### Chapter 2

```Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Prentice Hall
Slide 2- 1
Chapter 2
Limits and Continuity
2.1
Rates of Change and Limits
Quick Review
In Exercises 1 - 4, find f (2).
3
2
1.
f (x)= 2 x - 5 x + 4
3.
æ xö
f (x)= sin ççp ÷
÷
çè 2 ÷
ø
4.
ìï 3x - 1,
ïï
f (x)= í 1
ïï 2 ,
ïî x - 1
2.
4 x2 - 5
f (x )= 3
x +4
x< 2
x³ 2
Slide 2- 4
Quick Review
In Exercises 5 - 8, write the inequality in the form a < x < b.
5.
x<4
6.
x < c2
7.
x- 2 < 3
8.
x- c < d 2
In Exercises 9 and 10, write the fraction in reduced form.
9.
x 2 - 3x - 18
x+3
2x2 - x
10.
2 x2 + x - 1
Slide 2- 5
Quick Review Solutions
In Exercises 1 - 4, find f (2).
3
2
1.
f (x)= 2 x - 5 x + 4
3.
æ xö
f (x)= sin ççp ÷
÷
çè 2 ÷
ø
4.
ìï 3x - 1,
ïï
f (x)= í 1
ïï 2 ,
ïî x - 1
0
2.
4 x2 - 5
f (x )= 3
x +4
11
12
0
x< 2
x³ 2
1
3
Slide 2- 6
Quick Review Solutions
In Exercises 5 - 8, write the inequality in the form a < x < b.
5.
x<4
- 4< x < 4
6.
x < c2
- c2 < x < c2
7.
x- 2 < 3
- 1< x < 5
8.
x- c < d 2
c- d 2 < x< c + d 2
In Exercises 9 and 10, write the fraction in reduced form.
9.
x 2 - 3 x - 18
x+3
x- 6
2x2 - x
10.
2x2 + x - 1
x
x+ 1
Slide 2- 7





Average and Instantaneous Speed
Definition of Limit
Properties of Limits
One-Sided and Two-Sided Limits
Sandwich Theorem
…and why
Limits can be used to describe continuity, the derivative
and the integral: the ideas giving the foundation of
calculus.
Slide 2- 8
Average and Instantaneous Speed
A body's average speed during an interval of time is
found by dividing the distance covered by the elapsed
time.
Experiments show that a dense solid object dropped
from rest to fall freely near the surface of the earth will fall
y 16t 2 feet in the first t seconds.
Slide 2- 9
Definition of Limit
Let c and L be real numbers. The function f has limit L as x
approaches c if, given any positive number  , there is a positive
number  such that for all x,
0  x  c    f  x   L  .
We write
lim f  x   L
x c
Slide 2- 10
Definition of Limit continued
The sentence lim f  x   L is read, "The limit of f of x as x
x c
approaches c equals L". the notation means that the values f  x 
of the function f approach or equal L as the values of x approach
(but do not equal) c.
Figure 2.2 illustrates the fact that the existence of a limit as x  c never
depends on how the function may or may not be defined at c.
Slide 2- 11
Definition of Limit continued
The function f has limit 2 as x  1 even though f is not defined at 1.
The function g has limit 2 as x  1 even though g 1  2.
The function h is the only one whose limit as x  1 equals its value at x =1.
Slide 2- 12
Properties of Limits
If L, M , c, and k are real numbers and
lim f  x   L
x c
1.
Sum Rule :
and
lim g  x   M , then
x c
lim  f  x   g  x    L  M
x c
The limit of the sum of two functions is the sum of their limits.
2.
DifferenceRule :
lim  f  x   g  x    L  M
x c
The limit of the difference of two functions is the difference
of their limits.
Slide 2- 13
Properties of Limits continued
3.
Product Rule:
lim  f  x  g  x    L M
x c
The limit of the product of two functions is the product of their limits.
4.
Constant Multiple Rule:
lim  k f  x    k L
x c
The limit of a constant times a function is the constant times the limit
of the function.
5.
Quotient Rule :
lim
x c
f  x
g  x

L
, M 0
M
The limit of the quotient of two functions is the quotient
of their limits, provided the limit of the denominator is not zero.
Slide 2- 14
Properties of Limits continued
6.
If r and s are integers, s  0, then
Power Rule :
r
s
r
s
lim  f  x    L
x c
r
s
provided that L is a real number.
The limit of a rational power of a function is that power of the
limit of the function, provided the latter is a real number.
Other properties of limits:
lim  k   k
x c
lim  x   c
x c
Slide 2- 15
Example Properties of Limits
Use any of the properties of limits to find
lim  3 x3  2 x  9 
x c
lim  3x3  2 x  9   lim3x3  lim 2 x  lim9
x c
x c
x c
 3c3  2c  9
x c
sum and difference rules
product and multiple rules
Slide 2- 16
Polynomial and Rational Functions
1.
If f  x   an x n  an1 x n1  ...  a0 is any polynomial function and
c is any real number, then
lim f  x   f  c   an c n  an1c n1  ...  a0
x c
2.
If f  x  and g  x  are polynomials and c is any real number, then
lim
x c
f  x
g  x

f c
g c
, provided that g  c   0.
Slide 2- 17
Example Limits
Use Theorem 2 to find lim  4 x2  2 x  6 
x5
lim  4 x 2 - 2 x  6   4  5  2  5  6  4  25  10  6 100  10  6  96
2
x 5
Slide 2- 18
Evaluating Limits
As with polynomials, limits of many familiar
functions can be found by substitution at points
where they are defined. This includes
trigonometric functions, exponential and
logarithmic functions, and composites of these
functions.
Slide 2- 19
Example Limits
Find
1  sin x
x 0 cos x
lim
Solve graphically:
The graph of f  x  
1  sin x
suggests that the limit exists and is 1.
cos x
Confirm Analytically:
Find
1  sin x  1  sin 0 
1  sin x lim
x 0
lim


x  0 cos x
lim cos x
cos 0
x 0

1 0
1
1
Slide 2- 20
Example Limits
5
x 0 x
Find lim
5
Solve graphically: The graph of f  x   suggests that
x
the limit does not exist.
[-6,6] by [-10,10]
Confirm Analytically :
We can't use substitution in this example because when x is relaced by 0,
the denominator becomes 0 and the function is undefined.
This would suggest that we rely on the graph to see that the
limit does not exist.
Slide 2- 21
One-Sided and Two-Sided Limits
Sometimes the values of a function f tend to different limits as x approaches a
number c from opposite sides. When this happens, we call the limit of f as x
approaches c from the right the right-hand limit of f at c and the limit as x
approaches c from the left the left-hand limit.
right-hand: lim f  x  The limit of f as x approaches c from the right.
x c
left-hand: lim f  x  The limit of f as x approaches c from the left.
x c
Slide 2- 22
One-Sided and Two-Sided Limits continued
We sometimes call lim f  x  the two-sided limit of f at c to distinguish it from
x c
the one-sided right-hand and left-hand limits of f at c.
A function f  x  has a limit as x approaches c if and only if the right-hand
and left-hand limits at c exist and are equal. In symbols,
lim f  x   L  lim f  x   L and
x c
x c
lim f  x   L.
x c
Slide 2- 23
Example One-Sided and
Two-Sided Limits
Find the following limits from the given graph.
a.
4
o
b.
c.
1 2 3
d.
e.
lim f  x 
x  0
0
lim f  x 
 Does Not Exist
lim f  x   4
x2
lim f  x   Does Not Exist
x2
lim f  x   0
x 3
x  2


Slide 2- 24
Sandwich Theorem
If we cannot find a limit directly, we may be able to find
it indirectly with the Sandwich Theorem. The theorem
refers to a function f whose values are sandwiched between
the values of two other functions, g and h.
If g and h have the same limit as x  c then f has that limit too.
Slide 2- 25
Sandwich Theorem
If g  x   f  x   h  x  for all x  c in some interval about c, and
lim g  x  = lim h  x  = L,
x c
x c
then
lim f  x  = L
x c
Slide 2- 26
2.2
Limits Involving Infinity
Quick Review
In Exercises 1 - 4, find f - 1 and graph f , f - 1 and y = x in the
same viewing window.
1.
f (x)= 2 x - 3
2.
f (x )= e x
Slide 2- 28
Quick Review
3.
f (x)= tan- 1 x
4.
f (x)= cot- 1 x
Slide 2- 29
Quick Review
In Exercises 5 and 6, find the quotient q (x) and remainder r (x )
when f (x)is divided by g (x ).
5.
f (x)= 2 x 3 - 3x 2 + x - 1,
g (x )= 3x 3 + 4 x - 5
6.
f (x)= 2 x 5 - x 3 + x - 1,
g (x )= x 3 - x 2 + 1
Slide 2- 30
Quick Review
æ1 ÷
ö
ç
In Exercises 7 - 10, write a formula for (a ) f (- x ) and (b) f ç ÷
.
çè x ÷
ø
Simplify where possible.
7.
f (x)= cos x
9.
f (x)=
ln x
x
8.
f (x )= e-
x
æ
10. f (x)= ççx +
çè
1ö
÷
sin x
÷
÷
xø
Slide 2- 31
Quick Review Solutions
In Exercises 1 - 4, find f - 1 and graph f , f - 1 and y = x in the
same viewing window.
1.
f (x)= 2 x - 3
f
- 1
2.
x+ 3
(x)=
2
f (x)= e x
f - 1 (x)= ln x
[-12,12] by [-8,8]
[-6,6] by [-4,4]
Slide 2- 32
Quick Review Solutions
3.
f (x)= tan - 1 x
4.
f (x)= cot - 1 x
f - 1 (x)= tan x
f - 1 (x )= cot x
p
p
- < x<
2
2
0< x < p
  
  
  3 , 3  by   2 , 2 
 0,   by  1,  
Slide 2- 33
Quick Review Solutions
In Exercises 5 and 6, find the quotient q (x ) and remainder r (x )
when f (x ) is divided by g (x ).
5.
f (x )= 2 x 3 - 3 x 2 + x - 1,
2
q (x )= ,
3
6.
g (x )= 3x 3 + 4 x - 5
5
7
r (x )= - 3x - x +
3
3
f (x )= 2 x 5 - x 3 + x - 1,
q (x )= 2 x 2 + 2 x + 1,
2
g (x)= x3 - x 2 + 1
r (x )= - x 2 - x - 2
Slide 2- 34
Quick Review Solutions
æ1 ÷
ö
ç
In Exercises 7 - 10, write a formula for (a ) f (- x ) and (b) f ç ÷
.
çè x ÷
ø
Simplify where possible.
7.
f (x )= cos x
f (- x )= cos x,
ln x
9. f (x )=
x
æ
10. f (x )= ççx +
çè
8.
æ1 ÷
ö
1
ç
fç ÷
= cos
çè x ÷
ø
x
f (- x )= -
ö
1÷
sin x
÷
÷
ø
x
f (x )= e-
x
æ1 ö÷ - x1
f (- x )= e , f çç ÷
=e
èç x ø÷
x
ln (- x )
x
æ
f (- x )= ççx +
çè
,
æ1 ö
1
f çç ÷
=
x
l
n
÷
çè x ÷
ø
x
ö
1÷
sin x,
÷
÷
ø
x
æ1 ÷
ö æ
ç
fç ÷
= çx +
çè x ø÷ èçç
1 ö÷ 1
÷sin
÷
ø
x
x
Slide 2- 35





Finite Limits as x→±∞
Sandwich Theorem Revisited
Infinite Limits as x→a
End Behavior Models
Seeing Limits as x→±∞
…and why
Limits can be used to describe the behavior of functions
for numbers large in absolute value.
Slide 2- 36
Finite limits as x→±∞
The symbol for infinity (∞) does not represent a real number.
We use ∞ to describe the behavior of a function when the
values in its domain or range outgrow all finite bounds.
For example, when we say “the limit of f as x approaches
infinity” we mean the limit of f as x moves increasingly far to
the right on the number line.
When we say “the limit of f as x approaches negative
infinity (- ∞)” we mean the limit of f as x moves increasingly
far to the left on the number line.
Slide 2- 37
Horizontal Asymptote
The line y  b is a horizontal asymptote of the graph of a function
y  f  x  if either
lim f  x   b
x 
or
lim f  x   b
x 
Slide 2- 38
Example Horizontal Asymptote
Use a graph and tables to find  a  lim f  x  and
x 
 c  Identify all horizontal asymptotes.
f  x 
f  x .
 b  xlim

x 1
x
f  x  1
 a  lim
x 
f  x  1
 b  xlim

 c  Identify all horizontal asymptotes.
[-6,6] by [-5,5]
y 1
Slide 2- 39
Example Sandwich Theorem Revisited
The sandwich theorem also holds for limits as x  .
cos x
graphically and using a table of values.
x 
x
Find lim
The graph and table suggest that the function oscillates about the x-axis.
cos x
0
x 
x
Thus y  0 is the horizontal asymptote and lim
Slide 2- 40
Properties of Limits as x→±∞
If L, M and k are real numbers and
lim f  x   L
x 
1.
Sum Rule :
and
lim g  x   M , then
x 
lim  f  x   g  x    L  M
x 
The limit of the sum of two functions is the sum of their limits.
2.
Difference Rule :
lim  f  x   g  x    L  M
x 
The limit of the difference of two functions is the difference
of their limits
Slide 2- 41
Properties of Limits as x→±∞
3.
Product Rule:
lim  f  x  g  x    L M
x 
The limit of the product of two functions is the product of their limits.
4.
Constant Multiple Rule:
lim  k f  x    k L
x 
The limit of a constant times a function is the constant times the limit
of the function.
5.
Quotient Rule :
lim
x 
f  x
g  x

L
, M 0
M
The limit of the quotient of two functions is the quotient
of their limits, provided the limit of the denominator is not zero.
Slide 2- 42
Properties of Limits as x→±∞
6.
If r and s are integers, s  0, then
Power Rule :
r
s
r
s
lim  f  x    L
x 
r
s
provided that L is a real number.
The limit of a rational power of a function is that power of the
limit of the function, provided the latter is a real number.
Slide 2- 43
Infinite Limits as x→a
If the values of a function f ( x) outgrow all positive bounds as x approaches
a finite number a, we say that lim f  x   . If the values of f become large
xa
and negative, exceeding all negative bounds as x approaches a finite number a,
we say that lim f  x    .
xa
Slide 2- 44
Vertical Asymptote
The line x  a is a vertical asymptote of the graph of a function
y  f  x  if either
lim f  x     or lim f  x    
x a 
x a
Slide 2- 45
Example Vertical Asymptote
Find the vertical asymptotes of the graph of f ( x) and describe the behavior
of f ( x) to the right and left of each vertical asymptote.
8
f  x 
4  x2
The values of the function approach   to the left of x   2.
The values of the function approach + to the right of x   2.
The values of the function approach + to the left of x  2.
The values of the function approach   to the right of x  2.
8
8
lim



and
lim

2
2

x 2 4  x
x 2 4  x
8
8
lim



and
lim

2
2
x2 4  x
x  2 4  x
So, the vertical asymptotes are x  2 and x  2
[-6,6] by [-6,6]
Slide 2- 46
End Behavior Models
The function g is
 a  a right end behavior model for f if and only if lim
x 
 b  a left end behavior model for f if and only if lim
x 
f  x
g  x
f  x
g  x
1.
1.
Slide 2- 47
Example End Behavior Models
Find an end behavior model for
3x 2  2 x  5
f  x 
4 x2  7
Notice that 3 x 2 is an end behavior model for the numerator of f , and
4 x 2 is one for the denominator. This makes
3x 2 3
= an end behavior model for f .
2
4x 4
Slide 2- 48
End Behavior Models
If one function provides both a left and right end behavior model, it is simply called
an end behavior model.
In general, g  x   an x n is an end behavior model for the polynomial function
f  x   an x n  an 1 x n 1  ...  a0 , an  0
Overall, all polynomials behave like monomials.
Slide 2- 49
End Behavior Models
3
is also a horizontal
4
asymptote of the graph of f . We can use the end behavior model of a
rational function to identify any horizontal asymptote.
A rational function always has a simple power function as
In this example, the end behavior model for f , y 
an end behavior model.
Slide 2- 50
Example “Seeing” Limits as x→±∞
We can investigate the graph of y  f  x  as x   by investigating the
1
graph of y  f   as x  0.
 x
1
Use the graph of y  f   to find lim f  x  and lim f  x 
x 
x 
 x
1
for f  x   x cos .
x
cos x
1
The graph of y  f   =
is shown.
x
 x
1
lim f  x   lim f     
x 
x 0
 x
1
lim f  x   lim f     
x 
x 0
 x
Slide 2- 51
Quick Quiz Sections 2.1 and 2.2
You may use a graphing calculator to solve the following problems.
1.
A
 B
C
 D
E
x2  x  6
Find lim
if it exists
x 3
x3
1
1
2
5
does not exist
Slide 2- 52
Quick Quiz Sections 2.1 and 2.2
You may use a graphing calculator to solve the following problems.
1.
A
 B
C
 D
E
x2  x  6
Find lim
if it exists
x 3
x3
1
1
2
5
does not exist
Slide 2- 53
Quick Quiz Sections 2.1 and 2.2
2.
A
 B
C
D
E
3 x  1,

Find lim f  x  =  5
x2
 x  1 ,
5
3
13
3
7

does not exist
x2
x2
if it exists
Slide 2- 54
Quick Quiz Sections 2.1 and 2.2
2.
A
 B
C
D
E
3 x  1,

Find lim f  x  =  5
x2
 x  1 ,
5
3
13
3
7

does not exist
x2
x2
if it exists
Slide 2- 55
Quick Quiz Sections 2.1 and 2.2
3.
Which of the following lines is a horizontal asymptote for
3x3  x 2  x  7
f  x 
2 x3  4 x  5
3
A
y

x
 
2
 B y  0
C y 
D
E
2
3
7
5
3
y
2
y
Slide 2- 56
Quick Quiz Sections 2.1 and 2.2
3.
Which of the following lines is a horizontal asymptote for
3x3  x 2  x  7
f  x 
2 x3  4 x  5
3
A y  x
2
 B y  0
2
3
C
y
D
7
5
3
y
2
E
y
Slide 2- 57
2.3
Continuity
Quick Review
1.
3x 2 - 2 x + 1
Find lim
x® - 1
x3 + 4
2.
Let f (x )= int x. Find each limit.
f (x )
(a ) xlim
®-1
-
f (x )
(b) xlim
®-1
+
f (x )
(c) xlim
®-1
3.
ìï x 2 - 4 x + 5,
Let f (x )= ïí
ïïî 4 - x,
Find each limit.
(d ) f (- 1)
x< 2
x³ 2
f (x )
(a ) xlim
®2
f (x )
(b) xlim
®2
f (x )
(c) lim
x® 2
(d ) f (2)
-
+
Slide 2- 59
Quick Review
In Exercises 4 - 6, find the remaining functions in the list of functions:
f , g, f o g, g o f .
2x - 1
1
4. f (x)=
, g (x)= + 1
x+ 5
x
5.
f (x)= x 2 ,
(g o f )(x)= sin x 2 ,
domain of g = [0, ¥ )
Slide 2- 60
Quick Review
7.
1
g (x)= x - 1, (g o f )(x)= , x > 0
x
Use factoring to solve 2 x 2 + 9 x - 5 = 0
8.
Use graphing to solve
6.
x3 + 2 x - 1= 0
Slide 2- 61
Quick Review
ìï 5 - x,
x£ 3
ï
In Exercises 9 and 10, let f (x)= í
ïïî - x 2 + 6 x - 8, x > 3
9. Solve the equation f (x)= 4
10. Find a value of c for which the equation f (x)= c
has no solution.
Slide 2- 62
Quick Review Solutions
1.
3x 2 - 2 x + 1
Find lim
x® - 1
x3 + 4
2.
Let f (x )= int x. Find each limit.
f (x )
(a ) xlim
®-1
- 2
f (x ) - 1
(b) xlim
®-1
f (x )
(c) xlim
®-1
no limit
(d ) f (- 1) - 1
-
3.
2
+
ìï x 2 - 4 x + 5,
Let f (x )= ïí
ïîï 4 - x,
Find each limit.
f (x )
(a ) xlim
®2
-
f (x )
(c) lim
x® 2
x< 2
x³ 2
1
no limit
f (x )
(b) xlim
®2
+
(d ) f (2)
2
2
Slide 2- 63
Quick Review Solutions
In Exercises 4 - 6, find the remaining functions in the list of functions:
f , g, f o g, g o f .
4.
f (x )=
2x- 1
1
, g (x )= + 1
x+ 5
x
( f o g )(x) =
5.
x+ 2
,
6x + 1
x¹ 0
(g o f )(x ) =
3x + 4
,
2x- 1
x¹ 5
f (x )= x 2 ,
(g o f )(x )= sin x 2 , domain of g = [0, ¥ )
g (x )= sin x, x ³ 0
( f o g )(x)= sin 2 x, x ³ 0
Slide 2- 64
Quick Review Solutions
6.
g (x)=
x - 1,
f (x)=
1
+ 1,
2
x
1
(g o f )(x)= ,
x
x> 0
x> 0
( f o g )(x)=
7.
Use factoring to solve
2 x2 + 9 x - 5 = 0
8.
Use graphing to solve
x3 + 2 x - 1= 0
x
,
x- 1
x=
x> 1
1
, - 5
2
x » 0.453
Slide 2- 65
Quick Review Solutions
ìï 5 - x,
x£ 3
ï
In Exercises 9 and 10, let f (x)= í 2
ïïî - x + 6 x - 8, x > 3
9. Solve the equation f (x)= 4
x= 1
10. Find a value of c for which the equation f (x)= c
has no solution.
Any c in [1, 2)
Slide 2- 66





Continuity at a Point
Continuous Functions
Algebraic Combinations
Composites
Intermediate Value Theorem for Continuous
Functions
…and why
Continuous functions are used to describe how a body
moves through space and how the speed of a
chemical reaction changes with time.
Slide 2- 67
Continuity at a Point
Any function y  f  x  whose graph can be sketched in one continuous motion
without lifting the pencil is an example of a continuous function.
Slide 2- 68
Example Continuity at a Point
Find the points at which the given function is continuous and the points at
which it is discontinuous.
o
Points at which f is continuous
At x  0
At x  6
lim f  x   f  0 
x 0
lim f  x   f  6 
x  6
At 0 < c < 6 but not 2  c  3
lim f  x   f  c 
x c
Points at which f is discontinuous
At x  2
lim f  x  does not exist
x2
At c  0, 2  c  3, c  6
these points are not in the domain of f
Slide 2- 69
Continuity at a Point
Interior Point: A function y  f  x  is continuous at an interior point c of its
domain if lim f  x   f  c 
x c
Endpoint: A function y  f  x  is continuous at a left
endpoint a or is continuous at a right endpoint b of its domain if
lim f  x   f  a 
x a
or
lim f  x   f  b 
x b 
respectively.
Slide 2- 70
Continuity at a Point
If a function f is not continuous at a point c , we say
that f is discontinuous at c and c is a point of
discontinuity of f.
Note that c need not be in the domain of f.
Slide 2- 71
Continuity at a Point
The typical discontinuity types are:
a) Removable (2.21b and 2.21c)
b) Jump
(2.21d)
c) Infinite
(2.21e)
d) Oscillating (2.21f)
Slide 2- 72
Continuity at a Point
Slide 2- 73
Example Continuity at a Point
Find and identify the points of discontinuity of y 
3
 x  1
2
There is an infinite discontinuity at x 1.
[-5,5] by [-5,10]
Slide 2- 74
Continuous Functions
A function is continuous on an interval if and only
if it is continuous at every point of the interval. A
continuous function is one that is continuous at
every point of its domain. A continuous function
need not be continuous on every interval.
Slide 2- 75
Continuous Functions
The given function is a continuous function because it is
continuous at every point of its domain. It does have a
point of discontinuity at x   2 because it is not defined there.
y
2
 x  2
2
[-5,5] by [-5,10]
Slide 2- 76
Properties of Continuous Functions
If the functions f and g are continuous at x  c, then the
following combinations are continuous at x  c.
1.
Sums :
f g
2.
Differences:
f g
3.
4.
Products:
Constant multiples:
f g
k  f , for any number k
5.
Quotients:
f
, provided g  c   0
g
Slide 2- 77
Composite of Continuous Functions
If f is continuous at c and g is continuous at f  c  , then the
composite g f is continuous at c.
Slide 2- 78
Intermediate Value Theorem for Continuous
Functions
A function y  f  x  that is continuous on a closed interval [a, b]
takes on every value between f  a  and f  b  . In other words,
if y0 is between f  a  and f  b  , then y0  f  c  for some c in [a, b].
Slide 2- 79
Intermediate Value Theorem for
Continuous Functions
The Intermediate Value Theorem for Continuous Functions
is the reason why the graph of a function continuous on an
interval cannot have any breaks. The graph will be
connected, a single, unbroken curve. It will not have
jumps or separate branches.
Slide 2- 80
2.4
Rates of Change and Tangent Lines
Quick Review
In Exercises 1 and 2, find the increments D x and D y from point A
to point B.
1.
A(- 5, 2), B (3,5)
2.
A(1,3), B (a, b)
In Exercises 3 and 4, find the slope of the line determined by the
points.
3.
(- 2,3), (5, - 1)
4. (- 3, - 1),
(3,3)
Slide 2- 82
Quick Review
In Exercises 5 - 9, write an equation for the specified line.
3
2
5.
through (- 2,3) with slope =
6.
through (1,6) and (4, - 1)
7.
through (1, 4) and parallel to y = -
3
x+ 2
4
Slide 2- 83
Quick Review
3
x+ 2
4
8.
through (1, 4) and perpendicular to y = -
9.
through (- 1,3) and parallel to 2 x + 3 y = 5
10. For what value of b will the slope of the line through (2,3)
and (4, b) be
5
?
3
Slide 2- 84
Quick Review Solutions
In Exercises 1 and 2, find the increments D x and D y from point A
to point B.
1.
A(- 5, 2), B (3,5)
2.
D x = 8, D y = 3
D x = a - 1,
A(1,3), B (a, b)
D y = b- 3
In Exercises 3 and 4, find the slope of the line determined by the
points.
3.
(- 2,3), (5, - 1)
-
4
7
4. (- 3, - 1),
(3,3)
2
3
Slide 2- 85
Quick Review Solutions
In Exercises 5 - 9, write an equation for the specified line.
3
2
5.
through (- 2,3) with slope =
6.
through (1,6) and (4, - 1)
7.
through (1, 4) and parallel to y = -
y=
y= -
3
x+ 2
4
3
x+ 6
2
7
25
x+
3
3
y= -
3
19
x+
4
4
Slide 2- 86
Quick Review Solutions
8.
through (1, 4) and perpendicular to y = -
3
x+ 2
4
4
8
y= x+
3
3
9.
through (- 1,3) and parallel to 2 x + 3 y = 5
2
7
x+
3
3
10. For what value of b will the slope of the line through (2,3)
y= -
and (4, b) be
5
?
3
b=
19
3
Slide 2- 87





Average Rates of Change
Tangent to a Curve
Slope of a Curve
Normal to a Curve
Speed Revisited
…and why
The tangent line determines the direction of a body’s
motion at every point along its path.
Slide 2- 88
Average Rates of Change
The average rate of change of a quantity over a period
of time is the amount of change divided by the time it
takes.
In general, the average rate of change of a function over an
interval is the amount of change divided by the length of the
interval.
Also, the average rate of change can be thought of as the
slope of a secant line to a curve.
Slide 2- 89
Example Average Rates of Change
Find the average rate of change of f  x   2 x 2  3 x  7
over the interval  -2,4
f  2   f  4 



2  2   3  2   7  2  4   3  4   7
2  4
21  27 6

 1
6
6
2
2

2  4
Slide 2- 90
Tangent to a Curve
In calculus, we often want to define the rate at which the value
of a function y = f(x) is changing
with respect to x at any particular value x = a to be the slope of
the tangent to the curve y = f(x) at x = a.
The problem with this is that we only have one point and our
usual definition of slope requires two points.
Slide 2- 91
Tangent to a Curve
The process becomes:
1. Start with what can be calculated, namely, the slope of a
secant through P and a point Q nearby on the curve.
2. Find the limiting value of the secant slope (if it exists) as Q
approaches P along the curve.
3. Define the slope of the curve at P to be this number and
define the tangent to the curve at P to be the line through P
with this slope.
Slide 2- 92
Example Tangent to a Curve
Given y  x 2  2 at x   1 find:
the slope of the curve and an equation of the tangent line.
Then draw a graph of the curve and tangent line in the
same viewing window.
a 
Write an expression for the slope of the secant line and find the
limiting value of the slope as Q approaches P along the curve.
When x   1, y  x 2  2  3 so =P  1,3
 1  h 
2


2


1
 2


y  1  h   y  1


lim
 lim
h 0
h 0
h
h
h  h  2
3  2h  h 2  3
h 2  2h
lim
 lim
 lim
 lim  h  2    2
h 0
h

0
h

0
h 0
h
h
h
2
Slide 2- 93
Example Tangent to a Curve
 b  The tangent line has slope
 2 and passes through  1,3 .
The equation of the tangent line is
y  3   2  x   1 
y   2  x  1  3
curve
y   2x  2  3
y   2x 1
y  x2  2
tangent
y   2x  1
Slide 2- 94
Slope of a Curve
To find the tangent to a curve y = f(x) at a point P(a,f(a))
calculate the slope of the secant line through P and a point
Q(a+h, f(a+h)). Next, investigate the limit of the slope as
h→0.
If the limit exists, it is the slope of the curve at P and we
define the tangent at P to be the line through P with this
slope.
Slide 2- 95
Slope of a Curve
Slide 2- 96
Slope of a Curve at a Point
The slope of the curve y  f  x  at the point P  a, f  a   is the number
m  lim
h 0
f a  h  f a
h
provided the limit exists.
The tangent line to the curve at P is the line through P with this slope.
Slide 2- 97
Slope of a Curve
All of the following mean the same:
1. the slope of y  f ( x) at x  a
2. the slope of the tangent to y  f ( x ) at x  a
3. the (instantaneous) rate of change of f ( x) with respect to x at x  a
4. lim
h 0
f a  h  f a
h
The expression
f a  h  f a
h
is the difference quotient of f at a.
Slide 2- 98
Normal to a Curve
The normal line to a curve at a point is the line
perpendicular to the tangent at the point.
The slope of the normal line is the negative reciprocal of the
slope of the tangent line.
Slide 2- 99
Example Normal to a Curve
Given y  x 2  2 at x   1 write the equation of the normal line.
Draw a graph of the curve, the tangent line and the normal line in the
same viewing window.
From an earlier example, the slope of the tangent line was found
to be  2 so the slope of the normal is
y  3
1
x   1 

2
1
y   x  1  3
2
1
1 6
y x 
2
2 2
1
7
y x
2
2
1
.
2
tangent
y   2x  1
curve
y  x2  2
normal line
1
7
y x
2
2
Slide 2- 100
Speed Revisited
The function y 16t 2 is an object's position function. An object's average speed
along a coordinate axis for a given period of time is the average rate of change
of its position y  f (t ).
It's instantaneous speed at any time t is the instantaneous rate of change of
position with respect to time at time t , or lim
h 0
f t  h   f t 
h
.
Slide 2- 101
Quick Quiz Sections 2.3 and 2.4
You may use a graphing calculator to solve the following problems.
1. Which of the following values is the average rate of change
of f  x   x  1 over the interval  0,3 ?
A
 B
1
C

3
1
3
1
3
 D
E
3
Slide 2- 102
Quick Quiz Sections 2.3 and 2.4
You may use a graphing calculator to solve the following problems.
1. Which of the following values is the average rate of change
of f  x   x  1 over the interval  0,3 ?
A
 B
3
1
1
C 
3
1
D
 
3
E 3
Slide 2- 103
Quick Quiz Sections 2.3 and 2.4
2.
Which of the following statements is false for the function
3
0 x4
 4 x,

x4
f  x   2,
 x  7,
4 x6

6 x8 ?
1,
f  x  exists
 A  lim
x4
 B  f  4  exists
f  x  exists
 C  lim
x 6
f  x  exists
 D  xlim
8
 E  f is continuous at x  4

Slide 2- 104
Quick Quiz Sections 2.3 and 2.4
2.
Which of the following statements is false for the function
3
0 x4
 4 x,

x4
f  x   2,
 x  7,
4 x6

6 x8 ?
1,
f  x  exists
 A  lim
x4
 B  f  4  exists
f  x  exists
 C  lim
x 6
f  x  exists
 D  xlim
8
 E  f is continuous at x  4

Slide 2- 105
Quick Quiz Sections 2.3 and 2.4
3.
Which of the following is an equation for the tangent line
to f  x   9  x 2 at x  2?
1
9
A
y

x

 
4
2
 B y   4 x  13
C
 D
E
y   4x  3
y  4x  3
y  4 x  13
Slide 2- 106
Quick Quiz Sections 2.3 and 2.4
3.
Which of the following is an equation for the tangent line
to f  x   9  x 2 at x  2?
1
9
A
y

x

 
4
2
 B y   4 x  13
C
 D
E
y   4x  3
y  4x  3
y  4 x  13
Slide 2- 107
Chapter Test
In Exercises 1 and 2, find the limits.
1
1
2+ x 2
1. lim
2.
x® 0
x
lim
x® ¥
x + sin x
x + cos x
In Exercises 3 and 4, determine whether the limit exists on the basis of
the graph of y = f (x ). The domain of f is the set of real numbers.
3.
lim f (x )
x ® c-
4.
lim f (x )
x® b
Slide 2- 108
Chapter Test
In Exercise 5, use the graph of the function with domain - 1£ x £ 3.
5.
Determine
g (x )
(a ) xlim
®3
-
(b) g (3).
(c) whether g (x) is continuous at x = 3.
(d ) the points of discontinuity of g (x).
(e) Writing to Learn whether any points of discontinuity
are removable. If so, describe the extended function. If not,
explain why not.
Slide 2- 109
Chapter Test
In Exercise 6, (a ) find the vertical asymptotes of the graph of
y = f (x ), and (b) describe the behavior of f (x ) to the right
and left of any vertical asymptote.
6.
x- 1
f (x)= 2
x (x + 2)
Slide 2- 110
Chapter Test
7.
Given
ìï 1, x £ - 1
ïï
ïï - x, - 1 < x < 0
ï
f (x )= ïí 1, x = 0
ïï
ïï - x, 0 < x < 1
ïï
ïî 1, x ³ 1
(a ) Find the right-hand and left-hand limits of f at
x = - 1, 0 and 1.
(b) Does f have a limit as x approaches - 1? 0? 1?
If so, what is it? If not, why not?
(c) Is f continuous at x= - 1? 0? 1? Explain.
Slide 2- 111
Chapter Test
In Exercise 8, find (a ) a power function end behavior model and
(b) any horizontal asymptotes.
2x + 1
x2 - 2x + 1
8.
f (x)=
9.
Find the average rate of change of f (x)= 1+ sin x
over the interval [0,
p
].
2
Slide 2- 112
Chapter Test
10. Let f (x)= x 2 - 3x and P = (1, f (1)).
Find (a ) the slope of the curve y = f (x ) at P, (b) an equation
of the tangent at P and (c) an equation of the normal at P.
Slide 2- 113
Chapter Test Solutions
In Exercises 1 and 2, find the limits.
1.
1
1
2+ x 2
lim
x® 0
x
-
1
4
2.
lim
x® ¥
x + sin x
x + cos x
1
In Exercises 3 and 4, determine whether the limit exists on the basis of
the graph of y = f (x ). The domain of f is the set of real numbers.
3.
lim f (x )
x ® c-
Exists
4.
lim f (x )
x® b
Exists
Slide 2- 114
Chapter Test Solutions
In Exercise 5, use the graph of the function with domain - 1£ x £ 3.
5. Determine
g (x )
(a ) xlim
®3
-
1
(b) g (3).
(c) whether g (x) is continuous at x = 3.
(d ) the points of discontinuity of g (x).
1.5
No
at x = 3 and points
not in the domain
(e) Writing to Learn whether any points of discontinuity
are removable. If so, describe the extended function. If not,
explain why not.
Removable at x = 3 by assigning the value 1 to g (3).
Slide 2- 115
Chapter Test Solutions
In Exercise 6, (a ) find the vertical asymptotes of the graph of
y = f (x ), and (b) describe the behavior of f (x ) to the right
and left of any vertical asymptote.
x- 1
6. f (x )= 2
vertical asymptotes: x = 0, x = - 2
x (x + 2 )
At x = 0
Left-hand limit = limx® 0
x- 1
2
=- ¥
x ( x + 2)
Right-hand limit = lim+
x® 0
x- 1
2
=- ¥
x ( x + 2)
At x = - 2
Left-hand limit = limx® 2
x- 1
2
x ( x + 2)
=¥
Right-hand limit = lim+
x® 2
x- 1
2
=- ¥
x ( x + 2)
Slide 2- 116
Chapter Test Solutions
7.
Given
ìï 1, x £ - 1
ïï
ïï - x, - 1 < x < 0
ï
f (x )= ïí 1, x = 0
ïï
ïï - x, 0 < x < 1
ïï
ïî 1, x ³ 1
(a ) Find the right-hand and left-hand limits of f at
x = - 1, 0 and 1.
(b) Does f have a limit as x approaches - 1? 0? 1?
If so, what is it? If not, why not?
(c) Is f continuous at x= - 1? 0? 1? Explain.
Slide 2- 117
Chapter Test Solutions
7. a 
At x   1:
Left-hand limit = lim f  x   lim 1 1
x 1
x 1
Right-hand limit = lim f  x   lim   x  1
x 1
x 1
At x  0 :
Left-hand limit = lim f  x   lim   x   0
x 0
x 0
Right-hand limit = lim f  x   lim   x   0
x 0
x 0
At x 1
Left-hand limit = lim f  x   lim   x    1
x 1
x 1
Right-hand limit = lim f  x   lim 1 1
x 1
x 1
Slide 2- 118
Chapter Test Solutions
7. b 
At x   1: Yes, the limit is 1.
At x  0 : Yes, the limit is 0.
At x   1: No, the limit doesn't exist because the two
one-sided limits are different.
c
At x   1: Continuous because f  1  the limit.
At x  0 : Discontinuous because f  0   the limit.
At x 1: Discontinuous because the limit does not exist.
Slide 2- 119
Chapter Test Solutions
In Exercise 8, find (a ) a power function end behavior model and
(b) any horizontal asymptotes.
2x + 1
x2 - 2x + 1
(a )
2
x
(b) y = 0
8.
f (x)=
9.
Find the average rate of change of f (x )= 1+ sin x
é pù
over the interval ê0, ú.
êë 2 ú
û
2
p