Functions and Their Graphs

1
2
Functions, Limits and the Derivative

• Functions and Their Graphs
• The Algebra of Functions
• Functions and Mathematical Models
• Limits
• One-Sided Limits and Continuity
• The Derivative

2
2.1

• Functions and Their Graphs

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.

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).

5
Example

• Let the function f be defined by the rule
• Find f(1)
• Solution

Example 1, page 51
6
Example

• Let the function f be defined by the rule
• Find f( 2)
• Solution

Example 1, page 51
7
Example

• Let the function f be defined by the rule
• Find f(a)
• Solution

Example 1, page 51
8
Example

• Let the function f be defined by the rule
• Find f(a h)
• Solution

Example 1, page 51
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
• 1000 thermometers per week.
• 2000 thermometers per week.

Applied Example 2, page 51
10
Applied Example

• Solution
• We have
• The weekly profit by producing 1000 thermometers
  is
• or 2,000.
• The weekly profit by producing 2000 thermometers
  is
• or 7,000.

Applied Example 2, page 51
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.

12
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
16 2x
x
x
16
Applied Example 3, page 52
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.
• Find the expression that gives the volume V 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
14
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
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
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
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
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
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).

20
Example

• The graph of a function f is shown below

y
y
(x, y)
Range
x
x
Domain
Example 5, page 53
21
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
22
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
23
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
24
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
25
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

x y
3 10
2 5
1 2
0 1
1 2
2 5
3 10
Example 6, page 54
26
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 y
3 10
2 5
1 2
0 1
1 2
2 5
3 10
x
3 2 1 1 2 3
Example 6, page 54
27
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 y
3 10
2 5
1 2
0 1
1 2
2 5
3 10
x
3 2 1 1 2 3
Example 6, page 54
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
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

x y
3 3
2 2
1 1
0 0
1 1
2 1.41
3 1.73
Example 7, page 55
30
Example Sketching a Graph

• Sketch the graph of the function defined by the
  equation
• Solution
• Plotting these data and graphing we get

y
x y
3 3
2 2
1 1
0 0
1 1
2 1.41
3 1.73
3 2 1
x
3 2 1 1 2 3
Example 7, page 55
31
2.2
The Algebra of Functions
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.

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
33
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
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).

y
400 200 0 200 400
y D(t)
t
Billions of Dollars
t
1992 1994 1996 1998 2000
D(t)
Year
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.

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
  n 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
• The quotient f/g of f and g has domain A n B
  excluding all numbers x such that g(x) 0 and
  rule given by
• Quotient

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
  n B 1,?).
• The rules are as follows

Example 1, page 68
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 ? ½ .
• Thus, the domain is 1, ½) U ( ½,?).
• The rule is as follows

Example 1, page 68
39
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
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) 0.0001x2 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
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) 0.0005x2 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
42
Applied Example

• Solution
• 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)
• ( 0.0005x2 20x) ( 0.0001x2 10x
  10,000)
• 0.0004x2 10x 10,000
• 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
• or 50,000 per month.

Applied Example 3, page 69
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

44
The Composition of Two Functions

• Let f and g be functions.
• Then the composition of g and f is the function
  ggf (read g circle f ) defined by
• (ggf )(x) g(f(x))
• The domain of ggf is the set of all x in the
  domain of f such that f(x) lies in the domain of
  g.

45
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
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
  0.01x2/3 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 0.2t2 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 5 years from now.

Applied Example 5, page 70
47
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
48
Applied Example

• Solution
• Part (b)
• The level of CO five years from now is
• (ggf )(5) g(f(5)) 0.010.2(5)2 4(5)
  642/3
• (0.01)892/3 0.20
• or approximately 0.20 parts per million.

Applied Example 5, page 70
49
2.3
Functions and Mathematical Models
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)

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
Formulate
Solve
Test
Solution of real-world problem
Solution of mathematical model
Interpret
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
  a0, a1, . an are
  constants called the coefficients of the
  polynomial function.
• Examples
• The function below is polynomial function of
  degree 5

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
  a0, a1, . an are
  constants called the coefficients of the
  polynomial function.
• Examples
• The function below is polynomial function of
  degree 3

54
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.

Year 1999 2000 2001 2002 2003 2004
Market 12.07 14.07 16.21 18.28 20.00 21.72
Applied Example 1, page 76
55
Applied ExampleMarket for Cholesterol-Reducing
Drugs
Year 1999 2000 2001 2002 2003 2004
Market 12.07 14.07 16.21 18.28 20.00 21.72

• 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
56
Applied ExampleMarket for Cholesterol-Reducing
Drugs
Year 1999 2000 2001 2002 2003 2004
Market 12.07 14.07 16.21 18.28 20.00 21.72

• 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
57
Applied ExampleMarket for Cholesterol-Reducing
Drugs
Year 1999 2000 2001 2002 2003 2004
Market 12.07 14.07 16.21 18.28 20.00 21.72

• 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
58
Applied ExampleMarket for Cholesterol-Reducing
Drugs
Year 1999 2000 2001 2002 2003 2004
Market 12.07 14.07 16.21 18.28 20.00 21.72

• 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 1.95 billion per year.

Applied Example 1, page 76
59
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 gt 0
Opens downwards if a lt 0
y
y
x
x
60
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

Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380
Applied Example 2, page 78
61
Applied ExampleGlobal Warming
Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380

• 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
62
Applied ExampleGlobal Warming
Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380

• 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
63
Applied ExampleGlobal Warming
Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380

1. Use the model to estimate the average amount of
   atmospheric CO2 in 1980 (t 23).
2. Assume that the trend continued and use the model
   to predict the average amount of atmospheric CO2
   in 2010.

Applied Example 2, page 78
64
Applied ExampleGlobal Warming
Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380

• 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
65
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

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0
y (trillion)
6 4 2
t (years)
5 10 15 20 25 30
Applied Example 3, page 79
66
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

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0
y (trillion)
6 4 2
t (years)
5 10 15 20 25 30
Applied Example 3, page 79
67
Applied ExampleSocial Security Trust Fund Assets
Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0

1. The first baby boomers will turn 65 in 2011. What
   will be the assets of the Social Security trust
   fund at that time?
2. The last of the baby boomers will turn 65 in
   2029. What will the assets of the trust fund be
   at the time?
3. Use the graph of function A(t) to estimate the
   year in which the current Social Security system
   will go broke.

Applied Example 3, page 79
68
Applied ExampleSocial Security Trust Fund Assets
Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0

• 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
69
Applied ExampleSocial Security Trust Fund Assets
Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0

• Solution
• The graph shows that function A crosses 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
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)

71
Rational and Power Functions

• Examples of rational functions

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

73
Rational and Power Functions

• Many functions involve combinations of rational
  and power functions.
• Examples

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 5000 10,000 15,000 20,000
Cost/mile, y () 83.8 62.9 52.2 47.1
Applied Example 4, page 80
75
Applied Example Driving Costs
Miles/year, x 5000 10,000 15,000 20,000
Cost/mile, y () 83.8 62.9 52.2 47.1

• 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
76
Applied Example Driving Costs
Miles/year, x 5000 10,000 15,000 20,000
Cost/mile, y () 83.8 62.9 52.2 47.1

• 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.

Applied Example 4, page 80
77
Applied Example Driving Costs
Miles/year, x 5000 10,000 15,000 20,000
Cost/mile, y () 83.8 62.9 52.2 47.1

• 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.

Applied Example 4, page 80
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.

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.

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).

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
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
83
Applied Example Supply and Demand

• Solution
• The equilibrium price is given by
• or 40 per headset.

Applied Example 5, page 82
84
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 function f of
  one variable.
• Note any restrictions to be placed on the domain
  of f by the nature of the
  problem.

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
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 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
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
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)
• 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
89
2.4
Limits
90
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
91
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
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.

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) 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?

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.

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 gt 2. Then, the average
  velocity of the maglev over the time interval 2,
  t is given by

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
  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.

t 2.5 2.1 2.01 2.001 2.0001
Average Velocity 18 16.4 16.04 16.004 16.0004
97
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

t 2.5 2.1 2.01 2.001 2.0001
g(t) 18 16.4 16.04 16.004 16.0004
t 1.5 1.9 1.99 1.999 1.9999
g(t) 14 15.6 15.96 15.996 15.9996
98
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.

99
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.

100
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
  x sufficiently close to 2.
• Therefore,

f(x) x3
y
8 6 4 2 2
x
2 1 1 2 3
Example 1, page 101
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
  x sufficiently close to 1.
• Therefore,

y
g(x)
5 3 1
x
2 1 1 2 3
Example 2, page 101
102
Examples

• Let Evaluate
• Solution

• 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.

y
5
x
2 1 1 2
Example 3b, page 101
103
Theorem 1Properties of Limits

• Suppose and
• Then,
• r, a real number
• c, a real number
• Provided that M ? 0

104
Examples

• Use theorem 1 to evaluate the following limits

Example 4, page 102
105
Examples

• Use theorem 1 to evaluate the following limits

Example 4, page 102
106
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.

107
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.

108
Examples

• Evaluate
• Solution
• As weve seen, here we have an indeterminate form
  0/0.
• 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
109
Examples

• Evaluate
• Solution
• Notice in the graphs below that the two functions
  yield the same graphs, except for the value x 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
110
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
111
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.

y
400 300 200 100
x
10 20 30 40 50 60
112
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

x 1 2 5 10 100 1000
f(x) 1 1.6 1.92 1.98 1.9998 1.999998
113
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.

114
Examples

• Let
• Evaluate and
• Solution
• Graphing f(x) reveals that

y
1 1
x
3 3
Example 7, page 107
115
Examples

• Let
• Evaluate and
• Solution
• Graphing g(x) reveals that

y
x
3 2 1 1 2 3
Example 7, page 107
116
Theorem 2Properties of Limits

• 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

• For all n gt 0, and
• provided that is defined.

117
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 x3

Example 8, page 108
118
Examples

• Evaluate
• Solution
• Again, we see that property 5 does not apply.
• So we divide numerator and denominator by x2

Example 9, page 108
119
Examples

• Evaluate
• Solution
• Again, we see that property 5 does not apply.
• But dividing numerator and denominator by x2 does
  not help in this case
• In other words, the limit does not exist.
• We indicate this by writing

Example 10, page 109
120
2.5
One-Sided Limits and Continuity
121
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.

y
1 1
x
1 1
122
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

y
1 1
x
1 1
123
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

y
1 1
x
1 1
124
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.

125
Theorem 3Properties of Limits

• The connection between one-side limits and the
  two-sided limit defined earlier is given by the
  following theorem.

• 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

126
Examples

• Show that exists by studying the one-sided
• limits of f as x approaches 0
• Solution
• For x gt 0, we find
• And for x 0, we find
• Thus,

y
2 1
x
2 1 1 2
Example 1, page 118
127
Examples

• Show that does not exist.
• Solution
• For x lt 0, we find
• And for x ? 0, we find
• Thus, does
• not exist.

y
1 1
x
Example 1, page 118
128
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 ).

y
x
a
129
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.

y
x
a
b
130
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.

y
x
b
c
131
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.

y
x
c
d
132
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 at x a.
• Also, f is continuous on an interval if f is
  continuous at every number in the interval.

133
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
134
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
135
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
136
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.

y
1 1
x
Example 2, page 120
137
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.

y
1
x
Example 2, page 120
138
Properties of Continuous Functions

• The constant function f(x) c is continuous
  everywhere.
• The identity function f(x) x is continuous
  everywhere.
• If f and g are continuous at x 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.

139
Properties of Continuous Functions

• Using these properties, we can obtain the
  following additional properties.
• A polynomial function y P(x) is continuous at
  every value of x.
• A rational function R(x) p(x)/q(x) is
  continuous at every value of x where q(x) ? 0.

140
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
141
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
142
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
143
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.

144
Intermediate Value Theorem

• Mathematically, recall that the position of the
  maglev is a function of time given by f(t) 4t2
  for 0 ? t ? 30

y
s2 s3 s1
t
t1
t3
t2

• Suppose the position of the maglev is s1 at some
  time t1 and its position is s2 at some time t2.
• Then, if s3 is any number between s1 and s2,
  there must be at least one t3 between t1 and t2
  giving the time at which the maglev is at s3
  (f(t3) s3).

145
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.

y
y
f(b) M f(a)
f(b) M f(a)
x
x
c1
c2
c3
b
a
a
c
b
146
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).

y
y
f(b) f(a)
f(b) f(a)
x
x
c2
c3
b
a
c1
a
c
b
147
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 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 lt c
  lt 1 such that f(c) 0.

Example 5, page 124
148
2.6
The Derivative
149
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
150
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)
60 40 20
t (sec)
1 2 3 4
151
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.
A
The tangent line has the same slope as the curve
does at point A.
x
152
Slopes of Lines and of Curves

• The slope of a point in a curve is given by the
  slope of the tangent 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
153
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
Slope 1.8
A
x
154
Slopes of Lines and of Curves

• To calculate accurately the slope of a tangent to
  a curve, we must make the change in x as small as
  possible

y
Slope 1.8
Dy 3
As we let Dx get smaller, the slope of the secant
becomes more and more similar to the slope of the
tangent to the curve at that point.
A
Dx 4
x
155
Slopes of Lines and of Curves

• To calculate accurately the slope of a tangent to
  a curve, we must make the change in x as small as
  possible

y
Slope 1.8
Dy 2.4
As we let Dx get smaller, the slope of the secant
becomes more and more similar to the slope of the
tangent to the curve at that point.
A
Dx 3
x
156
Slopes of Lines and of Curves

• To calculate accurately the slope of a tangent to
  a curve, we must make the change in x as small as
  possible

y
Slope 1.8
Dy 2.2
As we let Dx get smaller, the slope of the secant
becomes more and more similar to the slope of the
tangent to the curve at that point.
A
Dx 2
x
157
Slopes of Lines and of Curves

• To calculate accurately the slope of a tangent to
  a curve, we must make the change in x as small as
  possible

y
Slope 1.8
Dy 1.5
As we let Dx get smaller, the slope of the secant
becomes more and more similar to the slope of the
tangent to the curve at that point.
A
Dx 1
x
158
Slopes of Lines and of Curves

• To calculate accurately the slope of a tangent to
  a curve, we must make the change in x as small as
  possible

y
Slope