎𝑎 |𝑏𝑏|, 𝑎𝑎 > 0, 𝑎𝑎 ≠ 1, 𝑏𝑏 ≠ 0, where 𝑛𝑛 is natural number. The properties 1 and 3 imply that 1
7. log 𝑎𝑎 = −log 𝑎𝑎 𝑏𝑏, 𝑎𝑎, 𝑏𝑏 > 0, 𝑎𝑎 ≠ 1. 𝑏𝑏
In order to write the change the base of logarithm one needs to use the following rule: 8. log 𝑎𝑎 𝑏𝑏 =
9. log 𝑎𝑎 𝑏𝑏 =
log 𝑐𝑐 𝑏𝑏
log 𝑐𝑐 𝑎𝑎 1
log 𝑏𝑏 𝑎𝑎
, 𝑎𝑎, 𝑏𝑏, 𝑐𝑐 > 0, 𝑎𝑎, 𝑐𝑐 ≠ 1.
, 𝑎𝑎, 𝑏𝑏 > 0, 𝑎𝑎, 𝑏𝑏 ≠ 1.
Exercise. To compute the value 5𝑙𝑙𝑙𝑙𝑙𝑙 5√5 15 .
Solution. 5 3
√225.
log 5 √5 15
=5
log 3 15 52
= |the rule 5. | = 5
2 log 5 15 3
=
2
�5log 5 15 �3
2
= 153 =
Example. It is known that log 7 2 = 𝑎𝑎. Find the value log 1 28. 2
Solution. Here one needs to apply formula 8 to change the base of logarithm: log 7 28 log 7 7 + log 7 4 1 + log 7 22 1 + 2𝑎𝑎 log 1 28 = = = − = − . −1 1 𝑎𝑎 log 2 log 2 2 7 7 log 7 2 The solution of all problems involving logarithms should begin with finding the domain where the formula has sense (is defined). Doing the simplifications of formulas one needs to control the equivalence of the steps. For example, the 44
formula 2 is not correct if at least one of the number 𝑎𝑎, 𝑏𝑏 or 𝑐𝑐 is negative. Simplifying the expression log 𝑎𝑎 (𝑓𝑓(𝑥𝑥)𝑔𝑔(𝑥𝑥)) with help of formula 2, one may lose those solution 𝑥𝑥 that make 𝑓𝑓(𝑥𝑥) and 𝑔𝑔(𝑥𝑥) to be negative. Therefore the following variants of formulas 2 and 3 are highly recommended to use: 8. log 𝑎𝑎 (𝑏𝑏c)= log 𝑎𝑎 |𝑏𝑏| + log 𝑎𝑎 |𝑐𝑐|, 𝑎𝑎 > 0, 𝑏𝑏 ∙ 𝑐𝑐 > 0, 𝑎𝑎 ≠ 1. 𝑏𝑏
9. log 𝑎𝑎 � �=log 𝑎𝑎 |𝑏𝑏| − log 𝑎𝑎 |𝑐𝑐|, 𝑎𝑎 > 0, 𝑏𝑏 ∙ 𝑐𝑐 > 0, 𝑎𝑎 ≠ 1. 𝑐𝑐
The simplest logarithmic equations and inequalities.
Let us discuss now the common algorithms to work with logarithmic equations and inequalities. In the sequel the symbol ⇔ means «equivalently» and the {𝐴𝐴
form �{𝐵𝐵 reads as «the union of systems A and B». 10. 𝑙𝑙𝑙𝑙𝑙𝑙𝑓𝑓(𝑥𝑥) 𝑔𝑔(𝑥𝑥) = 𝑙𝑙𝑙𝑙𝑙𝑙𝑓𝑓(𝑥𝑥) h(𝑥𝑥) ⇔
�
𝑔𝑔(𝑥𝑥 )=h (𝑥𝑥) 𝑓𝑓(𝑥𝑥)>0 𝑔𝑔(𝑥𝑥 )>0 𝑓𝑓(𝑥𝑥 )≠1
.
If 𝑓𝑓(𝑥𝑥) is a constant 𝑎𝑎, such that 𝑎𝑎 > 0, 𝑎𝑎 ≠ 1, then the formula 12 reduces to: 𝑔𝑔(𝑥𝑥) = ℎ(𝑥𝑥) 11. log 𝑎𝑎 𝑔𝑔(𝑥𝑥) = log 𝑎𝑎 h(𝑥𝑥) ⇔ � . 𝑔𝑔(𝑥𝑥) > 0 Example. To solve the equation log 2 (𝑥𝑥 2 − 1) = 2log 2 (𝑥𝑥 + 1).
Solution. Let us notice that the present equation is different from those given is 13., namely, the factor 2 is «extra». Thus, one can not use the direct equivalent step as given in formula 13. The first thing to do is to reduce the formula to the type given in 13: 1 2 1 − log �𝑥𝑥 � = log 2 (𝑥𝑥 + 1)2 ⇔ 2 2 log 2 �𝑥𝑥 − � = 2log 2 (𝑥𝑥 + 1) ⇔ � 2 2 𝑥𝑥 + 1 > 0 1
1 2 3 𝑥𝑥 2 − = (𝑥𝑥 + 1)2 − = 𝑥𝑥 2 + 2𝑥𝑥 + 1 𝑥𝑥 2𝑥𝑥 = − 2 2 2 ⇔ ⇔ � ⇔� � 𝑥𝑥+1>0 𝑥𝑥>−1 𝑥𝑥 > −1 (𝑥𝑥+1)2 >0 𝑥𝑥≠−1
3 3 𝑥𝑥 = − ⇔ 𝑥𝑥 = − . � 4 4 𝑥𝑥 > −1 45
Example. To solve the equation log 𝑥𝑥 (4𝑥𝑥 − 1) + 1 = 0.
Solution. Let us deduce the equation to the form 12. log 𝑥𝑥 (4𝑥𝑥 − 1) + 1 = 0 ⇔ log 𝑥𝑥 (4𝑥𝑥 − 1) = −log 𝑥𝑥 𝑥𝑥 ⇔
1 log 𝑥𝑥 (4𝑥𝑥 − 1) = log 𝑥𝑥 𝑥𝑥 −1 ⇔ log 𝑥𝑥 (4𝑥𝑥 − 1) = log 𝑥𝑥 . 𝑥𝑥 Now one can use the equivalent passage: 1 ⎧ 4𝑥𝑥 − 1 = 𝑥𝑥(4𝑥𝑥 − 1) − 1 ⎪ 𝑥𝑥 1 =0 ⇔ 𝑥𝑥 > 0, 𝑥𝑥 ≠ 1 ⇔ � log 𝑥𝑥 (4𝑥𝑥 − 1) = log 𝑥𝑥 𝑥𝑥 𝑥𝑥 ⎨ 𝑥𝑥 > 0, 𝑥𝑥 ≠ 1 1 ⎪ > 0 ⎩ 𝑥𝑥 𝑥𝑥(4𝑥𝑥 − 1) − 1 =0 ⇔ � 𝑥𝑥 𝑥𝑥 > 0, 𝑥𝑥 ≠ 1
4𝑥𝑥 2 − 𝑥𝑥 − 1 = 0 ⇔ � . 𝑥𝑥 > 0, 𝑥𝑥 ≠ 1
Let us solve the quadratic equation 4𝑥𝑥 2 − 𝑥𝑥 − 1 = 0 by using discriminant formula. 𝐷𝐷 = 1 − 4 ∙ 4 ∙ (−1) = 17. Then the roots are 𝑥𝑥1 =
1 + √17 1 − √17 , 𝑥𝑥2 = . 8 8
The first root is strictly positive and does not equal to1, while the second is less than zero and can not belong to the domain, therefore our solution is the only 𝑥𝑥1 =
1+√17 . 8
Finally, let us solve a «nontrivial equation». Exercise. Solve the equation (𝑥𝑥 2 )lg 𝑥𝑥 = 108 .
Solution. First, as soon as we see that the given equation contains the logarithmic function, one needs to find the domain: 𝑥𝑥 > 0 by the definition of logarithm. Further, we use the properties of logarithm as well as the identity (𝑥𝑥 2 )lg 𝑥𝑥 = 10lg�𝑥𝑥
2 lg 𝑥𝑥 �
to simplify the equation:
(𝑥𝑥 2 )lg 𝑥𝑥 = 108 ⇔
2 lg 𝑥𝑥 lg�𝑥𝑥 2 lg 𝑥𝑥 � �=8 ⇔ = 108 ⇔ �lg�𝑥𝑥 �10 𝑥𝑥 > 0 𝑥𝑥 > 0
46
2 𝑥𝑥 = 100 lg 𝑥𝑥 = ±2 2 lg x ∙ lg 𝑥𝑥 = 8 ( ⇔ � lg 𝑥𝑥) = 4 ⇔ � ⇔� � 1 𝑥𝑥 > 0 𝑥𝑥 > 0 𝑥𝑥 > 0 . 𝑥𝑥 = 100 1 Answer: 𝑥𝑥 = 100, . 100
The simplest logarithmic inequalities can be solved by using the fact that logarithm is decreasing when the base is less than 1 and increasing in the opposite case.
12.
log 𝑓𝑓(𝑥𝑥) 𝑔𝑔(𝑥𝑥) ≥ log 𝑓𝑓(𝑥𝑥) h(𝑥𝑥) ⇔
𝑔𝑔(𝑥𝑥 )≥h (𝑥𝑥)
� h (𝑥𝑥 )>0
𝑓𝑓(𝑥𝑥)>1 � h (𝑥𝑥)≥𝑔𝑔(𝑥𝑥) .
� 𝑔𝑔(𝑥𝑥 )>0
0 1 tains the second restriction (to fits the domain): both 4𝑥𝑥 + 5 and 𝑥𝑥 must be strictly positive. We need to solve the system: 5
4𝑥𝑥 + 5 > 0 𝑥𝑥 > − 𝑥𝑥 > 1 4 0 < 4𝑥𝑥 + 5 ≤ 𝑥𝑥 ⇔ �4𝑥𝑥 + 5 ≤ 𝑥𝑥 ⇔ �3𝑥𝑥 ≤ −5 ⇔ �𝑥𝑥 ≤ − 5 ⇔ ∅. � 𝑥𝑥 > 1 3 𝑥𝑥 > 1 𝑥𝑥 > 1
Thus, the system has no solutions for this case. 2. 0 < 𝑥𝑥 < 1. Here we use the fact that the function is decreasing, therefore the following system holds: 47
5 4𝑥𝑥 + 5 ≥ 𝑥𝑥 3𝑥𝑥 ≥ −5 𝑥𝑥 ≥ − ⇔ � ⇔ � � 3 ⇔ 0 < 𝑥𝑥 < 1. 0 < 𝑥𝑥 < 1 0 < 𝑥𝑥 < 1 0 < 𝑥𝑥 < 1 Thus, the answer is: 𝑥𝑥 ∈ (0,1).
Exercises. 1. To simplify: 𝑎𝑎) lg 10, 𝑏𝑏) ln√𝑒𝑒, 𝑐𝑐) log 5 1000 − log 5 40, 𝑑𝑑) 4
�2√2 � 2
√2
�2−3 3√4 7
2. To solve the solutions: 𝑎𝑎) lg 𝑥𝑥 = −1, 𝑏𝑏)3 ln 𝑥𝑥 = 2,
.
𝑐𝑐) 2𝑥𝑥 = 3, 𝑑𝑑) 3𝑥𝑥 + 2 ∙ 3𝑥𝑥−1 = 45, ln(𝑥𝑥 + 5) 𝑓𝑓) 𝑒𝑒) log 𝑥𝑥 (4𝑥𝑥 − 3) = 0, = 2. ln(𝑥𝑥 − 1) 3. To solve the inequalities: 𝑎𝑎) 52𝑥𝑥 ≤ 3 ∙ 5𝑥𝑥 , 𝑏𝑏) log 𝑥𝑥−1 (𝑥𝑥 + 1) ≥ 0.
48
V. TRIGONOMETRY The main formulas of trigonometry Let 𝑥𝑥 be the angle corresponding the point 𝑃𝑃�𝑥𝑥𝑝𝑝 , 𝑦𝑦𝑝𝑝 � on the unit circle. By the definition of cosine and sine of the angle x, the following relation holds: 𝑥𝑥𝑝𝑝 = cos 𝑥𝑥, 𝑦𝑦𝑝𝑝 = sin 𝑥𝑥.
By using the Pythagorian theorem and also the fact that the radius of circle equals 1, we obtain the main trigonometric identity: 1.
sin2 𝑥𝑥 + cos 2 𝑥𝑥 = 1.
2.
tg 𝑥𝑥 ∙ ctg 𝑥𝑥 = 1,
Moreover, it holds that
where by the definition, tg 𝑥𝑥 =
sin 𝑥𝑥
cos 𝑥𝑥
, ctg 𝑥𝑥 =
cos 𝑥𝑥 sin 𝑥𝑥
.
Assuming that cos 𝑥𝑥 ≠ 0 (or correspondingly, sin 𝑥𝑥 ≠ 0) one can divide the identity 1. by cos 𝑥𝑥, (correspondingly by sin 𝑥𝑥). This yields 3. 1 + tg 2 𝑥𝑥 =
1
cos 2 𝑥𝑥 1
4. 1 + ctg2 𝑥𝑥 =
sin 2 𝑥𝑥
.
Double angles formulas are: 5. sin 2𝑥𝑥 = 2 sin 𝑥𝑥 cos 𝑥𝑥 6. cos 2𝑥𝑥 = cos 2 𝑥𝑥 − sin2 𝑥𝑥 7. tg 2𝑥𝑥 =
2 tg 𝑥𝑥
1−tg 2 𝑥𝑥
Trigonometric functions of sum and difference are: 8. sin(𝑥𝑥 ± 𝑦𝑦) = sin 𝑥𝑥 cos 𝑦𝑦 ± cos 𝑥𝑥 sin 𝑦𝑦 9. cos (𝑥𝑥 ± 𝑦𝑦) = cos 𝑥𝑥 cos 𝑦𝑦 ∓ sin 𝑥𝑥 sin 𝑦𝑦 10. tg(𝑥𝑥 ± 𝑦𝑦) =
tg 𝑥𝑥±tg 𝑦𝑦
1∓tg 𝑥𝑥 tg 𝑦𝑦
Power reducing formulas are: 11. sin2 𝑥𝑥 =
1−cos 2𝑥𝑥 2
49
12. cos 2 𝑥𝑥 =
1+cos 2𝑥𝑥 2
The sum and difference of trigonometric functions can be reduced to product by formulas: 13. sin 𝑥𝑥 ± sin 𝑦𝑦 = 2 sin
𝑥𝑥±𝑦𝑦
14. cos 𝑥𝑥 + cos 𝑦𝑦 = 2 cos
𝑐𝑐𝑐𝑐𝑐𝑐
2 𝑥𝑥+𝑦𝑦
15. cos 𝑥𝑥 − cos 𝑦𝑦 = − 2 sin
cos
2 𝑥𝑥+𝑦𝑦 2
𝑥𝑥∓𝑦𝑦
2 𝑥𝑥−𝑦𝑦
sin
2 𝑥𝑥−𝑦𝑦 2
In the opposite way, to replace the product by the sum one can with help of formulas 1
16. sin 𝑥𝑥 sin 𝑦𝑦 = (cos(𝑥𝑥 − 𝑦𝑦) − cos(𝑥𝑥 + 𝑦𝑦)) 2 1
17. sin 𝑥𝑥 cos 𝑦𝑦 = (sin(𝑥𝑥 + 𝑦𝑦) + sin(𝑥𝑥 − 𝑦𝑦)) 2 1
18. cos 𝑥𝑥 cos 𝑦𝑦 = (cos(𝑥𝑥 + 𝑦𝑦) + cos(𝑥𝑥 − 𝑦𝑦)). 2
The angles can be measured in degrees as well as in radians. If 𝛼𝛼° is the degree measure of the angle and 𝛽𝛽 is radian one, then to convert them one needs to use the identity 𝛽𝛽 =
𝜋𝜋
180°
𝛼𝛼°, and vice-verse, 𝛼𝛼° =
180° 𝜋𝜋
𝛽𝛽.
Below we give the table with values of the main trigonometric function. This table one must know by heart!
0 sin x
0
cos x
1
tg x
0
ctg x
----
𝜋𝜋� 6
𝜋𝜋� 4
1� 2
√2� 2
1� √3
1
√3� 2 √3
√2� 2 1
𝜋𝜋� 3
√3� 2 1� 2 √3
1� √3
𝜋𝜋� 2 1
𝜋𝜋 0
3𝜋𝜋� 2 -1
2𝜋𝜋
0
-1
0
1
----
0
----
0
0
----
0
----
0
The symbol “----” means that the function is not defined for given value of the argument. Such situation can appear when the denominator equals zero and
50
formally this imply the division by zero. For example, tg
𝜋𝜋
2
=
𝜋𝜋 2 𝜋𝜋 cos 2
sin
1
= this oper0
ation is forbidden by rules. Therefore one says that the function 𝑡𝑡𝑡𝑡 𝑥𝑥 is not de𝜋𝜋 fined at point 𝑥𝑥 = . 2
Example. Evaluate
cos 2 45°−sin 60° tg 45°−sin 30°
.
Solution. Substitute the known values from the table: 2
√2 √3 2 √3 1 √3 1 − √3 �2� − 2 2 − 2 − 2 cos 45° − sin 60° 2 = =4 =2 = = 1 − √3. 1 1 1 1 tg 45° − sin 30° 1−2 1−2 2 2
Example. Given the formula sin 𝑥𝑥 − cos 𝑥𝑥 = 0,3. Evaluate sin4 𝑥𝑥 + cos 4 𝑥𝑥. Solution. Let us rewrite the expression we should evaluate:
1 sin4 𝑥𝑥 + cos 4 𝑥𝑥 = (sin2 𝑥𝑥 + cos 2 𝑥𝑥)2 − 2 sin2 𝑥𝑥 cos 2 𝑥𝑥 = 1 − (2 sin 𝑥𝑥 cos 𝑥𝑥)2 . 2
How can one evaluate 2 sin 𝑥𝑥 cos 𝑥𝑥, if sin 𝑥𝑥 − cos 𝑥𝑥 = 0,3? Let us square both sides of the last equation: 0,32 = (sin 𝑥𝑥 − cos 𝑥𝑥)2 = sin2 𝑥𝑥 + cos 2 𝑥𝑥 − 2 sin 𝑥𝑥 cos 𝑥𝑥,
this implies 2 sin 𝑥𝑥 cos 𝑥𝑥 = 1 − 0,09 = 0,91. Hence, 1
1
sin4 𝑥𝑥 + cos 4 𝑥𝑥 = 1 − (2 sin 𝑥𝑥 cos 𝑥𝑥)2 = 1 − ∙ 2
2
We leave the answer as it is. Example. Simplify
cos 2𝑥𝑥
ctg 2 𝑥𝑥−tg 2 𝑥𝑥
.
912
100 2
.
Solution. We start with simplifications at denominator:
51
cos 2 𝑥𝑥 sin2 𝑥𝑥 cos 2 𝑥𝑥 ∙ cos 2 𝑥𝑥 − sin2 𝑥𝑥 ∙ sin2 𝑥𝑥 ctg 𝑥𝑥 − tg 𝑥𝑥 = − = sin2 𝑥𝑥 cos 2 𝑥𝑥 sin2 𝑥𝑥cos 2 𝑥𝑥 cos 4 𝑥𝑥 − sin4 𝑥𝑥 (cos 2 𝑥𝑥 − sin2 𝑥𝑥)(cos 2 𝑥𝑥 + sin2 𝑥𝑥) = = 1 sin2 𝑥𝑥cos 2 𝑥𝑥 (sin 2𝑥𝑥)2 4 cos 2𝑥𝑥 ∙ 1 4 cos 2𝑥𝑥 = = . 2 1 (sin 2𝑥𝑥) 2 (sin 2𝑥𝑥) 4 2
2
Now we plug it into the denominator of the original fraction:
cos 2𝑥𝑥 cos 2𝑥𝑥 cos 2𝑥𝑥 ∙ (sin 2𝑥𝑥)2 1 2 = = = sin 2𝑥𝑥. 4 cos 2𝑥𝑥 4 ctg2 𝑥𝑥 − tg 2 𝑥𝑥 4 cos 2𝑥𝑥 2 (sin 2𝑥𝑥) 1
Answer: sin2 2𝑥𝑥. 4
Example. To evaluate a) cos 15° 𝑏𝑏) sin 10° sin 30° sin 50° sin 70°.
Solution. a) There exist two ways to evaluate cos 15°.
1. One can represent the angle 15° as the difference between 45° and 30°, and after that to apply formula 9. «the cosine of the difference» . cos 15° = cos(45° − 30°) = cos 45° cos 30° + sin 45° sin 30° = √2 √3 √2 1 √2(√3 + 1) ∙ + ∙ = . 2 2 2 2 4
2. The second way is to use the formula for cosine of half of the angle, since 15° =
30° 2
, and 30°is the table angle �corresponds to
Applying formula 12 one gets:
𝜋𝜋
6
in radian measure�.
√3 1+ 2 1 + cos 30° 2 + √3 cos2 15° = = = . 2 2 4
It remains only to take the square root with sign «+» because the cosine of 15°is positive: 2 + √3 �2 + √3 cos 15° = � = . 4 2
52
One can think for the moment that the answer is different from those we get by the first method of solution. However, it is not so. Simplifying the obtained value, we derive �2+√3 2
=
2�2+√3 4
=
√2�4+2√3 4
=
2
√2�1+√3 +2∙1∙√3 4
what coincides with the answer by first method.
=
√2�(1+√3)2 4
=
√2(√3+1) 4
b) Rewrite the product of functions as the sum: 1 sin 10° sin 50° = (cos (10° − 50°) − cos(10° + 50°)) = 2 1 1 1 = (cos (−40°) − cos 60°) = | cos 𝑥𝑥 is the even funtion| = �cos 40° − �. 2 2 2 Next, we substitute this product into the original expression: 1
1
sin 10° sin 30° sin 50° sin 70° = �cos 40° − � sin 30° sin 70° = 2 2
1
1
�cos 40° sin 70° − 2 sin 70°�. 4
Now replace the product by sum according to 17: 1
cos 40° sin 70° = (sin(70° + 40°) + sin(70° − 40°)) = 2
1 1 (sin 110° + sin30°) = (sin(180° − 70°) + sin30°) = 2 2 1 1 1 1 �sin 180° cos 70° − cos 180° sin 70°) + � = sin 70° + . 2 2 4 2 Substitute this into the previous calculations, we get: 1
1
sin 10° sin 30° sin 50° sin 70° = �cos 40° sin 70° − sin 70°� = 4 2
1 1
1
1
1
1
1
� sin 70° + 4 − 2 sin 70°�=4 ∙ 4 = 16 . 4 2 Answer:
1
16
53
Trigonometric equations. Inverse trigonometric functions Let us discuss first the definitions of the inverse trigonometric functions. For a given number 𝑎𝑎 we define the angle (argument) 𝑥𝑥 such that sin 𝑥𝑥 = 𝑎𝑎 . This angle 𝑥𝑥 is called arcisn of the value 𝑎𝑎 and denoted by 𝒙𝒙 = 𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚𝐚 𝒂𝒂. In other words the arcsin of the number is the angle whose sin is equal to the given number, see the figure. To define the arcsin uniquely it is assumed always that the codomain of the 𝜋𝜋 𝜋𝜋
arcsin is the segment �− , �. Since −1 ≤ sin 𝑥𝑥 ≤ 1, the condition 2 2
−1 ≤ 𝑎𝑎 ≤ 1 must be satisfied. Thus, the function arcsin 𝑎𝑎 has no sense for other values of 𝑎𝑎, different from those belonging to [-1,1].
Analogously one can define the functions arccos 𝑎𝑎, arctg 𝑎𝑎, arcctg 𝑎𝑎. The values of 𝑎𝑎 in arctg 𝑎𝑎 and arcctg 𝑎𝑎 can be an arbitrary from ℝ, because tan and cotan varies over the whole ℝ. Thus, 𝜋𝜋 𝜋𝜋 𝑥𝑥 ∈ �− , � . 𝑥𝑥 = arcsin 𝑎𝑎 ⇔ sin 𝑥𝑥 = 𝑎𝑎, −1 ≤ 𝑎𝑎 ≤ 1, 2 2 𝑥𝑥 = arccos 𝑎𝑎 ⇔ cos 𝑥𝑥 = 𝑎𝑎,
−1 ≤ 𝑎𝑎 ≤ 1,
𝑥𝑥 = arctg 𝑎𝑎 ⇔ tg 𝑥𝑥 = 𝑎𝑎,
𝑥𝑥 = arcctg 𝑎𝑎 ⇔ ctg 𝑥𝑥 = 𝑎𝑎,
𝑎𝑎 ∈ ℝ,
𝑎𝑎 ∈ ℝ,
𝑥𝑥 ∈ [0, 𝜋𝜋]. 𝜋𝜋 𝜋𝜋 𝑥𝑥 ∈ �− , � . 2 2 𝑥𝑥 ∈ (0, 𝜋𝜋).
The following formulas are valid for the values with the opposite sign: arcsin(−𝑎𝑎) = − arcsin 𝑎𝑎,
arccos(−𝑎𝑎) = 𝜋𝜋 − arccos 𝑎𝑎,
arctg(−𝑎𝑎) = − arctg 𝑎𝑎, arcctg (– 𝑎𝑎) = −arcctg 𝑎𝑎. The present identities can be verified by the definition of the inverse trigonome𝑡𝑡ℎ𝑒𝑒 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 sin 𝑦𝑦 𝑖𝑖𝑖𝑖 𝑜𝑜𝑜𝑜𝑜𝑜
tric functions. Indeed, 𝑦𝑦 = arcsin�– 𝑎𝑎� ⇔ sin 𝑦𝑦 = −𝑎𝑎 �������������������� sin(−𝑦𝑦) = 𝑎𝑎 ⇔ −𝑦𝑦 = arcsin 𝑎𝑎 ⇔ 𝑦𝑦 = − arcsin 𝑎𝑎 ⇔ 54
arcsin(−𝑎𝑎) = − arcsin 𝑎𝑎.
Another identities can be verified similarly. Example. Evaluate 𝜋𝜋
1
𝑎𝑎) arcsin 1, b) arccos , 𝑐𝑐) arccos �− � , 𝑑𝑑) arctg √3 . 2 2
𝜋𝜋 𝜋𝜋
Solution. 𝑎𝑎) By the definition, arcsin 1 is the angle belonging to �− , �, 2 2
𝜋𝜋
having the sine that equals 1, thus, the angle is . 2
𝜋𝜋
b) One needs to understand which angle has cosine that equals . This is 2
𝜋𝜋
tricky problem. Certainly, many give the answer that arccos = 0, and explain 𝜋𝜋
2
it that cos = 0. This is absolutely false, one should not mix the angle with the 2
𝜋𝜋
value of cosine! The formula arccos involves 2
𝜋𝜋
as the value of cosine and not
2
𝜋𝜋
the angle! The next attempt to answer the question is the following: arccos is 2
𝜋𝜋
the angle as it is written having cosine equal to , since this value is not listed 2
in the table. Such answer is also not correct: can it be so that cosine of an angel 𝜋𝜋 𝜋𝜋 is ? Of course not, because > 1: there is no an angle 𝛼𝛼, such that 2
cos 𝛼𝛼 =
(or, ∅).
𝜋𝜋
2
2
𝜋𝜋
> 1. Thus, the correct answer is: the angle arccos does not exist 2
c) One needs to replace the negative argument: 1 1 arccos �− � = 𝜋𝜋 − arccos . 2 2 1
How to evaluate arccos ? This value is the angle from [0, 𝜋𝜋] with cosine 1
𝜋𝜋
2
1
𝜋𝜋
equal to , i.e. . Then arccos �− � = 𝜋𝜋 − = 2 2 3 3
2𝜋𝜋 3
.
d) One needs to remember the table angles to get the value arctg √3: this is 𝜋𝜋 𝜋𝜋 the angle such that its tan equals √3. This is , therefore arctg √3 = . 3
3
The simplest trigonometric equations. All trigonometric equations can be solved by reduction to one of the following simplest types: 1.
𝑥𝑥 = (−1)𝑛𝑛 arcsin 𝑎𝑎 + 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ, 𝑖𝑖𝑖𝑖 |𝑎𝑎| ≤ 1 sin 𝑥𝑥 = 𝑎𝑎 ⇔ � . ∅, 𝑖𝑖𝑖𝑖 |𝑎𝑎| > 1 55
𝑥𝑥 = ± arccos 𝑎𝑎 + 2𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ, 𝑖𝑖𝑖𝑖 |𝑎𝑎| ≤ 1 2. cos 𝑥𝑥 = 𝑎𝑎 ⇔ � . ∅, 𝑖𝑖𝑖𝑖 |𝑎𝑎| > 1 3. tg 𝑥𝑥 = 𝑎𝑎 ⇔ 𝑥𝑥 = arctg 𝑎𝑎 + 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ. 4. ctg 𝑥𝑥 = 𝑎𝑎 ⇔ 𝑥𝑥 = arcctg 𝑎𝑎 + 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ. These formulas one should know by heart. Also it is useful to remember some partial cases of 1.and 2., written in more simple form: 𝜋𝜋 5. sin 𝑥𝑥 = 1 ⇔ 𝑥𝑥 = + 2𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ 2
6. sin 𝑥𝑥 = 0 ⇔ 𝑥𝑥 = 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ 𝜋𝜋 7. sin 𝑥𝑥 = −1 ⇔ 𝑥𝑥 = − + 2𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ 2
8. cos 𝑥𝑥 = 1 ⇔ 𝑥𝑥 = 2𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ 𝜋𝜋 9. cos 𝑥𝑥 = 0 ⇔ 𝑥𝑥 = + 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ 2
10. cos 𝑥𝑥 = −1 ⇔ 𝑥𝑥 = 𝜋𝜋 + 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ
Example. Solve the equation 2sin 3𝑥𝑥 = −1. Solution. 2sin 3𝑥𝑥 = −1 ⇔ sin 3𝑥𝑥 = −
1 2
⇔
1 arcsin 1 2 + 𝜋𝜋𝜋𝜋 ⇔ 3𝑥𝑥 = (−1)𝑛𝑛 arcsin �− � + 𝜋𝜋𝜋𝜋 ⇔ 𝑥𝑥 = (−1)𝑛𝑛+1 2 3 3 𝜋𝜋 𝜋𝜋𝜋𝜋 𝜋𝜋 𝜋𝜋𝜋𝜋 𝑥𝑥 = (−1)𝑛𝑛+1 6 + ⇔ 𝑥𝑥 = (−1)𝑛𝑛+1 + , 𝑛𝑛 ∈ ℤ. 3 18 3 3 Example. Solve the equation tg(3𝑥𝑥 + 10) = 50.
Solution. Apply the common formula for equations with tan: tg(3𝑥𝑥 + 10) = 50 ⇔ 3𝑥𝑥 + 10 = arctg 50 + 𝜋𝜋𝜋𝜋 ⇔
and here it is no problem that the argument of arctan is so large (much more than one), since the tan (as well as cotan) can take any values from ℝ ⇔ 𝑥𝑥 =
arctg 50 10 𝜋𝜋𝜋𝜋 − + , 3 3 3
56
𝑛𝑛 ∈ ℤ.
Homogeneous trigonometric equations. Let 𝐴𝐴, 𝐵𝐵, 𝐶𝐶 are some numbers. Equations of types 1. 𝐴𝐴 sin 𝑥𝑥 + 𝐵𝐵 cos 𝑥𝑥 = 0
2. 𝐴𝐴 sin2 𝑥𝑥 + 𝐵𝐵sin 𝑥𝑥 cos 𝑥𝑥 + 𝐶𝐶cos 2 𝑥𝑥 = 0
are called trigonometric homogeneous, of the first and the second order respectively. Equation 1. сan be solved as follows. If both 𝐴𝐴 and 𝐵𝐵 equals zero, then any 𝑥𝑥 is the solution, since it is always true that
𝐴𝐴 sin 𝑥𝑥 + 𝐵𝐵 cos 𝑥𝑥 = 0 ∙ sin 𝑥𝑥 + 0 ∙ cos 𝑥𝑥 = 0. Let now at least one of numbers 𝐴𝐴 or 𝐵𝐵 differs from zero (assume 𝐴𝐴 ≠ 0). Assume that cos 𝑥𝑥 = 0. Then the equation 1. implies 𝐴𝐴 sin 𝑥𝑥 + 𝐵𝐵 ∙ 0 = 0 ⇒ 𝐴𝐴 sin 𝑥𝑥 = 0 . Since 𝐴𝐴 ≠ 0, it yields that sin 𝑥𝑥 = 0. Consequently, our assumption gives that both sin 𝑥𝑥 = 0, cos 𝑥𝑥 = 0, what contradicts to the identity sin2 𝑥𝑥 + cos 2 𝑥𝑥 = 1. Therefore our assumption was not true, i.e. cos 𝑥𝑥 ≠ 0. Hence, one can divide the equation 1. by cos 𝑥𝑥. As a result we get 𝐵𝐵 𝐴𝐴 tg𝑥𝑥 + 𝐵𝐵 = 0 ⇒ tg𝑥𝑥 = − . 𝐴𝐴
The solution procedure for the obtained equation is known (see the formulas 3. from the previous section).
The second order homogeneous equation (equation number 2) can be solved analogously. First one needs to see that cos 𝑥𝑥 ≠ 0, then to divide the equation by cos 2 𝑥𝑥. After such a procedure the equation converts to the form 𝐴𝐴tg 2 𝑥𝑥 + 𝐵𝐵tg𝑥𝑥 + 𝐶𝐶 = 0.
This is the quadratic equation with respect to 𝑡𝑡 = tg 𝑥𝑥.
Equations 1. and 2. can be solved also by reducing to the equation with respect to ctg 𝑥𝑥, it is sufficient to prove that sin 𝑥𝑥 ≠ 0 and divide the equation by sin 𝑥𝑥 or respectively by sin2 𝑥𝑥 in case of the second order equation. Example. Solve the equation 3sin2 𝑥𝑥 + sin x cos 𝑥𝑥 − 1 = 0.
Solution. First, one can observe, that the considered equation is not homogeneous of second degree (even if it looks similar with those of type 2) since it contains the term +1. However, the equation can be reduced to homogeneous one by using the identity 1 = sin2 𝑥𝑥 + cos 2 𝑥𝑥. Indeed, 57
3sin2 𝑥𝑥 + sin x cos 𝑥𝑥 − 1 = 0 ⇔ 3sin2 𝑥𝑥 + sin x cos 𝑥𝑥 − sin2 𝑥𝑥 − cos 2 𝑥𝑥 = 0 ⇔
2sin2 𝑥𝑥 + sin x cos 𝑥𝑥 − cos 2 𝑥𝑥 = 0. If cos 𝑥𝑥 = 0 ⇒ sin 𝑥𝑥 = 0, what contra-
dicts the main trigonometric identity. Thus, cos 𝑥𝑥 ≠ 0, and therefore one can divide the equation by cos 𝑥𝑥. After the division one gets 2tg 2 𝑥𝑥 + tg 𝑥𝑥 − 1 = 0. 1
Denote 𝑡𝑡 = tg 𝑥𝑥. Then 2𝑡𝑡 2 + 𝑡𝑡 − 1 = 0 ⇔ 𝑡𝑡1 = , 𝑡𝑡2 = −1. Solving the equa2
tions
tg 𝑥𝑥 =
1 2
and tg 𝑥𝑥 = −1, one gets the answer: 1 π 𝑥𝑥 = arctg + πn, 𝑥𝑥 = − + πn, n ∈ ℤ. 2 4
Observe that equations of type 𝐴𝐴1 sin2 𝑥𝑥 + 𝐴𝐴2 cos 2 𝑥𝑥 + 𝐴𝐴3 sin 2𝑥𝑥 + 𝐴𝐴4 cos 2𝑥𝑥 + 𝐴𝐴5 sin 𝑥𝑥 cos 𝑥𝑥 + 𝐴𝐴6 = 0
can also be reduced to type 2. One should use the formula for double angles and the main trigonometric identity. Example. Solve the solution 12sin2 𝑥𝑥 + 3 sin 2𝑥𝑥 − 2 cos 2 𝑥𝑥 = 2. Solution. Replace sin 2𝑥𝑥 with 2 sin 𝑥𝑥 cos 𝑥𝑥 and the number 2 with 2sin2 𝑥𝑥 + 2cos 2 𝑥𝑥. Then we obtain the equation
10sin2 𝑥𝑥 + 6 sin 𝑥𝑥cos 𝑥𝑥 − 4 cos 2 𝑥𝑥 = 0. Analogously to the last example, one can show that cos 𝑥𝑥 ≠ 0. Dividing the equation by cos 2 𝑥𝑥, we obtain 10tg2 𝑥𝑥 + 10 tg 𝑥𝑥 − 4 = 0 ⇔
�
tg 𝑥𝑥=−1 2 5
tg 𝑥𝑥=
⇔ �
𝜋𝜋 4
𝑥𝑥=− +𝜋𝜋𝑛𝑛 2 5
𝑥𝑥=arctg +𝜋𝜋𝜋𝜋 ,
𝑛𝑛 ∈ ℤ.
Graphs for trigonometric functions. 1. 𝒚𝒚(𝒙𝒙) = 𝐬𝐬𝐬𝐬𝐬𝐬 𝒙𝒙, 𝒚𝒚(𝒙𝒙) = 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙. These functions are periodic with period 2𝜋𝜋, i.e. their values are repeated after equal intervals of argument 𝑥𝑥 (periodically with period 2𝜋𝜋): sin(𝑥𝑥 + 2𝜋𝜋) = sin 𝑥𝑥, cos(𝑥𝑥 + 2𝜋𝜋) = cos 𝑥𝑥. The domain is ℝ, the codomain is 𝐸𝐸 = [−1,1]. The graphs are illustrated below. 58
𝟐𝟐. 𝒚𝒚(𝒙𝒙) = 𝐭𝐭𝐭𝐭 𝒙𝒙, 𝒚𝒚(𝒙𝒙) = 𝐜𝐜𝐜𝐜𝐜𝐜 𝒙𝒙.
These functions have satisfy also periodicity property, the period is 𝜋𝜋: tg(𝑥𝑥 + 𝜋𝜋) = tg 𝑥𝑥, ctg(𝑥𝑥 + 𝜋𝜋) = ctg 𝑥𝑥. The domain of function 𝑦𝑦(𝑥𝑥) = tg 𝑥𝑥 = 𝜋𝜋
sin 𝑥𝑥
cos 𝑥𝑥
contains all points of real line except zeroes of denomina-
tor: 𝐷𝐷 = ℝ\ � + 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ�. Analogously, the domain for 𝑦𝑦(𝑥𝑥) = ctg 𝑥𝑥 = 2
must exclude the solutions of the equation sin 𝑥𝑥 = 0, therefore 𝐷𝐷 = ℝ\{𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ}. The codomains for both tg 𝑥𝑥 and ctg 𝑥𝑥 is `ℝ.
Exercise. Draw the graph of function 𝑦𝑦 = 2 sin(𝑥𝑥 + 𝜋𝜋) + 1. 59
cos 𝑥𝑥 sin 𝑥𝑥
Solution. We shall use the composition of the transformations. The algorithm is as follows: 1. Draw the graph of the function 𝑦𝑦 = sin 𝑥𝑥 2. Shift it 𝜋𝜋 units left along OX (since sin(𝑥𝑥 + 𝜋𝜋) = sin(𝑥𝑥 − (−𝜋𝜋)). 3. Stretch it vertically along OY by a factor of 2 (i.e. multiply every value of the function by 2) 4. Shift up 1 unit along OY.
Inverse trigonometric functions By the definition, 𝜋𝜋 𝜋𝜋
1. 𝑦𝑦(𝑥𝑥) = arcsin 𝑥𝑥 ⇔ 𝑦𝑦 ∈ �− , � , sin 𝑦𝑦 = 𝑥𝑥. 2 2
2. 𝑦𝑦(𝑥𝑥) = arccos 𝑥𝑥 ⇔ 𝑦𝑦 ∈ [0, 𝜋𝜋], cos 𝑦𝑦 = 𝑥𝑥. 𝜋𝜋 𝜋𝜋
3. 𝑦𝑦(𝑥𝑥) = arctg 𝑥𝑥 ⇔ 𝑦𝑦 ∈ �− , � , tg 𝑦𝑦 = 𝑥𝑥. 2 2 4. 𝑦𝑦(𝑥𝑥) = arcctg 𝑥𝑥 ⇔ 𝑦𝑦 ∈ (0, 𝜋𝜋), ctg 𝑦𝑦 = 𝑥𝑥.
Due to the periodicity and symmetry of trigonometric functions, the values of the inverse trigonometric functions are used to be considered on some period𝜋𝜋 𝜋𝜋
ic cell. This is the reason why the restrictions 𝑦𝑦 ∈ �− , � , 𝑦𝑦 ∈ [0, 𝜋𝜋], 2 2
𝜋𝜋 𝜋𝜋
𝑦𝑦 ∈ �− , � , 𝑦𝑦 ∈ (0, 𝜋𝜋) appear. Moreover, since −1 ≤ sin 𝑦𝑦, cos 𝑦𝑦 ≤ 1, the 2 2
domain D for arcsin and arccos is the segment [−1,1]. The graphs are illustrated on the figures below.
60
61
Exercises. 1. Compute:
𝑎𝑎)
𝜋𝜋 𝜋𝜋 cos2 6 + sin2 6 𝜋𝜋 5𝜋𝜋 tg 4 + cos 3
, 𝑏𝑏) cos 78° cos 18° + cos 12° cos 72°, 𝑐𝑐)
2. Fill in the tables. a
-1 -
arcsin a
√3 2
-
√2 2
-
1 2
tg 22° + tg 113° . 1 − tg 158° ∙ ctg23°
1 √2 √3 1 2 2 2
0
arccos a
A arctg a
√3 -1
−
√3 3
0 1 √3 √3 3
arcctg a
3. Solve the equations: 𝜋𝜋 𝜋𝜋 b) ctg(𝑥𝑥 + ) = −1, 𝑐𝑐) sin 2𝑥𝑥 = . 3 3 1 𝑑𝑑) 5sin2 𝑥𝑥 − sin 𝑥𝑥 cos 𝑥𝑥 − 1 = 0 , 𝑒𝑒)cos 2 2𝑥𝑥 − sin2 2𝑥𝑥 + = 0. 2 𝑎𝑎) cos 3𝑥𝑥 =
√3 , 2
4. Simplify the expression:
𝑎𝑎)(sin 𝛼𝛼 + cos 𝛼𝛼)2 + (sin 𝛼𝛼 − cos 𝛼𝛼)2 , 𝑏𝑏)(tg 𝛼𝛼 + ctg 𝛼𝛼)2 + (tg 𝛼𝛼 − ctg 𝛼𝛼)2 , 𝑐𝑐) sin 𝛼𝛼 (2 + ctg 𝛼𝛼)(2ctg 𝛼𝛼 + 1) − 5 cos 𝛼𝛼. 𝜋𝜋
5. Draw the graph of function 𝑦𝑦 = 2 cos(𝑥𝑥 + ) − 1. 4
62
Bibliography [1] V.V. Vavilov, I.I. Melnikov, S.N. Olekhnik, P.I. Pasichenko, M., Nauka, Gl. Red. Fiz.-Mat. Liter., 1988. − 432 p.[Russian] [2] S. N. Olekhnik, M.K. Potapov, M.: Nauka, UNC DO MGU Press, 1997. − 132 p. [Russian] [3] V.V. Kalinin. Matematika. Domashnjaja obsheobrazovatel’naja biblioteka. − M.: Astrel, ACT, 2000. – 255p.[Russian] [4] V. V. Tkachuk, Matematika abiturientu, M.: MCNMO, 2007. − 976p. [Russian] [5] G. Forsling, Övningar i matematik, Linköping University, 2003. − 46p.
63
Answer keys.
Section I. 1
2
3
1. a) 1 , b) 1 , c) 7 . 2 a)√5 >
3 a)
6
4 𝑏𝑏 5 𝑎𝑎
, b)
1
32 𝑎𝑎
3
. 4 a) 2, b) 5
𝑎𝑎 2
7
b) 𝜋𝜋 − 1 < �6√2.
5
𝑏𝑏 2
, c) � � . 5 a) 3, b) 34. 6 a) 11, b) 43. 𝑎𝑎 𝑎𝑎 2 +𝑎𝑎𝑎𝑎 +𝑏𝑏 2
7 a) 𝑎𝑎 = 2, 𝑎𝑎 = 8, b) 𝑎𝑎 = −1, 𝑎𝑎 = 5.
Section II. 1. a) 𝑥𝑥 ∈ (−∞, −5) ∪ (−5, −3) ∪ (3, +∞), b) 𝑥𝑥 ∈ (2, +∞), 1
c) 𝑥𝑥 ∈ �− , 0� ∪ (0, +∞). 3. a) 𝑓𝑓�𝑔𝑔(𝑥𝑥)� = 2
1
ln |𝑥𝑥|
b) �𝑔𝑔(𝑥𝑥)� = 2(𝑒𝑒 𝑥𝑥 + 1)2 + 3√𝑒𝑒 𝑥𝑥 + 1, 𝑔𝑔�𝑓𝑓(𝑥𝑥)� = �𝑒𝑒 2𝑥𝑥 𝑈𝑈
4. a) 𝐸𝐸0 =
𝑡𝑡 �1−𝑒𝑒 𝑅𝑅𝑅𝑅 �
𝑡𝑡
1
, 𝑔𝑔�𝑓𝑓(𝑥𝑥)� = ln � �,
, b) 𝑈𝑈 = 𝐸𝐸0 �1 − 𝑒𝑒 𝑅𝑅𝑅𝑅 �, c) 𝑅𝑅 =
−𝑡𝑡
4 +3𝑥𝑥
𝑈𝑈 � 𝐸𝐸 0
𝐶𝐶ln �1−
.
+ 1.
𝑥𝑥
Section III. 1. a) 2𝑥𝑥 + 2 +
10𝑥𝑥 −12
𝑥𝑥 2 −2𝑥𝑥+2
b) (𝑥𝑥 + 1) �𝑥𝑥 +
, b) 𝑥𝑥 2 + 1 +
1−√5 � �𝑥𝑥 2
+
1+√5 �. 2
1
𝑥𝑥 2 −1
. 2. a) (𝑥𝑥 − 1)(𝑥𝑥 + 1)(𝑥𝑥 2 + 1),
3. a) 𝑥𝑥 =
23 14
c) 𝑥𝑥1 = −2, 𝑥𝑥2 = 14, d) 𝑥𝑥1 = −3, 𝑥𝑥2 = −4. 4. a) 𝑥𝑥 ∈ (−∞, −2] ∪ [2, +∞),
, b) 𝑥𝑥1 = 0, 𝑥𝑥2 =
16 15
1
b) 𝑥𝑥 ∈ (−∞, −5) ∪ �− , 3�, c) 𝑥𝑥 ∈ (0,1) ∪ (2, +∞). 2
Section IV. 1
1. a) 1, b) , c) 2, d) 2. 2. a) 2
1
10
2
3. a) 𝑥𝑥 ∈ (−∞, log 5 3], b) 𝑥𝑥 ∈ (2, +∞). Section V. 2
1
1. a) , b) , c) -1. 3
3
, b) 𝑒𝑒 3 , c) log 2 3, d) 3, e) , f)
2
64
2
√17−1 . 2
,
2. A
−
-1
arcsin a −
𝜋𝜋 2
−
A arctg a
𝜋𝜋 3
5𝜋𝜋 6
𝜋𝜋
arccos a
1 1 √2 √3 √2 √3 0 1 − − 2 2 2 2 2 2
−√3 −
𝜋𝜋 3
lutions,
𝜋𝜋
18
d) 𝑥𝑥1,2 = arctg 4. a) 2, b)
4
𝜋𝜋 4
− −
𝜋𝜋 𝜋𝜋 0 𝜋𝜋 − 4 4 6 𝜋𝜋 4
−
sin 2 2𝛼𝛼
3
1
𝜋𝜋
7𝜋𝜋 12
+ 𝜋𝜋𝜋𝜋, e) 𝑥𝑥 = ± +
− , c) 4
𝜋𝜋 𝜋𝜋 2 4
𝜋𝜋 3
2
8
1
sin 𝛼𝛼
6
.
65
𝜋𝜋 3
𝜋𝜋 4
𝜋𝜋 6
𝜋𝜋 2 0
√3 0 1 √3 √3 3 3
+ 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ, b) 𝑥𝑥 = ± 1±√17
𝜋𝜋 𝜋𝜋 2 3
2𝜋𝜋 3
−
𝜋𝜋 4
𝜋𝜋 𝜋𝜋 0 6 6
−
3𝜋𝜋 4
-1
arcctg a − 𝜋𝜋 6 3. a) 𝑥𝑥 = ±
−
𝜋𝜋 6 𝜋𝜋 3
𝜋𝜋 3 𝜋𝜋 6
+ 𝜋𝜋𝜋𝜋, 𝑛𝑛 ∈ ℤ, c) there are no so-
𝜋𝜋𝜋𝜋 2
, 𝑛𝑛 ∈ ℤ.
EDUCATIONAL HANDBOOK
KOROLEVA YULIA OLEGOVNA KOROLEV ALEXANDER VLADIMIROVICH
BRIDGING COURSE IN MATH
EDITED BY AUTHOR
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