BASIC CALCULUS REVIEWER NOTES FOR GRADE 11- STEM STUDENTS FROM 1ST QUARTER AND 2ND QUARTER, Summaries of Mathematics

Download BASIC CALCULUS REVIEWER NOTES FOR GRADE 11- STEM STUDENTS FROM 1ST QUARTER AND 2ND QUARTER and more Summaries Mathematics in PDF only on Docsity! Understanding Basic Calculus S.K. Chung 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 1998-1999 through the second semester of 2006-2007. 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 Mathe- matics. Students who are familiar with the materials may skip this chapter. Chapter 1 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 2 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 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 10 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 soft- wares, 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 http://www.wolfram.com . Contents 0 Revision 1 0.1 Exponents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 0.2 Algebraic Identities and Algebraic Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 0.3 Solving Linear Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4 0.4 Solving Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 0.5 Remainder Theorem and Factor Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 0.6 Solving Linear Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 0.7 Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 0.8 Pythagoras Theorem, Distance Formula and Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 0.9 Parabola . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 0.10 Systems of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 1 Sets, Real Numbers and Inequalities 23 1.1 Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.1.2 Set Operations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 1.2 Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 1.2.1 The Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 1.2.2 Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 1.3 Solving Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 1.3.1 Quadratic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38 1.3.2 Polynomial Inequalities with degrees ≥ 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39 2 Functions and Graphs 43 2.1 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 2.2 Domains and Ranges of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45 2.3 Graphs of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 2.4 Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 2.5 Compositions of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 2.6 Inverse Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 2.7 More on Solving Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 3 Limits 73 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 3.2 Limits of Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 3.3 Limits of Functions at Infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 3.4 One-sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 3.5 Two-sided Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89 3.6 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94 4 Differentiation 103 4.1 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 4.2 Rules for Differentiation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 4.3 Higher-Order Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 2 Chapter 0. Revision 0.2 Algebraic Identities and Algebraic Expressions Identities Let a and b be real numbers. Then we have (1) (a + b)2 = a2 + 2ab + b2 (2) (a − b)2 = a2 − 2ab + b2 (3) (a + b)(a − b) = a2 − b2 Remark The above equalities are called identities because they are valid for all real numbers a and b. Caution In general, (a + b)2 , a2 + b2. Note: (a + b)2 = a2 + b2 if and only if a = 0 or b = 0. Example Expand the following: (1) (√ x + 2 )2 (2) ( x − 5 x )2 (3) (√ x2 + 1 + 7 ) (√ x2 + 1 − 7 ) Solution (1) (√ x + 2 )2 = (√ x )2 + 2 (√ x ) (2) + 22 = x + 4 √ x + 4 (2) ( x − 5 x )2 = x2 − 2(x) ( 5 x ) + ( 5 x )2 = x2 − 10 + 25 x2 (3) (√ x2 + 1 + 7 ) (√ x2 + 1 − 7 ) = (√ x2 + 1 )2 − 72 = ( x2 + 1 ) − 49 = x2 − 48  Example Simplify the following: (1) x2 − x − 6 x2 − 6x + 9 (2) x2 x2 − 1 − 1 (3) 2 x2 + 2x + 1 − 1 x2 − x − 2 (4) ( x − y−1 )−1 (5) 3 + 6 x x + x x + 1 0.2. Algebraic Identities and Algebraic Expressions 3 Solution (1) x2 − x − 6 x2 − 6x + 9 = (x − 3)(x + 2) (x − 3)2 = x + 2 x − 3 (2) x2 x2 − 1 − 1 = x2 − (x2 − 1) x2 − 1 = 1 x2 − 1 (3) 2 x2 + 2x + 1 − 1 x2 − x − 2 = 2 (x + 1)2 − 1 (x + 1)(x − 2) = 2(x − 2) − (x + 1) (x + 1)2(x − 2) = x − 5 (x + 1)2(x − 2) (4) ( x − y−1 )−1 = ( x − 1 y )−1 = ( xy − 1 y )−1 = y xy − 1 (5) 3 + 6 x x + x x + 1 = 3x + 6 x x (x + 1) + x x + 1 = 3x + 6 x x2 + 2x x + 1 = 3(x + 2) x · x + 1 x (x + 2) = 3(x + 1) x2  FAQ What is expected if we are asked to simplify an expression? For example, in (5), can we give 3x + 3 x2 as the answer? Answer There is no definite rule to tell which expression is simpler. For (5), both 3(x + 1) x2 and 3x + 3 x2 are acceptable. Use your own judgment.  4 Chapter 0. Revision Exercise 0.2 1. Expand the following: (a) (2x + 3)2 (b) (3x − y)2 (c) (x + 3y)(x − 3y) (d) (x + 3y)(x + 4y) (e) ( 2 √ x − 3 )2 (f) (√ x + 5 ) (√ x − 5 ) 2. Factorize the following: (a) x2 − 7x + 12 (b) x2 + x − 6 (c) x2 + 8x + 16 (d) 9x2 + 9x + 2 (e) 9x2 − 6x + 1 (f) 5x2 − 5 (g) 3x2 − 18x + 27 (h) 2x2 − 12x + 16 3. Simplify the following: (a) x2 − x − 6 x2 − 7x + 12 (b) x2 + 3x − 4 2 − x − x2 (c) 2x x2 − 1 ÷ 4x2 + 4x x − 1 (d) 1 x + h − 1 x h 0.3 Solving Linear Equations 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 (0.3.1) 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 (0.3.1) is a real number x0 such that F(x0) = 0. Example The equation 2x + 3 = 0 has exactly one solution, namely −3 2 . 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: (1) 2x + 3 = 0 (2) 2x = −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. 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 x2 + 2x − 15 = 0 ⇐⇒ (x + 5)(x − 3) = 0 ⇐⇒ x + 5 = 0 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 x1 = −5 and x2 = 3” which means “there are two solutions −5 and 3 and they are denoted by x1 and x2 respectively”. In Chapter 1, 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 (0.4.1) are given by x = −b ± √ b2 − 4ac 2a . Remark b2 − 4ac is called the discriminant of (0.4.1). (1) If b2 − 4ac > 0, then (0.4.1) has two distinct solutions. (2) If b2 − 4ac = 0, then (0.4.1) has one solution. (3) If b2 − 4ac < 0, then (0.4.1) has no (real) solution. FAQ Why is “(real)” added? Answer When the real number system is enlarged to the complex number system, (0.4.1) has two complex solutions if b2 − 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. (1) 2x2 − 9x + 10 = 0 (2) x2 + 2x + 3 = 0 Solution (1) Using the quadratic formula, we see that the equation has two solutions given by x = 9 ± √ (−9)2 − 4(2)(10) 2(2) = 9 ± 1 4 . 8 Chapter 0. Revision Thus the solutions are 5 2 and 2. (2) Since 22 − 4(1)(3) = −8 < 0, the equation has no solutions.  Example Solve the equation x (x + 2) = x (2x + 3). Solution Expanding both sides, we get x2 + 2x = 2x2 + 3x x2 + x = 0 x (x + 1) = 0 x = 0 or x = −1 The solutions are −1 and 0.  Remark If we cancel the factor x on both sides, we get x + 2 = 2x + 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 + 2) = x (2x + 3) ⇐⇒ x + 2 = 2x + 3 or x = 0 ... Example Find the value(s) of k such that the equation 3x2 + kx + 7 = 0 has only one solution. Solution The given equation has only one solution iff k2 − 4(3)(7) = 0. Solving, we get k = ±√84.  Exercise 0.4 1. Solve the following equations. (a) 4x − 4x2 = 0 (b) 2 + x − 3x2 = 0 (c) 4x (x − 4) = x − 15 (d) x2 + 2 √ 2x + 2 = 0 (e) x2 + 2 √ 2x + 3 = 0 (f) x3 − 7x2 + 3x = 0 2. Find the value(s) of k such that the equation x2 + kx + (k + 3) = 0 has only one solution. 3. Find the positive number such that sum of the number and its square is 210. 0.5 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) = x3 + 3x2 − 2x + 2. Find the remainder when p(x) is divided by x − 2. Solution The remainder is p(2) = 23 + 3(22) − 2(2) + 2 = 18.  Factor Theorem (x − c) is a factor of a polynomial p(x) if and only if p(c) = 0. 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) = x3 + kx2 + x − 6. Suppose that (x + 2) is a factor of p(x). (1) Find the value of k. (2) With the value of k found in (1), factorize p(x). Solution (1) Since ( x − (−2) ) is a factor of p(x), it follows from the Factor Theorem that p(−2) = 0, that is (−2)3 + k(−2)2 + (−2) − 6 = 0. Solving, we get k = 4. (2) Using long division, we get x3 + 4x2 + x − 6 = (x + 2)(x2 + 2x − 3). By inspection, we have p(x) = (x + 2)(x + 3)(x − 1).  FAQ Can we find the quotient (x2 + 2x − 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 (ax2 + bx + c). Thus we have x3 + 4x2 + x − 6 = (x + 2)(ax2 + bx + c) (0.5.1) Comparing the coefficient of x3, we see that a = 1. Similarly, comparing the constant term, we get c = −3. Hence we have x3 + 4x2 + x − 6 = (x + 2)(x2 + bx − 3). To find b, we may compare the x2 term (or the x term) to get 4 = 2 + b, which yields b = 2. Remark The compare coefficient method in fact consists of the following steps: (1) Expand the right side of (0.5.1) to get ax3 + (2a + b)x2 + (2b + c)x + 2c (2) Compare the coefficients of the given polynomial with that obtained in Step (1) to get 1 = a 4 = 2a + b 1 = 2b + c −6 = 2c 12 Chapter 0. Revision (2) 3(x − 2) + 5 > 3x + 7 Solution (1) Using rules for inequalities, we get 2x + 1 > 7(x + 3) 2x + 1 > 7x + 21 1 − 21 > 7x − 2x −20 > 5x −4 > x. The solutions are all the real numbers x such that x < −4, that is, all real numbers less than −4. (2) Expanding the left side, we get 3(x − 2) + 5 = 3x − 1 which is always less than the right side. Thus the inequality has no solution.  Exercise 0.6 1. Solve the following inequalities for x. (a) 1 − x 2 ≥ 3x − 7 3 (b) 2(3 − x) ≤ √3(1 − x) (c) 3x 1 − x + 3 < 0 (d) 2x 2x + 3 > 1 0.7 Lines A linear equation in two unknowns x and y is an equation that can be written in the form ax + by + c = 0 (0.7.1) 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 F(x, y) = 0, (0.7.2) where F is a function (from a collection of ordered pairs into R). Definition An ordered pair (of real numbers) is a pair of real numbers x0, y0 enclosed inside parenthesis: (x0, y0). Remark Two ordered pairs (x0, y0) and (x1, y1) are equal if and only if x0 = x1 and y0 = y1. For example, the ordered pairs (1, 2) and (2, 1) are not equal. Definition A solution to Equation (0.7.2) is an ordered pair (x0, y0) such that F(x0, y0) = 0. Example Consider the equation 2x + 3y − 4 = 0. By direct substitution, we see that (2, 0) is a solution whereas (1, 2) is not a solution. 0.7. Lines 13 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 per- pendicular 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 P1 and P2 are distinct points in the plane, then the ordered pairs corresponding to P1 and P2 are different. There- fore, 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. -1 1 2 3 4 -1 1 2 3 Figure 0.1 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 (x0, y0) 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 ` 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 (1)x + (−1)y + 0 = 0, that is, x − y = 0. Terminology If a line ` is represented by an equation in the form (0.7.3), we say that the equation is a general linear form for `. Remark In Equation (0.7.3), (1) if A = 0, then the equation reduces to y = −C B and its graph is a horizontal line; (2) if B = 0, then the equation reduces to x = −C A and its graph is a vertical line. 14 Chapter 0. Revision Example Consider the line ` given by 2x + 3y − 4 = 0 (0.7.4) For each of the following points, determine whether it lies on ` or not. (1) A = (4,−1) (2) B = (5,−2) Solution (1) Putting (x, y) = (4,−1) into (0.7.4), we get L.S . = 2(4) + 3(−1) − 4 = 1 , 0. Therefore A does not lie on `. (2) Putting (x, y) = (5,−2) into (0.7.4), we get L.S . = 2(5) + 3(−2) − 4 = 0 = R.S . Therefore B lies on `.  Example Consider the line ` given by x + 2y − 4 = 0 (0.7.5) Find the points of intersection of ` 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 ` with the x-axis is (4, 0). • Putting x = 0 into (0.7.5), we get 2y − 4 = 0 from which we obtain y = 2. The point of intersection of ` with the y-axis is (0, 2).  Remark The point (4, 0) and (0, 2) are called the x-intercept and y-intercept of ` 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 conven- tion, the x-intercept is 4 and the y-intercept is 2.  0.8. Pythagoras Theorem, Distance Formula and Circles 17 Example Find equations in general linear form for the two lines passing through the point (3,−2) such that one is parallel to the line y = 3x + 1 and the other is perpendicular to it. Solution Let `1 (respectively `2) be the line that passes through the point (3,−2) and parallel (respectively perpendicular) to the given line. It is clear that the slope of the given line is 3. Thus the slope of `1 is 3 and the slope of `2 is −1 3 . From these, we get the point-slope forms for `1 and `2: y − (−2) = 3(x − 3) and y − (−2) = −1 3 (x − 3) respectively. Expanding and rearranging terms, we get the following linear forms 3x − y − 11 = 0 and x + 3y + 3 = 0 for `1 and `2 respectively.  Exercise 0.7 1. 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 (−2, 3). (b) With slope 2 and passing through (5,−1). (c) With slope −3 and y-intercept (0, 7). (d) Passing through (−3, 2) and parallel to 2x − y − 3 = 0. (e) Passing through (1, 4) and perpendicular to x + 3y = 0. (f) Passing through (1,−1) 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 a2 + b2 = c2. a b c Figure 0.2 Distance Formula Let P = (x1, y1) and Q = (x2, y2). Then the distance PQ between P and Q is PQ = √ (x2 − x1)2 + (y2 − y1)2. P(x1, y1) Q(x2, y2) Figure 0.3 18 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)2 + (y − k)2 = r2. (0.8.1) 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)2 + (y − k)2 = r. Squaring both sides yields (0.8.1).  C(h, k) P(x, y) Figure 0.4 Example Find the center and radius of the circle given by x2 − 4x + y2 + 6y − 12 = 0. Solution Using the completing square method, the given equation can be written in the form (0.8.1). x2 − 4x + y2 + 6y = 12 (x2 − 4x + 4) + (y2 + 6y + 9) = 12 + 4 + 9 (x − 2)2 + (y + 3)2 = 25 (x − 2)2 + (y − (−3))2 = 52 The center is (2,−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 (y2 + 6y + a) is a complete square. That is, y2 + 6y + a = (y + b)2 (0.8.2) for some number b. Expanding the right-side of (0.8.2) (do this in your head) and comparing the coefficients of y on both sides, we get 2b = 6, that is, b = 3. Hence comparing the constant terms on both sides, we get a = b2 = 9. Summary a = square of half of the coefficient of y.  Exercise 0.8 1. 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 (12, 17) (d) (−2, 9) and (3,−1) 2. For each of the following circles, find its radius and center. (a) x2 + y2 − 4y + 1 = 0 (b) x2 + y2 + 4x − 2y − 4 = 0 (c) 2x2 + 2y2 + 4x − 2y + 1 = 0 3. For each of the following, find the distance from the given point to the given line. (a) (−2, 3) and the y-axis (b) the origin and x + y = 1 (c) (1, 2) and 2x + y − 6 = 0 0.9. Parabola 19 0.9 Parabola The graph of y = ax2 + bx + c where a , 0, is a parabola. The parabola intersects the x-axis at two distinct points if b2 − 4ac > 0. It touches the x-axis (one intersection point only) if b2 − 4ac = 0 and does not intersect the x-axis if b2 − 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 Figure 0.5(a) a < 0 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)2 + k (0.9.1) The vertex is (h, k) because (x − h)2 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 = x2 + 6x + 5. Find its vertex and axis of symmetry. Solution Using the completing square method, the given equation can be written in the form (0.9.1). y = x2 + 6x + 5 y = (x2 + 6x + 9) − 9 + 5 y = (x + 3)2 − 4 y = ( x − (−3) )2 − 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 x2 is 1, what should we do if it is not 1? 22 Chapter 0. Revision Solution From the equation of the line, we get y = 1 − x. Putting into the equation of the parabola, we get 1 − x = x2 + 2 0 = x2 + x + 1. Since ∆ = 12 − 4(1)(1) < 0, the above quadratic equation has no solution. Hence the system x + y − 1 = 0 y = x2 + 2 has no solution, that is, the line and the parabola do not intersect.  Exercise 0.10 1. Consider a rectangle with perimeter 28 cm and diagonal 10 cm. Find the length and width of the rectangle. Chapter 1 Sets, Real Numbers and Inequalities 1.1 Sets 1.1.1 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 collec- tion, 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, {1, 2, 3} is the set which has exactly three elements, namely 1, 2 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 {1, 2, 3, . . . , 99, 100}, or simply {1, 2, . . . , 100}. 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 {1, 2, . . . , 100}. We may have to refer to the set later many times. Instead of writing {1, 2, . . . , 100} 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 = {1, 2, . . . , 100}.” 24 Chapter 1. Sets, Real Numbers and Inequalities which means that the set {1, 2, . . . , 100} is given the “name” A. If we want to refer to the set later, we can just write A. For example, • “Let A = {1, 2, . . . , 100}. Then 100 is an element of A, but 101 is not an element of A.” If we consider another set, say {1, 2, 3, 4, 5} and want to give it a name, we must not use the symbol A again, because in the problem, A always means the set {1, 2, . . . , 100}. For example, • “Let A = {1, 2, . . . , 100}. Let B = {1, 2, 3, 4, 5}. Then every element of B is also an element of A. But there are elements of A that are not elements of B.” Remark The equality sign “=” can be used in several ways as the following examples illustrate. (1) 1 + 2 = 3. (2) x2 + 1 = 5. (3) Let A = {1, 2, 3}. The equality sign in (1) means equality of two quantities: the quantity on the left and the quantity on the right are equal. The equality sign in (2) is an equality in an equation. It is true when x = 2 (for example) and it is not true when x = 1 (for example). Instead of using the equality sign, some authors use “==”. The equation in (2) may be written as (2′) x2 + 1 == 5. The equality sign in (3) has a different meaning. The sentence in (3) means that the set {1, 2, 3} is denoted by A. The symbol “=” assigns a name to an object (a set is also an object). The name is written on the left side and the object on the right side. Instead of using the equality sign, some authors use the symbol “:=”. The sentence in (3) may be written as (3′) Let A := {1, 2, 3}. In this course, we will not use the notations “:=” and “==”. Readers can determine the meaning of “=” from the context. Notation Given an object x and a set A, either x is an element of A or x is not an element of A. (1) If x is an element of A, we write x ∈ A (read “x belongs to A”). (2) If x is not an element of A, we write x < A (read “x does not belong to A”). There is a set that has no element. It is called the empty set, denoted by ∅. This is a Scandinavian letter, a zero 0 together with a slash /. Definition The set that has no element is called the empty set and is denoted by ∅. Remark Because the empty set has no element, if we list all the elements of it and enclose “them” by braces, we get { }. This is an alternative notation for the empty set. [Description] Another way to denote a set is to describe a common property of the elements of the set, using the following notation: {x : P(x)} or {x | P(x)} 1.1. Sets 27 A and B are equal, denoted by A = B. Otherwise, we say that A and B are unequal, denoted by A , B. Example Let A = {1, 3, 5, 7, 9} and let B = {x ∈ Z : x is a positive odd number less than 10}. Then we have A = B, that is, A and B are equal. This is because every element of A is also an element of B and vice versa. Recall: Z is the set of all integers. Thus B is the set of all integers that are positive, odd and less than 10. Remark To prove that the sets A and B in the above example are equal, we check whether the condition given in the definition is satisfied. This is called proof by definition. Example Let A = {1, 3, 5, 7, 9} and let B = {x ∈ Z+ : x is a prime number less than 10}. Then we have A , B. Recall: Z+ is the set of all positive integers. Proof The number 9 is an element of A, but it is not an element of B. Therefore, it is not true that every element of A is also an element of B. Hence we have A , B.  FAQ In the above two examples, the assertions are quite obvious. Do we need to prove them? Answer Sometimes, mathematicians also write “obvious” in proofs of theorems. To some people, a result may be obvious; but, it may not be obvious to other people. If you say obvious, make sure that it is really obvious— if your classmates ask you why, you should be able to explain to them. It is impractical to explain everything. In proving theorems or giving solutions to examples, reasons that are “obvious” will not be given. When you answer questions, you should use your own judgment.  Remark Because it is impractical (in fact, impossible) to explain everything, discussion below will not be so detail as that above. If you don’t understand a concept, read the definition again. Try different ways to understand it. Relate it with what you have learnt. Guess what the meaning is. See whether your guess is correct if you apply it to examples . . . Example Let A = {1, 2, 3} and let B = {1, 3, 2}. Then we have A = B. Proof Obvious (use definition).  The above example shows that in listing elements of a set, order is not important. It should also be noted that in listing elements, there is no need to repeat the elements. For example, {1, 2, 3, 2, 1} and {1, 2, 3} are the same set. Definition Let A and B be sets. If every element of A is also an element of B, then we say that A is a subset of B, denoted by A ⊆ B. Otherwise, we say that A is not a subset of B, denoted by A * B. Note (1) A ⊆ A. (2) A = B if and only if A ⊆ B and B ⊆ A. (3) A * B means that there is at least one element of A that is not an element of B. 28 Chapter 1. Sets, Real Numbers and Inequalities Remark Instead of A ⊆ B, some authors use A ⊂ B to denote A is a subset of B. Example Let A = {1, 2, 3, 4, 5}, B = {1, 3, 5} and C = {2, 4, 6}. Then we have B ⊆ A and C * A. The relation between A, B and C can be described by the dia- gram shown in Figure 1.1. 1 2 3 6 4 5 Figure 1.1 FAQ For the given sets A, B and C, we also have the following: (1) A * B (2) A * C (3) C * B (4) B * C Why are they omitted? Answer Good and correct observation. Given three sets, there are six ways to pair them up. The example just illustrates the meaning of ⊆ and *.  1.1.2 Set Operations Definition Let A and B be sets. (1) The intersection of A and B, denoted by A ∩ B, is the set whose elements are those belonging to both A and B, that is, A ∩ B = {x : x ∈ A and x ∈ B}. (2) The union of A and B, denoted by A ∪ B, is the set whose elements are those belonging to either A or B or both A and B, that is A ∪ B = {x : x ∈ A or x ∈ B}. Remark In mathematics, “P or Q” means “either P or Q or both P and Q”. Example Let A = {2, 3, 5}, B = {2, 5, 6, 8} and C = {1, 2, 3}. Find the following sets. (1) A ∩ B (2) A ∪ B (3) (A ∩ B) ∩C (4) A ∩ (B ∩C) Solution (1) A ∩ B = {2, 5} (2) A ∪ B = {2, 3, 5, 6, 8} (3) (A ∩ B) ∩C = {2, 5} ∩ {1, 2, 3} = {2} 1.1. Sets 29 (4) A ∩ (B ∩C) = {2, 3, 5} ∩ {2} = {2}  Note Given any sets A, B and C, we always have (A ∩ B) ∩C = A ∩ (B ∩C) and (A ∪ B) ∪C = A ∪ (B ∪C). Thus we may write A ∩ B ∩C and A ∪ B ∪C without ambiguity. We say that set intersection and set union are associative. Definition Let A and B be sets. The relative complement of B in A, denoted by A\B or A−B (read “A setminus (or minus) B”), is the set whose elements are those belonging to A but not belonging to B, that is, A \ B = {x ∈ A : x < B}. Example Let A = {a, b, c} and B = {c, d, e}. Then we have A \ B = {a, b}. For each problem, we will consider a set that is “large” enough, containing all objects under consideration. Such a set is called a universal set and is usually denoted by U. In this case, all sets under consideration are subsets of U and they can be written in the form {x ∈ U : P(x)}. Example In considering addition and subtraction of whole numbers (0, 1, 2, 3, 4, . . .), we may use Z (the set of all integers) as a universal set. (1) The set of all positive even numbers can be written as {x ∈ Z : x > 0 and x is divisible by 2}. (2) The set of all prime numbers can be written as {x ∈ Z : x > 0 and x has exactly two divisors}. Definition Let U be a universal set and let B be a subset of U. Then the set U \ B is called the complement of B (in U) and is denoted by B′ (or Bc). Example Let U = Z+, the set of all positive integers. Let B be the set of all positive even numbers. Then B′ is the set of all positive odd numbers. Example Let U = {1, 2, 3, . . . , 12} and let A = {x ∈ U : x is a prime number} B = {x ∈ U : x is an even number} C = {x ∈ U : x is divisible by 3}. Find the following sets. (1) A ∪ B (2) A ∩C (3) B ∩C (4) (A ∪ B) ∩C (5) (A ∩C) ∪ (B ∩C) 32 Chapter 1. Sets, Real Numbers and Inequalities (a) A − B = A′ ∩ B (b) (A ∪ B) ∩C = A ∪ (B ∩C) (c) (A′ ∪ B′) ∩ B = B − A A statement above is true means that it is true for all possible choices of A, B, C and U. To show that the statement is false, it is enough to give a counterexample. To show that it is true, you can draw a Venn diagram to convince yourself; but to be more rigorous, you should use formal mathematical logic. 1.2 Real Numbers 1.2.1 The Number Systems (1) The numbers 0, 1, 2, 3, . . . are called natural numbers. The set of all natural numbers is denoted by N, that is, N = {0, 1, 2, 3, . . .}. Remark The three dots “. . .” means that the pattern is repeated indefinitely. FAQ In some books, N is defined to be {1, 2, 3, . . .}. Which one should we follow? Answer Some authors do not include 0 in N. This is just a convention; once we know the definition, it will not cause any problem.  (2) The numbers 0, 1,−1, 2,−2, . . . are called integers. The set of all integers is denoted by Z, that is, Z = {· · · ,−3,−2,−1, 0, 1, 2, 3, · · · }. (3) Numbers in the form p q where p, q ∈ Z and q , 0 are called rational numbers. The set of all rational numbers is denoted by Q, that is, Q = { p q : p, q are integers, and q , 0 } . Note Z ⊆ Q, that is, every integer is a rational number. For example, the integer 2 can be written as 2 1 and is therefore a rational number. All rational numbers can be represented by decimal numbers that terminate, such as 3 4 = 0.75, or by non-terminating but repeating decimals, such as 4 11 = 0.363636 · · · . Numbers that can be represented by non-terminating and non-repeating decimals are called irrational numbers. For example, π and √ 2 are irrational numbers. The following shows the first 50 decimals of π: π = 3.14159265358979323846264338327950288419716939937511 . . . Remark The proof for the fact that π is irrational is difficult. (4) Rational numbers together with irrational numbers are called real numbers. The set of all real numbers is denoted by R. 1.2. Real Numbers 33 In R, we have the algebraic operations +,× (and −,÷ also) as well as binary relations <,≤, >,≥. Numbers greater than (respectively smaller than) 0 are called positive (respectively negative). Real Number Line Real numbers can be represented by points on a line, called the real number line. >| −1 | 0 | 1 | 2 Figure 1.4 Notation The following nine types of subsets of R are called intervals: [a, b] = {x ∈ R : a ≤ x ≤ b} (1.2.1) (a, b) = {x ∈ R : a < x < b} (1.2.2) [a, b) = {x ∈ R : a ≤ x < b} (1.2.3) (a, b] = {x ∈ R : a < x ≤ b} (1.2.4) [a,∞) = {x ∈ R : a ≤ x} (1.2.5) (a,∞) = {x ∈ R : a < x} (1.2.6) (−∞, b] = {x ∈ R : x ≤ b} (1.2.7) (−∞, b) = {x ∈ R : x < b} (1.2.8) (−∞,∞) = R (1.2.9) where a and b are real numbers with a < b and∞ and −∞ (read “infinity” and “minus infinity”) are just symbols but not real numbers. FAQ What are the meaning of∞ and −∞? Answer Intuitively, you may imagine that there is a point, denoted by ∞, very far away on the right (and −∞ on the left). So (a,∞) is the set whose elements are the points between a and ∞, that is, real numbers greater than a.  Remark The notation (a, b), where a < b, has two different meanings. It denotes an ordered pair as well as an interval. To avoid ambiguity, some authors use ]a, b[ to denote the open interval {x ∈ R : a < x < b}. In this course, we will not use this notation. Readers can determine the meaning from the context. Terminology • Intervals in the form (a, b), [a, b], (a, b] and [a, b) are called bounded intervals and those in the form (−∞, b), (−∞, b], (a,∞), [a,∞) and (−∞,∞) are called unbounded intervals. • Intervals in the form (a, b), (−∞, b), (a,∞) and (−∞,∞) are called open intervals. For each of such intervals, the endpoint(s), if there is any, does not belong to the interval. • Intervals in the form [a, b], (−∞, b], [a,∞) and (−∞,∞) are called closed intervals. For each of such intervals, the endpoint(s), if there is any, belongs to the interval. • Intervals in the form [a, b] are called closed and bounded intervals. 34 Chapter 1. Sets, Real Numbers and Inequalities • A set {a} with exactly one element of R is called a degenerated interval (its length is 0). • Some authors also include ∅ as an interval (called the empty interval). In this course, an interval means a nonempty, non-degenerated interval, that is, an infinite subset of R that can be written in the form (1.2.1), (1.2.2), (1.2.3), (1.2.4), (1.2.5), (1.2.6), (1.2.7), (1.2.8) or (1.2.9). Example For each of the following pairs of intervals A and B, (1) A = [1, 5] and B = (3, 10] (2) A = [−2, 3] and B = (7, 11] (3) A = [−7,−2) and B = [−2,∞) • determine whether it is (i) an open interval, (ii) a closed interval , (iii) a bounded interval; • find A ∩ B and determine whether it is an interval. • find A ∪ B and determine whether it is an interval. Solution (1) Both A and B are not open intervals. A is a closed interval but B is not a closed interval. Both A and B are bounded intervals. A ∩ B = (3, 5]; it is an interval. A ∪ B = [1, 10]; it is an interval. (2) Both A and B are not open intervals. A is a closed interval but B is not a closed interval. Both A and B are bounded intervals. A ∩ B = ∅; it is not an interval. A ∪ B = [−2, 3] ∪ (7, 11]; it is not an interval. (3) Both A and B are not open intervals. B is a closed interval but A is not a closed interval. A is a bounded interval but B is not a bounded interval. A ∩ B = ∅; it is not an interval. A ∪ B = [−7,∞); it is an interval.  1.2.2 Radicals Definition (1) Let a and b be real numbers and let q be a positive integer. If aq = b, we say that a is a qth root of b. Example (a) −2 is the cube root of −8. (b) 3 and −3 are the square roots of 9. 1.3. Solving Inequalities 37 1.3 Solving Inequalities An inequality in one unknown x can be written in one of the following forms: (1) F(x) > 0 (2) F(x) ≥ 0 (3) F(x) < 0 (4) F(x) ≤ 0 where F is a function from a subset of R into R. Definition Consider an inequality in the form F(x) > 0 (the other cases can be treated similarly). (1) A real number x0 satisfying F(x0) > 0 is called a solution to the inequality. (2) The set of all solutions to the inequality is called the solution set to the inequality. To solve an inequality means to find all the solutions to the inequality, or equivalently, to find the solution set. In this section, we consider polynomial inequalities anxn + an−1xn−1 + · · · + a1x + a0 < 0 (or > 0, or ≤ 0, or ≥ 0) (1.3.1) where n ≥ 1 and an , 0. When n = 1, (1.3.1) is a linear inequality. A revision for solving linear inequalities is given in Chapter 0. In the following examples, we consider several linear inequalities simultaneously. Example Find the solution set to the following compound inequality: 1 ≤ 3 − 2x ≤ 9 Solution The inequality means 1 ≤ 3 − 2x and 3 − 2x ≤ 9. Solving them separately, we get 2x ≤ 2 x ≤ 1 and −6 ≤ x −3 ≤ x. The solution set is {x ∈ R : x ≤ 1 and − 3 ≤ x} = {x ∈ R : −3 ≤ x ≤ 1}.  Remark Using interval notation, the solution set can be written as [−3, 1]. Example Find the solution set to the following: 2x + 1 < 3 and 3x + 10 < 4. Give your answer using interval notation. Solution Solving the inequalities separately, we get 2x < 2 x < 1 and 3x < −6 x < −2. 38 Chapter 1. Sets, Real Numbers and Inequalities Therefore, we have solution set = {x ∈ R : x < 1 and x < −2} = {x ∈ R : x < −2} = (−∞,−2).  Example Find the solution set to the following: 2x + 1 > 9 and 3x + 4 < 10 Solution Solving the inequalities separately, we get 2x > 8 x > 4 and 3x < 6 x < 2. The solution set is {x ∈ R : x > 4 and x < 2} = ∅.  1.3.1 Quadratic Inequalities A quadratic inequality (in one unknown) is an inequality that can be written in the form ax2 + bx + c < 0 (or > 0, or ≤ 0, or ≥ 0) (1.3.2) where a , 0. This corresponds to n = 2 in (1.3.1). We use an example to describe three methods for solving quadratic inequalities. The first two methods make use of the following properties of real numbers. (1) α > 0 and β > 0 =⇒ α · β > 0 (2) α < 0 and β < 0 =⇒ α · β > 0 (3) α > 0 and β < 0 =⇒ α · β < 0 From these we get (4) α · β > 0 ⇐⇒ (α > 0 and β > 0) or (α < 0 and β < 0) (5) α · β < 0 ⇐⇒ (α > 0 and β < 0) or (α < 0 and β > 0) Example Find the solution set to the inequality x2 + 2x − 15 > 0. Solution (Method 1) First we factorize the quadratic polynomial: x2 + 2x − 15 > 0 (x + 5)(x − 3) > 0, and then apply Property (4): (x + 5 > 0 and x − 3 > 0) or (x + 5 < 0 and x − 3 < 0) (x > −5 and x > 3) or (x < −5 and x < 3) x > 3 or x < −5 The solution set is {x ∈ R : x < −5 or x > 3} = (−∞,−5) ∪ (3,∞). 1.3. Solving Inequalities 39 (Method 2) By factorization, we have L.S . = (x + 5)(x − 3). The left-side is zero when x = −5 or 3. These two points divide the real number line into three intervals: (−∞,−5), (−5, 3), (3,∞). In the following table, the first two rows give the signs of (x + 5) and (x − 3) on each of these intervals. Hence, using Properties (1), (2) and (3), we obtain the signs of (x + 5)(x − 3) in the third row. x < −5 x = −5 −5 < x < 3 x = 3 x > 3 x + 5 − 0 + + + x − 3 − − − 0 + (x + 5)(x − 3) + 0 − 0 + The solution set is (−∞,−5) ∪ (3,∞). Remark To determine the sign of (x + 5), first we note that it is 0 when x = −5. Since (x + 5) increases as x increases, it is positive when x > −5 and negative when x < −5. (Method 3) The graph of y = x2 +2x−15 is a parabola opening upward and it cuts the x-axis at x1 = −5 and x2 = 3. To solve the inequality x2 + 2x − 15 > 0 means to find all x such that the corresponding points on the parabola has y-coordinates greater than 0. From the graph, we see that the parabola is above the x-axis if and only if x < −5 or x > 3. Therefore, the solution set is (−∞,−5) ∪ (3,∞). -6 -4 -2 2 4 -15 -5 5 y = x2 + 2x − 15 Figure 1.5  1.3.2 Polynomial Inequalities with degrees ≥ 3 In this section, we consider polynomial inequalities (1.3.1) of degree n ≥ 3. To solve such polynomial inequal- ities, for example p(x) > 0, we can use methods similar to that for quadratic inequalities. The first step is to factorize p(x). Example Factorize the polynomial p(x) = x3 + 3x2 − 4x − 12. Solution First we try to find a factor of the form (x − c) where c is an integer. For this, we try c = ±1,±2,±3,±4,±6,±12. Direct substitution gives p(2) = 0 and so (x − 2) is a factor of p(x). Using long division, we obtain x3 + 3x2 − 4x − 12 = (x − 2)(x2 + 5x + 6) and then using inspection we get x3 + 3x2 − 4x − 12 = (x − 2)(x + 2)(x + 3).  In the above procedure, we make use of the following 42 Chapter 1. Sets, Real Numbers and Inequalities Chapter 2 Functions and Graphs 2.1 Functions Informal definition Let A and B be sets. A function from A into B, denoted by f : A −→ B, is a “rule” that assigns to each element of A exactly one element of B. Remark If the sets A and B are understood (or are not important for the problem under consideration), instead of saying “a function from A into B”, we simply say “a function” and instead of writing f : A −→ B, we simply write f . Terminology and Notation The sets A and B are called the domain and codomain of f respectively. The domain of f is denoted by dom ( f ). FAQ In the above informal definition, what is the meaning of a rule? Answer It is difficult to tell what a rule is. The above informal definition describes the idea of a function. There is a rigorous definition. However, it involves more definitions and notations. Interested readers may consult books on set theory or foundation of mathematics.  Notation & Terminology Let f be a function. For each x belonging to the domain of f , the corresponding element (in the codomain of f ) assigned by f is denoted by f (x) and is called the image of x under f . Remark Some people write f (x) to denote a function. This notation may be misleading because it also means an image. However, sometimes for convenience, such notations are used. For example, we write x2 to denote the square function, that is, the function f (from R into R) given by f (x) = x2. In this course, most of the functions we consider are functions whose domains and codomains are subsets of R. A variable that represents the “input numbers” for a function is called an independent variable. A variable that represents the “output numbers” is called a dependent variable because its value depends on the value of the independent variable. Example Consider the function f : R −→ R given by f (x) = x2 + 2. 44 Chapter 2. Functions and Graphs We may also write y = x2 + 2 to represent this function. For each input x, the function gives exactly one output x2 + 2, which is y. If x = 3, then y = 11; if x = 6, then y = 38 etc. The independent variable is x and the dependent variable is y. Example Let g(x) = x2 − 3x + 7. Find the following: (1) g(10) (2) g(a + 1) (3) g(r2) (4) g(x + h) (5) g(x + h) − g(x) h Solution (1) g(10) = 102 − 3(10) + 7 = 77 (2) g(a + 1) = (a + 1)2 − 3(a + 1) + 7 = (a2 + 2a + 1) − 3a − 3 + 7 = a2 − a + 5 (3) g(r2) = (r2)2 − 3(r2) + 7 = r4 − 3r2 + 7 (4) g(x + h) = (x + h)2 − 3(x + h) + 7 = x2 + 2xh + h2 − 3x − 3h + 7 (5) g(x + h) − g(x) h = [(x + h)2 − 3(x + h) + 7] − (x2 − 3x + 7) h = (x2 + 2xh + h2 − 3x − 3h + 7) − (x2 − 3x + 7) h = 2xh + h2 − 3h h = 2x + h − 3  Exercise 2.1 1. Let f (x) = x − 5 x2 + 4 . Find the following: (a) f (2) (b) f (3.5) (c) f (a + 1) (d) f ( √ a) (e) f (a2) (f) f (a) + f (1) 2. Let f (x) = x x + 1 and g(x) = √ x − 1. Find the following: (a) f (1) + g(1) (b) f (2)g(2) (c) f (3) g(3) (d) f (a − 1) + g(a + 1) (e) f (a2 + 1)g(a2 + 1) 2.2. Domains and Ranges of Functions 47 Solve for x. x2 = y − 2 x = ±√ y − 2. Note that x can be solved if and only if y − 2 ≥ 0. The range of f is {y ∈ R : y − 2 ≥ 0} = {y ∈ R : y ≥ 2} = [2,∞). Alternatively, to see that the range is [2,∞), we may use the graph of y = x2 + 2 which is a parabola. The lowest point (vertex) is (0, 2). For any y ≥ 2, we can always find x ∈ R such that f (x) = y. -3 -2 -1 1 2 3 2 4 6 8 10 Figure 2.1 (2) Put y = g(x) = 1 x − 2 . Solve for x. y = 1 x − 2 x − 2 = 1 y x = 1 y + 2. Note that x can be solved if and only if y , 0. The range of g is {y ∈ R : y , 0} = R \ {0}. -1 1 2 3 -4 -2 2 4 Figure 2.2 (3) Put y = h(x) = √ 1 + 5x. Note that y cannot be negative. Solve for x. y = √ 1 + 5x, y ≥ 0 y2 = 1 + 5x, y ≥ 0 x = y2 − 1 5 , y ≥ 0. Note that x can always be solved for every y ≥ 0. The range of h is {y ∈ R : y ≥ 0} = [0,∞). Remark y = √ 1 + 5x =⇒ y2 = 1 + 5x but the converse is true only if y ≥ 0. 1 2 3 1 2 3 4 Figure 2.3 Example Let f (x) = √ x + 7 − √ x2 + 2x − 15. Find the domain of f . Solution Note that f (x) is defined if and only if x + 7 ≥ 0 and x2 + 2x − 15 ≥ 0. Solve the two inequalities separately: • x + 7 ≥ 0 x ≥ −7; 48 Chapter 2. Functions and Graphs • x2 + 2x − 15 ≥ 0 (x + 5)(x − 3) ≥ 0 x < −5 x = −5 −5 < x < 3 x = 3 x > 3 x − 3 − − − 0 + x + 5 − 0 + + + (x − 3)(x + 5) + 0 − 0 + thus, x ≤ −5 or x ≥ 3. Therefore, we have dom ( f ) = {x ∈ R : x ≥ −7 and (x ≤ −5 or x ≥ 3)} = {x ∈ R : (x ≥ −7 and x ≤ −5) or (x ≥ −7 and x ≥ 3)} = {x ∈ R : −7 ≤ x ≤ −5 or x ≥ 3} = [−7,−5] ∪ [3,∞).  Example Let f (x) = 2x + 1 x2 + 1 . Find the range of f . Solution Put y = f (x) = 2x + 1 x2 + 1 . Solve for x. y = 2x + 1 x2 + 1 yx2 + y = 2x + 1 yx2 − 2x + (y − 1) = 0 x = 2 ± √ 4 − 4y(y − 1) 2y if y , 0, x = −1 2 if y = 0, = 1 ± √ 1 − y2 + y y if y , 0. Combining the two cases, we see that x can be solved if and only if 1 − y2 + y ≥ 0, that is, y2 − y − 1 ≤ 0. The range of f is {y ∈ R : y2 − y − 1 ≤ 0}. To solve the inequality y2 − y − 1 ≤ 0, first we find the zero of the left-side by quadratic formula to get 1 ± √5 2 . By the factor theorem and comparing coefficient of y2, we see that y2 − y − 1 = ( y − 1 − √5 2 ) ( y − 1 + √ 5 2 ) y < 1−√5 2 y = 1−√5 2 1−√5 2 < y < 1+ √ 5 2 y = 1+ √ 5 2 y > 1+ √ 5 2 y − 1−√5 2 − 0 + + + y − 1+ √ 5 2 − − − 0 + y2 − y − 1 + 0 − 0 + From the table, we see that ran ( f ) = { y ∈ R : 1 − √5 2 ≤ y ≤ 1 + √ 5 2 } = [1 − √5 2 , 1 + √ 5 2 ]  2.3. Graphs of Equations 49 Remark To solve the inequality y2 − y − 1 ≤ 0, we may also use graphical method: The figure shown is the graph of z = y2−y−1, where the horizontal and vertical axes are the y-axis and the z-axis respectively. Figure 2.4 Exercise 2.2 1. For each of the following functions f , find its domain. (a) f (x) = x2 − 5 (b) f (x) = 2 5x + 6 (c) f (x) = 1 x2 − 5 (d) f (x) = 1 x2 − 2x − 3 (e) f (x) = 1√ 2x − 3 (f) f (x) = 1 1 − 2x − √x + 3 (g) f (x) = 3 1 − x2 + √ 2x + 5 (h) f (x) = 1√ x2 + 3x − 10 2. For each of the following functions f , find its range. (a) f (x) = x2 − 5 (b) f (x) = x2 − 2x − 3 (c) f (x) = 2 5x + 6 (d) f (x) = 3 − 1 2x − 1 (e) f (x) = 1√ 2x − 3 (f) f (x) = 1 x2 − 5 (g) f (x) = 1 x2 − 2x − 3 3. Consider a rectangle with perimeter 28 (units). Let the width of the rectangle be w (units) and let the area of the region enclosed by the rectangle be A (square units). Express A as a function of w. State the domain of A and find the range of A. 2.3 Graphs of Equations Recall that an ordered pair of real numbers is denoted by (x0, y0) where x0 and y0 are real numbers. The set of all ordered pairs is denoted byR2 (read “R two”). The superscript 2 indicates that elements inR2 are represented by two real numbers. Since an ordered pair of real numbers represents a point in the coordinate plane, R2 can be identified with the plane. Let f : A −→ R be a function where A ⊆ R2. Each element in the domain of f is an ordered pair (x, y) of real numbers. Its image under f is denoted by f ( (x, y) ) , or simply f (x, y). Functions whose domains are subsets of R2 are called functions of two variables. Example Let f : R2 −→ R be the function given by f (x, y) = x + y2. Then we have (1) f (1, 2) = 1 + 22 = 5 (2) f (2, 1) = 2 + 12 = 3 52 Chapter 2. Functions and Graphs Example The graph of y = x3 is symmetric about the origin. Figure 2.9 We close this section with the following example of finding intersection of two curves (in fact, one is a line). This is the same as solving a system of two equations in two unknowns. Example Let ℰ and ℒ be the ellipse and the line given by 2x2 + y2 = 6 and x + 2y − 3 = 0 respectively. Find ℰ ∩ ℒ. Solution We need to solve the following system: 2x2 + y2 = 6 (2.3.4) x + 2y − 3 = 0 (2.3.5) From (2.3.5), we get x = 3 − 2y. Putting into (2.3.4) and solving 2(3 − 2y)2 + y2 = 6 2(9 − 12y + 4y2) + y2 = 6 9y2 − 24y + 12 = 0 3(y − 2)(3y − 2) = 0 we get y = 2 or y = 2 3 . Substitute back into (2.3.5), we get (x, y) = (−1, 2) or (5 3 , 2 3 ). Therefore we have ℰ ∩ ℒ = { (−1, 2), (5 3 , 2 3 ) } .  Exercise 2.3 1. Consider the graph of 2x2 + 3y2 = 4 (which is an ellipse). Find its x- and y-intercepts. 2. Suppose the graph of y = ax2 + bx + c has x-intercepts (2, 0) and (−3, 0) and y-intercept (0,−6). Find a, b and c. 3. Consider the graph of y = x2 + 4x + 5. (a) Find its x- and y-intercepts. (b) Show that the graph lies entirely above the x-axis. 4. Let C = {(x, y) ∈ R2 : x2 + y2 = 5}, E = {(x, y) ∈ R2 : x2 + 2y2 = 6} and L = {(x, y) ∈ R2 : 2x + y− 3 = 0}. Find the following: (a) L ∩C (b) L ∩ E (c) C ∩ E 5. Let C = {(x, y) ∈ R2 : x2 + y2 = 1} and L = {(x, y) ∈ R2 : ax + y = 2} where a is a constant. Find the values of a such that C ∩ L is a singleton (that is, a set with only one element). 2.4. Graphs of Functions 53 2.4 Graphs of Functions Let f : A −→ R be a function where A ⊆ R. The graph of f is the following subset of R2: { (x, y) ∈ R2 : x ∈ A and y = f (x) } . Example (1) Constant Functions A constant function is a function f that is given by f (x) = c, where c is a constant (a real number). The domain of every constant function is R. The range is a singleton: {c}. The graph is a horizontal line whose y-intercept is (0, c). c Figure 2.10 Remark Let f (x) = x0. Note that for all x , 0, we have f (x) = 1 and that f (0) is undefined. So there is a small difference between f and the constant function 1 whose domain is R. However, for convenience, we treat the function x0 as the constant function 1. In the above discussion, we use the symbol 1 to represent the function with domain and codomain equal to R and assigning every x ∈ R to the number 1. Thus the symbol 1 has two different meanings. It may be a function or a number. This abuse of notation is sometimes used in mathematics. Readers can determine the meaning from the context. (2) Linear Functions A linear function is a function f given by f (x) = ax + b, where a and b are constants and a , 0. The domain of every linear function is R. The range is also R (note that a is assumed to be non-zero). The graph is a line with slope a and y-intercept (0, b). b Figure 2.11 (3) Quadratic Functions A quadratic function is a function f given by f (x) = ax2 + bx + c, where a, b and c are constants and a , 0. The domain of every quadratic function is R. The range is [k,∞) if a > 0 and (−∞, k] if a < 0 where k is the y-coordinate of the vertex. The graph is a parabola which opens upward if a > 0 and downward if a < 0. 54 Chapter 2. Functions and Graphs a > 0 Figure 2.12(a) a < 0 Figure 2.12(b) Remark Besides using the completing square method to find the vertex, we can also use differentiation (see Chapter 5). (4) Polynomial Functions A function f given by f (x) = anxn + an−1xn−1 + · · · + a1x + a0, where a0, a1, . . . , an are constants with an , 0, is called a polynomial function of degree n. If n = 0, f is a constant function. If n = 1, f is a linear function. If n = 2, f is a quadratic function. Example Let f (x) = x3 − 3x2 + x − 1. The graph of f is shown in Figure 2.13. In Chapter 5, we will discuss how to sketch graphs of poly- nomial functions. -1 1 2 -6 -4 -2 2 Figure 2.13 The domain of every polynomial function f is R. There are three possibilities for the range. (a) If the degree is odd, then ran ( f ) = R. (b) If the degree is even and positive, then (i) ran ( f ) = [k,∞) if an > 0; (ii) ran ( f ) = (−∞, k] if an < 0, where k is the y-coordinate of the lowest point for case (i), or the highest point for case (ii), of the graph. Remark The constant function 0 is also considered to be a polynomial function. However, its degree is assigned to be −∞ (for convenience of a rule for degree of product of polynomials). (5) Rational Functions A rational function is a function f in the form f (x) = p(x) q(x) , where p and q are polynomial functions. 2.4. Graphs of Functions 57 Remark Let a be a positive constant. • The graph of y = f (x) + a can be obtained from that of y = f (x) by moving it a units up. • The graph of y = f (x) − a can be obtained from that of y = f (x) by moving it a units down. (b) Note that √ x − 2 is defined for x ≥ 2 only. The graph of y = √ x − 2 is obtained by moving that of y = √ x two units to the right. 1 2 3 4 5 6 1 2 Figure 2.19 Remark Let a be a positive constant. • The graph of y = f (x − a) can be obtained from that of y = f (x) by moving it a units to the right. • The graph of y = f (x + a) can be obtained from that of y = f (x) by moving it a units to the left. (c) Note that √ 2 − x is defined for x ≤ 2 only. The graph of y = √ 2 − x and that of y = √ x − 2 are symmetric with respect to the vertical line x = 2. 2 4 6 8-2-4 1 2 Figure 2.20 In general, two subsets of the plane are said to be symmetric about a line ` if for each point P belonging to any one of the two sets, there is a point Q belonging to the other set such that either P = Q belongs to ` or • the line segment PQ is perpendicular to `; • P and Q are equidistant from `.  (7) Exponential Functions Let b be a positive real number different from 1. The exponential function with base b, denoted by expb, is the function given by expb(x) = bx. The domain of every exponential function is R. The range of every exponential function is (0,∞). The y-intercept of the graph of every exponential function is (0, 1). This is because b0 = 1. Remark • Because there are infinitely many exponential functions, one for each base, the notation expb, where b is written as a subscript, indicates that the base is b. Thus, for example, exp2 and exp3 are the exponential functions with base 2 and 3 respectively. • Sometimes, for convenience, we also write bx to denote the exponential function with base b. 58 Chapter 2. Functions and Graphs Example Consider exp2, the exponential function with base 2. It is the function from R to R given by exp2(x) = 2x. The graph of exp2 goes up (as x increases) and the rate that the graph goes up increases as x increases. -3 -2 -1 1 2 3 2 4 6 8 Figure 2.21 Example Consider exp 1 3 , the exponential function with base 1 3 . It is the function from R to R given by exp 1 3 (x) = ( 1 3 )x . The graph of exp 1 3 goes down (as x increases). -3 -2 -1 1 2 3 5 10 15 20 25 Figure 2.22 In Chapter 8, exponential functions will be discussed in more detail. (8) Logarithmic Function Recall that for every positive real number x, there is a unique real number y such that 10y = x. Different positive x give different values of y. In this way, we obtain a function defined for all positive real numbers. This function, denoted by log, is called the common logarithmic function. For each positive real number x, log(x) is defined to be the unique real number such that 10log(x) = x. That is, log(x) = y if and only if y = 10x. For simplicity, log(x) is also written as log x. Remark Sometimes, for convenience, we also write log x to denote the common logarithmic function. So the notation log x has two different meanings. It can be a function (the log function) or a number (the image of x under the log function). The domain of log is (0,∞). The range is R. The graph of log is shown in Figure 2.23. As x increases, the graph goes up. The rate that the graph goes up de- creases as x increases. Note that the x-intercept is (1, 0). This is because log 1 = 0. 20 40 60 80 100 0.5 1 1.5 2 Figure 2.23 Remark In Chapter 8, logarithmic functions with bases other than 10 will be considered. Relation be- tween exponential functions and logarithmic functions will be discussed. (9) Trigonometric Functions Similar to the common logarithmic function, we use three letters sin, cos and tan to denote the sine, cosine and tangent functions respectively. Recall that sin(x) is defined to be the y-coordinate of the point on the unit circle x2 + y2 = 1 corresponding to the angle with measure x radians. More details on trigonometric functions can be found in Chapter 7. 2.4. Graphs of Functions 59 For simplicity, we write sin(x) = sin x etc. • The sine function: sin The domain of the sine function is R. The range is [−1, 1]. The graph of the sine function has a waveform as shown in Figure 2.24 (the symbol p stands for the number π). The graph is symmetric about the origin. This is because sin(−x) = − sin x. The graph crosses the x-axis infinitely often, at points with x-coordinates 0,±π,±2π, . . . 2p 4p-2p-4p -1 1 Figure 2.24 The sine function is periodic with period 2π, that is, sin(x + 2π) = sin x for all x ∈ R. Definition If f is a function such that f (x + p) = f (x) for all x ∈ dom ( f ), where p is a positive constant, then we say that f is periodic with period p. • The cosine function: cos The domain of the cosine function is R. The range is [−1, 1]. The graph of the cosine function has a waveform. It is symmetric about the y-axis. This is be- cause cos(−x) = cos x. The graph crosses the x-axis infinitely often, at points with x-coordinates ±π 2 ,±3π 2 , . . . 2p 4p-2p-4p -1 1 Figure 2.25 The cosine function is periodic with period 2π, that is, cos(x + 2π) = cos x for all x ∈ R. Remark The graph of the cosine function can be obtained by shifting the graph of the sine function π 2 units to the left. This is because cos x = sin ( x + π 2 ) for all x ∈ R. 62 Chapter 2. Functions and Graphs Remark Alternatively, the graph can be obtained as follows: • The graph of y = −|x| and that of y = |x| are symmetric about the x-axis. So the graph of y = −|x| is an inverted V-shape figure. • Move the inverted V-shape figure 1 unit up. 1 2-1-2 1 -1 -2 Figure 2.28 (b) The graph of y = |x − 1| is a V-shape figure. It can be obtained by moving the graph of y = |x| one unit to the right. 1 2 3-1-2 1 2 Figure 2.29 (11) Piecewise-defined Functions Below we give more examples of piecewise-defined functions. Example Let f : [−2, 6] −→ R be the function given by f (x) =  x2 if − 2 ≤ x < 0 2x if 0 ≤ x < 2 4 − x if 2 ≤ x ≤ 6. For each of the following, find its value: (a) f (−1) (b) f (1 2 ) (c) f (3) Sketch the graph of f . Solution (a) f (−1) = (−1)2 = 1 (b) f (1 2 ) = 2 · 1 2 = 1 (c) f (3) = 4 − 3 = 1 The graph of f consists of three parts: • the curve y = x2, −2 ≤ x < 0 (part of a parabola); • the line segment y = 2x, 0 ≤ x < 2 (excluding the right endpoint); • the line segment y = 4 − x, 2 ≤ x ≤ 6. 2 4 6-2 -2 2 4 Figure 2.30 Remark In the figure, the little circle indicates that the point (2, 4) is not included in the graph. The little dot (which can be omitted) emphasizes that the point (2, 2) is included.  2.4. Graphs of Functions 63 The next example shows that piecewise-defined functions can be used in daily life. The piecewise-defined function in the example is called a step function. It jumps from one value to another. Example Suppose the long-distance rate for a telephone call from City A to City B is $1.4 for the first minute and $0.9 for each additional minute or fraction thereof. If y = f (t) is a function that indicates the total charge y for a call of t minutes’ duration, sketch the graph of f for 0 < t ≤ 4 1 2 . Solution Note that f (t) =  1.4 if 0 < t ≤ 1 2.3 if 1 < t ≤ 2 3.2 if 2 < t ≤ 3 4.1 if 3 < t ≤ 4 5.0 if 4 < t ≤ 4 1 2 . The graph of y = f (t) is shown in Figure 2.31. 1 2 3 4 4.5 1.4 2.3 3.2 4.1 5 Figure 2.31  Remark The ceiling of a real number t, denoted by dte, is defined to be the smallest integer greater than or equal to t. Using this notation, we have f (t) = 1.4 + 0.9 (dte − 1 ) . Exercise 2.4 1. For each of the following equations, sketch its graph. (a) y = 2x − 3 (b) y + 3 = 2(x − 5) (c) 7x − 5y + 4 = 0 (d) y = |2x − 1| + 5 (e) x2 = y2 (f) y = −x2 (g) y − 2 = −x2 (h) y − 2 = −(x − 3)2 (i) y = x2 + 2x − 3 (j) y = √ 1 − x2 (k) x = √ y 2. For each of the following equations, use a computer software to sketch its graph. (a) y = x3 (b) y = x3 − 2x2 − 3x + 4 (c) y = 2x3 − x + 5 (d) y = x3 − 3x2 + 3x − 1 (e) y = −x3 (f) y = −x3 + 2x2 + 3x − 4 (g) y = x4 (h) y = x4 − x3 − x2 + x + 1 (i) y = x4 − 3x3 + 2x2 + x − 1 (j) y = x4 − 4x3 + 6x2 − 4x (k) y = −x4 (l) y = −x4 − 2x3 + 3x 64 Chapter 2. Functions and Graphs Can you generalize the results for graphs of polynomial functions of degree 3, 4, . . . ? 3. Let f (x) = 2x − 1 x2 + 3 . The graph of f is shown on page 55. Note that there is a highest point and a lowest point. Find the coordinates of these two points. Hint: consider the range of f The points are called relative extremum points. An easy way to find their coordinates is to use differentiation, see Chapter 5. 4. An object is thrown upward and its height h(t) in meters after t seconds is given by h(t) = 1 + 4t − 5t2. (a) When will the object hit the ground? (b) Find the maximum height attained by the object. 5. The manager of an 80-unit apartment complex is trying to decide what rent to charge. Experience has shown that at a rent of $20000, all the units will be full. On the average, one additional unit will remain vacant for each $500 increase in rent. (a) Let n represent the number of $500 increases. Find an expression for the total revenue R from all the rented apartments. What is the domain of R? (b) What value of n leads to maximum revenue? What is the maximum revenue? 2.5 Compositions of Functions Consider the function f given by f (x) = sin2x. Recall that sin2x = (sin x)2. For each input x, to find the output y = f (x), (1) first calculate sin x, call the resulted value u; (2) and then calculate u2. These two steps correspond to two functions: (1) u = sin x; (2) y = u2. Given two functions, we can “combine” them by letting one function acting on the output of the other. Definition Let f and g be functions such that the codomain of f is a subset of the domain of g. The composition of g with f , denoted by g ◦ f , is the function given by (g ◦ f )(x) = g ( f (x) ) . (2.5.1) The right-side of (2.5.1) is read “g of f of x”. Figure 2.32 indicates that f is a function from A to B and g is a function from C to D where B ⊆ C. 2.6. Inverse Functions 67 Example The following two figures show the graphs of f and g in the last two examples. It is easy to see from the Horizontal Line Test that f is injective whereas g is not injective. 1 2-1-2 1 2 3 4 1 2-1-2 1 2 3 4f (x) = 2x g(x) = x2 injective not injective Figure 2.33 Figure 2.34 Let f be an injective function. Then given any element y of ran ( f ), there is exactly one element x of dom ( f ) such that f (x) = y. This means that if we use an element y of ran ( f ) as input, we get one and only one output x. The function obtained in this way is called the inverse of f . Definition Let f : X −→ Y be an injective function and let Y1 be the range of f . The inverse (function) of f , denoted by f −1, is the function from Y1 to X such that for every y ∈ Y1, f −1(y) is the unique element of X satisfying f ( f −1(y) ) = y. The following figure indicates a function f from a set X to a set Y . Assuming that f is injective, for each y belonging to the range of f , there is one and only one element x of X such that f (x) = y. This element x is defined to be f −1(y). That is, f −1(y) = x if and only if f (x) = y. Remark (1) For every x ∈ X, we have ( f −1 ◦ f )(x) = x. For every y ∈ Y1, we have ( f ◦ f −1)(y) = y. (2) f −1 is injective and ( f −1)−1(x) = f (x) for all x ∈ dom ( f ). X x • • y > < f f −1 Y Figure 2.35 Steps to find inverse functions Let f : X −→ R be an injective function where X ⊆ R. To find the inverse function of f means to find the domain of f −1 as well as a formula for f −1(y). If the formula for f (x) is not very complicated, dom ( f −1) and f −1(y) can be found by solving the equation y = f (x) for x. (Step 1) Put y = f (x). (Step 2) Solve x in terms of y. The result will be in the form x = an expression in y. (Step 3) From the expression in y obtained in Step 2, the range of f can be determined. This is the domain of f −1. The required formula is f −1(y) = the expression in y obtained in Step 2. Remark Steps 1 and 2 can be used to find range of a function. If the function is not injective, the expression in y obtained in Step 2 does not give a function; some y give more than one values of x. 68 Chapter 2. Functions and Graphs Example Let f (x) = 2x3 + 1. Find the inverse of f . Solution The domain of f is R. It is not difficult to show that f is injective and that the range of f is R. These two facts can also be seen from the following steps: Put y = f (x). That is, y = 2x3 + 1. Solve for x: y − 1 = 2x3 y − 1 2 = x3 3 √ y − 1 2 = x (x can be solved for all real numbers y) Thus we have dom ( f −1) = R and f −1(y) = 3 √ y − 1 2 .  Example Let g : [0,∞) −→ R be the function given by g(x) = x2. Find the inverse of g. Solution Because the domain of g is [0,∞), the function g is injective . Moreover, the range of g is [0,∞). These two facts can also be seen from the following steps: Put y = g(x). That is, y = x2. Note that y ≥ 0 and that x ≥ 0 since x ∈ dom ( f ). Solve for x: y = x2, y ≥ 0, x ≥ 0 √ y = x (x can be solved if and only if y ≥ 0, x = −√y is rejected) Thus we have dom (g−1) = [0,∞) and g−1(y) = √ y.  Remark Usually, we use x to denote the independent variable of a function. For the above examples, we may write f −1(x) = 3 √ x − 1 2 and g−1(x) = √ x. Caution f −1(x) , 1 f (x) Remark We use sin−1 or arcsin to denote the inverse of sin etc. Although the sine function is not injective, we can make it injec- tive by restricting the domain to [−π2 , π2 ]. x = sin−1 y means sin x = y and −π2 ≤ x ≤ π 2 . The domain of sin−1 is [−1, 1] because −1 ≤ sin x ≤ 1. - p 2 p 2 1 -1 y = sin x Figure 2.36 FAQ Why do we use the notation f −1? Answer The following example gives a reason why we use such a notation. Let f (x) = 2x. Then f is injective and its inverse is given by f −1(x) = 1 2 x. The multiplicative constant 1 2 is 2−1. Another reason is to have the “index law” (details omitted): f m ◦ f n = f m+n for m, n ∈ Z.  2.7. More on Solving Equations 69 Graph of the inverse function Let f : X −→ R be a function, where X ⊆ R. Then its graph is a subset of the plane. If, in addition, f is injective, then f has an inverse and dom ( f −1) ⊆ R. Hence the graph of f −1 is also a subset of the plane. There is a nice relationship between the graph of f and that of f −1: (∗) The graph of f and the graph of f −1 are symmetric about the line x = y. Reason Suppose P(a, b) belongs to the graph of f . This means that b = f (a) or equivalently, a = f −1(b). Thus Q(b, a) belongs to the graph of f −1. It is straightforward to show that the line segment PQ is perpendicular to the line y = x (denoted by `) and that P and Q are equidistant from `.  1 2 1 2 y = x2 y = √ x Figure 2.37 Figure 2.37 is an illustration for (∗). The function f : [0,∞) −→ R given by f (x) = x2 is injective. Its range is [0,∞). The domain of f −1 is [0,∞) and f −1(x) = √ x. Exercise 2.6 1. For each of the following functions f , determine whether it is injective or not. (a) f (x) = x3 + 2x (b) f (x) = x2 − 5 2. For each of the following functions f , find its inverse. (a) f (x) = 3x − 2 (b) f (x) = x5 + 3 (c) f (x) = 1 + 2x 1 7 (d) f (x) = 3√ 2x3 − 1 2.7 More on Solving Equations In this section, we will consider fractional equations and radical equations. In solving equations, if there is a one-sided implication (=⇒) in any one of the steps, we have to check solution. If all the steps are two-sided implications (⇐⇒), there is no need to check solution. Example For each of the following equations, find its solution set. (1) 5 x − 2 = 10 x + 3 (2) x x − 1 + 2 x = 1 x2 − x Solution 72 Chapter 2. Functions and Graphs Adding this condition, each step below is a two-sided implication: √ x − √x − 3 = 3 ⇐⇒ √ x − 3 = √ x − 3 ⇐⇒ (√ x − 3 )2 = x − 3 and x ≥ 9 ⇐⇒ x − 6 √ x + 9 = x − 3 and x ≥ 9 ... ⇐⇒ √ x = 2 and x ≥ 9 ⇐⇒ x = 4 and x ≥ 9 From this we see that there is no solution.  Exercise 2.7 1. For each of the following equations, find its solution set. (a) (2x + 1)(x − 2) = x(x + 2) (b) (2x + 1)(x − 2) = x(x − 2) (c) 1 x + 1 = 2 x + 2 (d) x x + 2 − x x − 2 = −4x x2 − 4 (e) 3 − √2x + 5 = 0 (f) √ x2 − 9 + x = 9 (g) √ x2 − 9 + 9 = x (h) √ x + 5 + 1 = 2 √ x (i) x6 − 9x3 + 8 = 0 2. Let C(q) = 2q + 12 be the cost to produce q units of a product and let R(q) = 10q − q2 be the revenue. (a) Find the profit (function). (b) Find the break-even quantity. 3. The stopping distance y in feet of a car traveling at x mph is described by the equation y = 0.056057x2 + 1.06657x. (a) Find the stopping distance for a car traveling at 35 mph. (b) How fast can one drive if one needs to be certain of stopping within 200 ft? 4. Find the right-angle triangle such that the sides adjacent to the right angle differ by 1 unit and the perime- ter is 12 units. Chapter 3 Limits Calculus is the study of differentiation and integration (this is indicated by the Chinese translation of “calcu- lus”). Both concepts of differentiation and integration are based on the idea of limit. In this chapter, we use an intuitive approach to consider limits, omitting the more difficult ε-δ definition. 3.1 Introduction In this section, we introduce the idea of limit by considering two problems. The first problem is to “find” the velocity of an object at a particular instant. The idea is related to differentiation. The second problem is to “find” the area under the graph of a curve (and above the x-axis). The idea is related to integration. Problem 1 Suppose an object moves along the x-axis and its displacement (in meters) s at time t (in seconds) is given by s(t) = t2, t ≥ 0. We want to consider its velocity at a certain time instant, say at t = 2. Idea Velocity (or speed) is defined by velocity = distance traveled time elapsed (3.1.1) (3.1.1) can only be applied to find average velocities over time intervals. We (still) don’t have a definition for velocity at t = 2. time distance Figure 3.1 To define the velocity at t = 2, we consider short time intervals about t = 2, say from t = 2 to t = 2 + 1 2n . Using (3.1.1), we can compute the average velocity over the time intervals [2, 2.5], [2, 2.25] etc. n Time interval Velocity 1 [2,2.5] 4.5 m/s 2 [2,2.25] 4.25 m/s 3 [2,2.125] 4.125 m/s 4 [2,2.0625] 4.0625 m/s ... 74 Chapter 3. Limits In general, the velocity vn over the time interval [ 2, 2 + 1 2n ] is vn = (2 + 1 2n )2 − 22 1 2n = 4 + 2 · 2 · 1 2n + ( 1 2n )2 − 4 1 2n = 4 + 1 2n . It is clear that if n is very large (that is, if the time interval is very short), vn is very close to 4. The velocity, called the instantaneous velocity, at t = 2 is (defined to be) 4. Problem 2 Find the area of the region that lies under the curve y = x2 and above the x-axis for x between 0 and 1. 1 1 Figure 3.2 Idea Similar to the idea in Problem 1, we use approximation to find/define area. First we divide the interval [0, 1] into finitely many subintervals of equal lengths: [ 0, 1 n ] , [ 1 n , 2 n ] , [ 2 n , 3 n ] , . . . , [ n − 1 n , 1 ] . For each subinterval [ i − 1 n , i n ] , we consider the rectangular region with base on the subinterval and height ( i − 1 n )2 (the largest region that lies under the curve). If we add the area of these rectangular regions, the sum is smaller than that of the required region. However, if n is very large, the error is very small and we get a good approximation for the required area. 1 1 Figure 3.3 The following table gives the sum S n of the areas of the rectangular regions (correct to 3 decimal places) for several values of n. In general, if there are n subintervals, the sum S n is S n = 1 n · 02 + 1 n · ( 1 n )2 + 1 n · ( 2 n )2 + · · · + 1 n · ( n − 1 n )2 = 12 + 22 + · · · + (n − 1)2 n3 = n(n − 1)(2n − 1) 6n3 By Sum of Squares Formula = 2n3 − 3n2 + n 6n3 = 1 3 − 1 2n + 1 6n2 n Sum of areas 2 0.125 3 0.185 4 0.219 ... 10 0.285 ... 100 0.328 ... 500 0.332 It is clear that if n is very large (so that the error is small), S n is very close to 1 3 . 3.2. Limits of Sequences 77 Rules for Limits of Sequences (L1) lim n→∞ k = k (where k is a constant) (L2) lim n→∞ 1 np = 0 (where p is a positive constant) (L3) lim n→∞ 1 bn = 0 (where b is a constant greater than 1) (L4) lim n→∞(an + bn) = lim n→∞ an + lim n→∞ bn (L5) lim n→∞ anbn = lim n→∞ an · lim n→∞ bn (L6) lim n→∞ an bn = lim n→∞ an lim n→∞ bn provided that lim n→∞ bn , 0. Remark • The meaning of (L1) is that if an = k for all n where k is a constant, then the sequence (an)∞n=1 is convergent and its limit is k. • The meaning of (L4) is that if both (an)∞n=1 and (bn)∞n=1 are convergent and their limits are L and M respectively, then (an + bn)∞n=1 is also convergent and its limit is L + M. • The following is a special case of (L5). It can be obtained by putting an = k for all n and applying (L1). (L5s) lim n→∞ kbn = k lim n→∞ bn • Using (L4) and (L5s), we get (L4′) lim n→∞(an − bn) = lim n→∞ an − lim n→∞ bn In fact, Rule (L4) is valid for sum and difference of finitely many sequences. This general result will be referred to as Rule (L4). Similarly, the result for product of finitely many sequences will be referred to as Rule (L5). In Problem 1 in the last section, the sequence obtained can be represented by the formula an = 4 + 1 2n . Our intuition tells us that the limit of the sequence is 4. Below we use rules for limits to justify this result. Example Find lim n→∞ ( 4 + 1 2n ) , if it exists. Explanation The sequence under consideration is given by an = 4 + 1 2n . The question asks for the following (1) Does the limit of (an)∞n=1 exist or not (or equivalently, is the sequence convergent)? (2) If the answer to (1) is affirmative, find the limit. Solution lim n→∞ ( 4 + 1 2n ) = lim n→∞ 4 + lim n→∞ 1 2n Rule (L4) = 4 + 0 Rules (L1) and (L3) = 4  78 Chapter 3. Limits Remark Below is the logic in the above calculation: (1) In the first step, because the constant sequence (4)∞n=1 and the sequence ( 1 2n )∞ n=1 are convergent, we can apply Rule (L4). (2) The limits of the two sequences are found by Rule (L1) and Rule (L3) respectively in the second step. The sequence in the next example is the one obtained in Problem 2 in the last section. Example Find lim n→∞ 2n3 − 3n2 + n 6n3 , if it exists. Solution lim n→∞ 2n3 − 3n2 + n 6n3 = lim n→∞ ( 1 3 − 1 2n + 1 6n2 ) Rewrite the expression = lim n→∞ 1 3 − lim n→∞ ( 1 2 · 1 n ) + lim n→∞ ( 1 6 · 1 n2 ) Rule (L4), rewrite 2nd and 3rd terms = 1 3 − 1 2 · lim n→∞ 1 n + 1 6 · lim n→∞ 1 n2 Rules (L1) and (L5s) = 1 3 − 1 2 · 0 + 1 6 · 0 Rule (L2) = 1 3  Example Find lim n→∞ (1 + 2n), if it exists. Solution Limit does not exist. This is because we can’t find any real number L satisfying the condition that 2n + 1 is close to L if n is large.  Remark If we apply rules for limits, we get lim n→∞(2n + 1) = lim n→∞ 2n + lim n→∞ 1 Rule (L4) = 2 lim n→∞ n + 1 Rules (L1) and (L5s) However, we can’t proceed because lim n→∞ n does not exist. From this, we see that the given limit does not exist. FAQ Can we say that lim n→∞(1 + 2n) is∞? Answer Limit of a sequence is a real number satisfying Condition (∗) given in the definition on page 76. Because∞ is not a real number, we should say that the limit does not exist. In the next section, we will discuss the meaning of lim x→∞ f (x) = ∞ etc.  Example Find lim n→∞ n + 1 2n + 1 , if it exists. Explanation We can’t use Rule (L6) because limits of the numerator and the denominator do not exist. However, we can’t conclude from this that the given limit does not exist. To find the limit, we use a trick: divide the numerator and the denominator by n. 3.2. Limits of Sequences 79 Solution lim n→∞ n + 1 2n + 1 = lim n→∞ n + 1 n 2n + 1 n Divide numerator and denominator by n = lim n→∞ ( 1 + 1 n ) lim n→∞ ( 2 + 1 n ) Rule (L6), rewrite numerator and denominator = lim n→∞ 1 + lim n→∞ 1 n lim n→∞ 2 + lim n→∞ 1 n Rule (L4) = 1 + 0 2 + 0 Rules (L1) and (L2) = 1 2  Remark We can apply the following shortcut (called the Leading Terms Rule). The method is to throw away the constant term 1 in the numerator and the denominator (note that if n is very large, compared with n or 2n, 1 is very small). lim n→∞ n + 1 2n + 1 = lim n→∞ n 2n = lim n→∞ 1 2 = 1 2 . The Leading Terms Rule for limits of functions at infinity will be discussed in more details in the next section (see page 83). Exercise 3.2 1. For each of the following, find the limit if it exists. (a) lim n→∞ 1√ n (b) lim n→∞(7 − 5 n2 ) (c) lim n→∞ 3n2 − 4000 2n2 + 10000 (d) lim n→∞ n2 − 12345 n + 1 (e) lim n→∞ 5n2 + 4 2n3 + 3 (f) lim n→∞ (1 n + (−1)n ) 2. Suppose $50,000 is deposited at a bank and the annual interests rate is 2%. (a) What amount (correct to the nearest cent) will the account have after one year if interests is (i) compounded quarterly; (ii) compounded monthly? (b) If interest is compounded n times a year, express the amount An after one year in terms of n. ∗(c) Does lim n→∞ An exist? What is the value? ∗3. For each of the following sequences (an)∞n=1, use computer to find the first 100 (or more) terms. Does lim n→∞ an exists? If yes, what is the value? (a) an = ( 1 + 1 n )n (b) an = ( 1 + 2 n )n (c) an = n sin 1 n (angles are in radians) 82 Chapter 3. Limits Example Find lim x→∞ x2 + 1 3x3 − 4x + 5 , if it exists. Explanation Because limits of the numerator and denominator do not exist, we can’t apply Rule (6). The first step is to divide the numerator and denominator by x3 so that the limits at infinity of the new numerator and denominator exist. Solution lim x→∞ x2 + 1 3x3 − 4x + 5 = lim x→∞ x2 + 1 x3 3x3 − 4x + 5 x3 Divide numerator and denominator by x3 = lim x→∞ ( 1 x + 1 x3 ) lim x→∞ ( 3 − 4 x2 + 5 x3 ) Rule (L6), rewrite numerator and denominator = lim x→∞ 1 x + lim x→∞ 1 x3 lim x→∞ 3 − lim x→∞ 4 x2 + lim x→∞ 5 x3 Rule (L4) = 0 + 0 3 − 0 + 0 Rules (L1), (L2) and (L5s) = 0  The next example is similar to the last one. To find limits at infinity for rational functions, we can divide the numerator and denominator by a suitable power of x. Example Find lim x→∞ x3 + 1 3x3 − 4x + 5 , if it exists. Solution lim x→∞ x3 + 1 3x3 − 4x + 5 = lim x→∞ x3 + 1 x3 3x3 − 4x + 5 x3 Divide numerator and denominator by x3 = lim x→∞ ( 1 + 1 x3 ) lim x→∞ ( 3 − 4 x2 + 5 x3 ) Rule (L6), rewrite numerator and denominator = lim x→∞ 1 + lim x→∞ 1 x3 lim x→∞ 3 − lim x→∞ 4 x2 + 5 x3 Rule (L4) = 1 + 0 3 − 0 + 0 Rules (L1), (L2) and (L5s) = 1 3  To find limits at infinity for rational functions, we can also use the following shortcut. 3.3. Limits of Functions at Infinity 83 Leading Terms Rule Let f (x) = anxn + an−1xn−1 + · · ·+ a1x + a0 and g(x) = bmxm + bm−1xm−1 + · · ·+ b1x + b0, where an , 0 and bm , 0. Then we have lim x→∞ f (x) g(x) = lim x→∞ anxn + an−1xn−1 + · · · + a1x + a0 bmxm + bm−1xm−1 + · · · + b1x + b0 = lim x→∞ anxn bmxm Proof The idea is to extract factor anxn in the numerator and bmxm in the denominator. Putting ϕ(x) = 1 + an−1 an · 1 x + an−2 an · 1 x2 + · · · + a1 an · 1 xn−1 + a0 an · 1 xn 1 + bm−1 bm · 1 x + bm−2 bm · 1 x2 + · · · + b1 bm · 1 xm−1 + b0 bm · 1 xm we have f (x) g(x) = anxn bmxm · ϕ(x). It is straightforward to check that lim x→∞ϕ(x) = 1. Hence by Rule (5), we obtain the required result.  Remark • (a) If n = m, the limit is an bn . (b) If n < m, the limit is lim x→∞ ( an bm · 1 xm−n ) = 0. (c) If n > m, the limit is lim x→∞ ( an bm · xn−m ) which does not exist because as x increases indefinitely, xn−m increases indefinitely. • The Leading Terms Rule can also be applied to “functions similar to rational functions”, for example, for f (x) = x + 2 √ x + 3 and g(x) = 5x + 6 √ x + 7, we have lim x→∞ x + 2 √ x + 3 5x + 6 √ x + 7 = lim x→∞ x 5x • The Leading Terms Rule can’t be applied to limits of rational functions at a point: lim x→a f (x) g(x) , where a ∈ R. Below we re-do the last two examples using the Leading Terms Rule. Example lim x→∞ x2 + 1 3x3 − 4x + 5 = lim x→∞ x2 3x3 Leading Terms Rule = lim x→∞ ( 1 3 · 1 x ) Simplify and rewrite expression = 1 3 · 0 Rules (L2) and (L5s) = 0 Example lim x→∞ x3 + 1 3x3 − 4x + 5 = lim x→∞ x3 3x3 Leading Terms Rule = lim x→∞ 1 3 Simplify expression = 1 3 Rule (L1) 84 Chapter 3. Limits In the next example, the function can be considered as a product or a quotient of two functions. However, we can’t apply Rule (5) or (6) because limit at infinity of one of the functions does not exist. To find the limit, we need the following result. Sandwich Theorem Let f , g and h be functions such that f (x), g(x) and h(x) are defined for sufficiently large x. Suppose that f (x) ≤ g(x) ≤ h(x) if x is sufficiently large and that both lim x→∞ f (x) and lim x→∞ h(x) exist and are equal (with common limit denoted by L). Then we have lim x→∞ g(x) = L. Remark The condition “ f (x) ≤ g(x) ≤ h(x) if x is sufficiently large” means that there is a real number r such that the inequalities are true for all x > r. Example Find lim x→∞ sin x x , if it exists. Explanation The given function can be written as a product of two functions: sin x and 1 x . For the second function, its limit at infinity is 0. However, for the first function, its limit at infinity does not exist. Thus we can’t apply Rule (5). Solution Since −1 ≤ sin x ≤ 1 for all real numbers x, it follows that −1 x ≤ sin x x ≤ 1 x for all x > 0. Note that lim x→∞ −1 x = lim x→∞ 1 x = 0. Thus by the Sandwich Theorem, we have lim x→∞ sin x x = 0.  Example Find lim x→∞(1 + log x), if it exists. Remark Since lim x→∞ log x does not exist, we can’t apply Rule (L4). Solution lim x→∞(1 + log x) does not exist. This is because if x increases without bound, so does 1 + log x.  Infinite Limits In the last example, although limit does not exist, we know that if x increases indefinitely, so does (1 + log x). In the limit notation lim x→∞, the symbol x → ∞ indicates that “x increases indefinitely”, or “x approaches ∞”. Using the same idea, we also write 1 + log x → ∞ which indicates that the value increases indefinitely (as x increases indefinitely). Putting y = 1 + log x, we write y → ∞ as x → ∞. Concerning the graph of y = f (x) in the coordinate plane, x→ ∞ means that x goes to the right indefinitely, approaching the point∞ (an imaginary point on the right) and y → ∞ means that y goes up indefinitely, approaching the point ∞ (an imaginary point at the top). Notation Let f be a function such that f (x) is defined for sufficiently large x. Suppose that (∗) f (x) is arbitrarily large if x is sufficiently large. Then we write lim x→∞ f (x) = ∞. Remark • Because ∞ is not a real number, lim x→∞ f (x) = ∞ does not mean the limit exists. In fact, it indicates that the limit does not exist and explains why it does not exist. 3.4. One-sided Limits 87 Right-side Limits Definition Let a ∈ R and let f be a function such that f (x) is defined for x sufficiently close to and greater than a. Suppose L is a real number satisfying (∗) f (x) is arbitrarily close to L if x is sufficiently close to and greater than a. Then we say that L is the right-side limit of f at a and we write lim x→a+ f (x) = L. Remark • The condition “ f (x) is defined for x sufficiently close to and greater than a” means that there is a positive real number δ such that f (x) is defined for all x ∈ (a, a+δ). For simplicity, instead of saying the condition, we will say “ f is defined on the right-side of a”. • Instead of saying Condition (∗), we will say (∗∗) f (x) is close to L if x is close to and greater than a. • In the definition, it doesn’t matter whether f is defined at a or not. If f (a) is defined, its value has no effect on the existence and the value of lim x→a+ f (x). This is because right-side limit depends on the values of f (x) for x close to and greater than a. Example Let f (x) = 1 − 2− 1√ x . The domain of f is (0,∞) and so f is defined on the right-side of 0. The graph of f is shown in the following figure. Imagine a small creature living on the curve. Suppose it moves to the left so that the x-coordinate of its position approaches 0 (from the right). From the graph, we see that the y-coordinate will approach 1. In other words, the right-side limit of f at 0 is 1, that is, lim x→0+ (1 − 2− 1√ x ) = 1. 0.5 1 1.5 2 0.2 0.4 0.6 0.8 1 Figure 3.9 FAQ How can we know the graph of f ? Answer This example is chosen to illustrate the idea of right-side limit. The graph is generated by computer. • To see why lim x→0+ (1−2− 1√ x ) = 1, note that if x is small positive, then so is √ x and hence 1√ x is large positive; the expression 2− 1√ x = 2−large positive = 1 2large positive is small positive and so f (x) = (1 − small positive) is close to 1. • To see why the graph goes down (as x increases), note that if x increases, then so does √ x and so 1√ x decreases; hence − 1√ x increases; therefore 2− 1√ x increases and thus f (x) decreases. An alternative way to see this is to use differentiation (see Chapter 5 and Chapter 9).  Example Let f (x) = √ x. The domain of f is [0,∞) and so f is defined on the right-side of 0. Note that if x is close to and greater than 0, then f (x) is close to 0. This means that lim x→0+ √ x = 0. 0.2 0.4 0.6 0.8 1 0.2 0.4 0.6 0.8 1 Figure 3.10 88 Chapter 3. Limits Example Let f (x) = sin 1 x . The domain of f is R \ {0}. Thus f is defined on the right-side of 0 and we can consider the right-side limit of f at 0. Remark In fact, f (x) is also defined for x < 0 and so we can consider its left-side limit. See the definition for left-side limit below. The graph of f (for 0 < x ≤ 2) is shown in Figure 3.11. Note that when x approaches 0 from the right-side, f (x) oscillates between −1 and 1. Thus lim x→0+ sin 1 x does not exist. 0.5 1 1.5 2 -1 -0.5 0.5 1 Figure 3.11 Example Let f (x) = sin x x2 − x . The domain of f is R \ {0, 1}. Thus f (x) is defined for 0 < x < 1 and we can consider lim x→0+ f (x). The graph of f (for 0 < x ≤ 0.8) is shown in Figure 3.12. From the graph, we see that lim x→0+ sin x x2 − x = −1. Remark The limit can be calculated using the following fact: lim x→0 sin x x = 1 It is a two-sided limit. See next section for more details. 0.2 0.4 0.6 0.8 -5 -4 -3 -2 -1 Figure 3.12 Left-side Limits If a function f is defined on the left-side of a, we can consider its left-side limit. The notation lim x→a− f (x) = L means that (∗) f (x) is arbitrarily close to L if x is sufficiently close to and less than a. Example Let f (x) = sin x x2 − x . The domain of f is R \ {0, 1}. Thus f (x) is defined for x < 0 and we can consider lim x→0− f (x). The graph of f (for x close to and less than 0) is shown in the following figure. From the graph, we see that lim x→0− sin x x2 − x = −1. -2 -1.5 -1 -0.5 -2 -1.5 -1 -0.5 Figure 3.13 Similar to lim x→∞ f (x) = ∞ etc., we have the following notations: (1) lim x→a+ f (x) = ∞ (2) lim x→a+ f (x) = −∞ (3) lim x→a− f (x) = ∞ (4) lim x→a− f (x) = −∞ 3.5. Two-sided Limits 89 Readers can figure out the meaning of the notations themselves. Geometrically, if any one of these notations is true, then the line x = a is a vertical asymptote for the graph of f . Example lim x→0+ 1 x = ∞ lim x→1− 1 x − 1 = −∞ 0.2 0.4 0.6 0.8 1 10 20 30 40 50 Figure 3.14 0.2 0.4 0.6 0.8 1 -50 -40 -30 -20 -10 Figure 3.15 Exercise 3.4 1. For each of the following, find the limit if it exists. (a) lim x→1+ √ x − 1 (b) lim x→0+ 1√ x (c) lim x→0− 1 x3 (d) lim x→0− sin 1 x (e) lim x→0+ (2 − 3 1 x ) (f) lim x→0− (2 − 3 1 x ) 3.5 Two-sided Limits Definition Let a ∈ R and let f be a function that is defined on the left-side and right-side of a. Suppose that both lim x→a− f (x) and lim x→a+ f (x) exist and are equal (with the common limit denoted by L which is a real number). Then the two-sided limit, or more simply, the limit of f at a is defined to be L, written lim x→a f (x) = L. Remark • lim x→a f (x) = L means that (∗) f (x) is arbitrarily close to L if x is sufficiently close to (but not equal to) a. • The condition “ f be a function that is defined on the left-side and the right-side of a” means that there is a positive real number δ such that f (x) is defined for all x ∈ (a − δ) ∪ (a, a + δ). • In considering lim x→a f (x), it doesn’t matter whether f is defined at a or not. Example Let f (x) = sin x x2 − x . The domain of f is R \ {0, 1}. Thus f is defined on the left-side and right-side of 0. In Section 3.4, the left-side and right-side limits of f at 0 were found to be lim x→0+ sin x x2 − x = −1 and lim x→0− sin x x2 − x = −1. Thus, by definition, the limit of f at 0 exists and lim x→0 sin x x2 − x = −1. -2 -1.5 -1 -0.5 0.5 -2 -1.5 -1 -0.5 Figure 3.16