Functions and Their Graphs The Algebra of Functions Functions and Mathematical Models Limits

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Functions and Their Graphs The Algebra of Functions Functions and Mathematical Models Limits One-Sided Limits and Continuity The Derivative. 2. Functions, Limits and the Derivative. 2.1. Functions and Their Graphs. Functions.

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Functions and Their Graphs

The Algebra of Functions

Functions and Mathematical Models

Limits

One-Sided Limits and Continuity

The Derivative

2

Functions, Limits and the Derivative

### 2.1

Functions and Their Graphs

Functions
• Function: A function is a rule that assigns to each element in a set Aone 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.
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 xin its domain, is called the range of f(x).
• The output f(x) associated with an input x is unique:
• Eachxmust correspond to one and onlyone value of f(x).
Example
• Let the function fbe defined by the rule
• Find: f(1)
• Solution:

Example 1, page 51

Example
• Let the function fbe defined by the rule
• Find: f(–2)
• Solution:

Example 1, page 51

Example
• Let the function fbe defined by the rule
• Find: f(a)
• Solution:

Example 1, page 51

Example
• Let the function fbe defined by the rule
• Find: f(a + h)
• Solution:

Example 1, page 51

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 ThermoMaster’s weekly profit if its level of production is:
• 1000 thermometers per week.
• 2000 thermometers per week.

Applied Example 2, page 51

Applied Example

Solution

• We have
• The weekly profit by producing1000 thermometers is

or \$2,000.

• The weekly profit by producing2000 thermometers is

or \$7,000.

Applied Example 2, page 51

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.
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.

x

10 10 – 2x

x

x

16 – 2x

x

16

Applied Example 3, page 52

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.
• Find the expression that gives the volumeV of the box as a function of x.
• What is the domain of the function?
• The dimensions of the resulting box are:

x

10 – 2x

16 – 2x

Applied Example 3, page 52

Applied Example: Packaging

Solution

a. The volume of the box is given by multiplying its dimensions(length ☓ width ☓ height), so:

x

10 – 2x

16 – 2x

Applied Example 3, page 52

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

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

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

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

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 functionf is the set of all points (x, y) in the xy-plane such that x is the domain of f and y = f(x).
Example
• The graph of a function f is shown below:

y

y

(x, y)

Range

x

x

Domain

Example 5, page 53

Example
• The graph of a function f is shown below:
• What is the value of f(2)?

y

4

3

2

1

–1

–2

x

1 2 3 4 5 6 7 8

(2,–2)

Example 5, page 53

Example
• The graph of a function f is shown below:
• What is the value of f(5)?

y

4

3

2

1

–1

–2

(5,3)

x

1 2 3 4 5 6 7 8

Example 5, page 53

Example
• The graph of a function f is shown below:
• What is the domain of f(x)?

y

4

3

2

1

–1

–2

x

1 2 3 4 5 6 7 8

Domain: [1,8]

Example 5, page 53

Example
• The graph of a function f is shown below:
• What is the range of f(x)?

y

4

3

2

1

–1

–2

Range:

[–2,4]

x

1 2 3 4 5 6 7 8

Example 5, page 53

Example: Sketching a Graph
• Sketch the graph of the function defined by the equation

y = x2 + 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:

Example 6, page 54

Example: Sketching a Graph
• Sketch the graph of the function defined by the equation

y = x2 + 1

Solution

• The domain of the function is the set of all real numbers.
• Then plot these values in a graph:

y

10

8

6

4

2

x

– 3 – 2 – 1 1 2 3

Example 6, page 54

Example: Sketching a Graph
• Sketch the graph of the function defined by the equation

y = x2 + 1

Solution

• The domain of the function is the set of all real numbers.
• And finally, connect the dots:

y

10

8

6

4

2

x

– 3 – 2 – 1 1 2 3

Example 6, page 54

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

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:

Example 7, page 55

Example: Sketching a Graph
• Sketch the graph of the function defined by the equation

Solution

• Plotting these data and graphing we get:

y

3

2

1

x

– 3 – 2 – 1 1 2 3

Example 7, page 55

2.2

The Algebra of Functions

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.

y

2000

1800

1600

1400

1200

1000

y = R(t)

y = S(t)

S(t)

Billions of Dollars

R(t)

t

1990 1992 1994 1996 1998 2000

t

Year

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.

y

2000

1800

1600

1400

1200

1000

y = R(t)

y = S(t)

S(t)

Billions of Dollars

D(t) = R(t) – S(t)

R(t)

t

1990 1992 1994 1996 1998 2000

t

Year

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 asR(t) and S(t).

y

400

200

0

–200

–400

y = D(t)

t

Billions of Dollars

t

1992 1994 1996 1998 2000

D(t)

Year

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.
The Sum, Difference, Product and Quotient of Functions
• Let f and g be functions with domains A and B, respectively.
• The sumf + g, the differencef – g, and the productfg of f and g are functions with domain A ∩ Band 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

• The quotientf/g of f and g has domain A ∩ Bexcluding all numbers x such that g(x)= 0 and rule given by

Quotient

Example
• Let and g(x) = 2x + 1.
• Find the sums, the differenced, the productp, and the quotientq 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

Example
• Let and g(x) = 2x + 1.
• Find the sums, the differenced, the productp, and the quotientq of the functions f and g.

Solution

• The domain of the quotient function is [–1,) together with the restriction x≠–½.
• Thus, the domain is [–1, –½) U (–½,).
• The rule is as follows:

Example 1, page 68

Applied Example
• Suppose Puritron, a manufacturer of water filters, has a monthly fixed cost of \$10,000 and a variable cost of

–0.0001x2 + 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

Applied Example

Solution

• Puritron’s 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) = –0.0001x2 + 10x

• The total cost is the sum of the fixed cost F and the variable cost V:

C(x) = V(x) + F(x)

= –0.0001x2 + 10x + 10,000 (0 x 40,000)

Applied Example 2, page 68

Applied Example

Lets now consider profits

• Suppose that the total revenueR realized by Puritron from the sale of x water filters is given by

R(x) = –0.0005x2 + 20x (0 ≤x≤ 40,000)

• Find
• The total profit function for Puritron.
• The total profit when Puritron produces 10,000 filters per month.

Applied Example 3, page 69

Applied Example

Solution

• The totalprofitP realized by the firm is the difference between the total revenueR and the total costC:

P(x) = R(x) – C(x)

= (–0.0005x2 + 20x)– (–0.0001x2 + 10x + 10,000)

= –0.0004x2 + 10x – 10,000

• The totalprofit realized by Puritron when producing10,000 filters per month is

P(x) = –0.0004(10,000)2 + 10(10,000)– 10,000

= 50,000

or \$50,000 per month.

Applied Example 3, page 69

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 gat the pointf(x), we find that:
• This is an entirely new function, which we could call h:
The Composition of Two Functions
• Let f and g be functions.
• Then the composition of g and f is the function ggf(read “gcirclef ”) defined by

(ggf)(x) = g(f(x))

• The domain of ggfis the set of all x in the domain of f such that f(x) lies in the domain of g.
Example
• Let
• Find:
• The rule for the composite function ggf.
• The rule for the composite function fgg.

Solution

• To find ggf, evaluate the function g at f(x):
• To find fgg, evaluate the function f at g(x):

Example 4, page 70

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 0.01x2/3 parts per million when the number ofmotor vehiclesisx 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 0.2t2 + 4t + 64 thousand.
• Find:
• An expression for the concentration of CO in the air due to automobile exhaust t years from now.
• The level of concentration 5years from now.

Applied Example 5, page 70

Applied Example

Solution

• Part (a):
• The level of CO is described by the function

g(x) =0.01x2/3

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) = 0.2t2 + 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

(ggf)(t) = g(f(t)) = 0.01(0.2t2 + 4t + 64)2/3

Applied Example 5, page 70

Applied Example

Solution

• Part (b):
• The level of CO five years from now is:

(ggf)(5) = g(f(5)) = 0.01[0.2(5)2 + 4(5) + 64]2/3

= (0.01)892/3≈ 0.20

or approximately 0.20 parts per million.

Applied Example 5, page 70

2.3

Functions and Mathematical Models

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)
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

Formulate

Mathematical model

Solve

Test

Solution of real-world problem

Solution of mathematical model

Interpret

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 a0, a1, …. an are constants called the coefficients of the polynomial function.

• Examples:
• The function below is polynomial function of degree 5:
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 a0, a1, …. an are constants called the coefficients of the polynomial function.

• Examples:
• The function below is polynomial function of degree 3:
Applied ExampleMarket 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.

Applied Example 1, page 76

Applied ExampleMarket for Cholesterol-Reducing Drugs

M(t) = 1.95t + 12.19

• Sketch the graph of the function M and the given data on the same set of axes.
• Assuming that the projection held and the trend continued, what was the market for cholesterol-reducing drugs in 2005(t = 6)?
• What was the rate of increase of the market for cholesterol-reducing drugs over the period in question?

Applied Example 1, page 76

Applied ExampleMarket for Cholesterol-Reducing Drugs

M(t) = 1.95t + 12.19

Solution

• Graph:

y

25

20

15

M(t)

Billions of Dollars

t

1 2 3 4 5

Applied Example 1, page 76

Year

Applied ExampleMarket for Cholesterol-Reducing Drugs

M(t) = 1.95t + 12.19

Solution

• The projected market in 2005 for cholesterol-reducing drugs was

M(6) = 1.95(6) + 12.19 = 23.89

or \$23.89 billion.

Applied Example 1, page 76

Applied ExampleMarket for Cholesterol-Reducing Drugs

M(t) = 1.95t + 12.19

Solution

• 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 \$1.95billion per year.

Applied Example 1, page 76

Modeling a Polynomial Function of Degree 2
• A polynomial function of degree 2 has the form
• Or more simply, y = ax2 + bx + c, and is called a quadratic function.
• The graph of a quadratic function is a parabola:

Opens upwards if a > 0

Opens downwards if a < 0

y

y

x

x

Applied ExampleGlobal Warming
• The increase in carbon dioxide (CO2) in the atmosphere is a major cause of global warming.
• Below is a table showing the average amount of CO2, measured in parts per million volume (ppmv) for various years from 1958 through 2007:

Applied Example 2, page 78

Applied ExampleGlobal Warming
• Below is a scatter plot associated with these data:

y (ppmv)

380

360

340

320

t (years)

10 20 30 40 50

Applied Example 2, page 78

Applied ExampleGlobal Warming
• A mathematical model giving the approximate amount of CO2 is given by:

y (ppmv)

380

360

340

320

t (years)

10 20 30 40 50

Applied Example 2, page 78

Applied ExampleGlobal Warming
• Use the model to estimate the average amount of atmospheric CO2 in 1980(t= 23).
• Assume that the trend continued and use the model to predict the average amount of atmospheric CO2 in 2010.

Applied Example 2, page 78

Applied ExampleGlobal Warming

Solution

• The average amount of atmospheric CO2 in 1980 is given by

or approximately 338 ppmv.

• Assuming that the trend will continue, the average amount of atmospheric CO2 in 2010 will be

Applied Example 2, page 78

Applied ExampleSocial 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:

y (\$trillion)

6

4

2

t (years)

5 10 15 20 25 30

Applied Example 3, page 79

Applied ExampleSocial 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:

y (\$trillion)

6

4

2

t (years)

5 10 15 20 25 30

Applied Example 3, page 79

Applied ExampleSocial Security Trust Fund Assets
• The first baby boomers will turn 65 in 2011. What will be the assets of the Social Security trust fund at that time?
• The last of the baby boomers will turn 65 in 2029. What will the assets of the trust fund be at the time?
• Use the graph of function A(t) to estimatethe year in which the current Social Security system will go broke.

Applied Example 3, page 79

Applied ExampleSocial Security Trust Fund Assets

Solution

• 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.

Applied Example 3, page 79

Applied ExampleSocial Security Trust Fund Assets

Solution

• The graph shows that function Acrosses the t-axis at about t = 32, suggesting the system will go broke by 2040:

y (\$trillion)

6

4

2

t (years)

5 10 15 20 25 30

Applied Example 3, page 79

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)
Rational and Power Functions
• Examples of rational functions:
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:
Rational and Power Functions
• Many functions involve combinations of rational and power functions.
• Examples:
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.

Applied Example 4, page 80

Applied Example: Driving Costs
• Below is the scatter plot associated with this data:

y (¢)

140

120

100

80

60

40

20

C(x)

t (years)

5 10 15 20 25

Applied Example 4, page 80

Applied Example: Driving Costs
• Using this model, estimate the average cost of driving a 2007 medium-sized sedan 8,000 miles per year and 18,000miles per year.

Solution

• The average cost for driving a car 8,000 miles per year is

or approximately 68.8¢/mile.

Applied Example 4, page 80

Applied Example: Driving Costs
• Using this model, estimate the average cost of driving a 2007 medium-sized sedan 8,000 miles per year and 18,000miles per year.

Solution

• The average cost for driving a car 18,000 miles per year is

or approximately 48.95¢/mile.

Applied Example 4, page 80

Some Economic Models
• People’s decision on how much to demand or purchase of a given product depends on the price of the product:
• Thehigherthe price thelessthey want to buy of it.
• A demand functionp = d(x) can be used to describe this.
Some Economic Models
• Similarly, firms’ decision on how much to supply or produce of a product depends on the price of the product:
• Thehigherthe price, themorethey want to produce of it.
• A supply functionp = s(x) can be used to describe this.
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 curveintersect: where d(x)= s(x).
Applied Example: Supply and Demand
• The demandfunction 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

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

Applied Example: Supply and Demand

Solution

• The equilibrium price is given by:

Applied Example 5, page 82

Constructing Mathematical Models
• Some mathematical models can be constructed using elementary geometric and algebraic arguments.

Guidelines for constructing mathematical models:

• Assign a letter to each variable mentioned in the problem. If appropriate, draw and label a figure.
• Find an expression for the quantity sought.
• Use the conditions given in the problem to write the quantity sought as a functionf of one variable.

Note any restrictions to be placed on the domain of f by the nature of the problem.

Applied Example: Enclosing an Area
• The owner of the Rancho Los Feliz has 3000yards 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 functionf in the variablex giving the area of the grazing land if she uses all of the fencing.

Applied Example 6, page 84

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 – 2x2

• 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 – 2x2(0 x 1500)

Applied Example 6, page 84

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 \$1for each additional person.
• Letting x denote the number of passengers above200, find a function giving the revenue realized by the company.

Applied Example 7, page 84

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)

= –x2 + 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) =–x2 + 100x + 60,000 (0 x 300)

Applied Example 7, page 84

2.4

Limits

Introduction to Calculus
• Historically, the development of calculus by Isaac Newton and Gottfried W. Leibniz resulted from the investigation of the following problems:
• Finding the tangent line to a curve at a given point on the curve:

y

T

t

Introduction to Calculus
• Historically, the development of calculus by Isaac Newton and Gottfried W. Leibniz resulted from the investigation of the following problems:
• Finding the area of planar region bounded by an arbitrary curve.

y

R

t

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.
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 timet is given by

s = f(t) = 4t2 (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

• But what if we want to find the velocity of the maglev at any given point in time?
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 exactlyt = 2, but it is a useful approximation.
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
Example: A Speeding Maglev
• By choosing the values of tcloser 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 average velocity becomes to the instantaneous velocity of the train at t = 2, as the table below demonstrates:
• The closert 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.
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 approaches2.
• We take values of tapproaching2from the right (as we did before), and we find that g(t)approaches16:
• Similarly, we take values of t approaching2from the left, and we find that g(t) also approaches16:
Intuitive Definition of a Limit
• We have found that as tapproaches2from either side, g(t)approaches16.
• In this situation, we say that the limit of g(t) as tapproaches2 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.
Limit of a Function
• The functionf has a limitLas xapproachesa, written
• If the value of f(x) can be made as close to the number L as we please by taking x valuessufficiently close to (but not equal to) a.
Examples
• Let f(x) = x3. Evaluate

Solution

• You can see in the graph that f(x) can be as close to 8 as we please by taking xsufficiently close to 2.
• Therefore,

f(x) = x3

y

8

6

4

2

–2

x

–2 –1 1 2 3

Example 1, page 101

Examples
• Let Evaluate

Solution

• You can see in the graph that g(x) can be as close to 3 as we please by taking xsufficiently close to 1.
• Therefore,

y

g(x)

5

3

1

x

–2 –1 1 2 3

Example 2, page 101

Examples
• Let Evaluate

Solution

• The graph shows us that as xapproaches0 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 xapproaches0.

y

5

x

–2 –1 1 2

Example 3b, page 101

Theorem 1Properties of Limits

Suppose and

Then,

• r, a real number
• c, a real number
• Provided thatM≠ 0
Examples
• Use theorem 1 to evaluate the following limits:

Example 4, page 102

Examples
• Use theorem 1 to evaluate the following limits:

Example 4, page 102

Indeterminate Forms
• Let’s consider

which we evaluated earlier for the maglev example by looking at values for xnearx = 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 quotientf(x)/g(x) as xapproaches2 has the indeterminate form 0/0.
• This expression does not provide us with a solution to our problem.
Strategy for Evaluating Indeterminate Forms
• Replace the given function with an appropriate one that takes on the same values as the original function everywhereexcept at x = a.
• Evaluate the limit of this function as xapproachesa.
Examples
• Evaluate

Solution

• As we’ve seen, here we have an indeterminate form0/0.
• We can rewrite

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

Examples
• Evaluate

Solution

• Notice in the graphs below that the two functions yield the same graphs, except for the valuex = 2:

y

y

20

16

12

8

4

20

16

12

8

4

x

x

–3 –2 –1 1 2 3

–3 –2 –1 1 2 3

Example 5, page 104

Examples
• Evaluate

Solution

• As we’ve seen, here we have an indeterminate form0/0.
• We can rewrite (with the constraint that h≠ 0):
• Thus, we can say that

Example 6, page 105

Limits at Infinity
• There are occasions when we want to know whether f(x)approachesa 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.

y

400

300

200

100

x

10 20 30 40 50 60

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
Limit of a Function at Infinity
• The function f has the limitL as xincreases without bound (as xapproaches infinity), written

if f(x) can be made arbitrarily close toL by taking x large enough.

• Similarly, the function f has the limitM as xdecreases without bound (as xapproaches negative infinity), written

if f(x) can be made arbitrarily close toM by taking x large enough in absolute value.

Examples
• Let
• Evaluate and

Solution

• Graphing f(x) reveals that

y

1

–1

x

–3 3

Example 7, page 107

Examples
• Let
• Evaluate and

Solution

• Graphing g(x) reveals that

y

x

–3 –2 –1 1 2 3

Example 7, page 107

Theorem 2Properties 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 forlimits to infinity:
Examples
• Evaluate

Solution

• The limits of both the numerator and denominatordo not exist as x approaches infinity, so property 5 is not applicable.
• We can find the solution instead by dividingnumerator and denominatorbyx3:

Example 8, page 108

Examples
• Evaluate

Solution

• Again, we see that property 5 does not apply.
• So we dividenumerator and denominatorbyx2:

Example 9, page 108

Examples
• Evaluate

Solution

• Again, we see that property 5 does not apply.
• But dividingnumerator and denominatorbyx2 does not help in this case:
• In other words, the limit does not exist.
• We indicate this by writing

Example 10, page 109

2.5

One-Sided Limits and Continuity

One-Sided Limits
• Consider the function
• Its graph shows that fdoesnothave a limit as xapproaches zero, because approaching from each side results in different values.

y

1

–1

x

–1 1

One-Sided Limits
• Consider the function
• If we restrictx to be greater than zero (to the right of zero), we see that f(x)approaches1 as close to as we please as xapproaches 0.
• In this case we say that the right-hand limit of f as xapproaches0 is 1, written

y

1

–1

x

–1 1

One-Sided Limits
• Consider the function
• Similarly, if we restrictx to be less than zero (to the left of zero), we see that f(x)approaches–1 as close to as we please as xapproaches 0.
• In this case we say that the left-hand limit of f as xapproaches0 is –1, written

y

1

–1

x

–1 1

One-Sided Limits
• The function f has the right-hand limitL as xapproaches 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 limitL as xapproaches 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.
Theorem 3Properties 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.
Examples
• Show that exists by studying the one-sided

limits of f as xapproaches0:

Solution

• For x > 0, we find
• And for x≤ 0, we find
• Thus,

y

2

1

x

– 2 –1 1 2

Example 1, page 118

Examples
• Show that does not exist.

Solution

• For x< 0, we find
• And for x 0, we find
• Thus, does

not exist.

y

1

–1

x

Example 1, page 118

Continuous Functions
• Loosely speaking, a function is continuous at a given point if its graph at that point has noholes, 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 ).

y

x

a

Continuous Functions
• Loosely speaking, a function is continuous at a given point if its graph at that point has noholes, 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 limitof f(x)as x approachesb.

y

x

a

b

Continuous Functions
• Loosely speaking, a function is continuous at a given point if its graph at that point has noholes, 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.

y

x

b

c

Continuous Functions
• Loosely speaking, a function is continuous at a given point if its graph at that point has noholes, 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.

y

x

c

d

Continuity of a Function at a Number
• A function f is continuous at a number x = a if the following conditions are satisfied:
• f(a) is defined.
• If f is not continuous at x = a, then f is said to be discontinuous atx = a.
• Also, f is continuous on an interval if f is continuous at every number in the interval.
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.

y

5

4

3

2

1

x

– 2 –1 1 2

Example 2, page 120

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.

y

5

4

3

2

1

x

– 2 –1 1 2

Example 2, page 120

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

y

5

4

3

2

1

x

– 2 –1 1 2

Example 2, page 120

Examples
• Find the values of x for which the function is continuous:

Solution

• The function F is discontinuous atx = 0 because the limit of Ffails to exist as xapproaches0. It is continuous everywhere else.

y

1

–1

x

Example 2, page 120

Examples
• Find the values of x for which the function is continuous:

Solution

• The function G is discontinuous atx = 0 because the limit of Gfails to exist as xapproaches0. It is continuous everywhere else.

y

–1

x

Example 2, page 120

Properties of Continuous Functions
• The constant function f(x) = c is continuous everywhere.
• The identity function f(x) = x is continuous everywhere.

Iff andg are continuous atx = a, then

• [f(x)]n, where n is a real number, is continuous at x = a whenever it is defined at that number.
• f ± g is continuous at x = a.
• fg is continuous at x = a.
• f/g is continuous at g(a) ≠ 0.
Properties of Continuous Functions
• Using these properties, we can obtain the following additional properties.
• A polynomial functiony = P(x) is continuous at every value of x.
• A rational functionR(x)= p(x)/q(x) is continuous at every value of x where q(x)≠ 0.
Examples
• Find the values ofx 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 ofx.

Example 3, page 121

Examples
• Find the values ofx 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 ofx.

Example 3, page 121

Examples
• Find the values ofx 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

Intermediate Value Theorem
• Let’s look again at the maglev example.
• The train cannot vanish at any instant of time and cannot skip portions of track and reappear elsewhere.
Intermediate Value Theorem
• Mathematically, recall that the position of the maglev is a function of time given by f(t) = 4t2for 0 t 30:

y

s2

s3

s1

t

t1

t3

t2

• Suppose the position of the maglev is s1 at some timet1 and its position is s2 at some timet2.
• Then, if s3is any number betweens1 and s2, there must be at least onet3 between t1 and t2 giving the time at which the maglev is at s3 (f(t3) = s3).
Theorem 4Intermediate 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 numberc in [a, b] such that f(c) = M.

y

y

f(b)

M

f(a)

f(b)

M

f(a)

x

x

a

c1

c2

c3

b

a

c

b

Theorem 5 Existence of Zeros of a Continuous Function
• A special case of this theorem is when a continuous function crosses thexaxis.
• 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 equationf(x) = 0 in the interval (a, b).

y

y

f(b)

f(a)

f(b)

f(a)

x

x

a

c1

c2

c3

b

a

c

b

Example
• Let f(x) = x3 + x + 1.
• Show that f is continuous for all values of x.
• 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

• The function f is a polynomial function of degree 3 and is therefore continuous everywhere.
• f (–1) = (–1)3 + (–1)+ 1 = –1 andf (1) = (1)3 + (1)+ 1 = 3

Since f (–1) andf (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

2.6

The Derivative

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) = 4t2 (0 t 30)

where s is measured in feet and t in seconds.

• Its graph is:

s (ft)

60

40

20

t (sec)

1 2 3 4

An Intuitive Example
• The graph rises slowlyat first but more rapidlyover 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)

60

40

20

t (sec)

1 2 3 4

Slopes of Lines and of Curves
• The slope at a point of a curve is given by the slope of thetangent to the curve at that point:

y

Suppose we want to find the slope at point A.

A

The tangent line has the same slope as the curve does at point A.

x

Slopes of Lines and of Curves
• The slope of a point in a curve is given by the slope of thetangent to the curve at that point:

y

Slope = 1.8

Dy = 1.8

The slope of the tangent in this case is 1.8:

A

Dx = 1

x

Slopes of Lines and of Curves
• To calculate accurately the slope of a tangent to a curve, we must make the change inx assmallas possible:

y

Slope = 1.8

Dy = 3

As we let Dxget smaller, the slopeof thesecant becomes more and more similar to the slopeof thetangent to the curve at that point.

A

Dx = 4

x

Slopes of Lines and of Curves
• To calculate accurately the slope of a tangent to a curve, we must make the change inx assmallas possible:

y

Slope = 1.8

Dy = 2.4

As we let Dxget smaller, the slope of thesecant becomes more and more similar to the slope of thetangent to the curve at that point.

A

Dx = 3

x

Slopes of Lines and of Curves
• To calculate accurately the slope of a tangent to a curve, we must make the change inx assmallas possible:

y

Slope = 1.8

Dy = 2.2

As we let Dxget smaller, the slope of thesecant becomes more and more similar to the slope of thetangent to the curve at that point.

A

Dx = 2

x

Slopes of Lines and of Curves
• To calculate accurately the slope of a tangent to a curve, we must make the change inx assmallas possible:

y

Slope = 1.8

As we let Dxget smaller, the slope of thesecant becomes more and more similar to the slope of thetangent to the curve at that point.

Dy = 1.5

A

Dx = 1

x

Slopes of Lines and of Curves
• To calculate accurately the slope of a tangent to a curve, we must make the change inx assmallas possible:

y

Slope = 1.8

As we let Dxget smaller, the slope of thesecant becomes more and more similar to the slope of thetangent to the curve at that point.

A

Dy = 0.00179

Dx = 0.001

x

Slopes of Lines and of Curves
• In general, we can express the slope of the secant as follows:
Slopes of Lines and of Curves
• Thus, as happroaches zero, the slopeof thesecantapproaches the slopeof thetangent to the curve at that point:
Slopes of Lines and of Curves
• Thus, as happroaches zero, the slopeof thesecantapproaches the slopeof thetangent to the curve at that point:
Slopes of Lines and of Curves
• Thus, as happroaches zero, the slopeof thesecantapproaches the slopeof thetangent to the curve at that point:
Slopes of Lines and of Curves
• Thus, as happroaches zero, the slopeof thesecantapproaches the slopeof thetangent to the curve at that point:
Slopes of Lines and of Curves
• Thus, as happroaches zero, the slopeof thesecantapproaches the slopeof thetangent to the curve at that point:
Slopes of Lines and of Curves
• Thus, as happroaches zero, the slopeof thesecantapproaches the slopeof thetangent to the curve at that point.
• Expressed in limits notation:

The slopeof the tangent line to the graph

of f at the point P(x, f(x)) is given by

if it exists.

Average Rates of Change
• We can see that measuring theslopeof thetangent line to a graph is mathematically equivalent to finding therate of change of f at x.
• The number f(x + h) – f(x) measures the change in y that corresponds to a changeh 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].
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
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 instantaneousrate of change of f at x

(as opposed to the averagerate 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.
Instantaneous Rates of Change
• The instantaneousrate 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 .
The Derivative of a Function
• The derivative of a function f with respect tox is the function f′′(read “f prime”).
• The domain of f′is the set of allx where the limit exists.
• Thus, the derivative of function fis a functionf ′that gives the slopeof 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.
The Derivative of a Function
• Four Step Process for Finding f′(x)
• Compute f(x + h).
• Form the difference f(x + h) –f(x).
• Form the quotient
• Compute
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

Examples
• Find the slope of the tangent line to the graph f(x) = x2at 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 = x2 + 2xh + h2.

Step 2.f(x + h) – f(x) = x2 + 2xh + h2 – x2 = h(2x + h).

Step 3.

Step 4.

Example 3, page 138

Examples
• Find the slope of the tangent line to the graph f(x) = x2at any point (x, f(x)).
• The slope of the tangent line is given byf ′(x) = 2x.
• Now, find and interpretf ′(2).

Solution

f ′(2) = 2(2) = 4.

y

• This means that, at the point(2, 4)…

… the slope of the tangent line to the graph is 4.

(2, 4)

Example 3, page 138

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

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

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

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

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 theslopeis at that point)

y

x

a

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.

y

x

a

b

Applied Example: Wages
• Mary works at the B&O department store, where, on a weekday, she is paid\$8 an hour for the first8 hours and \$12 an hour of overtime.
• The function

gives Mary’s earnings on a weekday in which she worked xhours.

• Sketch the graph of the function f and explain why it is not differentiable at x = 8.

Applied Example 8, page 143

Applied Example: Wages

Solution

• The graph of f has a corner at x = 8 and so is not differentiable at that point.

y

130

110

90

70

50

30

10

(8, 64)

x

2 4 6 8 10 12

Applied Example 8, page 143