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

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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

Contents

• 0 Revision
  • 0.1 Exponents
  • 0.2 Algebraic Identities and Algebraic Expressions
  • 0.3 Solving Linear Equations
  • 0.4 Solving Quadratic Equations
  • 0.5 Remainder Theorem and Factor Theorem
  • 0.6 Solving Linear Inequalities
  • 0.7 Lines
  • 0.8 Pythagoras Theorem, Distance Formula and Circles
  • 0.9 Parabola
  • 0.10 Systems of Equations
• 1 Sets, Real Numbers and Inequalities
  • 1.1 Sets
    • 1.1.1 Introduction
    • 1.1.2 Set Operations
  • 1.2 Real Numbers
    • 1.2.1 The Number Systems
    • 1.2.2 Radicals
  • 1.3 Solving Inequalities
    • 1.3.1 Quadratic Inequalities
    • 1.3.2 Polynomial Inequalities with degrees ≥
• 2 Functions and Graphs
  • 2.1 Functions
  • 2.2 Domains and Ranges of Functions
  • 2.3 Graphs of Equations
  • 2.4 Graphs of Functions
  • 2.5 Compositions of Functions
  • 2.6 Inverse Functions
  • 2.7 More on Solving Equations
• 3 Limits
  • 3.1 Introduction
  • 3.2 Limits of Sequences
  • 3.3 Limits of Functions at Infinity
  • 3.4 One-sided Limits
  • 3.5 Two-sided Limits
  • 3.6 Continuous Functions
• 4 Differentiation
  • 4.1 Derivatives
  • 4.2 Rules for Differentiation
  • 4.3 Higher-Order Derivatives
• 5 Applications of Differentiation iv CONTENTS
  • 5.1 Curve Sketching
    • 5.1.1 Increasing and Decreasing Functions
    • 5.1.2 Relative Extrema
    • 5.1.3 Convexity
    • 5.1.4 Curve Sketching
  • 5.2 Applied Extremum Problems
    • 5.2.1 Absolute Extrema
    • 5.2.2 Applied Maxima and Minima
    • 5.2.3 Applications to Economics
• 6 Integration
  • 6.1 Definite Integrals
  • 6.2 Fundamental Theorem of Calculus
  • 6.3 Indefinite Integrals
  • 6.4 Application of Integration
• 7 Trigonometric Functions
  • 7.1 Angles
  • 7.2 Trigonometric Functions
  • 7.3 Differentiation of Trigonometric Functions
• 8 Exponential and Logarithmic Functions
  • 8.1 Exponential Functions
  • 8.2 Logarithmic Functions
  • 8.3 Differentiation of Exp and Log Functions
• 9 More Differentiation
  • 9.1 Chain rule
  • 9.2 Implicit Differentiation
  • 9.3 More Curve Sketching
  • 9.4 More Extremum Problems
• 10 More Integration
  • 10.1 More Formulas
  • 10.2 Substitution Method
  • 10.3 Integration of Rational Functions
  • 10.4 Integration by Parts
  • 10.5 More Applications of Definite Integrals
• A Answers
• B Supplementary Notes
  • B.1 Mathematical Induction
  • B.2 Binomial Theorem
  • B.3 Mean Value Theorem
  • B.4 Fundamental Theorem of Calculus

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 = a^2 + 2 ab + b^2 (2) (a − b)^2 = a^2 − 2 ab + b^2 (3) (a + b)(a − b) = a^2 − b^2

Remark The above equalities are called identities because they are valid for all real numbers a and b.

Caution In general, (a + b)^2 , a^2 + b^2. Note: (a + b)^2 = a^2 + b^2 if and only if a = 0 or b = 0.

Example Expand the following:

(1)

x + 2

x − (^5) x

x^2 + 1 + 7

x^2 + 1 − 7

Solution

(1)

x + 2

x

x

= x + 4 √x + 4

x − (^5) x

= x^2 − 2(x)

x

x

= x^2 − 10 + (^25) x 2

(3)

x^2 + 1 + 7

x^2 + 1 − 7

x^2 + 1

x^2 + 1

= x^2 − 48 

Example Simplify the following:

(1) x

(^2) − x − 6 x^2 − 6 x + 9 (2) x

2 x^2 − 1 −^1 (3) (^) x (^2) + 22 x + 1 − (^) x (^2) −^1 x − 2

(4)

x − y−^1

3 + (^6) x x + (^) x +x 1

0.2. Algebraic Identities and Algebraic Expressions 3

Solution

(1) x

(^2) − x − 6 x^2 − 6 x + 9 =^

(x − 3)(x + 2) (x − 3)^2 = x x^ +−^23

(2) x

2 x^2 − 1 −^1 =^

x^2 − (x^2 − 1) x^2 − 1 = (^) x (^2 1) − 1

(3) (^) x (^2) + 22 x + 1 − (^) x (^2) − 1 x − 2 = (^) (x +^2 1) 2 − (^) (x + 1)(^1 x − 2)

= 2((xx^ −+^ 2)1)^2 −(^ x( x−^ + 2)^ 1)

= (^) (x + x1)^ − (^2) (^5 x − 2)

x − y−^1

x − (^1) y

( (^) xy − 1 y

= (^) xy y− 1

3 + (^6) x x + (^) x +x 1

3 x + 6 x x (x + 1) + x x + 1

=

3 x + 6 x x^2 + 2 x x + 1 = 3(x^ x+ 2)· (^) x (xx^ + +^1 2)

= 3(x x^ + 2 1) 

FAQ What is expected if we are asked to simplify an expression? For example, in (5), can we give 3 x x^ + 2 3 as the answer?

Answer There is no definite rule to tell which expression is simpler. For (5), both 3(x x^ + 2 1) and 3 x x^ + 2 3 are acceptable. Use your own judgment. 

0.3. Solving Linear Equations 5

Properties of real numbers Let a, b and c be real numbers. Then we have

(1) a = b ⇐⇒ a + c = b + c (2) a = b =⇒ ac = bc and ac = bc =⇒ a = b if c , 0

Remark

• =⇒ is the symbol for “implies”. The first part of Property (2) means that if a = b, then ac = bc.
• ⇐⇒ is the symbol for “=⇒ and ⇐=”. Property (1) means that if a = b, then a + c = b + c and vice versa, that is, a = b iff a + c = b + c. In mathematics, we use the shorthand “iff ” to stand for “if and only if ”.

Example Solve the following equations for x.

(1) 3 x − 5 = 2(7 − x) (2) a(b + x) = c − dx, where a, b, c and d are real numbers with a + d , 0.

Solution

(1) Using properties of real numbers, we get

3 x − 5 = 2(7 − x) 3 x − 5 = 14 − 2 x 3 x + 2 x = 14 + 5 5 x = 19 x = 195.

The solution is 195.

FAQ Can we omit the last sentence? Answer The steps above means that a real number x satisfies 3x − 5 = 2(7 − x) if and only if x = 195. It’s alright if you stop at the last line in the equation array because it tells that given equation has one and only one solution, namely 195. 

FAQ What is the difference between the word “solution” after the question and the word “solution” in the last sentence? Answer They refer to different things. The first “solution” is solution (answer) to the problem (how to solve the problem) whereas the second “solution” means solution to the given equation. Sometimes, an equation may have no solution, for example, x^2 + 1 = 0 but the procedures (explanations) to get this information is a solution to the problem. 

FAQ Can we use other symbols for the unknown? Answer In the given equation, if x is replaced by another symbol, for example, t, we get the equation 3 t − 5 = 2(7 − t) in one unknown t. Solution to this equation is also 195. In writing an equation, the symbol

6 Chapter 0. Revision

for the unknown is not important. However, if the unknown is expressed in t, all the intermediate steps should use t as unknown: 3 t − 5 = 2(7 − t) ... t = (^195) 

(2) Using properties of real numbers, we get a(b + x) = c − dx ab + ax = c − dx ax + dx = c − ab (a + d)x = c − ab x = ca^ − +^ ab d. 

Exercise 0.

1. Solve the following equations for x. (a) 2(x + 4) = 7 x + 2 (b) 5 x^2 + 3 − 5 = 5 x^4 −^4

(c) (a + b)x + x^2 = (x + b)^2 (d) (^) ax − xb = c where a, b and c are constants with a , b.

0.4 Solving Quadratic Equations

A quadratic equation (in one unknown) is an equation that can be written in the form

ax^2 + bx + c = 0 (0.4.1)

where a, b, and c are constants and a , 0. To solve (0.4.1), we can use the Factorization Method or the Quadratic Formula.

Factorization Method The method makes use of the following result on product of real numbers:

Fact Let a and b be real numbers. Then we have

ab = 0 ⇐⇒ a = 0 or b = 0.

Example Solve x^2 + 2 x − 15 = 0.

Solution Factorizing the left side, we obtain

(x + 5)(x − 3) = 0.

Thus x + 5 = 0 or x − 3 = 0. Hence x = −5 or x = 3. 

8 Chapter 0. Revision

Thus the solutions are 52 and 2.

(2) Since 2^2 − 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

x^2 + 2 x = 2 x^2 + 3 x x^2 + 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 = 2 x + 3 which has only one solution. In canceling the factor x, it is assumed that x , 0. However, 0 is a solution and so this solution is lost. To use cancellation, we should write x (x + 2) = x (2x + 3) ⇐⇒ x + 2 = 2 x + 3 or x = 0 ...

Example Find the value(s) of k such that the equation 3x^2 + kx + 7 = 0 has only one solution.

Solution The given equation has only one solution iff

k^2 − 4(3)(7) = 0.

Solving, we get k = ±

Exercise 0.

1. Solve the following equations. (a) 4 x − 4 x^2 = 0 (b) 2 + x − 3 x^2 = 0 (c) 4 x (x − 4) = x − 15 (d) x^2 + 2

2 x + 2 = 0 (e) x^2 + 2

2 x + 3 = 0 (f) x^3 − 7 x^2 + 3 x = 0

2. Find the value(s) of k such that the equation x^2 + 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) = x^3 + 3 x^2 − 2 x + 2. Find the remainder when p(x) is divided by x − 2.

Solution The remainder is p(2) = 23 + 3(2^2 ) − 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) = x^3 + kx^2 + 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

x^3 + 4 x^2 + x − 6 = (x + 2)(x^2 + 2 x − 3).

By inspection, we have p(x) = (x + 2)(x + 3)(x − 1). 

FAQ Can we find the quotient (x^2 + 2 x − 3) by inspection (without using long division)?

Answer The “inspection method” that some students use is called the compare coefficient method. Since the quotient is quadratic, it is in the form (ax^2 + bx + c). Thus we have

x^3 + 4 x^2 + x − 6 = (x + 2)(ax^2 + bx + c) (0.5.1)

Comparing the coefficient of x^3 , we see that a = 1. Similarly, comparing the constant term, we get c = −3. Hence we have x^3 + 4 x^2 + x − 6 = (x + 2)(x^2 + bx − 3).

To find b, we may compare the x^2 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

ax^3 + (2a + b)x^2 + (2b + c)x + 2 c

(2) Compare the coefficients of the given polynomial with that obtained in Step (1) to get 1 = a 4 = 2 a + b 1 = 2 b + c − 6 = 2 c

0.6. Solving Linear Inequalities 11

(2) We write b > a to denote that b is greater than a and we write a < b to denote that a is less than b. (3) We write b ≥ a to denote that b is greater than or equal to a and we write a ≤ b to denote that a is less than or equal to b.

A linear inequality in one unknown x is an inequality that can be written in one of the following forms: (1) ax + b < 0 (2) ax + b ≤ 0 (3) ax + b > 0 (4) ax + b ≥ 0

where a and b are constants with a , 0. More generally, an inequality in one unknown x is an inequality that can be written in one of the following forms:

(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 A solution to an inequality F(x) < 0 is a real number x 0 such that F(x 0 ) < 0. The definition also applies to other types of inequalities.

Example Consider the inequality 2x + 3 ≥ 0. By direct substitution, we see that 1 is a solution and −2 is not a solution.

To solve an inequality means to find all solutions to the inequality.

Rules for Inequalities Let a, b and c be real numbers. Then the following holds.

(1) If a < b, then a + c < b + c. (2) If a < b and c > 0, then ac < bc. (3) If a < b and c < 0, then ac > bc. Note: The inequality is reversed. (4) If a < b and b ≤ c, then a < c. (5) If a < b and a and b have the same sign, then (^1) a > (^1) b. (6) If 0 < a < b and n is a positive integer, then an^ < bn^ and √na < n

b.

Terminology Two numbers have the same sign means that both of them are positive or both of them are negative.

Remark One common mistake in solving inequalities is to apply a rule with the wrong sign (positive or negative). For example, if c is negative, it would be wrong to apply Rule (2).

Example Solve the following inequalities.

(1) 2 x + 1 > 7(x + 3)

12 Chapter 0. Revision

(2) 3(x − 2) + 5 > 3 x + 7

Solution

(1) Using rules for inequalities, we get 2 x + 1 > 7(x + 3) 2 x + 1 > 7 x + 21 1 − 21 > 7 x − 2 x − 20 > 5 x − 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 = 3 x − 1 which is always less than the right side. Thus the inequality has no solution. 

Exercise 0.

1. Solve the following inequalities for x. (a) 1 − 2 x≥ 3 x^3 − 7 (b) 2(3 − x) ≤

3(1 − x)

(c) (^1 3) −x x + 3 < 0 (d) (^2) x^2 +x 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 x 0 , y 0 enclosed inside parenthesis: (x 0 , y 0 ).

Remark Two ordered pairs (x 0 , y 0 ) and (x 1 , y 1 ) are equal if and only if x 0 = x 1 and y 0 = y 1. 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 (x 0 , y 0 ) such that F(x 0 , y 0 ) = 0.

Example Consider the equation 2 x + 3 y − 4 = 0.

By direct substitution, we see that (2, 0) is a solution whereas (1, 2) is not a solution.