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- Slide 1
- 2 Functions and Their Graphs The Algebra of Functions Functions and Mathematical Models Limits One-Sided Limits and Continuity The Derivative Functions, Limits and the Derivative
- Slide 2
- 2.1 Functions and Their Graphs
- Slide 3
- Functions Function: A function is a rule that assigns to each element in a set A one and only one element in a set B. The set A is called the domain of the function. It is customary to denote a function by a letter of the alphabet, such as the letter f.
- Slide 4
- Functions The element in B that f associates with x is written f(x) and is called the value of f at x. The set of all the possible values of f(x) resulting from all the possible values of x in its domain, is called the range of f(x). The output f(x) associated with an input x is unique: Each x must correspond to one and only one value of f(x).
- Slide 5
- Example Let the function f be defined by the rule Find: f(1) Solution: Example 1, page 51
- Slide 6
- Example Let the function f be defined by the rule Find: f( 2) Find: f( 2) Solution: Example 1, page 51
- Slide 7
- Example Let the function f be defined by the rule Find: f() Find: f(a) Solution: Example 1, page 51
- Slide 8
- Example Let the function f be defined by the rule Find: f() Find: f(a + h) Solution: Example 1, page 51
- Slide 9
- Applied Example ThermoMaster manufactures an indoor-outdoor thermometer at its Mexican subsidiary. Management estimates that the profit (in dollars) realizable by ThermoMaster in the manufacture and sale of x thermometers per week is Find ThermoMasters weekly profit if its level of production is: a.1000 thermometers per week. b.2000 thermometers per week. Applied Example 2, page 51
- Slide 10
- Applied Example Solution We have a.The weekly profit by producing 1000 thermometers is or $2,000. or $2,000. b.The weekly profit by producing 2000 thermometers is or $7,000. or $7,000. Applied Example 2, page 51
- Slide 11
- Determining the Domain of a Function Suppose we are given the function y = f(x). Then, the variable x is called the independent variable. The variable y, whose value depends on x, is called the dependent variable. To determine the domain of a function, we need to find what restrictions, if any, are to be placed on the independent variable x. In many practical problems, the domain of a function is dictated by the nature of the problem.
- Slide 12
- x 16 2x Applied Example: Packaging An open box is to be made from a rectangular piece of cardboard 16 inches wide by cutting away identical squares (x inches by x inches) from each corner and folding up the resulting flaps. 10 10 2x 16 x xx Applied Example 3, page 52
- Slide 13
- Applied Example: Packaging An open box is to be made from a rectangular piece of cardboard 16 inches wide by cutting away identical squares (x inches by x inches) from each corner and folding up the resulting flaps. a.Find the expression that gives the volume V of the box as a function of x. b.What is the domain of the function? The dimensions of the resulting box are: x 16 2x 10 2x Applied Example 3, page 52
- Slide 14
- x 16 2x Applied Example: Packaging Solution a.The volume of the box is given by multiplying its dimensions (length width height), so: 10 2x Applied Example 3, page 52
- Slide 15
- Applied Example: Packaging Solution b.Since the length of each side of the box must be greater than or equal to zero, we see that must be satisfied simultaneously. Simplified: All three are satisfied simultaneously provided that: Thus, the domain of the function f is the interval [0, 5]. Applied Example 3, page 52
- Slide 16
- More Examples Find the domain of the function: Solution Since the square root of a negative number is undefined, it is necessary that x 1 0. Thus the domain of the function is [1, ). Example 4, page 52
- Slide 17
- More Examples Find the domain of the function: Solution Our only constraint is that you cannot divide by zero, so Which means that Or more specifically x 2 and x 2. Thus the domain of f consists of the intervals ( , 2), (2, 2), (2, ). Example 4, page 52
- Slide 18
- More Examples Find the domain of the function: Solution Here, any real number satisfies the equation, so the domain of f is the set of all real numbers. Example 4, page 52
- Slide 19
- Graphs of Functions If f is a function with domain A, then corresponding to each real number x in A there is precisely one real number f(x). Thus, a function f with domain A can also be defined as the set of all ordered pairs (x, f(x)) where x belongs to A. The graph of a function f is the set of all points (x, y) in the xy-plane such that x is the domain of f and y = f(x).
- Slide 20
- Example The graph of a function f is shown below: x y Domain Range x y (x, y) Example 5, page 53
- Slide 21
- Example The graph of a function f is shown below: What is the value of f(2)? x y 4 3 2 1 1 2 (2, 2) 1234 5678 1234 5678 Example 5, page 53
- Slide 22
- Example The graph of a function f is shown below: What is the value of f(5)? x y (5, 3) 4 3 2 1 1 2 1234 5678 1234 5678 Example 5, page 53
- Slide 23
- Example The graph of a function f is shown below: What is the domain of f(x)? x y 4 3 2 1 1 2 Domain: [1,8] 1234 5678 1234 5678 Example 5, page 53
- Slide 24
- Example The graph of a function f is shown below: What is the range of f(x)? x y 4 3 2 1 1 2 Range: [2,4] 1234 5678 1234 5678 Example 5, page 53
- Slide 25
- Example: Sketching a Graph Sketch the graph of the function defined by the equation y = x 2 + 1 Solution The domain of the function is the set of all real numbers. Assign several values to the variable x and compute the corresponding values for y: xy 310 25 12 01 12 25 310 Example 6, page 54
- Slide 26
- 3 2 11 23 3 2 11 23 108642 Example: Sketching a Graph Sketch the graph of the function defined by the equation y = x 2 + 1 Solution The domain of the function is the set of all real numbers. Then plot these values in a graph: x y xy 310 25 12 01 12 25 310 Example 6, page 54
- Slide 27
- 3 2 11 23 3 2 11 23 108642 Example: Sketching a Graph Sketch the graph of the function defined by the equation y = x 2 + 1 Solution The domain of the function is the set of all real numbers. And finally, connect the dots: x y xy 310 25 12 01 12 25 310 Example 6, page 54
- Slide 28
- Example: Sketching a Graph Sketch the graph of the function defined by the equation Solution The function f is defined in a piecewise fashion on the set of all real numbers. In the subdomain ( , 0), the rule for f is given by In the subdomain [0, ), the rule for f is given by Example 7, page 55
- Slide 29
- Example: Sketching a Graph Sketch the graph of the function defined by the equation Solution Substituting negative values for x into, while substituting zero and positive values into we get: xy33 22 11 00 11 21.41 31.73 Example 7, page 55
- Slide 30
- Example: Sketching a Graph Sketch the graph of the function defined by the equation Solution Plotting these data and graphing we get: 3 2 11 23 3 2 11 23 321 x yxy33 22 11 00 11 21.41 31.73 Example 7, page 55
- Slide 31
- 2.2 The Algebra of Functions
- Slide 32
- The Sum, Difference, Product and Quotient of Functions Consider the graph below: R(t) denotes the federal government revenue at any time t. S(t) denotes the federal government spending at any time t. 199019921994199619982000 200018001600140012001000 t y Billions of Dollars Year y = S(t) y = R(t) t R(t)R(t)R(t)R(t) S(t)S(t)S(t)S(t)
- Slide 33
- D(t) = R(t) S(t) D(t) = R(t) S(t) The Sum, Difference, Product and Quotient of Functions Consider the graph below: The difference R(t) S(t) gives the budget deficit (if negative) or surplus (if positive) in billions of dollars at any time t. 199019921994199619982000 200018001600140012001000 t y Billions of Dollars Year y = S(t) y = R(t) t R(t)R(t)R(t)R(t) S(t)S(t)S(t)S(t)
- Slide 34
- The Sum, Difference, Product and Quotient of Functions The budget balance D(t) is shown below: D(t) is also a function that denotes the federal government deficit (surplus) at any time t. This function is the difference of the two function R and S. D(t) has the same domain as R(t) and S(t). 19921994199619982000 4002000 200 400 t y Billions of Dollars Year t D(t) D(t) D(t) D(t) y = D(t)
- Slide 35
- The Sum, Difference, Product and Quotient of Functions Most functions are built up from other, generally simpler functions. For example, we may view the function f(x) = 2x + 4 as the sum of the two functions g(x) = 2x and h(x) = 4.
- Slide 36
- The Sum, Difference, Product and Quotient of Functions Let f and g be functions with domains A and B, respectively. The sum f + g, the difference f g, and the product fg of f and g are functions with domain A B and rule given by (f + g)(x) = f(x) + g(x) Sum (f g)(x) = f(x) g(x) Difference (fg)(x) = f(x)g(x) Product (fg)(x) = f(x)g(x) Product The quotient f/g of f and g has domain A B excluding all numbers x such that g(x) = 0 and rule given by Quotient
- Slide 37
- Example Let and g(x) = 2x + 1. Find the sum s, the difference d, the product p, and the quotient q of the functions f and g. Solution Since the domain of f is A = [1, ) and the domain of g is B = ( , ), we see that the domain of s, d, and p is A B = [1, ). The rules are as follows: Example 1, page 68
- Slide 38
- Example Let and g(x) = 2x + 1. Find the sum s, the difference d, the product p, and the quotient q of the functions f and g. Solution The domain of the quotient function is [1, ) together with the restriction x . Thus, the domain is [1, ) U ( , ). The rule is as follows: Example 1, page 68
- Slide 39
- Applied Example Suppose Puritron, a manufacturer of water filters, has a monthly fixed cost of $10,000 and a variable cost of 0.0001x 2 + 10x(0 x 40,000) dollars, where x denotes the number of filters manufactured per month. Find a function C that gives the total monthly cost incurred by Puritron in the manufacture of x filters. Applied Example 2, page 68
- Slide 40
- Applied Example Solution Puritrons monthly fixed cost is always $10,000, so it can be described by the constant function: F(x)= 10,000 The variable cost can be described by the function: V(x)= 0.0001x 2 + 10x The total cost is the sum of the fixed cost F and the variable cost V: C(x)= V(x) + F(x) C(x)= V(x) + F(x) = 0.0001x 2 + 10x + 10,000 (0 x 40,000) Applied Example 2, page 68
- Slide 41
- Applied Example Lets now consider profits Suppose that the total revenue R realized by Puritron from the sale of x water filters is given by R(x)= 0.0005x 2 + 20x (0 x 40,000) Find a.The total profit function for Puritron. b.The total profit when Puritron produces 10,000 filters per month. Applied Example 3, page 69
- Slide 42
- Applied Example Solution a.The total profit P realized by the firm is the difference between the total revenue R and the total cost C: P(x)= R(x) C(x) P(x)= R(x) C(x) = ( 0.0005x 2 + 20x) ( 0.0001x 2 + 10x + 10,000) = ( 0.0005x 2 + 20x) ( 0.0001x 2 + 10x + 10,000) = 0.0004x 2 + 10x 10,000 b.The total profit realized by Puritron when producing 10,000 filters per month is P(x)= 0.0004(10,000) 2 + 10(10,000) 10,000 = 50,000 = 50,000 or $50,000 per month. Applied Example 3, page 69
- Slide 43
- The Composition of Two Functions Another way to build a function from other functions is through a process known as the composition of functions. Consider the functions f and g: Evaluating the function g at the point f(x), we find that: This is an entirely new function, which we could call h:
- Slide 44
- The Composition of Two Functions Let f and g be functions. Then the composition of g and f is the function g f (read g circle f ) defined by (g f )(x) = g(f(x)) The domain of g f is the set of all x in the domain of f such that f(x) lies in the domain of g.
- Slide 45
- Example Let Find: a.The rule for the composite function g f. b.The rule for the composite function f g. Solution To find g f, evaluate the function g at f(x): To find f g, evaluate the function f at g(x): Example 4, page 70
- Slide 46
- Applied Example An environmental impact study conducted for the city of Oxnard indicates that, under existing environmental protection laws, the level of carbon monoxide (CO) present in the air due to pollution from automobile exhaust will be 0.01x 2/3 parts per million when the number of motor vehicles is x thousand. A separate study conducted by a state government agency estimates that t years from now the number of motor vehicles in Oxnard will be 0.2t 2 + 4t + 64 thousand. Find: a.An expression for the concentration of CO in the air due to automobile exhaust t years from now. b.The level of concentration 5 years from now. Applied Example 5, page 70
- Slide 47
- Applied Example Solution Part (a): The level of CO is described by the function g(x) = 0.01x 2/3 where x is the number (in thousands) of motor vehicles. where x is the number (in thousands) of motor vehicles. In turn, the number (in thousands) of motor vehicles is described by the function f(x) = 0.2t 2 + 4t + 64 where t is the number of years from now. Therefore, the concentration of CO due to automobile exhaust t years from now is given by (g f )(t) = g(f(t)) = 0.01(0.2t 2 + 4t + 64) 2/3 Applied Example 5, page 70
- Slide 48
- Applied Example Solution Part (b): The level of CO five years from now is: (g f )(5)= g(f(5)) = 0.01[0.2(5) 2 + 4(5) + 64] 2/3 = (0.01)89 2/3 0.20 or approximately 0.20 parts per million. or approximately 0.20 parts per million. Applied Example 5, page 70
- Slide 49
- 2.3 Functions and Mathematical Models
- Slide 50
- Mathematical Models As we have seen, mathematics can be used to solve real- world problems. We will now discuss a few more examples of real-world phenomena, such as: The solvency of the U.S. Social Security trust fund (p.79) Global warming (p. 78)
- Slide 51
- Mathematical Modeling Regardless of the field from which the real-world problem is drawn, the problem is analyzed using a process called mathematical modeling. The four steps in this process are: Real-world problem Mathematical model Solution of real- world problem Solution of mathematical model Formulate Interpret Solve Test
- Slide 52
- Modeling With Polynomial Functions A polynomial function of degree n is a function of the form where n is a nonnegative integer and the numbers a 0, a 1, . a n are constants called the coefficients of the polynomial function. Examples: The function below is polynomial function of degree 5:
- Slide 53
- Modeling With Polynomial Functions A polynomial function of degree n is a function of the form where n is a nonnegative integer and the numbers a 0, a 1, . a n are constants called the coefficients of the polynomial function. Examples: The function below is polynomial function of degree 3:
- Slide 54
- Applied Example Market for Cholesterol-Reducing Drugs In a study conducted in early 2000, experts projected a rise in the market for cholesterol-reducing drugs. The U.S. market (in billions of dollars) for such drugs from 1999 through 2004 was A mathematical model giving the approximate U.S. market over the period in question is given by M(t) = 1.95t + 12.19 where t is measured in years, with t = 0 for 1999. Year199920002001200220032004 Market12.0714.0716.2118.2820.0021.72 Applied Example 1, page 76
- Slide 55
- Applied Example Market for Cholesterol-Reducing Drugs M(t) = 1.95t + 12.19 a.Sketch the graph of the function M and the given data on the same set of axes. b.Assuming that the projection held and the trend continued, what was the market for cholesterol- reducing drugs in 2005 (t = 6)? c.What was the rate of increase of the market for cholesterol-reducing drugs over the period in question? Year199920002001200220032004 Market12.0714.0716.2118.2820.0021.72 Applied Example 1, page 76
- Slide 56
- Applied Example Market for Cholesterol-Reducing Drugs M(t) = 1.95t + 12.19 Solution a.Graph: Year199920002001200220032004 Market12.0714.0716.2118.2820.0021.72 1234512345 25 20 15 t y Billions of Dollars Year M(t)M(t)M(t)M(t) Applied Example 1, page 76
- Slide 57
- Applied Example Market for Cholesterol-Reducing Drugs M(t) = 1.95t + 12.19 Solution b.The projected market in 2005 for cholesterol-reducing drugs was M(6) = 1.95(6) + 12.19 = 23.89 or $23.89 billion. Year199920002001200220032004 Market12.0714.0716.2118.2820.0021.72 Applied Example 1, page 76
- Slide 58
- Applied Example Market for Cholesterol-Reducing Drugs M(t) = 1.95t + 12.19 Solution c.The function M is linear, and so we see that the rate of increase of the market for cholesterol-reducing drugs is given by the slope of the straight line represented by M, which is approximately $1.95 billion per year. Year199920002001200220032004Market12.0714.0716.2118.2820.0021.72 Applied Example 1, page 76
- Slide 59
- Modeling a Polynomial Function of Degree 2 A polynomial function of degree 2 has the form Or more simply, y = ax 2 + bx + c, and is called a quadratic function. The graph of a quadratic function is a parabola: x y x y Opens upwards if a > 0 Opens downwards if a < 0
- Slide 60
- Applied Example Global Warming The increase in carbon dioxide (CO 2 ) in the atmosphere is a major cause of global warming. Below is a table showing the average amount of CO 2, measured in parts per million volume (ppmv) for various years from 1958 through 2007: Year195819701974197819851991199820032007 Amount315325330335345355365375380 Applied Example 2, page 78
- Slide 61
- Applied Example Global Warming Below is a scatter plot associated with these data: Year195819701974197819851991199820032007 Amount315325330335345355365375380 1020304050 380360340320 t (years) y (ppmv) Applied Example 2, page 78
- Slide 62
- Applied Example Global Warming A mathematical model giving the approximate amount of CO 2 is given by: Year195819701974197819851991199820032007 Amount315325330335345355365375380 1020304050 380360340320 t (years) y (ppmv) Applied Example 2, page 78
- Slide 63
- Applied Example Global Warming a.Use the model to estimate the average amount of atmospheric CO 2 in 1980 (t = 23). b.Assume that the trend continued and use the model to predict the average amount of atmospheric CO 2 in 2010. Year195819701974197819851991199820032007 Amount315325330335345355365375380 Applied Example 2, page 78
- Slide 64
- Applied Example Global Warming Solution a.The average amount of atmospheric CO 2 in 1980 is given by or approximately 338 ppmv. b.Assuming that the trend will continue, the average amount of atmospheric CO 2 in 2010 will be Year195819701974197819851991199820032007Amount315325330335345355365375380 Applied Example 2, page 78
- Slide 65
- Applied Example Social Security Trust Fund Assets The projected assets of the Social Security trust fund (in trillions of dollars) from 2008 through 2040 are given by: The scatter plot associated with these data is: Year Year200820112014201720202023202620292032203520382040 Assets Assets2.43.24.04.75.35.75.95.64.93.61.70 51015202530 642 t (years) y ($trillion) Applied Example 3, page 79
- Slide 66
- Applied Example Social Security Trust Fund Assets The projected assets of the Social Security trust fund (in trillions of dollars) from 2008 through 2040 are given by: A mathematical model giving the approximate value of assets in the trust fund (in trillions of dollars) is: 51015202530 642 t (years) y ($trillion) Year Year200820112014201720202023202620292032203520382040 Assets Assets2.43.24.04.75.35.75.95.64.93.61.70 Applied Example 3, page 79
- Slide 67
- Applied Example Social Security Trust Fund Assets a.The first baby boomers will turn 65 in 2011. What will be the assets of the Social Security trust fund at that time? b.The last of the baby boomers will turn 65 in 2029. What will the assets of the trust fund be at the time? c.Use the graph of function A(t) to estimate the year in which the current Social Security system will go broke. Year Year200820112014201720202023202620292032203520382040 Assets Assets2.43.24.04.75.35.75.95.64.93.61.70 Applied Example 3, page 79
- Slide 68
- Applied Example Social Security Trust Fund Assets Solution a.The assets of the Social Security fund in 2011 (t = 3) will be: or approximately $3.18 trillion. The assets of the Social Security fund in 2029 (t = 21) will be: or approximately $5.59 trillion. Year Year200820112014201720202023202620292032203520382040 Assets Assets2.43.24.04.75.35.75.95.64.93.61.70 Applied Example 3, page 79
- Slide 69
- Applied Example Social Security Trust Fund Assets Solution b.The graph shows that function A crosses the t-axis at about t = 32, suggesting the system will go broke by 2040: 51015202530 642 y ($trillion) Year Year200820112014201720202023202620292032203520382040 Assets Assets2.43.24.04.75.35.75.95.64.93.61.70 t (years) Applied Example 3, page 79
- Slide 70
- Rational and Power Functions A rational function is simply the quotient of two polynomials. In general, a rational function has the form where f(x) and g(x) are polynomial functions. Since the division by zero is not allowed, we conclude that the domain of a rational function is the set of all real numbers except the zeros of g (the roots of the equation g(x) = 0)
- Slide 71
- Rational and Power Functions Examples of rational functions:
- Slide 72
- Rational and Power Functions Functions of the form where r is any real number, are called power functions. We encountered examples of power functions earlier in our work. Examples of power functions:
- Slide 73
- Rational and Power Functions Many functions involve combinations of rational and power functions. Examples:
- Slide 74
- Applied Example: Driving Costs A study of driving costs based on a 2007 medium-sized sedan found the following average costs (car payments, gas, insurance, upkeep, and depreciation), measured in cents per mile: A mathematical model giving the average cost in cents per mile is: where x (in thousands) denotes the number of miles the car is driven in 1 year. Miles/year, x 500010,00015,00020,000 Cost/mile, y () 83.862.952.247.1 Applied Example 4, page 80
- Slide 75
- Applied Example: Driving Costs Below is the scatter plot associated with this data: 510152025 14012010080604020 t (years) y ( ) C(x)C(x)C(x)C(x) Miles/year, x 500010,00015,00020,000 Cost/mile, y () 83.862.952.247.1 Applied Example 4, page 80
- Slide 76
- Applied Example: Driving Costs Using this model, estimate the average cost of driving a 2007 medium-sized sedan 8,000 miles per year and 18,000 miles per year. Solution The average cost for driving a car 8,000 miles per year is or approximately 68.8/mile. Miles/year, x 500010,00015,00020,000 Cost/mile, y () 83.862.952.247.1 Applied Example 4, page 80
- Slide 77
- Applied Example: Driving Costs Using this model, estimate the average cost of driving a 2007 medium-sized sedan 8,000 miles per year and 18,000 miles per year. Solution The average cost for driving a car 18,000 miles per year is or approximately 48.95/mile. Miles/year, x 500010,00015,00020,000 Cost/mile, y () 83.862.952.247.1 Applied Example 4, page 80
- Slide 78
- Some Economic Models Peoples decision on how much to demand or purchase of a given product depends on the price of the product: The higher the price the less they want to buy of it. A demand function p = d(x) can be used to describe this.
- Slide 79
- Some Economic Models Similarly, firms decision on how much to supply or produce of a product depends on the price of the product: The higher the price, the more they want to produce of it. A supply function p = s(x) can be used to describe this.
- Slide 80
- Some Economic Models The interaction between demand and supply will ensure the market settles to a market equilibrium: This is the situation at which quantity demanded equals quantity supplied. Graphically, this situation occurs when the demand curve and the supply curve intersect: where d(x) = s(x).
- Slide 81
- Applied Example: Supply and Demand The demand function for a certain brand of bluetooth wireless headset is given by The corresponding supply function is given by where p is the expressed in dollars and x is measured in units of a thousand. Find the equilibrium quantity and price. Applied Example 5, page 82
- Slide 82
- Applied Example: Supply and Demand Solution We solve the following system of equations: Substituting the second equation into the first yields: Thus, either x = 400/9 (but this is not possible), or x = 20. So, the equilibrium quantity must be 20,000 headsets. Applied Example 5, page 82
- Slide 83
- Applied Example: Supply and Demand Solution The equilibrium price is given by: or $40 per headset. Applied Example 5, page 82
- Slide 84
- Constructing Mathematical Models Some mathematical models can be constructed using elementary geometric and algebraic arguments. Guidelines for constructing mathematical models: Guidelines for constructing mathematical models: 1.Assign a letter to each variable mentioned in the problem. If appropriate, draw and label a figure. 2.Find an expression for the quantity sought. 3.Use the conditions given in the problem to write the quantity sought as a function f of one variable. Note any restrictions to be placed on the domain of f by the nature of the problem.
- Slide 85
- Applied Example: Enclosing an Area The owner of the Rancho Los Feliz has 3000 yards of fencing with which to enclose a rectangular piece of grazing land along the straight portion of a river. Fencing is not required along the river. Letting x denote the width of the rectangle, find a function f in the variable x giving the area of the grazing land if she uses all of the fencing. Applied Example 6, page 84
- Slide 86
- Applied Example: Enclosing an Area Solution This information was given: The area of the rectangular grazing land is A = xy. The amount of fencing is 2x + y which must equal 3000 (to use all the fencing), so: 2x + y = 3000 Solving for y we get: y = 3000 2x Substituting this value of y into the expression for A gives: A = x(3000 2x) = 3000x 2x 2 Finally, x and y represent distances, so they must be nonnegative, so x 0 and y = 3000 2x 0 (or x 1500). Thus, the required function is: f(x) = 3000x 2x 2 (0 x 1500) Applied Example 6, page 84
- Slide 87
- Applied Example: Charter-Flight Revenue If exactly 200 people sign up for a charter flight, Leasure World Travel Agency charges $300 per person. However, if more than 200 people sign up for the flight (assume this is the case), then each fare is reduced by $1 for each additional person. Letting x denote the number of passengers above 200, find a function giving the revenue realized by the company. Applied Example 7, page 84
- Slide 88
- Applied Example: Charter-Flight Revenue Solution This information was given. If there are x passengers above 200, then the number of passengers signing up for the flight is 200 + x. The fare will be (300 x) dollars per passenger. The revenue will be R= (200 + x)(300 x) = x 2 + 100x + 60,000 The quantities must be positive, so x 0 and 300 x 0 (or x 300). So the required function is: f(x) = x 2 + 100x + 60,000 (0 x 300) Applied Example 7, page 84
- Slide 89
- 2.4 Limits
- Slide 90
- Introduction to Calculus Historically, the development of calculus by Isaac Newton and Gottfried W. Leibniz resulted from the investigation of the following problems: 1.Finding the tangent line to a curve at a given point on the curve: t y T
- Slide 91
- Introduction to Calculus Historically, the development of calculus by Isaac Newton and Gottfried W. Leibniz resulted from the investigation of the following problems: 2.Finding the area of planar region bounded by an arbitrary curve. t y R
- Slide 92
- Introduction to Calculus The study of the tangent-line problem led to the creation of differential calculus, which relies on the concept of the derivative of a function. The study of the area problem led to the creation of integral calculus, which relies on the concept of the anti-derivative, or integral, of a function.
- Slide 93
- Example: A Speeding Maglev From data obtained in a test run conducted on a prototype of maglev, which moves along a straight monorail track, engineers have determined that the position of the maglev (in feet) from the origin at time t is given by s = f(t) = 4t 2 (0 t 30) s = f(t) = 4t 2 (0 t 30) Where f is called the position function of the maglev. The position of the maglev at time t = 0, 1, 2, 3, , 10 is f(0) = 0 f(1) = 4 f(2) = 16 f(3) = 36 f(10) = 400 f(0) = 0 f(1) = 4 f(2) = 16 f(3) = 36 f(10) = 400 But what if we want to find the velocity of the maglev at any given point in time?
- Slide 94
- Example: A Speeding Maglev Say we want to find the velocity of the maglev at t = 2. We may compute the average velocity of the maglev over an interval of time, such as [2, 4] as follows: or 24 feet/second. This is not the velocity of the maglev at exactly t = 2, but it is a useful approximation.
- Slide 95
- Example: A Speeding Maglev We can find a better approximation by choosing a smaller interval to compute the speed, say [2, 3]. More generally, let t > 2. Then, the average velocity of the maglev over the time interval [2, t] is given by
- Slide 96
- Example: A Speeding Maglev By choosing the values of t closer and closer to 2, we obtain average velocities of the maglev over smaller and smaller time intervals. The smaller the time interval, the closer the becomes to the instantaneous velocity of the train at t = 2, as the table below demonstrates: The smaller the time interval, the closer the average velocity becomes to the instantaneous velocity of the train at t = 2, as the table below demonstrates: The closer t gets to 2, the closer the average velocity gets to 16 feet/second. Thus, the instantaneous velocity at t = 2 seems to be 16 feet/second. t2.52.12.012.0012.0001 Average Velocity 1816.416.0416.00416.0004
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- Intuitive Definition of a Limit Consider the function g, which gives the average velocity of the maglev: Suppose we want to find the value that g(t) approaches as t approaches 2. We take values of t approaching 2 from the right (as we did before), and we find that g(t) approaches 16: Similarly, we take values of t approaching 2 from the left, and we find that g(t) also approaches 16: t2.52.12.012.0012.0001 g(t)g(t)g(t)g(t)1816.416.0416.00416.0004 t1.51.91.991.9991.9999 g(t)g(t)g(t)g(t)1415.615.9615.99615.9996
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- Intuitive Definition of a Limit We have found that as t approaches 2 from either side, g(t) approaches 16. In this situation, we say that the limit of g(t) as t approaches 2 is 16. This is written as Observe that t = 2 is not in the domain of g(t). But this does not matter, since t = 2 does not play any role in computing this limit.
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- Limit of a Function The function f has a limit L as x approaches a, written If the value of f(x) can be made as close to the number L as we please by taking x values sufficiently close to (but not equal to) a.
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- Examples Let f(x) = x 3. Evaluate Solution You can see in the graph that f(x) can be as close to 8 as we please by taking x sufficiently close to 2. Therefore, 2 1123 2 1123 8642 88664422 2 288664422 2 2 x y f(x) = x 3 Example 1, page 101
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- Examples Let Evaluate Solution 2 1123 2 1123 531 x y g(x)g(x)g(x)g(x) You can see in the graph that g(x) can be as close to 3 as we please by taking x sufficiently close to 1. Therefore, Example 2, page 101
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- Examples Let Evaluate Solution 2 112 2 112 5 x y The graph shows us that as x approaches 0 from either side, f(x) increases without bound and thus does not approach any specific real number. Thus, the limit of f(x) does not exist as x approaches 0. Example 3b, page 101
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- Theorem 1 Properties of Limits Supposeand Then, 1. r, a real number 2. c, a real number 3. 3. 4. 4. 5. Provided that M 0 Supposeand Then, 1. r, a real number 2. c, a real number 3. 3. 4. 4. 5. Provided that M 0
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- Examples Use theorem 1 to evaluate the following limits: Example 4, page 102
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- Examples Use theorem 1 to evaluate the following limits: Example 4, page 102
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- Indeterminate Forms Lets consider which we evaluated earlier for the maglev example by looking at values for x near x = 2. If we attempt to evaluate this expression by applying Property 5 of limits, we get In this case we say that the limit of the quotient f(x)/g(x) as x approaches 2 has the indeterminate form 0/0. This expression does not provide us with a solution to our problem.
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- Strategy for Evaluating Indeterminate Forms 1. Replace the given function with an appropriate one that takes on the same values as the original function everywhere except at x = a. 2. Evaluate the limit of this function as x approaches a.
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- Examples Evaluate Solution As weve seen, here we have an indeterminate form 0/0. We can rewrite x 2 x 2 Thus, we can say that Note that 16 is the same value we obtained for the maglev example through approximation. Example 5, page 104
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- Examples Evaluate Solution Notice in the graphs below that the two functions yield the same graphs, except for the value x = 2: 20161284 x y 3 2 1123 3 2 1123 20161284 x y Example 5, page 104
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- Examples Evaluate Solution As weve seen, here we have an indeterminate form 0/0. We can rewrite (with the constraint that h 0): Thus, we can say that Example 6, page 105
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- Limits at Infinity There are occasions when we want to know whether f(x) approaches a unique number as x increases without bound. In the graph below, as x increases without bound, f(x) approaches the number 400. We call the line y = 400 a horizontal asymptote. In this case, we can say that and we call this a limit of a function at infinity. 400 300 200 100 x y 102030405060
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- Example Consider the function Determine what happens to f(x) as x gets larger and larger. Solution We can pick a sequence of values of x and substitute them in the function to obtain the following values: As x gets larger and larger, f(x) gets closer and closer to 2. Thus, we can say that x125101001000 f(x)f(x)f(x)f(x)11.61.921.981.99981.999998
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- Limit of a Function at Infinity The function f has the limit L as x increases without bound (as x approaches infinity), written if f(x) can be made arbitrarily close to L by taking x large enough. Similarly, the function f has the limit M as x decreases without bound (as x approaches negative infinity), written if f(x) can be made arbitrarily close to M by taking x large enough in absolute value.
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- Examples Let Evaluateand Solution Graphing f(x) reveals that 1 11 1 111 1 1 x y 3 3 3 3 Example 7, page 107
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- Examples Let Evaluateand Solution Graphing g(x) reveals that x y 3 2 1 123 3 2 1 123 Example 7, page 107
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- Theorem 2 Properties of Limits For all n > 0,and provided that is defined. All properties of limits listed in Theorem 1 are valid when a is replaced by or . In addition, we have the following properties for limits to infinity:
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- Examples Evaluate Solution The limits of both the numerator and denominator do not exist as x approaches infinity, so property 5 is not applicable. We can find the solution instead by dividing numerator and denominator by x 3 : Example 8, page 108
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- Examples Evaluate Solution Again, we see that property 5 does not apply. So we divide numerator and denominator by x 2 : Example 9, page 108
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- Examples Evaluate Solution Again, we see that property 5 does not apply. But dividing numerator and denominator by x 2 does not help in this case: In other words, the limit does not exist. We indicate this by writing Example 10, page 109
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- 2.5 One-Sided Limits and Continuity
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- One-Sided Limits Consider the function Its graph shows that f does not have a limit as x approaches zero, because approaching from each side results in different values. 1 11 1 111 1 1 x y 1 1 1 1
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- One-Sided Limits Consider the function If we restrict x to be greater than zero (to the right of zero), we see that f(x) approaches 1 as close to as we please as x approaches 0. In this case we say that the right-hand limit of f as x approaches 0 is 1, written 1 11 1 111 1 1 x y 1 1 1 1
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- One-Sided Limits Consider the function Similarly, if we restrict x to be less than zero (to the left of zero), we see that f(x) approaches 1 as close to as we please as x approaches 0. In this case we say that the left-hand limit of f as x approaches 0 is 1, written 1 11 1 111 1 1 x y 1 1 1 1
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- One-Sided Limits The function f has the right-hand limit L as x approaches from the right, written If the values of f(x) can be made as close to L as we please by taking x sufficiently close to (but not equal to) a and to the right of a. Similarly, the function f has the left-hand limit L as x approaches from the left, written If the values of f(x) can be made as close to L as we please by taking x sufficiently close to (but not equal to) a and to the left of a.
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- Theorem 3 Properties of Limits Let f be a function that is defined for all values of x close to x = a with the possible exception of a itself. Then The connection between one-side limits and the two- sided limit defined earlier is given by the following theorem.
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- 2 1 12 2 1 12Examples Show that exists by studying the one-sided limits of f as x approaches 0: limits of f as x approaches 0:Solution For x > 0, we find And for x 0, we find Thus, 21 x y Example 1, page 118
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- Show that does not exist. Solution For x < 0, we find And for x 0, we find Thus, does not exist. 1 11 1 111 1 1Examplesx y Example 1, page 118
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- Continuous Functions Loosely speaking, a function is continuous at a given point if its graph at that point has no holes, gaps, jumps, or breaks. Consider, for example, the graph of f This function is discontinuous at the following points: At x = a, f is not defined (x = a is not in the domain of f ). a x y
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- Continuous Functions Loosely speaking, a function is continuous at a given point if its graph at that point has no holes, gaps, jumps, or breaks. Consider, for example, the graph of f This function is discontinuous at the following points: At x = b, f(b) is not equal to the limit of f(x) as x approaches b. a x y b
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- Continuous Functions Loosely speaking, a function is continuous at a given point if its graph at that point has no holes, gaps, jumps, or breaks. Consider, for example, the graph of f This function is discontinuous at the following points: At x = c, the function does not have a limit, since the left-hand and right-hand limits are not equal. x y bc
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- Continuous Functions Loosely speaking, a function is continuous at a given point if its graph at that point has no holes, gaps, jumps, or breaks. Consider, for example, the graph of f This function is discontinuous at the following points: At x = d, the limit of the function does not exist, resulting in a break in the graph. x y cd
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- Continuity of a Function at a Number A function f is continuous at a number x = a if the following conditions are satisfied: 1. f(a) is defined. 2. 2. 3. 3. If f is not continuous at x = a, then f is said to be discontinuous at x = a. Also, f is continuous on an interval if f is continuous at every number in the interval.
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- Examples Find the values of x for which the function is continuous: Solution The function f is continuous everywhere because the three conditions for continuity are satisfied for all values of x. 2 1 12 2 1 12 54321 x y Example 2, page 120
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- Examples Find the values of x for which the function is continuous: Solution The function g is discontinuous at x = 2 because g is not defined at that number. It is continuous everywhere else. 2 1 12 2 1 12 54321 x y Example 2, page 120
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- Examples Find the values of x for which the function is continuous: Solution The function h is continuous everywhere except at x = 2 where it is discontinuous because 2 1 12 2 1 12 54321 x y Example 2, page 120
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- Examples Find the values of x for which the function is continuous: Solution The function F is discontinuous at x = 0 because the limit of F fails to exist as x approaches 0. It is continuous everywhere else. 1 11 1 111 1 1 x y Example 2, page 120
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- Examples Find the values of x for which the function is continuous: Solution The function G is discontinuous at x = 0 because the limit of G fails to exist as x approaches 0. It is continuous everywhere else. 1 1 x y Example 2, page 120
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- Properties of Continuous Functions 1. The constant function f(x) = c is continuous everywhere. 2. The identity function f(x) = x is continuous everywhere. If f and g are continuous at x = a, then 3. [f(x)] n, where n is a real number, is continuous at x = a whenever it is defined at that number. 4. f g is continuous at x = a. 5. fg is continuous at x = a. 6. f /g is continuous at g(a) 0.
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- Properties of Continuous Functions Using these properties, we can obtain the following additional properties. 1. A polynomial function y = P(x) is continuous at every value of x. 2. A rational function R(x) = p(x)/q(x) is continuous at every value of x where q(x) 0.
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- Examples Find the values of x for which the function is continuous. Solution The function f is a polynomial function of degree 3, so f(x) is continuous for all values of x. Example 3, page 121
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- Examples Find the values of x for which the function is continuous. Solution The function g is a rational function. Observe that the denominator of g is never equal to zero. Therefore, we conclude that g(x) is continuous for all values of x. Example 3, page 121
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- Examples Find the values of x for which the function is continuous. Solution The function h is a rational function. In this case, however, the denominator of h is equal to zero at x = 1 and x = 2, which we can see by factoring. Therefore, we conclude that h(x) is continuous everywhere except at x = 1 and x = 2. Example 3, page 121
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- Intermediate Value Theorem Lets look again at the maglev example. The train cannot vanish at any instant of time and cannot skip portions of track and reappear elsewhere.
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- Intermediate Value Theorem Mathematically, recall that the position of the maglev is a function of time given by f(t) = 4t 2 for 0 t 30: t1t1t1t1 s2s2s2s2 s3s3s3s3 s1s1s1s1 t y Suppose the position of the maglev is s 1 at some time t 1 and its position is s 2 at some time t 2. Then, if s 3 is any number between s 1 and s 2, there must be at least one t 3 between t 1 and t 2 giving the time at which the maglev is at s 3 (f(t 3 ) = s 3 ). t3t3t3t3 t2t2t2t2
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- Theorem 4 Intermediate Value Theorem The Maglev example carries the gist of the intermediate value theorem: If f is a continuous function on a closed interval [a, b] and M is any number between f(a) and f(b), then there is at least one number c in [a, b] such that f(c) = M. a f(b)f(b)MMf(a)f(a)f(b)f(b)MMf(a)f(a)M x y c b x y c1c1c1c1 c2c2c2c2 c3c3c3c3ba f(b)f(b)MMf(a)f(a)f(b)f(b)MMf(a)f(a)M
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- Theorem 5 Existence of Zeros of a Continuous Function A special case of this theorem is when a continuous function crosses the x axis. If f is a continuous function on a closed interval [a, b], and if f(a) and f(b) have opposite signs, then there is at least one solution of the equation f(x) = 0 in the interval (a, b). f(b)f(b)f(a)f(a)f(b)f(b)f(a)f(a) x y c b x y c2c2c2c2 c3c3c3c3ba f(b)f(b)f(a)f(a)f(b)f(b)f(a)f(a) a c1c1c1c1
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- Example Let f(x) = x 3 + x + 1. a.Show that f is continuous for all values of x. b.Compute f(1) and f(1) and use the results to deduce that there must be at least one number x = c, where c lies in the interval (1, 1) and f(c) = 0. Solution a.The function f is a polynomial function of degree 3 and is therefore continuous everywhere. b. f (1) = (1) 3 + (1) + 1 = 1 and f (1) = (1) 3 + (1) + 1 = 3 Since f (1) and f (1) have opposite signs, Theorem 5 tells us that there must be at least one number x = c with 1 < c < 1 such that f(c) = 0. Example 5, page 124
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- 2.6 The Derivative
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- An Intuitive Example Consider the maglev example from Section 2.4. The position of the maglev is a function of time given by s = f(t) = 4t 2 (0 t 30) where s is measured in feet and t in seconds. Its graph is: s (ft) 1234123412341234 604020 t (sec)
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- An Intuitive Example The graph rises slowly at first but more rapidly over time. This suggests the steepness of f(t) is related to the speed of the maglev, which also increases over time. If so, we might be able to find the speed of the maglev at any given time by finding the steepness of f at that time. But how do we find the steepness of a point in a curve? s (ft) 1234123412341234 604020 t (sec)
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- A Slopes of Lines and of Curves The slope at a point of a curve is given by the slope of the tangent to the curve at that point: y Suppose we want to find the slope at point A. The tangent line has the same slope as the curve does at point A. x
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- The slope of a point in a curve is given by the slope of the tangent to the curve at that point: y = 1.8 x = 1 A Slopes of Lines and of Curves The slope of the tangent in this case is 1.8: Slope = 1.8 x y
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- The slope at a point of a curve is given by the slope of the tangent to the curve at that point: x A Slopes of Lines and of Curves Slope = 1.8 y
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- To calculate accurately the slope of a tangent to a curve, we must make the change in x as small as possible: y = 3 x = 4 As we let x get smaller, the slope of the secant becomes more and more similar to the slope of the tangent to the curve at that point. Slopes of Lines and of Curves Slope = 1.8 A x y
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- To calculate accurately the slope of a tangent to a curve, we must make the change in x as small as possible: y = 2.4 x = 3 A Slope = 1.8 Slopes of Lines and of Curves As we let x get smaller, the slope of the secant becomes more and more similar to the slope of the tangent to the curve at that point. x y
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- To calculate accurately the slope of a tangent to a curve, we must make the change in x as small as possible: y = 2.2 x = 2 A Slope = 1.8 Slopes of Lines and of Curves As we let x get smaller, the slope of the secant becomes more and more similar to the slope of the tangent to the curve at that point. x y
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- y = 1.5 x = 1 To calculate accurately the slope of a tangent to a curve, we must make the change in x as small as possible: A Slope = 1.8 Slopes of Lines and of Curves As we let x get smaller, the slope of the secant becomes more and more similar to the slope of the tangent to the curve at that point. x y
- Slide 158
- To calculate accurately the slope of a tangent to a curve, we must make the change in x as small as possible: y = 0.00179 x = 0.001 A Slope = 1.8 Slopes of Lines and of Curves As we let x get smaller, the slope of the secant becomes more and more similar to the slope of the tangent to the curve at that point. x y
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- In general, we can express the slope of the secant as follows: Slopes of Lines and of Curves
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- Thus, as h approaches zero, the slope of the secant approaches the slope of the tangent to the curve at that point: Slopes of Lines and of Curves
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- Thus, as h approaches zero, the slope of the secant approaches the slope of the tangent to the curve at that point: Slopes of Lines and of Curves
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- Thus, as h approaches zero, the slope of the secant approaches the slope of the tangent to the curve at that point: Slopes of Lines and of Curves
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- Slide 164
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- Thus, as h approaches zero, the slope of the secant approaches the slope of the tangent to the curve at that point. Expressed in limits notation: The slope of the tangent line to the graph The slope of the tangent line to the graph of f at the point P(x, f(x)) is given by if it exists. Slopes of Lines and of Curves
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- Average Rates of Change We can see that measuring the slope of the tangent line to a graph is mathematically equivalent to finding the rate of change of f at x. The number f(x + h) f(x) measures the change in y that corresponds to a change h in x. Then the difference quotient measures the average rate of change of y with respect to x over the interval [x, x + h]. In the maglev example, if y measures the position the train at time x, then the quotient give the average velocity of the train over the time interval [x, x + h].
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- Average Rates of Change The average rate of change of f over the interval [x, x + h] or slope of the secant line to the graph of f through the points (x, f(x)) and (x + h, f(x + h)) is
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- Instantaneous Rates of Change By taking the limit of the difference quotient as h goes to zero, evaluating we obtain the rate of change of f at x. This is known as the instantaneous rate of change of f at x (as opposed to the average rate of change). In the maglev example, if y measures the position of a train at time x, then the limit gives the velocity of the train at time x.
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- Instantaneous Rates of Change The instantaneous rate of change of f at x or slope of the tangent line to the graph of f at (x, f(x)) is This limit is called the derivative of f at x.
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- The Derivative of a Function The derivative of a function with respect to is the function (read prime). The derivative of a function f with respect to x is the function f (read f prime). The domain of is the set of all x where the limit exists. The domain of f is the set of all x where the limit exists. Thus, the derivative of function f is a function f that gives the slope of the tangent to the line to the graph of f at any point (x, f(x)) and also the rate of change of f at x.
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- The Derivative of a Function Four Step Process for Finding f (x) 1.Compute f(x + h). 2.Form the difference f(x + h) f(x). 3.Form the quotient 4.Compute
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- Examples Find the slope of the tangent line to the graph f(x) = 3x + 5 at any point (x, f(x)). Solution The required slope is given by the derivative of f at x. To find the derivative, we use the four-step process: Step 1. f(x + h) = 3(x + h) + 5 = 3x + 3h + 5. Step 2. f(x + h) f(x) = 3x + 3h + 5 (3x + 5) = 3h. Step 3. Step 4. Example 2, page 138
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- Examples Find the slope of the tangent line to the graph f(x) = x 2 at any point (x, f(x)). Solution The required slope is given by the derivative of f at x. To find the derivative, we use the four-step process: Step 1. f(x + h) = (x + h) 2 = x 2 + 2xh + h 2. Step 2. f(x + h) f(x) = x 2 + 2xh + h 2 x 2 = h(2x + h). Step 3. Step 4. Example 3, page 138
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- Examples Find the slope of the tangent line to the graph f(x) = x 2 at any point (x, f(x)). The slope of the tangent line is given by . The slope of the tangent line is given by f (x) = 2x. Now, find and interpret . Now, find and interpret f (2).Solution . f (2) = 2(2) = 4. This means that, at the point (2, 4) the slope of the tangent line to the graph is 4. (2, 4) y Example 3, page 138
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- Applied Example: Demand for Tires The management of Titan Tire Company has determined that the weekly demand function of their Super Titan tires is given by where p is measured in dollars and x is measured in thousands of tires. Find the average rate of change in the unit price of a tire if the quantity demanded is between 5000 and 6000 tires; between 5000 and 5100 tires; and between 5000 and 5010 tires. What is the instantaneous rate of change of the unit price when the quantity demanded is 5000 tires? Applied Example 7, page 141
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- Applied Example: Demand for Tires Solution The average rate of change of the unit price of a tire if the quantity demanded is between x and x + h is Applied Example 7, page 141
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- Applied Example: Demand for Tires Solution The average rate of change is given by 2x h. To find the average rate of change of the unit price of a tire when the quantity demanded is between 5000 and 6000 tires [5, 6], we take x = 5 and h = 1, obtaining or $11 per 1000 tires. Similarly, with x = 5, and h = 0.1, we obtain or $10.10 per 1000 tires. Finally, with x = 5, and h = 0.01, we get or $10.01 per 1000 tires. Applied Example 7, page 141
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- Applied Example: Demand for Tires Solution The instantaneous rate of change of the unit price of a tire when the quantity demanded is x tires is given by In particular, the instantaneous rate of change of the price per tire when quantity demanded is 5000 is given by 2(5), or $10 per tire. Applied Example 7, page 141
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- Differentiability and Continuity Sometimes, one encounters continuous functions that fail to be differentiable at certain values in the domain of the function f. For example, consider the continuous function f below: It fails to be differentiable at x = a, because the graph makes an abrupt change (a corner) at that point. (It is not clear what the slope is at that point) a x y
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- Differentiability and Continuity Sometimes, one encounters continuous functions that fail to be differentiable at certain values in the domain of the function f. For example, consider the continuous function f below: It also fails to be differentiable and x = b because the slope is not defined at that point. a x y b
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- Applied Example: Wages Mary works at the B&O department store, where, on a weekday, she is paid $8 an hour for the first 8 hours and $12 an hour of overtime. The function gives Marys earnings on a weekday in which she worked x hours. Sketch the graph of the function f and explain why it is not differentiable at x = 8. Applied Example 8, page 143
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- Applied Example: Wages Solution The graph of f has a corner at x = 8 and so is not differentiable at that point. y 1301109070503010 24681012 x (8, 64) Applied Example 8, page 143
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- End of Chapter

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