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
2.1
Functions and Their Graphs
y
4
3
(5, 3)
2
1
1
–1
–2
2
3
4
5
6
7
8
x
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.
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).
Example
 Let the function f be defined by the rule
f  x   2 x2  x  1
✦ Find: f(1)
✦ Solution:
f 1  2 1  1  1  2  1  1  2
2
Example 1, page 51
Example
 Let the function f be defined by the rule
f  x   2 x2  x  1
✦ Find: f( – 2)
✦ Solution:
f  2   2  2    2   1  8  2  1  11
2
Example 1, page 51
Example
 Let the function f be defined by the rule
f  x   2 x2  x  1
✦ Find: f(a)
✦ Solution:
f  a   2  a    a   1  2a 2  a  1
2
Example 1, page 51
Example
 Let the function f be defined by the rule
f  x   2 x2  x  1
✦ Find: f(a + h)
✦ Solution:
f  a  h   2  a  h    a  h   1  2a 2  4ah  2h 2  a  h  1
2
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
P  x   0.001x 2  8x  5000
 Find ThermoMaster’s weekly profit if its level of
production is:
a. 1000 thermometers per week.
b. 2000 thermometers per week.
Applied Example 2, page 51
Applied Example
Solution
 We have
P  x   0.001x 2  8x  5000
a. The weekly profit by producing 1000 thermometers is
P 1000   0.001 1000   8 1000   5000  2000 =
2
or $2,000.
b. The weekly profit by producing 2000 thermometers is
P  2000   0.001  2000   8  2000   5000  7000 =
2
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
16
Applied Example 3, page 52
x
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.
 The dimensions of the
resulting box are:
x
10 – 2x
16 – 2x
a. Find the expression that gives the volume V of the box as
a function of x.
b. What is the domain of the function?
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:
V  f  x   16  2 x   10  2 x   x
 160  52 x  4 x 2  x
 4 x 3  52 x 2  160 x
x
10 – 2x
Applied Example 3, page 52
16 – 2x
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
16  2 x  0
10  2 x  0
x0
must be satisfied simultaneously. Simplified:
x 8
x 5
x  0=
All three are satisfied simultaneously provided that:
0 x5
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:
f  x  x 1
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:
1
f  x  2
x 4
Solution
 Our only constraint is that you cannot divide by zero, so
x2  4  0
 Which means that
x 2  4   x  2 x  2  0
 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:
f  x   x2  3
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 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).
Example
 The graph of a function f is shown below:
y
y
(x, y)
Range
x
Domain
Example 5, page 53
x
Example
 The graph of a function f is shown below:
✦ What is the value of f(2)?
y
4
3
2
1
1
2
3
–1
–2
Example 5, page 53
(2, –2)
4
5
6
7
8
x
Example
 The graph of a function f is shown below:
✦ What is the value of f(5)?
y
4
3
(5, 3)
2
1
1
–1
–2
Example 5, page 53
2
3
4
5
6
7
8
x
Example
 The graph of a function f is shown below:
✦ What is the domain of f(x)?
y
4
3
2
1
1
2
3
4
5
6
–1
–2
Domain: [1,8]
Example 5, page 53
7
8
x
Example
 The graph of a function f is shown below:
✦ What is the range of f(x)?
y
4
3
2
Range:
[–2,4]
1
1
–1
–2
Example 5, page 53
2
3
4
5
6
7
8
x
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
–3
–2
–1
0
1
2
3
Example 6, page 54
y
10
5
2
1
2
5
10
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
x
–3
–2
–1
0
1
2
3
Example 6, page 54
y
10
5
2
1
2
5
10
10
8
6
4
2
–3
–2
–1
1
2
3 x
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
x
–3
–2
–1
0
1
2
3
Example 6, page 54
y
10
5
2
1
2
5
10
10
8
6
4
2
–3
–2
–1
1
2
3 x
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
  x
f  x  
 x
if x  0
if x  0
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
f  x  x
 In the subdomain [0, ), the rule for f is given by
f  x  x
Example 7, page 55
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
  x
f  x  
 x
if x  0
if x  0
Solution
 Substituting negative values for x into f  x    x, while
substituting zero and positive values into f  x   x we get:
x
–3
–2
–1
0
1
2
3
y
3
2
1
0
1
1.41
1.73
Example 7, page 55
Example: Sketching a Graph
 Sketch the graph of the function defined by the equation
  x
f  x  
 x
if x  0
if x  0
Solution
 Plotting these data and graphing we get:
x
–3
–2
–1
0
1
2
3
y
3
2
1
0
1
1.41
1.73
Example 7, page 55
y
f  x  x
3
f  x  x
2
1
–3
–2
–1
1
2
3
x
2.2
The Algebra of Functions
y
Billions of Dollars
2000
y = R(t)
1800
y = S(t)
1600
S(t)
1400
R(t)
1200
1000
1990
1992
1994
t 1996
Year
1998
2000
t
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
Billions of Dollars
2000
y = R(t)
1800
y = S(t)
1600
S(t)
1400
R(t)
1200
1000
1990
1992
1994
t
Year
1996
1998
2000
t
The Sum, Difference, Product and Quotient
of Functions
 Consider the graph below:
✦ The difference R(t) – S(t) gives the budget deficit (if negative)
or surplus (if positive) in billions of dollars at any time t.
y
Billions of Dollars
2000
y = R(t)
1800
y = S(t)
1600
S(t)
D(t) = R(t) – S(t)
1400
R(t)
1200
1000
1990
1992
1994
t
Year
1996
1998
2000
t
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
Billions of Dollars
400
200
0
D(t)
–200
–400
y = D(t)
1992
1994
t 1996
Year
1998
2000
t
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 sum f + g, the difference f – g, and the product fg of f
and g are functions with domain A ∩ B and rule given by
(f + g)(x) = f(x) + g(x)
Sum
(f – g)(x) = f(x) – g(x)
(fg)(x) = f(x)g(x)
Difference
Product
 The quotient f/g of f and g has domain A ∩ B excluding all
numbers x such that g(x) = 0 and rule given by
 f 
f ( x)
 g   x   g ( x)
 
Quotient
Example
 Let f ( x ) 
x  1 and g(x) = 2x + 1.
 Find the sum s, the difference d, the product p, and the
quotient q of the functions f and g.
Solution
 Since the domain of f is A = [–1,) and the domain of g
is B = (– , ), we see that the domain of s, d, and p is
A ∩ B = [–1,).
 The rules are as follows:
s( x )  ( f  g )( x )  f ( x )  g ( x )  x  1  2 x  1
d ( x )  ( f  g )( x )  f ( x )  g ( x )  x  1  2 x  1
p( x )  ( fg )( x )  f ( x ) g ( x )   2 x  1 x  1
Example 1, page 68
Example
 Let f ( x ) 
x  1 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:
 f 
f ( x)
x 1
q( x )    ( x ) 

g ( x) 2 x  1
g
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 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
a. The total profit function for Puritron.
b. The total profit when Puritron produces 10,000 filters per
month.
Applied Example 3, page 69
Applied Example
Solution
a. The total profit P realized by the firm is the difference
between the total revenue R and the total cost C:
P(x) = R(x) – C(x)
= (– 0.0005x2 + 20x) – (– 0.0001x2 + 10x + 10,000)
= – 0.0004x2 + 10x – 10,000
b. The total profit realized by Puritron when producing 10,000
filters per month is
P(x) = – 0.0004(10,000)2 + 10(10,000) – 10,000
= 50,000
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:
g ( x)  x
f ( x)  x 2  1
 Evaluating the function g at the point f(x), we find that:
g  f ( x)  
f ( x)  x 2  1
 This is an entirely new function, which we could call h:
h( x )  x 2  1
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.
Example
 Let f ( x )  x  1 and g ( x ) 
2
x  1.
 Find:
a. The rule for the composite function ggf.
b. The rule for the composite function fgg.
Solution
 To find ggf, evaluate the function g at f(x):
( g f )( x )  g ( f ( x )) 
f ( x)  1  x 2  1  1
 To find fgg, evaluate the function f at g(x):
( f g )( x )  f ( g ( x ))  ( g ( x ))2  1  ( x  1)2  1
 x  2 x 11  x  2 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 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:
a. An expression for the concentration of CO in the air due
to automobile exhaust t years from now.
b. The level of concentration 5 years from now.
Applied Example 5, page 70
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
y ($trillion)
6
4
2
5
10
15
20
25
30
t (years)
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
Solve
Test
Solution of realworld problem
Mathematical
model
Interpret
Solution of
mathematical model
Modeling With Polynomial Functions
 A polynomial function of degree n is a function of the form
f ( x )  an x n  an 1 x n 1    a2 x 2  a1 x  a0
(an  0)
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:
f ( x)  2 x5  3x 4  12 x 3  2 x 2  6
Modeling With Polynomial Functions
 A polynomial function of degree n is a function of the form
f ( x )  an x n  an 1 x n 1    a2 x 2  a1 x  a0
(an  0)
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:
g ( x )  0.001x 3  0.2 x 2  10 x  200
Applied Example
Market for Cholesterol-Reducing Drugs
 In a study conducted in early 2000, experts projected a
rise in the market for cholesterol-reducing drugs.
 The U.S. market (in billions of dollars) for such drugs
from 1999 through 2004 was
Year
1999
2000
2001
2002
2003
2004
Market
12.07
14.07
16.21
18.28
20.00
21.72
 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 Example
Market 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
a. Sketch the graph of the function M and the given data
on the same set of axes.
b. Assuming that the projection held and the trend
continued, what was the market for cholesterolreducing drugs in 2005 (t = 6)?
c. What was the rate of increase of the market for
cholesterol-reducing drugs over the period in question?
Applied Example 1, page 76
Applied Example
Market 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
a. Graph:
y
Billions of Dollars
25
M(t)
20
15
1
Applied Example 1, page 76
2
3
Year
4
5
t
Applied Example
Market 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
b. 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 Example
Market 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
c. The function M is linear, and so we see that the rate of
increase of the market for cholesterol-reducing drugs is
given by the slope of the straight line represented by M,
which is approximately $1.95 billion per year.
Applied Example 1, page 76
Modeling a Polynomial Function of Degree 2
 A polynomial function of degree 2 has the form
f ( x )  a2 x 2  a1 x  a0
( a2  0)
 Or more simply, y = ax2 + bx + c, and is called a
quadratic function.
 The graph of a quadratic function is a parabola:
y
Opens upwards if
a>0
y
x
Opens downwards if
a<0
x
Applied Example
Global 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
Applied Example 2, page 78
325
330
335
345
355
365
375
380
Applied Example
Global 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:
380
y (ppmv)
360
340
320
Applied Example 2, page 78
10
20
30
40
50
t (years)
Applied Example
Global 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:
380
y (ppmv)
A(t )  0.01076t 2  0.8212t  313.4
360
340
320
Applied Example 2, page 78
10
20
30
40
50
t (years)
Applied Example
Global Warming
Year
1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount
315
325
330
335
345
355
A(t )  0.010716t 2  0.8212t  313.4
365
375
380
(1  t  50)
a. Use the model to estimate the average amount of atmospheric
CO2 in 1980 (t = 23).
b. 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 Example
Global Warming
Year
1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount
315
325
330
335
345
355
A(t )  0.010716t 2  0.8212t  313.4
365
375
380
(1  t  50)
Solution
a. The average amount of atmospheric CO2 in 1980 is given by
A(23)  0.010716  23  0.8212  23  313.4  337.96
2
or approximately 338 ppmv.
b. Assuming that the trend will continue, the average amount of
atmospheric CO2 in 2010 will be
A(53)  0.010716  53  0.8212  53  313.4  387.03
2
Applied Example 2, page 78
Applied Example
Social Security Trust Fund Assets
 The projected assets of the Social Security trust fund (in
trillions of dollars) from 2008 through 2040 are given by:
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
 The scatter plot associated with these data is:
y ($trillion)
6
4
2
Applied Example 3, page 79
5
10
15
20
25
30
t (years)
Applied Example
Social Security Trust Fund Assets
 The projected assets of the Social Security trust fund (in
trillions of dollars) from 2008 through 2040 are given by:
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
 A mathematical model giving the approximate value of assets
in the trust fund (in trillions of dollars) is:
y ($trillion)
A(t )  0.00000324t 4  0.000326t 3  0.00342t 2  0.254t  2.4
6
4
2
Applied Example 3, page 79
5
10
15
20
25
30
t (years)
Applied Example
Social Security Trust Fund Assets
Year
Assets
2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
2.4
3.2
4.0
4.7
5.3
5.7
5.9
5.6
4.9
3.6
1.7
0
A(t )  0.00000324t 4  0.000326t 3  0.00342t 2  0.254t  2.4
a. The first baby boomers will turn 65 in 2011. What will be the
assets of the Social Security trust fund at that time?
b. The last of the baby boomers will turn 65 in 2029. What will
the assets of the trust fund be at the time?
c. Use the graph of function A(t) to estimate the year in which
the current Social Security system will go broke.
Applied Example 3, page 79
Applied Example
Social Security Trust Fund Assets
Year
Assets
2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
2.4
3.2
4.0
4.7
5.3
5.7
5.9
5.6
4.9
3.6
1.7
0
A(t )  0.00000324t 4  0.000326t 3  0.00342t 2  0.254t  2.4
Solution
a. The assets of the Social Security fund in 2011 (t = 3) will be:
A(3)  0.00000324  3  0.000326  3  0.00342  3  0.254  3  2.4  3.18
4
3
2
or approximately $3.18 trillion.
The assets of the Social Security fund in 2029 (t = 21) will be:
A(21)  0.00000324  21  0.000326  21  0.00342  21  0.254  21  2.4  5.59
4
3
or approximately $5.59 trillion.
Applied Example 3, page 79
2
Applied Example
Social Security Trust Fund Assets
Year
2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets
2.4
3.2
4.0
4.7
5.3
5.7
5.9
5.6
4.9
3.6
1.7
0
A(t )  0.00000324t 4  0.000326t 3  0.00342t 2  0.254t  2.4
Solution
b. The graph shows that function A crosses the t-axis at about
t = 32, suggesting the system will go broke by 2040:
y ($trillion)
6
4
Trust runs
out of funds
2
Applied Example 3, page 79
5
10
15
20
25
30
t (years)
Rational and Power Functions
 A rational function is simply the quotient of two
polynomials.
 In general, a rational function has the form
R( x) 
f ( x)
g ( x)
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:
3x 3  x 2  x  1
F ( x) 
x2
x2  1
G( x)  2
x 1
Rational and Power Functions
 Functions of the form
f ( x)  x r
where r is any real number, are called power functions.
 We encountered examples of power functions earlier in
our work.
 Examples of power functions:
f ( x)  x  x
1/2
and
1
g ( x )  2  x 2
x
Rational and Power Functions
 Many functions involve combinations of rational and
power functions.
 Examples:
1  x2
f ( x) 
1  x2
g ( x )  x 2  3x  4
h( x )  (1  2 x )
1/2
1
 2
( x  2)3/2
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:
Miles/year, x
5000
10,000
15,000
20,000
Cost/mile, y (¢)
83.8
62.9
52.2
47.1
 A mathematical model giving the average cost in cents per
mile is:
C ( x) 
164.8
x 0.42
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
Miles/year, x
5000
10,000
15,000
20,000
Cost/mile, y (¢)
83.8
62.9
52.2
47.1
164.8
C ( x )  0.42
x
 Below is the scatter plot associated with this data:
140
120
100
80
60
40
20
y (¢)
Applied Example 4, page 80
C(x)
5
10
15
20
25
t (years)
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
164.8
C ( x )  0.42
x
 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
C (8) 
164.8
 8
0.42
or approximately 68.8¢/mile.
Applied Example 4, page 80
 68.81
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
164.8
C ( x )  0.42
x
 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
C (18) 
164.8
18
0.42
 48.95
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:
✦ The higher the price the less they want to buy of it.
✦ A demand function p = 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:
✦ The higher the price, the more they want to produce of it.
✦ A supply function p = 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 curve intersect: where d(x) = s(x).
Applied Example: Supply and Demand
 The demand function for a certain brand of bluetooth
wireless headset is given by
p  d ( x )  0.025x 2  0.5x  60
 The corresponding supply function is given by
p  s( x )  0.02 x 2  0.6 x  20
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:
p  0.025x 2  0.5x  60
p  0.02 x 2  0.6 x  20
 Substituting the second equation into the first yields:
0.02 x 2  0.6 x  20  0.025 x 2  0.5 x  60
0.045 x 2  1.1x  40  0
45 x 2  1100 x  40,000  0
9 x 2  220 x  8,000  0
 9 x  400  x  20  0
 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:
p  0.02  20   0.6  20   20  40
2
or $40 per headset.
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:
1. Assign a letter to each variable mentioned in the
problem. If appropriate, draw and label a figure.
2. Find an expression for the quantity sought.
3. Use the conditions given in the problem to write
the quantity sought as a function f of one variable.
Note any restrictions to be placed on the domain
of f by the nature of the problem.
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
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 $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
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
y
400
f ( x)
300
200
100
10 20 30 40 50 60
x
Introduction to Calculus
 Historically, the development of calculus by Isaac Newton
and Gottfried W. Leibniz resulted from the investigation
of the following problems:
1. Finding the tangent line to a curve at a given point on the
curve:
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:
2. 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 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?
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:
Distance covered f (4)  f (2)

Time elapsed
42
4(42 )  4(2 2 )

2
64  16

2
 24
or 24 feet/second.
 This is not the velocity of the maglev at exactly t = 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
Distance covered f (t )  f (2)

Time elapsed
t2
4(t 2 )  4(22 )

t2
4(t 2  4)

t2
Example: A Speeding Maglev
4(t 2  4)
Average velocity 
t2
 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:
t
2.5
2.1
2.01
2.001
2.0001
Average Velocity
18
16.4
16.04
16.004
16.0004
 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.
Intuitive Definition of a Limit
 Consider the function g, which gives the average velocity
of the maglev:
4(t 2  4)
g (t ) 
t 2
 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:
t
2.5
2.1
2.01
2.001
2.0001
g(t)
18
16.4
16.04
16.004
16.0004
✦ Similarly, we take values of t approaching 2 from the left,
and we find that g(t) also approaches 16:
t
1.5
1.9
1.99
1.999
1.9999
g(t)
14
15.6
15.96
15.996
15.9996
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
4(t 2  4)
lim
g (t )  lim
 16
t 2
t 2
t2
 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 function f has a limit L as x approaches a, written
lim
f ( x)  L
x a
 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.
Examples
 Let f(x) = x3. Evaluate
lim f ( x ).
x 2
Solution
 You can see in the graph that
f(x) can be as close
8 as we please by taking
x sufficiently close to 2.
 Therefore,
f(x) = x3
y
to
8
6
4
2
lim x 3  8
x2
–2
–1
1
–2
Example 1, page 101
2
3
x
Examples
x  2
 Let g ( x )  
1
if x  1
Evaluate lim g ( x ).
x1
if x  1
Solution
 You can see in the graph
y
that g(x) can be as close
to 3 as we please by taking
x sufficiently close to 1.
 Therefore,
g(x)
5
3
1
lim g ( x )  3
x1
–2
Example 2, page 101
–1
1
2
3
x
Examples
1
 Let f ( x )  2
x
Evaluate lim f ( x ).
x0
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.
Example 3b, page 101
y
5
f ( x) 
–2
–1
1
2
1
x2
x
Theorem 1
Properties of Limits
Suppose
Then,
1.
lim
f ( x)  L
x a
and
lim g ( x )  M
xa
r
lim
f ( x)  lim f ( x)   Lr

xa
 xa

r
r, a real number
2. lim cf ( x )  c lim f ( x )  cL
xa
c, a real number
xa
3. lim  f ( x )  g ( x )  lim f ( x )  lim g ( x )  L  M
xa
xa
xa
4.
lim
f ( x) g ( x)  lim f ( x)  lim g ( x)  LM

xa
 xa
  xa

5.
f ( x) L
f ( x ) lim
x

a
lim


x a g ( x )
lim g ( x ) M
x a
Provided that M ≠ 0
Examples
 Use theorem 1 to evaluate the following limits:
3
lim
x   lim x   23  8
x2
x2
3


3/2
lim
5x  5  lim x   5(4)3/2  40
x4
x4
3/2


4
lim
(5x  2)  5 lim x   lim 2  5(1)4  2  3
x1
4
Example 4, page 102
 x1 
x1
Examples
 Use theorem 1 to evaluate the following limits:
lim
2x x  7  2  lim x 
 x3 
x3
3
2
3
lim x 2  7  2(3)3 (3)2  7  216
x3
2
lim
(2
x
 1) 2(2)2  1 9
2x  1
 x2

 3
lim
x2
lim ( x  1)
2 1
3
x 1
2
x2
Example 4, page 102
Indeterminate Forms
4( x 2  4)
 Let’s consider lim
x2 x  2
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
2
lim
4(
x
 4)
4( x  4) x2
0
lim


x 2 x  2
lim
x2
0
x 2
2
 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.
Strategy for Evaluating Indeterminate Forms
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.
1.
Examples
4( x 2  4)
 Evaluate lim
x2 x  2
Solution
 As we’ve seen, here we have an indeterminate form 0/0.
 We can rewrite
4( x 2  4) 4( x  2)( x  2)

 4( x  2)
x2
x2
 Thus, we can say that
x≠2
4( x 2  4)
lim
 lim 4( x  2)  16
x2 x  2
x2
 Note that 16 is the same value we obtained for the maglev
example through approximation.
Example 5, page 104
Examples
4( x 2  4)
 Evaluate lim
x2 x  2
Solution
 Notice in the graphs below that the two functions yield the
same graphs, except for the value x = 2:
4( x 2  4)
f ( x) 
x2
y
20
–3
–2
–1
Example 5, page 104
20
16
16
12
12
8
8
4
4
1
2
3
x
g ( x )  4( x  2)
y
–3
–2
–1
1
2
3
x
Examples
 Evaluate
1 h 1
lim
h 0
h
Solution
 As we’ve seen, here we have an indeterminate form 0/0.
 We can rewrite (with the constraint that h ≠ 0):
1 h 1
1 h 1 1 h 1
h
1




h
h
1  h  1 h( 1  h  1)
1 h 1
 Thus, we can say that
1 h 1
1
1
1
lim
 lim


h0
h

0
h
1 h 1
1 1 2
Example 6, page 105
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
y
a horizontal asymptote.
400
 In this case, we can say
f ( x)
that
300
lim f ( x )  400
x
and we call this a limit
of a function at infinity.
200
100
10
20
30 40
50
60
x
Example
2x2
 Consider the function f ( x ) 
1  x2
 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:
x
1
2
5
10
100
1000
f(x)
1
1.6
1.92
1.98
1.9998
1.999998
 As x gets larger and larger, f(x) gets closer and closer to 2.
 Thus, we can say that
2 x2
lim
2
x 1  x 2
Limit of a Function at Infinity
 The function f has the limit L as x increases without
bound (as x approaches infinity), written
lim f ( x )  L
x
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
lim f ( x )  M
x
if f(x) can be made arbitrarily close to M by taking x
large enough in absolute value.
Examples
 1
 Let f ( x )  
 1
 Evaluate
if x  0
if x  0
lim f ( x )
lim f ( x ) and
x
x 
Solution
 Graphing f(x) reveals that
y
lim f ( x )  1
x 
f ( x)
1
lim f ( x )  1
x
–3
3
–1
Example 7, page 107
x
Examples
1
 Let g ( x )  2
x
 Evaluate
lim g ( x ) and
lim g ( x )
x 
x 
Solution
y
 Graphing g(x) reveals that
1
g ( x)  2
x
lim g ( x )  0
x
lim g ( x )  0
x 
–3
Example 7, page 107
–2
–1
1
2
3
x
Theorem 2
Properties 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 > 0,
lim
x 
1
0
n
x
and
1
provided that n is defined.
x
lim
x
1
0
n
x
Examples
x2  x  3
 Evaluate lim
x
2 x3  1
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:
1 1
3


2
3
( x 2  x  3) / x 3
000 0
x
x
x
lim
 lim

 0
x (2 x 3  1) / x 3
x
1
20
2
2 3
x
Example 8, page 108
Examples
3x 2  8 x  4
 Evaluate lim
x 2 x 2  4 x  5
Solution
 Again, we see that property 5 does not apply.
 So we divide numerator and denominator by x2:
8 4
 2
(3x  8 x  4) / x
x x  3 00  3
lim

lim
x (2 x 2  4 x  5) / x 2
x
4 5
2  2 200 2
x x
2
Example 9, page 108
2
3
Examples
2 x 3  3x 2  1
 Evaluate lim
x x 2  2 x  4
Solution
 Again, we see that property 5 does not apply.
 But dividing numerator and denominator by x2 does not
help in this case:
1
2
(2 x  3x  1) / x
x
lim
 lim
x ( x 2  2 x  4) / x 2
x
2 4
1  2
x x
3
2
2
2x  3 
 In other words, the limit does not exist.
 We indicate this by writing
2 x 3  3x 2  1
lim

2
x x  2 x  4
Example 10, page 109
2.5
One-Sided Limits and Continuity
y
a
b
c
d
x
One-Sided Limits
x 1
x 1
 Consider the function f ( x )  
if x  0
if x  0
 Its graph shows that
y
f does not have a limit
as x approaches zero,
because approaching
from each side results
in different values.
f ( x)
1
–1
1
–1
x
One-Sided Limits
x 1
x 1
 Consider the function f ( x )  
if x  0
if x  0
 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
lim f ( x )  1
x0
y
f ( x)
1
–1
1
–1
x
One-Sided Limits
x 1
x 1
 Consider the function f ( x )  
if x  0
if x  0
 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
lim f ( x )  1
x0
y
f ( x)
1
–1
1
–1
x
One-Sided Limits
 The function f has the right-hand limit L as x
approaches from the right, written
lim f ( x )  L
x a 
 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
lim f ( x )  L
x a 
 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 3
Properties 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
lim
f ( x)  L
xa
if and only if
lim f ( x )  lim f ( x )  L
xa 
xa 
Examples
 Show that lim f ( x ) exists by studying the one-sided
x 0
limits of f as x approaches 0:
 x
f ( x)  
  x
if x  0
if x  0
Solution
 For x > 0, we find
y
lim f ( x )  0
x0
 And for x ≤ 0, we find
2
lim f ( x )  0
1
x0
 Thus,
lim
f ( x)  0
x 0
Example 1, page 118
f ( x)
–2
–1
1
2
x
Examples
 Show that lim g ( x) does not exist.
x0
1
g ( x)  
 1
if x  0
if x  0
Solution
 For x < 0, we find
y
lim g ( x)  1
x 0
 And for x  0, we find
g ( x)
1
lim
g ( x)  1
x0
 Thus,
lim
g ( x) does
x0
not exist.
Example 1, page 118
x
–1
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
y
a
x
 This function is discontinuous at the following points:
✦ At x = a, f is not defined (x = a is not in the domain of f ).
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
y
a
b
x
 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.
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
y
b
c
x
 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.
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
y
c
d
x
 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.
Continuity of a Function at a Number
 A function f is continuous at a number x = a
if the following conditions are satisfied:
1. f(a) is defined.
2. lim f ( x ) exists.
xa
3. lim f ( x )  f ( a )
x a
 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.
Examples
 Find the values of x for which the function is continuous:
f ( x)  x  2
Solution
 The function f is continuous everywhere because the three
conditions for continuity are satisfied for all values of x.
y
5
f ( x)  x  2
4
3
2
1
–2
Example 2, page 120
–1
x
1
2
Examples
 Find the values of x for which the function is continuous:
x2  4
g ( x) 
x2
Solution
 The function g is discontinuous at x = 2 because g is not
defined at that number. It is continuous everywhere else.
y
x2  4
g ( x) 
x2
5
4
3
2
1
–2
Example 2, page 120
–1
x
1
2
Examples
 Find the values of x for which the function is continuous:
x  2
h( x )  
1
if x  2
if x  2
Solution
 The function h is continuous everywhere except at x = 2
where it is discontinuous because
h(2)  1  lim h( x )  4
x 2
y
y  h( x )
5
4
3
2
1
Example 2, page 120
–2
–1
x
1
2
Examples
 Find the values of x for which the function is continuous:
 1 if x  0
F ( x)  
 1 if x  0
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
y  F ( x)
1
x
–1
Example 2, page 120
Examples
 Find the values of x for which the function is continuous:
 1
if x  0

G( x )   x

 1 if x  0
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
y  G ( x)
x
–1
Example 2, page 120
Properties of Continuous Functions
1. The constant function f(x) = c is continuous everywhere.
2. The identity function f(x) = x is continuous everywhere.
If f and g are continuous at x = a, then
3. [f(x)]n, where n is a real number, is continuous at
x = a whenever it is defined at that number.
4. f ± g is continuous at x = a.
5. fg is continuous at x = a.
6. f /g is continuous at g(a) ≠ 0.
Properties of Continuous Functions
 Using these properties, we can obtain the following
additional properties.
1. A polynomial function y = P(x) is continuous at every
value of x.
2. A rational function R(x) = p(x)/q(x) is continuous at
every value of x where q(x) ≠ 0.
Examples
 Find the values of x for which the function is continuous.
f ( x)  3x 3  2 x 2  x  10
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
Examples
 Find the values of x for which the function is continuous.
8 x10  4 x 2  1
g ( x) 
x2  1
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
Examples
 Find the values of x for which the function is continuous.
4 x 3  3x 2  1
h( x )  2
x  3x  2
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) = 4t2 for 0  t  30:
y
y  4t 2
s2
s3
s1
t1
t3
t2
t
 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).
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)
f(b)
y  f ( x)
y  f ( x)
M
M
f(a)
f(a)
a
c
b
x
a c1
c2
c3
b
x
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(b)
y  f ( x)
y  f ( x)
a
f(a)
c
b
x
a c1
f(a)
c2
c3
b
x
Example
 Let f(x) = x3 + x + 1.
a. Show that f is continuous for all values of x.
b. Compute f(–1) and f(1) and use the results to deduce that
there must be at least one number x = c, where c lies in the
interval (–1, 1) and f(c) = 0.
Solution
a. The function f is a polynomial function of degree 3 and is
therefore continuous everywhere.
b. f (–1) = (–1)3 + (–1) + 1 = –1 and f (1) = (1)3 + (1) + 1 = 3
Since f (–1) and f (1) have opposite signs, Theorem 5 tells
us that there must be at least one number x = c with
–1 < c < 1 such that f(c) = 0.
Example 5, page 124
2.6
The Derivative
y
f(x + h)
f(x + h) – f(x)
f(x)
A
h
x
x+h
x
An Intuitive Example
 Consider the maglev example from Section 2.4.
 The position of the maglev is a function of time given by
(0  t  30)
s = f(t) = 4t2
where s is measured in feet and t in seconds.
 Its graph is:
s (ft)
s  f (t )
60
40
20
1
2
3
4
t (sec)
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)
s  f (t )
60
40
20
1
2
3
4
t (sec)
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
A
Suppose we want to find the
slope at point A.
The tangent line has the
same slope as the curve does
at point A.
x
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
y = 1.8
A
x = 1
The slope of the tangent in
this case is 1.8:
y 1.8
Slope 

 1.8
x
1
x
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
Slope 
y 3
  0.75
x 4
y = 3
A
x = 4
As we let x get smaller, the
slope of the secant becomes
more and more similar to the
slope of the tangent to the
curve at that point.
x
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
Slope 
y 2.4

 0.8
x
3
y = 2.4
A
x = 3
As we let x get smaller, the
slope of the secant becomes
more and more similar to the
slope of the tangent to the
curve at that point.
x
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
Slope 
y 2.2

 1.1
x
2
y = 2.2
A
x = 2
As we let x get smaller, the
slope of the secant becomes
more and more similar to the
slope of the tangent to the
curve at that point.
x
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
A
y = 1.5
x = 1
Slope 
y 1.5

 1.5
x
1
As we let x get smaller, the
slope of the secant becomes
more and more similar to the
slope of the tangent to the
curve at that point.
x
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
A y = 0.00179
x = 0.001
Slope 
y 0.00179

 1.8
x
0.001
As we let x get smaller, the
slope of the secant becomes
more and more similar to the
slope of the tangent to the
curve at that point.
x
Slopes of Lines and of Curves
 In general, we can express the slope of the secant as follows:
Slope 
y f ( x  h )  f ( x ) f ( x  h )  f ( x )


x
( x  h)  x
h
y
f(x + h)
f(x + h) – f(x)
f(x)
A
h
x
x+h
x
Slopes of Lines and of Curves
 Thus, as h approaches zero, the slope of the secant approaches
the slope of the tangent to the curve at that point:
y
f(x + h)
f(x + h) – f(x)
f(x)
A
h
x
x+h
x
Slopes of Lines and of Curves
 Thus, as h approaches zero, the slope of the secant approaches
the slope of the tangent to the curve at that point:
y
f(x + h)
f(x + h) – f(x)
f(x)
A
h
x
x+h
x
Slopes of Lines and of Curves
 Thus, as h approaches zero, the slope of the secant approaches
the slope of the tangent to the curve at that point:
y
f(x + h)
f(x + h) – f(x)
f(x)
A
h
x
x+h
x
Slopes of Lines and of Curves
 Thus, as h approaches zero, the slope of the secant approaches
the slope of the tangent to the curve at that point:
y
f(x + h)
f(x + h) – f(x)
f(x)
A
h
x
x+h
x
Slopes of Lines and of Curves
 Thus, as h approaches zero, the slope of the secant approaches
the slope of the tangent to the curve at that point:
y
f(x + h)
f(x)
A
f(x + h) – f(x)
h
xx + h
x
Slopes of Lines and of Curves
 Thus, as h approaches zero, the slope of the secant approaches
the slope of the tangent to the curve at that point.
 Expressed in limits notation:
The slope of the tangent line to the graph
of f at the point P(x, f(x)) is given by
lim
h0
if it exists.
f ( x  h)  f ( x)
h
Average Rates of Change
 We can see that measuring the slope of the tangent line to
a graph is mathematically equivalent to finding the
rate of change of f at x.
 The number f(x + h) – f(x) measures the change in y that
corresponds to a change h in x.
 Then the difference quotient
f ( x  h)  f ( x)
h
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
f ( x  h)  f ( x)
h
Instantaneous Rates of Change
 By taking the limit of the difference quotient as h goes to
zero, evaluating
f ( x  h)  f ( x)
lim
h0
h
we obtain the rate of change of f at x.
 This is known as the instantaneous rate of change of f at x
(as opposed to the average rate of change).
 In the maglev example, if y measures the position of a
train at time x, then the limit gives the velocity of the train
at time x.
Instantaneous Rates of Change
 The instantaneous rate of change of f at x or slope
of the tangent line to the graph of f at (x, f(x)) is
f ( x  h)  f ( x)
lim
h0
h
 This limit is called the derivative of f at x .
The Derivative of a Function
 The derivative of a function f with respect to x is the
function f′′ (read “f prime”).
f ( x  h)  f ( x)
f ( x )  lim
h0
h
 The domain of f ′ is the set of all x where the limit exists.
 Thus, the derivative of function f is a function f ′ that
gives the slope of the tangent to the line to the graph of f
at any point (x, f(x)) and also the rate of change of f at x.
The Derivative of a Function
 Four Step Process for Finding f ′(x)
1. Compute f(x + h).
2. Form the difference f(x + h) – f(x).
f ( x  h)  f ( x)
.
h
f ( x  h)  f ( x)
4. Compute f ( x )  lim
.
h0
h
3. Form the quotient
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.
f ( x  h )  f ( x ) 3h

 3.
h
h
Step 4. f ( x )  lim f ( x  h )  f ( x )  lim 3  3.
h0
h0
h
Example 2, page 138
Examples
 Find the slope of the tangent line to the graph f(x) = x2 at
any point (x, f(x)).
Solution
 The required slope is given by the derivative of f at x.
 To find the derivative, we use the four-step process:
Step 1. f(x + h) = (x + h)2 = x2 + 2xh + h2.
Step 2. f(x + h) – f(x) = x2 + 2xh + h2 – x2 = h(2x + h).
Step 3.
f ( x  h )  f ( x ) h(2 x  h )

 2 x  h.
h
h
Step 4. f ( x )  lim
h0
Example 3, page 138
f ( x  h)  f ( x)
 lim(2 x  h )  2 x .
h0
h
Examples
 Find the slope of the tangent line to the graph f(x) = x2 at
any point (x, f(x)).
 The slope of the tangent line is given by f ′(x) = 2x.
 Now, find and interpret f ′(2).
y
Solution
f ′(2) = 2(2) = 4.
5
 This means that, at the point
(2, 4)
4
(2, 4)…
3
… the slope of the tangent
line to the graph is 4.
2
4
1
1
–2
Example 3, page 138
–1
0
1
2
x
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
p  f ( x )  144  x 2
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
f ( x  h )  f ( x ) [144  ( x  h) 2 ]  (144  x 2 )

h
h
144  x 2  2 xh  h 2  144  x 2

h
2xh  h 2 h( 2 x  h )


h
h
 2x  h
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
2(5)  1  11
or –$11 per 1000 tires.
 Similarly, with x = 5, and h = 0.1, we obtain
2(5)  0.1  10.1
or –$10.10 per 1000 tires.
 Finally, with x = 5, and h = 0.01, we get
2(5)  0.01  10.01
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
f ( x )  lim
h0
f ( x  h)  f ( x)
 lim( 2 x  h )  2 x
h0
h
 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 the slope is at that point)
y
a
x
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
a
b
x
Applied Example: Wages
 Mary works at the B&O department store, where, on a
weekday, she is paid $8 an hour for the first 8 hours and
$12 an hour of overtime.
 The function
if 0  x  8
8 x
f ( x)  
12 x  32 if x  8
gives Mary’s earnings on a weekday in which she worked x
hours.
 Sketch the graph of the function f and explain why it is not
differentiable at x = 8.
Applied Example 8, page 143
Applied Example: Wages
Solution
y
130
110
90
70
50
30
10
if 0  x  8
8 x
f ( x)  
12 x  32 if x  8
(8, 64)
x
2
4
6
8
10
12
 The graph of f has a corner at x = 8 and so is not
differentiable at that point.
Applied Example 8, page 143
End of
Chapter