The student will become familiar with a beginning library of. Calculus for Business & Economics 1

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1 Learning Objectives for Section Elementary Functions: - Graphs &Transformations The student will become familiar with a beginning library of elementary functions. The student will be able to transform functions using vertical and horizontal shifts. The student t will be able to transform functions using stretches and shrinks. The student will be able to graph piecewise defined functions. 1

2 Problem Complete the table and graph the function f(x) = x 2 : x y = x

3 Solution Complete the table and graph the function f(x) = x 2 : x y = x

4 Problem Sketch the graph of f(x) = (x 2) 2 and explain, in words, how it is related to the graph of ff( f(x) = x 2 : x y = (x-2)

5 Comparison of f(x) = x 2 and f(x) = (x 2) 2 5

6 Solution (continued) The graph has the same shape as the original function. The difference is that the original graph has been translated two units to the right on the x axis. Conclusion: The graph of the function (x 2) 2 is the graph of shifted horizontally two units to the right on the x axis. Notice that replacing x by x 2shifts the graph horizontally to the right and not the left. Correspondingly, replacing x by x + 2 would shift the graph horizontally to the left. 6

7 Problem Sketch the graphs of f(x) = x 3 and f(x)=x x

8 Solution When we shift the graph of f(x) = x 3 by 5 units upward, we get the graph of f(x)=x

9 Problem Sketch the graphs of f(x) = x and f(x) = x and find the domains. 9

10 Solution The domain is all non negativenegative real numbers. By reflecting one of the graphs about the x-axis, we get the graph of the other one. 10

11 Problem 3 3 Sketch the graphs of f(x) = x and f(x) = x

12 Solution 3 When we shift the graph of f(x) = x by 1 unit horizontally 3 to the left, we get the graph of f(x) = x+ 1 12

13 Absolute Value Function a( x) = x Various graphs involved with the absolute function. 13

14 Absolute Value Function (continued) a( x ) = x Various graphs involved with the absolute function. 14

15 Summary of Graph Transformations 1. Vertical Translation: at y = f(x) + k (1) k > 0: Shift graph of y = f (x) up k units. (2) k < 0: Shift graph of y = f (x) down k units. 2. Horizontal Translation: y = f (x+h) (1) h > 0: Shift graph of y = f (x) left h units. (2)h<0:Shiftgraph of y = f (x)right h units units. 3. Reflection: y = f (x). Reflect the graph of y = f (x) in the x axis. 4. Vertical Stretch and Shrink: y =Af(x) (1) A > 1: Shrink graph of y = f (x) vertically by multiplying each coordinate value by A. (2) 0 < A < 1: Stretch graph of y = f (x) vertically by multiplying each coordinate value by A. 15

16 Piecewise-Defined Functions Earlier we noted that the absolute value of a real number x can be defined as x, if x > 0 x = x, if x < 0 Notice that this function is defined by different rules for different parts of its domain. Functions whose definitions involve more than one rule are called piecewise-defined functions. Graphing one of these functions involves graphing each rule over the appropriate portion of the domain. 16

17 Example Sketch the graph of the function: f(x) 2 2x, if x < 2 = x 2, if x 2 17

18 Solution We have the following graph of f(x) 2 2x, if x < 2 = x 2, if x 2 18

19 End of Slides Please, solve the exercise problems, especially the applications given in the textbook. If you don t solve a problem by your hands, you will have a difficult time on the exam. Studying this course consists of understanding the concepts/techniques and solving the problems, especially the applications. 19

20 Learning Objectives for Section 2.3 Solving Quadratic Functions The student will be able to identify and define quadratic functions, equations, and inequalities. The student will be able to identify and use properties of quadratic functions and their graphs. The student will be able to solve applications of quadratic functions. The student will be able to graph and identify properties of polynomial and rational functions. 1

21 Quadratic Functions If a, b, c are real numbers with a 0, then the function 2 f(x) = ax + bx+ c is a quadratic function and its graph is called a parabola. 2

22 Vertex Form of the Quadratic Functions It is convenient to convert the general form of a quadratic function 2 f(x) = ax + bx+ c to what is known as the vertex form: 2 f(x) = a(x h) + k This vertex form gives the vertex vertex of f(x): (h, k) 3

23 Completing the Square To find the Vertex of a Quadratic Function The example below illustrates the procedure: Consider 2 f(x) = 3x + 6x 1 vertex. Complete the square to find the vertex 4

24 Solution Step 1. Factor the coefficient of x 2 outofthefirsttwoterms: of the first two terms: 2 2 f(x) = 3x + 6x 1= 3[x 2x] 1 2 Step 2. Recall the formula x 2 + 2ax+ a 2 = (x+ a) By the formula, we have 2 f(x) = 3[x 2 2 2x + ( 1) ( 1) ] 1 = [ x 2x ( 1) ] 3( 1) 1 2 = 3( x 1) + 2 Step 3. From the vertex form, we deduce the vertex (1,2) 5

25 Intercepts sofaquadratic Function The graph of any function has two types of special points: ( something, 0) and ( 0, something) The first one ( something,0) is called the x intercept (or root) and the second one ( 0,some thing) is called the y intercept of the function. 3 Example) f(x) = 2x 3 has the x intercept (,0) and the y intercept (0, 3) 2 6

26 Example Find the xandy intercepts of the quadratic function 2 f(x) = 3x + 6x 1 7

27 Solution 1. x intercept : We solve f(x)=0, i.e., find x satisfying 2 0 = 3x + 6x 1 By the Quadratic Formula, we have x 2 2 b b 4ac 6 6 4( 3)( 1) ± ± = = 2a 2( 3) 6± 24 = 0.18, Therefore, we have two x intercepts: (0.18,0) and (1.82,0) 8

28 Solution (continued) 2 2. y intercept : Given function f(x) = 3x + 6x 1 implies f(0) = 3(0) + 6(0) 1= 1 Therefore, we have only one y intercept: (0, 1) 9

29 Generalization 2 For f(x) = a(x h) + k, 1. If a 0, then the graph of f(x) is a parabola. (1) If a > 0, then the graph is concave upward. (2) If a< 0, then the graph is concave downward. In any case, we have the vertex at ( h,k) 2. Axis of symmetry: x = h 3. (1) If a > 0, then f(x) has the minimum f(h) = k (2) If a< 0, then f(x) has the maximum f(h) = k 4. Domain of f(x) is the set of all real numbers. 5. Range: (1) If a> 0, then f(x) has the range {y y k} (2) If a < 0, then f(x) has the range {y y k} 10

30 Solving Quadratic Inequalities Solve the quadratic inequality 3 x x

31 Solving Quadratic Inequalities Solve the quadratic inequality 3 x x 1 0 Solution This inequality holds for those values of x for which the graph of f(x) () is at or above the x axis. This happens for x between the two x intercepts, including the intercepts. (If you look at the graph of the function, you can understand d this very easily.) Thus, the solution set for the quadratic inequality is 0.18 x

32 Please, take a look at the Mth Mathematicati file Section

33 Application of Quadratic Functions A Macon, Georgia, peach orchard farmer now has 20 trees per acre. Each tree produces, on the average, 300 peaches. For each additional tree that the farmer plants, the number of peaches per tree is reduced by 10. How many more trees should the farmer plant to achieve the maximum yield of peaches? What is the maximum yield? 14

34 Solution Yield = (number of peaches per tree)(number of trees) Yield = (300)(20) = 6000 (currently) Plant t one more tree: Yield = (300 1(10))(20 + 1) = (290)(21) = 6090 peaches. Plant two more trees: Yield = (300 2(10))(20 + 2) = (280)(22) = 6160 Plant x more trees: Yield = (300 x(10))(20 + x) = -10x x

35 Solution (continued) From the result above, we define 2 Y( x ) = 10x + 100x To find the maximum yield, note that the function Y(x) is a quadratic function of which graph is concave downward. Hence, the vertex of the function will be the maximum value of the yield. Graph is below, with the y value in thousands. To find the vertex, we complete the square. 16

36 Solution (continued) We complete the square as follows: 2 Y( x ) = 10( x 10x ) ( x 10x ( 5 ) ( 5 ) ) 6000 = ( x 5) 10( 25 ) 60 = = 10( x 5 ) It implies that the graph has the vertex (5,6250) So, the farmer should plant 5 additional trees and obtain a 2 yield of 6250 peaches. The term 10x in the quadratic function implies that the graph is concave downward, so the vertex must be the maximum

37 Break Even Analysis The financial department of a company that produces digital cameras has the revenue and cost functions for x million cameras are as follows: R(x) = x(94.8 5x) C( x ) = x Both have domain 1 x 15 Break even points are the production levels at which R(x) = C(x) Find the break-even points algebraically to the nearest thousand cameras. 18

38 Solution (continued) From R(x) = C(x), we deduce x( x ) = x x 5 x = x 2 5x 75.1x = By the Quadratic Formula, we get x = ± 2 b± b 4ac 2a ± ( 75.1) 4( 5 )(156 ) = 2( 5 ) 2.49,

39 Solution (continued) Thus, the break eveneven points are x 2.49 and x If we graph the cost and revenue functions on a graphing utility, we obtain the following graphs, showing the two intersection points: 20

40 Quadratic Regression A visual inspection of the plot of a data set might indicate that a parabola would be a better model of the data than a straight line. In that case, rather than using linear regression to fit a linear model to the data, we would use quadratic regression on a graphing calculator to find the function of the form 2 y( x ) = ax + bx + c that best fits the data. We skip the example. 21

41 Polynomial Functions A polynomial function of degree n is a function that can be written in the form n n 1 2 P(x) = a x + a x + + a x + ax + a n n n The domain of a polynomial function is the set of all real numbers. A polynomial of degree 0 is a constant. A polynomial of degree 1 is a linear function. A polynomial of degree 2 is a quadratic function. A polynomial of degree 3 is a cubic function. 22

42 Shape of Polynomials AA polynomial is called odd if it only contains odd powers of x. AA polynomial is called even if it only contains even powers of x. Let s look at the shapes of some even and odd polynomials. Look for some of the following properties: Symmetry Number of x intercepts Number of local maxima/minima What happens as x goes to + + or -? 23

43 Examples Please, confer the Mathematica file Section

44 Observation of Odd Polynomials For an odd polynomial, the graph is symmetric about the origin the graphs starts negative, ends positive, or vice versa, depending on whether h the leading coefficient is positive or negative either way, a polynomial of degree n crosses the x axis at least once, at most n times. For an even polynomial, the graph is symmetric about the y axisy the graphs starts negative, ends negative, or starts and ends positive, depending on whether the leading coefficient is positive or negative either way, a polynomial of degree n crosses the x axis at most n times. It may or may not cross at all. 25

45 Characteristics of Polynomials Graphs of polynomials are continuous. One can sketch the graph without lifting up the pencil. Graphs p of polynomials have no sharp corners. Graphs of polynomials usually have turning points, which is a point that separates an increasing portion of the graph from a decreasing portion. A polynomial of degree n can have at most n linear factors. Therefore, the graph of a polynomial l function of positive i degree n can intersect the x axis at most n times. A polynomial of degree n may intersect the x axis fewer than n times. 26

46 Rational Functions AA rational function is a quotient of two polynomials, P(x) and Q(x), for all x such that Q(x) is not equal to zero. Example: Let P(x) = x + 5 and Q(x) = x 2, then x + 5 R( x ) = x 2 is a rational function whose domain is all real values of x with the exception of 2 (Why?) 27

47 Vertical Asymptotes of Rational Functions x values at which the function is undefined represent vertical asymptotes to the graph of the function. A vertical asymptote is a line of the form x = k which the graph of the function approaches but does not cross. In the figure below, which is the graph of x + 5 R( x ) = x 2 the line x = 2 is a vertical asymptote. 28

48 Horizontal Asymptotes of Rational Functions A horizontal asymptote of a function is a line of the form y = k which the graph of the function approaches as x approaches ± For example, in the graph of x+ 5 R( x ) = x 2 the line y = 1 is a horizontal asymptote. 29

49 Generalization about Asymptotes of Rational Functions The number of vertical asymptotes of a rational function f (x) = n(x)/d(x) ( ( ) is at most equal to the degree of d(x). A rational function has at most one horizontal asymptote. The graph of a rational function approaches the horizontal asymptote (when one exists) both as x increases and decreases without bound. 30

50 End of Slides Please, solve the exercise problems, especially the applications given in the textbook. If you don t solve a problem by your hands, you will have a difficult time on the exam. Studying this course consists of understanding the concepts/techniques and solving the problems, especially the applications. 31

51 Objectives for Section E Exponential Functions The student will be able to graph and identify the properties of exponential functions. The student will be able to apply base e exponential functions, including growth and decay applications. The student will be able to solve compound interest problems. 1

52 Exponential Functions The equation f(x ) = b x for b> 0 defines the exponential function with base b. The domain is the set of all real numbers, while the range is the set of all positive real numbers. 2

53 Riddle Here is a problem related to exponential functions: Suppose you received a penny on the first day of December, two pennies on the second day of December, four pennies on the third day, eight pennies on the fourth day and so on. How many pennies would you receive on December 31 if this pattern continues? Would you rather take this amount of money or receive a lump sum payment of $10,000,000? 3

54 Solution Day No of Pennies Currently Total Dec. 1 1= Dec. 2 2= Dec. 3 4= Dec. 4 8= Dec. n 2 n n-1 Dec n

55 Solution (continued) According to the table, if this pattern continued, you would have 2 30 pennies on Dec. 31. The exponent on two is one less than the day of the month. Since 2 30 =10,737,418.24, which is bigger than $10,000,000, thus on Dec. 31, you should get $10,737, This example shows how an exponential lf function grows extremely rapidly. In this case, the exponential function f(x) = 2 is used to model this problem. x 5

56 Graph of f(x) = 2 x When we extend the previous table for a real number x, we have the following graph: 6

57 Characteristics of the Graph of f(x) = b x when b> 1 1. All graphs will approach the x axis as x becomes unbounded and negative. 2. All graphs will pass through (0,1)(y intercept). 3. There are no x intercepts. 4. Domain is all real numbers. 5. Range is all positive real numbers. 6. The graph is always increasing on its domain. 7. All graphs are continuous curves. 7

58 Characteristics of the Graph of f(x) = b x when 0< b< 1 1. All graphs will approach the x axis as x gets large. 2. All graphs will pass through (0,1)(y intercept). 3. There are no x intercepts. 4. Domain is all real numbers. 5. Range is all positive real numbers. 6. The graph is always decreasing on its domain. 7. All graphs are continuous curves. 8

59 Graphs of Exponential Functions with Various Basis Please, take a look at the Mathematica file Section 2.4 9

60 Exponential Function with Base e e is an irrational number called the Euler constant. It represents e º e can be approximated as closely as we like by evaluating the x expression x 10

61 l sttopics/e.h.ac nd /Hi s/e.h.uk uk/history stor y Leonhard Euler ( ) Leonhard Euler first demonstrated that as x gets positively large, x will approach a fixed constant we now call e. So much of our mathematical notation is due to Euler that it will come as no surprise to find that the notation e for this number is due to him. The claim which has sometimes been made, however, that Euler used the letter e because it was the first letter of his name is ridiculous. It is probably not even the case that the e comes from exponential but it may have just be the next vowel after a and Euler was already using the notation a in his work. Whatever the reason is, the notation e made its first appearance in a letter Euler wrote to Goldbach in ( ac. uk/ ~ history/ HistTopics/ e. html# s19) x 11

62 Leonhard Euler (continued) He made various discoveries regarding g e in the following years, but it was not until 1748 when Euler published Introductio in analysin infinitorum that he gave a full treatment of the ideas surrounding e. He showed that e = 1 + 1/1! + 1/2! + 1/3! +... and that e is the limit of (1 + 1/n) n as n tends to infinity. Euler gave an approximation for e to 18 decimal places, e =

63 e Graph of f(x) = e x Since 2 < e < 3, we have 2 x < e x < 3 x for positive x, 2 x > e3 x > 3 x for negative x. e 13

64 Relative Growth Rates Functions of the form y(t) = c e kt, where c and k are constants and the independent variable t represents time, are often used to model population growth and radioactive decay. Note that if t = 0, then y = c. So, the constant c represents the initial population pp (or initial amount.) The constant k is called the relative growth rate. If the relative growth rate is k = 0.02, then at any time t, the population p is growing at a rate of 0.02y persons (2% of the population) per year. We say that population is growing continuously at relative growth rate k to mean that the population y is given by the model y(t) () = c e kt kt. 14

65 Growth & Decay Applications - Atmospheric Pressure The atmospheric pressure P decreases with increasing height. The pressure is related to the number of kilometers h above the sea level by the formula: P(h)=760e h Question 1. Find the pressure at sea level (h=0). Question 2. Find the pressure at height of 7 kilometers. 15

66 Solution Answer 1. Pressure at sea level: P(0)=760e 0 =760 Answer 2. Pressure at a height of 7 kilometers: P(7)=760e (7) =

67 Depreciation of a Machine A machine is initially worth V 0 dollars but loses 10% of its value each year. Its value after t years is given by the formula V(t)= V t 0 (0.9 ) Find the value after 8 years of a machine whose initial value is $30,000. Solution) V(8)=30000(0.9 8 )=$12,914 17

68 Compound Interest The compound interest formula is A= P 1+ r n Here, A is the future value of the investment, P is the initial amount (principal), r is the annual interest rate as a decimal, n represents the number of compounding periods per year, and t is the number of years. nt 18

69 Example Find the amount to which $1500 will grow if deposited in a bank at 5.75% interest compounded quarterly for 5 years. 19

70 Solution Find the amount to which $1500 will grow if deposited in a bank at 5.75% interest compounded quarterly for 5 years. Solution) Use the compound interest formula: r A= P 1 + n Putting P=1500, r=0.0575, n=4, t=5, we obtain 4( 5 ) A = = $ nt 20

71 End of Slides Please, solve the exercise problems, especially the applications given in the textbook. If you don t solve a problem by your hands, you will have a difficult time on the exam. Studying this course consists of understanding the concepts/techniques and solving the problems, especially the applications. 21

72 Learning Objectives for Section 2.5 Logarithmic Functions The student will be able to use and apply inverse functions. The student will be able to use and apply logarithmic functions and properties of logarithmic functions. The student will be able to evaluate logarithms. 1

73 Logarithmic Functions There is a close connection between a logarithmic function and an exponential function inverse relation. We will study the concept of inverse functions as a prerequisite for our study of logarithmic function. 2

74 One to One Functions We wish to define an inverse of a function. Before we do so, it is necessary to discuss the topic of one to one functions. First of all, only certain functions are one to one one. Definition: A function is said to be one to to one one if distinct inputs of a function correspond to distinct outputs. In notation, For x x, f(x ) f(x )

75 Graph of One to to One Functions If we choose two different x values, the corresponding y values of the one-to-one function should be different. Lots of graphs will be given in class. 4

76 Horizontal Line Test Recall that for an equation to be a function, its graph must pass the vertical line test. That is, a vertical line that sweeps across the graph of a function from left to right will intersect the graph only once at each x value. There is a similar geometric test to determine if a function is one-to-one. one. It is called the horizontal line test.any horizontal line drawn through the graph of a one-to-one function will cross the graph only once. If a horizontal line crosses a graph more than once, then the function that is graphed is not one-to-one. one. 5

77 Definition of Inverse Function Given a one-to to-one one function, the inverse function is found by interchanging the x and y values of the original function. That is to say, if an ordered pair (a, b) belongs to the original function, then the ordered pair (b, a) belongs to the inverse function. Note: If a function is not one-to to-one one (fails the horizontal line test), then the inverse of such a function does not exist. 6

78 Logarithmic Functions The logarithmic function with base 2 is defined to be the inverse of the exponential function with base 2, y=2 x. Notice that the exponential function y=2 x is one-to-one and therefore has an inverse. Generally, the logarithmic function with base n > 0 is defined to be the inverse of the exponential function with base n > 0. 7

79 Inverse of an Exponential Function Start with y=2 x Now, interchange x and y coordinates : x=2 y There are no algebraic techniques that can be used to solve for y, so we simply call this function y the logarithmic function with base 2. The definition of this new function is: log x = y if and only if x = 2 y 2 8

80 Graph, Domain, Range of Logarithmic Functions The domain of the logarithmic function y = log 2 x is the same as the range of the exponential function y= 2 x. Why? The range of the logarithmic function is the same as the domain of the exponential function (Again, why?) Another fact: If one graphs any one-to-one function and its inverse on the same grid, the two graphs will always be symmetric with respect to the line y = x. 9

81 Logarithmic Exponential Conversions Study the examples below. You should be able to convert a logarithmic into an exponential expression and vice versa. x log416 = x = 16 4 x = log3 = log3 = log 3 33 = = 9 81 = 9 log81 9 = = 5 log = 3 10

82 Solving Equations Using the definition of a logarithm, you can solve equations involving logarithms. Examples: 3 3 logb1000 = 3 b = 10 b = 10 5 log6 x = 5 6 = x 7776 = x In each of the above, we converted from log form to exponential form and solved the resulting equation. 11

83 Properties of Logarithms (Must Memorize) These are the properties of logarithms. M and N are positive real numbers, b not equal to 1, and p and x are real numbers. (For 4, we need ed x > 0). x log b x 1. logb1 = 0 2. logbb = 1 3. logbb = x 4. b = x 5. logb( MN) = logbm + logbn M 6. logb = logb M logb N N p 7logM 7. b = plogm b 8. log M = log N if and only if M = N b b 12

84 Example Solve for x: log 4 (x + 6) + l og 4 ( x 6 ) = 3 13

85 Solution Solve for x: lo g 4 (x + 6) + log 4 (x 6) = 3 log ( x + 6 )( x 6 ) = log 4( x 36 ) = = x = x = x ± 10 = x 2 x = 10 Question: why is the answer the positive 10 only? 14

86 Example Solve for x: log 10 π l og π = x 15

87 Solution Solve for x: log 10 π log π = x π log 10 = x π 1 log 10 = x log 10 = x 10 x = 4 16

88 Common Logs & Natural Logs Common log: log x = log 10 x If no base is indicated, the logarithm is assumed to be base 10. Natural log: ln x = loge x Here e represents the Euler constant, e º

89 Example Solve for x: ln( x + 1) ln x = 1 18

90 Solution Solve for x: ln( x + 1) ln x = 1 x+ 1 ln = 1 x x+ 1 e = x ex = x + 1 (e 1)x = 1 x 1 = e 1 19

91 Application How long will it take money to double if compounded monthly at 4 % interest? 20

92 Solution By the compound interest formula, A= P 1+ 2P = P 1 + r n nt = ( ) 12t t 12t ( ) ( ) ln 2 = ln = 12t ln ln2 12 ln ( ) = t t =

93 Logarithmic Regression Among increasing functions, the logarithmic functions with bases b > 1 increase much more slowly for large values of x than either exponential or polynomial functions. When a visual inspection of the plot of a data set indicates a slowly increasing function, a logarithmic function provides a good model. We use logarithmic regression on a graphing calculator to find the function of the form y = a+ blnx that best fits the data. Again, since the regression subject depends on the calculator, we skip this subject. 22

94 End of Slides Please, solve the exercise problems, especially the applications given in the textbook. If you don t solve a problem by your hands, you will have a difficult time on the exam. Studying this course consists of understanding the concepts/techniques and solving the problems, especially the applications. 23

95 Chapter 2 Review - Important Terms, Symbols, Concepts 2.1. Functions Point-by-point plotting may be used to sketch the graph of an equation in two variables: plot enough points from its solution set in a rectangular coordinate system so that the total graph is apparent and then connect these points with a smooth curve. A function is a correspondence between two sets of elements such that to each element in the first set there corresponds one and only one element in the second set. The first set is called the domain and the second set is called the range. 1

96 Section 2.1 Functions (continued) If x represents the elements in the domain of a function, then x is the independent variable or input. If y represents the elements in the range, then y is the dependent variable or output. If in an equation in two variables we get exactly one output for each input, then the equation specifies a function. The graph of such a function is just the graph of the equation. If we get more than one output for a given input, then the equation does not specify a function. 2

97 Section 2.1 Functions (continued) The vertical line test can be used to determine whether or not an equation in two variables specifies a function. The functions specified by equations of the form y = mx + b, where m is not equal to 0, are called linear functions. Functions specified by equations of the form y = b are called constant functions. 3

98 Section 2.1 Functions (continued) If a function is specified by an equation and the domain is not indicated, we agree to assume that the domain is the set of all inputs that produce outputs that are real numbers. The symbol f(x) represents the element in the range of f that corresponds to the element x of the domain. Break-eveneven and profit-loss analysis uses a cost function C and a revenue function R to determine when a company will have a loss (R < C), break even (R = C) or a profit (R > C). 4

99 Section 2.2 Elementary Functions Graphs & Transformations The six basic elementary functions are the identity function, the square and cube functions, the square root and cube root functions and the absolute value function. Performing an operation on a function produces a transformation of the graph of the function. The basic graph transformations ti are: vertical and horizontal translations (shifts), reflection in the x-axis, and vertical stretches t and shrinks. A piecewise-defined function is a function whose definition involves more than one formula. 5

100 Section 2.3 Quadratic Functions If a, b, and c are real numbers with a not equal to 0, then the function f(x) = ax 2 + bx + c is a quadratic function in standard form, and its graph is a parabola. The Quadratic Formula, 2 b b 4ac 2 x ± =, when b 4ac 0 2a can be used to find the x-intercepts. 6

101 Section 2.3 Quadratic Functions (continued) Completing the square in the standard form of a quadratic function produces the vertex form, f(x) = a(x - h) 2 +k From the vertex form of a quadratic function, we can read off the vertex, axis of symmetry, intercepts, maximum or minimum, and range, and sketch the graph. If a revenue function R(x) and a cost function C(x) intersect at apoint(x 0, y 0 ), then both this point and its coordinate x 0 are referred to as break-even even points. 7

102 Section 2.3 Quadratic Functions (continued) Quadratic regression on a graphing calculator produces the function of the form y = ax 2 + bx + c that best fits a data set. A quadratic function is a special case of a polynomial function, that is, a function that can be written in the form f(x) = a n x n + a n-1 x n a 1 x + a 0 Unlike polynomial functions, a rational function can have vertical asymptotes (but not more than the degree of the denominator) and at most one horizontal asymptote. 8

103 Section 2.4 Exponential Functions An exponential function is a function of the form f (x) = b x, where b is not equal to 1, but is a positive constant called the base. Thedomain of f(x) is the set of all real numbers and the range is the set of positive real numbers. The graph of an exponential function is continuous, passes through (0,1), and has the x-axis as a horizontal asymptote. Exponential functions obey the familiar laws of exponents and satisfy additional properties. 9

104 Section 2.4 Exponential Functions (continued) The base that is used most frequently in mathematics is the irrational number e Exponential functions can be used to model population growth and radioactive decay. Exponential regression on a graphing calculator produces the function of the form y = a(b x ) that best fits a data set. Exponential functions are used in computations of compound nt interest: r A= P 1 + n 10

105 Section 2.5 Logarithmic Functions A function is said to be one-to to-one one if each range value corresponds to exactly one domain value. The inverse of a one-to-one function f is the function formed by interchanging the independent and dependent variables of f. That is, (a, b) is a point on the graph of f if and only if (b, a) is a point on the graph of the inverse of f. A function that is not one-to to-one one does not have an inverse. 11

106 Section 2.5 Logarithmic Functions (continued) The inverse of the exponential function with base b is called the logarithmic function with base b, denoted y = log b x. The domain of log b x is the set of all positive real numbers and the range is the set of all real numbers. Because y = log b xis the inverse of the function y = b x, y = log b x is equivalent to x = b y. Properties of logarithmic functions can be obtained from corresponding properties of exponential functions. 12

107 Section 2.5 Logarithmic Functions (continued) Logarithms with base 10 are called common logarithms, denoted by log x. Logarithms with base e are called natural logarithms, denoted by ln x. Logarithms can be used to find an investment s s doubling time - the length of time it takes for the value of an investment to double. Logarithmic regression on a graphing calculator produces the function of the form y = a + b ln x that best fits a data set. 13

108 End of Slides Please, solve the exercise problems, especially the applications given in the textbook. If you don t solve a problem by your hands, you will have a difficult time on the exam. Studying this course consists of understanding the concepts/techniques and solving the problems, especially the applications. 14

109 Learning Objectives for Section 3.1 Introduction to Limits The student will learn about: Functions and graphs Limits from a graphic approach Limits from an algebraic approach Limits of difference quotients. 1

110 Functions & Graphs Brief Overview The graph of a function is the graph of the set of all ordered d pairs that t satisfy the function. Example: f(x) = 2x 1 2

111 Analyzing a Limit (Graphical Approach) Based on the graph, we observe as x goes to 3, f(x) = 2x 1 goes to 5 In fact, without using the graph, we observe as x goes to 1000, f(x) = 2x 1 goes to 1999 We introduce a notation and express as follows: as x ö 1000, f(x) = 2x 1 ö

112 Limit DEFINITION Using the limit notation, as x ö c, f(x) ö L is compacted into the form: lim f( x ) L x c = It reads as x goes to c, the limit of f(x) is L. The meaning of the equation is whenever x is close (either side) to c, but not equal to c, f(x) is close to th e single real number L. 4

113 One Sided Limit We write lim f( x ) K = x c and call K the limit from the left (or left hand limit) if f (x) is close to K whenever x is close to c, but to th e left of c on the real number line. We write lim f( x ) L + = x c and call L the limit from the right (or right hand hand li mit) if f (x) is close to L whenever x is close to c, but t o the right of c on the real number line. In order for a limit to exist, the limit from the left and the limit from the right must exist and be equal. 5

114 Example (confer the Mathematica file Sec 3.1) lim f( x ) = 2 x 2 but lim f(x ) = 3 x 2 So, lim f( x) :notexists exists. x 2 + lim f( x ) = 5 x 4 andlim f(x) = 5 x 4 x 4 f(x) = So, l im f(x) 5 + 6

115 Limit Properties (Algebraic Approach) Let f and g be two functions, and assume that the following two limits exist and are finite: lim f( x ) = L and lim g( x ) = K x c x c Then (1) the limit of the sum of the functions is e qual to the sum of the limits and the limit of the difference of the functions is equal to the differenc e of the limits: lim x c[ f(x) + g(x)] lim x c[ f(x) g(x)] = limx c f( x ) + limx c g( x) = limx c f( x ) limx c g( x ) = L + K = L K 7

116 Limit Properties (Algebraic Approach) (2) The limit of a constant times a function is equ al to the constant times the limit of the function: lim [ af( x )] = a[lim f( x )] = al, a: constant x c x c (3) The limit of the product of the functions is the product of the limits of the functions: lim [ f(x)g(x)] = [lim f(x)][lim g(x)] = LK x c x c x c 8

117 Limit Properties (Algebraic Approach) (4) The limit of the quotient of the function is the quotient of the limit of the function, provided tha tki is not equal lto 0: f(x ) lim f(x ) L x c lim x c = =, if lim x c g( x ) 0 g( x ) limx c g( x ) K (5) The limit of the n-th root of a function is the n -th root of the limit of that function: n x c = x c n = n lim [ f( x )] [lim f( x )] L 9

118 Example 2 2 lim( x 3x ) = lim x 3 lim x = 4 3( 2) = 2 x 2 x 2 x 2 lim( 2x) x 4 lim 2x 8 = = x 4 3x + 1 lim( 3x + 1) 13 x 4 From these examples, we conclude that (1) for any polynomial f(x), lim f(x ) = f(c ) x c (2) for any rational function R(x) with a nonzero de nominator at x = c, lim R( x ) = R( c ) x c 10

119 Indeterminate Forms If then, lim f( x) = 0 and lim g( x) = 0 x c x c lim f(x) g(x) or lim g( x ) f( x ) x c x c is said to be indeterminate. The term indetermina te is used because the limit may or may not exist. How to make the indeterminate form to be dete rminate? Simplify the quotient! 11

120 Examples 2 x 4 (x + 2)(x 2) lim = lim = lim(x+ 2) = 4 x 2 x 2 x 2 x 2 x (x 1) (x 1) x 1 lim = lim = lim = x x 1 x 1 x 1 ( x + 1)( x 1) x

121 Difference Quotients For f(x) = 3x 1, find f(a + h) f(a) lim h 0 h 13

122 Solution For f(x) = 3x 1, find f(a + h) f(a) lim h 0 h Solution: f(a+ h) = 3(a+ h) 1= 3a+ 3h 1 f(a) = 3a 1 f(a+ + h) f(a) = 3h l im f(a + h) f(a) 3h = lim = h h h 0 h

123 Summary We started by using a table to investigate the i dea of a limit. This was an intuitive way to approa ch hlimits. i We saw that if the left and right limits at a poi nt were the same, we had a limit at that point. We saw that we could add, subtract, multiply, and divide limits. We now have some very powerful tools for deali ng with limits it and can go on to our study of calcul l us. 15

124 3.7 Marginal Analysis in Business and Economics Definition. Let x be the number of units of a product. (1) For the total cost function C(x), the marginal cost function is dened by its derivative C 0 (x). (2) For the total revenue function R(x), the marginal revenue function is dened by its derivative R 0 (x). (3) For the total prot function P (x), the marginal profit function is dened by its derivative P 0 (x). Remark. Total cost of producing (x + 1) items is C(x + 1). Total cost of producing x items is C(x). Thus, we deduce that the total cost of producing (x + 1) th item is C(x + 1) C(x). Similar statements can be made for the revenue and prot. Example. Let the total cost function C(x) = 90 0:1x is given. (1) Find the marginal cost function. (2) Find the marginal cost when x = 500 and interpret the result. (3) Find the exact cost of producing 501 th item. Answer. (1) C 0 (x) = 90 0:1x. (2) C 0 (500) = 40. At production level of 500 units, the total cost increases at the rate of 40. (3) C(501) C(500) = 39:95. Theorem. The marginal cost function approximates the exact cost of producing the (x + 1) th item, i.e., C 0 (x) C(x + 1) C(x): Similar statements can be made for total revenue function and total prot function. Example. For C(x) = x item. 0:05x 2, use the marginal cost to approximate the cost of the 335 th Answer. C 0 (x) = 90 0:1x implies C 0 (334) = 56:6. Thus the approximated cost is 56:6. For your information, the exact cost is C(335) C(334) = 56:55. We observe the approximated cost is very close to the exact cost. So it is good to use the approximated cost instead of the exact cost. More Examples? Please solve the problems given in the textbook. 1

125 4.1 The Constant e and Continuous Compound Interest Theorem (Continuous Compound Interest Formula). A = P e rt where P is the principal, r is the annual nominal interest rate compounded continuously, t is the time in years and A is the amount at time t and e is the Euler constant e 2:71. Example. Let 1000 DHS be invested at 5% compounded continuously. (1) What amount will be in the account after 3 years? How much interest will be earned? (2) How long will it take for the account to be worth 2000 DHS? Answer. (1) By the formula, we get A = P e rt = 1000e (0:05)(3) 1161:83 DHS: The interest earned is 1161: = 161:83 DHS: (2) We solve the following equation: 2000 = 1000e 0:05t ; i:e:; 2 = e 0:05t ; i:e:; ln 2 = 0:05t; i:e:; t = ln 2 0:05 13:8629: Thus, it takes about 13:8629 years. Example. A bank note will pay DHS at maturity y 5 years from now. How much should you be willing to pay for the note now if money is worth 7% compounded continuously? Answer. The problem asks to nd P in the following equation: = P e 0:07(5) ; i:e:; = P e 0:35 ; i:e:; P = e 0: :6 a promissory note issued by a bank payable to bearer on demand but without interest and circulating as money y the time when a note or bill of exchange becomes due 2

126 4.2 Derivatives of Exponential and Logarithmic Functions Theorem (Derivative of e x ). The derivative of f(x) = e x is the function itself, i.e., d dx ex = e x ; (e x ) 0 = e x : Example. Find f 0 (x) for f(x) = 5e x 3x 4 7x e + e 2. Answer. f 0 (x) = 5(e x ) 0 3(x 4 ) 0 7(x e ) 0 + (e 2 ) 0 = 5e x 12x 3 7ex e = 5e x 12x 3 7ex e 1 : Theorem (Derivative of ln x). The derivative of f(x) = ln x is 1=x, i.e., d dx ln x = 1 x ; (ln x)0 = 1 x : Be careful! \ln x" means \the natural logarithmic function", i.e., ln x = log e x, where e is the Euler constant. Example. Find y 0 for y = 3e x + 5 ln x + x 4 ln x 6. Answer. First we simplify y and then dierentiate it: y = 3e x + 5 ln x + x 4 6 ln x = 3e x ln x + x 4 ; y 0 = 3(e x ) 0 (ln x) 0 + (x 4 ) 0 = 3e x 1 x + 4x3 : Theorem (Other Exponential and Logarithmic Functions). For b > 0, b 6= 1, d dx ex = e x ; d dx bx = b x ln b; d dx ln x = 1 x ; d 1 1 dx log b x = ln b x : Example. Find f 0 (x) for f(x) = 2 x 3 x + x x + log 2 x 6 log 5 x + log 4 x 5 + log 3 (10x 2 ). Answer. First we simplify f(x) and then dierentiate it: f(x) = 2 x 3 x + x x + log 2 x 6 log 5 x + log 4 x 5 + log 3 (10x 2 ) = 2 x 3 x + x x + log 2 x 6 log 5 x + 5 log 4 x + log log 3 x 2 = 2 x 3 x + x x + log 2 x 6 log 5 x + 5 log 4 x + log log 3 x f 0 (x) = (2 x ) 0 (3 x ) 0 + (x 10 ) 0 + (10 x ) 0 + (log 2 x) 0 6(log 5 x) 0 + 5(log 4 x) 0 + (log 3 10) 0 + 2(log 3 x) 0 = 2 x ln 2 3 x ln x x ln ln 2 x ln 5 x ln 4 x ln 3 x 1 = 2 x ln 2 3 x ln x x 6 ln 10 + ln 2 ln ln ln 3 x : 3

127 4.3 Derivatives of Products and Quotients Theorem (Product Rule). If y = f(x) = F (x)s(x), then f 0 (x) = F 0 (x)s(x) + F (x)s 0 (x): Also, y 0 = F 0 S + F S 0 ; dy dx =! df S + F dx! ds : dx Example. Find f 0 (x) for f(x) = (3x + 2)(4x 2 Answer. Let F (x) = 3x + 2 and S(x) = 4x 2 5x). 5x. Then f(x) = F (x)s(x) and by the formula above, f 0 (x) = F 0 (x)s(x) + F (x)s 0 (x) = (3x + 2) 0 (4x 2 5x) + (3x + 2)(4x 2 5x) 0 = 3(4x 2 5x) + (3x + 2)(8x 5) = 2(18x 2 7x 5): Theorem (Quotient Rule). If y = f(x) = F (x) S(x), then f 0 (x) = F 0 (x)s(x) F (x)s 0 (x) S 2 : (x) Also, y 0 = F 0 S F S 0 S 2 ; dy dx = df dx S F ds dx S 2 : Example. Find f 0 (x) for f(x) = x3 2x 1. Answer. Let F (x) = x 3 and S(x) = 2x 1. Then f(x) = F (x) S(x) f 0 (x) = F 0 (x)s(x) F (x)s 0 (x) S 2 (x) = (x3 ) 0 (2x 1) (x 3 )(2x 1) 0 (2x 1) 2 and by the formula above, = 3x2 (2x 1) 2x 3 (2x 1) 2 = 4x3 3x 2 (4x 3)x2 = (2x 1) 2 (2x 1) 2 : 4

128 4.4 The Chain Rule Definition. A function m is a composite of functions f and g if m(x) = f[g(x)]: Example. Let f(u) = e u and g(x) = 3x. Find f[g(x)] and g[f(u)]. Answer. f[g(x)] = f[ 3x] = e 3x : g[f(x)] = g[e u ] = 3e u : Exercise. (1) Let f(u) = e u and g(x) = e x. Find f[g(x)] and g[f(u)]. (2) Let f(u) = e u and g(x) = ln x. Find f[g(x)] and g[f(u)]. Theorem (Chain Rule). If y = f(u) and u = g(x) dene the composite function y = m(x) = f[g(x)] then y = f 0 [g(x)]g 0 (x); or dy dx = dy du du dx : Example. Find dy=du, du=dx and dy=dx (express dy=dx as a function of x). (1) y = u 3=2 and u = 3x (2) y = e u and u = 2x Answer. (1) (2) dy du = d du u3=2 = (u 3=2 ) 0 = 3 2 u1=2 : du dx = d dx (3x2 + 1) = (3x 2 + 1) 0 = 6x: dy dx = dy dy du dx = 3 2 u1=2 (6x) = 9x(3x 2 + 1) 1=2 : dy du = d du eu = (e u ) 0 = e u : du dx = d dx (2x3 + 5) = (2x 3 + 5) 0 = 6x 2 : dy dx = dy dy du dx = eu (6x 2 ) = 6x 2 e 2x3 +5 : Exercise. Find dy=dx for y = e 3x + 3 ex + e ex + e ln(x+2) + log 1:8 3 x + log 4 (x 2 + 3) + log 5 (e x ) + ln(ln x). 1

129 4.5 Implicit Differentiation Problem 1: Let us consider the problem: nd dy=dx from 3x 2 + y 2 = 0. The given equation implies y = 2 3x 2. Dierentiating it with respect to x, we get dy dx = d 2 3x 2 = (2 3x 2 ) 0 = 6x; i:e:; dx dy dx = 6x: We say that the equation y = 2 3x 2 gives explicitly y as a function of x, while the equation 3x 2 + y 2 = 0 gives implicitly y. The technique discussed here in this section is about nding dy=dx without changing the implicit form of an equation into the explicit one. Problem 2: Let us consider the problem: nd dy=dx from y 2 x = 0. Solution 1 (Using Explicit Form). The equation is given in the implicit form. explicit form: y 2 = x; i:e:; y = p x = x 1=2 : We change it into the Dierentiating it with respect to x, we get dy dx = d x 1=2 = x 1=2 0 1 = dx 2 x 1=2 = 1 2 p x : ( ) Solution 2 (Implicit Dierentiation). We just dierentiate the whole equation with respect to x: d y 2 dx x = d dx (0) ; i:e:; dy 2 dx dx dx = 0; i:e:; dy 2 dx 1 = 0: Here we have the problem: how do we dierentiate y 2 with respect to x? That is, what is dy2 dx? This is the place where we must use the CHAIN RULE and this is the reason why we learned the chain rule just before this section. Using the chain rule, we have dy 2 dx = dy2 dy dy dx = y 2 0 dy dx dy = 2y dx ; dy 2 dx = 2y dy dx : Thus, we have dy 2 dx 1 = 0; i:e:; 2y dy dx 1 = 0; i:e:; dy dx = 1 2y : Since y = p x, we have dy dx = 1 2y = 1 2 p x ; which is exactly same as the result in ( ). 2

130 Example. Use implicit dierentiation to nd y 0 and evaluate y 0 at the indicated point: x 2 + y 3 = 31 at (x; y) = (2; 3). Solution. Dierentiating the whole equation with respect to x, we get d x 2 + y 3 = d dx dx 31; dx 2 dx + dy3 dx = 0; 2x + 3y 2 dy dx = 0; dy 3y2 dx = 2x; dy dx = Therefore, the value of y 0 at (x; y) = (2; 3) is obtained as follows: Exercise. Find y 0 from 2x + 6y 4 = 0. y 0 j (2;3) = dy dx (2;3) = 2(2) 3(3 2 ) = 4 27 : dy3 2x + dy 2x 3y 2 : dy dx = 0; Exercise. Use implicit dierentiation to nd y 0 and evaluate y 0 at the indicated point: 5x 3 y 1 = 0 at (1; 4). Exercise. Use implicit dierentiation to nd y 0 and evaluate y 0 at the indicated point. (1) x 3 y = ln y at (1; 1). (2) ln y = 2y 2 x at (2; 1). (3) e y = x 3 + y 4 at (1; 0). 3

131 5.1 First Derivative and Graphs In this section, we discuss four topics: Increasing and Decreasing Functions, Local Extrema, First{Derivative Test, Applications to Economics. Increasing and Decreasing Functions Let us think about the escalator. We have two kinds of escalator: one always going upward and the other one always going downward. When we take the escalator going upward, obviously we can reach upstairs. If we take the other escalator, we can reach downstairs. Now we consider the graph of y = x 2. On the interval [0; 1), as x goes to the right{ hand side, the graph is going upward. So the graph of y = x 2 on the interval [0; 1) looks like the escalator always going upward. On the other hand, the graph of y = x 2 on the interval ( 1; 0] looks like the escalator always going downward, because as x goes to the right{hand side, the graph is going downward. If a graph of a function looks like the escalator always going upward on an interval [a; b], we say that the graph/function is increasing on [a; b]. If a graph of a function looks like the escalator always going downward on an interval [a; b], we say that the graph/function is decreasing on [a; b]. We observe the followings: (1) On the interval where the graph/function f is increasing, the slope of the tangent line is positive, which implies f 0 (x) > 0 on the interval. (2) On the interval where the graph/function f is decreasing, the slope of the tangent line is negative, which implies f 0 (x) < 0 on the interval. Then what is the slope of the tangent line at the point where the graph/function is neither increasing nor decreasing? The slope should be between positiveness and negativeness, that is, the slope should be zero, which implies f 0 (x) = 0 on the interval. We can summarize this argument in the Theorem. 1

132 Theorem 1 (Increasing and Decreasing Functions). For the interval (a; b), f 0 (x) f(x) Graph of f Examples + Increases Rises Given in Class Decreases Falls Given in Class Discussion 2. For the graphs of f(x) = x 2 and g(x) = jxj, discuss the relationship between the graph of each function at x = 0 and the derivative of the function at x = 0. (Please confer the gures below.) f x x 2 g x x x x Example 3. Given the function f(x) = x 2 6x + 10, (1) Which values of x correspond to horizontal tangent lines? (2) For which values of x is f(x) increasing? Decreasing? (3) Sketch a graph of f. Add any horizontal tangent lines. Solution. (1) A horizontal tangent line at x = a is equivalent to f 0 (a) = 0. So we nd x = a at which f 0 (x) = 0: f 0 (x) = 2x 6 = 2(x 3) = 0 at x = 3: Therefore, f 0 (3) = 0 and so we have the horizontal tangent line at x = 3. (2) f(x) is increasing on (a; b) if and only if f 0 (x) > 0 on (a; b). Since f 0 (x) = 2(x 3) > 0 for x > 3, i.e., on (3; 1), thus f(x) is increasing on (3; 1). 2

133 f(x) is decreasing on (c; d) if and only if f 0 (x) < 0 on (c; d). Since f 0 (x) = 2(x 3) < 0 for x < 3, i.e., on ( 1; 3), thus f(x) is increasing on ( 1; 3). When we use the sign chart, we can easily get the sign of f 0 (x): x Less than 3 3 Bigger than 3 f 0 (x) = 2(x 3) 0 + f(x) Decreases 1 Increases (3) Please confer the gure below. X f x x 2 6x x The point at which f has the horizontal (f 0 (x) = 0) or vertical tangent line (f 0 (x) does not exist) is very important. Definition 4 (Critical Values). The values of x in the domain of f where f 0 (x) = 0 or where f 0 (x) does not exist are called the critical values of f. Example 5. Find (1) the critical values of f, (2) the intervals on which f is increasing, and (3) those on which f is decreasing, for f(x) = 1 x 3. Solution. We start with the derivative: f 0 (x) = 3x 2. (1) Since f 0 (x) = 3x 2 = 0 at x = 0 which is in the domain of f, so f has the only critical value 0. (2) We use the sign chart: 3

134 x Less than 0 0 Bigger than 0 f 0 (x) = 3x 2 0 f(x) Decreases 1 Decreases Thus, there is no interval on which f is increasing. (3) By the sign chart in (2), f is decreasing on ( 1; 0) and (0; 1). Since f is continuous at x = 0, it follows that f is decreasing for all x. Please confer the gure below. f x x 3 X x Example 6. Find (1) the critical values of f, (2) the intervals on which f is increasing, and (3) those on which f is decreasing, for f(x) = (1 + x) 1=3. Solution. We start with the derivative: f 0 (x) = 1 3(1 + x) 2=3. (1) Since f 0 (x) does not exist at x = 1 which is in the domain of f, so f has the only critical value 1. (2) We use the sign chart: f is increasing on ( f is increasing for all x. x Less than 1 1 Bigger than 1 f 0 (x) + Does Not Exist + f(x) Increases 0 Increases 1; 0) and (0; 1). Since f is continuous at x = 0, it follows that 4

135 (3) By the sign chart in (2), there is no interval where f is decreasing. Please confer the gure below. f x 1 x 1 3 X x 1 Example 7. Find (1) the critical values of f, (2) the intervals on which f is increasing, and (3) those on which f is decreasing, for f(x) = 1 x. Solution. We start with the derivative: f 0 (x) = 1 x 2. (1) Since f 0 (x) does not exist at x = 0 which is not in the domain of f, so f does not have any critical value. However, we use the value x = 0 in the sign chart. (2) We use the sign chart: x Less than 0 0 Bigger than 0 f 0 (x) Does Not Exist f(x) Decreases Does Not Exist Decreases Thus, there is no interval on which f is increasing. (3) By the sign chart in (2), f is decreasing on ( 1; 0) and (0; 1). Please confer the gure below. X 5