1.

1 Limits: A numerical and Graphical Approach (Page 1 of 40)

1.1 Limits: A Numerical and Graphical Approach

Example 1
The graph of f (x) = 2x + 3 is
shown.

What happens to the value of

f (x) as x approaches 3?
x
f(x)

2
7

2.6
8.2

2.9
8.8

2.99
8.98

2.999
8.998

x ! 3"
f (x) ! ?

x
f(x)

3.4

3.1

3.01

3.001

x ! 3+
f (x) ! ?

As x ! 3 , f (x) ! _______ . That is, as x approaches 3 from the

left, f (x) = 2x + 3 approaches 9. The number 9 is said to be the
limit of 2x + 3 as x approaches 3 from either side. It is written in
limit notation as

lim(2x + 3) = 9
x!3

Definition of a Limit
As x approaches a, the limit of f (x) is L, written

lim f (x) = L ,
x!a

if all values of f (x) are arbitrarily close to L for values of x that

are sufficiently close, but not equal, to a.

1.1 Limits: A numerical and Graphical Approach (Page 2 of 40)

Theorem
As x approaches a, the limit of f (x) is L if the limit from the left
exists and the limit from the right exists and both limits are L. That
is,
if lim" f (x) = lim+ f (x) = L ,then lim f (x) = L .
x!a

x!a

x!a

Example 2

# 2x + 2, x < 1
The graph of g(x) = $
is shown.
2x
!
4,
x
"
1
%
a.

Find lim g(x) .

b.

Find lim g(x) .

x!1

x!3

x ! 1" g(x)
(x < 1)
0
2
0.7
3.4
0.9
3.8
0.99
3.98
0.999
3.998
+
x !1
g(x)
(x > 1)
2
0
1.5
-1
1.1
-1.8
1.01
-1.98
1.001 -1.998

x ! 3"
(x < 3)
2
2.7
2.9
2.99
2.999
x ! 3+
(x > 3)
4
3.5
3.1
3.01
3.001

g(x)

g(x)

1.1 Limits: A numerical and Graphical Approach (Page 3 of 40)

Example 3
a.

Limits when there is a Hole in the Graph

x =1
" 5,
Graph of G(x) = #
$ x + 1, x ! 1

b.

Find lim G(x)

c.

Find lim G(x)

x!1

x!"2

1.1 Limits: A numerical and Graphical Approach (Page 4 of 40)

Example 4
a.
x
f(x)

Limits at Infinity
1
Graph of f (x) =
+3
x!2
-3
2.8

x
2
f(x) VA

-2
-1
0
2.75 2.67 2.5
3
4

1
2

4
6
8
3.5 3.25 3.17

b.

Find lim f (x)

c.

Find lim" f (x) =

d.

Find lim+ f (x) =

e.

Find lim f (x)

f.

Find lim f (x)

g.

Find lim f (x)

x!3

x!2

x!2

x
1.5 1.9 1.99 1.999 x ! 2 "
f(x) 1
-7 -97 -997 f (x) !
x
2.5 2.1 2.01 2.001 x ! 2 +
f(x) 5
13 103 1003 f (x) !

x!2

x!"

x!" #

x
5
10 100 1000 x ! 2 +
f(x) 3.33 3.13 3.01 3.001 f (x) !
x
-5 -10 -100 -1000 x ! 2 +
f(x) 2.86 2.92 2.990 2.999 f (x) !

1.2 Algebraic Limit and Continuity (Page 5 of 40)

1.2 Algebraic Limits and Continuity

Limit Principles
Let lim f (x) = L , lim g(x) = M and n be a positive integer. Then
x!a

1.1

x!a

lim c = c
x!a

The limit of a constant is the constant.

lim x = a
x!a

1.2

lim [ f (x)] = " lim f (x) $ = Ln

# x!a
%
x!a
The limit of a power is the power of the limit.
n

lim n f (x) = n lim f (x) = n L ,

x!a

x!a

provided L > 0 if n is even.

The limit of a root is the root of the limit.

1.3

lim[ f (x) g(x)] = lim f (x) lim g(x) = L M

x!a

x!a

x!a

The limit of a sum [difference] is the sum [difference] of the

limits.
1.4

lim[ f (x)" g(x)] = lim f (x)" lim g(x) = L " M

x!a

x!a

x!a

The limit of a product is the product of the limits.

1.5

f (x) L
f (x) lim
x!a
lim
=
=
x!a g(x)
lim g(x) M

provided M ! 0

x!a

The limit of a quotient is the quotient of the limits.

1.6

lim[c " f (x)] = c " lim f (x) = c " L

x!a

x!a

The limit of a constant times a function is the constant times

the limit.

1.2 Algebraic Limit and Continuity (Page 6 of 40)

Limit Principles
Let lim f (x) = L , lim g(x) = M and n be a positive integer. Then
x!a

1.1
1.2

lim c = c

x!a

lim x = a

x!a

x!a
n

n
lim [ f (x)] = " lim f (x) $ = Ln
# x!a
%
x!a
lim n f (x) = n lim f (x) = n L ,
x!a

x!a

provided L > 0 if n is even.

1.3

lim[ f (x) g(x)] = lim f (x) lim g(x) = L + M

1.4

lim[ f (x)" g(x)] = lim f (x)" lim g(x) = L " M

1.5

f (x) L
f (x) lim
x!a
lim
=
=
x!a g(x)
lim g(x) M

x!a

x!a

x!a

x!a

x!a

x!a

provided M ! 0

x!a

1.6

lim[c " f (x)] = c " lim f (x) = c " L

x!a

x!a

Example 1
Use the Limit Principles to find lim(x 2 " 3x + 7)
x!4

Theorem on Limit of Rational Functions

For any rational function F, with a in the domain of F,
lim F(x) = F(a) .
x!a

That is, the limit of a rational function can be evaluated by direct

substitution.

1.2 Algebraic Limit and Continuity (Page 7 of 40)

Example 2
Limit of a Polynomial
Find lim(x 4 " 5x 3 + x + 7)
x!2

Example 3
Limit of a Radical
Find lim x 2 " 3x + 7
x!4

Example 4

Limit of a Rational Function where the

numerator and denominator have a
common factor.

x2 " 9
Find lim
x!"3 x + 3

Example 5
Limit of a Polynomial
2
Find lim(3x + 3xh + h 2 )
h!0

1.2 Algebraic Limit and Continuity (Page 8 of 40)

Definitions of Continuity
1.
A function f is continuous at the point x = a if:
i.
the output at a exists
f (a) exists
ii.
the limit exists
lim f (x) exists
x!a

iii.
2.

lim f (x) = f (a)

x!a

the limit equals the output

Otherwise, f has a discontinuity at x = a.

A function is continuous over an interval I, if it continuous
at each point in I.

Example 6
Discuss the continuity of each function over (! ", ") . For each
discontinuity, state which condition of continuity failed.
x2 ! 4
1.
f (x) =
x+2
Continuous Intervals:
Discontinuities:

2.

G(x) =

1
x

Continuous Intervals:

Discontinuities:

1.2 Algebraic Limit and Continuity (Page 9 of 40)

Definitions of Continuity
1.
A function f is continuous at the point x = a if:
i.
the output at a exists
f (a) exists
ii.
the limit exists
lim f (x) exists
x!a

iii.
2.

lim f (x) = f (a)

x!a

the limit equals the output

Otherwise, f has a discontinuity at x = a.

A function is continuous over an interval I, if it continuous
at each point in I.

Example 7
Discuss the continuity of each function over (! ", ") . For each
discontinuity, state which condition of continuity failed.
# x + 2, x ! 1
H (x) = $
1.
%x"2 x >1
Continuous Intervals:

Discontinuities:

2.

# x ! 2, x " !2
k(x) = $
x = !2
% 5,
Continuous Intervals:

Discontinuities:

1.2 Algebraic Limit and Continuity (Page 10 of 40)

Example 8
Is the function f continuous at x = 4?
Why or why not?

# x 2 ! 16
, x"4
%
f (x) = $ x ! 4
%& 2
x=4

Example 10
Is the function f continuous at x = 2?
Why or why not?

x<2
# 3 ! x,
f (x) = $ 2
% x ! 4x + 5 x " 2

1.3 Average Rate of Change (Page 11 of 40)

1.3 Average Rate of Change

Average Rate of Change
The average rate of change of a
function f from x1 to x2 is the ratio of
the change in output to the change in
input:

! output f (x2 ) " f (x1 )

ARC =
=
! input
x2 " x1

( x2 , f (x2 ))
( x1, f (x1 ))
x

Example 1
The graph shows the total
number of suits produced
by Raggs, Ltd., during one
morning of work.
a. What was the number
of suits produced from
9 am to 10 am?

b.

What was the average rate of change from 9 am to 11 am?

Include the units and explain its meaning in this application.

1.3 Average Rate of Change (Page 12 of 40)

Example 3
For f (x) = x 2 , find the average rate of change as
a) x changes from 1 to 3.

b)

x changes from 1 to 2.

c)

x changes from 2 to 3.

The Difference Quotient

The average rate of change of a
function f with respect to x is called
the difference quotient. It is given by

DQ =

f (x + h) ! f (x)
h

Notice that the difference quotient is

also the slope of the secant line
connecting the two points.

( x + h, f (x + h))

( x, f (x))
x

1.3 Average Rate of Change (Page 13 of 40)

The Difference Quotient

The average rate of change of a
function f with respect to x is called
the difference quotient. It is given by

DQ =

f (x + h) ! f (x)
h

( x + h, f (x + h))
( x, f (x))

Notice that the difference quotient is

also the slope of the secant line
connecting the two points.
Exercise 2
For f (x) = 5x 2 .
a. Find the simplified form of the difference quotient.

b.

Evaluate the difference quotient at x = 5 and h = {2, 1, 0.1,

0.01}.

1.3 Average Rate of Change (Page 14 of 40)

Exercise 14
For f (x) = x 2 ! 4x .
a. Find the simplified form of the
difference quotient.

b.

Evaluate the difference quotient at

x = 5 and h = {2, 1, 0.1, 0.01}.

Exercise 30
Let A(t) = 2000(1.015)4t be the amount of money in a savings
account that pays 6% interest, compounded quarterly for t years,
when an initial investment of $2000 is made.
a.
Find A(3)
b.

Find A(5)

c.

Find A(5) ! A(3)

d.

Find

A(5) ! A(3)
and interpret its meaning in this application.
5!3

1.3 Average Rate of Change (Page 15 of 40)

Example 6
For f (x) = x 3 , find the simplified form of the difference quotient.

Example 7
For f (x) = 3 / x , find the simplified form of the difference
quotient.

1.4 Limit Definition of the Derivative (Page 16 of 40)

1.4 Differentiation Using the Limit Definition

The Slope of the Secant Line
The slope of the secant line to f
between points ( x, f (x)) and

( x, +h f (x + h)) is given by
msec =

( x + h, f (x + h))

f (x + h) ! f (x)
h

( x, f (x))
x

The Slope of the Tangent Line

The slope of the tangent line to f at
( x, f (x)) is given by

mtan

f (x + h) " f (x)
.
= lim
h!0
h

This limit is called the instantaneous

rate of change of f (x) at x.

( x + h, f (x + h))

( x, f (x))
x

1.4 Limit Definition of the Derivative (Page 17 of 40)

Definition of a Derivative
For a function y = f (x) , the derivative
of f at x is the function f ! (read fprime) defined by

f !(x) = lim
h"0

f (x + h) # f (x)
.
h

( x + h, f (x + h))
( x, f (x))
x

Three Steps to Find the Derivative

f (x + h) ! f (x)
1.
Write the difference quotient,
h
2.
Simplify the difference quotient.
3.

Find the limit as h approaches zero.

Exercise 8
Let f (x) = !2x + 5 .
a) Graph the function.
b)

Find f !(x) using the limit

definition.

d)

Find f !("2) , f !(0) , and f !(1)

1.4 Limit Definition of the Derivative (Page 18 of 40)

Exercise 2
Let f (x) = 12 x 2 .
a) Graph the function.

b)

Draw the tangent line to the

graph at the points whose xcoordinates are -2, 0, and 1.

c)

Find f !(x) using the limit definition.

1.

Write the difference

quotient.

2.

Simplify the
difference quotient.

3.

Find the limit as h

approaches zero.

d)

Find f !("2) , f !(0) , and f !(1)

f (x + h) ! f (x)
=
h

1.4 Limit Definition of the Derivative (Page 19 of 40)

Exercise 6
Let f (x) = !x 3 .
a) Graph the function.

b)

Draw the tangent line to the

graph at the points whose xcoordinates are -2, 0, and 1.

c)

Find f !(x) using the limit

definition.

1.

Write the difference

quotient.

2.

Simplify the
difference quotient.

3.

Find the limit as h

approaches zero.

d)

Find f !("2) , f !(0) , and f !(1)

f (x + h) ! f (x)
=
h

1.4 Limit Definition of the Derivative (Page 20 of 40)

Exercise 16
Let f (x) = 2 / x .
a) Graph the function.

b)

Draw the tangent line to the

graph at the points whose xcoordinates are -2, 0, and 1.

c)

Find f !(x) using the limit

definition.

1.

Write the difference

quotient.

2.

Simplify the
difference quotient.

3.

Find the limit as h

approaches zero.

d)

Find f !("2) , f !(0) , and f !(1)

f (x + h) ! f (x)
=
h

1.4 Limit Definition of the Derivative (Page 21 of 40)

Exercise 20
Find the equation of the tangent
line to the graph of f (x) = 2 / x at
given point. Recall f !(x) = "2 / x 2 .
(a) (3, 2/3)

(b) (-1, -2)

(c) (1, 2)

f (x) = 2 / x

1.4 Limit Definition of the Derivative (Page 22 of 40)

Points where f is NOT Differentiable

The function y = f (x) is not differentiable at x = a if
1.
f (a) is undefined.
2.
f (x) is not continuous at x = a.
3.
x = a is where f has a sharp point or corner.
4.
f has a vertical tangent line at x = a.

f (x) =

x2 ! 4

f (x) =

x+2

1
x!2

G(x) =

1
x

, G !(x) = "

1
x2

f (x) = x

# x + 2, x ! 1

H (x) = $

%x " 2

f (x) = (x ! 2)2 / 3 + 1
f "(x) =

1
f "(x) = !
(x ! 2)2

f (x) = (x ! 3)1/ 3 + 1
f "(x) =

x >1

1
3(x ! 3)2 / 3

2
3 x!2
3

1.5 Power Rule and Sum Rule (Page 23 of 40)

1.5 Differentiation Rules: The Power Rule and the Sum

& Difference Rule
Derivative Notations
If y = f (x) , the derivative of y with respect to x is expressed as:

y!

y-prime

f !(x)

f-prime of x

dy
dx

the derivative of y with respect to x

d
f (x)
dx

the derivative of f with respect to x

Theorem 1

The Power Rule

d n
For any real number n,
x = n ! x n"1
dx
Example 1
Find the derivative of each funciton.
y = x4
a.
b.

y=x

c.

y=

1
x

1.5 Power Rule and Sum Rule (Page 24 of 40)

Example 2
Find the derivative of each funciton.
y = x1/3
a.

b.

y= x

c.

y = x 2.35

Theorem 2

The Constant Rule

d
For any real number c,
c=0
dx

Theorem 3

Constant Times a Function

d
d
For any real number k,
c ! f (x)] = c !
f (x)
[
dx
dx
Example 3
Find each of the following derivatives.
d
d
a.
b.
9x 5
( !6x )
dx
dx

( )

c.

d ! 1 $
#
&
dx " 5x 2 %

1.5 Power Rule and Sum Rule (Page 25 of 40)

Theorem 4
Sum and Difference Rules
Sum
The derivative of the
d
d
d
f
(x)
+
g(x)
=
f
(x)
+
g(x)
[
]
sum is the sum of
dx
dx
dx
the derivatives.
Difference The derivative of the
difference is the
difference of the
derivatives.

Example 4
Find each of the following derivatives.
d
a.
!3x 4 ! 2x + 7
dx

b.

d "
1 3
2
!17x
!
+
$
dx #
x

c.

d " 1
%
! x 3/5 '
$#
&
dx 2x

%
x'
&

d
d
d
f (x) ! g(x)] =
f (x) ! g(x)
[
dx
dx
dx

1.5 Power Rule and Sum Rule (Page 26 of 40)

Example 5
The volume of a sperical tumor can be approximated by
V (r) = 43 ! r 3 , where r is the radius of the tumor.
a.
Find the rate change of volume with respect to r. That is, find
V !(r) .

b.

Find the rate change of volume at r = 1.2. That is, find

V !(1.2) .

Example 6
Find the point(s) on the graph of
f (x) = !x 3 + 6x 2 (shown) at which
a.
the tangent line is horizontal.

b.

the tangent line has slope 6. Find

the exact value of the x- and ycoordinates.

1.6 Product Rule and Quotient Rule (Page 27 of 40)

1.6 Differentiation Rules: The Product Rule and

Quotient Rule
Theorem 5
The Product Rule
Let P(x) = f (x)! g(x) . Then

P !(x) =

d
d
d
f (x)! g(x)] = f (x)! g(x) + g(x)!
f (x)
[
dx
dx
dx
= f (x)! g "(x) + g(x)! f "(x)

Example 1
Find the derivative of each funciton.
y = (x 4 ! 2x 2 + 7)(!2x 2 + 7x)
a.

b.

y = (x 4 ! 3 x )(x 2 + 5x ! 3)

1.6 Product Rule and Quotient Rule (Page 28 of 40)

Theorem 6
The Quotient Rule
D(x)" N !(x) # N(x)" D !(x)
N(x)
Let Q(x) =
.
Then Q !(x) =
D(x)
[ D(x)]2

Q=

hi
ho

Example 3

x6
Find the derivative of y = 2 .
x

Example 4

1+ x 2
Find the derivative of y =
.
x3

Q! =

ho "D hi # hi "D ho
ho " ho

1.6 Product Rule and Quotient Rule (Page 29 of 40)

Example 5

x 2 + 3x
Find the derivative of f (x) =
.
x+5

Exercise 98
Find the equation of the tangent line to the graph of y =

x = 1 and x = 1 / 4 .

x
at
x +1

1.6 Product Rule and Quotient Rule (Page 30 of 40)

Average Cost, Average Revenue and Average Profit

Average Cost
If C(x) is the cost of producing x items,
then the average cost of producing x items is
C(x)
.
x
Average Revenue If R(x) is the revenue from selling x items,
then the average revenue from selling x
R(x)
items is
.
x
Average Profit

If P(x) is the profit selling x items, then the

P(x)
average profit from selling x items is
.
x

Example 6
Paulsens Greenhouse finds that the cost, in dollars, of growing x
geraniums is given by

C(x) = 200 + 100 4 x ,

and the revenue from the sale of x geraniums is given by

R(x) = 120 + 90 x
a.
b.

Find the average cost, average revenue and average profit

when x geraniums are grown and sold.
Find the rate at which average profit is changing when 300
geraniums are being grown and sold.

1.7 The Chain Rule (Page 31 of 40)

1.7 Differentiation Rules: The Extended Power Rule and

the Chain Rule
Theorem 7
The Extended Power Rule
n
Let P(x) = [ f (x)] . Then

P !(x) =

d
k
k"1 d
f (x)] = k ! [ f (x)] !
f (x)
[
dx
dx
= k ! [ f (x)]

k"1

Example 1
Find the derivative of each funciton.
y = (x 4 ! 2x 2 + 7)2
a.

b.

f (x) = 1+ x 3

c.

f (x) = (7 ! x)55

! f #(x)

1.7 The Chain Rule (Page 32 of 40)

Example 2
Find the derivative of each funciton.
y = (1! x 2 )2 + (5 + 4x)2
a.

b.

f (x) = (3x ! 5)4 (12 ! x)8

c.

f (x) =

x+5
x!5

1.7 The Chain Rule (Page 33 of 40)

Composition of Functions
The composition of functions f and g, written f ! g is defined as
( f ! g ) (x) = f ( g(x))
Example 5
Let f (x) = x 3 and g(x) = 1! x 2 .
a.
Find ( f ! g ) (x)

b.

Find ( g ! f ) (x)

Example 6
Let f (x) = x and g(x) = 1! x .
a.
Find ( f ! g ) (x)

b.

Find ( g ! f ) (x)

1.7 The Chain Rule (Page 34 of 40)

Theorem 8

The Chain Rule

d
The derivative f ! g is given by
( f ! g ) (x) = f ! ( g(x)) " g!(x)
dx
If y = f (u) and u = g(x) , then

dy dy du
=
!
dx du dx

Example 7
Let y = u and u = 1! x 2 .
dy
a.
Find
using the chain rule.
dx

b.

Express y as a function of x and find

dy
.
dx

1.7 The Chain Rule (Page 35 of 40)

Example 8
1
Let y = 2
and u = 5 ! 3t .
u +u
dy
a.
Find
using the chain rule.
dt

b.

Express y as a function of t and find

dy
.
dt

Example 9
Let h(x) = ( f ! g ) (x) = (3x 3 ! 2x)4 . Find f (x) and g(x) . Then find
h '(x) and h !(1) .

1.7 The Chain Rule (Page 36 of 40)

Example 10
Let h(x) = ( f ! g ) (x) =

h '(x) and h !(1) .

1
. Find f (x) and g(x) . Then find
7x + 2

Exercise 74
A company determines that its total cost in thousands of dollars,
for producing x items is

C(x) = 5x 2 + 60
and it plans to boost production t months from now according to
the function

x(t) = 20t + 40 .
How fast will the costs be rising 4 months from now?

1.8 Higher Order Derivatives (Page 37 of 40)

1.8 Higher Order Derivatives

Example 1
Let y = f (x) = x 5 ! 5x 3 ! x . The first, second, third, forth, fifth and sixth
derivatives are shown. Note the five various notations for the higher
order derivative.
1st
Derivative
y! = f !(x)

dy
=
dx
d
=
f (x)
dx
d 5
#$ x " 5x 3 " x %&
=
dx
= 5x 4 " 15x 2

y(4 )

4th
Derivative
= f (4 ) (x)

d4y
= 4
dx
d4
= 4 f (x)
dx
d4
= 4 "# 6x 2 ! 30 $%
dx
= 12x

2nd
Derivative
y!! = f !!(x)

d2y
= 2
dx
d2
= 2 f (x)
dx
d2
= 2 #$ 5x 4 " 15x 2 %&
dx
= 20x 3 " 30x
5th
Derivative
(5)
y = f (5) (x)

d5y
= 5
dx
d5
= 5 f (x)
dx
d5
= 5 [12x ]
dx
= 12

3rd
Derivative
y!!! = f !!!(x)

d 3y
= 3
dx
d3
= 3 f (x)
dx
d3
= 3 #$ 20x 3 " 30x %&
dx
= 60x 2 " 30
6th
Derivative
(6)
y = f (6) (x)

d6y
= 6
dx
d6
= 6 f (x)
dx
d6
= 6 [12 ]
dx
=0

1.8 Higher Order Derivatives (Page 38 of 40)

Example 2
Let y = 1 / x . Find d 2 y / dx 2 .

Exercise 22
Let f (x) = (3x 2 + 2x + 1)5 . Find f !(x) and f !!(x) .

1.8 Higher Order Derivatives (Page 39 of 40)

Exercise 34
Let y = (x 3 ! 2)(5x + 1) . Find f !(x) and f !!(x) .

Exercise 42
Let f (x) = x !3 + 2x1/3 . Find f !(x) and f (5) (x) .

1.8 Higher Order Derivatives (Page 40 of 40)

Velocity and Acceleration

Let s(t) be the distance an object travels from its starting point at time t.
Then,
1. The velocity of the
ds
v(t)
=
= s !(t)
object is given by
dt
2. The acceleration of
the object is given by

dv
= v!(t)
dt
d2s
= 2 = s !!(t)
dt

a(t) =

Exercise 48

1
Let s(t) = t 2 ! t + 3 , where s(t) is the number of meters traveled in t
2
seconds.
a.
Find v(t) and a(t) .

b.

Find the velocity and acceleration when t = 1 sec.