# Understanding Basic Calculus

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1 Understanding Basic Calculus S.K. Chung

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3 Dedicated to all the people who have helped me in my life.

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5 i Preface This book is a revised and expanded version of the lecture notes for Basic Calculus and other similar courses offered by the Department of Mathematics, University of Hong Kong, from the first semester of the academic year through the second semester of It can be used as a textbook or a reference book for an introductory course on one variable calculus. In this book, much emphasis is put on explanations of concepts and solutions to examples. By reading the book carefully, students should be able to understand the concepts introduced and know how to answer questions with justification. At the end of each section (except the last few), there is an exercise. Students are advised to do as many questions as possible. Most of the exercises are simple drills. Such exercises may not help students understand the concepts; however, without practices, students may find it difficult to continue reading the subsequent sections. Chapter 0 is written for students who have forgotten the materials that they have learnt for HKCEE Mathematics. Students who are familiar with the materials may skip this chapter. Chapter is on sets, real numbers and inequalities. Since the concept of sets is new to most students, detail explanations and elaborations are given. For the real number system, notations and terminologies that will be used in the rest of the book are introduced. For solving polynomial inequalities, the method will be used later when we consider where a function is increasing or decreasing as well as where a function is convex or concave. Students should note that there is a shortcut for solving inequalities, using the Intermediate Value Theorem discussed in Chapter 3. Chapter is on functions and graphs. Some materials are covered by HKCEE Mathematics. New concepts introduced include domain and range (which are fundamental concepts related to functions); composition of functions (which will be needed when we consider the Chain Rule for differentiation) and inverse functions (which will be needed when we consider exponential functions and logarithmic functions). In Chapter 3, intuitive idea of limit is introduced. Limit is a fundamental concept in calculus. It is used when we consider differentiation (to define derivatives) and integration (to define definite integrals). There are many types of limits. Students should notice that their definitions are similar. To help students understand such similarities, a summary is given at the end of the section on two-sided limits. The section of continuous functions is rather conceptual. Students should understand the statements of the Intermediate Value Theorem (several versions) and the Extreme Value Theorem. In Chapters 4 and 5, basic concepts and applications of differentiation are discussed. Students who know how to work on limits of functions at a point should be able to apply definition to find derivatives of simple functions. For more complicated ones (polynomial and rational functions), students are advised not to use definition; instead, they can use rules for differentiation. For application to curve sketching, related concepts like critical numbers, local extremizers, convex or concave functions etc. are introduced. There are many easily confused terminologies. Students should distinguish whether a concept or terminology is related to a function, to the x-coordinate of a point or to a point in the coordinate plane. For applied extremum problems, students

6 ii should note that the questions ask for global extremum. In most of the examples for such problems, more than one solutions are given. In Chapter 6, basic concepts and applications of integration are discussed. We use limit of sums in a specific form to define the definite integral of a continuous function over a closed and bounded interval. This is to make the definition easier to handle (compared with the more subtle concept of limit of Riemann sums). Since definite integrals work on closed intervals and indefinite integrals work on open intervals, we give different definitions for primitives and antiderivatives. Students should notice how we can obtain antiderivatives from primitives and vice versa. The Fundamental Theorem of Calculus (several versions) tells that differentiation and integration are reverse process of each other. Using rules for integration, students should be able to find indefinite integrals of polynomials as well as to evaluate definite integrals of polynomials over closed and bounded intervals. Chapters 7 and 8 give more formulas for differentiation. More specifically, formulas for the derivatives of the sine, cosine and tangent functions as well as that of the logarithmic and exponential functions are given. For that, revision of properties of the functions together with relevant limit results are discussed. Chapter 9 is on the Chain Rule which is the most important rule for differentiation. To make the rule easier to handle, formulas obtained from combining the rule with simple differentiation formulas are given. Students should notice that the Chain Rule is used in the process of logarithmic differentiation as well as that of implicit differentiation. To close the discussion on differentiation, more examples on curve sketching and applied extremum problems are given. Chapter 0 is on formulas and techniques of integration. First, a list of formulas for integration is given. Students should notice that they are obtained from the corresponding formulas for differentiation. Next, several techniques of integration are discussed. The substitution method for integration corresponds to the Chain Rule for differentiation. Since the method is used very often, detail discussions are given. The method of Integration by Parts corresponds to the Product Rule for differentiation. For integration of rational functions, only some special cases are discussed. Complete discussion for the general case is rather complicated. Since Integration by Parts and integration of rational functions are not covered in the course Basic Calculus, the discussion on these two techniques are brief and exercises are not given. Students who want to know more about techniques of integration may consult other books on calculus. To close the discussion on integration, application of definite integrals to probability (which is a vast field in mathematics) is given. Students should bear in mind that the main purpose of learning calculus is not just knowing how to perform differentiation and integration but also knowing how to apply differentiation and integration to solve problems. For that, one must understand the concepts. To perform calculation, we can use calculators or computer softwares, like Mathematica, Maple or Matlab. Accompanying the pdf file of this book is a set of Mathematica notebook files (with extension.nb, one for each chapter) which give the answers to most of the questions in the exercises. Information on how to read the notebook files as well as trial version of Mathematica can be found at

7 Contents 0 Revision 0. Exponents Algebraic Identities and Algebraic Expressions Solving Linear Equations Solving Quadratic Equations Remainder Theorem and Factor Theorem Solving Linear Inequalities Lines Pythagoras Theorem, Distance Formula and Circles Parabola Systems of Equations Sets, Real Numbers and Inequalities 3. Sets Introduction Set Operations Real Numbers The Number Systems Radicals Solving Inequalities Quadratic Inequalities Polynomial Inequalities with degrees Functions and Graphs 43. Functions Domains and Ranges of Functions Graphs of Equations Graphs of Functions Compositions of Functions Inverse Functions More on Solving Equations Limits Introduction Limits of Sequences Limits of Functions at Infinity One-sided Limits Two-sided Limits Continuous Functions Differentiation Derivatives Rules for Differentiation Higher-Order Derivatives

8 iv CONTENTS 5 Applications of Differentiation 7 5. Curve Sketching Increasing and Decreasing Functions Relative Extrema Convexity Curve Sketching Applied Extremum Problems Absolute Extrema Applied Maxima and Minima Applications to Economics Integration Definite Integrals Fundamental Theorem of Calculus Indefinite Integrals Application of Integration Trigonometric Functions Angles Trigonometric Functions Differentiation of Trigonometric Functions Exponential and Logarithmic Functions 9 8. Exponential Functions Logarithmic Functions Differentiation of Exp and Log Functions More Differentiation Chain rule Implicit Differentiation More Curve Sketching More Extremum Problems More Integration 9 0. More Formulas Substitution Method Integration of Rational Functions Integration by Parts More Applications of Definite Integrals A Answers 55 B Supplementary Notes 69 B. Mathematical Induction B. Binomial Theorem B.3 Mean Value Theorem B.4 Fundamental Theorem of Calculus

9 Chapter 0 Revision 0. Exponents Definition () Let n be a positive integer and let a be a real number. We define a n to be the real number given by a n a } {{ } a a. n factors () Let n be a negative integer n, that is, n k where k is a positive integer, and let a be a real number different from 0. We define a k to be the real number given by a k a k. (3) (i) Let a be a real number different from 0. We define a 0. (ii) We do not define 0 0 (thus the notation 0 0 is meaningless). Terminology In the notation a n, the numbers n and a are called the exponent and base respectively. Rules for Exponents Let a and b be real numbers and let m and n be integers (when a 0 or b 0, we have to add the condition: m, n different from 0). Then we have () a m a n a m+n () a m a n am n provided that a 0 (3) (a m ) n a mn (4) (ab) n a n b n (5) ( a n a b) n b n provided that b 0 Exercise 0.. Simplify the following; give your answers without negative exponents. (a) x 6 x 3 (b) (c) ( x y 3) 4 (d) x y z 3 ( x ) 3 y 4 ( x y )

10 Chapter 0. Revision 0. Algebraic Identities and Algebraic Expressions Identities Let a and b be real numbers. Then we have () (a + b) a + ab + b () (a b) a ab + b (3) (a + b)(a b) a b Remark The above equalities are called identities because they are valid for all real numbers a and b. Caution In general, (a + b) a + b. Note: (a + b) a + b if and only if a 0 or b 0. Example Expand the following: () ( ) x + () ( x 5 ) x (3) ( x ) ( x + 7 ) Solution () () (3) ( x + ) ( x ) + ( x ) () + x + 4 x + 4 ( x 5 x) x (x) ( ) 5 + x x x ( ) 5 x ( x ) ( x + 7 ) ( x + ) 7 ( x + ) 49 x 48 Example Simplify the following: () () (3) (4) x x 6 x 6x + 9 x x x + x + x x ( ) x y (5) x x + x x +

11 0.. Algebraic Identities and Algebraic Expressions 3 Solution () () (3) (4) x x 6 x 6x + 9 (x 3)(x + ) (x 3) x + x 3 x x x (x ) x x x + x + x x ( x y ) ( x ) y (x + ) (x + )(x ) (x ) (x + ) (x + ) (x ) x 5 (x + ) (x ) ( xy y y xy ) (5) x x + x x + 3x + 6 x x (x + ) + x x + 3x + 6 x x + x x + 3(x + ) x x + x (x + ) 3(x + ) x FAQ What is expected if we are asked to simplify an expression? For example, in (5), can we give 3x + 3 as the answer? Answer There is no definite rule to tell which expression is simpler. For (5), both acceptable. Use your own judgment. x 3(x + ) and 3x + 3 x x are

12 4 Chapter 0. Revision Exercise 0.. Expand the following: (a) (x + 3) (b) (3x y) (c) (x + 3y)(x 3y) (d) (x + 3y)(x + 4y) ( ) ( ) ( ) (e) x 3 (f) x + 5 x 5. Factorize the following: (a) x 7x + (b) x + x 6 (c) x + 8x + 6 (d) 9x + 9x + (e) 9x 6x + (f) 5x 5 (g) 3x 8x + 7 (h) x x Simplify the following: (a) (c) x x 6 x 7x + x x 4x + 4x x (b) (d) 0.3 Solving Linear Equations x + 3x 4 x x x + h x h A linear equation in one (real) unknown x is an equation that can be written in the form ax + b 0, where a and b are constants with a 0 (in this course, we consider real numbers only; thus a constant means a real number that is fixed or given). More generally, an equation in one unknown x is an equation that can be written in the form F(x) 0 (0.3.) Remark To be more precise, F should be a function from a subset of R into R. See later chapters for the meanings of function and R. Definition A solution to Equation (0.3.) is a real number x 0 such that F(x 0 ) 0. Example The equation x has exactly one solution, namely 3. To solve an equation (in one unknown) means to find all solutions to the equation. Definition We say that two equations are equivalent if the have the same solution(s). Example The following two equations are equivalent: () x () x 3 To solve an equation, we use properties of real numbers to transform the given equation to equivalent ones until we obtain an equation whose solutions can be found easily.

13 0.3. Solving Linear Equations 5 Properties of real numbers Let a, b and c be real numbers. Then we have () a b a + c b + c () a b ac bc and ac bc a b if c 0 Remark is the symbol for implies. The first part of Property () means that if a b, then ac bc. is the symbol for and. Property () means that if a b, then a + c b + c and vice versa, that is, a b iff a + c b + c. In mathematics, we use the shorthand iff to stand for if and only if. Example Solve the following equations for x. () 3x 5 (7 x) () a(b + x) c, where a, b, c and d are real numbers with a + d 0. Solution () Using properties of real numbers, we get The solution is x 5 (7 x) 3x 5 4 x 3x + x x 9 x 9 5. FAQ Can we omit the last sentence? Answer The steps above means that a real number x satisfies 3x 5 (7 x) if and only if x 9 5. It s alright if you stop at the last line in the equation array because it tells that given equation has one and only one solution, namely 9 5. FAQ What is the difference between the word solution after the question and the word solution in the last sentence? Answer They refer to different things. The first solution is solution (answer) to the problem (how to solve the problem) whereas the second solution means solution to the given equation. Sometimes, an equation may have no solution, for example, x + 0 but the procedures (explanations) to get this information is a solution to the problem. FAQ Can we use other symbols for the unknown? Answer In the given equation, if x is replaced by another symbol, for example, t, we get the equation 3t 5 (7 t) in one unknown t. Solution to this equation is also 9. In writing an equation, the symbol 5

14 6 Chapter 0. Revision for the unknown is not important. However, if the unknown is expressed in t, all the intermediate steps should use t as unknown: 3t 5 (7 t). t 9 5 () Using properties of real numbers, we get a(b + x) c ab + ax c ax + c ab (a + d)x c ab x c ab a + d. Exercise 0.3. Solve the following equations for x. (a) (x + 4) 7x + (b) (c) (a + b)x + x (x + b) (d) where a, b and c are constants with a b. 5x x 4 4 x a x b c 0.4 Solving Quadratic Equations A quadratic equation (in one unknown) is an equation that can be written in the form ax + bx + c 0 (0.4.) where a, b, and c are constants and a 0. To solve (0.4.), we can use the Factorization Method or the Quadratic Formula. Factorization Method The method makes use of the following result on product of real numbers: Fact Let a and b be real numbers. Then we have ab 0 a 0 or b 0. Example Solve x + x 5 0. Solution Factorizing the left side, we obtain (x + 5)(x 3) 0. Thus x or x 3 0. Hence x 5 or x 3.

15 0.4. Solving Quadratic Equations 7 FAQ Can we write x 5 and x 3? Answer The logic in solving the above equation is as follows x + x 5 0 (x + 5)(x 3) 0 x or x 3 0 x 5 or x 3 It means that a (real) number x satisfies the given equation if and only if x 5 or x 3. The statement x 5 or x 3 cannot be replaced by x 5 and x 3. To say that there are two solutions, you may write the solutions are 5 and 3. Sometimes, we also write the solutions are x 5 and x 3 which means there are two solutions 5 and 3 and they are denoted by x and x respectively. In Chapter, you will learn the concept of sets. To specify a set, we may use listing or description. The solution set to an equation is the set consisting of all the solutions to the equation. For the above example, we may write the solution set is { 5, 3} (listing); the solution set is {x : x 5 or x 3} (description). When we use and, we mean the listing method. Quadratic Formula Solutions to Equation (0.4.) are given by Remark b 4ac is called the discriminant of (0.4.). x b ± b 4ac. a () If b 4ac > 0, then (0.4.) has two distinct solutions. () If b 4ac 0, then (0.4.) has one solution. (3) If b 4ac < 0, then (0.4.) has no (real) solution. FAQ Why is (real) added? Answer When the real number system is enlarged to the complex number system, (0.4.) has two complex solutions if b 4ac < 0. However, these solutions are not real numbers. In this course, we consider real numbers only. So you may simply say that there is no solution. Example Solve the following quadratic equations. () x 9x () x + x Solution () Using the quadratic formula, we see that the equation has two solutions given by x 9 ± ( 9) 4()(0) () 9 ± 4.

16 8 Chapter 0. Revision Thus the solutions are 5 and. () Since 4()(3) 8 < 0, the equation has no solutions. Example Solve the equation x (x + ) x (x + 3). Solution Expanding both sides, we get x + x x + 3x x + x 0 x (x + ) 0 x 0 or x The solutions are and 0. Remark If we cancel the factor x on both sides, we get x + x + 3 which has only one solution. In canceling the factor x, it is assumed that x 0. However, 0 is a solution and so this solution is lost. To use cancellation, we should write x (x + ) x (x + 3) x + x + 3 or x 0. Example Find the value(s) of k such that the equation 3x + kx has only one solution. Solution The given equation has only one solution iff k 4(3)(7) 0. Solving, we get k ± 84. Exercise 0.4. Solve the following equations. (a) 4x 4x 0 (b) + x 3x 0 (c) 4x (x 4) x 5 (d) x + x + 0 (e) x + x (f) x 3 7x + 3x 0. Find the value(s) of k such that the equation x + kx + (k + 3) 0 has only one solution. 3. Find the positive number such that sum of the number and its square is Remainder Theorem and Factor Theorem Remainder Theorem If a polynomial p(x) is divided by x c, where c is a constant, the remainder is p(c). Example Let p(x) x 3 + 3x x +. Find the remainder when p(x) is divided by x. Solution The remainder is p() 3 + 3( ) () + 8. Factor Theorem (x c) is a factor of a polynomial p(x) if and only if p(c) 0.

17 0.5. Remainder Theorem and Factor Theorem 9 Proof This follows immediately from the remainder theorem because (x c) is a factor means that the remainder is 0. Example Let p(x) x 3 + kx + x 6. Suppose that (x + ) is a factor of p(x). () Find the value of k. () With the value of k found in (), factorize p(x). Solution () Since ( x ( ) ) is a factor of p(x), it follows from the Factor Theorem that p( ) 0, that is ( ) 3 + k( ) + ( ) 6 0. Solving, we get k 4. () Using long division, we get x 3 + 4x + x 6 (x + )(x + x 3). By inspection, we have p(x) (x + )(x + 3)(x ). FAQ Can we find the quotient (x + x 3) by inspection (without using long division)? Answer The inspection method that some students use is called the compare coefficient method. Since the quotient is quadratic, it is in the form (ax + bx + c). Thus we have x 3 + 4x + x 6 (x + )(ax + bx + c) (0.5.) Comparing the coefficient of x 3, we see that a. Similarly, comparing the constant term, we get c 3. Hence we have x 3 + 4x + x 6 (x + )(x + bx 3). To find b, we may compare the x term (or the x term) to get 4 + b, which yields b. Remark The compare coefficient method in fact consists of the following steps: () Expand the right side of (0.5.) to get ax 3 + (a + b)x + (b + c)x + c () Compare the coefficients of the given polynomial with that obtained in Step () to get a 4 a + b b + c 6 c

18 0 Chapter 0. Revision (3) Solve the above system to find a, b and c. Example Factorize p(x) x 3x. Solution Solving p(x) 0 by the quadratic formula, we get x ( 3) ± ( 3) 4()( ) () 3 ± 7. 4 By the Factor Theorem, both ( x ) ( 3 7) and x are factors of p(x). Therefore, we have 4 p(x) ( x ) ( x 3 ) 7 4 where the factor is obtained by comparing the leading term (that is, the x term). FAQ Can we say that p(x) can t be factorized? Answer Although p(x) does not have factors in the form (x c) where c is an integer, it has linear factors as given above. If the question asks for factors with integer coefficients, then p(x) cannot be factorized as product of linear factors. FAQ Can we use the above method to factorize, for example, p(x) 6x + x? Answer If you don t know how to factorize p(x) by inspection, you can solve p(x) 0 using the quadratic formula (or calculators) to get x or x. Therefore (by the Factor Theorem and comparing the leading 3 term), we have ( p(x) 6 x ) ( x + ) 3 (x )(3x + ). Exercise 0.5. For each of the following expressions, use the factor theorem to find a linear factor (x c) and hence factorize it completely (using integer coefficients). (a) x 3 3x + (b) x 3 7x + x + 3 (c) x 3 x 4x + 3 (d) x 3 5x + x 7. Solve the following equation for x. (a) x 3 9x 8x (b) x 3 x + 0 (c) x 3 5x + x Solving Linear Inequalities Notation and Terminology Let a and b be real numbers. () We say that b is greater than a, or equivalently, that a is less than b to mean that b a is a positive number.

19 0.6. Solving Linear Inequalities () We write b > a to denote that b is greater than a and we write a < b to denote that a is less than b. (3) We write b a to denote that b is greater than or equal to a and we write a b to denote that a is less than or equal to b. A linear inequality in one unknown x is an inequality that can be written in one of the following forms: () ax + b < 0 () ax + b 0 (3) ax + b > 0 (4) ax + b 0 where a and b are constants with a 0. More generally, an inequality in one unknown x is an inequality that can be written in one of the following forms: () F(x) < 0 () F(x) 0 (3) F(x) > 0 (4) F(x) 0 where F is a function from a subset of R into R. Definition A solution to an inequality F(x) < 0 is a real number x 0 such that F(x 0 ) < 0. The definition also applies to other types of inequalities. Example Consider the inequality x By direct substitution, we see that is a solution and is not a solution. To solve an inequality means to find all solutions to the inequality. Rules for Inequalities Let a, b and c be real numbers. Then the following holds. () If a < b, then a + c < b + c. () If a < b and c > 0, then ac < bc. (3) If a < b and c < 0, then ac > bc. Note: The inequality is reversed. (4) If a < b and b c, then a < c. (5) (6) If a < b and a and b have the same sign, then a > b. If 0 < a < b and n is a positive integer, then a n < b n and n a < n b. Terminology Two numbers have the same sign means that both of them are positive or both of them are negative. Remark One common mistake in solving inequalities is to apply a rule with the wrong sign (positive or negative). For example, if c is negative, it would be wrong to apply Rule (). Example Solve the following inequalities. () x + > 7(x + 3)

20 Chapter 0. Revision () 3(x ) + 5 > 3x + 7 Solution () Using rules for inequalities, we get x + > 7(x + 3) x + > 7x + > 7x x 0 > 5x 4 > x. The solutions are all the real numbers x such that x < 4, that is, all real numbers less than 4. () Expanding the left side, we get 3(x ) + 5 3x which is always less than the right side. Thus the inequality has no solution. Exercise 0.6. Solve the following inequalities for x. (a) (c) 0.7 Lines x 3x 7 3 (b) (3 x) 3( x) 3x x + 3 < 0 (d) x x + 3 > A linear equation in two unknowns x and y is an equation that can be written in the form ax + by + c 0 (0.7.) where a, b and c are constants with a, b not both 0. More generally, an equation in two unknowns x and y is an equation that can be written in the form where F is a function (from a collection of ordered pairs into R). F(x, y) 0, (0.7.) Definition An ordered pair (of real numbers) is a pair of real numbers x 0, y 0 enclosed inside parenthesis: (x 0, y 0 ). Remark Two ordered pairs (x 0, y 0 ) and (x, y ) are equal if and only if x 0 x and y 0 y. For example, the ordered pairs (, ) and (, ) are not equal. Definition A solution to Equation (0.7.) is an ordered pair (x 0, y 0 ) such that F(x 0, y 0 ) 0. Example Consider the equation x + 3y 4 0. By direct substitution, we see that (, 0) is a solution whereas (, ) is not a solution.

21 0.7. Lines 3 Rectangular Coordinate System Given a plane, there is a one-to-one correspondence between points in the plane and ordered pairs of real numbers (see the construction below). The plane described in this way is called the Cartesian plane or the rectangular coordinate plane. First we construct a horizontal line and a vertical line on the plane. Their point of intersection is called the origin. The horizontal line is called the x-axis and the vertical line y-axis. For each point P in the plane we can label it by two real numbers. To this ends, we draw perpendiculars from P to the x-axis and y-axis. The first perpendicular meets the x-axis at a point which can be represented by a real number a. Similarly, the second perpendicular meets the y-axis at a point which can be represented by a real number b. Moreover, the ordered pair of numbers a and b determines P uniquely, that is, if P and P are distinct points in the plane, then the ordered pairs corresponding to P and P are different. Therefore, we may identify the point P with the ordered pair (a, b) and we write P (a, b) or P(a, b). The numbers a and b are called the x-coordinate and y-coordinate of P respectively Figure 0. The x- and y-axes divide the (rectangular) coordinate plane into 4 regions (called quadrants): Quadrant I {(a, b) : a > 0 and b > 0}, Quadrant II {(a, b) : a < 0 and b > 0}, Quadrant III {(a, b) : a < 0 and b < 0}, Quadrant IV {(a, b) : a > 0 and b < 0}. Lines in the Coordinate Plane Consider the following equation Ax + By + C 0 (0.7.3) where A, B and C are constants with A, B not both zero. It is not difficult to see that the equation has infinitely many solutions. Each solution (x 0, y 0 ) represents a point in the (rectangular) coordinate plane. The collection of all solutions (points) form a line, called the graph of Equation (0.7.3). Moreover, every line in the plane can be represented in this way. For example, if l is the line passing through the origin and making an angle of 45 degrees with the positive x-axis, then it is the graph of the equation y x. Although this equation is not in the form (0.7.3), it can be written as ()x + ( )y + 0 0, that is, x y 0. Terminology If a line l is represented by an equation in the form (0.7.3), we say that the equation is a general linear form for l. Remark In Equation (0.7.3), () if A 0, then the equation reduces to y C B and its graph is a horizontal line; () if B 0, then the equation reduces to x C and its graph is a vertical line. A

22 4 Chapter 0. Revision Example Consider the line l given by x + 3y 4 0 (0.7.4) For each of the following points, determine whether it lies on l or not. () A (4, ) () B (5, ) Solution () Putting (x, y) (4, ) into (0.7.4), we get L.S. (4) + 3( ) 4 0. Therefore A does not lie on l. () Putting (x, y) (5, ) into (0.7.4), we get L.S. (5) + 3( ) 4 0 R.S. Therefore B lies on l. Example Consider the line l given by x + y 4 0 (0.7.5) Find the points of intersection of l with the x-axis and the y-axis. Solution Putting y 0 into (0.7.5), we get x 4 0 from which we obtain x 4. The point of intersection of l with the x-axis is (4, 0). Putting x 0 into (0.7.5), we get y 4 0 from which we obtain y. The point of intersection of l with the y-axis is (0, ). Remark The point (4, 0) and (0, ) are called the x-intercept and y-intercept of l respectively. FAQ Can we say that the x-intercept is 4 etc? Answer Some authors define x-intercept to be the x-coordinate of point of intersection etc. Using this convention, the x-intercept is 4 and the y-intercept is.

23 0.7. Lines 5 Definition For a non-vertical line l, its slope (denoted by m l or simply m) is defined to be m l y y x x where P (x, y ) and P (x, y ) are any two distinct points lying on l. Remark The number m l is well-defined, that is, its value is independent of the choice of P and P. FAQ What is the slope of a vertical line? Answer The slope of a vertical line is undefined because if P (x, y ) and P (x, y ) lie on a vertical line, then x x and so which is undefined. y y y y x x 0 Some students say that the slope is infinity, denoted by. However, is not a number; it is just a notation. Moreover, infinity is ambiguous does it mean positive infinity (going up, very steep) or negative infinity (going down, very steep)? Example Find the slope of the line (given by) x 5y Solution Take any two points on the line, for example, take P (, ) and P (3, 3). The slope m of the line is FAQ Can we take other points on the line? m 3 3 ( ) 5. Answer You can take any two points. For example, taking A ( 0, 9 5 m ( 9 ) ) and B ( 9, 0), we get 5. Equations for Lines Let l be a non-vertical line in the coordinate plane. Suppose P (x, y ) is a point lying on l and m is the slope of l. Then an equation for l can be written in the form called a point-slope form for l. y y m(x x ) (0.7.6) Remark Since there are infinitely many points on a line, l has infinitely many point-slope forms. However, we also say that (0.7.6) is the point-slope form of l. FAQ Can we write the equation in the following form? y y x x m (0.7.7) Answer Equation (0.7.7) represents a line minus one point. If you put (x, y) (x, y ) into (0.7.7), the left-side is 0 0 which is undefined. This means that the point (x, y ) does not lie on L. However, once you get (0.7.7), you can obtain the point-slope form (0.7.6) easily.

24 6 Chapter 0. Revision Suppose the y-intercept of l is (0, b) and the slope of l is m. Then a point-slope form for l is y b m(x 0) which can be written as y mx + b called the slope-intercept form for l. Example Find the slope of the line having general linear form x + 3y 4 0. Solution Rewrite the given equation in slope-intercept form: x + 3y 4 0 3y x + 4 y 3 x The slope of the line is 3. Example Let l be the line that passes through the points A(, 3) and B(, 4). Find an equation in general linear form for l. Solution Using the points A and B, we get the slope m of l m 3 ( 4) 7. Using the slope m and the point A (or B), we get the point slope form y 3 7(x ). (0.7.8) Expanding and rearranging terms, (0.7.8) can be written in the following general linear form 7x + y 0 0. Parallel and Perpendicular Lines Let l and l be (non-vertical) lines with slopes m and m respectively. Then () l and l are parallel if and only if m m ; () l and l are perpendicular to each other if and only if m m. Note If l and l are vertical, then they are parallel. If l is vertical and l is horizontal (or the other way round), then they are perpendicular to each other.

25 0.8. Pythagoras Theorem, Distance Formula and Circles 7 Example Find equations in general linear form for the two lines passing through the point (3, ) such that one is parallel to the line y 3x + and the other is perpendicular to it. Solution Let l (respectively l ) be the line that passes through the point (3, ) and parallel (respectively perpendicular) to the given line. It is clear that the slope of the given line is 3. Thus the slope of l is 3 and the slope of l is 3. From these, we get the point-slope forms for l and l : y ( ) 3(x 3) and y ( ) (x 3) 3 respectively. Expanding and rearranging terms, we get the following linear forms 3x y 0 and x + 3y for l and l respectively. Exercise 0.7. For each of the following, find an equation of the line satisfying the given conditions. Give your answer in general linear form. (a) Passing through the origin and (, 3). (b) With slope and passing through (5, ). (c) With slope 3 and y-intercept (0, 7). (d) Passing through ( 3, ) and parallel to x y 3 0. (e) Passing through (, 4) and perpendicular to x + 3y 0. (f) Passing through (, ) and perpendicular to the y-axis. 0.8 Pythagoras Theorem, Distance Formula and Circles Pythagoras Theorem Let a, b and c be the (lengths of the) sides of a right-angled triangle where c is the hypotenuse. Then we have a + b c. c b a Figure 0. Distance Formula Let P (x, y ) and Q (x, y ). Then the distance PQ between P and Q is PQ (x x ) + (y y ). Q(x, y ) P(x, y ) Figure 0.3

26 8 Chapter 0. Revision Equation of Circles Let C be the circle with center at C(h, k) and radius r. Then an equation for C is (x h) + (y k) r. (0.8.) Proof Let P(x, y) be any point on the circle. Since the distance from P to the center C is r, using the distance formula, we get (x h) + (y k) r. Squaring both sides yields (0.8.). C(h, k) P(x, y) Figure 0.4 Example Find the center and radius of the circle given by x 4x + y + 6y 0. Solution Using the completing square method, the given equation can be written in the form (0.8.). x 4x + y + 6y (x 4x + 4) + (y + 6y + 9) (x ) + (y + 3) 5 (x ) + (y ( 3)) 5 The center is (, 3) and the radius is 5. FAQ How do we get the number 9 etc (the numbers added to both sides)? Answer We want to find a number (denoted by a) such that (y + 6y + a) is a complete square. That is, y + 6y + a (y + b) (0.8.) for some number b. Expanding the right-side of (0.8.) (do this in your head) and comparing the coefficients of y on both sides, we get b 6, that is, b 3. Hence comparing the constant terms on both sides, we get a b 9. Summary a square of half of the coefficient of y. Exercise 0.8. For each of the following pairs of points, find the distance between them. (a) ( 3, 4) and the origin (b) (4, 0) and (0, 7) (c) (7, 5) and (, 7) (d) (, 9) and (3, ). For each of the following circles, find its radius and center. (a) x + y 4y + 0 (b) x + y + 4x y 4 0 (c) x + y + 4x y For each of the following, find the distance from the given point to the given line. (a) (, 3) and the y-axis (b) the origin and x + y (c) (, ) and x + y 6 0

27 0.9. Parabola Parabola The graph of y ax + bx + c where a 0, is a parabola. The parabola intersects the x-axis at two distinct points if b 4ac > 0. It touches the x-axis (one intersection point only) if b 4ac 0 and does not intersect the x-axis if b 4ac < 0. If a > 0, the parabola opens upward and there is a lowest point (called the vertex of the parabola). If a < 0, the parabola opens downward and there is a highest point (vertex). a > 0 a < 0 Figure 0.5(a) Figure 0.5(b) The vertical line that passes through the vertex is called the axis of symmetry because the parabola is symmetric about this line. To find the vertex, we can use the completing square method to write the equation in the form y a(x h) + k (0.9.) The vertex is (h, k) because (x h) is always non-negative and so if a > 0, then y k and thus (h, k) is the lowest point; if a < 0, then y k and thus (h, k) is the highest point. Example Consider the parabola given by y x + 6x + 5. Find its vertex and axis of symmetry. Solution Using the completing square method, the given equation can be written in the form (0.9.). y x + 6x + 5 y (x + 6x + 9) y (x + 3) 4 y ( x ( 3) ) 4. The vertex is ( 3, 4) and the axis of symmetry is the line given by x 3 (the vertical line that passes through the vertex). FAQ In the above example, the coefficient of x is, what should we do if it is not?

28 0 Chapter 0. Revision Answer To illustrate the procedure, let s consider y x + 3x 4. To rewrite the equation in the form (0.9.), consider the first two terms and rewrite it in the form a ( x + b a x). Exercise 0.9 y ( x + 3 x) 4 ( y x + 3 x + ( 3 ( 3 4) 4 ( y x + 3 x + ( 3 4 y ( x + 3 ) y ( x ) 9 6 ) ) 4 ) 4 ) 4 8. For each of the following parabolas, find its x-intercept(s), y-intercept and vertex. (a) y x + 4x (b) y x + 6x 7 (c) y x + x Systems of Equations A system of two equations in two unknowns x and y can be written as F (x, y) 0 F (x, y) 0. Usually, each equation represents a curve in the coordinate plane. Solving the system means to find all ordered pairs (x 0, y 0 ) such that F (x 0, y 0 ) 0 and F (x 0, y 0 ) 0, that is, to find all points P(x 0, y 0 ) that lies on the intersection of the two curves. To solve a system of two linear equations (with two unknowns x and y) we can use elimination or substitution. Example Solve the following system of equations Solution ax + by + c 0 + ey + f 0, (Elimination) Multiply (0.0.) and (0.0.) by 3 and respectively, we get x + 3y 7 (0.0.) 3x + 5y (0.0.) 6x + 9y (0.0.3) 6x + 0y (0.0.4)

29 0.0. Systems of Equations Subtracting (0.0.3) from (0.0.4), we get y. Substituting y back into (0.0.) or (0.0.) and solving, we get x. The solution to the system is (, ). Remark The point (, ) is the intersection point of the lines given by x + 3y 7 and 3x + 5y. (Substitution) From (0.0.), we get x 7 3y. Substituting into (0.0.), we get ( ) 7 3y 3 + 5y 3(7 3y) + 0y y and we can proceed as in the elimination method. To solve a system in two unknowns, with one linear equation and one quadratic equation ax + by + c 0 + exy + f y + gx + hy + k 0 we can use substitution. From the linear equation, we can express x in terms of y (or vice versa). Substituting into the quadratic equation, we get a quadratic equation in y which can be solved by factorization or by formula. Substituting the value(s) of y back into the linear equation, we get the corresponding value(s) of x. Example Solve the following system of equations x y 4 (0.0.5) x + y 5 (0.0.6) Solution From (0.0.5), we get x 4 + y. Substituting into (0.0.6), we get (4 + y) + y 5 5y + 6y + 0. Solving we get y or y 5. Substituting y into (0.0.5), we get x ; substituting y 5 into (0.0.5), we get x 5. The solutions to the system are (, ) and ( 5, ). 5 Remark If we substitute y into (0.0.6), we get two values of x, one of which should be rejected. The solutions are the intersection points of the line x y 4 and the circle x + y 5. Example Find the point(s) of intersection, if any, of the line and the parabola given by x + y 0 and y x + respectively.

30 Chapter 0. Revision Solution From the equation of the line, we get y x. Putting into the equation of the parabola, we get x x + 0 x + x +. Since 4()() < 0, the above quadratic equation has no solution. Hence the system x + y 0 y x + has no solution, that is, the line and the parabola do not intersect. Exercise 0.0. Consider a rectangle with perimeter 8 cm and diagonal 0 cm. Find the length and width of the rectangle.

31 Chapter Sets, Real Numbers and Inequalities. Sets.. Introduction Idea of definition A set is a collection of objects. This is not a definition because we have not defined what a collection is. If we give a definition for collection, it must involve something that have not been defined. It is impossible to define everything. In mathematics, set is a fundamental concept that cannot be defined. The idea of definition given above describes what a set is using daily language. This helps us understand the meaning of a set. Terminology An object in a set is called an element or a member of the set. To describe sets, we can use listing or description. [Listing] To denote a set with finitely many elements, we can list all the elements of the set and enclose them by braces. For example, {,, 3} is the set which has exactly three elements, namely, and 3. If we want to denote the set whose elements are the first one hundred positive integers, it is impractical to write down all the elements. Instead, we write {,, 3,..., 99, 00}, or simply {,,..., 00}. The three dots... (read and so on ) means that the pattern is repeated, up to the number(s) listed at the end. Suppose in a problem, we consider a set, say {,,..., 00}. We may have to refer to the set later many times. Instead of writing {,,..., 00} repeatedly, we can give it a name by using a symbol to represent the set. Usually, we use small letters (eg. a, b,...) to denote objects and capital letters (eg. A, B,...) to denote sets. For example, we may write Let A {,,..., 00}.

33 .. Sets 5 read the set of all x such that P(x) (is true). For example, the set whose elements are the first one hundred positive integers can be expressed as ( ) {x : x is a positive integer less than 0} In considering property, it is understood that the property applies to a certain collection of objects only. For example, when we say an old person (a person is said to be old if his or her age is 65 or above), the property of being old is applied to people. It is meaningless to say this is an old atom (unless we have a definition which tells whether an atom is old or not). The property of being a positive integer less than 0 is applied to numbers. In this course, we consider real numbers only. The set of all real numbers is denoted by R. In considering the set given in ( ), it is understood that x is a real number. To make this explicit, we write ( ) {x R : x is a positive integer less than 0} read the set of all x belonging to R such that x is a positive integer less than 0. Notation () The set of all real numbers is denoted by R. () The set of all rational numbers is denoted by Q. (3) The set of all integers is denoted by Z. (4) The set of all positive integers is denoted by Z +. (5) The set of all natural numbers is denoted by N. Definition () A rational number is a number that can be written in the form p q () Positive integers together with 0 are called natural numbers. where p and q are integers and q 0. Remark Some authors do not include 0 as natural number. In that case, N means the set of all positive integers. Example () To say that is a natural number, we may write N. () To say that is a rational number, we may write Q. Note: The number is a rational number because it can be written as or 6 3 etc. (3) To say that π is not a rational number, we may write π Q. Note: π ; the rational number is only an approximation to π. 7 7 Definition Let A and B be sets. If every element of A is also an element of B and vice versa, then we say that A and B are equal, denoted by A B. Remark In mathematics, definitions are important. Students who want to take more courses in mathematics must pay attention to definitions. Understand the meaning, give examples, give nonexamples.

34 6 Chapter. Sets, Real Numbers and Inequalities In the definition, the first sentence Let A and B be sets describes the setting. The definition for equality applies to sets only and does not apply to other objects. Of course, we can consider equality of other objects, but it is another definition. In the first sentence Let A and B be sets, the use of plural sets does not mean that A and B are two different sets. It also includes the case where A and B are the same set. The following are alternative ways to say this: Let A and B be set(s). Let A be a set and let B be a set. However, these alternative ways are rather cumbersome and will not be used in most situations. Some students may not be familiar with the use of the word let. It is used very often in mathematics. Consider the following sentences: Let A {,, 3, 4, 5}. Let A be a set. The word let appears in both sentences. However, the meanings of let in the two sentences are quite different. In the first sentence, let means denote whereas in the second sentence, it means suppose. The definition for equality of sets can also be stated in the following ways: Suppose A and B are sets. If every element of A is also an element of B and vice versa, then we say that A and B are equal. If A and B are sets and if every element of A is also an element of B and vice versa, then we say that A and B are equal. The definition can also be stated in a way that the assumption that A and B are sets is combined with the condition for equality of A and B. If every element of a set A is also an element of a set B and vice versa, then we say that A and B are equal. The definition tells that if A and B are sets having the same elements, then A B. Conversely, it also tells that if A and B are sets and A B, then A and B have the same elements because this is the condition to check whether A and B are equal. Some mathematicians give the definition using iff: Let A and B be sets. We say that A and B are equal if and only if every element of A is also an element of B and vice versa. Sometimes, we also give definition of a concept together with its opposite. The following is a definition of equality of sets together with its opposite. In this course, we will use the following format (a) describe the setting; (b) give condition(s) for the concept; (c) give condition(s) for the opposite concept, whenever it is appropriate. Definition Let A and B be sets. If every element of A is also an element of B and vice versa, then we say that

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