Functions Modeling Change: A Precalculus Course. Marcel B. Finan Arkansas Tech University c All Rights Reserved

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1 Functions Modeling Change: A Precalculus Course Marcel B. Finan Arkansas Tech University c All Rights Reserved

2 PREFACE This supplement consists of my lectures of a freshmen-level mathematics class offered at Arkansas Tech University. The lectures are designed to accompany the textbook Functions Modeling Change:A preperation for Calculus written by the Harvard Consortium. This book has been written in a way that can be read by students. That is, the text represents a serious effort to produce exposition that is accessible to a student at the freshmen or high school levels. The lectures cover the Appendices A - H of the book as well as various sections from Chapters,, 3, 4, 5, 6, 7, 8, and 9. These chapters are well suited for a 4-hour one semester course in Precalculus Marcel B. Finan April 003

3 Contents Review of Basic Algebra 5. Exponents Multiplying Algebraic Expressions Factoring Algebraic Expressions Fractions of Algebraic Expressions Solving Equations Solving Inequalities The Concept of a Function. Functions The Concept of Proportionality The Rate of Change Evaluating Functions Changes in Input and Output Domain and Range of a Function Piecewise Defined Functions Linear Functions What is a Linear Function? Formulas for Linear Functions Solving Systems of Equations Geometric Properties of Linear Functions Fitting Linear Functions to Data Exponential and Logarithmic Functions Introduction to Exponential Functions Exponential Functions Versus Linear Functions Graphs of Exponential Functions Logarithms and Their Properties The Logarithmic Function Logarithmic and Exponential Equations Continuous Growth Rate and the Number e Compound Interest

4 4 CONTENTS 5 Building New Functions from Old Ones 7 5. Composition of Functions Inverse Functions Combinations of Functions Vertical and Horizontal Translations Reflections and Symmetry Vertical Stretches and Compressions Horizontal Stretches and Compressions Polynomial and Rational Functions Completing the Square Procedure and the Quadratic Formula The Family of Quadratic Functions Polynomial Functions Rational Functions Trigonometric Functions Angles and Arcs The Trigonometric Functions Periodicity of the Trigonometric Functions Graphs of the Sine and Cosine Functions Sinusoidal Functions Other Trigonometric Functions Inverse Trigonometric Functions Inverse Sine and Cosine Functions Inverse Tangent, Cotangent, Secant, and Cosecant Functions Trigonometric Equations The Trigonometry of Triangles Trigonometric Functions of Acute Angles The Law of Sines The Law of Cosines Trigonometric Identities Verification of Trigonometric Identities Sum and Difference Identities The Double-Angle and Half-Angle Identities

5 Chapter Review of Basic Algebra Many of the content of this chapter were assumed to be covered in either highschool or in remedial mathematics classes. It may be good to review some basic rules and definitions of algebra. As you work through the problems, try to make the vocabulary and the manipulations second nature so that you can use them quickly and appropriately as we progress later on in the course.. Exponents Let a be any real number. For any positive integer n we define a n = a a a a }{{} n factors where we call a the base and n the exponent. For a 0 we define a 0 =. For negative integers, we define a n = a. Note that any number a is a power with n exponent equals to. That is, a = a because here we have only one factor of a. Also, it is worth noticing that a = n = a n. a n Example (i) Simplify π 0 and ( 3) 0. (ii) Use a calculator to compute 0 0. (iii) Write the following with positive exponent: (0.75) 6 and 3 4. (i) π 0 = ( 3) 0 = since any nonzero number raised to the power 0 is. (ii) Using a calculator we get an error message. This means that 0 0 is not a number, like 0. (iii) (0.75) = (0.75) 6 and 3 4 = 6 3 = 4 8. Remark. Be aware of the following notational conventions: 5

6 6 CHAPTER. REVIEW OF BASIC ALGEBRA (i) a n = ( )a n and a n ( a) n. (ii) ab n = ( a)(b n ). Thus, the notation 4 means 4 multiplied by, that is, 4 = 6. On the other hand, ( ) 4 means the product of ( ) by itself four times. That is, ( ) 4 = ( )( )( )( ) = 6. Example Calculate 3 3 ( 4). 3 3 ( 4) = 7 6 = 43. It is important to keep in mind the following Rules of Exponentiation: (i) Product Rule: a m a n = a m+n. This is obvious, since a m a n is the product of m-factors of a and n-factors of a, a total of (m + n) factors. a (ii) Quotient Rule: m a = a m n, a 0. n To see this, suppose first that m > n. Then m = n + k for some positive integer k. In this case, a m a n = n factors of a {}}{ a a a a a a }{{} n factors of a Now, if m < n then n = m + k and a m a n = a a a }{{} m factors of a m factors of a {}}{ a a a (iii) Power Rule: (a m ) n = a mn. Applying rule (i) we see that k factors of a {}}{ a a a a a a }{{} k factors of a = = a } a {{ a } = a k = a m n k factors of a = a } a {{ a } a k = a k = a m n k factors of a n factors of m {}}{ (a m ) n = a } m a m {{ a m } = a m + m + m = a m.n n factors of a m (iv) Raising a product to a power: (ab) n = a n b n. To see this, (ab) n = (ab)(ab) (ab) = }{{} n factors of ab n factors of a n factors of b {}}{ b b b {}}{} a a a {{} Since is commutative = a n b n

7 .. EXPONENTS 7 (v) Raising a quotient to a power: ( ) a n b = a n b. n To see this, ( a b ) n = ( a b ) ( a ) b ( a ) b }{{} n factors of a b = n factors of a {}}{ a a a = an b } b {{ } b b n. n factors of b Exercise Is it true that the power of a sum is always equal to the sum of powers? That is, do we always have (a + b) n = a n + b n? If not, give an example. Let a =, b =, and n =. Then (a + b) n = ( + ( )) = 0 and a n + b n = + ( ) = so that (a + b) n a n + b n. Scientific notation for a number is an expression of the form N 0 n where N < 0, N is a decimal number; n is positive integer if the number is greater than 0 and a negative integer if the number is between 0 and. Example 3 Convert each of the following to decimal notation: (a) (b) (a) = (b) = Exercise The mass of a neutron is Express this number in scientific notation = Next, we would like to define a power with fractional exponent. A number c is said to be an nth root of a if c n = a. We write c = n a. We call a the radicand and the symbol n the radical. Note that if n is even and a < 0 then a = c n > 0 a contradiction so that n a is undefined for n even and a < 0.

8 8 CHAPTER. REVIEW OF BASIC ALGEBRA Rules of Radicals (i) If n is even then n a n = a for if c = n a n 0 then c n = a n. For n even and a 0 we must have c = a. For n even and a 0 we must have c = a. Thus, c = a. (ii) If n is odd then n a n = a. For if c = n a then only solution to the equation c n = a n is c = a. (iii) n ab = n a n b For if c = n ab, d = n a and e = n b then c n = ab = d n e n = (d e) n That is, c = d e. (iv) n a b = n a n b This result is established using a similar argument to that of (iii) (Try to prove it!) Remark Formula (iii) requires that a, b 0 for n even and formula (iv) requires that a 0 and b > 0 for n even. Removing the radicals in a denominator or a numerator is called rationalizing the denominator or rationalizing the numerator. Pairs of expressions of the form a + b and a b are called conjugates. Rationalizing is done by multiplying an expression by its conjugate. Recall that (a b)(a + b) = a b. Exercise 3 (a) Rationalize the denominator: 3+ (b) Rationalize the numerator: x + h x (a) (b) 3 + = 3 (3 + )(3 ) = 3. ( x + h x)( x + h + x) x + h x = = x + h x h =. x + h + x x + h + x x + h + x The expression n a m can be written in terms of a radical exponent such as n am = a m n. Note that for n even we require that a m 0. Example 4 Complete the following: (i) a n = (ii) a m n =

9 .. MULTIPLYING ALGEBRAIC EXPRESSIONS 9 (i) a n = n a. (ii) a m n = n a m Exercise 4 Simplify using no negative exponents in the final answer: (x y) 3 x 5 y 4 (xy ) (x y) 3 x 5 y 4 (xy ) = x 3 y 3 x 5 y 4 x y 4 = x 4 y 7 y x = y 4 x Exercise 5 Compute without using a calculator: ( 8 7 ) 3. Exercise 6 Simplify 3 6. ( 8 7 ) 3 = [ ( 3 ) 3 ] 3 = ( ) = = 3 3 = 3. Recommended Problems (pp ): 7,, 3, 4, 6, 9, 3, 30, 3, 33, 37, 4, 45, 47, 48, 50, 5, 56, 57, 59, 6, 68, 69, 73, 78, 79, 8, 83, 85, 88, 90, 9, 93, 95.. Multiplying Algebraic Expressions The multiplication of algebraic expressions is based on the distributive property for real numbers: a(b + c) = ab + ac (b + c)a = ba + ca To multiply (or expand) algebraic expressions, multiply each term of the first expression by each term of the second and then combine like terms. To multiply two algebraic expressions, each is the sum of two terms, one can use the so-called FOIL procedure: Multiply the first terms, then the outer terms, then the inner terms, then the last terms. Then collect like terms, if possible.

10 0 CHAPTER. REVIEW OF BASIC ALGEBRA Example 5 f (x-7)(3x+4) i o l F O I L = 6x + 8x - x - 8 = 6x - 3x - 8 Example 6 Expand the following and simplify by gathering like terms: (i) (a b)(a + b) (ii) (a + b) (iii) (a b). (i) (a b)(a + b) = a + ab ba b = a b since ab = ba by the commutative property of multiplication. (ii) (a + b) = (a + b)(a + b) = a + ab + ba + b = a + ab + b. (iii) (a b) = (a b)(a b) = a ab ba + b = a ab + b. Exercise 7 Expand the following and simplify by gathering like terms: (i) (3x )(x + x) (ii) (3 x) (i) (3x )(x + x) = 3x 3 + 6x x x = 3x 3 + 5x x. (ii) (3 x) = (3 x)(3 x) = 9 3 x 3 x + 4 x = 9 3x + 4 x. Recommended Problems (p. 500):,, 8,, 3, 6, 7, 9,, 5, 6, 9..3 Factoring Algebraic Expressions Factoring is the reverse process of multiplying; that is, we write the given expression as a product. This can be done in using some techniques that we discuss in this section. Factoring by removing a common factor. When an algebraic expression is to be factored, we should always look first to factor out a factor that is common to all the terms using the distributive property. Example 7 Factor

11 .3. FACTORING ALGEBRAIC EXPRESSIONS (i) a b (ii) x y 0x 3 y (iii) a(a b) (b a) (i) a b = (a + b) (ii) x y 0x 3 y = 4x y(3y 5x) (iii) a(a b) (b a) = a(a b) + (a b) = (a + )(a b) Grouping Terms Another way to factor algebraic expressions is to group terms and then remove common factors. Example 8 Factor (i) x hx x + h (ii) hx + 4hx 3x (iii) x 3 x 8x + 4 (i) x hx x + h = x(x h) (x h) = (x )(x h). (ii) hx + 4hx 3x = hx 4hx 3x + = hx(x 4) 3(x 4) = (hx 3)(x 4). (iii) x 3 x 8x + 4 = x (x ) 4(x ) = (x 4)(x ) = (x )(x + )(x ) Factoring Quadratics To factor a quadratic expression of the form x +bx+c, we look for two numbers whose product is c and sum is b. The two numbers are found by trial and error. Example 9 Factor (i) x 6x + 9 (ii) x + x (iii) x 4x. (i) x 6x + 9 = (x 3)(x 3) = (x 3) (ii) x + x = (x 3)(x + 4) (iii) x 4x = (x 6)(x + ). Now, to factor an expression of the form ax + bx + c with a 0, we use FOIL method in reverse. We look for two factors of the form (ux + v)(wx + t)) such that uw = a, vt = c, and ut + vw = b. Example 0 Factor (i) 3x 0x 8 (ii) 3x x

12 CHAPTER. REVIEW OF BASIC ALGEBRA (i) 3x 0x 8 = (3x + )(x 4) (ii) 3x x = (3x + )(x ) Special Factorizations The following identities are important in factoring: Example Factor a b = (a b)(a + b) (difference of two sqaures) (a + b) = a + ab + b (square of a sum) (a b) = a ab + b (square of a difference) a 3 b 3 = (a b)(a + ab + b ) (difference of cubes) a 3 + b 3 = (a + b)(a ab + b ) (sum of cubes) (i) 4x 9 (ii) x 4 + 8x (iii) x + 8x + 6 (iv) 5x 30x + 9 (v) x 3 7 (i) 4x 9 = (x 3)(x + 3) (ii) x 4 + 8x = x(x 3 + 8) = x(x + )(x x + 4) (iii) x + 8x + 6 = (x + 4) (iv) 5x 30x + 9 = (5x 3) (v) x 3 7 = (x 3)(x + 3x + 9) Recommended Problems (p. 50): 5, 8, 0,, 4, 6, 9,, 3, 6, 9, 30, 3, 35, 38, Fractions of Algebraic Expressions By a an algebraic fraction we mean a ratio of the form a b or a b where a and b are two algebraic expressions. We call a the numerator and b the denominator. Like numeric fractions, algebraic fractions obey the following rules: (i) a c + b c = a+b c, c 0. (ii) a b + c d = ad+bc bd, b 0, d 0. (iii) a b c d = ac bd, b 0, d 0. = a b d, b 0, d 0. (iv) a b c d c = ad bc (v) a b = a b = a b, b 0. Example Simplify: x x+ (a) 3x (b) x x+. 3x

13 .4. FRACTIONS OF ALGEBRAIC EXPRESSIONS 3 (a) (b) x x+ 3x x x+ 3x = x x+ 3x = x 3(x+). = x 3x x+ = 3x3 x+. Finding A Common Denominator Property (iii) above shows that for any nonzero number c we have a b = ac bc That is, the value of a fraction is unchanged if both terms are multiplied by the same nonzero number. When adding or subtracting fractions with different denominators, it is recommended that the denominator of the result be the least common denominator or LCD. To find the LCD of two expressions a and b, decompose each expression as a product of irreducible factors( i.e. factors that can not be written as product themselves such as x 4 = (x )(x+) or (x +4x+4) = (x+) ), and then take all the factors that appear in each decomposition. For common factors take the one with highest exponent. For example, to find the LCD of x 4 and (x 3 8) we decompose x 4 = (x )(x + ) and (x 3 8) = (x ) (x + x + 4). So LCD(x 4, (x 3 8) ) = (x ) (x + )(x + x + 4). Reduced Form of a Fraction It is recommended that when dealing with algebraic fractions, to write the final answer in reduced form. That is, the top and the bottom have no common factors. For example, x 4 x = (x )(x+) x = x +. Example 3 Perform the following operations. Express your answers in reduced form. (i) 4 x + 3 x. (ii) 9 x +5x+6 + x+3. 4 (i) x + 3 x = 8 (x ) + 3 (x ) = 4 (x ). 9 (ii) x +5x+6 + x+3 = 9 (x+)(x+3) + (x+) (x+)(x+3) = 9+(x+) (x+)(x+3) = x+33 (x+)(x+3). Warning. You can always multiply the top and bottom of a fraction by the same number without changing the value of the fraction. On the contrary, the value of a fraction is changed if you add or subtract the same number to the top and bottom of a fraction. Also, squaring the top and the bottom of a fraction will result in a different fraction. Be careful with cancellation. Cancellation occurs only when you have a common factor in both the numerator and the denominator. For example, x + 4 x + 5x + 6 = (x + ) (x + )(x + 3) = x + 3.

14 4 CHAPTER. REVIEW OF BASIC ALGEBRA You cannot cancel for example the x in the expression because x is not a common factor. x x+ to obtain + = Splitting Fractions By reversing the rule for adding or subtracting fractions, we can split up an algebraic fraction into two fractions: a + b c = a c + b c or a b c Example 4 Split 3x + x into two reduced fractions. 3 = a c b c 3x + x 3 = 3x x 3 + x 3 = 3 x + x 3. Recommended Problems (p ):, 5, 8, 0,, 5, 8, 9, 3, 6, 8, 35, 37, 4, 44, 5, 53, 55, 58, 59, 6, 64, Solving Equations An equation is an equality between two algebraic expressions. A number that satisfies an equation is called a solution, zero or a root. Example 5 Which of the following numbers is a solution to the equation 3x + 7 = 5 (A) 4 (B) 7 (C) 4 (D) 5 3 (E) None of the above. Since 3( 4) + 7 = 5 then the answer is (A). The collection of all solutions to an equation is called the solution set. To solve an equation is to find the solution set. Four important principles for solving equations: () The Addition Principle: Adding or subtracting the same number to both sides of an equation does not change the solution set of the original equation. () The Multiplication Principle: Multiplying (respectively dividing) both sides of an equation by the same number (respectively nonzero number) does not change the solution set of the original equation.

15 .5. SOLVING EQUATIONS 5 (3) The Principle of Zero Products: If ab = 0 then a = 0 or b = 0. (4) The Principle of Square Roots: If x = k with k 0 then x = ± k. Example 6 Solve 3x 5 = 4 Adding five to both sides to obtain 3x = 9 and now dividing both sides by 3 to obtain x = 3. Example 7 Solve x 5 = 3x + 7 Subtract 7 + x from both sides to obtain x =. Exercise 8 Sometimes to solve an equation, you must simplify the equation before using the properties listed above. Solve (x + )(x + 3) = x + 6 Multiplying the left-hand side to obtain x + 5x + 6 = x + 6. Subtract x + 6 from both sides to otbain 5x = 0 or x =. A radical equation is an equation in which variables appear in one or more radicals. This type of equations is usually solved by using the so called Principle of Powers: If a = b then a n = b n where n is a positive integer. When solving this type of equations we must always check for extraneous solutions at the end. Example 8 Solve: x + 3 = 7. Squaring both sides to obtain x + 3 = 49. Solving this equation to get x = 3. Note that = 49 = 7 so x = 3 is indeed a solution. Exercise 9 Solve: x = x 3. Squaring both sides to obtain 4x = (x 3) or 4x = x 6x + 9. Rearranging this equation to get x 0x + 9 = 0. Factoring, we have (x )(x 9) = 0.

16 6 CHAPTER. REVIEW OF BASIC ALGEBRA Solving we find x = or x = 9. Substituting into the original equation we get = so that x = is not a solution. Now, substituting x = 9 to obtain 6 = 6 so that x = 9 is the only solution. A rational equation is an equation that contains rational expressions. This type of equations is solved by multiplying through by the Least Common Denominator. However, when solving this type of equations one must check for extraneous solutions at the end. Example 9 Solve x + x x = 0 The least common denominator is x. Multiplying through x and cancelling to obtain + x(x + ) = 0 or x + x = 0. Factoring, we find (x )(x + ) = 0 and solving this equation to obtain x = or x =. The value x = is not a solution since it leads to a division by zero. Exercise 0 Solve x + x(x + 3) + x = 3 x + 3 Multiplying through by x(x + 3) to obtain or (x + ) + x + 3 = 6x 6x + 5 = 6x. Subtracting 6x from both sides we find 5 = 0 which is impossible so the given equation has no real solutions. Another way to solve equations is by means of factoring to obtain a product of the form ab = 0 and then applying the Principle of Zero Products to get a = 0 or b = 0. Example 0 Solve the equation: x 3 = 4x. Rearranging to obtain x 3 4x = 0 or x(x )(x + ) = 0. The roots are then x = 0, x =, x =. Exercise Solve: (x + )(x + 3) = 5.

17 .6. SOLVING INEQUALITIES 7 Multiplying on the left x + 4x + 3 = 5 and subtracting 5 from both sides to obtain x + 4x = 0. Factoring we find (x + 6)(x ) = 0. Solving we find x = or x = 6. Recommeded Problems (pp. 55-6): 7,, 3, 0, 3, 7, 30, 33, 4, 5, 5, 54, 65, 66, 68, 75, Solving Inequalities Sets of real numbers can be expressed as intervals. For example, if a and b are real numbers such that a < b then the open interval (a, b) consists of all real numbers between, but not including, a and b. In set-builder notation we write (a, b) = {x a < x < b}. The points a and b are the endpoints of the interval. The parentheses indicate that the endpoints are not included in the interval. If an endpoint is to be included then we use a bracket instead of a parenthesis. Exercise Find the set-builder notation of the following intervals and then graph on the real line: (a) (a, b] (b) [a, b) (c) [a, b] (a) (a, b] = {x a < x b}. (b) [a, b) = {x a x < b}. (c) [a, b] = {x a x b}. a a a b b b (a,b] [a,b) [a,b] Some intervals extend without bound in one or both directions. For example, the interval [a, ) begins at a and extends to the right without bound. In set-builder notation we have [a, ) = {x x a}.

18 8 CHAPTER. REVIEW OF BASIC ALGEBRA Exercise 3 Write the set-builder notation of the following intervals: (i) (a, ) (ii) (, b) (iii) (, b] (iv) (, ) (i) (a, ) = {x a < x}. (ii) (, b) = {x x < b}. (iii) (, b] = {x x b}. (iv) (, ) = {x x is any real number} Exercise 4 Write interval notation for each set: (i) {x 0 x < 5}. (ii) {x x > }. (iii) {x x 7}. (i) {x 0 x < 5} = [ 0, 5). (ii) {x x > } = (, ). (iii) {x x 7} = (, 7) (7, ). A linear inequality is a linear equation with the equal sign replaced by an inequality symbol such as <, >,, or. For example, ax + by c. Two important principles for solving inequalities: Addition Principle: If a < b then a + c < b + c and a c < b c. Multiplication Principle: If a < b and c > 0 then ac < bc. If c < 0 then ca > cb. Thus when multiplying by a negative number, we must reverse the inequality sign. Exercise 5 Solve the inequality: 5x + 6 > 0. Write the answer in interval notation. 5x + 6 > 0 5x > 0 6 5x > 4 ( 5 In interval notation, (, 4 5 ). ) ( 5x) < Exercise 6 Solve the inequality: 6 x x 9 x ( 5 x < 4 5 ) (4)

19 .6. SOLVING INEQUALITIES 9 In interval notation, (, 3). 6 x + x < x + x < 9 x + x 6 < 3x < < 3 3x < 3 9 < x < 3 A polynomial inequality is a polynomial equation with the equality sign replaced by one of the inequality symbols. To solve a polynomial inequality one first finds the zeros of the polynomial equation. These zeros divide the x-axis into intervals in which the sign of the polynomial is tested using test values. Finally, determine the intervals for which the inequality is satisfied and write the solution set in interval notation. Remember to use brackets when endpoints are in the solution set( this usually occurs with inequalities such as or ). Exercise 7 Solve (x )x (x )x. Rewriting to find (x )x(x ) 0. The zeros of the left-hand side are x =, x =, or x = 0. The figure below shows the sign of the product on each subintervals. (-)(-)(-) (-)(+)(-) (-)(+)(+) (+)(+)(+) 0 (-) (+) (-) (+) The interval of solution is [0, ] [, ). Exercise 8 Solve : x(x ) > 6. Rewriting to find (x 3)(x + ) > 0. The zeros of the left-hand side are x =, or x = 3. The figure below shows the sign of the product on each subintervals. (-)(-) (-)(+) (+)(+) - 3 (+) (-) (+) The interval of solution is (, ) (3, ).

20 0 CHAPTER. REVIEW OF BASIC ALGEBRA Exercise 9 Solve: x 3 + 5x 0x. Rewriting to find x(x 5) 0. Since (x 5) 0 for all x then the inequality is valid for all x 0. The interval of solution is (, 0]. Rational inequalities are inequalities that involve rational expressions. To solve a rational inequality we start by simplifying the inequality so that we get 0 on the right side and a fraction on the left side. Next find the values of x that make the numerator zero or the denominator undefined. Now proceed with these values in a similar fashion as the procedure of solving a polynomial inequality. Be careful not to use brackets at endpoints where the rational expression is undefined. Exercise 0 x Solve: x Taking common denominator we find 3x+6 x+3 0. The figure below shows that the interval of solution is the interval ( 3, ]. (-) (-) (-) (+) -3 - (+) (+) Exercise x Solve: x 5 x+5. (+) (-) (+) x Taking common denominator we find +5 (x )(x+5) 0. The figure below shows that the interval of solution is the interval (, 5) (, ). (+) (-)(-) (+) (-)(+) (+) (+)(+) -5 (+) (-) (+) Recommended Problems (pp. 54-5): 3, 5, 6, 8,,, 3, 6, 8, 0, 5, 6, 9, 3, 3, 39, 4, 45, 49, 53, 56, 57.

21 Chapter The Concept of a Function In this chapter, we will introduce a concept that is fundamental to many areas of mathematics-the concept of a function. We will see that functions can be represented by graphs, charts, and formulas that can be used to model real-life situations mathematically.. Functions Functions play a crucial role in mathematics. The goal of this section is to introduce this important concept. Before giving the definition of a function we first consider an application. Example The sales tax on an item is 6%. So if p denotes the price of the item and C the total cost of buying the item then if the item is sold at $ then the cost is + (0.06)() = $.06. If the item is sold at $ then the cost of buying the item is + (0.06)() = $. and so on. Thus we have a relationship between the quantities C and p such that each value of p determines exactly one value of C. In this case, we say that C is a function of p. The relationship between C and p can be expressed by a table, a graph, or a formula. The chart below gives the total cost of buying an item at price p as a function of p. p C The graph of the function C is obtained by plotting the data in the above table to obtain

22 CHAPTER. THE CONCEPT OF A FUNCTION The formula that describes the relationship between C and p is given by C(p) =.06p. By a function we mean a relationship f between two quantities x and y such that each value of x yields a unique value y. We call x the input and y the output. In function notation we write y = f(x) Note that y depends on x so that we call x the independent variable and y the dependent variable. Exercise Using the following table, determine which variable is a function of what variable. x y Since each value of x determines exactly one value of y then y is a function of x. On the contrary, x is not a function of y since the input 73 has two output values 7 and 0. So x is not a function of y. Now, most of the functions that we will encounter in this course have formulas. For example, the area A of a circle is a function of its radius r. In function notation, we write A(r) = πr. However, there are functions that can not be represented by a formula. For example, if D is a day in July and T is lowest temperature in that day, then T is a function of D. Note that T can not be expressed as a formula in D. Functions of this nature, are mostly represented by either a graph or a table of numerical data. Exercise 3 The table below shows the daily low temperature for a one-week period in New

23 .. FUNCTIONS 3 York City during July. (a) What was the low temperature on July 9? (b) When was the low temperature 73 F? (c) Is the daily low temperature a function of the date?explain. (d) Can you express T as a formula? D T (a) The low temperature on July 9 was 69 F. (b) On July 7 and July 0 the low temperature was 73 F. (c) T is a function of D since each value of D determines exactly one value of T. (d) T can not be expressed as a formula in terms of D otherwise there would be no need for meteorologists. Next, we present a relationship between two quantities x and y that is not a function. Exercise 4 Let x and y be two quantities related by the equation (a) Is x a function of y? Explain. (b) Is y a function of x? Explain. x + y = 4. (a) For y = 0 we have two values of x, namely, x = and x =. So x is not a function of y. (b) For x = 0 we have two values of y, namely, y = and y =. So y is not a function of x. Suppose that the graph of a relationship between two quantities x and y is given. To say that y is a function of x means that for each value of x there is exactly one value of y. Graphically, this means that each vertical line must intersect the graph at most once. Hence, to determine if a graph is a function one uses the following test: Vertical Line Test: A graph is a function if and only if every vertical line crosses the graph at most once. According to the vertical line test and the definition of a function, if a vertical line cuts the graph for example twice, the graph could not be the graph of a function since we have two y values for the same x value and this violates the definition of a function.

24 4 CHAPTER. THE CONCEPT OF A FUNCTION Exercise 5 Which of the graphs (a) through (i) represent y as a function of x? (Note that an open circle indicates a point that is not included in the graph; a solid dot indicates a point that is included in the graph.) c a b d e f g h j (a), (d), and (g) are functions. The rest are not by the vertical line test. Recommended Problems (pp. 6-0):, 3, 5, 6, 7, 0, 3, 5,, 5.. The Concept of Proportionality Suppose you live in a city where the sales tax is 7.87%. Let p be your total purchases in dollars from a grocery store. Then what will you be actually paying at the cash register? If we denote by C the amount you are supposed to pay to the cashier then C consists of the amount p dollars plus the sales tax 7.87%p = p dollars, a total of C =.0787p. In this case, we say that the cost C is directly proportional to p. In general, a quantity y is said to be directly proportional or simply proportional to the nth power of a quantity x if there is a constant k such that y = kx n, n > 0. (.) We call k the constant of proportionality. Note that for x = 0, y = 0. That is, if y is proportional to a power of x then the graph must definitely cross the origin.

25 .. THE CONCEPT OF PROPORTIONALITY 5 Example In the formula y = π x, y is proportional to x with constant of proportionality π. Similarly, the area of a circle is directly proportional to the square of the radius with constant of proportionality π. On the other hand, in the formula y = + 5x, y is not proportional to x since the graph does not cross the origin. k x = k n In (.), we required that n > 0. If n < 0 then we can write y = x. In m this case, we say that y is inversely proportional to the mth power of x with constant of proportionality k. Note that for inverse proportionality, when x increases, y decreases (i.e. graph falls to the right) and when x decreases, y increases (i.e. graph rises to the right). Moreover, the graph will never cross the x-axis since there is no x value where y = 0. Example 3 In making a 500-mile trip by car, the time it takes depends on the speed of the car. The greater the speed, the less the time it will take. If you decrease the speed, the time increases. Thus, the time t is inversely proportional to the speed S. From Physics, the constant of proportionality is the distance traveled. Therefore, t = 500 S. Exercise 6 Suppose that a is inversely proportional to b, and when b = 5, a =. Find a when b =. Since a is inversely proportional to b then a = k b. Since b = 5 when a = then k 5 = or k = 5. In this case, a = 5 5 b. So for b =, a = 4. Inverse propor- Exercise 7 Which of the following could describe direct proportionality? tionality? Explain. (a) (b) (c) (d)

26 6 CHAPTER. THE CONCEPT OF A FUNCTION (a) Neither since the graph does not cross the origin and as x increases y increases. (b) Direct proportionality since graph crosses the origin. (c) Inverse proportionality since y decreases as x increases. (d) Neither since the graph crosses the x-axis and does not cross the origin. In many applications, when a quantity y is proportional to a quantity x and both quantities have units then the constant of proportionality k has also units which is the units of y divided by the units of x. In this case, we refer to k as a rate. For example, the distance d traveled by a car is proportional to the time t spent traveling. That is, d = kt. In this case, k = d t. If d is expressed in miles and t is hours then k is expressed as miles/hour. That is, k is the speed of the car. Exercise 8 The distance a car travels on the highway is proportional to the quantity of gas consumed. A car travels 5 miles on 5 gallons of gas. Find the constant of proportionality and explain its meaning. Since distance traveled = k quantity of gas then k = 5 5 = 45 miles/gallon. The car consumed on average one gallon for every 45 miles traveled. Recommeded Problems (pp. 7-0):, 3, 4, 9,, 4, 7, 0,,, 4, 6, 9, 34, The Rate of Change In this section we will introduce the concept of the average rate of change of a function. We start with an example. Example 4 (Average Speed) During a typical trip to school, your car will undergo a series of changes in its speed. If you were to inspect the speedometer readings at regular intervals, you would notice that it changes often. The speedometer of a car reveals information about the instantaneous speed of your car; that is, it shows your speed at a particular instant in time. The instantaneous speed of an object is not to be confused with the average speed. Average speed is a measure of the distance traveled in a given period of time. That is, Average Speed = Distance traveled Time elapsed

27 .3. THE RATE OF CHANGE 7 So if the trip to school takes 0. hours (i.e. minutes) and the distance traveled is 5 miles then the average speed of your car is Ave. Speed = 5 miles 0. hours = 5miles/hour This says that on the average, your car was moving with a speed of 5 miles per hour. During your trip, there may have been times that you were stopped and other times that your speedometer was reading 50 miles per hour; yet on the average you were moving with a speed of 5 miles per hour. Looking at the above example, we see that the average speed over the interval [0,0.] is s(0.) s(0) Average Speed = 0. 0 where s(t) is the position function of the car at time t hours. Example 5 An object moves a distance of s(t) = 6t feet from its starting point in t seconds. Complete the following table time interval [.8,] [.9,] [.99,] [.999,] [,.000] [,.00] [,.0] Average velocity time interval [.8,] [.9,] [.99,] [.999,] [,.000] [,.00] [,.0] Average velocity The average speed is a example of the average rate of a function. In general, the average rate of change of a function f(x) from x = a to x = b is the difference quotient f(b) f(a). b a Geometrically, this quantity represents the slope of the secant line going through the points (a, f(a)) and (b, f(b)) on the graph of f(x). Exercise 9 Calculate the average rate of change of f(x) = x between x = and x =. Ave. rate of change = f() f( ) ( ) = 3 3 = Average Rate of Change and Monotone Functions We say that a function is increasing if its graph climbs as x moves from left to right. That is, the function values increase as x increases. It is said to be

28 8 CHAPTER. THE CONCEPT OF A FUNCTION decreasing if its graph falls as x moves from left to right. This means that the function values decrease as x increases. A function is monotone if it is either increasing or decreasing. As an application of the average rate of change, we can use such quantity to decide whether a function is increasing or decreasing. If a function f is increasing on an interval I then by taking any two points in the interval I, say a < b, we see that f(a) < f(b) and in this case f(b) f(a) b a > 0. Thus, if the average rate of change is positive in an interval then the function is increasing in that interval. Similarly, if the average rate of change is negative in an interval I then the function is decreasing. Exercise 30 The table below gives values of a function w = f(t). Is this function increasing or decreasing? t w The average of w over the interval [0, 4] is w(4) w(0) 4 0 = = 0.5 The average rate of change of the remaining intervals are given in the chart below time interval [0,4] [4,8] [8,] [,6] [6, 0] [0,4] Average Since the average rate of change is always negative on [0, 4] then the function is decreasing over there. Exercise 3 Determine the intervals where the function is increasing and decreasing y x

29 .4. EVALUATING FUNCTIONS 9 The function is increasing on (, ) (, ) and decreasing on the interval (, ). Recommended Problems (pp. 7-9): 4, 5, 6, 7, 9, 0, 4, 5..4 Evaluating Functions By evaluating a function, we mean figuring out the output value corresponding to a given input value. Thus, notation like f(0) = 4 means that the function s output, corresponding to the input 0, is equal to 4. If the function is given by a formula, say of the form y = f(x), then to find the output value corresponding to an input value a we replace the letter x in the formula of f by the input a and then perform the necessary algebraic operations to find the output value. Exercise 3 Let g(x) = x + 5+x. Evaluate the following expressions: (a) g() (b) g(a) (c) g(a) (d) g(a) g(). (a) g() = + 5+ = 5 7 (b) g(a) = a + 5+a (c) g(a) = a + 5+a 5+a 5+a = a a 9 5+a (d) g(a) g() = a + 5+a 5 7 = 7(a +) 7(5+a) 5 5+a 7 5+a = 7a 5a 8 7a+35. Finding Output and Input Values from Tables and Graphs Next, suppose that a function is given by a table of numeric data. For example, the table below shows the daily low temperature T for a one-week period in New York City during July. D T Then T (8) = 77 F. This means, that the low temperature on July 8, was 77 F. Remark. Note that, from the above table one can find the value of an input value given an output value listed in the table. For example, there are two values of D such that T (D) = 75, namely, D = and D =. Next, to evaluate the output value of a function from its graph, we locate the input value on the horizontal line and then we draw a line perpendicular to the horizontal line at the input value. This line will cross the graph of the function at a point whose y-value is the function s output value.

30 30 CHAPTER. THE CONCEPT OF A FUNCTION Exercise 33 (a) Using the graph below, evaluate f(.5). (b) For what value of x, f(x) = 4? 0 8 y x (a) f(.5) 6 (b) f() = 4. Finding Input Values by Solving Equations So far we have seen how to find the output value of a function given an input value. In some cases, this situation is reversed. That is, given an output value we want to find the corresponding input value(s). We have seen above how to find an input value, given the output value, when the function is defined by a table or its graph is given. When the function is given by a formula, then the input values are obtained by solving an equation. That is, the input values are solutions to the equation f(x) = b, where b is a given output value. Exercise 34 Consider the function y = x 4. (a) Find an x-value that results in y =. (b) Is there an x-value that results in y =? Explain. (a) Letting y = to obtain x 4 = or x 4 = 0.5. Squaring both sides to obtain x 4 = 0.5 and adding 4 to otbain x = 4.5. (b) Since y = x 4 then the right-hand side is always positive so that y > 0. The equation y = has no solutions. Recommended Problems (pp ):, 3, 6, 7, 9, 0,, 4, 6..5 Changes in Input and Output Consider a function f(x). In this section we would like to understand the effect of changes to the input and output of f such as f(x + k), f(x k), f(x) +

31 .5. CHANGES IN INPUT AND OUTPUT 3 k, f(x) k, f(kx), and kf(x) where k is a positive constant. The expression f(x + k) means that each input of the function f(x) is being increased (if k > 0) or decreased (if k < 0) by a constant value k. The output f(x + k) is obtained by replacing the letter x by the sum x + k in the formula of f(x). The expression f(x) + k means that the output of the function f(x) is increased (if k > 0) or decreased (if k < 0) by a constant value k. For example, if A is the area of a circle of radius r cm, i.e. A = πr, then the expression A + 5 means that the area in increased by 5 cm whereas the expression A 5 means that the area is decreased by 5 cm. The quantity f(x) + k is obtained by adding the number k to the expression of f(x). An increase or decrease in the input of f(x) by a constant k is represented symbolically by either f(x + k) or f(x k). For example, A(r + ) is the area of the circle when the radius in increased by cm. To find f(x + k) we replace the letter x by the quantity x + k in the fromula of f. For example, since A = πr then A(r + ) = π(r + ). The expression f(kx) means that each input of the function f(x) is multiplied by a constant k. For example, if we double the radius of the circle of radius r then the area of the larger circle is A(r) = π(r) = 4πr. The output f(kx) is obtained by replacing the letter x in the formula of f(x) by the quantity kx. The expression kf(x) means that the output values of the function f(x) are multiplied by a constant k. The expression of kf(x) is obtained by multiplying the expression of f(x) by the constant k. Thus, twice the area of the circle of radius r is represented symbolically by the expression A = πr. Exercise 35 Suppose that f(x) = x. Find the following functions: (i) f (x) = f(x + ) (ii) f (x) = f(x) 3 (iii) f 3 (x) = 3f(x) (iv) f 4 (x) = f(x). (i) f (x) = f(x + ) = (x + ) = x. (ii) f (x) = f(x) 3 = x 3 = x. (iii) f 3 (x) = 3f(x) = 3( x) = 3 3x. (iv) f 4 (x) = f(x) = x. Exercise 36 Suppose every day I take the same taxi over the same route from home to the train station. The trip is x miles, so the cost for the trip is C = f(x). Match each story in (a)-(d) to a function in (i) - (iv) which represents the amount paid to the taxi driver. (a) I received a raise yesterday, so today I gave my driver a five dollar tip. (b) I had a new driver today and he got lost. He drove five extra miles and

32 3 CHAPTER. THE CONCEPT OF A FUNCTION charged me for it. (c) I haven t paid my driver all week. Today is Friday and I ll pay what I owe for the week. (d) The meter in the taxi went crazy and showed five times the number of miles I actually traveled. (i) C = 5f(x) (ii) C = f(x) + 5 (iii) C = f(5x) (iv) C = f(x + 5). (a) - (ii), (b) - (iv), (c) - (i), (d) - (iii). Now, given two quantities x and y. Suppose that y is a function of x. That is, y = f(x) in function notation. We have seen from Exercise 30 in Section., that x need not be a function of y. This is not always the case. That is, sometimes x can also be a function of y, i.e. x = g(y). In this case, we call f and g inverses of each other and we write g = f. Thus, if y = f(x) and x = g(y) then we will write x = f (y). Remarks. The notation f is just a symbol to represent the inverse function. So be aware not to treat f as f(x) being raised to the power. That is, f (x) f(x).. Note that f takes x as input and y as output whereas f goes the opposite direction. That is, f takes y as input and x as output. 3. Inverse functions will be considered again in Section 8. of the textbook. Process for finding f :. Replace f(x) with y.. Interchange the letters x and y. 3. Solve for y in terms of x. 4. Replace y with f (x). Example 6 The function f(x) = x has an inverse. Find the formula for the inverse function. We find its inverse as follows:. Replace f(x) with y to obtain y = x Interchange x and y to obtain x = y Solve for y to obtain y 3 = x 7 or y = 3 x Replace y with f (x) to obtain f (x) = 3 x 7. Recommended Problems (pp. 84-6):, 3, 4, 6, 8, 9, 3, 5, 6, 8.

33 .6. DOMAIN AND RANGE OF A FUNCTION 33.6 Domain and Range of a Function If we try to find the possible input values that can be used in the function y = x we see that we must restrict x to the interval [, ), that is x. Similarly, the function y = x takes only certain values for the output, namely, y > 0. The above discussion leads to the following definitions: By the domain of a function we mean all possible input values. Graphically, the domain is part of the horizontal axis. The range of a function is the collection of all possible output values. The range is part of the vertical axis. The domain and range of a function can be found either algebraically or graphically. Finding the Domain and the Range Algebraically When finding the domain of a function, ask yourself what values can t be used. Your domain is everything else. There are simple basic rules to consider: The domain of all polynomial functions, i.e. functions of the form f(x) = a n x n + a n x n + + a x + a 0, is the Real numbers IR. Square root functions can not contain a negative underneath the radical. Set the expression under the radical greater than or equal to zero and solve for the variable. This will be your domain. Rational functions, i.e. ratios of polynomials, can not have zeros in the denominator. Determine which values of the input cause the denominator to equal zero, and set your domain to be everything else. Exercise 37 Find, algebraically, the domain and the range of each of the following functions. Write your answers in interval notation: (a) y = πx (b) y = x 4 (c) y = + x. (a) Since the function is a polynomial then its domain is the interval (, ). To find the range, note that y 0 for all x. So the range is the interval [0, ). (b) The domain of y = x 4 consists of all numbers such that x 4 > 0 or x > 4. That is, the interval (4, ). To find the range, notice that y > 0 for all values of x in the domain of y. Thus, the range is the interval (0, ). (c) The domain of y = + x is the interval (, 0) (0, ). To find the range, write x in terms of y to obtain x = y. The values of y for which this later formula is defined is the range of the given function, that is, (, ) (, ). Remark. Note that the domain of the function y = πx of the previous problem consists of all real numbers. If this function is used to model a real-world situation, that is, if the x stands for the radius of a circle and y is the corresponding area then

34 34 CHAPTER. THE CONCEPT OF A FUNCTION the domain of y in this case consists of all numbers x 0. In general, for a word problem the domain is the set of all x values such that the problem makes sense. Finding the Domain and the Range Graphically We often use a graphing calculator to find the domain and range of functions. In general, the domain will be the set of all x values that has corresponding points on the graph. We note that if there is an asymptote (shown as a vertical line on the TI series) we do not include that x value in the domain. To find the range, we seek the top and bottom of the graph. The range will be all points from the top to the bottom (minus the breaks in the graph). Exercise 38 Use a graphing calculator to find the domain and the range of each of the following functions. Write your answers in interval notation: (a) y = πx (b) y = x 4 (c) y = + x. (a) The graph of y = πx is given below y x The domain is the set (, ) and the range is [0, ). (b) The graph of y = x 4 is given below

35 .7. PIECEWISE DEFINED FUNCTIONS y x The domain is the set (4, ) and the range is (0, ). (c) The graph of y = + x is given below 0 y 0 0 x 0 0 The domain is the set (, 0) (0, ) and the range is (, ) (, ). Recommended Problems (pp ):, 4, 7,, 6,, 3, 5, 6..7 Piecewise Defined Functions Let s consider the following real-world situation: The charge for a taxi ride is $.50 for the first 5 of a mile, and $0.5 for each additional 5 of a mile (rounded up to the nearest 5 mile). (a) Sketch a graph of the cost function C as a function of the distance traveled x, assuming that 0 x. The graph is given below

36 36 CHAPTER. THE CONCEPT OF A FUNCTION x (b) Find a formula for C in terms of x on the interval [0, ]. A formula of C(x) is.50 if 0 x 5.75 if 5 < x 5 C(x) =.00 if 5 < x if 3 5 < x if 4 5 < x (c) What is the cost for a 4 5 mile ride? The cost for a 4 5 ride is C( 4 5 ) = $.5. Note that the function C(x) is defined by different outputs on different parts of the interval [0, ]. A function of this type is called a piecewise defined function. More generally, a piecewise defined function is a function with different output formulas on different pieces of the domain of the function. Example 7 The absolute value function x is a piecewise defined function since x = { x for x 0 x for x < x

37 .7. PIECEWISE DEFINED FUNCTIONS 37 Example 8 Sketch the graph of the piecewise defined function given by x + 4 for x f(x) = for < x < 4 x for x The following table gives values of f(x). x f(x) The graph of the function is given below x Recommended Problems (pp ):, 3, 4, 5, 7, 8, 0,, 3.

38 38 CHAPTER. THE CONCEPT OF A FUNCTION

39 Chapter 3 Linear Functions In this chapter we discuss a family of functions known as linear functions. In the next chapter we discuss the so-called exponential and logarithmic functions. 3. What is a Linear Function? This section is designed to introduce students to the concept of linear functions. So, the first question that comes to mind is the question of recognizing a linear function. In general, the average rate of change of a function is different on different intervals. A linear function is a function with the property that the average rate of change on any interval is the same. We say that y is changing at a constant rate with respect to x. In this case, the graph is a straight line ( and thus the term linear). It follows, that if f is linear and if [a, b] and [c, d] are two intervals of equal length, i.e. b a = d c then f(b) f(a) b a = f(d) f(c) d c and this implies that f(b) f(a) = f(d) f(c). That is, if a function y is such that equal increments in x corresponds to equal increments in y then y is a linear function of x. Exercise 39 Which of the following tables could represent a linear function? x f(x) x g(x) Since equal increments in x yield equal increments in y then f(x) is a linear 39

40 40 CHAPTER 3. LINEAR FUNCTIONS function. On the contrary, since then g(x) is not linear. Now, suppose that f(x) is a linear function of x. Then f changes at a constant rate m. That is, if we pick two points (0, f(0)) and (x, f(x)) then m = f(x) f(0). x 0 That is, f(x) = mx + f(0). This is the function notation of the linear function f(x). Another notation is the equation notation, y = mx + f(0). We will denote the number f(0) by b. In this case, the linear function will be written as f(x) = mx + b or y = mx + b. Since b = f(0) then the point (0, b) is the point where the line crosses the vertical line. We call it the y-intercept. So the y-intercept is the output corresponding to the input x = 0, sometimes known as the initial value of y. If we pick any two points (x, y ) and (x, y ) on the graph of f(x) = mx + b then we must have m = y y x x. We call m the slope of the line. Exercise 40 The value of a new computer equipment is $0,000 and the value drops at a constant rate so that it is worth $ 0 after five years. Let V (t) be the value of the computer equipment t years after the equipment is purchased. (a) Find the slope m and the y-intercept b. (b) Find a formula for V (t). (a) Since V (0) = 0, 000 and V (5) = 0 then the slope of V (t) is m = 0 0, = 4, 000 and the vertical intercept is V (0) = 0, 000. (b) A formula of V (t) is V (t) = 4, 000t + 0, 000. Exercise 4 The table below shows the cost of selling various amounts of coffee per day from a cart. (a) Using the table, show that the relationship appears to be linear. (b) Plot the data in the table. (c) Find the slope of the line. Explain what this means in the context of the given situation. x(cups per day) C(dollars)

41 3.. FORMULAS FOR LINEAR FUNCTIONS 4 (a) Since an increment of 5 in x yields an increment of.5 in y then the table shows that equal increments in x yield equal increments in y so that y is linear. (b) The graph is given below (c) The slope of the line is m = = 0.5. This number means that an increase in production by one cup will increase cost by 5 cents. Recommended Problems (pp. 43-5):,, 3, 8, 9,,, 9, Formulas for Linear Functions In this section we will discuss ways for finding the formulas for linear functions. Recall that f is linear if and only if f(x) can be written in the form f(x) = mx + b. So the problem of finding the formula of f is equivalent to finding the quantities m and b. Suppose that we know two points on the graph of f(x), say (x, f(x )) and (x, f(x )). Since the slope m is just the average rate of change of f(x) on the interval [x, x ] then m = f(x ) f(x ) x x. To find b, we use one of the points in the formula of f(x); say we use the first point. Then f(x ) = mx + b. Solving for b we find b = f(x ) mx. Example 9 Let s find the formula of a linear function given by a table of data values. The table below gives data for a linear function. Find the formula. x f(x)

42 4 CHAPTER 3. LINEAR FUNCTIONS We use the first two points to find the value of m : m = f(.3) f(.).3. To find b we can use the first point to obtain Solving for b we find b =. Thus, = =.(.) + b. f(x) =.x +. =. Example 30 Suppose that the graph of a linear function is given and two points on the graph are known. For example, the graph below is the graph of a linear function going through the points (00, ) and (60,.6). Find the formula. (60,6) (00,) The slope m is found as follows: m = = To find b we use the first point to obtain = 0.083(00) + b. Solving for b we find b = 7.3. So the formula for f(x) = x. Example 3 Sometimes a linear function is given by a verbal description as in the following problem: In a college meal plan you pay a membership fee; then all your meals are at a fixed price per meal. If 30 meals cost $5.50 and 60 meals cost $50 then find the formula for the cost C of a meal plan in terms of the number of meals n. We find m first: m = = $3.5/meal.

43 3.3. SOLVING SYSTEMS OF EQUATIONS 43 To find b or the membership fee we use the point (30, 5.50) in the formula C = mn+b to obtain 5.50 = 3.5(30)+b. Solving for b we find b = $55. Thus, C = 3.5n So far we have represented a linear function by the expression y = mx + b. This is known as the slope-intercept form of the equation of a line. Now, if the slope m of a line is known and one point (x 0, y 0 ) is given then by taking any point (x, y) on the line and using the definition of m we find y y 0 x x 0 = m. Cross multiply to obtain: y y 0 = m(x x 0 ). This is known as the point-slope form of a line. Example 3 Find the equation of the line passing through the point (00, ) and with slope m = 0.0. Using the above formula we have: y = 0.0(x 00) or y = 0.0x. Note that the form y = mx + b can be rewritten in the form Ax + By = C (3.) where A = m, B =, and C = b. The form (3.) is known as the standard form of a linear equation. Example 33 Rewrite in standard form: 3x + y + 40 = x y. Subtracting x y + 40 from both sides to obtain x + 3y = 40. Recommended Problems (pp ): 3, 5, 7, 0,,, 3, 4, 6, 9,, 5, 6, 7, Solving Systems of Equations This section deals with the question of solving two equations in two unknowns. That is, we want to find an ordered pair that satisfy the two equations simultaneously. We will discuss two methods of solutions: graphical method and algebraic method (method of substitution). If the system consists of two equations of the form ax + by = c then we call the system a linear system. A system which is not linear is called a non linear system.

44 44 CHAPTER 3. LINEAR FUNCTIONS Geometrically, solving a linear system is equivalent to finding the coordinates of the point of intersection of two lines. Because the graph of a linear equation is a straight line, points that satisfy both equations lie on both lines. For some systems these points can be found by graphing. Example 34 Solve the following system by graphing,i.e using a graphing calculator: { 8x + 3y = 4 x + y = The graph of each line and the point of intersecrion are shown below 5 0 [, 0] 5 3 x 5 Exercise 4 Solve the system by graphing: { x 3y = 6 x + 3y = 3 The graph shows that the system has no solutions or inconsistent.

45 3.3. SOLVING SYSTEMS OF EQUATIONS x Exercise 43 Solve the system by graphing: { 4x 6y = 30 x 3y = 5 The graph shows that the system has infinite number of solutions x A second way for solving a linear system is by substitution. In this method, we use one equation to write one variable in terms of the other and then substitute the resulting expression in the unused equation. This way one eliminates a variable and gets an equation involving only one variable. Example 35 Solve the following system by substitution { 4x + y = 5 x + y = 4 Using the second equation, we find x = 4 y. Replacing this expression in the first equation to obtain 4( 4 y) + y = 5 or 7y =. Solving for y we find y = 3. Thus, x = 4 ( 3) =. Exercise 44 Solve the system by substitution: { x y = 3 x 4y = 7

46 46 CHAPTER 3. LINEAR FUNCTIONS Using the first equation, we find x = 3 + y. Replacing this expression in the second equation to obtain (3 + y) 4y = 7 or 0y =. Since this is impossible we conclude that the system has no solutions. Exercise 45 Solve the system by substitution: { x + 3y = 5 + x + 4y x y = 5 Using the second equation, we find x = 5 + y. Replacing this expression in the first equation to obtain (5 + y) + 3y = y + 4y or 5y + 0 = 5y + 0. Since this equation is valid for any y then the system has an infinite number of solutions given by (y + 5, y). Exercise 46 A third way for solving linear systems is by the method of elimination. In this method we eliminate a variable by adding equations. Solve the system by elimination: { x 3y = 7 3x + y = 5 Multiplying the second equation by 3 and adding to obtain x + 9x = 7 5. Solving for x we find x =. Substituting this value for x in one of the original equations and solving for y we find y =. Exercise 47 (Non linear Systems) Solve the nonlinear system { x + y = 36 y = x 3 Replacing the expression for y, given in the second equation, into the first equation to obtain x + (x 3) = 36 or after simplifying x 6x 5 = 0. Solving this quadratic equation by the quadratic formula we find x = 6 ± 36 4 = 3 ± 59. Now, replacing this into the second equation to obtain y = 3 ± 59.

47 3.4. GEOMETRIC PROPERTIES OF LINEAR FUNCTIONS 47 Exercise 48 Solve the nonlinear system Taking ratios, { Q0 a 0 = 90.7 Q 0 a 3 = 9 Q 0 a 3 9 = Q 0 a or a 3 = Taking the cube root of both sides to obtain Now, use the first equation to find Q 0, or Q 0 = a = Q 0 (.00) 0 = (.00) 0 Recommended Problmes (p. 59):, 3, 5, 6, 8, 9, 0, 4, Geometric Properties of Linear Functions In this section we discuss four geometric related questions of linear functions. The first question considers the importance of the slope of a straight line. We have seen that the graph of a linear function f(x) = mx + b is a straight line. But a line can be horizontal, vertical, rising to the right or falling to the right. The slope is the parameter that provides information about the steepness of a straight line. If m = 0 then f(x) = b is a constant function whose graph is a horizontal line at (0, b). For a vertical line, the slope is undefined since any two points on the line have the same x-value and this leads to a division by zero in the formula for the slope. The equation of a vertical line has the form x = a. Suppose that the line is neither horizontal nor vertical. If m > 0 then by Section.3 (Section.3 of the textbook), f(x) is increasing. That is, the line is rising to the right. If m < 0 then f(x) is decreasing. That is, the line is falling to the right. It follows that the larger the value of m is the more the line rises and the smaller the value of m is the more the line falls. The parameter b tells us where the line crosses the vertical axis.

48 48 CHAPTER 3. LINEAR FUNCTIONS Example 36 Arrange the slopes of the lines in the figure from largest to smallest. E B C G F D A According to the figure we have m G > m F > m D > m A > m E > m B > m C. Next, we consider the following question: How can we examine a pair of linear equations to determine whether their graphs are parallel- that is, they do not intersect? Let s first try and solve the following exercise. Example 37 Use a graphing calculator and complete the following table: Pairs of Equations m m b b Parallel? y = 0.38x + 4., y = 0.38x 5. y = 0.38x + 4., 50y + 9x = x + = y, y + 3x = 4x 6y = 30, x 3y = 5 y = 3, y = 4. x = 5, x = Pairs of Equations m m b b Parallel? y = 0.38x + 4., y = 0.38x YES y = 0.38x + 4., 50y + 9x = YES x + = y, y + 3x = -3 NO 4x 6y = 30, x 3y = NO y = 3, y = YES x = 5, x = UNDEFINED UNDEFINED NONE NONE YES From this table, we see that two lines are parallel if and only if:

49 3.4. GEOMETRIC PROPERTIES OF LINEAR FUNCTIONS 49 (i) the two lines are vertical; (ii) if nonvertical, then the two lines have the same slope and different y- intercepts. Now, the third question of this section is: How can we examine a pair of linear equations to determine whether they are perpendicular? Example 38 Use a graphing calculator and complete the following table: Pairs of Equations m m Perpendicular? y = 5 x + 3, y = 5 x + 3 y = 0.875x, y = 6 3 x 5. 3y + 4x =, 4y 3x = 8 y = 3x + 4, y = x 7 x = 5, y = x = 5, x = Pairs of Equations m m Perpendicular? y = 5 x + 3, y = 5 x YES y = 0.875x, y = 6 3 x YES 3y + 4x =, 4y 3x = YES y = 3x + 4, y = x 7-3 NO x = 5, y = UNDEFINED 0 YES Note that two lines are perpendicular if and only if: (i) the product of their slopes is -. That is, the slope of one is the negative reciprocal of the slope of the other; (ii) if one of the lines is vertical and the other is horizontal. The final question is the question of finding the point of intersection of two lines. The point of intersection of two lines is basically the solution to a system of two linear equations. This system can be solved by the method of substitution which we discussed in Section 3.3 (Appendix G of the textbook). Exercise 49 Find the point of intersection of the two lines y = 3 3 x and y = x. Equating the two equations to obtain 3 3 x = x. Multiply through by 6 to obtain 8 4x = 4 + 9x. Solving for x we find x = 4 3. To find y, replace the value of x in the first equation obtaining y = 3. Recommended Problems (pp. 58-6):, 3, 6, 7, 8, 3, 4, 5, 7, 9, 3.

50 50 CHAPTER 3. LINEAR FUNCTIONS 3.5 Fitting Linear Functions to Data In general, data obtained from real life events, do not match perfectly simple functions. Very often, scientists, engineers, mathematicians and business experts can model the data obtained from their studies, with simple linear functions. Even if the function does not reproduce the data exactly, it is possible to use this modeling for further analysis and predictions. This makes the linear modeling extremely valuable. Let s try to fit a set of data points from a crankcase motor oil producing company. They want to study the correlation between the number of minutes of TV advertisement per day for their product, and the total number of oil cases sold per month for each of the different advertising campaigns. The information is given in the following table : x:tv ads(min/day) y:units sold(in millions) Using TI-83 we obtain the following scatter plot of this given data. steps of getting the graph are discussed later in this section. The The figure below shows the plot and the optimum linear function that describes the data. That line is called the best fitting line and has been derived with a very commonly used statistical technique called the method of least squares. The measure of how well this linear function fits the experimental points, is called regression analysis.

51 3.5. FITTING LINEAR FUNCTIONS TO DATA 5 Graphic calculators, as the TI-83, have built in programs which allow us to find the slope and the y intercept of the best fitting line to a set of data points. That is, the equation of the best linear fit. The calculators also give as a result of their procedure, a very important value called the correlation coefficient. This value is in general represented by the letter r and it is a measure of how well the best fitting line fits the data points. Its value varies from - to. The TI-83 prompts the correlation coefficient r as a result of the linear regression. If it is negative, it is telling us that the line obtained has negative slope. Conversely, positive values of r indicate a positive slope in the best fitting line. If r is close to 0 then the data may be completely scattered, or there may be a non-linear relationship between the variables. The square value of the correlation coefficient r is generally used to determine if the best fitting line can be used as a model for the data. For that reason, r is called the coefficient of determination. In most cases, a function is accepted as the model of the data, if this coefficient of determination is greater than 0.5. A coefficient of determination tells us which percent of the variation on the real data is explained by the best fitting line. An r = 0.9 means that 9% of the variation on the data points is described by the best fitting line. The closer the coefficient of determination is to the better the fit. The following are the steps required to find the best linear fit using a TI-83 graphing calculator.. Enter the following data into two lists L and L. a. Push the STAT key and select the Edit option. b. Up arrow to move to the Use the top of the list L. c. Clear the list by hitting CLEAR ENTER d. Type in the x values of the data. Type in the number and hit enter.

52 5 CHAPTER 3. LINEAR FUNCTIONS e. Move to the list L, clear it and enter the y data in this list.. Graph the data as a scatterplot. a. Hit nd STAT PLOT. (upper left) b. Move to plot and hit enter c. Turn the plot on by hitting enter on the ON option. d. Move to the TYPE option and select the dot graph type. Hit enter to select it e. Move to the Xlist and enter in nd L. f. Move to the Xlist and enter in nd L. g. Move to Mark and select the small box option h. Go to the window button and hit enter (top center). Set the Xmin=0, Xmax=00, Ymin=50 and Ymax=00, or HIT ZOOM and select ZOOMSTAT. i. Clear all other functions. Go to Y= and if there are equations entered for any of the functions Y, Y,., you must delete those functions. j. Hit GRAPH button and a plot of the points should show up 3. Fit a line to the data. a. Turn on the option to display the correlation coefficient, r. This is accomplished by hitting nd Catalog (lower left). Scroll down the list until you find Diagnostic On, hit enter for this option and hit enter a second time to activate this option. The correlation coefficient will be displayed when you do the linear regression b. Hit STAT, CALC, and select option 8 LinREg. c. Enter nd, nd, with a comma in between. d. Press ENTER. The equation for a line through the data is shown. The slope is b, the intercept is a, the correlation coefficient is r, and the coefficient of determination is r. 4. Graph the best fit line with the data a. Press Y=, then press VARS to open the Variables window. b. Arrow down to select 5: Statistics... then press ENTER c. Right arrow over to select EQ and press ENTER. This places the formula for the regression equation into the Y= window. d. Press GRAPH to graph the equation. Your window should now show the graph of the regression equation as well as each of the data points. Recommended Problems (pp. 65-7):, 3, 5, 7.

53 Chapter 4 Exponential and Logarithmic Functions In the previous chapter we discussed the family of linear functions. Recall that a linear function is a function for which the dependent variable (or output) increases by a constant amount for each one unit increase in the independent variable (or the input). In contrast, an exponential function is a function where the dependent variable increases at a constant factor for each one unit increase in the independent variable. In this chapter, we study the family of exponential functions and their inverses, the logarithmic functions. 4. Introduction to Exponential Functions Linear functions are functions that change at a constant rate. For example, if f(x) = mx + b then f(x + ) = m(x + ) + b = f(x) + m. Exponential functions are functions that change at a constant percent rate. That is, if f(x) is an exponential function then f(x + ) = f(x) + r%f(x). Before we go on to discuss exponential problems, we should pause and discuss some of the terms that are frequently used in these problems. First let s talk about the terms factor and percent. For example, suppose you invested $00, and after a time, your investment was worth $300. The final value ($300) would be three (3) times the initial value. We would say that your investment had increased by a factor of 3. The general equation for percentage changes is N final N initial N initial 00 = % change Thus, a $00 investment that increased to $300 (increased by a factor of 3), had a percentage increase of: = 00% 53

54 54 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 39 A quantity increases from 0 to. By what percent has it increased? The increase is 0 =. = 0%. Exercise 50 If we multiply a quantity A by.5 then A is increased by what percent? Since.5A = A + 0.5A = A + 5%A then multiplying A by.5 is equivalent to increasing A by 5%. Exponential functions are used to model increasing quantities such as population problems. Example 40 Suppose that you are observing the behavior of cell duplication in a lab. In one experiment, you started with one cell and the cells doubled every minute. That is, the population cell is increasing at the constant rate of 00%. Write an equation to determine the number (population) of cells after one hour. The table below shows the number of cells for the first 5 minutes. Let P (t) be the number of cells after t minutes. t P(t) At time 0, i.e t=0, the number of cells is or 0 =. After minute, when t =, there are two cells or =. After minutes, when t =, there are 4 cells or = 4. Therefore, one formula to estimate the number of cells (size of population) after t minutes is the equation (model) f(t) = t. Now, to determine the number of cells after one hour we convert to minutes to obtain t = 60 minutes so that f(60) = 60 = cells. Exponential functions can also model decreasing quantities. Example 4 If you start a biology experiment with 5,000,000 cells and 45% of the cells are dying every minute, how long will it take to have less than 50,000 cells?

55 4.. INTRODUCTION TO EXPONENTIAL FUNCTIONS 55 Let P (t) is the number of cells after t minutes. Then P (t+) = P (t) 45%P (t) or P (t + ) = 0.55P (t). By constructing a table data we find t P(t) 0 5,000,000,750,000,5, , , , , , , So it takes 8 minutes for the population to reduce to less than 50,000 cells. A formula of P (t) is P (t) = 5, 000, 000(0.55) t. From the previous two examples, we see that an exponential function has the general form P (t) = b a t, a > 0 a. Since b = P (0) then we call b the initial value. Since P (t + ) = ka t+ = a(ka t ) = ap (t) then we call a the growth factor. Also, since P (t) = P (t) + r%p (t) where r = a then we call r the percent growth rate. Note that for a > (or r < ) the exponential function represents a growth model whereas for 0 < a < ( or r > ) the function represents a decay model. Remark. Why a is restricted to a > 0 and a? Since t is allowed to have any value then a negative a will create meaningless expressions such as a (if t = ). Also, for a = the function P (t) = b is called a constant function and its graph is a horizontal line. Exercise 5 Suppose you are offered a job at a starting salary of $40,000 per year. To strengthen the offer, the company promises annual raises of 6% per year for the first 0 years. Let P (t) be your salary after t years. Find a formula for P (t) and then compute your projected salary after 4 years from now. A formula of P (t) is P (t) = 40, 000(.06) t. After four years, the projected salary is P (3) = 40, 000(.06) 4 50, Exercise 5 The amount in milligrams of a drug in the body t hours after taking a pill is given by A(t) = 5(0.85) t. (a) What is the initial dose given?

56 56 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS (b) What percent of the drug leaves the body each hour? (c) What is the amount of drug left after 0 hours? (a) Initial dose give is A(0) = 5 mg. (b) r = a = 0.85 =.5 so that 5% of the drug leaves the body each hour. (c) A(0) = 5(0.85) mg. Recommended Problems (pp ):,, 3, 4, 7, 9, 3, 5, 6, 9, Exponential Functions Versus Linear Functions The first question in this section is the question of recognizing whether a function given by a table of values is exponential or linear. We know that for a linear function, equal increments in x correspond to equal increments in y. For an exponential function let us first assume that we have a formula for the function, say f(x) = ba x. Then f(x+) f(x) = a. Thus, if the ratio of consecutive y-values of a function is the same then the function is an exponential function. Example 4 Decide if the function is linear or exponential?find a formula for each case. x f(x) x g(x) Since then f(x) is an exponential function. To find a formula for f(x) = ba x we use the first two points obtaining.5 = f(0) = b and 3.75 = f() = ba =.5a. Hence, a = so that f(x) =.5(.) x. On the other hand, equal increments in x corresponds to equal increments in the g-values so that g(x) is linear, say g(x) = mx + b. Since g(0) = 0 then b = 0. Also, = g() = m so that g(x) = x. The next question of this section is the question of finding a formula for an exponential function. The above example shows how to find linear and exponential functions using two data points. The next example illustrates how to find the formula of an exponential function given two points on its graph.

57 4.3. GRAPHS OF EXPONENTIAL FUNCTIONS 57 Example 43 Find a formula for the exponential function whose graph is given below (-,.5) (,.6) Write f(x) = ba x. Since f( ) =.5 then ba =.5. Similarly, ba =.6. Taking the ratio we find ba ba =.6.5 or a =.64 or a = 0.8. From ba =.6 we find that b = = so that f(x) = (0.8)x. Later on in this chapter we will try to solve exponential equations, that is, equations involving exponential functions. Usually, the process requires the use of the so-called logarithm function which we will discuss in Section 4.4. For the time being, we will exhibit a graphical method for solving an exponential equation. Example 44 Estimate to two decimal places the solutions to the exponential equation x + = x. Let f(x) = + x and g(x) = x. The solutions to the given equation are the x-values of the points of intersection of the graphs of f(x) and g(x). Using a graphing calculator we see that the two graphs intersect at two points one in the first quadrant and one in the second quadrant. Using the INTERSECT key we find x = and x.69. Recommended Problems (pp. 7-9):,, 3, 5, 8, 9, 3, 5, 6, 7,, 3, Graphs of Exponential Functions Recall that an exponential function with base a and initial value b is a function of the form f(x) = b a x. Since b = f(0) then (0, b) is the vertical intercept of f(x). In this section we consider graphs of exponential functions. Let s see the effect of the parameter b on the graph of f(x) = ba x.

58 58 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 45 Graph, on the same axes, the exponential functions f (x) = (.) x, f (x) = (.) x, and f 3 (x) = 0.75(.) x. Using a graphing calculator, we find x where f is in red, f in green and f 3 in yellow. Note that these functions have the same growth factor but different b and therefore different vertical intercepts. We know that the slope of a linear function measures the steepness of the graph. Similarly, the parameter a measures the steepness of the graph of an exponential function. Example 46 Graph, on the same axes, the exponential functions f ( x) = x, f (x) = 3 x, and f 3 (x) = 4 x. Using a graphing calculator we find x

59 4.3. GRAPHS OF EXPONENTIAL FUNCTIONS 59 where f is in red, f in green, and f 3 in yellow. Observations. (i) Note that the greater the value of a, the more rapidly the graph rises. That is, the growth factor a affects the steepness of an exponential function. (ii) Since a > then the graphs represent exponential growths. (iii) Note that as x decreases, the function values approach the x-axis. Symbolically, as x, y 0. Example 47 Graph, on the same axes, the exponential functions f (x) = x =, f x (x) = 3 x, and f 3 (x) = 4 x = 3. x Using a graphing calculator we find x where f is in red, f in green, and f 3 in yellow. Observations. (i) Note that the smaller the value of a, the more rapidly the graph falls. (ii) Since 0 < a < then the graphs represent exponential decays. (iii) Note that as x increases, the function values approach the x-axis. Symbolically, as x, y 0. General Observations (i) For a >, as x decreases, the function values get closer and closer to 0. Symbolically, as x, y 0. For 0 < a <, as x increases, the function values gets closer and closer to the x-axis. That is, as x, y 0. We call the x-axis, a horizontal asymptote. (ii) The domain of an exponential function consists of the set of all real numbers whereas the range consists of the set of all positive real numbers.

60 60 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS (iii) Since the average rate of change is the slope of a secant line connecting two points on a graph then the average rate of change of the exponential functions discussed above is always increasing. This means that the graph opens up. We also say that the graph is concave up ( If the average rate of change of a function is decreasing then the graph opens down and we say that the graph is concave down). Example 48 Show that the graph of the function defined by the table below is concave down. x f(x) Average Rate of Change Finding the average rate of change we find x f(x) Average Rate of Change Since the average rate of change is decreasing then the graph of f(x) is concave down. Recommended Problems (pp. 4-7):,, 3, 4, 6, 7, 9, 0,, 4, 8, 5, 6, Logarithms and Their Properties In this section we find an algebraic way to solve equations of the form a x = b where a and b are positive constants. For this, we introduce two functions that are found in today s calculators, namely, the functions log x and ln x. If x > 0 then we define log x to be a number y that satisfies the equality 0 y = x. For example, log 00 = since 0 = 00. Similarly, log 0.0 = since 0 = 0.0. We call log x the common logarithm of x. Thus, y = log x if and only if 0 y = x.

61 4.4. LOGARITHMS AND THEIR PROPERTIES 6 Example 49 (a) Rewrite log 30 =.477 using exponents instead of logarithms. (b) Rewrite = using logarithms instead of exponents. (a) log30 =.477 is equivalent to = 30. (b) = is equivalent to log = 0.8. Example 50 Without a calculator evaluate the following expressions: (a) log (b) log 0 0 (c) log ( 0 ) (d) 0 log 00 log (0.0) (e) 0 (a) log = 0 since 0 0 =. (b) log 0 0 = log = 0 by (a). (c) log ( 0 ) = log 0 =. (d) 0 log 00 = 0 = 00. (e) 0 log (0.0) = 0 = 0.0. Properties of Logarithms (i) Since 0 x = 0 x we can write (ii) Since log x = log x then log 0 x = x 0 log x = x (iii) log = 0 since 0 0 =. (iv) log 0 = since 0 = 0. (v) Suppose that m = log a and n = log b. Then a = 0 m and b = 0 n. Thus, a b = 0 m 0 n = 0 m+n. Rewriting this using logs instead of exponents, we see that log (a b) = m + n = log a + log b. (vi) If, in (v), instead of multiplying we divide, that is a b = 0m 0 = 0 m n then n using logs again we find ( a ) log = log a log b. b (vii) It follows from (vi) that if a = b then log a log b = log = 0 that is log a = log b. (viii) Now, if n = log b then b = 0 n. Taking both sides to the power k we find b k = (0 n ) k = 0 nk. Using logs instead of exponents we see that log b k = nk = k log b that is log b k = k log b.

62 6 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 5 Solve the equation: 4(.7) x = 7(.088) x. Rewriting the equation into the form ( ).7 x.088 = 7 4 (vii) and (viii) to obtain ( ).7 x log = log Thus, log 7 4 x = log (.7 )..088 Exercise 53 Solve the equation log (x + ) + 3 = 0. and then using properties Subtract 3 from both sides to obtain log (x + ) = 3. Switch to exponential form to get x + = 0 3 = Subtract and then divide by to obtain x = Remarks. All of the above arguments are valid for the function ln x for which we replace the number 0 by the number e =.78. That is, y = ln x is equivalent to x = e y, ln (a b) = ln a + ln b, etc. We call ln x the natural logarithm of x. Keep in mind the following: log (a + b) log a + log b log (a b) log a log b log (ab) log a log b log ( ) a b log a log b log ( ) a log a (Give an example for each case!) Recommended Problems (pp. 33-5):, 4, 6, 7, 8, 9, 3, 5, 8, 3, 4, 30, 3, 33, The Logarithmic Function We have seen that the notation y = log x is equivalent to 0 y = x. Since 0 raised to any power is always positive then the domain of the function log x consists of all positive numbers. That is, log x cannot be used with negative numbers. Now, let us sketch the graph of this function by first constructing the following chart:

63 4.5. THE LOGARITHMIC FUNCTION 63 x log x Average Rate of Change 0 underfined From the chart we see that the graph is always increasing. Since the average rate of change is decreasing then the graph is always concave down. Now plotting these points and connecting them with a smooth curve to obtain x We observe the following: (a) The range of log x consists of all real numbers. (b) The graph never crosses the y-axis since a positive number raised to any power is always positive. (c) The graph crosses the x-axis at x =. (d) As x gets closer and closer to 0 from the right the function log x decreases without bound. That is, as x 0 +, x. We call the y-axis a vertical asymptote. In general, if a function increases or decreases without bound as x gets closer to a number a then we say that the line x = a is a vertical asymptote. Next, let s graph the function y = 0 x by using the above process: x 0 x Average Rate of Change

64 64 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Note that this chart can be obtained from the chart of log x discussed above by interchanging the variables x and y. This means, that the graph of y = 0 x (in red) is a reflection of the graph of y = log x (in blue) about the line y = x (in green) as seen in the picture below. 3 y 0 x 3 Example 5 Sketch the graphs of the functions y = ln x and y = e x on the same axes. The functions y = ln x and y = e x are inverses of each like the functions y = log x and y = 0 x. So their graphs are reflections of one another across the line y = x as shown below Recommended Problems (pp. 37-9):,, 3, 4, 6, 7, 9, 0,, 3, 5, 6, 0, 3.

65 4.6. LOGARITHMIC AND EXPONENTIAL EQUATIONS Logarithmic and Exponential Equations We have seen how to solve an equation such as 00(0.886) x = 5 graphically by means of the operation INTERSECTION found in the latest versions of the TI graphing calculators. Equations that involve exponential functions are referred to as exponential equations. Equations involving logarithmic functions are called logarithmic equations. The purpose of this section is to study ways for solving these equations. In order to solve an exponential equation, we use algebra to reduce the equation into the form a x = b where a and b > 0 are constants and x in the unknown variable. Taking the common logarithm of both sides and using property the log (a x ) = x we find x = log b log a. Example 53 Solve the equation 00(0.886) x = 5 algebraically. Dividing both sides by 00 to obtain (0.886) x = 0.5. Take the log of both log (0.886) sides to obtain x log (0.886) = log 0.5. Thus, x = log (0.5) 7.8. Exercise 54 Solve the equation 50, 000(.035) x = 50, 000(.06) x. Divide both sides by 50, 000(.06) x to obtain Take log of both sides to obtain x log ( ) x.035 = 5.06 ( ).035 = log 5.06 Divide both sides by the coefficient of x to obtain log 5 x = log ( ) Doubling Time In some exponential models one is interested in finding the time for an exponential growing quantity to double. We call this time the doubling time. To find it, we start with the equation b a t = b or a t =. Solving for t we find t = log log a. Example 54 Find the doubling time of a population growing according to P = P 0 e 0.t.

66 66 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Setting the equation P 0 0.t = P 0 and dividing both sides by P 0 to obtain 0.t =. Take ln of both sides to obtain 0.t = ln. Thus, t = ln Half-Life On the other hand, if a quantity is decaying exponentially then the time required for the quantity to reduce into half is called the half-life. To find it, we start with the equation ba t = b and we divide both sides by b to obtain a t = 0.5. Take the log of both sides to obtain t log a = log (0.5). Solving for t we find t = log (0.5) log a. Example 55 The half-life of Iodine-3 is about 3 hours. You begin with 50 grams of this substance. What is a formula for the amount of Iodine-3 remaining after t hours? Since the problem involves exponential decay then if Q(t) is the quantity remaining after t hours then Q(t) = 50a t with 0 < t <. But Q(3) = 5. That is, 50a 3 = 5 or a 3 = 0.5. Thus a = (0.5) and Q(t) = 50(0.95) t. Can all exponential equations be solved using logarithms? The answer is no. For example, the only way to solve the equation x + = x is by graphical methods which give the solutions x.69 and x =. Exercise 55 Solve the equation (.0) t = t. Using a graphing calculator and the key INTERSECTION one finds t 99.4 We end this section by describing a method for solving logarithmic equations. The method consists of rewriting the equation into the form log x = a or ln x = a and then find the exponential form to obtain x = 0 a or x = e a. Also, you must check these values in the original equation for extraneous solutions. Example 56 Solve the equation: log (x ) log (x + ) = log (x ). Using the property of the logarithm ( ) of a quotient we can rewrite the given x x equation into the form log x+ = log (x ). Thus, x+ = x. Cross multiply and then foil to obtain (x+)(x ) = x or x = 0. Solving we find x = 0. However, this is not a solution because it yields logarithms of negative numbers when plugged into the original equation. Exercise 56 Solve the equation: ln (x ) + ln (x 3) = ln x.

67 4.7. CONTINUOUS GROWTH RATE AND THE NUMBER E 67 Using the property ln (ab) = ln a + ln b we can rewrite the given equation into the form ln (x )(x 3) = ln x. Thus,(x )(x 3) = x or x 7x+6 = 0. Factoring to obtain (x )(x 6) = 0. Solving we find x = or x = 6. The value x = must be discarded since it yields a logarithm of a negative number. Recommended Problems (pp. 44-7):, 3, 5, 6, 7, 9,, 3, 5, 7, 0,. 4.7 Continuous Growth Rate and the Number e For any positive number c, e ln c = c. Thus, any positive number a can be written in the form a = e k where k = ln a. If a > then ln a > 0 and therefore k > 0. If 0 < a < then ln a < 0 and thus k < 0. Now, suppose that Q(t) = ba t. Then Q(t) = b(e k ) t = be kt. Note that for k > 0 e k > so that Q(t) represents an exponential growth and if k < 0 then e k < so that Q(t) is an exponential decay. We call the constant k the continuous growth rate. Example 57 If f(t) = 3(.07) t is rewritten as f(t) = 3e kt, find k. By comparison of the two functions we find e k =.07. Take the ln of both sides to obtain k = ln (.07). Example 58 A population increases from its initial level of 7.3 million at the continuous rate of.% per year. Find a formula for the population P (t) as a function of the yeat t. When does the population reach 0 million? We are given the intial value 7.3 million and the continuous growth rate k = 0.0. Therefore, P (t) = 7.3e 0.0t. Next,we want to find the time when P (t) = 0. That is, 7.3e 0.0t = 0. Divide both sides by 7.3 and then take the ln of ln 0 both sides to obtain 0.0t = ln 7.3. Thus, t = ln ln Next, in order to convert from Q(t) = be kt to Q(t) = ba t we let a = e k. For example, to convert the formula Q(t) = 7e 0.3t to the form Q(t) = ba t we let b = 7 and a = e Thus, Q(t) = 7(.35) t.

68 68 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Example 59 Find the annual percent rate and the continuous percent growth rate of Q(t) = 00(0.886) t. The annual percent of decrease is r = a = = 0.4 =.4%. To find the continuous percent growth rate we let e k = and solve for k by taking the ln of both sides and obtain k = ln (0.886) 0. =.%. Why k is called the continuous growth rate (Optional) Suppose that an amount of money P dollars is invested into an account that pays an annual interest r compounded annually, that is after the first year the interest will be used for the amount P + rp rather than just P. Thus, the balance at the end of the first year is P + rp = P ( + r). At the end of the second year the balance will be P ( + r) + rp ( + r) = P ( + r). Continuing, we find that at the end of the t-th year the balance will be P ( + r) t. Now if the interest is compounded every six month then the balance at the end of the first year will be P ( + r ) + r P ( + r ) = P ( + r ). At the end of the second year, the balance will be P ( + r )4 = P ( + r ). Continuing we find that the balance at the end of the t years is P ( + r )t. Similarly, if the interest is compounded quarterly we will find that the balance will be P ( + r 4 )4t and in general, if the interest is compounded n times a year than the balance in the account after t years will be P ( + r n )nt. What happens in the long run or in the continuous case? That is, when n? In the vocabulary of finance, the interest is said to be compounded continuously. Example 60 Suppose that r = 5%. Complete the following table n ( n )n It follows that in the long run, ( + 5% n )n e 5%. In general, ( + r n )n e r when n. Recommended Problems (pp. 5 - ):, 3, 5, 7, 9, 0,, 5, 7, Compound Interest The term compound interest refers to a procedure for computing interest whereby the interest for a specified interest period is added to the original

69 4.8. COMPOUND INTEREST 69 principal. The resulting sum becomes a new principal for the next interest period. The interest earned in the earlier interest periods earn interest in the future interest periods. The future value is the total amount due on the maturity date of the loan or the investment. The maturity value is calculated by using the compound amount formula which is given by A = P ( + r n )nt where P is the present value or principal, r is the nominal annual interest rate, n is the number of periods in a year, and t is the time in years. Possible values of n are: (annually), (semiannually), 4 (quarterly), (monthly), and 365 (daily). When interest is compounded more frequently than once a year, the account effectively earns more than the nominal rate. Thus, we distinguish between nominal rate and effective rate. The effective annual rate tells how much interest the investment actually earns. The quantity ( + r n )n is known as the effective interest rate. Example 6 Translating a value to the future is referred to as compounding. What will be the maturity value of an investment of $5, 000 invested for four years at 9.5% compounded semi-annually? Using the formula for compound interest with P = $5, 000, t = 4, n =, and r =.095 we obtain ( A = 5, ) 8 $, Example 6 Translating a value to the present is referred to as discounting. We call ( + r n ) nt the discount factor. What principal invested today will amount to $8, 000 in 4 years if it is invested at 8% compounded quarterly? The present value is found using the formula ( P = A + r ) ( nt = 8, ) 6 $5, n 4 Example 63 What is the effective rate of interest corresponding to a nominal interest rate of 5% compounded quarterly? effective rate = ( ) = 5.% 4

70 70 CHAPTER 4. EXPONENTIAL AND LOGARITHMIC FUNCTIONS Continuous Compound Interest When the compound formula is used over smaller time periods the interest becomes slightly larger and larger. That is, frequent compounding earns a higher effective rate, though the increase is small. This suggests compounding more and more, or equivalently, finding the value of A in the long run. The example at the end of the previous section shows that as n the expression ( + n) r n approaches e r so that A = P e rt. This formula is known as the continuous compound formula. In this case, the annual effective interest rate is found using the formula e r. Example 64 Find the effective rate if $000 is deposited at 5% annual interest rate compounded continuously. The effective interest rate is e = 5.7% Example 65 Which is better: An account that pays 8% annual interest rate compounded quarterly or an account that pays 7.95% compounded continuously? The effective rate corresponding to the first option is That of the second option ( ) 4 8.4% 4 e % Thus, we see that 7.95% compounded continuously is better than 8% compounded quarterly. Recommended Problems (pp. 56-7):, 5, 7, 9,, 4, 6, 7, 0,.

71 Chapter 5 Building New Functions from Old Ones In this chapter we discuss various ways for building new functions from old ones. New functions can be obtained by composiing functions, finding the inverse of a function, using arithmetic combinations, and finally by making changes to either the input or the output of a function. 5. Composition of Functions In this section we will discuss a procedure for building new functions from old ones known as the composition of functions. We start with an example of a real-life situation where composite functions are applied. Example 66 You have two money machines, both of which increase any money inserted into them. The first machine doubles your money. The second adds five dollars. The money that comes out is described by f(x) = x in the first case, and g(x) = x + 5 in the second case, where x is the number of dollars inserted. The machines can be hooked up so that the money coming out of one machine goes into the other. Find formulas for each of the two possible composition machines. Suppose first that x dollars enters the first machine. Then the amount of money that comes out is f(x) = x. This amount enters the second machine. The final amount coming out is given by g(f(x)) = f(x) + 5 = x + 5. Now, if x dollars enters the second machine first, then the amount that comes out is g(x) = x + 5. If this amount enters the second machine then the final amount coming out is f(g(x)) = (x + 5) = x + 0. The function f(g(x)) is called the composition of the functions f and g; the 7

72 7 CHAPTER 5. BUILDING NEW FUNCTIONS FROM OLD ONES function g(f(x)) is called the composition of the functions g and f. In general, suppose we are given two functions f and g such that the range of g is contained in the domain of f so that the output of g can be used as input for f. We define a new function, called the composition of f and g, by the formula (f g)(x) = f(g(x)). In a similar way, we can define the composition of g and f to be the function (g f)(x) = g(f(x)) so that the output of f is the input of g. Using a Venn diagram we have Composition of Functions Defined by Tables Example 67 Complete the following table x f(x) g(x) f(g(x))

73 5.. COMPOSITION OF FUNCTIONS 73 x f(x) g(x) f(g(x)) undefined Composition of Functions Defined by Formulas Example 68 Suppose that f(x) = x + and g(x) = x 3. (a) Find f g and g f. (b) Calculate (f g)(5) and (g f)( 3). (c) Are f g and g f equal? (a) (f g)(x) = f(g(x)) = f(x 3) = (x 3) + = x 5. Similarly, (g f)(x) = g(f(x)) = g(x + ) = (x + ) 3 = 4x + 4x. (b) (f g)(5) = (5) 5 = 45 and (g f)( 3) = 4( 3) + 4( 3) =. (c) f g g f. With only one function you can build new functions using composition of the function with itself. Also, there is no limit on the number of functions that can be composed. Example 69 Suppose that f(x) = x + and g(x) = x 3. (a) Find (f f)(x). (b) Find [f (f g)](x). (a) (f f)(x) = f(f(x)) = f(x + ) = (x + ) + = 4x + 3. (b) [f (f g)](x) = f(f(g(x))) = f(f(x 3)) = f(x 5) = (x 5) + = 4x 9. Decomposition of Functions If a formula for (f g)(x) is given then the process of finding the formulas for f and g is called decomposition. Example 70 Decompose (f g)(x) = 4x. One possible answer is f(x) = x and g(x) = 4x. Another possible answer is f(x) = x and g(x) = x. Recommended Problems (pp ):, 3, 5, 6, 7, 9,, 3, 7,, 9, 4, 43, 45, 49, 5, 53, 57, 6, 63.

74 74 CHAPTER 5. BUILDING NEW FUNCTIONS FROM OLD ONES 5. Inverse Functions We have seen that when every vertical line crosses a curve at most once then the curve is the graph of a function f. We called this procedure the vertical line test. Now, if every horizontal line crosses the graph at most once then the function can be used to build a new function, called the inverse function and is denoted by f, such that if f takes an input x to an output y then f takes y as its input and x as its output. That is f(x) = y if and only if f (y) = x. Remark. The test used to identify invertible functions which we discussed above is referred to as the horizontal line test. Example 7 Use a graphing calculator to decide whether or not the function is invertible, that is, has an inverse function: (a) f(x) = x (b) g(x) = x. (a) Using a graphing calculator we find the graph 30 y x 0 0 We see that every horizontal line crosses the graph once so the function is invertible. (b) The graph of g(x) = x show that there are horizontal lines that cross the graph twice so that g is not invertible.

75 5.. INVERSE FUNCTIONS y 0 x Warning It is important not to confuse between f (x) and (f(x)). The later is just the reciprocal of f(x), that is, (f(x)) = f(x) whereas the former is how the inverse function is represented. How do you find the formula for f from the formula of f? The procedure consists of the following steps:. Replace f(x) with y.. Interchange the letters x and y. 3. Solve for y in terms of x. 4. Replace y with f (x). Example 7 Use a graphing calculator to show that the function f(x) = x 3 +7 has an inverse. Find the formula for the inverse function. As seen in Example, f(x) is invertible. We find its inverse as follows:. Replace f(x) with y to obtain y = x Interchange x and y to obtain x = y Solve for y to obtain y 3 = x 7 or y = 3 x Replace y with f (x) to obtain f (x) = 3 x 7. Sometimes it is difficult to find a formula for the inverse function. In this case, a graphical method is used to evaluate the inverse of a function at a given point. To be more precise, let f(x) = x 3 + x +. Using a graphing calculator, one can easily check that the graph satisfies the horizontal line test and consequently f(x) is invertible. Finding a formula of f (x) is difficult. So if we want for example to evaluate f (4) then we write x = f (4) and set the equation or x 3 + x + = 4 x 3 + x 3 = 0.

76 76 CHAPTER 5. BUILDING NEW FUNCTIONS FROM OLD ONES Now, using a graphing calculator and the INTERSCTION key one looks for the x-intercepts which is found to be x.3. Thus, f (4).3. Exercise 57 Show that f(x) = x is invertible and find its inverse. Graph on the same axes both f(x) and f (x). What is the relationship between the graphs? Graphing f(x) we find y 3 0 x Thus, the horizontal test applies and the function is invertible. To find a formula for the inverse function, we follow the four steps discussed above:. Replace f(x) with y to obtain y = x.. Interchange x and y to obtain x = y. 3. Solve for y by taking log of both sides to obtain y = log x log. 4. Replace y with f (x) to obtain f (x) = log x log. Graphing both f(x) and f (x) on the same axes we find

77 5.. INVERSE FUNCTIONS 77 3 y 0 x 3 So the graphs are reflections of one another across the line y = x as shown above. Domain and Range of an Inverse Function Using a Venn diagram the relationship between f and f is shown below This shows that the outputs of f are the inputs of f and the outputs of f are the inputs of f. It follows that Domain of f = Range of f and Range of f = Domain of f. Compositions of f and its Inverse Suppose that f is an invertible function. Then the expressions y = f(x) and x = f (y) are equivalent. So if x is in the domain of f then f (f(x)) = f (y) = x and for y in the domain of f we have f(f (y)) = f(x) = y It follows that for two functions f and g to be inverses of each other we must have f(g(x)) = x and g(f(x)) = x for x in the appropriate domain.