the german university in cairo (guc) dr. tarek emam

1

The German University in Cairo (GUC) Dr. Tarek Emam

Mathematics Department Winter 2010

Math101 for Management Students

Lecture Notes

c©Tarek Emam 2010 [email protected]

Contents

0.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 6

1 Pre-Calculus 7

1.1 Numbers . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Order and Intervals on the Real Number Line . . . . 10

1.3 Solving Inequalities . . . . . . . . . . . . . . . . . . . 11

1.4 Absolute Value and Distance on the real number line 14

1.4.1 Intervals Defined by Absolute Values . . . . . 15

1.4.2 Basic types of Inequalities Involving Absolute

Value . . . . . . . . . . . . . . . . . . . . . . . 15

2 Functions and graphs 19

2.1 The Cartesian plane . . . . . . . . . . . . . . . . . . 19

2.2 Functions and Graphs . . . . . . . . . . . . . . . . . 22

2.3 Domain of a function . . . . . . . . . . . . . . . . . 24

2.3.1 Domain of a Polynomial . . . . . . . . . . . . 24

2.3.2 Domain of a Rational Function . . . . . . . . 24

2.4 Range of a function . . . . . . . . . . . . . . . . . . 26

2.5 Graphs of Functions and Relations . . . . . . . . . . 26

2.6 The Vertical Line Test . . . . . . . . . . . . . . . . . 28

2.7 Lines in the Plane and Slope . . . . . . . . . . . . . 30

2.7.1 Finding the slope of a Line . . . . . . . . . . . 32

2.7.2 Point-Slope Form of the equation of a Line . . 33

2.7.3 Equation of a Line passing through two points 34

3

4 CONTENTS

3 Limit and Continuity 37

3.1 The Limit of a Function . . . . . . . . . . . . . . . . 37

3.1.1 Rules for Limits . . . . . . . . . . . . . . . . . 39

3.2 Limit of a polynomial Function . . . . . . . . . . . . 40

3.3 The replacement Theorem . . . . . . . . . . . . . . . 41

3.3.1 Limit of a rational function . . . . . . . . . . 41

3.3.2 The Root Trick . . . . . . . . . . . . . . . . . 43

3.4 One-Sided Limits . . . . . . . . . . . . . . . . . . . . 44

3.5 Continuity . . . . . . . . . . . . . . . . . . . . . . . . 46

3.5.1 Continuity of a polynomial . . . . . . . . . . 48

3.5.2 Continuity of a Rational Function . . . . . . . 48

4 Derivatives and Applications 51

4.1 The Derivative of a Function . . . . . . . . . . . . . . 51

4.2 Geometrical Meaning of f ′(x) . . . . . . . . . . . . . 52

4.3 Some Rules of Differentiation . . . . . . . . . . . . . 55

4.4 Equation of Tangent Line . . . . . . . . . . . . . . . 56

4.5 Rate of Change . . . . . . . . . . . . . . . . . . . . . 57

4.6 Rates of Change in Economics: Marginals . . . . . . 58

4.7 The product Rule . . . . . . . . . . . . . . . . . . . . 60

4.8 The Quotient Rule . . . . . . . . . . . . . . . . . . . 61

4.9 The Chain Rule . . . . . . . . . . . . . . . . . . . . . 62

4.9.1 The General Power Rule . . . . . . . . . . . . 62

4.10 Higher Order derivatives . . . . . . . . . . . . . . . . 64

4.11 Increasing and Decreasing Functions . . . . . . . . . 64

4.11.1 Test for Increasing and Decreasing Functions . 65

4.12 Critical Numbers and their Use . . . . . . . . . . . . 66

4.13 Extrema and the First Derivative Test . . . . . . . . 68

4.14 Absolute Extrema . . . . . . . . . . . . . . . . . . . . 69

4.15 The Second Derivative Test . . . . . . . . . . . . . . 70

4.16 Optimization in Business and Economics . . . . . . . 71

CONTENTS 5

5 Exponential and Logarithmic Functions 75

5.1 Exponential Functions . . . . . . . . . . . . . . . . . 75

5.1.1 Properties of Exponents . . . . . . . . . . . . 75

5.1.2 Graphs of Exponential Functions . . . . . . . 76

5.2 Natural Exponential Function . . . . . . . . . . . . . 76

5.3 Business Application: Compound Interest . . . . . . 78

5.4 Finding present value . . . . . . . . . . . . . . . . . . 79

5.5 Derivatives of Exponential Functions . . . . . . . . . 79

5.6 Logarithmic Functions . . . . . . . . . . . . . . . . . 80

5.6.1 The Natural Logarithm Function . . . . . . . 81

5.6.2 Properties of Logarithms . . . . . . . . . . . . 81

5.6.3 Solving Exponential and Logarithmic Equations 82

5.7 Derivatives of Logarithmic Functions . . . . . . . . . 83

6 Function of Two Variables 85

6.1 Partial Differentiation . . . . . . . . . . . . . . . . . 86

6.1.1 Partial Derivatives of a Function of Two Vari-

ables . . . . . . . . . . . . . . . . . . . . . . . 86

6.1.2 Higher order Partial derivatives . . . . . . . . 88

6.2 Relative Extrema . . . . . . . . . . . . . . . . . . . . 88

6.2.1 Critical points . . . . . . . . . . . . . . . . . . 89

6.2.2 The Second -Partial Test for Relative Extrema 90

7 Sequences and Series 93

7.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . 93

7.1.1 The Limit of a Sequence . . . . . . . . . . . . 94

7.1.2 Pattern Recognition for Sequences . . . . . . . 94

7.1.3 Application . . . . . . . . . . . . . . . . . . . 95

7.2 Series and Convergence . . . . . . . . . . . . . . . . . 95

7.2.1 Infinite Series . . . . . . . . . . . . . . . . . . 95

7.3 Geometric Series . . . . . . . . . . . . . . . . . . . . 96

7.3.1 Convergence of an Infinite Geometric Series . 98

6 CONTENTS

0.1 Introduction

You may ask : Why should I study Mathematics, while I am a Man-

agement Student? The answer is : you need Mathematics to under-

stand many concepts in Management sciences. There are plenty

of applications of Mathematics, such as finding the optimal profit

or minimal cost. Mathematics is also an essential background for

many further courses in Management sciences such as operational

research and finance. This course is planed in order to enable you

to use Mathematics to solve real life problems.

Chapter 1

Pre-Calculus

1.1 Numbers

Natural Numbers

The natural numbers 0, 1, 2, 3, 4, .. (we denote the Natural numbers

by N ) are used for counting. These numbers are not enough to deal

with daily life.

Integers

If you work in trading and you gained 200 LE on Saturday, but you

lost 50 LE On Sunday then the net income of you is 200− 50 = 150

LE. If Ahmad is working also in trading, and if He gained 100 LE

on Saturday and lost 200 LE on Sunday, then the net income of

Ahmad is 100− 200 which is not a natural number. In fact, the net

income of Ahmad is −100 LE. −100 is a negative number. negative

numbers, such as −1,−2,−3, ... form together with natural numbers

the Integers.., −3,−2,−1, 0, 1, 2, 3, .... The integers are denoted by

Z

Rational numbers

Although Integers solved some problems but we still need more num-

bers. For example if you know that the price of 4 pencils is 3 LE.

What is the price of a pencil? If we assume that the price of a pencil

is x, hence

4.x = 3

7

8 CHAPTER 1. PRE-CALCULUS

To get the value of x, we divide the equation by 4, then we write

x = 34, which is called a rational number

Definition

A number x is called a rational number if it can be written as x = ab,

where a and b are integers and b 6= 0. We denote the rational

numbers by Q

Properties of Rational Numbers

Let a, b, c, and d be integers, with b 6= 0 and d 6= 0. Then

(1) ab

= a.db.d

(2) −ab

= a−b

= −ab

(3) a1

= a

(4) ab

+ cd

= a.d+c.bb.d

(5) ab. cd

= a.cb.d

(6) ( bd)−1 = d

b

Example 1

a- 1015

= 23

b- 23

+ 35

= 2.5+3.33.5

= 1615

c- 27.35

= 635

d- (23)−1 = 3

2

Remark

You may noted that: the Natural Numbers are part of Integers, The

Integers are part of Rational Numbers.

1.1. NUMBERS 9

Irrational Numbers and Real numbers

The Real Numbers consists of Rational Numbers, and Irrational

numbers

The real numbers are denoted by R

Irrational Number is a real number which is not Rational

e.g.√

2 , π, 3√

5

No one of these numbers can be written in the form of ab, where a

and b are integers, and b 6= 0

The real numbers are usually represented by drawing the so-called

real number line

every point on this line represents a real number and

every real number can be found on the real line.

Remark

every real number has a decimal representation. Rational numbers

have either terminating or infinitely repeating decimal representa-

tion

for example, 25

= 0.4, a terminating decimal

but 13

= 0.333..., infinitely repeating decimal

Some irrational numbers occurs so frequently such as√

2, π, and e

(Euler number)

these number have decimal approximations as follows:√

2 ≈ 1.41424135623

π ≈ 3.1415926535

e ≈ 2.7182818284


1.2 Order and Intervals on the Real Number

Line

The real numbers are ordered: 0 is less than 1, −2 is less than −1.5,

and so on. a is less than b (symbolically a < b)if and only if a lies

to the left of b on the real number line.

Open Interval

The set of real numbers between a and b is called the open interval

between a and b and is denoted by ]a, b[ or (a, b)

for example the open interval ] − 1, 2[ is the set of all real numbers

between −1 and 2 and you have to note that the two end points −1

and 2 are not included in the interval.

a number x belongs to this interval satisfies the inequality

a < x < b

Closed Interval

The closed interval [a, b] is the set of all real numbers between a and

b including the end points a and b.

Intervals of the form [a, b[ and ]a, b] are neither open nor closed.

We can represent the intervals graphically as follows:

1.3. SOLVING INEQUALITIES 11

Note that the square bracket is used to denote ”less than or equal to”

(≤)” or greater than or equal to” (≥). Furthermore , the symbols

∞ and −∞ denote Positive and negative infinity. These symbols do

not denote real numbers; they merely let you describe unbounded

conditions more concisely.

1.3 Solving Inequalities

In Calculus, you are frequently required to solve inequalities involv-

ing variable expressions such as 3x − 4 < 5.

Properties of Inequalities

Let a, b, c, and d be real numbers.

1. a < b and b < c ⇒ a < c

2. a < b, and c < d ⇒ a + c < b + d

3. a < b ⇒ ac < bc , c > 0

4. a < b ⇒ ac > bc , c < 0

5. a < b ⇒ a + c < b + c


6. a < b ⇒ a − c < b − c

Note that similar properties hold if < is replaced by ≤

Example 2 Solving an Inequality

Find the solution set of the inequality 3x − 4 < 5

Solution

3x − 4 < 5

3x − 4 + 4 < 5 + 4 Add 4 to each side

3x < 9 Simplify

13(3x) < 1

3(9) Multiply each side by 1

3

x < 3 Simplify

So The Solution Set is the interval ] −∞, 3[, which can be repre-

sented graphically as follows

Example 3 Solving Inequality

Solve the inequality

a- 2x − 5 > 7

b- 4x + 1 < 2x

Solution

a- 2x − 5 > 7

2x > 7 + 5 Move 5 to the RH side

2x > 12 Simplify

x > 6 Divide by 2

Then the solution set is the interval ]6,∞[

Example 4 Production levels

In addition to fixed overhead costs of $500 per day, the cost of pro-

ducing x units of an item is $2.5 per unit. during the month of

1.3. SOLVING INEQUALITIES 13

August, the total cost of production varied from a high of $1325 to

a low of $1200 per day. Find the high and low production levels

during the month.

Solution

Since it costs $2.5 to produce a unit, then it costs $2.5x to produce

x units, but the fixed cost per day is $500, then the total daily cost

of producing x units is

C = 2.5x + 500

The cost ranged from $1200 to $1325, then we can write

1200 ≤ 2.5x + 500 ≤ 1325

Adding −500 to all sides, we get

1200 − 500 ≤ 2.5x + 500 − 500 ≤ 1325 − 500

⇒ 700 ≤ 2.5x ≤ 825

Divide each side by 2.5, we get

7002.5

≤ 2.5x2.5

≤ 8252.5

⇒ 280 ≤ x ≤ 330

So, the daily production levels during the month of August varied

from the low of 280 units to 330 units.

Example 5

The revenue for selling x units of a product is

R = 30x

and the cost of producing x units is

C = 23x + 300

To obtain a profit, the revenue must be greater than the cost. For

what values of x will this product return profit?

Solution

Profit P is defined as


P = R − C, so to obtain profit we should have:

R > C

⇒ 30x > 23x + 300

⇒ 7x > 300

x > 42.857

So this product will return profit if the number of units is 43 or

greater

1.4 Absolute Value and Distance on the real

number line

Definition of Absolute Value

The Absolute Value of a real number x is

|x| =

x, ifx ≥ 0

−x, ifx < 0.

For example,

to find |3|, since 3 > 0, then we apply the first rule to get |3| = 3.

To find | − 5|, since −5 < 0, then we apply the second rule to get

| − 5| = −(−5) = 5.

You can conclude that |a| can not be negative.

Properties of Absolute Value

1. |ab| = |a||b|2. |a

b| = |a|

|b| , b 6= 0

3. |an| = |a|n

4.√

a2 = |a|Distance on the real line

Consider two points a and b on the real line

1.4. ABSOLUTE VALUE AND DISTANCE ON THE REAL NUMBER LINE 15

The distance can not be negative. So the distance between a and b

is |a − b| or |b − a|Example 6

Determine the distance between −2 and 5 on the real line

Solution

| − 2 − 5| = | − 7| = 7

1.4.1 Intervals Defined by Absolute Values

Example 7

Find the interval on the real number line that contains all numbers

that lie no more than two units from 3

Solution

If x is any point in the interval, we need to find x such that the

distance between x and 3 is less than or equal 2. This implies that

|x − 3| ≤ 2.

then x − 3 lie between −2 and 2, then

− 2 ≤ x − 3 ≤ 2

Adding 3 to the sides, we get

1 ≤ x ≤ 5

So the interval is [1, 5]

1.4.2 Basic types of Inequalities Involving Absolute Value

There are two basic types of Inequalities Involving Absolute Value:

1- |x| ≤ d if and only if −d ≤ x ≤ d.

2- |x| ≥ d if and only if x ≤ −d or x ≥ d


Example 8

Solve the Following Inequalities

a- |x − 2| ≤ 4

b- |x + 1| ≥ 3

c- |x − 5| ≤ 2

d- |x − 1| ≥ 5

Solution

a-

|x − 2| ≤ 4

⇒ −4 ≤ x − 2 ≤ 4

⇒ −4 + 2 ≤ x ≤ 4 + 2

⇒ −2 ≤ x ≤ 6

hence the solution set is the interval [−2, 6]

b-

|x + 1| ≥ 3

⇒ x + 1 ≤ −3 or x + 1 ≥ 3

⇒ x ≤ −4 or x ≥ 2

x ≤ −4 means x belongs to the interval ]−∞,−4] , x ≥ 2 means x

belongs to the interval [2,∞[

hence , the solution set is the union of the two intervals ] −∞,−4]

and [2,∞[, which can be written as

] −∞,−4] ∪ [2,∞[

c-

|x − 5| ≤ 2

⇒ −2 ≤ x − 5 ≤ 2

⇒ −2 + 5 ≤ x ≤ 2 + 5

⇒ 3 ≤ x ≤ 7

1.4. ABSOLUTE VALUE AND DISTANCE ON THE REAL NUMBER LINE 17

hence the solution set is the interval [3, 7]

d-

|x − 1| ≥ 5

⇒ x − 1 ≤ −5 or x − 1 ≥ 5

⇒ x ≤ −4 or x ≥ 6

x ≤ −4 means x belongs to the interval ]−∞,−4] , x ≥ 6 means x

belongs to the interval [6,∞[

hence , the solution set is the union of the two intervals ] −∞,−4]

and [6,∞[, which can be written as

] −∞,−4] ∪ [6,∞[

Example 9

The estimated daily production x at a refinery is given by:

|x − 200, 000| ≤ 25, 000

Where x is measured in barrels of oil. Determine the high and low

production levels

Solution

|x − 200, 000| ≤ 25, 000

⇒ −25, 000 ≤ x − 200, 000 ≤ 25, 000

⇒ −25, 000 + 200, 000 ≤ x ≤ 25, 000 + 200, 000

⇒ 175, 000 ≤ x ≤ 225, 000

hence the high production level is 225, 000 barrels , and the low pro-

duction level is 175, 000 barrels.

Solve the following example by yourself

Solve the inequality

a- |x − 1| < 9

b- |x + 2| > 4

Chapter 2

Functions and graphs

2.1 The Cartesian plane

The Cartesian plane is formed by using two real number lines inter-

secting at right angles. The horizontal line is usually called x-axis,

and the vertical line is usually called the y-axis. The point of inter-

section of these two axes is the origin, and the two axes divide the

plane into four parts called quadrants.

Each point in the plane corresponds to an ordered pair (x, y) of real

numbers x and y, called coordinates of the point. The x-coordinate

19

20 CHAPTER 2. FUNCTIONS AND GRAPHS

represents the directed distance from the y-axis to the point, and

the y-coordinate represents the directed distance from the x-axis to

the point.

Example 1

Plot the points (−1, 2), (3, 4), (3, 0), (−2,−3)

Solution

-2 -1 1 2 3

-3

-2

-1

1

2

3

4

H3,0L

H3,4L

H-1,2L

H-2,-3L

2.1. THE CARTESIAN PLANE 21

Distance between two points

Consider the two points in the cartesian plane (a, b) and (c, d).

The distance between these two points can be calculated using the

Pythagorean theorem

S2 = (c − a)2 + (d − b)2

S =√

(c − a)2 + (d − b)2

The Distance formulaThe distance S between the points (a, b) and (c, d) in the plane is

S =√

(c − a)2 + (d − b)2

Example 2

Find the distance between the two points (3,−1) and (2, 4)

Solution

S =√

(2 − 3)2 + (4 − (−1))2 =√

1 + 25 =√

26


2.2 Functions and Graphs

Function

Consider that D is a subset of the real numbers R , then we define

a function on D as

Definition

A function on D is a rule which assigns to each x in D exactly one

real number y, and we write f(x) = y. the function itself is denoted

by

f : D → R

Example 3

1- The rule given by f(x) = 3x is a function, since each x is assigned

to a triple of x which is unique. For example f(1) = 3, f(3) = 9

2- The rule f(x) = ±√x is not a function since it assigns two values

√x and −√

x for each x

Definition

A polynomial is a function of the form

f(x) = anxn + an−1xn−1 + ... + a1x + a0

where n is a natural number, a,is are real numbers, with an 6= 0

n is called the degree of the polynomial

where n is a natural number, a,is are real numbers, with an 6= 0

n is called the degree of the polynomial

an is called the leading coefficient

a0 is called the absolute coefficient

Example 4

For each of the following polynomials, determine the degree, the

leading coefficient, and the absolute coefficient

2.2. FUNCTIONS AND GRAPHS 23

(a) f(x) = 4x3 − 2x + 3

(b) f(x) = −3x5 + 3x4 + 1.2

(c) f(x) = 9x2 + 4x

Solution

(a) Degree, n = 3, Leading coefficient = 4, Absolute

coefficient = 3

(b) Degree, n = 5, Leading coefficient = −3, Absolute

coefficient = 1.2

(c) Degree, n = 2, Leading coefficient = 9, Absolute

coefficient = 0

Notes

1- A linear function f(x) = mx + c is a polynomial of degree 1

2- A constant function f(x) = c, where c is constant is a polynomial

of degree 0

Rational Function Definition

A function f(x) is said to be rational if f(x) = g(x)h(x)

, where g(x) and

h(x) are polynomials, and h(x) 6= 0

For Example, The following are rational functions

f(x) = 2x2−3x3−2

K(x) = x−9x2+3x

Consider the function f(x) = 43−x

. To get the value of this function

at x = 1, we simply substitute in the rule of the function to get

f(1) = 43−1

= 42

= 2. Similarly we can find that f(2) = 42−1

= 4.

But we can not find a value for f(x) at x = 3, since in this case

f(3) = 40, but it is not allowed to divide by 0, which means that the

function f is not defined at x = 3. In this case we say that x = 3 is


not in the domain of f(x)

2.3 Domain of a function

The domain of f(x) is the set of real numbers for which the function

f(x) is defined

2.3.1 Domain of a Polynomial

A polynomial is a function of the form

f(x) = anxn + an−1xn−1 + ... + a1x + a0

From the definition of a polynomial, it is easy to realize that the

domain of a polynomial is the set of all Real numbers R

2.3.2 Domain of a Rational Function

Consider the rational function f(x) = x−1x2−9

, you find that this func-

tion is not defined for values of x that make the denominator= 0;

we can get these values by solving the equation x2 − 9 = 0 ⇒ x2 =

9 ⇒ x = ±3. This means that domain of the function is all real

numbers except for zeroes of the denominator.

This means that the Domain of f(x) = R − {3,−3}

We can generalize this method to find the domain of a Ra-

tional Function f(x) = g(x)h(x)

2.3. DOMAIN OF A FUNCTION 25

Step 1 Find the zeroes of the denominator by solving the equation

h(x) = 0

Step 2 The Domain of the function f(x) = R − { Zeroes of the

denominator}

Example 5

Find the domain of the function f(x) = x3

x2−25

Solution

Step 1

To find the zeroes of the denominator,we solve the equation x2−25 =

0 ⇒ x2 = 25 ⇒ x = ±5

Step 2

The domain of f(x) is R − {5,−5}

Example 6

Find the domain of the following functions

(a)- f(x) = 1x2−9

(b)- g(x) =√

x − 2

Solution

(a)- The function f is defined for all real numbers except for the

numbers which make the denominator equal zero. But the denom-

inator is 0 if x2 − 9 = 0 ⇒ x2 = 9 ⇒ x = 3 or −3 . Hence the

domain of f is R − {3,−3}(b)- The square root is defined only for non-negative numbers, the

g(x) is defined for values of x satisfying x − 2 ≥ 0 ⇒ x ≥ 2, hence

the domain of g(x) is the interval [2,∞)


2.4 Range of a function

The range of the function f(x) is the all real numbers y for which

there is some x with y = f(x)

Example 7

From the graph, find the range of the function

f(x) = x2 − 1

-2 -1 1 2

-1

1

2

3

Solution

From the figure, we can find that the values of f(x) are in the in-

terval [−1,∞)

2.5 Graphs of Functions and Relations

Relation

An expression which involves the two variables x and y is called a

relation between x and y.

Example 8

(a) the expression x2 + y2 = 25 defines a relation between x and y.

(b) the expression y = 2x − 3 defines a relation between x and y .

(c) the expression x.y − 3 = 6y defines a relation between x and y.

You may also note that every function defines a relation. If f(x) is

a function, then the corresponding relation is given by

2.5. GRAPHS OF FUNCTIONS AND RELATIONS 27

f(x) = y

The Graph of a Relation

The graph of a relation is given by all points (a, b) which satisfy the

relation. The graph of a function f(x) is the graph of the relation

f(x) = y.

Example 9

(a) Consider the function f(x) = x2 − 4, the corresponding graph is

given by

-4 -2 2 4

5

10

15

20

(b) the graph of the relation x2 + y2 = 25 is given by


-4 -2 2 4

-4

-2

2

4

2.6 The Vertical Line Test

From the graph of a relation we can determine if this relation is a

function or not.

As we know a function assigns every x to exactly one y. So a graph

of a relation is a graph of a function if on every vertical line there

is at most one point of the graph.

For Example, if we consider the two graphes of Example 9. The

first graph is a graph of a function , since on every vertical line there

is at most one point of the graph.

The second graph is not a graph of a function , if you draw a vertical

line at x = 1, you will find that two points of the graph lie on this

line.

The Vertical Line TestThe graph of a relation is the graph of a function if on every verticalline there is at most one point of the graph

2.6. THE VERTICAL LINE TEST 29

Example 10

Decide whether the following graphs are graphs of functions or not.

(a) y = x3 − 3x + 4

-4 -2 2 4

-2.5

2.5

5

7.5

10

If you draw any vertical line, you find that the line intersects the

graph of the relation in at most one point.

(b ) x =√

|y|

0.5 1 1.5 2

-4

-2

2

4

you can note that there is a vertical line intersects the graph of the

relation in two points, which means that the relation is not a func-

tion.


2.7 Lines in the Plane and Slope

The simplest mathematical model for relating two variables is the

linear equation y = mx + c. The equation is called linear because

its graph is a straight line

for example the relation y = 2x + 1 represents a straight line

-2 -1 1 2x

-2

2

4

y

Returning to the straight line equation y = mx + c, if you let x = 0

you see that the line intersects the y-axis at y = c, in other words

the y-intercept is (0, c).

The Slope of a line is the number of units rises (or falls) vertically

for each unit of horizontal change from left to right.

Example 11

Find the slope and y-intercept of the straight line then graph this

line

(a) 2y − 4x = 8

(b) x + y = 4

Solution

(a) we first write the equation in the standard form y = mx + c

2y − 4x = 8, divided by 2, we get, y − 2x = 4 ⇒ y = 2x + 4

comparing with the standard form, we find that the y-intercept

c = 4, and the slope m = 2

The graph will take the form

2.7. LINES IN THE PLANE AND SLOPE 31

-3 -2 -1 1 2 3

-2

2

4

6

8

10

Example 21 Using Slope as a Rate of Change

A manufacturing company determines that the total cost in dollars

of producing x units of a product is C = 25x + 3500. Decide the

practical significance of the y-intercept and slope of the line given

by the equation.

Solution

The y-intercept (0, 3500) tells you that the cost of producing zero

units is $3500. This is the Fixed cost of production-it includes

costs that must be paid even if there is no production.

The slope of m = 25 tells you that the additional cost of producing

each unit is $25, as shown in figure.

Economists call the cost per unit the Marginal cost. If the

production increases by one unit, then the ”margin” of extra amount

of cost is $25


20 40 60 80 100x

1000

2000

3000

3500

4000

5000

6000C

2.7.1 Finding the slope of a Line

The slope m of the line passing through (x1, y1) and (x2, y2) is

m =∆y

∆x=

y1 − y2

x1 − x2

, where x1 6= x2

Example 13

Find the slope of the line passing through each pair of points.

(a) (−2, 0) and (3, 1) (b) (−1, 2) and (2, 2)

(c) (2, 5) and (2, 1)

Solution

(a) Let (x1, y1) = (3, 1) and (x2, y2) = (−2, 0) ,then the slope is

given by

m = y2−y1

x2−x1= 1−0

3−(−2)= 1

5

(b) the slope can be calculated as

m = 2−22−(−1)

= 03

= 0


(c) the slope is calculated as

m = 1−52−2

= −40

this means that the slope is not defined

The slope of a horizontal line is zero,while the slope of a vertical line is not defined

2.7.2 Point-Slope Form of the equation of a Line

The equation of the line with slope m passing through the point

(x1, y1) is given by

y − y1 = m(x − x1)

Example 14

Find the equation of the line that has a slope of 3 and passes through

the point (1,−2)

Solution

Using the point-slope form with m = 3 and (x1, y1) = (1,−2) we

get

y − y1 = m(x − x1)


y − (−2) = 3(x − 1)

y + 2 = 3x − 3

y = 3x − 5

2.7.3 Equation of a Line passing through two points

Consider a line that passes through the two given points (x1, y1) and

(x2, y2). To get the equation of this line we first calculate the slope

as:

m =y1 − y2

x1 − x2

then we use the point-slope form to write the equation of the line

using one of the points and the slope m

Example 15

Find the equation of the line that passes through the points (1,−3)

and (2, 3)

Solution

the slope is

m =3 − (−3)

2 − 1=

6

1= 6

x1 = 1 , y1 = −3

Substituting in the formula

y − y1 = m(x − x1)

, we get

y − (−3) = 6(x − 1)

⇒ y + 3 = 6x − 6

⇒ y = 6x − 9

Example 16

A company constructs a warehouse for $825, 000. The warehouse


has an estimated useful life of 25 years, after which its value is ex-

pected to be $75, 000. Write a linear equation giving the value y of

warehouse during its 25 years of useful life. (Let t represents the

time in years)

Solution

In the beginning (t = 0), the value y = 825000. When t = 25, the

value y = 75000. Since the relation between the value y and the

time t is assumed to be linear, then the two points (0, 825000) and

(25, 75000) lie on the line . Hence the equation of the line is given by

y − y1 = m(x − x1)

, where

m = 75000−82500025−0

= −75000025

= −30000

hence

y − 825000 = −30000(t − 0)

⇒ y = −30000t + 825000

Example 17

Your annual salary was $26, 300 in 2002 and $29, 700 in 2004. As-

sume your salary can be modeled by a Linear Equation.

(a) Write a linear equation giving your salary S in terms of the year.

Let t = 2 represent 2002.

(b) Use the linear model to predict your salary in 2008.

Solution

Chapter 3

Limit and Continuity

3.1 The Limit of a Function

To understand the concept of a limit of a function, consider the fol-

lowing two examples

We have a function f(x) = x2 − x + 2, we need to investigate the

behavior of this function for values of x near 2. The following table

gives the values of f(x) for values of x close to 2. BUT NOT EQUAL TO 2

x f(x) x f(x)1.0 2.000000 3.0 8.0000001.5 2.750000 2.5 5.7500001.8 3.440000 2.2 4.6400001.9 3.710000 2.1 4.3110001.95 3.710000 2.05 4.1525001.99 3.970100 2.01 4.0301001.995 3.985025 2.005 4.0150251.999 3.997001 2.001 4.003001

37

38 CHAPTER 3. LIMIT AND CONTINUITY

-1 1 2 3 4

2

4

6

8

10y=x2-x+2

From the table and the graph of f shown in figure we see that

when x is close to 2 (on either side of 2), f(x) is close to 4. We

express this by saying ”the limit of the function f(x) = x2 − x + 2

as x approaches 2 is equal to 4.”

The notation of this is

limx→2

(x2 − x + 2) = 4

Now consider another example f(x) = x2−1x−1

Let us study the limit of f as x approaches 1. It is clear that f(x)

is undefined at x = 1. But the limit depends only on values of f(x)

near 1, not at 1.

Now let x 6= 1, then f(x) = (x−1)(x+1)x−1

⇒ f(x) = x + 1, then for

x 6= 1, the function behaves as the linear function f(x) = x + 1

represented graphically by the line y = x + 1

3.1. THE LIMIT OF A FUNCTION 39

It is clear that as x approaches 1, the value of f(x) approaches 2.

This can be checked numerically by taking values of x close to 1,

and we find the value of f(x) approaches 2.

Definition

If f(x) becomes arbitrary close to a single number L as x approaches

a from either side, then

limx→a

f(x) = L

which is read as ”the limit of f(x) as x approaches c is L”

3.1.1 Rules for Limits

Let c and a are real numbers, and let n be a natural number,then

Rule 1- limx→a

c = c

this means that the limit of a constant is the same constant

For example limx→5

7 = 7, limx→3

7 = 7,

Rule 2- limx→a

xn = an


this means that to find the limit of a monomial you just replace

x with a


x4 = 54 = 625, limx→3

x2 = 32 = 9,

Rule 3- limx→a

(f(x) ± g(x)) = limx→a

f(x) ± limx→a

g(x)

this means that the limit is distributed on plus or minus signs

For example

limx→4

(x2 + x4) = limx→4

x2 + limx→4

x4 = 42 + 44 = 16 + 256 = 262

Rule 4- limx→a

cf(x) = c limx→a

f(x)


8x2 = 8 limx→3

x2 = 8(32) = 72

Rule 5- limx→a

[f(x).g(x)] = [limx→a

f(x)].[limx→a

g(x)]

Rule 6- limx→a

f(x)

g(x)=

limx→a f(x)

limx→a g(x), if lim

x→ag(x) 6= 0

Rule 7- limx→a

n√

x = n√

a

3.2 Limit of a polynomial Function

To find the limit of a polynomial function we follow the rule:

limit of a polynomial functionif f(x) is a polynomial then lim

x→af(x) = f(a)

3.3. THE REPLACEMENT THEOREM 41

This means that to find the limit of a polynomial as x approaches

a, replace x with a

Then the limit of a polynomial as x approaches a is equal

to the value of the polynomial at x = a

Example 18

If f(x) = x2 − x + 2, find limx→2

f(x)

Solution

since f(x) is a polynomial then we find the limit of it by direct sub-

stitution

limx→2

f(x) = limx→2

(x2 − x + 2) = 22 − 2 + 2 = 4

3.3 The replacement Theorem

Let c be a real number and let f(x) = g(x) for all x 6= c. If the limit

of g(x) exists as x → c, then the limit of f(x) also exists and

limx→c

f(x) = limx→c

g(x)

3.3.1 Limit of a rational function

Let f(x) = g(x)h(x)

be a rational function, then to find limx→a

f(x) =

limx→a

g(x)

h(x)we proceed as follows:

Check whether h(a) = 0 or not :

Case 1 if h(a) 6= 0, then the limit can be obtained by direct substi-

tution, and we have

limx→a

f(x) = limx→a

g(x)

h(x)=

g(a)

h(a)

Case 2 If h(a) = 0, then factorize g(x) and h(x) as f(x) = (x −a)f̃(x), and g(x) = (x − a)g̃(x)


then

limx→a

f(x) = limx→a

g(x)

h(x)= lim

x→a

(x − a)g̃(x)

(x − a)h̃(x)= lim

x→a

g̃(x)

h̃(x)

Example 19

Find the following limits

(a) limx→1

x + 3

x2 − 2

(b) limx→2

x2 − 3x + 2

x − 2Solution

(a) If we replace x with 1, then we find that the denominator

12 − 2 = −1 6= 0, then

limx→1

x + 3

x2 − 2=

1 + 3

12 − 2=

4

−1= −4

(b) If we replace x with 2, we find that the denominator will be

2 − 2 = 0. Then to find the limit we factorize the numerator

limx→2

x2 − 3x + 2

x − 2= lim

x→2

��(x − 2)(x − 1)

��(x − 2)=

limx→2

(x − 1) = 2 − 1 = 1

Example 20

Find the limit

limx→−3

x2 + x − 6

x + 3solution

Direct substitution fails since both numerator and denominator are

zero when x = −3. Then we have to factorize numerator

limx→−3

x2 + x − 6

x + 3= lim

x→−3

��(x + 3)(x − 2)

��x + 3= lim

x→−3(x − 2) = −5

3.3. THE REPLACEMENT THEOREM 43

3.3.2 The Root Trick

This technique is used to rationalize the numerator , which is useful

in calculating limits involving roots.

Example 21

Find the limit: limx→0

√x + 1 − 1

xSolution


zero when x = 0. We rationalize the numerator through multiply-

ing numerator and denominator by the conjugate of the numerator

which is√

x + 1 + 1

then

limx→0

√x + 1 − 1

x= lim

x→0

√x + 1 − 1

x(

√x + 1 + 1√x + 1 + 1

)

= limx→0

(x + 1) − 1

x(√

x + 1 + 1)

= limx→0

�x

�x(√

x + 1 + 1)

=limx→0

1√x + 1 + 1

Replacing x with 0, we get

limx→0

√x + 1 − 1

x=

1

1 + 1=

1

2

Example 22

Find the limit: limx→0

√x + 4 − 2

xSolution


zero when x = 0. We rationalize the numerator through multiply-

ing numerator and denominator by the conjugate of the numerator

which is√

x + 4 + 2

then

limx→0

√x + 4 − 2

x= lim

x→0

√x + 4 − 2

x(

√x + 4 + 2√x + 4 + 2

)

= limx→0

(x + 4) − 4

x(√

x + 4 + 2)


= limx→0

�x

�x(√

x + 4 + 2)

=limx→0

1√x + 4 + 2

Replacing x with 0, we get

limx→0

√x + 4 − 2

x=

1

2 + 2=

1

4

3.4 One-Sided Limits

Definition

Let f(x) be a function and a is a real number.

1- If there is a real number L , such that f(x) approaches L as x

approaches a from left, we write

limx→a−

f(x) = L

and we say L is the left-sided limit of f(x)

2-If there is a real number K , such that f(x) approaches K as x

approaches a from right, we write

limx→a+

f(x) = K

and we say K is the left-sided limit of f(x)

If the left limit is equal to the right limit is equal to L for example,

then the limit of the function f(x) as x → a exists and it is equal

to L.

But if the left limit is not equal to the right limit, then the limit of

f(x) as x → a does not exist.

Example 23

Find the limit of f(x) as x approaches 1.

f(x) =

{

4 − x, x < 1

4x − x2, x > 1

3.4. ONE-SIDED LIMITS 45

Solution

We have to remember that we are concerned with the value of f(x)

near x = 1 rather than x = 1.

To get the left-sided limit limx→1−

f(x) we note that for x < 1, f(x) =

4 − x. Then

limx→1−

f(x) = limx→1−

(4 − x) = 4 − 1 = 3.

For x > 1, f(x) = 4x − x2, then the right-sided limit is given by:

limx→1+

f(x) = limx→1+

(4x − x2) = 4(1) − 12 = 3

One can note that limx→1−

f(x) = limx→1+

f(x) = 3.

In this case we say that the limit of the function f(x) as x → 1

exists and in fact

limx→1

f(x) = 3

Example 24

Find the limit as x → 0 from the left and the limit as x → 0 from

the right for the function

f(x) =

{

2x + 1, x < 0

4x2 − 2, x > 0

Solution

To get the left-sided limit limx→0−

f(x) we note that for x < 0, f(x) =

2x + 1. Then

limx→0−

f(x) = limx→0−

2x + 1 = 1.

For x > 0, f(x) = 4x2 − 2, then the right-sided limit is given by:

limx→0+

f(x) = limx→0+

(4x2 − 2) = 4(0) − 2 = −2

One can note that limx→0−

f(x) 6= limx→0+

f(x).

In this case we say that the limit of the function does not exist


3.5 Continuity

Definition

A function f(x) is said to be continuous at a number a if

limx→a

f(x) = f(a)

Note that this definition requires three things if f is continuous at

a:

1- f(a) is defined (that is a is in the domain of f)

2- limx→a

exists

3- limx→a

= f(a)

If one of these three things is not satisfied for some number a, then

f(x) is not continuous (discontinuous) at a.

Study carefully the following examples

Example 25

Given f(x) = x2−1x−1

.

Is f(x) continuous at 1?

Solution

f(x) is not defined at x = 1. That is 1 is not in the domain of f(x).

Hence f(x) is not continuous at x = 1.

One can observe this from the graph of f(x)

3.5. CONTINUITY 47

From the graph, you note that, there is a discontinuity when x = 1

because the graph has a break there.

Example 26

f(x) =

{

x2− 1, x ≤ 2

x + 1, x > 2

Is f(x) continuous at x = 2

Solution

From the definition of f(x), we find that f(2) = 22− 1 = 1 − 1 = 0.

That is f(2) is defined.

Now we proceed to find the limx→2

f(x). To find such limit we observe

that for x ≤ 2 the function is defined by a rule different from the

rule of definition for x > 2. So we have to find the left limit and the

right limit of the function as x approaches 2

The right-sided limit limx→2+

f(x) = limx→2+

(x + 1) = 2 + 1 = 3. You

note that to get the right limit we used the rule of the function for

x > 2.

The left-sided limit limx→2−

f(x) = limx→2−

(x

2− 1) = 1− 1 = 0. You note

that to get the left limit we used the rule of the function for x < 2.

Since limx→2+

f(x) 6= limx→2−

f(x), then limx→2

f(x) does not exist.

Then we deduce that the function is not continuous at x = 2, which

is clear from the graph of the function, there is a jump at x = 2


-4 -2 2 4

-2

2

4

6

Example 27

f(x) =

{

2, x = 1

4x + 1, x 6= 1

Is f(x) continuous at x = 1 ?

Solution

f(1) = 2 That is f(1) is defined

limx→1

f(x) = limx→1

(4x + 1) = 5

Since limx→1

f(x) 6= f(1), then f(x) is not continuous at x = 1.

3.5.1 Continuity of a polynomial

Every polynomial is continuous on R, i.e. continuous at any real

number

for example the function f(x) = x4 − 2x3 + x − 4 is a polynomial .

Then f(x) is continuous for any real number.

3.5.2 Continuity of a Rational Function

Every rational function is continuous on its domain.

for example f(x) = x2+2xx2−25

. f(x) is continuous on its domain, i.e.

f(x) is continuous on R − {5,−5}Example 28

Describe the interval(s) for which the function is continuous

(a) f(x) = x2−1x

3.5. CONTINUITY 49

(b) g(x) =

{

x3 − 2x, x ≤ 1

x + 1, x > 1

Solution

(a) f(x) is a rational function which is continuous on its domain.

The domain of f(x) is R−{0}. So f(x) is continuous on (−∞, 0)⋃

(0,∞).

(b) For x 6= 1, g(x) is a polynomial which is continuous. So it is

enough to investigate the continuity of g(x) at x = 1

g(1) = 13 − 2(1) = −1, hence g(1) is defined.

The right-sided limit is given by:

limx→1+

f(x) = limx→1+

(x + 1) = 1 + 1 = 2

The left-sided limit is given by:

limx→1−

f(x) = limx→1−

(x3 − 2x) = 1 − 2(1) = −1

since the right limit does not equal the left limit , then limx→1

f(x)

does not exist. So g(x) is not continuous at x = 1. Hence g(x) is

continuous on (−∞, 1)⋃

(1,∞)

Chapter 4

Derivatives andApplications

4.1 The Derivative of a Function

The derivative of f(x) at x is given by

f ′(x) = limh→0

f(x + h) − f(x)

h

Provided this limit exists. A function at x is differentiable at x if

its derivative exists at x. The process of finding derivatives is called

differentiation

Example 29

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

Solution

f ′(x) = limh→0

f(x + h) − f(x)

h

= limh→0

2(x + h) + 5 − (2x + 5)

h

= limh→0

2x + 2h + 5 − 2x − 5

h

= limh→0

2h

h= lim

h→02

= 2

51

52 CHAPTER 4. DERIVATIVES AND APPLICATIONS

Note

In addition to f ′(x), other notation can be used to denote the deriva-

tive of y = f(x). The most common are:dydx

, y′, and ddx

[f(x)]

Example 30

Find the derivative of f(x) = 3x2 − 2x

Solution

f ′(x) = limh→0

f(x + h) − f(x)

h

= limh→0

3(x + h)2 − 2(x + h) − (3x2 − 2x)

h

= limh→0

3x2 + 6xh + 3h2 − 2x − 2h − (3x2 − 2x)

h

= limh→0

3x2 + 6xh + 3h2 − 2x − 2h − 3x2 + 2x

h

= limh→0

6xh + 3h2 − 2h

h

= limh→0

(6x + 3h − 2)h

h= lim

h→06x + 3h − 2

= 6x − 2

4.2 Geometrical Meaning of f ′(x)

To understand the geometrical meaning of f ′(x), consider two points

on the graph of f(x).

4.2. GEOMETRICAL MEANING OF F′(X) 53

As shown in figure , the coordinates of the two points are P (x, f(x)),

and Q(x + h, f(x + h)). The line connecting the two points P and

Q is called the secant line . The slope of the secant line is given by:

msec =f(x + h) − f(x)

h

As h approaches zero, the point Q becomes closer to the point p,

and the secant line tends to be the tangent line to the graph of f(x)

at the point p, as shown in figure.


In this case limh→0

f(x + h) − f(x)

h= m the slope of the tangent

line to the graph of f(x) at x

To summarizeThe derivative of f(x) at x is given by

f ′(x) = limh→0

f(x + h) − f(x)

hwhich is the slope of the tangent

line to the graph of f(x) at x .

Example 31

Find a formula for the slope of the graph of f(x) = x2 + 1. What

are the slopes at the points (−1, 2) and (2, 5) ?

Solution

The slope of the graph is given by

4.3. SOME RULES OF DIFFERENTIATION 55

m = f ′(x) = limh→0

f(x + h) − f(x)

h

= limh→0

(x + h)2 + 1 − (x2 + 1)

h

= limh→0

x2 + 2xh + h2 + 1 − (x2 + 1)

h

= limh→0

2xh + h2

h

= limh→0

(2x + h)h

h= lim

h→02x + h

= 2x

Using the formula for the slope m = 2x we can find the slopes at

the specified points. At (−1, 2) the slope is m = 2(−1) = −2, and

at (2, 5), the slope is m = 2(2) = 4.

Differentiability implies continuity

If the function f(x) is differentiable at x = a, then f(x) is continu-

ous at x = a.

This means that if the function is discontinuous at some point , then

it has no derivative at that point.

4.3 Some Rules of Differentiation

Rule 1- (The constant Rule )If f(x) = c where c is constant, then f ′(x) = 0

this means that the derivative of a constant is zero

For Example Let f(x) = 5, then f ′(x) = 0


Rule 2- (The simple power rule)If f(x) = xn, then f ′(x) = nxn−1, where n is any real

number

For Example Let f(x) = x4, then f ′(x) = 4x3

Rule 3- (The constant multiple rule)ddx

[cf(x)] = c ddx

f(x)

For Example

Let f(x) = 4x7, then f ′(x) = 4(7x6) = 28x6

Rule 4- (The sum rule )If f(x) and g(x) are differentiable thenddx

[f(x) + g(x)] = df(x)dx

+ dg(x)dx

For Example Let f(x) = 3x4 + 5x−3

then f ′(x) = 3(4x3) + 5(−3x−4) = 12x3 − 15x−4

4.4 Equation of Tangent Line

To find an equation of the tangent line to the graph of f(x) at some

point (x1, y1), we use the formula

y − y1 = m(x − x1)

where m is the slope of the tangent to the graph of f(x) at the point

(x1, y1), in other words , m = f ′(x) at (x1, y1)

Example 32:

Find an equation of the tangent line to the graph of f(x) = x2 − 2

at the point (1,−1)

Solution

The equation of a straight line is given by:

4.5. RATE OF CHANGE 57

y − y1 = m(x − x1).

as we know f ′(x) gives the slope of the tangent

since f ′(x) = 2x, then the slope at the point (1,−1) is given by

m = 2(1) = 2

then the equation of the tangent line is given by

y − (−1) = 2(x − 1)

y + 1 = 2(x − 1)

y + 1 = 2x − 2

y = 2x − 3

Example 33

Find an equation of the tangent line to the graph of the function

f(x) = x3 + x at the point (−1,−2)

Solution

To get the slope of the line we find f ′(x)

f ′(x) = 3x2 + 1, then the slope of the tangent line at the point

(−1,−2) is given by substitution of x = −1 in f ′(x)

Slope = m = 3(−1)2 + 1 = 3 + 1 = 4

then the equation of the tangent line is given by

y − y1 = m(x − x1)

y − (−2) = 4(x − (−1))

y + 2 = 4x + 4

y = 4x + 2

4.5 Rate of Change

In the previous section we have studied one of the applications of

Derivatives: That is; f ′(x) gives the slope of the graph of f(x) at a

point (x, f(x)).

In this lecture we study another application of derivatives which is:

Rate of Change: f ′(x) gives the rate of change of f(x) with re-

spect to x at the point (x, f(x)).


Example 34

From 1998 through 2003, the revenue R (in millions of dollars per

year) for Microsoft Corporation can be modeled by:

R = 174.343t3 − 5630.45t2 + 63, 029.8t− 218, 635,

8 ≤ t ≤ 13

where t = 8 represent 1998. At what rate was Microsoft’s revenue

changing in 1999 ?

Solution

The Rate of change of R with respect to time is dRdt

. So we begin

with finding the derivative of R..

dRdt

= R′(t) = 174.343(3t2) − 5630.45(2t) + 63, 029.8 = 523.029t2 −11, 260.90t + 63, 029.

1999 corresponds to t = 9. Then

Rate of change of revenue in 1999 is given by 523.029(9)2−11, 260.90(9)+

63, 029 = $4047 millions per year

4.6 Rates of Change in Economics: Marginals

There are three important functions in Economics: The cost func-

tion C, the Revenue function R, and the Profit function P . An

equation relates these functions is

P = R − C

Economists refer to marginal profit, marginal revenue,and marginal

cost as the rates of change of the profit, revenue , and cost with

respect to the number x of units produced or sold.

In other words

dPdx

= marginal Profit

dRdx

= marginal Revenue

dCdx

= marginal Cost

4.6. RATES OF CHANGE IN ECONOMICS: MARGINALS 59

Example 35

The profit of selling x units of an alarm clock is given by

P = 0.0002x3 + 10x.

Find the marginal Profit for a production level of 50 units.

Solution

The marginal Profit is given by

dPdx

= 0.0002(3x2) + 10 = 0.0006x2 + 10

when x = 50, the marginal Profit is

0.0006(50)2 + 10 = $11.50 per unit

Demand Function

In practice, it is more common to encounter situations in which

sales can be increased only by lowering the price per item. The

relation between the price per unit p and the number of units x

that consumers are willing to purchase is given by the Demand

Function

p = f(x)

The total Revenue R is related to the price per unit p and the

quantity demanded (or Sold) x is given by

R = xp

Example 36

A fast food restaurant has determined that the monthly demand for

its hamburgers is given by

p = 60,000−x20,000

, Find the marginal revenue when x = 20, 000.

Solution

The price per unit is given from

p = 60,000−x20,000

,

and the revenue function is given by R = xp, then

R = xp = x(60,000−x20,000

) = 60,000x−x2

20,000

then R = 120,000

(60, 000x− x2).

Then the marginal revenue is given by

dRdx

= 120,000

(60, 000 − 2x)

when x = 20, 000, the marginal revenue is given by


120,000

(60, 000 − 2(20, 000)) = $1 per unit.

Example 37

Suppose in Example 36 that the cost of producing x hamburgers is

C = 5000 + 0.56x, 0 ≤ x ≤ 50, 000.

Find the profit and the marginal profit for each production level

(a) x = 20, 000 (b) x = 24, 400 (c)x = 30, 000

solution

From Example 36, you know that the total revenue from selling x

hamburgers is

R = 120,000

(60, 000x−x2). But the total profit is given by P = R−C,

you have

P = 120,000

(60, 000x− x2) − 5000 − 0.56x = 2.44x − x2

20,000− 5000

So the marginal profit is

dPdx

= 2.44 − x10,000

for x = 20, 000, profit P = $23, 800, Marginal Profit dPdx

= $0.44

per unit.


= $0.00

per unit.


= $−0.56

per unit.

If you are the manager of this restaurant, how much would

you charge for hamburgers ? Explain your answer

4.7 The product Rule

If f(x) and g(x) are two differentiable functions, then

d

dx[f(x)g(x)] = f ′(x)g(x) + f(x)g′(x)

Example 38

Find the derivative of y = (3x − 2x2)(5 + 4x).

Solution

4.8. THE QUOTIENT RULE 61

dy

dx= (3 − 4x)(5 + 4x) + (3x − 2x2)(0 + 4)

= 15 + 12x − 20x − 16x2 + 12x − 8x2

= 15 + 4x − 24x2

Example 39

Find the derivative of f(x) = ( 1x

+ 1)(x − 1)

Solution

We rewrite f(x) in the following form

f(x) = (x−1 + 1)(x − 1)

f ′(x) = (−x−2 + 0)(x − 1) + (x−1 + 1)(1)

= −x−1 + x−2 + x−1 + 1

= 1 + x−2

4.8 The Quotient Rule

If f(x) and g(x) are differentiable, then

d

dx

[f(x)

g(x)

]

=f ′(x)g(x) − f(x)g′(x)

[g(x)]2

Example 40

Find the derivative of y = x2+2x3x−5

Solution

dy

dx= y′ =

(2x + 2)(3x − 5) − (x2 + 2x)(3)

(3x − 5)2

=6x2 − 10x + 6x − 10 − 3x2 − 6x

(3x − 5)2

=3x2 − 10x − 10

(3x − 5)2

Example 41

Find the derivative of the following function


(a)- f(x) = 3√

x − 4x2 + 7

Solution

we first rewrite f(x) in the form

f(x) = 3x12 − 4x−2 + 7

then

f ′(x) = 3(1

2)x

−12 − 4(−2)x−3 + 0

=3

2x

−12 + 8x−3

4.9 The Chain Rule

If y = f(x) is a differentiable function of u, and u = g(x) is a differ-

entiable function of x, then y is a differentiable function of x, anddydx

= dydu

.dudx

Example 42

If y =√

u, and u = 3x2 − x + 1 , use the chain rule to find dydx

Solution

Let = udydx

= dydu

.dudx

= (1/2)u−1/2(6x − 1)

= (1/2)(3x2 − x + 1)−1/2(6x − 1) = 6x−12√

3x2−x+1

4.9.1 The General Power Rule

Using the Chain rule we can find a rule to find the derivative of

functions in the form y = [u(x)]n, this rule is called the general

power rule given by:

If y = [u(x)]n, where u is a differentiable function of x and n is a

real number. Thendydx

= n[u(x)]n−1u′(x)

Example 43

Find the derivative of

f(x) = (3x − 2x2)3

4.9. THE CHAIN RULE 63

Solution

f ′(x) = 3(3x − 2x2)2(3 − 4x) = (9 − 12x)(3x − 2x2)2

Example 44

Find the tangent line to the graph of

y = 3√

(x2 + 4)2

when x = 2

Solution

We rewrite the function in the form

y = (x2 + 4)2/3

then y′ = (2/3)(x2 + 4)−1/3(2x)

= 4x(x2+4)−1/3

3

= 4x

3 3√x2+4

when x = 2, y = 3√

(22 + 4)2 = 4 and the slope of the line tangent

to the graph at (2, 4) is given by4(2)

3 3√22+4= 8

3(2)= 4

3, using the formula for straight line equation

y − y1 = m(x − x1), such that x1 = 2, y1 = 4, m = 43, then

y − 4 = 43(x − 2)

y = 4 + 43x − 8

3

y = 43x + 4

3

Example 45

Find the Derivative of

(a) y = x2√

1 − x2

(b) f(x) = (3x−1x2+3

)2

Solution

(a) rewrite y

y = x2(1 − x2)1/2

using the product rule ( ddx

[f(x)g(x)] = f ′(x)g(x) + f(x)g′(x)) , we

get

y′ = 2x(1 − x2)1/2 + x2(1/2)(1 − x2)−1/2(−2x)

= 2x√

1 − x2 − x3√

1−x2

(b) f ′(x) = 2(3x−1x2+3

)(3(x2+3)−(3x−1)(2x)(x2+3)2

)

= 2(3x−1x2+3

)(3x2+9−6x2+2x(x2+3)2

)


= 2(3x−1x2+3

)(−3x2+2x+9(x2+3)2

)

4.10 Higher Order derivatives

The derivative of f ′(x) is the second derivative of f(x) and is de-

noted by f ′′(x)

The derivative of f ′′(x) is the third derivative of f(x) and is de-

noted by f ′′′(x)

’By continuing this process, you obtain Higher-order derivatives

of f(x)

Example 46

Find the first three derivatives of f(x) = 2x4 − 3x2

Solution

f(x) = 2x4 − 3x2

f ′(x) = 8x3 − 6x

f ′′(x) = 24x2 − 6

f ′′′(x) = 48x

4.11 Increasing and Decreasing Functions

A function f is increasing on an interval if for any x1 and x2 in the

interval

x2 > x1 implies f(x2) > f(x1)

A function f is decreasing on an interval if for any x1 and x2 in the

interval

x2 > x1 implies f(x2) < f(x1)

Consider the following graph of function

4.11. INCREASING AND DECREASING FUNCTIONS 65

From the figure you can find that the function is increasing on the

interval 1 and decreasing on the interval 2. The graph also shows

that if the function is increasing the slope of tangent rises which

means that f ′(x) > 0, while if the function is decreasing the slope

of tangent falls which means that f ′(x) < 0

4.11.1 Test for Increasing and Decreasing Functions

Let f be differentiable on the interval (a, b).

1. If f ′(x) > 0 for all x in (a, b), then f is increasing on (a, b)

2. If f ′(x) < 0 for all x in (a, b), then f is decreasing on (a, b)

3. If f ′(x) = 0 for all x in (a, b), then f is constant on (a, b)

Example 47

Show that the function

f(x) = x2

is decreasing on the interval (−∞, 0) and increasing on the interval

(0,∞)

solution

The derivative of f is given by


f ′(x) = 2x.

If x belongs to the interval (−∞, 0), then 2x < 0, which implies that

f ′(x) < 0, then f is decreasing on the interval (−∞, 0)

If x belongs to the interval (0,∞), then 2x > 0, which implies that

f ′(x) > 0, then f is increasing on the interval (0,∞)

4.12 Critical Numbers and their Use

If f is defined at c, then c is a critical number of f if f ′(c) = 0 or if

f ′ is undefined at c.

The critical numbers are used to find the open intervals on which the

function is increasing or decreasing using the following procedure

Guidelines for applying Increasing/decreasing Test

1. Find the derivative of f

2. Locate the critical numbers of f and use these numbers to deter-

mine test intervals. That is, find all x for which f ′(x) = 0 or f, (x)

is undefined.

3. Test the sign of f ′(x) at an arbitrary number in each test inter-

vals.

4. Use the test for increasing and decreasing functions to decide

whether f is increasing or decreasing on each interval.

Example 48

Find the open intervals on which the function is increasing or de-

creasing

f(x) = x3 − 32x2

Solution

First of all we note that the function is a polynomial, so the domain

of it is R

We find f ′(x)

f ′(x) = 3x2 − 3x, to get the critical numbers, we set f ′(x) = 0 and

solve for x

4.12. CRITICAL NUMBERS AND THEIR USE 67

3x2 − 3x = 0

3x(x − 1) = 0

x = 0 or x = 1.

There is no values for which f ′ is undefined. Then x = 0 and x = 1

are the only critical numbers. So, the intervals we need to test are

(−∞, 0), (0, 1), (1,∞).

The table summarize the test of these three intervals

Interval (−∞, 0) (0, 1) (1,∞)Test Value x = −1 x = 1

2x = 2

Sign of f ′(x) f ′(−1) = 6 > 0 f ′(12) = −3

4< 0 f ′(2) = 6 > 0

Conclusion Increasing Decreasing IncreasingExample 49

Find the open interval on which the function

f(x) = (x2 − 4)2/3

is increasing or decreasing.

solution

It is clear that the domain of f is R

We find f ′

f ′(x) = 23(x2 − 4)−1/3(2x) = 4x

3(x2−4)1/3

Setting f ′(x) = 0 and solving for x, we get

4x = 0, implies x = 0.

f ′(x) is undefined when the denominator = 0, which results in

x = −2 or x = 2

the three numbers −2, 0, 2 are the critical numbers.

This implies that the test intervals are


(−∞,−2), (−2, 0), (0, 2), (2,∞)

The table summarize the test of these three intervals

Interval (−∞,−2) (−2, 0) (0, 2) (2,∞)

Test Value x = −3 x = −1 x = 1 x = 3

Sign of f ′(x) f ′(−3) = −2.34 < 0 f ′(−1) = 0.92 > 0 f ′(1) = −0.92 < 0 f ′(3) = 2.34 > 0

Conclusion Decreasing Increasing Decreasing Increasing

4.13 Extrema and the First Derivative Test

Let f be continuous on the interval (a, b) in which c is the only crit-

ical number. If f is differentiable on the interval (except possibly

at c), then f(c) can be classified as a relative minimum , a relative

maximum, or neither. As shown.

1. On the interval (a, b), if f ′(x) is negative to the left of x = c and

positive to the right of x = c, then f(c) is a relative minimum .

2- On the interval (a, b), if f ′(x) is positive to the left of x = c and

negative to the right of x = c, then f(c) is a relative maximum.

3. On the interval (a, b), if f ′(x) has the same sign to the left and

right of x = c, then f(c) is not a relative extremum of f .

Example 50

Find the relative extrema of the function

f(x) = 2x3 − 3x2 − 36x + 14

Solution

4.14. ABSOLUTE EXTREMA 69

Begin by finding the critical numbers of f

f ′(x) = 6x2 − 6x − 36

6x2 − 6x − 36 = 0

6(x − 3)(x + 2) = 0

x = 3 ,x = −2

Because f ′(x) is defined for any real number, then the only critical

numbers of f are −2, 3

Using these numbers,you can form the three test intervals (−∞,−2), (−2, 3), (3,∞).

The testing of these intervals is shown in the table

Interval (−∞,−2) (−2, 3) (3,∞)Test Value x = −3 x = 0 x = 4Sign of f ′(x) f ′(−3) = 36 > 0 f ′(0) = −36 < 0 f ′(4) = 36 > 0Conclusion Increasing Decreasing Increasing

Using the first derivative test, you can conclude that the critical

number −2 yields a relative maximum [f ′(x) changes sign from pos-

itive to negative] , and the critical number 3 yields a relative mini-

mum [f ′(x) changes sign from negative to positive]

4.14 Absolute Extrema

Let f be defined on an interval I containing c.

1. f(c) is an absolute minimum of f on I if f(c) ≤ f(x) for every x

in I.

2. f(c) is an absolute maximum of f on I if f(c) ≥ f(x) for every

x in I

The absolute minimum and absolute maximum values of a function

on an interval are sometimes simply called the minimum and maxi-

mum.

Extreme value theorem

If f is continuous on [a, b], then f has both a minimum and a max-

imum value on [a, b].

How to find the extrema on a closed interval


To find the extrema of a continuous function on a closed interval

[a, b], use the steps below.

1. Evaluate f at each of the critical numbers in (a, b).

2. Evaluate f at each end point, a, and b.

3. The least of these values is the minimum, and the greatest is the

maximum.

Example 51

Find the absolute minimum and absolute maximum values of the

function f(x) = x2 − 6x + 2 on [0, 5]

Solution

We obtain the critical numbers

f ′(x) = 2x − 6

Solving f ′(x) = 0

2x − 6 = 0

x = 3

since f ′(x) is defined for any real number. Then 3 is the only critical

number, it belongs to the interval [0, 5]

We calculate the value of the function at the three points 3, 0, and5

f(3) = −7, f(0) = 2, and f(5) = −3

So, the maximum is 2 and the minimum is −7

4.15 The Second Derivative Test

Let f ′(c) = 0, and let f ′′(x) exist on an open interval containing c.

1. If f ′′(c) > 0, then f(c) is a relative minimum 2. If f ′′(c) < 0, then

f(c) is a relative maximum 3. If f ′′(c) = 0, then test fails. In such

cases, you can use the first derivative test to determine whether f(c)

is a relative minimum, a relative maximum, or neither.

Example 52

Using the second derivative test, find the relative extrema of

f(x) = x3 − 5x2 + 7x.

Solution

4.16. OPTIMIZATION IN BUSINESS AND ECONOMICS 71

We begin by finding the critical numbers of f

f ′(x) = 3x2 − 10x + 7

solving f ′(x) = 0, we get

3x2 − 10x + 7 = 0

(3x − 7)(x − 1) = 0

then x = 73

or x = 1. Sine f ′(x) is defined for all real numbers, then

the only critical numbers are 73

and 1.

Now we find f ′′(x)

f ′′(x) = 6x − 10

f ′′(73) = 14− 10 = 4 > 0, then f has a relative minimum at 7

3, such

that f(73) = (7

3)3 − 5(7

3)2 + 7(7

3) = 49

27

f ′′(1) = 6−10 = −4 < 0, then f has a relative maximum at 1, such

that f(1) = (1)3 − 5(1)2 + 7(1) = 3

4.16 Optimization in Business and Economics

One of the most common applications of calculus is the determina-

tion of optimum (minimum or maximum ) values. In fact it is a

goal of our course to learn how to deal with real life optimization

problems .

Example 53

A company has determined that its total revenue (in dollars) for a

product can be modeled by

R = −x3 + 450x2 + 52, 500x

where x is the number of units produced (and sold). What produc-

tion level will yield a maximum Revenue?

Solution

To maximize the revenue, we find the critical numbers of R

R′ = −3x2 + 900x + 52, 500

solving the equation R′ = 0, we get


−3x2 + 900x + 52, 500 = 0

−3(x2 − 300x − 17500) = 0

−3(x + 50)(x − 350) = 0

hence , the critical numbers are x = −50,x = 350

x = −50 is refused, since the production level can not be negative.

To be sure that x = 350 corresponds to a maximum value of Rev-

enue we find f ′′(350)

R′′(x) = −6x + 900

R′′(350) = −1200 < 0.So 350 is the production level that yields

maximum profit.

Example 54

A company estimates that the cost (in dollars) of producing x units

of a product can be modeled by

C = 800 + 0.04x + 0.0002x2. Find the production level that mini-

mizes the average cost per unit.

Solution

The average cost per unit is defined as A = Cx,

where C is the total cost of producing x units, and x is the number

of units produced

To minimize A, we find the critical numbers

A′ = −800x2 + 0.0002

Solving A′ = 0, we get

−800x2 + 0.0002 = 0

800x2 = 0.0002

x2 = 8000.0002

= 4000000

the critical numbers are x = ±2000

but −2000 is refused, since the production level can not be negative

To be sure that x = 2000 corresponds to a minimum value of A we

find A′′(2000)

A′′ = 1600x3

A′′ = 1600(2000)3

> 0

Hence 2000 is the production level that minimizes the average cost

4.16. OPTIMIZATION IN BUSINESS AND ECONOMICS 73

per unit.

Example 55

A marketing department of a business has determined that the de-

mand for a product can be modeled by

p = 50√x.

The cost of producing x units is given by

C = 0.5x + 500.

What price will yield a maximum profit?

Solution

Let R represents the revenue, P the profit, p the price per unit, x

the number of units, and C the total cost of producing x units.

The Revenue R = xp = 50x√x

= 50√

x, then the profit is given by

P = R − C = 50√

x − 0.5x − 500

To maximize profit we get the critical numbers

P ′ = 25√x− 0.5 = 0

√x = 50

x = 2500

you can check that P ′′(2500) < 0. This means that x = 2500 cor-

responds to maximum value of Profit and the price of one unit will

be p = 50√2500

= $1

Chapter 5

Exponential andLogarithmic Functions

5.1 Exponential Functions

If a > 0 and a 6= 1, then the exponential function with base a is

given by

f(x) = ax

For Example: f(x) = 2x is exponential function with base 2

g(x) = 3x is exponential function with base 3

h(x) = (1.4)x is exponential function with base 1.4

5.1.1 Properties of Exponents

Let a and b be positive numbers.

1. a0 = 1.

For example 20 = 1, 30 = 1 (1.5)0 = 1

2. axay = ax+y.

For Example 2x2y = 2x+y, 3x3y = 3x+y

3. ax

ay = ax−y.

For Example 4x

4y = 4x−y

4. (ax)y = axy.

For Example (5x)y = 5xy

5. (ab)x = axbx.

For Example ((2)(3))x = 2x3x

75

76 CHAPTER 5. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

6. (ab)x = ax

bx .

For Example (23)x = 2x

3x

7. a−x = 1ax .

For Example 5−x = 15x

5.1.2 Graphs of Exponential Functions

consider the function f(x) = 3x

x -4 -3 -2 -1 0 1 2 3 4f(x) 1/81 1/27 1/9 1/3 1 3 9 27 81

-2 -1 1 2

2

4

6

8

fHxL=3x

From the table and the graph of f(x) = 3x, we can find the fol-

lowing:

1. The domain of f(x) is R (the set of all real numbers)

2. The Range is (0,∞)

3. as x approaches ∞, f(x) approaches ∞.

4. as x approaches −∞, f(x) approaches 0.

These four properties are followed by any exponential function f(x) =

ax, where a > 0 , and a 6= 1.

5.2 Natural Exponential Function

A very important exponential function is the natural exponential

function defined as f(x) = ex, where e is an irrational number,

5.2. NATURAL EXPONENTIAL FUNCTION 77

whose decimal approximation is

e ≈ 2.71828182846.

Limit Definition of e

The irrational number e is defined to be the limit of (1 + x)1/x as

x → 0. That is

limx→0

(1 + x)1/x = e.

Or Equivalency limn→∞

(1 +1

n)n.

The behavior of f(x) = ex is similar to f(x) = ax, where a > 0 ,

and a 6= 1, see below the graph of f(x) = ex

-2 -1 1 2

1

2

3

4

5

6

7

fHxL=ex

Example 1

A bacterial culture is growing according to the model

y = 1.251+0.25e−0.4t , t ≥ 0

where y is the culture weight (in grams) and t is the time (in hours).

find the weight of the culture after 0 hours, 1 hour, and 10 hours.

What is the limit of this model as t increases without bound ?

solution

when t = 0

y = 1.251+0.25e−0.4(0) = 1 grams

when t = 1

y = 1.251+0.25e−0.4(1) = 1.071 grams

when t = 10

y = 1.251+0.25e−0.4(10) = 1.244 grams

As t approaches infinity, the limit of y is


limt→∞

1.25

1 + 0.25e−0.4t= lim

t→∞

1.25

1 + (0.25/e0.4t)=

1.25

1 + 0= 1.25 grams

5.3 Business Application: Compound Interest

If P dollars are deposited in an account at an annual interest rate

of r in (decimals form ), what is the balance after one year? The

answer depends on the number of times the interest is compounded,

according to the formula

A = P (1 + rn)n

where n is the number of compounding per year.

Of course A increases if n increases. But you may be surprised to

discover that as n increases, the balance A approaches a limit, and

we have

A = limn→∞

P (1 +r

n)n

= P limn→∞

[(1 +r

n)n/r]r

= Per

This limit is the balance after 1 year of continuous compounding

Summary of Compound Interest Formulas

Let p be the amount deposited, t the number of years, A the balance

, and r the annual interest rate (in decimal form).

1. Compounded n times per year : A = P (1 + rn)nt

2. Compounded Continuously : A = Pert.

Example 2

Steven deposits $12, 000 in an account for 25 years. Compare the

balances for each situation

(a) the interest rate is 7% compounded quarterly

(b) the interest rate is 7% compounded continuously.

Solution

(a) n = 4, r = 7% = 0.07, t = 25, P = 12, 000

The balance is given by A = 12, 000(1 + 0.074

)4(25) = 68, 017.87

(b) The balance is given by A = 12, 000e0.07(25) = 69, 055.23

5.4. FINDING PRESENT VALUE 79

It is clear that compounding continuously results in a balance greater

than that obtained from compounding quarterly

5.4 Finding present value

In planning for the future, this problem arises: ” how much money

P should be deposited now, at a fixed rate of interest r , in order

to have a balance of A , t years from now ? ” The answer to this

question is given by the present value of A.

To find the present value of a future investment, use the formula for

compound interest as shown.

A = P (1 + rn)nt

Solving for P gives a present value of

P = A(1+ r

n)nt

Example 3

An investor is purchasing a 12− year certificate of deposit that pays

an annual percentage rate of 8%, compounded monthly. How much

should the person invest in order to obtain a balance of $15, 000 at

maturity ?

Solution

Here, A = 15000, r = 0.08, n = 12, and t = 12. Using the formula

for present value P = A(1+ r

n)nt , we obtain

P = 15000(1+ 0.08

12)(12)(12)

= $5761.72

5.5 Derivatives of Exponential Functions

The derivative of ex is given by

ddx

(ex) = ex.

This rule is seems to be strange !! (The derivative of ex is ex). The

proof of this rule using the limit definition of differentiation is given

below. Before we start the proof, we note that for small value of h

we have eh ≈ 1 + h


Now let f(x) = ex, then

f ′(x) = limh→0

f(x + h) − f(x)

h

f ′(x) = limh→0

ex+h − ex

h

f ′(x) = limh→0

exeh − ex

h

f ′(x) = limh→0

ex(eh − 1)

husing eh ≈ 1 + h, which is true as h is very small, we get

f ′(x) = limh→0

ex(1 + h − 1)

h

f ′(x) = ex limh→0

h

h= ex(1) = ex

What about f(x) = eu(x) ?

The answer is that ddx

eu(x) = eu(x) dudx

For example ddx

ex2= ex2

(2x)

Example 4

(a) f(x) = e2x (b) f(x) = e−3x2

(c) f(x) = 6ex3(d) f(x) = e−x

(e) f(x) = x2ex (f)f(x) = ex

x

Solution

(a) f ′(x) = e2x(2) = 2e2x

(b) f ′(x) = e−3x2(−6x) = −6xe−3x2

(c) f ′(x) = 6ex3(3x2) = 18x2ex3

(d) f ′(x) = e−x(−1) = −e−x = − 1ex

(e) f ′(x) = 2xex + x2ex = ex(2x + x2)

(f) f ′(x) = xex−ex(1)x2 = ex(x−1)

x2

5.6 Logarithmic Functions

The Logarithm Function loga x is defined as

loga x = y if and only if ay = x. a is called the base of the logarithm.

In calculus the most useful base for logarithms is the number e

5.6. LOGARITHMIC FUNCTIONS 81

5.6.1 The Natural Logarithm Function

The natural logarithm function, denoted by lnx, is defined as

lnx = y if and only if ey = x.

where ln x is loge x, that is the base is e

The definition implies that the natural logarithmic function and the

natural exponential function are inverse functions. For Example

ln 1 = 0 and e0 = 1

ln e = 1 and e1 = e

The graph of the function y = lnx is shown below

2 4 6 8 10

-2

-1

1

2

y= ln x

Inverse properties of Logarithms and Exponents

1. ln ex = x 2. eln x = x

For Example

ln e3 = 3

eln 5 = 5

5.6.2 Properties of Logarithms

1. ln(xy) = ln x + ln y

2. ln xy

= lnx − ln y

3. ln xn = n ln x


We can use these properties to rewrite the logarithm of single quan-

tity as a sum, difference, or multiple of logarithms and vice versa

Example 5

Use the properties of logarithms to rewrite each expression as a sum,

difference, or multiple of logarithms .

(a) ln 109

(b) ln√

x2 + 1 (c) ln[x2(x + 1)]

Solution

(a) ln 109

= ln 10 − ln 9

(b) ln√

x2 + 1 = ln(x2 + 1)1/2 = 12ln(x2 + 1)

(c) ln[x2(x + 1)] = ln x2 + ln(x + 1) = 2 ln x + ln(x + 1)

Example 6

Use the properties of logarithms to rewrite each expression as the

logarithm of single quantity

(a) ln x + 2 ln y (b) 2 ln(x + 2) − 3 lnx

Solution

(a) ln x + 2 ln y = ln x + ln y2 = ln(xy2)

(b) 2 ln(x + 2) − 3 ln x = ln(x + 2)2 − ln x3 = ln (x+2)2

x3

5.6.3 Solving Exponential and Logarithmic Equations

Example 7

Solve each equation

(a) 10 + e0.1x = 14 (b) 3 + 2 lnx = 7

Solution

(a) Moving 10 to the right side , we get,

e0.1x = 14 − 10 = 4

Take ln for each side, we get

ln e0.1x = ln 4

From the property ln ex = x , we get ,

0.1x = ln 4

x = ln 40.1

= 10 ln 4

5.7. DERIVATIVES OF LOGARITHMIC FUNCTIONS 83

(b) Moving 3 to the right side, we get

2 lnx = 4

lnx = 2

Exponentiate each side, we get

elnx = e2

x = e2

5.7 Derivatives of Logarithmic Functions

1- If f(x) = ln x, then f ′(x) = 1x

2- If f(x) = ln u(x), then f ′(x) = 1u(x)

u′(x)

Example 8

Find the derivative of

(a) f(x) = ln 2x

(b) f(x) = ln(2x2 + 4)

(c) f(x) = x2 ln x

(d) f(x) = ln(√

x2 + 1)

Solution

(a) f ′(x) = 12x

(2) = 22x

= 1x

(b) f ′(x) = 12x2+4

(4x) = 4x2x2+4

= 2xx2+2

(c) f ′(x) = (2x)(ln x) + (x2)( 1x) = 2x lnx + x

(d) f(x) = ln(√

x2 + 1) = ln(x2 + 1)1/2 = 12ln(x2 + 1)

f ′(x) = 12( 1

x2+1)(2x) = x

x2+1

2 4 6 8 10

-2

-1

1

2

y= ln x

Chapter 6

Function of Two Variables

Introduction It is very seldom to find phenomenon that depends

only on one variable. In Real life applications, phenomenon always

depends on several variables.

For example the growth of a tree depends on the quantity of water,

the quantity of light, and the quantity of useful minerals a tree gets.

Another Example

The monthly payment M for an installment loan of P dollars taken

over t years at an annual interest rate r is given by

M = f(P, r, t) =Pr12

1−[ 11+(r/12)

]12t

You note that in this case we have a function of three variables p, r

, and t.

Evaluating function of several variables

Example 1

Given the function of two variables

f(x, y) = x2 − 3xy

Find

(a) f(2, 1)

(b) f(5, 3)

Solution

(a) f(2, 1) = 22 − 3(2)(1) = 4 − 6 = −2

(b) f(5, 3) = 52 − 3(5)(3) = 25 − 45 = −20

85

86 CHAPTER 6. FUNCTION OF TWO VARIABLES

6.1 Partial Differentiation

Real-life applications of functions of several variables are often con-

cerned with how changes in one of the variables will affect the values

of the functions. For instance, an economist who wants to determine

the effect of a tax increase on the economy might make calculations

using different tax rates while holding all other variables, such as

unemployment, constant.

You can follow a similar procedure to find the rate of change of a

function f with respect to one of its independent variables. That

is, you find the derivative of f with respect to one independent

variable, while holding the other variables constant. This process

is called Partial Differentiation, and each derivative is called a

Partial Derivative.

6.1.1 Partial Derivatives of a Function of Two Variables

If z = f(x, y) is, then the first partial derivative of f with re-

spect to x and y are the functions ∂z∂x

and ∂z∂y

, defined a shown.

∂z

∂x= lim

h→0

f(x + h, y) − f(x, y)

h∂z

∂y= lim

h→0

f(x, y + h) − f(x, y)

hExample 2

Find ∂z∂x

and ∂z∂y

for the function z = 3x − x2y2 + 2x3y

solution

to get ∂z∂x

, we hold y constant and differentiate with respect to x.

That is

∂z∂x

= 3(1) − (2x)y2 + 2(3x2)y = 3 − 2xy2 + 6x2y

∂z∂y

= 0 − x2(2y) + 2x3(1) = −2x2y + 2x3

Example 11

Find The first partial derivatives of

f(x, y) = 4x2 − xy2 + 2y2 + 8

Solution

6.1. PARTIAL DIFFERENTIATION 87

∂f∂x

= 8x − y2 + 0 + 0 = 8x − y2

∂f∂y

= 0 − x(2y) + 4y + 0 = −2xy + 4y

Note :

The partial derivative of f with respect to x can be written as ∂f∂x

or fx.

The partial derivative of f with respect to y can be written as ∂f∂y

or fy.

Example 3

Find the first partial derivatives of f(x, y) = xex2y and evaluate each

at the point (1, ln 2)

Solution

using the product rule

fx(x, y) = ∂∂x

(x)e(x2y) + x ∂∂x

(e(x2y)) = e(x2y) + xe(x2y)(2xy)

= e(x2y)(1 + 2x2y)

at the point (1, ln 2), the value of the derivative is

fx(1, ln 2) = e((1)2(ln 2))(1 + 2)((1)2(ln 2)) = 2(2 ln 2 + 1) ≈ 4.773

To find the first partial derivative with respect to y, hold x constant

and differentiate to obtain

fy(x, y) = x(x2)e(x2y) = x3e(x2y)

at the point (1, ln 2), the value of the derivative is

fy(1, ln 2) = (1)3e((1)2 ln 2) = 2

Example 4

A company produces two models of bicycles : a mountain bike,and

a racing bike. The cost function for producing x mountain bikes

and y racing bikes is given by

C = 10√

xy + 149x + 189y + 657.

Find the marginal costs (∂C∂x

and ∂C∂y

) when x = 120 and y = 160.

Solution

we rewrite C in the form

C = 10x1/2y1/2 + 149x + 189y + 657

∂C∂x

= 10(1/2)x−1/2y1/2 + 149 + 0 + 0 = 10(1/2)x−1/2y1/2 + 149

when x = 120 and y = 180, we find


∂C∂x

= 10(1/2)(120)−1/2(180)1/2 + 149 ≈ 155.1

∂C∂y

= 10(1/2)x1/2y−1/2 + 189 = 10(1/2)x1/2y−1/2 + 189

When x = 120 an dy = 180

∂C∂y

= 10(1/2)(120)1/2(180)−1/2 + 189 ≈ 193.1

6.1.2 Higher order Partial derivatives

As with ordinary derivative, it is possible to find partial derivatives

of second order or higher order. For instance, there are four ways

to find a second partial derivative of z = f(x, y)

∂∂x

(∂f∂x

) = fxx

∂∂y

(∂f∂y

) = fyy

∂∂y

(∂f∂x

) = fxy

∂∂x

(∂f∂y

) = fyx

Example 5

Find the second partial derivatives of

f(x, y) = 3xy2 − 2y + 5x2y2

Solution

Begin by finding the first partial derivatives

∂∂x

(f) = fx = 3y2 + 10xy2 ∂∂y

(f) = fy = 6xy − 2 + 10x2y

The differentiate with respect to x and y produces

fxx = 10y2 fyy = 6x + 10x2

fxy = 6y + 20xy fyx = 6y + 20xy

6.2 Relative Extrema

Earlier in this course, you learned how to use derivatives to find

the relative minimum and relative maximum values of a function

of single variable. In this section we will study how to use partial

derivatives to find the relative minimum and relative maximum val-

ues of a function of two variables.

6.2. RELATIVE EXTREMA 89

Relative extrema of a Function of Two Variables

Let f be a function defined on the region containing (x0, y0). The

number f(x0, y0) is a relative maximum of f if there is circular re-

gion R centered at (x0, y0) such that

f(x0, y0) ≥ f(x, y) for all (x, y) in this region.

The number f(x0, y0) is a relative minimum of f if there is circular

region R centered at (x0, y0) such that

f(x0, y0) ≤ f(x, y) for all (x, y) in this region.

6.2.1 Critical points

A point (x0, y0) is a critical point of f if fx(x0, y0) or fy(x0, y0) is

undefined or if

fx(x0, y0) = 0 and fy(x0, y0) = 0


6.2.2 The Second -Partial Test for Relative Extrema

To Find the relative extrema of a function of two variables, follow

the following steps

Step1: Find the first partial derivatives fx(x, y), and fy(x, y)

Step 2: Solve the two equations fx(x, y) = 0, and fy(x, y) = 0 si-

multaneously. The solutions will be in the form of the ordered pair

(a, b)

Step 3 Compute the value of the quantity

D = fxx(a, b)fyy(a, b) − [fxy(a, b)]2

Step 4: There are 4 cases:

1. if D > 0, and fxx(a, b) > 0, then f(a, b) is a relative minimum

2. if D > 0, and fxx(a, b) < 0, then f(a, b) is a relative maximum

3. if D < 0, the function has a saddle point at the point (a, b)

4. if D = 0, the test gives no information.

Example 6 Find the relative extrema and saddle points of

f(x, y) = xy − 14x4 − 1

4y4.

Solution

Step1: We find the first partial derivatives and the second partial

derivatives

fx(x, y) = y − x3

fy(x, y) = x − y3

fxx(x, y) = −3x2

fxy(x, y) = 1

fyy(x, y) = −3y2

Step 2

We solve the two equations fx(x, y) = 0, and fy(x, y) = 0 simulta-

neously.

the two equations are

y−x3 = 0 and x−y3 = 0 , from the first equation y = x3, substitute

in the second to get x − x9 = 0,

6.2. RELATIVE EXTREMA 91

then x(1 − x8) = 0, then x = 0 or x = ±1. substituting in the first

equation with values of x, we get y = 0 when x = 0, y = −1 when

x = −1, and y = 1 when x = 1.

Then we have three critical points (0, 0), (−1,−1), and (1, 1).

step 3 for the point (0, 0)

D = fxx(0, 0)fyy(0, 0) − [fxy(0, 0)]2 = 0 − 1 = −1

since D < 0 then f has a saddle point at (0, 0)

For the point (−1,−1)

D = fxx(−1,−1)fyy(−1,−1) − [fxy(−1,−1)]2 = (−3)(−3) − 1 =

8 > 0

fxx(−1,−1) = −3(−1)2 = −3 < 0

then f has a relative maximum at the point (−1,−1)

For the point (1, 1)

D = fxx(1, 1)fyy(1, 1) − [fxy(1, 1)]2 = (−3)(−3) − 1 = 8 > 0

fxx(1, 1) = −3(1)2 = −3 < 0

then f has a relative maximum at the point (1, 1)

Example 7

Find the relative extrema of

f(x, y) = 2x2 + y2 + 8x − 6y + 20

Solution

Step 1

fx(x, y) = 4x + 8

fy(x, y) = 2y − 6

fxx(x, y) = 4

fyy(x, y) = 2

fxy(x, y) = 0

Step 2

Solving the two equations 4x + 8 = 0 and 2y − 6 = 0, we get the

point (−2, 3) the only critical point of f

Step 3

D = fxx(−2, 3)fyy(−2, 3) − [fxy(−2, 3)]2 = 8 > 0, but fxx(−2, 3) =


4 > 0, then f has a relative minimum at (−2, 3)

Example 8

A company manufactures two competitive products, the prices of

which are p1 and p2. Find p1 and p2 so as to maximize the total

revenue

R = 500p1 + 800p2 + 1.5p1p2 − 1.5p21 − p2

2

solution

Follow the same procedure to solve this example.

Rp1 = 500 + 1.5p2 − 3p1

Rp2 = 800 + 1.5p1 − 2p2

Rp1p1 = 3

Rp1p2 = 1.5

Rp2p2 = −2

To get the critical points, we solve

Rp1 = 0 and Rp2 = 0

500 + 1.5p2 − 3p1 = 0

800 + 1.5p1 − 2p2 = 0

multiply the second equation by 2 then add to the first equation ,

we get

−2.5p2 + 2100 = 0

then p2 = 840

substitute in the first equation we get p1 = 586.67.

Then the only critical point is (586.67, 840)

Now:

D = Rp1p1Rp2p2 − Rp1p2 = (−3)(−2) − (1.5)2 = 3.75 > 0

But Rp1p1 = −3 < 0

hence the profit is maximum when p1 = 586.67 and p2 = 840

Chapter 7

Sequences and Series

7.1 Sequences

When we say that a collection of objects or events is ”in sequence,”

we mean that the collection is ordered so that it has a first member,

a second member, a third member, and so on.

Definition of a sequence

A Sequence is a function whose domain is the set of positive in-

tegers. The function values a1, a2, a3, ..., an, ... are the terms of the

sequence.

Example 1

Find the first four terms of each sequence.

(a) an = 2n + 1

(b) bn = 3n+1

Solution

(a) a1 = 2(1) + 1 = 3, a2 = 2(2) + 1 = 5, a3 = 2(3) + 1 = 7,

a4 = 2(4) + 1 = 9.

(b) b1 = 31+1

= 32, b2 = 3

2+1= 3

3= 1, b3 = 3

3+1= 3

4, b4 = 3

4+1= 3

5

93

94 CHAPTER 7. SEQUENCES AND SERIES

7.1.1 The Limit of a Sequence

We are concerned with sequences whose terms approach a limiting

value. Such sequences are said to converge; sequences that have no

limit are said to diverge.

For example: The terms of the sequence 12, 1

4, 1

8, 1

16, ..., 1

2n , ... ap-

proach 0 as n increases. You can write the limit as

limn→∞

an = limn→∞

1

2n= 0.

Example 2

Find the limit of each sequence (if it exists) as n approaches infinity

(a) an = 2nn+1

(b) an = n1−2n

(c) an = 2n

2n−1

Solution

(a) limn→∞

an = limn→∞

2n

n + 1= lim

n→∞

2

1 + (1/n)=

2

1 + 0= 2.

(b) limn→∞

an = limn→∞

n

1 − 2n= lim

n→∞

1

(1/n) − 2=

1

0 − 2= −1

2.

(c) limn→∞

an = limn→∞

2n

2n − 1= lim

n→∞

1

1 − (1/2n)=

1

1 − 0= 1.

7.1.2 Pattern Recognition for Sequences

Often, the terms of a sequence are generated by a procedure that

does not explicitly identify the nth term of the sequence. In such

cases , you need to discover a pattern in the sequence and find a

formula for the nth term.

N.B: If n is a positive integer, then n factorial is defined

as

n! = n.(n − 1).(n − 2)...3.2.1

Example 3

Determine an nth term of the sequence

12, 4

6, 9

24, 16

120, ...

Solution

You may observe that the numerators are n2, and the denominators

are (n + 1)!. So an = n2

(n+1)!

7.2. SERIES AND CONVERGENCE 95

7.1.3 Application

A deposit of $1000 is made in an account that earns 6% interest

compounded monthly. Find a sequence that represent the monthly

balances.

Solution

After 1 month, the balance will be

A1 = 1000 + 1000(0.0612

) = 1000 + 1000(0.05) = 1000(1.005)

After 2 months , the balance will be

A2 = 1000(1.005) + 1000(1.005)(0.005) = 1000(1.005)2

Continuing this pattern, you can determine that the balances after

n months is An = 1000(1.005)n

The sequence will be

1000(1.005), 1000(1.005)2, 1000(1.005)3, 1000(1.005)4, 1000(1.005)5,

1000(1.005)6, 1000(1.005)7, ...

7.2 Series and Convergence

The finite sum a1 + a2 + a3 + ... + aN can be written asN

∑

i=1

ai

For Example:

1 + 4 + 9 + 16 + 25 + ... + 100 can be written as10

∑

i=1

i2

7.2.1 Infinite Series

The infinite summation∞∑

n=1

an = a1 + a2 + a3 + a4 + ...

is called an infinite series

The nth-Term Test for Divergence

Consider the infinite series∞

∑

n=1

an. If


limn→∞

an 6= 0, then the series diverges

Example 4

Use the nth-Term Test to determine whether each series diverges.

(a)∞

∑

n=1

2n (b)∞

∑

n=1

n

1 + 2n

Solution

(a) limn→∞

an = limn→∞

2n = ∞ 6= 0

hence the series diverges

(b) limn→∞

an = limn→∞

n

1 + 2n= lim

n→∞

1

(1/n) + 2=

1

0 + 2=

1

26= 0

hence the series diverges

7.3 Geometric Series

If a is a non-zero number, then the infinite series∞

∑

n=0

arn

is called a geometric series with ratio r.

nth partial sum of a geometric series

The nth partial sum of the geometric series∞

∑

n=0

arn is

sn = a(1−rn+1)1−r

, r 6= 1

Example 5

Find the third, fifth, and tenth partial sum of the geometric series∞

∑

n=0

3(1

4)n = 3 +

3

4+

3

42+

3

43+ . . .

Solution

a = 3, r = 14. The nth partial sum is given by

sn = a(1−rn+1)1−r

= 3(1−(1/4)n+1)1−(1/4)

= 3(1−(1/4)n+1)3/4

7.3. GEOMETRIC SERIES 97

Using this formula, we can find the third, fifth, and tenth partial

sums as shown:

S3 = 3(1−(1/4)4)3/4

≃ 3.984 S5 = 3(1−(1/4)6)3/4

≃ 3.999

S10 = 3(1−(1/4)11)3/4

≃ 4.000

Note that

M∑

n=1

arn = −a +

M∑

n=0

arn = −a +a(1 − rM+1)

1 − r

Example 6

A deposit of $50 is made every month for 2 years in a saving ac-

counts that pays 6% compounded monthly. What is the balance at

the end of 2 years?

Solution

Using the formula for compound interest

A = P (1 + rn)nt

A is the balance, P is the deposit, n = 12 r = 0.06, and t is the

time in years.

The money that was first deposited the first month will have become

A24 = 50(1 + 0.0612

)(12)(2) = 50(1.005)24

Similarly, after 23 months, the money deposited the second month

will have become

A23 = 50(1 + 0.0612

)(12)(23/12) = 50(1.005)23

Continuing this process, you will find that the total balance result-

ing from the 24 deposits will be

A = A1 + A2 + A3 + · · ·+ A24 =

24∑

n=1

An =

24∑

n=1

50(1.005)n

Noting that the index begins with n = 1. Now

A =

24∑

n=1

50(1.005)n = −50+

24∑

n=0

50(1.005)n = −50+50(1 − 1.00525)

1 − 1.005≃

$1277.96


7.3.1 Convergence of an Infinite Geometric Series

An infinite geometric series given by∞

∑

n=0

arn

diverges if |r| ≥ 1. If |r| < 1, then the series converges to the sum∞

∑

n=0

arn =a

1 − r

Example 7

Decide whether each series converges or diverges

(a)

∞∑

n=0

(−1

2)n (b)

∞∑

n=0

(3

2)n (c)

∞∑

n=1

4

3n

Solution

(a) a = 1, r = −12. Since |r| = 1

2< 1, then the series converges.

Moreover∞∑

n=0

(−1

2)n =

a

1 − r=

1

1 − (−1/2)=

1

3/2=

2

3.

(b) r = 3/2 > 1 then the series diverges.

(c)

∞∑

n=1

4

3n=

∞∑

n=1

4(1

3)n

a = 1, r = 13

< 1. The series converges. Moreover∞

∑

n=1

4

3n= −4 +

∞∑

n=0

4(1

3)n = −4 +

4

1 − (1/3)= −4 + 6 = 2

Example 8

A manufacturer sells 10, 000 units of a product each year. In any

given year, each unit has a 10% chance of breaking. That is after

1 year you expect that only 9000 of the previous year’s 10000 units

will still be in use. During the next year, this number will drop by

an additional 10% t0 8100, and so on. How many units will be in

use after 20 years?

Solution

The situation is

By the end of year 0,

the number of units in use is 10000.


the number of units in use is 10000 + 10000(0.90).

7.3. GEOMETRIC SERIES 99


the number of units in use is 10000 + 10000(0.90) + 10000(0.90)2.


the number of units in use is 10000 + 10000(0.90) + 10000(0.90)2 +

10000(0.90)3.

A geometric series

After 20 years, the number of units in use will be20

∑

n=0

10000(0.09)n =10000(1 − (0.90)21)

1 − 0.9≃ 89000

the german university in cairo (guc) dr. tarek emam

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