CHAPTER 2 Review of Algebra

Transcription

1 CHAPTER 2 Review of Algebra 2.1 Real Numbers The text assumes that you know these sets of numbers. You are asked to characterize each set in Problem 1 of Chapter 1 review. Name Symbol Set Examples Counting numbers f1, 2, 3, 4, ÐÐÐg 72; 2,345, 950; p 25; p 1 or natural numbers Whole numbers f0, 1, 2, 3, ÐÐÐg 0; 72; 10 16, 425 ; 2 25 Integers fððð, 2, 1, 0, 1, 2, ÐÐÐg 5; 0; 10 ; 9,562; 82 Rational numbers Irrational numbers 0 Numbers that can be written in the form p q, where p and q are integers with q 6D 0 Numbers whose decimal representations do not terminate and do not repeat Real numbers The collection of all rational and irrational numbers ; 4 ; ;.8666ÐÐÐ; 17 5; 0; ; 15; 4 ; ; ÐÐÐ; p 2; p 3; ÐÐÐ; ; 2 All examples listed on this page are real numbers. Not all numbers are real numbers, however. There is another set of numbers (not considered in this handbook) called complex numbers, which includes the real numbers as a subset. 2.2 Algebraic Processes In order to review the basic algebraic processes with which you need to be familiar in calculus, we begin with a review of some terminology. 24

2 Chapter 2 25 TERM DEFINITION EXAMPLES Numerical expression Algebraic expression Term Polynomial Rational expression Radical expression A number or several numbers connected by defined mathematical operations. A numerical expression with at least one variable. A number, a variable, or a product of numbers and variables. Atermorasumof terms. (Note: Since x y D x C y, differences are included.) A polynomial divided by a nonzero polynomial. (Note: Includes polynomials, because if P is a polynomial, it can be written as P D P 1.) An algebraic expression with a variable expression as a radicand (the expression under a radical sign). 6; 5 C 2; 5 p 2 6x; x C y;2x 2 3x C 4 6; x;6x;2x 2 ; 1 2 y3 6; x C y;4x 3 3x C 1 x C y x y ; 1 x C 1 y p x; 3p y; x 2 C y 2 2 Many of the processes in algebra are based on an agreement, called the order-of-operations agreement, that tells us how to deal with expressions with more than one operation: First, carry out those operations enclosed in parentheses. Second, carry out all multiplications and divisions as they occur, left to right.

3 26 Chapter 2 Third, carry out all additions and subtractions as they occur, left to right. If there are exponents, use what is sometimes called the extended order-of-operations agreement which instructs us to do the operations involving exponents just before those involving multiplications and divisions. There are four principal processes used in algebra: Factor Simplify To factor an expression means to write the expression as a product. An expression is called completely factored if all fractions are eliminated by common factoring and if no further factoring is possible over the set of integers. To simplify a numerical expression means to carry out all the operations, according to their order, and write your answer as a single number. To simplify a polynomial means to carry out all the operations, according to their order, combine similar terms, and write your answer by arranging the terms in decreasing degree. If there are different terms of the same degree, then they are arranged alphabetically. To simplify a rational expression means to carry out all the operations, according to their order, factor the numerator and denominator, and make sure that there is no common factor (other than 1 or 1). If there is a negative, then the expression is written in the form p q and q. or p q for positive p To simplify a radical expression means to carry out all the operations, according to their order, and to make sure that the following conditions are satisfied: 1. When the radicand is written in completely factored form, there is no factor raised to a power greater than or equal to the index of the radical. 2. No radical appears in a denominator. 3. No fraction (or negative exponent) appears within a radical. 4. There is no common factor (other than 1) between the power of the radicand and the index of the radical. Sometimes the word expand is used instead of simplify. For example, to expand the expression x C y 2 means to

4 Chapter 2 27 simplify it; that is, write x C y 2 D x 2 C 2xy C y 2 Evaluate To evaluate an algebraic expression means to replace the variable or variables with a given number and then simplify the expression. In calculus, we often evaluate a function at two values and then we subtract those values. For example, if the function f x D x 2 C x C 1 is evaluated at x D 2, we have f 2 D 2 2 C 2 C 1 D 7, and if f is evaluated at x D 1, we have f 1 D 1 2 C 1 C 1 D 3. Finally, if we subtract these evaluations, we get f 2 f 1 D 7 3 D 4. Since this is a frequent calculation, a compact notation is sometimes used: f x j 2 1 D f 2 f 1 For example, 2.3 Powers and roots EXPONENTS Definition of Exponent: [x 2 C x C 1]j 2 1 D [22 C 2 C 1] [1 2 C 1 C 1] D 7 3 D 4 In the expression x n, the number n is called the exponent, the number b is called the base, andx n is called a power. Ifn is a positive integer, then x n D x} Ð x Ð{{ x ÐÐÐx},andx 0 D 1; x n D 1 x n If m and n are integers, then and n factors x 1 n D np x whenever np x is defined x m n D x 1 n m D np x m whenever np x is defined Factorial numbers: n! D n n 1 n 2 ÐÐÐ3 Ð 2 Ð 1 0! D 1 for n a nonnegative integer

5 28 Chapter 2 ROOTS Recall that for any positive even integer n (called the index of the radical) and any positive number x (called the radicand), y D np x if and only if y>0andy n D x We call y the positive nth proot of x. For example, the positive fourth root of 16 is denoted by 4 16; we write 4p 16 D 2, since 2 4 D 16. For any positive odd integer n and any number x (positive or negative), y D np x and only if y n D x and y is called the nth root of x. For example, 2 3 D 8. Note that p x 2 Djxjfor any number x. 3p 8 D 2, since LAWS OF EXPONENTS If r and s are real numbers, then x r Ð x s D x rcs x r s D x rs whenever x r is meaningful xy r D x r y r whenever x r and y r are meaningful ( ) x r D xr y y r whenever xr and y r are meaningful and y r 6D 0 x r x D s xr s whenever x r and x s are meaningful and x s 6D 0 FACTORS AND EXPANSIONS Difference of squares: a 2 b 2 D a b a C b Perfect square: a C b 2 D a 2 C 2ab C b 2 Difference of cubes: a 3 b 3 D a b a 2 C ab C b 2 Sum of cubes: a 3 C b 3 D a C b a 2 ab C b 2 Perfect cube: a C b 3 D a( 3 C) 3a 2 b C 3ab ( 2 ) C b 3 Binomial theorem: a C b n n D a 0 b 0 n C a 1 b C ( ) ( ) n a 2 b 2 n CÐÐÐC a n b n

6 Chapter 2 29 where ( n 0 ) ( ) n D r ( ) n D 1, D n ( ) n 1 1, D 2 ( ) n! n r! n r!, ÐÐÐ, D 1 n n n 1, ÐÐÐ, 1 Ð 2 The following factoring procedure will work for most of the factoring problems you encounter in calculus: To factor an expression: First, look for the greatest common factor. Next, check to see if the expression is a special type: Difference of squares: x 2 y 2 D x y x C y Difference of cubes: x 3 y 3 D x y x 2 C xy C y 2 Sum of cubes: x 3 C y 3 D x C y x 2 xy C y 2 Perfect square: x 2 C 2xy C y 2 D x C y 2 x 2 2xy C y 2 D x y 2 Finally, if the expression is a trinomial, factor it into two binomials. Sometimes, grouping the terms will help you to factor. EXAMPLE 2.1 Common factoring Factor the following expressions: a. Common monomial factors: a 2 b C 5a 3 b 2 C 7a 2 b 3 b. Common binomial factors: 5x 3a 5b C 9y 3a 5b 1 c. Common factoring to provide integral coefficients: 36 x y d. Multiple common factoring: 2x x 1 C x 1 1 C x 3 3 a. The common factor is a 2 b: a 2 b C 5a 3 b 2 C 7a 2 b 3 D a 2 b 1 C 5ab C 7b 2 b. The common factor is 3a 5b : 5x 3a 5b C 9y 3a 5b D 5x C 9y 3a 5b

7 30 Chapter 2 c. Treat the fraction as a common factor: d. The common factor is 3 1 C x : 1 36 x y D 1 36 x y D 1 x 36y 36 2x x 1 C x 1 1 C x 3 3 D 3 1 C x [ 2x 3 1 x 1 1 C x 2 1 ] D 3 1 C x [ 2x 2x 2 3 C 3x 1 C 1 C 2x C x 2 ] D 3 1 C x [3x 2 3x 2] EXAMPLE 2.2 Factoring special types Factor the following expressions: a. 3x 2 75 b. x C 3y 3 C 8 c. 9a 2 a C 3b 2 b 2 d. x 6 1 a. 3x 2 75 D 3 x 2 25 Common factor first D 3 x 5 x C 5 Difference of squares b. Recognize this as a sum of cubes: x C 3y 3 C 8 D [ x C 3y C 2][ x C 3y 2 x C 3y 2 C 2 2 ] D x C 3y C 2 x 2 C 6xy C 9y 2 2x 6y C 4 c. Use common factoring to provide integral coefficients: 9a 2 b a C 2 3b 2 D 1 b 2 [9a2 b 2 a C 3b 2 ] Common factor D 1 [3a b a C 3b ][3a C b a C 3b ] b2 Difference of squares

8 Chapter 2 31 D 1 b 2 3a ab 3b2 3a C ab C 3b 2 d. Treat this as a difference of squares. x 6 1 D x D x 3 1 x 3 C 1 D x 1 x 2 C x C 1 x C 1 x 2 x C 1 EXAMPLE 2.3 Factoring trinomials Factor the following expressions: a. x 2 8x C 15 b. 6w 2 9w 15 c. 6 x C y 2 9 x C y 15 d. 4x 4 13x 2 y 2 C 9y 4 a. x 2 8x C 15 D x 5 x 3 b. 6w 2 9w 15 D 3 2w 2 3w 5 Common factor first D 3 2w 5 w C 1 c. 6 x C y 2 9 x C y 15 D 3[2 x C y 2 3 x C y 5] D 3[2 x C y 5][ x C y C 1] D 3 2x C 2y 5 x C y C 1 d. 4x 4 13x 2 y 2 C 9y 4 D x 2 y 2 4x 2 9y 2 D x y x C y 2x 3y 2x C 3y EXAMPLE 2.4 Factoring using negative exponents x 1 2 y 1 2 x 1 2 y 1 2 D x 1 2 y 1 2 y x D y x p xy RADICALS An algebraic expression containing radicals is simplified if all four of the following conditions are satisfied:

9 32 Chapter 2 1. When the radicand is written in completely factored form, there is no factor raised to a power greater than or equal to the index of the radical. 2. No radical appears in a denominator. 3. No fraction (or negative exponent) appears within a radical. 4. There is no common factor (other than 1) between the exponent of the radicand and the index of the radical. LOGARITHMS In y D log b x, y is called a logarithm and b is called the base. The logarithm (y) is defined as the exponent on a base b that equals the number x. Properties of logarithms log MN D log M C log N log M/N D log M log N log M n D n log M log np M D 1 n log M log b b D 1 log n 1 D 0 Special bases: log x D log 10 x;thisiscalleda common logarithm ln x D log e x,where e ³ ; is called the natural logarithm EXAMPLE 2.5 Using laws of exponents a. 2x 2 y 3 4 D 2 4 x 2Ð4 y 3Ð4 D 16x 8 y 12 b. 2x 2 C y 3 4 D 2x 2 4 C 4 2x 2 3 y 3 C 6 2x 2 2 y 3 2 C 4 2x 2 y 3 3 C y 3 4 D 16x 8 C 32x 6 y 3 C 24x 4 y 6 C 8x 2 y 9 C y 12 EXAMPLE 2.6 Simplifying radical expressions x 6 y 7 3 z 9 ( x 6 y 7 D z 9 ) 1/3 D x6 1/3 y 7 1/3 z 9 1/3 D x2 y 7/3 z 3 D x2 y 2 3p y z 3 Note that y 7/3 D y 6/3C1/3 D y 2C1/3 D y 2 y 1/3 D y 2 3 p y

10 Chapter 2 33 EXAMPLE 2.7 Simplifying expressions with negative exponents a. ( 1 x 1 C y 1 1 D x C 1 ) 1 ( ) x C y 1 D D xy y xy x C y b. x 1 y 1 1 D xy EXAMPLE 2.8 theorem Expanding an expression by using the binomial Expand 2x 3y 5. In the binomial theorem, replace a by 2x and b by 3y. With n D 5, we have 2x 3y 5 D 1 2x 5 3y 0 C 5 2x 4 3y 1 C 5 Ð 4 2 2x 3 3y 2 C 5 Ð 4 Ð 3 2 Ð 3 2x 2 3y 3 C 5 Ð 4 Ð 3 Ð 2 2 Ð 3 Ð 4 2x 1 3y 4 C 1 2x 0 3y 5 D 32x 5 240x 4 y C 720x 3 y x 2 y 3 C 810xy 4 243y Sequences and Series This material is required in Chapter 8 of the text. A sequence is a list, as in fs 0,s 1,s 2, ÐÐÐ,s n g; a series is a sum: s 0 C s 1 C s 2 CÐÐÐCa n. Arithmetic sequence If a n denotes the nth term, d the common difference, and n thenumberofterms,then a n D a 1 C n 1 d Arithmetic series If A n denotes the nth partial sum of an arithmetic sequence, then A n D n 2 a 1 C a n or A n D n 2 [2a 1 C n 1 d] Geometric sequence If g n denotes the nth term, r the common ratio, and n the number of terms, then g n D g 1 r n 1 for n ½ 1

11 34 Chapter 2 Geometric series If G n denotes the nth partial sum of a geometric sequence, then G n D g 1 1 r n,r 6D 1 1 r Infinite geometric series If G is the sum of an infinite geometric series, then G D g 1 1 r 0 for jrj < 1 SUMS OF POWERS OF THE FIRST n INTEGERS You can use these formulas to help find Riemann sums in Chapter 5 of the text. n 1 D n kd1 n k D 1 C 2 C 3 CÐÐÐCn D kd1 n k 2 D 1 2 C 2 2 C 3 2 CÐÐÐCn 2 D kd1 n n C 1 2 n n C 1 2n C 1 6 n k 3 D 1 3 C 2 3 C 3 3 CÐÐÐCn 3 D n2 n C 1 2 kd1 n kd1 k 4 D1 4 C2 4 C3 4 CÐÐÐCn 4 D n n C 1 2n C 1 3n2 C 3n SOLVING EQUATIONS Even though there are many aspects of algebra that are important to the scientist and mathematician, the ability to solve simple equations is important to the layperson and can be used in a variety of everyday applications. Pay attention to the difference between an expression and an equation. Also, note that to solve is not the same as to simplify. An equation is a statement of equality. There are three types of equations: true, false,andopen.anopen equation is an equation with a variable. 4

12 Chapter 2 35 A true equation is an equation that lacks variables and that is true, such as 2 C 3 D 5 A false equation is an equation that lacks variables and that is false, such as 2 C 3 D 15 Our focus is on open equations those equations with a variable. The values that make an open equation true are said to satisfy the equation and are called the solutions or roots of the equation. To solve an open equation is to find all replacements for the variable(s) that make the equation true. There are three types of open equations. Those that are always true, such as x C 3 D 3 C x are called identities. Those that are always false, such as x C 3 D 4 C x are called contradictions. Most open equations, such as 2 C x D 15 are true for some replacements of the variable and false for other replacements. These are called conditional equations. Generally, when we speak of equations, we mean conditional equations. Our concern in solving equations is to find the numbers that satisfy a given equation, so we look for things to do to equations to make the solutions or roots more obvious. Two equations with the same solutions are called equivalent equations. An equivalent equation may be easier to solve than the original equation, so we try to get successively simpler equivalent equations until the solution is obvious. There are certain procedures you can use to create equivalent equations. In this section, we will discuss solving the two most common types of equations you will encounter: linear and quadratic. Linear equations : ax C b D 0 a 6D 0 Quadratic equations : ax 2 C bx C c D 0 a 6D 0 Linear Equations To solve the first-degree (or linear) equation, isolate the variable on one side. That is, the solution of ax C b D 0is x D b, where a 6D 0 a

13 36 Chapter 2 To solve a linear equation, use the following steps: Step 1: Use the distributive property to clear the equation of parentheses. If the equation is a rational expression, multiply both sides by the appropriate expression to eliminate the denominator. Be sure to check the solutions obtained by plugging the values back into the original equation. Step 2: Add the same number to both sides of the equality to obtain an equation in which all of the terms involving the variable are on one side and all of the other terms are on the other side. Step 3: Multiply (or divide) both sides of the equation by the same nonzero number to isolate the variable on one side. EXAMPLE 2.9 Solving a linear equation Solve 4 x 3 C 5x D 5 8 C x. 4 x 3 C 5x D 5 8 C x 4x 12 C 5x D 40 C 5x 9x 12 D 40 C 5x 4x 12 D 40 4x D 52 x D 13 Check: C 5 13 D 145, and 5 8 C 13 D 145. EXAMPLE 2.10 Solving a rational equation Solve x C 1 x 2 D x C 2 x 2. x C 1 x 2 D x C 2 x 2 x 2 2x C x 2 D x 2 4 x 2 D 4

14 Chapter 2 37 x D 2 x D 2 Notice that x D 2 causes division by 0, so the solution set is empty. EXAMPLE 2.11 Solving a literal equation Solve 4x C 5xy C 3y 2 D 10 for x. 4x C 5xy D 10 3y 2 4 C 5y x D 10 3y 2 x D 10 3y2 4 C 5y,y 6D 4 5 Quadratic Equations: To solve a second-degree (or quadratic) equation, first obtain a zero on one side. Next, try to factor the quadratic. If it is factorable, set each factor equal to zero and solve. If it is not factorable, use the quadratic formula: HANDBOOK THEOREM 1 Quadratic Formula The solution of ax 2 C bx C c D 0is x D b š p b 2 4ac, where a 6D 0 2a Examples here are similar to Problems 7 18 in Problem Set 1.1. EXAMPLE 2.12 Solving a quadratic equation by factoring Solve 2x 2 x 3 D 0. Factoring yeilds 2x 3 x C 1 D 0. If 2x 3 D 0, then x D 3 2 ;ifx C 1 D 0, then x D 1. The solution is x D 3 2, 1. EXAMPLE 2.13 Solving a quadratic equation by using the quadratic formula Solve 5x 2 3x 4 D 0.

15 38 Chapter 2 a D 5,bD 3, and c D 4; thus (since the equation does not factor), x D 3 š p D 3 š p EXAMPLE 2.14 formula Solving a literal equation by using the quadratic Solve x 2 C 2xy C 3y 2 4 D 0forx. a D 1,bD 2y, c D 3y 2 4; thus, x D 2y š 4y y D 2y š 16 8y 2 2 D yš 4 2y 2 EXAMPLE 2.15 Solving an absolute-value equation Solve j5x C 2j D 12. Use Property 7 of Table 1.1 in the text to write the absolute-value equation as two equations: 5x C 2 D 12 5x D 10 x D 2 5x C 2 D 12 5x D 14 x D 14 5 The solution is x D 2, Exponential Equations: See Examples 4, 7, and 9 of Section 2.4 of the text. Logarithmic Equations: See Examples 5 and 8 of Section 2.4 of the text. Higher Degree Equations: See Examples 2.36, 2.37, and 2.38 on pp Trigonometric Equations: See Section 4.7, pp

16 Chapter 2 39 EXAMPLE 2.16 Solving absolute-value equations a. Solve j5x C 2j D12. b. Solve j5 3wj D 3. a. Use property 7 of Table 1.1 in the text to write two equations: 5x C 2 D 12 5x D 10 x D 2 5x C 2 D 12 5x D 14 x D 14 5 The solution is x D 2, b. Use property 1 of Table 1.1 of the text, jaj ½a, to see that there are no real values for w that make this equation true. 2.6 Completing the Square In Section 1.1 of the text, this procedure of completing the square is used to find the equation of a circle. In algebra, the method of completing the square was first introduced as a technique for solving quadratic equations. The method leads to a proof of the quadratic formula. However, when we are working with conic sections, we need to complete the square whenever the conic is not centered at the origin. (See Chapters 6 and 7 of this Handbook.) Consider Ax 2 C Bx C C D 0 We wish to put this into the form xc? 2 D some number To do this, we perform the following steps below: Step 1: Subtract C (the constant term) from both sides: Ax 2 C Bx D C Step 2: Divide both sides by A.(A 6D 0; if A D 0, the equation would not be quadratic.) That is, we want the coefficient of the squared term to be 1. The result of dividing by A is x 2 C B A x D C A

17 40 Chapter 2 ( ) B 2 Step 3: Add to both sides. That is, take one-half of the 2A coefficient of the first-degree term, square it, and add it to both sides: ) 2 D B2 x 2 C B A x C ( B 2A 4A 2 C A Step 4: The expression on the left is now a perfect square and can be factored: ( x C B ) 2 D B2 2A 4A 2 C A EXAMPLE 2.17 Completing the square Complete the square for x 2 C 2x 5 D 0. x 2 C 2x 5 D 0 x 2 C 2x D 5 x 2 C 2x C 1 D 5 C 1 x C 1 2 D 6 EXAMPLE 2.18 involves fractions Completing the square when the expression Complete the square for 3x 2 C 5x 4 D 0. 3x 2 C 5x 4 D 0 3x 2 C 5x D 4 x 2 C 5 3 x D 4 3 x 2 C 5 ( ) x C D C ( x C 5 ) 2 D

18 Chapter 2 41 EXAMPLE 2.19 Completing the square in two variables Complete the square in both the x and the y terms: x 2 C y 2 C 4x 9y 13 D 0 Associate the x and y terms: x 2 C 4x C y 2 9y D 13 ( x 2 C 4x C 4 C y 2 9y C 81 ) D 13 C 4 C ( x C 2 2 C y 9 ) 2 D EXAMPLE 2.20 Completing the square in two variables Complete the square in both the x and the y terms: 2x 2 3y 2 12x C 6y C 7 D 0 Associate the x and y terms: 2x 2 12x C 3y 2 C 6y D 7 In this example, we cannot divide by the coefficient of the squared terms, because we need to make both the x 2 and the y 2 coefficients equal to 1. Instead, we factor 2 x 2 6x 3 y 2 2y D 7 Now we complete the square; be sure you add the same number to both sides. 2 x 2 6x C 9 3 y }{{} 2 2y C 1 D 7C18 3 }{{} add 18 to both sides add 3 to both sides 2.7 Solving Inequalities 2 x y 1 2 D 8 Linear Inequalities The first-degree inequality is solved following the steps outlined in the previous section for solving first-degree equations. The only difference is that, when we multiply or divide both sides of an inequality by

19 42 Chapter 2 a negative value, the order of the inequality is reversed. That is, if a<b,then a C c<bccfor any number c a c<b c for any number c ac < bc for any positive number c ac > bc for any negative number c A similar result holds if we use,>,or½. Answers to inequality problems are often intervals on a real-number line. We use the following notation for intervals: Interval notation is first introduced in Section 1.1 of the text. Interval notation is used in stating the solutions to most inequalities. You might wish to review in this section how interval notation is used. closed interval (endpoints included) : [a, b] a b open interval (endpoints excluded) : half-open (or half-closed) interval : a a b a, b a, 1 a b a, b] a b [a, b a 1,a] To denote two intervals that are not connected, we use the notation for the union of two sets: a b c d [a, b [ c, d] a b c [a, b [ b, c] EXAMPLE 2.21 Solving a linear inequality Solve 5x C 3 < 3x 15.

20 Chapter x C 3 < 15 2x < 18 x< 9 The solution is 1, 9. EXAMPLE 2.22 Solving a linear inequality Solve 2x C 1 x 5 2x 3 x C 2 2x 2 9x 5 2x 2 C x 6 9x 5 x 6 10x 1 x ½ 1 10 The solution set is [0.1, 1). The word between is often used with a double inequality. We say that x is between a and b if a<x<band that x is between a and b, inclusive if a x b. EXAMPLE 2.23 Solving a between relationship Solve 5 x C x C x C x 1 The solution set is [ 9, 1]. We say that x is between 9 and 1, inclusive. Quadratic Inequalities The method of solving quadratic inequalities is similar to the method of solving quadratic equations:

21 44 Chapter 2 Step 1: Obtain a zero on one side of the inequality. Step 2: Factor if possible. If the inequality is a rational expression, factor the numerator and the denominator separately. Step 3: Set each factor equal to zero. These values are not necessarily the solution of the inequality. If the inequality is not factorable, we treat the entire expression as a single factor and solve by the quadratic formula. Values for which the factors are zero are called the critical values of x. Plot these values on a number line. The points determine one or more intervals on the line. Step 4: Choose some value in each interval. It will make the inequality true or false in that interval; we accordingly include or exclude that interval from the solution set. EXAMPLE 2.24 Solving a quadratic inequality by factoring Solve x 2 < 6 x. x 2 C x 6 < 0 x 2 x C 3 < 0 }{{} Factors Product of factors < 0 means that the product of factors is negative; product of factors > 0 means that the product of factors is positive. For this example, we are seeking values of x that make the product of the factors negative. Signs of factors: Solve: x 2 = 0 x = 2 This is a critical value. x 2: x < 2 x > 2 Write: x 2 is neg x 2 is pos x + 3 = 0 x = 3 This is a critical value. x + 3: x < 3 x > 3 x + 3 is neg x + 3 is pos Write:

22 We summarize these steps by writing neg.neg pos critical values neg.pos neg 3 2 pos.pos pos Chapter 2 45 product of factors We see from the number line that the segment labeled negative is 3, 2 ; this is the solution (except for a consideration of the endpoints). In this example, the endpoints are not included, so the solution is an open interval. These steps are summarized as signs of factors pos a neg b pos sign of product (negative because of <) EXAMPLE 2.25 Solving an inequality by examining the factors Solve 2 x x C 3 x 1 ½ 0. Plotting critical values and checking the signs of the factors, we have Endpoints included because of pos neg pos neg The solution is 1, 3] [ [1, 2]. EXAMPLE 2.26 Solving a rational inequality Solve x C 3 x 2 < 0. Be careful not to multiply both sides by x 2, since we do not know whether x 2 is positive or negative. We could consider separate cases, but instead we solve the inequality as if it were quadratic. We set the numerator and denominator each equal to zero to obtain the critical values x D 3,x D 2. We then plot these values on a number line and check a value in each interval to determine the solution. The solution is 3,

23 46 Chapter 2 EXAMPLE 2.27 Solving a rational inequality Solve x 3 x > 1. Since 3 < 0for 3 x,wehave x 3 1 > 0 x x 3 x > 0 x 3 x > 0 x < 0 x > 0 neg neg neg pos + pos 0 neg The solution is 1, 0. EXAMPLE 2.28 Solving a rational inequality Solve x C 2 2x ½ 5. x C 2 5 ½ 0 2x x C 2 10x ½ 0 2x 2 9x ½ 0 2x neg 0 2 pos 9 neg The solution is 0, 2 9 ]. The endpoints are included when we have intervals with ½ or. However, values of the variable that cause division by zero are excluded.

24 Chapter 2 47 EXAMPLE 2.29 factor Solving a quadratic inequality that does not Solve x 2 C 2x 4 < 0. The left-hand expression is in simplified form and cannot be factored. Therefore, we proceed by considering x 2 C 2x 4 as a single factor. To find the critical values, we find the values for which the factor is zero: x 2 C 2x 4 D 0 x D 2 š p D 1š p 5 Next, we plot the critical values and check the sign of the expression in each of the intervals: The solution is 1 p 5, 1 C p Determinants Determinants are used in calculus in several places. We first see them in Section 9.4 of the text, when they are used as memory devices of an operation called the cross product. In Sections 12.3 and 12.8, we use determinants to calculate what is called the Jacobian of a transformation. Finally, in Chapter 14, the Wronskian is defined in terms of an n ð n determinant. The following arrays of numbers are examples of matrices: ( ) ( ) a a1 b 1 b 1 ( ) 1 a1 b a a 2 b 2 b 1 c 1 d a a 3 b 2 b 2 c 2 d 2 3 The first of these matrices has two rows and two columns and is called a two-by-two matrix (written 2 ð 2 matrix ), the second is a 3 ð 2 matrix, since it has three rows and two columns, and the last one is a 2 ð 4matrix.Thea s, b s, c s, etc., that appear are called the entries of the matrix. An n ð n matrix (i.e., a matrix with n rows and n columns, and hence a matrix with the same number of rows and columns) is called a square matrix. Associated with each square matrix is a certain number called the determinant of the matrix. We shall show how to find the determinant of a 2 ð 2matrixanda3ð 3matrix.

25 48 Chapter 2 DETERMINANT [ ] a b If A is the 2 ð 2matrix, then the determinant of A c d is defined to be the number ad bc. Some notations for this determinant are det A, a c will generally use jaj. b d,andjaj. We WARNING Note that determinants always have the same number of rows and columns; that is, they are square: 2ð 2, 3 ð 3, 4 ð 4, ÐÐÐ. EXAMPLE 2.30 Evaluating determinants a D 4 Ð D 12 2 D 10 b D 2 Ð 2 2 Ð 2 D 0 c D 1 Ð 1 0 Ð 0 D 1 d D 1 Ð 4 2 Ð 2 D 0 We next define the determinant of a 3 ð 3 matrix. We need a preliminary observation. If we delete the first row and first column of the matrix [ ] a1 b 1 c 1 A D a 2 b 2 c 2 a 3 b 3 c 3 [ ] b2 c We obtain the 2 ð 2matrix 2.This2ð2matrixisreferredto b 3 c 3 as the minor associated with the entry in the first row and first column. Similarly, the minor associated with b 1 isobtainedbydeletingthefirst [ ] a2 c row and second column of A, and hence this minor is 2.The [ ] a 3 c 3 a2 b minor associated with c 1 is 2. The determinant of the 3 ð 3 a 3 b 3 matrix A is obtained by using these minors as follows: b jaj Da 2 c 2 1 b 3 c 3 b 1 a 2 c 2 a 3 c 3 C c 1 a 2 b 2 a 3 b 3 ( ) ( ) ( ) minor minor minor D a 1 associated b 1 associated C c 1 associated with a 1 with b 1 with c 1

26 Chapter 2 49 EXAMPLE 2.31 Evaluating a determinant D C D 2C4 10 C 2 D 40 a 0 0 It is not difficult to show that 0 b 0 D abc 0 0 c We found the determinant for a 3 ð 3 matrix by using the entries in the first row and their minors. In fact, it is possible to find the determinant using any row (or column); the formulas are similar and use the minors of the terms in the given row (or column), but adjustments must be made for the signs of the minors. The interested reader can find these results in any standard college algebra book. The following four properties of determinants are worth noting: HANDBOOK THEOREM 2 Properties of determinants Property 1 If B is the matrix obtained from A by multiplying a row (or column) of A by a real number k,thenjbj DkjAj. Property 2 If B is the matrix obtained from A by adding to a row (or column) of A a multiple of some other row (or column) of A, thenjbj DjAj. Property 3 If the matrix B is obtained from A by interchanging two rows (or columns), then jbj D jaj. Property 4 If two rows (or columns) of a matrix A are equal or proportional, then jaj D0. EXAMPLE 2.32 Properties of determinants illustrated a. Property 1: ka 1 kb 1 a 2 b 2 D a 1 b 1 ka 2 kb 2 D k a 1 b 1 a 2 b 2 b. ( Property) 2: Multiplying each entry in the first row of the matrix 1 2 D A by 2, and adding the result to the corresponding 2 5 ( ) 1 2 entry in the second row, we obtain the matrix B D,and 0 1 property 2 asserts that jaj DjBj. (Check for yourself to see that this is correct.) c. Property 3 (interchange the first and third rows): D

27 50 Chapter 2 d Note that 2 times an entry in row 1 is equal to the corresponding entryinrow2. EXAMPLE 2.33 Multiplicative property of determinants Verify that 2 ð 2 determinants have the following multiplicative property: a 1 b 1 c 1 d 1 a 2 b 2 c 2 d 2 D a 1 a 2 C b 1 c 2 a 1 b 2 C b 1 d 2 c 1 a 2 C d 1 c 2 c 1 b 2 C d 1 d 2 a 1 b 1 c 1 d 1 a 2 b 2 c 2 d 2 D a 1d 1 b 1 c 1 a 2 d 2 b 2 c 2 D a 1 a 2 d 1 d 2 b 1 c 1 a 2 d 2 a 1 d 1 b 2 c 2 C b 1 b 2 c 1 c 2 D a 1 a 2 C b 1 c 2 c 1 b 2 C d 1 d 2 a 1 b 2 C b 1 d 2 c 1 a 2 C d 1 c 2 D a 1a 2 C b 1 c 2 a 1 b 2 C b 1 d 2 c 1 a 2 C d 1 c 2 c 1 b 2 C d 1 d Functions Definition: A function f is a rule that assigns, to each element x of a set X, a unique element y of a set Y. (See Section 1.3 of the text.) In this handbook, we will simply list and categorize some of the more common types of functions you will encounter in calculus. ALGEBRAIC FUNCTIONS: Definition: An algebraic function is a function that satisfies an equation of the form a n x y n C a n 1 x y n 1 CÐÐÐCa 0 x D 0 where the coefficient functions a k x are polynomials in x. Polynomial Functions Definition: f x D a n x n C a n 1 x n 1 C a n 2 x n 2 CÐÐÐCa 2 x 2 C a 1 x C a 0,a n 6D 0 Constant function: f x D a

28 Linear function: f x D ax C b Standard form: Ax C By C C D 0 Point-slope form: y k D m x h Slope-intercept form: y D mx C b ( ) y2 y 1 Two-point form: y y 1 D x x 1 or x 2 x 1 x y 1 x 1 y 1 1 D 0 x 2 y 2 1 x Intercept form: a C y b D 1 Horizontal line: y D k Vertical line: x D h Quadratic function: f x D ax 2 C bx C c, a 6D 0 Cubic function: f x D ax 3 C bx 2 C cx C d, a 6D 0 Miscellaneous Algebraic Functions Absolute value function: f x Djxj D { x if x ½ 0 x if x<0 Chapter 2 51 Greatest integer function: f x D [[x]] This is the unique integer [[x]] satisfying [[x]] x<[[x]] C 1; this means that [[x]] is the largest integer not exceeding x. For example, [[2.3]] D 2, [[ 2.3]] D 3, and [[2]] D 2. Power function: f x D x r for any real number r. Rational function: f x D P x,wherep x and D x are two D x polynomial functions, with D x 6D 0. TRANSCENDENTAL FUNCTIONS: Definition: Functions that are not algebraic are called transcendental. Exponential function: f x D b x b > 0,b6D 1 Logarithmic function: f x D log b x b > 0,b6D 1 If b D 10, then f is a common logarithm: f x D log x If b D e,thenf is a natural logarithm: f x D ln x

29 52 Chapter 2 Trigonometric functions: Let be any angle in standard position, and let P x, y be any point on the terminal side of the angle a distance of r from the origin r 6D 0. Then cos D x r,sind y r,tand y x, sec D r x,csc D r y,andcot D x y Polynomials THEOREMS Let P x and Q x be polynomial functions. Zero factor theorem: Q x D 0 (or both). If P x Q x D 0, then either P x D 0or Remainder theorem: If P x is divided by x r until a constant is obtained, then the remainder is equal to P r. Intermediate-value theorem for a polynomial function: If P x is a polynomial function on [a, b] such that P a 6D P b,thenptakes on every value between P a and P b over the interval [a, b]. Factor theorem: If r is a root of the polynomial equation P x D 0, then x r is a factor of P x. Also, if x r is a factor of P(x), then r is a root of the polynomial equation P x D 0. Root limitation theorem: at most, n distinct roots. A polynomial function of degree n has, Location theorem: If P is a polynomial function such that P a and P b are opposite in sign, then there is at least one real root on the interval [a, b]. Rational root theorem: If P x has integer coefficients and has a rational root p/q (where p/q is reduced), then p is a factor of the constant term, a 0,andq is a factor of the leading coefficient, a n. Upper and lower bound theorem: If a>0 and, in the synthetic division of P x by x a, all the numbers in the last row are either positive or negative, then a is an upper bound for the roots of P x D 0. If b<0 and, in the synthetic division of P x by x b, the numbers in the last row alternate in sign, then b is a lower bound for the roots of P x D 0. Descartes s rule of signs: of x. Then LetP x be written in descending powers 1. The number of positive real zeros is equal to the number of sign changes or is equal to that number decreased by an even integer.

30 Chapter The number of negative real zeros is equal to the number of sign changes in P x or is equal to that number decreased by an even integer. Fundamental theorem of algebra: If P x is a polynomial of degree n ½ 1 with complex coefficients, then P x D 0 has at least one complex root. Number-of-roots theorem: If P x is a polynomial of degree n ½ 1 with complex coefficients, then P x D 0 has exactly n roots (if roots are counted according to their multiplicity). EXAMPLE 2.34 Synthetic division Divide x 4 C 3x 3 12x 2 C 5x 2byx 2. x b. To do synthetic division, the divisor must be of the form This is the number b. # Coefficients of 2j 2 Ł given polynomial " " Add " Remainder Bring down leading coefficient Ł The numbers in this row are found as follows: 2 Ð 1 D 2; 2 Ð 5 D 10; 2 Ð 2 D 4; 2 Ð 1 D 2 The degree of the result is one less than the degree of the given polynomial and has coefficients given by the last row (1, 5, 2, and 1 for this example; the last entry is the remainder). Thus, x 4 C 3x 3 12x 2 C 5x 2 x 2 D x 3 C 5x 2 2x C 1 EXAMPLE 2.35 remainder Synthetic division with zero coefficients and a Divide x 5 3x 2 C 1byx 2.

31 54 Chapter 2 Note: x 5 3x 2 C 1 D x 5 C 0x 4 C 0x 3 3x 2 C 0x C j R x 5 3x 2 C 1 x 2 EXAMPLE 2.36 D x 4 C 2x 3 C 4x 2 C 5x C 10 C 21 x 2 Solving a polynomial equation Solve 12x 3 6x 2 24x C 18 D 0. Using the rational root theorem, p Dš1, š2, š3, š6, š9, š18 and q D 1, 2, 3, 4, 6, 12. There are many possible rational roots of the form p. We try these values using synthetic division, until we find q a root that, in turn, allows us to write a factorization of polynomials of a degree lower than that of the original polynomial. We obtain The root is 1 (zero remainder), so x 1 and 12x 2 C 6x 18 are factors. Solving thus yields 12x 3 6x 2 24x C 18 D 0 x 1 12x 2 C 6x 18 D 0 6 x 1 2x 2 C x 3 D 0 6 x 1 2x C 3 x 1 D 0 The roots are 3 2, 1, where 1 has multiplicity 2. EXAMPLE 2.37 roots Solving a polynomial equation with repeated Solve r 4 5r 3 C 6r 2 C 4r 8 D 0. The possible rational roots are š1, š2, š4, and š Coefficients of 1j given polynomial j

32 Chapter 2 55 Solving the remaining quadratic: x 2 4x C 4 D 0 x 2 2 D 0 x D 2 The roots are 1 and 2. Note: The root 2 has multiplicity 3. EXAMPLE 2.38 rational roots Solving a polynomial equation that has no Solve x 4 3x 2 6x 2 D 0. p: š1, š2 andq: 1,so p q using synthetic division: : š1, š2. We try these values, Lower bound There are no rational roots. (We have tried all the numbers on our list.) Next, we verify the type of roots by using Descartes s rule of signs. First, f x D x 4 3x 2 6x 2 There is one sign change, so there is one positive root. Next, f x D x 4 3x 2 C 6x 2 There are three sign changes, so we have three negative roots or one negative root. Using synthetic division to find some additional points yields 0, 2 for the y-intercept; 1.5, ; 3, 34 ; there are no roots larger than 3 Using the intermediate-value theorem and synthetic division, we can find the roots to any desired degree of accuracy. We find the following:

33 56 Chapter Remember that there is a root between 0.41 and 0.42, since the remainder is positive for 0.42 and negative for We continue in this fashion to approximate the root to any degree of accuracy desired. This is a good problem for a calculator or a computer. We repeat the procedure for the other roots to find (correct to the nearest tenth) 0.4, 2.4. The graph can be used to verify that these are the only real roots. You may need to consult a precalculus textbook for a review of how to solve polynomial equations PROBLEM SET 2 Perform the indicated operations in Problems Ð Ð Ð /3 Ð 4 1/ /4 Ð 8 1/ /3 4 1/2 Perform the indicated operations Problems Assume that a, b, and c are positive real numbers. 13. a 2 b 3 c 5 Ð a 3 b 5 c a 2 b 3 c 5 Ð a 2 b 4 c 2 ( a 15. ab 2 c b 3 ) c ( 2 a 2 b 2 ) 1 ( a 2 b 2 ) c c ( 2 a 2 b 2 ) x 2 C y 2 c x 2 C y a 1/2 b 2/3 c 1/5 15 Factor the expressions in Problems a 1/2 b 1/2 C 3a 1/2 b 1/ a 1/2 b C 4a 1/2 25. a 1/2 b C a 1/2 b a 1/2 b 1/3 C a 1/2 b 1/3 27. x x 12 (Adapted from Problem 20, Section 4.3, of text.)

34 28. 12u 3 6u 2 24u (Adapted from Problem 22, Section 4.3, of text.) 29. 2x 2x C 3 2 C 2x 2 2x C 3 2 (Adapted from Problem 42, Section 3.5, of text.) 30. [2x] 4x C 5 3 C x 2 [3 4x C ] (From Example 9, Section 3.5, of text.) 31. x C x C 3 2 C 2x C x C x 2 C 1 3 4x x C 5 2x 2 C 1 4 2x x In Problems 33 58, solve each equation or inequality for x x 9 ½ x > x x > 3x x x< < 3x C x < x 2 C 5x 6 D x 2 C 5x C 6 D x 2 D 7x 44. 7x 2 D x D 3 4x x 2 D 12x x 2 C x w D x 2 C wx C 5 D x 2 4x C 1 t 2 D y D 2x 2 C x C x 2 3t C 10 x C 6t C 4 D 0,t>2 52. x 3 2 C y 2 2 D x C 1 2x C 5 7 3x > x 2 3x C 2 3 2x < 0 x 2x 1 x 55. > x 2x C 3 x 2 < x 2 C 4x C 5 ½ x 2 2x 6 0 In Problems 59 70, find the value of the given determinant. Chapter i j k (From Example 5, For variables i, j, and k. Section 9.4, of text.) (From Example 1, Section 9.4, of text.)

35 58 Chapter In Problems 71 76, expand the given quantity. 71. a C b a C 2b x C 2y x C 2y x C y p a C 2b 5 Solve the polynomial equations in Problems x 3 x 2 4x C 4 D x 3 x 2 18x C 9 D x 3 2x 2 9x C 18 D x 3 C 2x 2 5x 6 D x 3 C 3x 2 4x 12 D x 3 C x 2 13x C 6 D x 3 3x 2 32x 15 D x 4 12x 3 C 54x 2 108x C 81 D x 4 C 3x 3 19x 2 3x C 18 D x 4 13x 2 C 36 D x 3 C 15x 2 C 71x C 105 D x 5 C 6x 4 C x 3 48x 2 92x 48 D x 6 3x 4 C 3x 2 1 D x 7 C 3x 6 4x 5 16x 4 13x 3 3x 2 D Evaluate the determinant sin cos sin sin cos cos sin sin sin cos cos sin cos 0 sin This problem is from Example 6 of Section 12.8