CALCULUS in MAM101
Lecture Notes
2025
By Dr Mbekezeli Nxumalo
Qualifications: Bsc Physical Sciences; Bsc Honours Physical Sciences (Cum Laude);
Msc Pure Mathematics (Cum Laude); PhD Pure Mathematics
Research Areas: General Topology; Pointfree Topology; Categorical Topology
Book: Calculus by James Stewart, 5th Edition
Chapter 1
Introduction
The purpose of this course is to introduce students to Mathematics as a language of communicating science ideas. The course aims to equip students with logical reasoning skills
and problem solving skills. It prepares students for careers in science fields such as Physics,
Mathematics, Applied Mathematics, Chemistry, Computer Sciences and Biological Sciences.
Prior knowledge required:
1. No prior knowledge is assumed.
You are expected to:
1. Attend all lectures.
2. Attend and write all tutorial tests.
3. Try tutorial questions before coming to the tut venue.
4. Come for consultation if you need help with some content.
5. Engage in class.
6. Have fun.
7. Challenge the lecturer.
8. Show respect.
1
By Dr. M. Nxumalo
9. Show level of independence in Tuts.
10. Write all assessments.
11. Solve all results that were left as exercises in class.
12. Make use of the ADP Tutorial service.
Lectures and tutorial worskhops are a key element in this course.
This course uses teaching, learning and assessment approaches such as tutorial tests,
quizzes, ungraded tasks, class tests, projects/assignments and exam in the design and delivery of the curriculum.
Assessments:
1. Class Test 1: (The Number System + Functions), Date: 07 March 2025, Venue: Great
Hall, Time: 18:30 - 21:30
2. Class Test 2: (Limits + Continuity + Derivatives), Date: 11 April 2025, Venue: Great
Hall, Time: 18:30 - 21:30
3. Class Test 3: (Differentiation Rules + Applications of Differentiation), Date: 05 May
2025, Venue: Great Hall, Time: 18:30 - 21:30
4. Class Record(40%): Class Tests= 30%; Assignments = 5%; Tuts = 5%.
5. Exam: 60%.
This course equips one with the following skills:
1. Problem-solving skills.
2. Logical Reasoning skills.
Remember:
“The only way to do Mathematics is to do it.”
2
Contents
1 Introduction
1
2 PART 1: The Number System
6
2.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2.2
Absolute values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
13
3 Functions
15
3.1
Definition of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
3.2
Domain and range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
3.3
Linear models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
3.4
Piecewise functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
3.5
Even and odd functions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
3.6
Increasing and decreasing functions . . . . . . . . . . . . . . . . . . . . . . . .
25
3.7
Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
25
3.8
Power functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
26
3.9
Rational functions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
3.10 Algebraic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
3.11 Trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
28
3.12 Transformations of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
3
CONTENTS
By Dr. M. Nxumalo
3.13 Combinations of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
33
3.14 Composition of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
35
3.15 Exponential functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
3.16 Inverse functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
3.17 Logarithmic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
3.18 Inverse trigonometric functions . . . . . . . . . . . . . . . . . . . . . . . . . .
49
4 Limits
50
4.1
The tangent problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
50
4.2
The velocity problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
4.3
The limit of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
52
4.4
One-sided limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
4.5
Infinite limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
54
4.6
Calculating limits using the limit laws
. . . . . . . . . . . . . . . . . . . . . .
56
4.7
The precise definition of a limit . . . . . . . . . . . . . . . . . . . . . . . . . .
61
4.8
Limits at infinity: horizontal asymptotes . . . . . . . . . . . . . . . . . . . . .
63
4.9
Infinite limits at infinity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
4.10 Tangents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
67
4.11 Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
69
4.12 Other rates of change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71
5 Continuity
74
6 Derivatives
82
7 PART 2: Differentiation Rules
86
4
CONTENTS
By Dr. M. Nxumalo
8 Applications of Differentiation
104
8.1
Maximum and Minimum Values . . . . . . . . . . . . . . . . . . . . . . . . . . 104
8.2
The Mean Value Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107
8.3
How Derivatives Affect the Shape of a Graph
. . . . . . . . . . . . . . . . . . 110
8.3.1
What does f 0 say about f ? . . . . . . . . . . . . . . . . . . . . . . . . . 110
8.3.2
What does f 00 say about f ? . . . . . . . . . . . . . . . . . . . . . . . . 112
8.4
Indeterminate Forms and L’Hospital’s Rule . . . . . . . . . . . . . . . . . . . . 117
8.5
Summary of Curve Sketching . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
8.5.1
Slant Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122
8.6
Optimization Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
8.7
Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
8.8
Antiderivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132
9 Introduction to Integration
136
9.1
Areas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
9.2
The Definite Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
5
Chapter 2
PART 1: The Number System
The purpose of this chapter is to give a brief introduction of some properties of the set of real
numbers. You are most likely to encounter these properties in most Mathematics courses.
Throughout, we denote by R (or (−∞, ∞)) the set of real numbers. The symbol −∞ is
not a real number, but an expression indicating that the number in question is a very small
negative number. While ∞ is an expression indicating that the number in question is a very
big positive number.
2.1
Introduction
A set is a collection of things whose members are called elements. We usually use capital
letters A, B, C, .. to represent sets and small letters a, b, x, y, ... to represent elements of a set.
We write a ∈ A to mean that a is an element of a set A, and a ∈
/ A to say that a is not an
element of A. A set with no elements is denoted by ∅ or {} and is called the empty set. A
set is nonempty if it has atleast one element.
A set A is a subset of a set B, denoted by A ⊆ B, if every element of A is also an element
of B. Two sets are equal if they are subsets of each other. Otherwise they are not equal.
Example 2.1.1. Let X represent a big square bowl with small square bowls placed inside it.
The circles represent beads of certain colours inside each bowl:
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2.1. INTRODUCTION
By Dr. M. Nxumalo
X
A
green
pink
yellow
red
B
red
white
cyan
blue
Here,
1. green, red and yellow beads are elements of A. That is, for example, green ∈ A
2. white, red and blue beads are elements of B. That is, for example, white ∈ B.
3. pink and cyan beads is not elements of A. That is, for example, pink ∈
/ A.
4. Bowl A is not equal to bowl B. This is because green ∈ A but green ∈
/ B.
5. Both A and B are non-empty sets.
We explore the algebraic structure of the real number system. We shall assume that the
set of real numbers R is a field (i.e., it satisfies the following postulate).
Postulate 1. [FIELD AXIOMS]. For every a, b, c ∈ R:
1. Closure Property: a + b, a · b ∈ R.
2. Associative Property: a + (b + c) = (a + b) + c and a · (b · c) = (a · b) · c.
7
2.1. INTRODUCTION
By Dr. M. Nxumalo
3. Commutative Property: a + b = b + a and a · b = b · a.
4. Distributive Law: a · (b + c) = (a · b) + (a · c).
5. Existence of the Additive Identity: There is a unique 0 ∈ R such that 0 + a = a.
6. Existence of the Multiplicative Identity: There is a unique 1 ∈ R such that 1 6= 0 and
1 · a = a.
7. Existence of the Additive Inverses: There is a unique −a ∈ R such that a + (−a) = 0.
8. Existence of the Additive Inverses: If a 6= 0, then there is a unique a−1 ∈ R such that
a · (a−1 ) = 1.
The set of all real numbers also possesses an ordering relation.
Postulate 2. [ORDER AXIOMS]. For every a, b, c ∈ R:
1. Trichotomy Property: a < b, b < a or a = b.
2. Transitive Property: a < b and b < c imply a < c.
3. The additive Property: a < b implies a + c < b + c.
4. The Multiplicative Property:
a < b and c > 0 imply ac < bc
and
a < b and c < 0 imply bc < ac.
By b > a we shall mean a < b. By a ≤ b we shall mean a < b or a = b. By a < b < c we shall
mean a < b and b < c.
We shall read a < b as a is lesser than b or b is greater than a. A number a ∈ R is
nonnegative if a ≥ 0 and positive if a > 0.
In the set of real numbers, for any x, y ∈ R:
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2.1. INTRODUCTION
By Dr. M. Nxumalo
1. (x, y) := {a ∈ R : x < a < y}. This is read as “all real numbers that are strictly
greater than x and at the same time strictly lesser than y”. Elements x and y ARE
NOT members of this set. We represent this set on a number line as follows:
This solid line indicates all elements included in the set
x
y
2. [x, y] := {a ∈ R : x ≤ a ≤ y}. This is read as “all real numbers that are greater or equal
to x and at the same time lesser or equal to y”. Elements x and y ARE members of this
set. We represent this set on a number line as follows:
x
y
3. (x, y] := {a ∈ R : x < a ≤ y}. This is read as “all real numbers that are strictly greater
than x and at the same time lesser or equal to y”. The element x is NOT a member of
this set, and the element y is a member of this set. We represent this set on a number
line as follows:
x
y
4. [x, y) := {a ∈ R : x ≤ a < y}. This is read as “all real numbers that are greater or equal
to x and at the same time strictly lesser than y”. The element x is a member of this set,
and the element y is NOT a member of this set. We represent this set on a number line
as follows:
x
y
5. A set of the form {x1 , x2 , x3 , ...} consists of listed elements. Every listed element belongs
9
2.1. INTRODUCTION
By Dr. M. Nxumalo
to the set. We represent this set on a number line as follows:
Elements in this region are not part of the set
•
•
•
•
x1
x2
x3
x4 ..................................
Remark 2.1.2.
..................................
1. We call sets of the form (x, y), [x, y], (x, y], [x, y) intervals. The set
(x, y) is called an open interval, the set [x, y] is called a closed interval and the sets
(x, y] and [x, y) are called half-open and half-closed intervals. Expressions of the
form {a ∈ R : x ≤ a < y} are referred to as set-builder notations.
2. The symbol := means “equal by definition” or “is defined to be”.
Example 2.1.3. For each of the following sets: (1) represent them in interval form, (2) draw
each number line.
1. A = {a ∈ R : 2a < 6}.
2. B = {x ∈ R : x2 − 1 = 0}.
Solution: 1.
A = {a ∈ R : 2a < 6}
= {a ∈ R : a < 3}
= (−∞, 3).
Number line:
3
2. The question is: Which real numbers can you insert in the equation x2 − 1 = 0 so that
the left side and the right side are equal?
B = {x ∈ R : x2 − 1 = 0}
= {−1, 1}.
10
2.1. INTRODUCTION
Number Line:
•
−1
By Dr. M. Nxumalo
•
0
1
The union of two subsets A and B of R is defined by
A ∪ B := {x ∈ R : x ∈ A or x ∈ B}
and is the set of all real numbers that belong to A or B or both. The intersection of A and
B is defined by
A ∩ B := {x ∈ R : x ∈ A and x ∈ B}
and is the set of all real numbers that belong to both A and B.
The complement of a set B relative to a set A is defined by
A r B := {x ∈ R : x ∈ A and x ∈
/ B}
and is the set of all real numbers that are elements of A bot not elements of B.
Example 2.1.4.
1. (−1, 2) ∩ [0, 3) = [0, 2). Let’s us represent this on a number line:
This is the intersection of these sets
−1
0
1
2
3
2. (−1, 2) ∪ (2, 3) = (−1, 2) ∪ (2, 3). Let’s us represent this on a number line:
This is the union of these sets
−1
0
1
2
11
3
2.1. INTRODUCTION
By Dr. M. Nxumalo
3. R r {1, 2} = (−∞, 1) ∪ (1, 2) ∪ (2, ∞). Let’s us represent this on a number line:
This is the complement of {1, 2}
•
•
1
2
The set of real numbers contain:
1. The set of natural numbers
N := {1, 2, 3, 4, 5, 6, .......}.
2. The set of integers
Z := {......, −3, −2, −1, 0, 1, 2, 3, ......}
3. The set of rational numbers
Q :=
nn
m
o
: n, m ∈ Z and n 6= 0 .
4. The set of irrational numbers
Qc = R r Q.
Equality in Q is defined by
n
p
= if and only if nq = mp.
m
q
Example 2.1.5.
1. R r Z = ... ∪ (−3, −2) ∪ (−2, −1) ∪ (−1, 0) ∪ (0, 1) ∪ (1, 2) ∪ ....
Number line:
•
•
•
−1
0
1
12
2.2. ABSOLUTE VALUES
2.2
By Dr. M. Nxumalo
Absolute values
Definition 2.2.1. The absolute value of a number a ∈ R is the number
(
a
if a ≥ 0
|a| =
−a
if a < 0.
Briefly, the absolute value of a number is the distance of the number from 0.
Example 2.2.2.
-1
0
1. |5| = 5. Let’s see this on a number line:
1
2
3
4
5
6
2. | − 10| = 10. Let’s see this on a number line:
-11
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
1
Proposition 2.2.3. For all a, b ∈ R and nonegative M ∈ R:
1. |ab| = |a||b|.
2. Fundamental Theorem of Absolute Values: |a| ≤ M if and only if −M ≤ a ≤ M .
3. Positive Definite: |a| ≥ 0 with |a| = 0 if and only if a = 0.
4. Symmetric: |a − b| = |b − a|.
5. Triangle Inequality: |a + b| ≤ |a| + |b| and |a| − |b| ≤ |a − b|.
Proposition 2.2.4. Let a, x, y ∈ R. Then
1. x < y + for all > 0 if and only if x ≤ y.
2. x > y − for all > 0 if and only if x ≥ y.
3. |a| < for all > 0 if and only if a = 0.
13
2.2. ABSOLUTE VALUES
By Dr. M. Nxumalo
Proof. 1. We want to show two things: (1) If every random > 0 satisfies x < y + , then
x ≤ y. (2) If x ≤ y, then any > 0 gives x < y + .
To prove (1), we assume that every > 0 satisfies x < y + but x y. That is x > y.
Then x − y > 0. The expression x − y is part of the > 0, so the assumption says that
x < y + (x − y). Therefore, x < y + x − y = x. This means that x < x. This cannot be true.
So, the assumption that x y is false. Thus x ≤ y.
To prove (2), we assume that x ≤ y and pick a random > 0. (we must verify that x <
y +). By definition of x ≤ y, we have that x < y or x = y. If x < y, then x+0 < y +0 < y +.
Therefore x < y + . If x = y, then x < y + . Thus x < y + for any > 0.
2. We want to show two things: (1) If every random > 0 satisfies x > y − , then x ≥ y.
(2) If x ≥ y, then any > 0 gives x > y − .
To prove (1), we assume that every > 0 satisfies x > y − but x y. That is x < y.
Then y − x > 0. The expression y − x is part of the > 0, so the assumption says that
x > y − (y − x). Therefore, x > y − y + x = x. This means that x > x. This cannot be true.
So, the assumption that x y is false. Thus x ≥ y.
To prove (2), we assume that x ≥ y and pick a random > 0. (we must verify that
x > y − ). By definition of x ≥ y, we have that x > y or x = y. If x > y, then x + 0 > y + 0 >
y + (−). Therefore x > y − . If x = y, then x > y − . Thus x > y − for any > 0.
3. We want to show two things: (1) If every random > 0 satisfies |a| < , then a = 0. (2)
If a = 0, then any > 0 gives |a| < .
To prove (1), we assume that every > 0 satisfies |a| < . Then |a| < = 0 + . By 1.
above, |a| ≤ 0. But it is always true that |a| ≥ 0, so |a| = 0.
To prove (2), we assume that a = 0. It follows that |a| = 0. Therefore every > 0 satisfies
|a| < .
14
Chapter 3
Functions
Functions appear in most areas of science, as long as there is a process of inserting some input
and expect an output.
3.1
Definition of a function
Definition 3.1.1. A function f : X → Y is a rule that assigns to each element x in a set X,
exactly one element, called f (x) (read f at x), in a set Y .
Remark 3.1.2. The above definition of a function says that:
1. Every element of X gets assigned by f .
2. An element of X gets assigned to exactly one element of Y .
Here is an algebraic definition of a function:
Definition 3.1.3. An assignment f : X → Y from a set X to a set Y is a function if for
every x1 , x2 ∈ X, x1 = x2 implies f (x1 ) = f (x2 ).
The above definition tells us that for a curve to be a function, there must not be a vertical
line that intersects the curve more than once. The method of using a vertical line to check
whether a curve is a function is called the vertical line test.
Example 3.1.4.
1. The assignment
15
3.1. DEFINITION OF A FUNCTION
By Dr. M. Nxumalo
f
X
Y
a
c
p
b
is not a function: This is because b ∈ X is not assigned by f . This says that not every
element of X gets assigned by f .
2. The assignment
g
X
Y
x
w
z
d
is not a function: This is because d ∈ X is mapped to more than one element of Y .
3. The curve
y
x
is not a function. This is because it fails the vertical line test.
We usually consider functions for which the sets X and Y are subsets of real numbers. A
function defined by a table of values is called a tabular function.
16
3.2. DOMAIN AND RANGE
3.2
By Dr. M. Nxumalo
Domain and range
Given a function f : X → Y , the set X is called the domain of f and Y is called the
codomain of f . By the range of f , we mean the set
f (X) := {f (x) : x ∈ X}.
This is the set of all elements of y that are mapped to some element of X. We shall write
Domf for the domain of f , and Rangef for the range of f .
Remark 3.2.1. It is helpful to think of a function as a machine:
x
input
f
f (x)
output
If x is in the domain of the function f , then when x enters the machine, it is accepted as an
input and the machine processes an output f (x) according to the rule of the function. The
collection of all the inputs forms the domain of f . The range is the collection of all the outputs.
The variable x representing our choice of inputs is called an “independent variable” of the
function f and f (x) is a“dependent variable”.
Example 3.2.2.
1. Find the domain and range of the function: f (x) =
√
x + 2.
Solution: For the domain of f , we are looking for inputs x which will make f (x) a valid
output (a real number in this case). We know that a square root of a number is only
valid when the number is nonnegative. So, the domain of f consists of all values of x
such that x + 2 ≥ 0. A simplification gives x ≥ −2 which tells us that the domain of f
consists of all real numbers x ≥ 2.
Let us write this formally: Domf = {x ∈ R : x ≥ −2}. Equivalently, Domf = [−2, ∞).
For the range of f , we are looking for all real numbers that are outputs of numbers from
√
Domf . Set y = f (x). Then y = x + 2 which implies that y 2 = x + 2. Therefore
x = y 2 − 2. Which values of y can we insert so that x ∈ [−2, ∞)? Obviously, any
nonnegative real number y will work. Therefore Rangef = [0, ∞).
Another way of getting both the domain and range of a function is through drawing a
graph:
17
3.2. DOMAIN AND RANGE
By Dr. M. Nxumalo
y
f (x) =
√
x+2
x
−2
0
The horizontal region covered by the curve forms the domain of f , and the vertical region
forms the range of f .
2. Find the domain and range of the function: g(x) =
1
x2 − x
.
Solution: Domain: For g(x) to be valid (or well-defined under the set of real numbers),
x2 − x must not be zero. That is,
x2 − x 6= 0 =⇒ x(x − 1) 6= 0
=⇒ x 6= 0 and x 6= 1.
Therefore
Domf = {x ∈ R : x 6= 0 and x 6= 1}.
Equivalently
Domf = (−∞, 0) ∪ (0, 1) ∪ (1, ∞).
Range:
(a) In the region (−∞, 0), we get that f (x) ∈ (0, ∞).
(b) In the region (0, 1), we get that f (x) ∈ (−∞, 4].
(c) In the region (1, ∞), we get that f (x) ∈ (0, ∞).
Therefore
Rangef = {y ∈ R : −∞ < y ≤ 4 or 0 < y < ∞}.
Equivalently,
Rangef = (−∞, −4] ∪ (0, ∞).
We did not get the above answer through guessing. We used a graph:
18
3.2. DOMAIN AND RANGE
By Dr. M. Nxumalo
y
f (x) =
1
x2 − x
x
0
−4
3. A rectangular storage container with an open top has a volume of 10 m3 . The length of
its base is twice its width. Material for the base costs R 10 per square meter; material
for the sides costs R 6 per square meter. Express the cost of materials as a function of
the width of the base.
Solution:
Consider the following daigram:
h
2w
w
Let w be the width of the box, 2w be the length of the box and h be the height of the
box.
The area of the base is w(2w) = 2w2 . In rands, it’s 10(2w2 ). Two of the sides have area
wh. and the other two have area 2wh. Therefore, the total area is given by
A = 2w2 + 2wh + 2(2wh) = 2w2 + 2wh + 4wh = 2w2 + 6wh.
The total cost is
C = 10(2w2 ) + 6(6wh) = 20w2 + 36wh.
19
3.3. LINEAR MODELS
By Dr. M. Nxumalo
The volume of the box is given by V = w(2w)h = 10. Therefore
h=
10
5
= 2.
2
2w
w
We get that
2
C = 20w + 36w
Therefore C(w) = 20w2 +
3.3
5
w2
= 20w2 +
180
.
w
180
, w > 0.
w
Linear models
Definition 3.3.1. A mathematical model is a mathematical description (often by means
of a function or an equation) of a real-world phenomenon such as the size of a population,
the demand for a product, the speed of a falling object, the concentration of a product in a
chemical reaction, the life expectancy of a person at birth, or the cost of emission reductions.
The purpose of the model is to understand the phenomenon and perhaps to make predictions about future behavior.
Here is an illustration of a mathematical modeling:
Real-world problem
Formulate
Test
Real-world
predictions
Mathematical model
Solve
Interpret
20
Mathematical
conclusions
3.3. LINEAR MODELS
By Dr. M. Nxumalo
There are many functions that can be used to model relationships observed in the real
world.
Definition 3.3.2. A given function is called linear if it is of the form f (x) = mx + c, where
1. m is called the gradient or slope of f . The gradient of a line indicates how steep is
the line. It measures the rate of change of one variable to another. That is, for any two
f (x2 ) − f (x1 )
points (x1 , f (x1 )) and (x2 , f (x2 )) on the graph of f , m =
.
x2 − x1
2. c is a constant. This number represents the value cut by the graph (resulting from the
function) along the y-axis. To get this value, we usually solve for c when x = 0.
A characteristic of linear functions is that they grow at a constant rate.
Example 3.3.3. As dry air moves upward, it expands and cools. If the ground temperature
is 20◦ C and the temperature at a height of 1 km is 10◦ C:
1. Express the temperature T (in ◦ C) as a function of the height h (in kilometers), assuming
that a linear model is appropriate.
2. Draw the graph of the function in part 1. What does the slope represent?
3. What is the temperature at a height of 2.5 km?
Solution:
1. Height h is our independent variable and temperature T is our dependent variable. So,
the scenario is represented by the equation T (h) = mh + c.
We have the following two points form the given scenario: (0, 20) and (1, 10). So,
m=
10 − 20
= −10.
1−0
Now, at h = 0, we have that T (0) = 20 which is our value c. Therefore,
T (h) = −10h + 20.
2.
21
3.4. PIECEWISE FUNCTIONS
By Dr. M. Nxumalo
T (◦ C)
20
h(km)
0
2
The slope represents the change of temperature with respect to height.
3. T (2.5) = −10(2.5) + 20 = −5◦ C.
3.4
Piecewise functions
Certain functions are separated into cases. They are called piecewise functions. Below, we
explore such functions.
Example 3.4.1.
1. A function f is defined by
(
1−x
f (x) =
x2
if x ≤ 1
if x > 1.
Evaluate f (0), f (1) and f (2).
Solution: This is how we read this function: f (x) = 1 − x for all x ≤ 1 and f (x) = x2
for all x > 1. That is, for any x ≤ 1, output is computed using f (x) = 1 − x, and for
x > 0, the output is computed using f (x) = x2 .
Since 0 ≤ 1, we have that f (0) = 1 − (0) = 1.
Since 1 ≤ 1, we have that f (1) = 1 − (1) = 0.
Since 2 > 1, we have that f (2) = (2)2 = 4.
Let us represent this function graphically:
y
1
0
22
x
1
3.4. PIECEWISE FUNCTIONS
By Dr. M. Nxumalo
2. Sketch the graph of the absolute value function f (x) = |x + 2|.
Solution: By definition of the absolute value of a number:
Therefore
(
x+2
f (x) = |x + 2| =
−(x + 2)
if x + 2 ≥ 0
if x + 2 < 0.
(
x+2
f (x) = |x + 2| =
−x − 2
if x ≥ −2
if x < −2.
Let us draw the graph:
y
x
−2 0
3. Find a formula for the function f graphed below:
y
5
2
x
−4−3
0
Solution: The graph on the left has slope of 2 and the y-intercept of (0, 8). So, f is
given by f (x) = 2x + 8, for x ≤ −3.
The other graph is given by f (x) = 5, for x > −3.
So,
(
2x + 8
f (x) =
5
23
if x ≤ −3
if x > −3.
3.5. EVEN AND ODD FUNCTIONS
3.5
By Dr. M. Nxumalo
Even and odd functions
Definition 3.5.1. An even function is a function f satisfying the condition that f (−x) = f (x)
for every x ∈ Domf .
The significance of an even function is that its graph is symmetric with respect to the
y-axis (it is cut in two equal halves).
Example 3.5.2.
1. The function f (x) = x2 − 1 is an even function. This is because
f (−x) = (−x)2 − 1 = x2 − 1 = f (x).
Here is the graph of f showing the symmetry with respect to the y-axis:
y
0
x
−1
2. The function f (x) = x − 1 is not an even function. This is because
f (−x) = (−x) − 1 6= f (x).
Definition 3.5.3. An odd function is a function f satisfying the condition that f (−x) =
−f (x) for every x ∈ Domf .
The graph of an odd function is symmetric about the origin (0, 0).
Example 3.5.4.
1. The function f (x) = x5 + x is an odd function. This is because
f (−x) = (−x)5 + (−x) = −x5 − x = −(x5 + x) = −f (x).
Here is the graph of f showing the symmetry about the origin:
y
x
0
24
3.6. INCREASING AND DECREASING FUNCTIONS
By Dr. M. Nxumalo
2. The function f (x) = 1 − x is not an odd function. This is because
f (−x) = 1 − (−x) = 1 + x 6= −f (x).
3.6
Increasing and decreasing functions
Definition 3.6.1. A function f is called increasing on an interval I if for all x1 , x2 ∈ I,
x1 < x2 =⇒ f (x1 ) < f (x2 ).
It is called decreasing on I if for all x1 , x2 ∈ I,
x1 < x2 =⇒ f (x1 ) > f (x2 ).
Example 3.6.2.
1. The function f : [0, 4] → R defined by f (x) = sin(x) increases on the
following intervals: I1 = [0, π2 ) and I2 = ( 3π
, 2π).
2
).
It decreases on the interval: I3 = ( π2 , 3π
2
See the following graph of the function f :
y
x
0
3.7
1
2
3
4
Polynomials
Unlike linear functions, some functions do not necessarily grow at a constant rate. We consider
some of these functions in this section.
Definition 3.7.1. A function P is a polynomial if
P (x) = an xn + an−1 xn−1 + ... + a2 x2 + a1 x + a0
where n ∈ N and the numbers a0 , a1 , a2 , ...., an are constants called the coefficients of the
polynomial.
25
3.8. POWER FUNCTIONS
By Dr. M. Nxumalo
The domain of any polynomial is R.
A polynomial is a constant function if it of the form P (x) = a for a ∈ R.
If the leading coefficient an 6= 0, then the degree of the polynomial is n.
A polynomial of degree 1 is of the form P (x) = mx + b and so it is called a linear function.
We alternatively use y − x1 = m(x − x1 ) for linear functions.
A polynomial of degree 2 is of the form P (x) = ax2 + bx + c and so it is called a quadratic
function.
A polynomial of degree 3 is of the form P (x) = ax3 + bx2 + cx + d and so it is called a
cubic function.
y
x
0 f (x) = x3 + 2x
3.8
Power functions
Definition 3.8.1. A function is a power function if is of the form f (x) = xa , where a ∈ R.
When a ∈ N, the resulting function is a polynomial with only one term. For instance,
f (x) = x, f (x) = x2 , f (x) = x3 , etc., are all polynomials.
The general shape of the graph of f (x) = xn depends on whether n is even or odd. If n is
even, then f (x) = xn is an even function and its graph is similar to that of y = x2 . Here is
the graph of y = x2 :
26
3.8. POWER FUNCTIONS
By Dr. M. Nxumalo
y
x
0
If n is odd, then f (x) = xn is an odd function and its graph is similar to that of y = x3 .
Here is the graph of y = x3 :
y
x
0
1
When a = n1 where n ∈ N, the function f (x) = x n is a root function.
1
1
If n is even, then the graph of f (x) = x n is similar to the graph of f (x) = x 2 whose domain
1
is [0, ∞). Here is the graph of y = x 2 :
y
x
0
1
1
If n is odd, then the graph of f (x) = x n is similar to the graph of f (x) = x 3 whose domain
1
is R. Here is the graph of y = x 3 :
y
x
0
When a = −1, the function f (x) = x−1 is the reciprocal of the function f (x) = x, and is
a hyperbolic function with the coordinates axes as its asymptotes.
27
3.9. RATIONAL FUNCTIONS
3.9
By Dr. M. Nxumalo
Rational functions
Some functions are ratios of polynomials.
Definition 3.9.1. A rational function is a function of the form:
f (x) =
P (x)
Q(x)
where P (x) and Q(x) are polynomials. Its domain is given by the set of all x such that
Q(x) 6= 0.
3.10
Algebraic functions
Definition 3.10.1. A function is an algebraic function if it can be constructed using algebraic operations (such as addition, substraction, multiplication, division, and taking roots).
3.11
Trigonometric functions
In addition to the functions discussed above, we have trigonometric functions.
It is understood that sin x means the sine of the angle whose radian measure is x.
y
1
−2π −π
g(x) = cos x
π
0
−1
28
x
2π
f (x) = sin x
3.12. TRANSFORMATIONS OF FUNCTIONS
By Dr. M. Nxumalo
From the above two graphs, we can see that
Domf = Domg = R
and
Rangef = Rangeg = [−1, 1].
The tangent function is related to the sine and cosine functions by the equation:
tan x =
sin x
.
cos x
y
h(x) = tan x
− 3π
2
−π
− π2
0
π
2
π
3π
2
x
Clearly,
Domh = R r
nπ
2
o
+ nπ : n ∈ Z
and Rangeh = R.
3.12
Transformations of functions
By applying certain transformations to the graph of a given function, we can obtain the
graphs of certain related functions. In this section, we discuss functions obtained after certain
functions have been transformed.
29
3.12. TRANSFORMATIONS OF FUNCTIONS
By Dr. M. Nxumalo
Let us consider translations.
Vertical and Horizontal Shifts: Suppose c > 0. To obtain the graph of:
1. y = f (x) + c, shift the graph of y = f (x) a distance c units upward
2. y = f (x) − c, shift the graph of y = f (x) a distance c units downward
3. y = f (x − c), shift the graph of y = f (x) a distance c units to the right
4. y = f (x + c), shift the graph of y = f (x) a distance c units to the left.
Now, let us consider stretching and reflecting transformations.
Vertical and Horizontal Stretching and Reflecting: Suppose c > 1. To obtain the
graph of:
1. y = cf (x), stretch the graph of y = f (x) vertically by a factor of c
2. y = 1c f (x), compress the graph of y = f (x) vertically by a factor of c
3. y = f (cx), compress the graph of y = f (x) horizontally by a factor of c
4. y = f ( xc ), stretch the graph of y = f (x) horizontally by a factor of c
5. y = −f (x), reflect the graph of y = f (x) about the x-axis
6. y = f (−x), reflect the graph of y = f (x) about the y-axis.
Example 3.12.1. Describe step by step transformations of the following functions (Each step
must be accompanied by a graph):
1. f (x) = x2 + 6x + 10.
Solution:
By completing the squares, we get that f (x) = (x + 3)2 + 1.
STEP 1: The original graph is the graph of f0 (x) = x2 :
30
3.12. TRANSFORMATIONS OF FUNCTIONS
By Dr. M. Nxumalo
y
x
0
STEP 2: Shift the graph of f0 3 units to the left to get f1 (x) = (x + 3):
y
x
−3
0
STEP 3: Shift the graph of f1 1 unit up to get f (x) = (x + 3)2 + 1:
y
1
x
−3
2. f (x) =
√
2x − 1.
Solution:
We have that f (x) =
q
1
2(x − 21 ) = 2(x − 21 ) 2 .
1
STEP 1: The original graph is the graph of f0 (x) = x 2 :
y
x
0
1
STEP 2: Shift the graph of f0 21 units to the right to get f1 (x) = (x − 21 ) 2 :
31
3.12. TRANSFORMATIONS OF FUNCTIONS
By Dr. M. Nxumalo
y
x
0 0.5
1
STEP 3: Compress the graph of f1 horizontally by a factor of 2 to get f (x) = 2(x − 21 ) 2 :
y
x
0 0.5
3. f (x) = 2| 12 x2 − 1| + 2.
Solution:
STEP 1: The original graph is the graph of f0 (x) = x2 :
y
x
0
STEP 2: Compress the graph of f0 vertically by the factor of 2 to get f1 (x) = 12 x2 :
y
x
0
STEP 3: Shift the graph of f1 a distance of 1 unit downward to get f2 (x) = 21 x2 + 1:
y
x
0
−1
32
3.13. COMBINATIONS OF FUNCTIONS
By Dr. M. Nxumalo
STEP 4: Apply absolute values on the graph of f2 to get f3 (x) = | 12 x2 − 1|:
y
1
x
0
STEP 5: Stretch the graph of f3 vertically by a factor of 2 to get f4 (x) = 2| 21 x2 − 1|:
y
2
x
0
STEP 6: Shift the graph of f4 a distance of 2 units upward to get f5 (x) = 2| 12 x2 − 1| + 2:
y
4
x
0
3.13
Combinations of functions
In this section, we discuss how different functions can be combined to form one function.
Algebra of functions: Let f : A → R and g : B → R be functions. Then the functions
f + g, f − g, f g and f /g are defined as follows:
1. (f + g)(x) = f (x) + g(x); the domain of f + g is A ∩ B.
2. (f − g)(x) = f (x) − g(x); the domain of f − g is A ∩ B.
33
3.13. COMBINATIONS OF FUNCTIONS
By Dr. M. Nxumalo
3. (f g)(x) = f (x)g(x); the domain of f g is A ∩ B.
4.
f
g
(x)
(x) = fg(x)
; the domain of f /g is {x ∈ A ∩ B : g(x) 6= 0}.
The graph of the function f + g is obtained from the graphs of f and g by graphical
addition.
√
Example 3.13.1. Let f : [0, ∞] → R be a function defined by f (x) = x and g : [−2, 2] → R
√
a function defined by g(x) = 9 − x2 . Find the functions (with their domains) f + g, f − g,
f g and f /g.
Solution:
(a) (f + g)(x) =
√
9 − x2 +
√
x:
y
f+g
f (x) =
√
9 − x2
g(x) =
√
x
x
0
The domain of f + g is Domf +g = Domf ∩ Domg = [0, 2].
(b) (f − g)(x) =
√
√
9 − x2 − x:
y
f (x) =
√
9 − x2
g(x) =
√
x
x
0
f-g
The domain of f − g is Domf −g = Domf ∩ Domg = [0, 2].
(c) (f g)(x) =
√
9 − x2 ·
√
x=
√
9x − x3 :
34
3.14. COMPOSITION OF FUNCTIONS
By Dr. M. Nxumalo
y
f (x) =
√
fg
9 − x2
g(x) =
√
x
x
0
The domain of f g is Domf g = Domf ∩ Domg = [0, 2].
q
√
√
2
(d) (f /g)(x) = 9 − x / x = x9 − x :
y
f (x) =
√
f/g
9 − x2
g(x) =
√
x
x
0
The domain of f /g is
Domf /g = {x ∈ Domf ∩ Domg : g(x) 6= 0} = (0, 2].
3.14
Composition of functions
We explore another way of combining two functions to get a new function.
Definition 3.14.1. Given two functions f and g, the composite function f ◦ g (also called
the composition of f and g) is defined by
(f ◦ g)(x) = f (g(x)).
Notes:
1. The domain of f ◦ g is the set of all x in the domain of g such that g(x) is in the domain
of f .
2. (f ◦ g)(x) is defined whenever g(x) and f (g(x)) are defined.
35
3.14. COMPOSITION OF FUNCTIONS
By Dr. M. Nxumalo
To find the domain of f ◦ g, we must:
1. Compute both domains of g and write down the domain of f ◦ g that directly results
from the computation of f ◦ g. We will call this function h at this stage.
2. Find the intersection between the domain of g and h.
Example 3.14.2.
1. If f (x) =
√
x and g(x) =
√
2 − x, find each function and its domain.
(a) f ◦ g
(b) g ◦ f
(c) f ◦ f
(d) g ◦ g
Solution:
p√
√
(a) (f ◦ g)(x) = f ( 2 − x) =
2 − x. For the domain:
p√
The domain of g is (−∞, 2] and the function h(x) =
2 − x is well-defined on (−∞, 2].
So,
Domf ◦g = Domg ∩(−∞, 2] = (−∞, 2].
p
√
√
(b) (g◦f )(x) = g( x) = 2 − x. For the domain: Domf = [0, ∞) and for the function
p
√
h(x) = 2 − x,
Domh = {x ∈ R : 2 −
√
x ≥ 0}
= {x ∈ R : x ≤ 4}
= (−∞, 4].
Therefore, Domg◦f = Domf ∩(−∞, 4] = [0, 4].
p√
√
1
1
(c) (f ◦ f )(x) = f ( x) =
x = x 4 . For the domain: Recall that the graph of y = x 4
1
is similar to that of y = x 2 . So, Domf ◦f = Domf = [0, ∞).
36
3.15. EXPONENTIAL FUNCTIONS
By Dr. M. Nxumalo
p
√
√
(d) (g ◦ g)(x) = g( 2 − x) = 2 − 2 − x. For the domain: Domg = (−∞, 2] and for
p
√
the function h(x) = 2 − 2 − x,
√
Domh = {x ∈ R : 2 − 2 − x ≥ 0}
√
= {x ∈ R : 2 − x ≤ 2}
= {x ∈ R : 2 − x ≤ 4}
= {x ∈ R : x ≥ 2}
= [−2, ∞).
Therefore, Domg◦g = Domg ∩[−2, ∞) = [−2, 2]. So, we exclude all values of x that are
greater than 4. Now, Domg◦f = [0, 4].
x
, g(x) = x10 and h(x) = x + 3.
2. Find f ◦ g ◦ h if f (x) = x+1
Solution:
(f ◦ g ◦ h)(x) = f (g(x + 3)) = f ((x + 3)10 ) =
3.15
(x + 3)10
.
(x + 3)10 + 1
Exponential functions
Some functions increase exponentially.
Definition 3.15.1. A function is an exponential function if it is of the form f (x) = ax ,
where a is a positive real number and is called the base of f (x).
y
( 14 )x
( 12 )x
10x
2x
1.5x
x
We can see from the above diagram that:
37
3.15. EXPONENTIAL FUNCTIONS
By Dr. M. Nxumalo
1. If 0 < a < 1, then the exponential function decreases.
2. If a > 1, then the exponential function increases.
We have the following laws of exponents: If a and b are positive numbers and x and y are any
real numbers, then
1. ax+y = ax ay .
2. ax−y =
ax
.
ay
3. (ax )y = axy .
4. (ab)x = ax bx .
The exponential function occurs very frequently in mathematical models of nature and
society. Here we indicate briefly how it arises in the description of population growth and
radioactive decay.
Example 3.15.2. The half-life of Strontium-90, 90 Sr, is 25 years. This means that half of
any given quantity of 90 Sr will disintegrate in 25 years.
1. If a sample of 90 Sr has a mass of 24 mg, find an expression for the mass m(t) that
remains after t years.
2. Find the mass remaining after 40 years, correct to the nearest milligram.
Solution:
1. The mass is initially 24 mg and is halved during each 25-year period, so
m(0) = 24 =
1
0
2 25
1
1
m(25) = (24) = 25
2
2 25
1 1
1
1
m(50) = · (24) = 2 (24) = 50 (24)
2 2
2
2 25
1 1 1
1
1
m(75) = · · (24) = 3 (24) = 75 (24)
2 2 2
2
2 25
38
3.15. EXPONENTIAL FUNCTIONS
By Dr. M. Nxumalo
1
1 1 1 1
1
· · · (24) = 4 (24) = 100 (24).
2 2 2 2
2
2 25
m(100) =
From this pattern, it appears that the mass remaining after t years is
m(t) =
1
2
t
25
t
(24) = 24 · 2− 25 .
40
2. m(40) = 24 · 2− 25 mg.
Consider the following graph:
y
f (x) = 2x
g(x) = 0.7x + 1
(0, 1)
x
The line g, with gradient 0.7, touches the graph of f at only one point: (0, 1). We are
interested to find a number a such that the line h(x) = x + 1 with gradient 1 will only touch
the graph g(x) = ax at the point (0, 1). This leads to the introduction of the constant e with
a value
e ≈ 2.71828.
This letter was chosen by Leonard Euler in 1727.
The graph of f (x) = ex is as follows:
y
(0, 1)
0
39
x
3.15. EXPONENTIAL FUNCTIONS
By Dr. M. Nxumalo
Example 3.15.3. Describe step by step transformations of the function f (x) = 12 e−x − 1
(Each step must be accompanied by a graph, domain and range).
Solution:
STEP 1: The original graph is the graph of f0 (x) = ex :
y
(0, 1)
x
0
Domf0 = R.
Rangef0 = (0, ∞).
STEP 2: Reflect the graph of f0 about the y-axis to get f1 (x) = e−x :
y
(0, 1)
x
0
Domf1 = R.
Rangef1 = (0, ∞).
STEP 3: Compress the graph of f1 vertically by a factor of 2 to get f2 (x) = 12 e−x :
y
0.5
0
Domf2 = R.
Rangef2 = (0, ∞).
40
x
3.16. INVERSE FUNCTIONS
By Dr. M. Nxumalo
STEP 4: Shift the graph of f2 1 unit down to get f (x) = 12 e−x − 1:
y
x
−1
Domf = R.
Rangef = (−1, ∞).
3.16
Inverse functions
For a function f (x), we usually speak about finding the output f (x) for an input x. What if
we want want to treat f (x) as an input and x an output? Would we still have a function?
We answer these questions in this section. In short, we want to study those functions which
return the original value for which a function has given the output.
By the inverse function, denoted by f −1 , of a function f , we refer to a function that is
obtained by reversing the function f .
Not every function has an inverse. Consider the following function:
f
X
Y
a
c
b
If we reverse f , we get:
41
p
3.16. INVERSE FUNCTIONS
By Dr. M. Nxumalo
f −1
Y
X
a
b
c
p
Clearly, f −1 is not a function.
Let us introduce a class of functions that are guaranteed to have inverse.
Definition 3.16.1. A function f : X → Y is called one to one (or injective) if no two
different elements of X gets mapped to the same value of Y . Algrebraically, all two elements
x1 , x2 of X satisfy the condition that
If x1 6= x2 , then f (x1 ) 6= f (x2 ).
Equivalently,
If f (x1 ) = f (x2 ), then x1 = x2 .
A function is injective if it passes the horizontal line test. That is, no horizontal line
will cut two different points of the graph.
Example 3.16.2. The function f (x) = ex is injective, as shown below:
y
f (x) = ex
x
Example 3.16.3. Use the algebraic definition of injectivity to show that the function f :
√
[5, ∞) → R, defined by, f (x) = x − 5 is injective.
42
3.16. INVERSE FUNCTIONS
By Dr. M. Nxumalo
Solution:
Proof. We pick two random elements a, b ∈ [5, ∞) and assume that f (a) = f (b). (We must
show that a=b).
f (a) = f (b) =⇒
√
a−5=
√
b−5
=⇒ a − 5 = b − 5
=⇒ a = b.
Thus f is injective.
Definition 3.16.4. A function f : X → Y is surjective (or onto) if every element of the
codomain Y belongs to the range f (X). That is f (X) = Y . Algrebraically,
For every y ∈ Y, there is x ∈ X such that f (x) = y.
Example 3.16.5.
1. The following function is surjective:
f
X
Y
a
c
p
b
2. The function f : R → R defined by f (x) = 2x , is no surjective. This is because y = −1
is in the codomain of f , but there is not x ∈ R such that f (x) = −1.
3. Use the algebraic definition of surjectivity to show that the function f : R → R, defined
by f (x) = x3 , is surjective.
Solution:
Proof. We pick a random y ∈ R. (We must show that there is x ∈ R satisfying f (x) = y).
Let us construct the x in question, we work backwards:
43
3.16. INVERSE FUNCTIONS
By Dr. M. Nxumalo
f (x) = y =⇒ x3 = y
√
=⇒ x = 3 y.
√
Set x = 3 y. It is clear that x ∈ R.
Then
√
√
f (x) = f ( 3 y) = ( 3 y)3 = y.
Therefore f is surjective.
Definition 3.16.6. A function is bijective if it is both injective and surjective.
Definition 3.16.7. Let f be a bijective function with domain A and codomain B. Then its
inverse function f −1 has domain B and range A and is defined by
f −1 (y) = x ⇐⇒ f (x) = y
for all y ∈ B.
Do not mistake the −1 in f −1 for an exponent. Thus
f −1 (x) 6=
1
.
f (x)
For a function f : X → Y with inverse f −1 , we have the following cancellation equations:
1. f −1 (f (x)) = x for all x ∈ X.
2. f (f −1 (y)) = y for all y ∈ Y .
Here are steps on how to find the inverse of a bijective function f :
Step 1 Write y = f (x).
Step 2 Solve this equation for x in terms of y (if possible).
Step 3 To express f −1 as a function of x, interchange x and y. The resulting equation is
y = f −1 (x).
44
3.17. LOGARITHMIC FUNCTIONS
By Dr. M. Nxumalo
The graph of f −1 is obtained by reflecting the graph of f about the line y = x.
Example 3.16.8. Sketch the graphs of f (x) =
√
−1 − x and its inverse function using the
same coordinate axes.
Solution:
y
y=x
x
f
f −1
3.17
Logarithmic functions
If a > 0 and a 6= 1, then the exponential function f (x) = ax has an inverse function f −1 ,
which is called the logarithmic function with base a and is denoted by loga .
Clearly,
loga (x) = y ⇐⇒ ay = x.
Here is the graph of f (x) = loga x, a > 1:
45
3.17. LOGARITHMIC FUNCTIONS
By Dr. M. Nxumalo
y
f −1
x
1
y=x
f
Here are some laws of logarithms:
1. loga (a) = 1
2. loga (xy) = loga x + loga y
x
3. loga
= loga x − loga y
y
4. loga (xr ) = r loga x, where r ∈ R.
Example 3.17.1. Evaluate log2 80 − log2 5.
Solution:
80
log2 80 − log2 5 = log2
5
= log2 (16)
= log2 (24 )
= 4 log2 2
= 4(1) = 4.
Certain logarithmic functions have base e. These are called natural logarithms and have
a special notation:
loge x = ln x.
Clearly,
ln x = y ⇐⇒ ey = x
46
3.17. LOGARITHMIC FUNCTIONS
By Dr. M. Nxumalo
We also have that ln e = 1.
Change of base formula: For any positive number a (a 6= 1), we have
loga x =
ln x
.
ln a
The graph of ln x is similar to that of loge x. Can you convice yourself that this is
true?
Example 3.17.2.
1. Find x if ln x = 5.
2. Solve the equation e5−3x = 10.
3. Sketch the graph of the function f (x) = 2 ln(x − 2) − 1.
Solution:
1. We have that e5 = x.
10
2. We have that ln 10 = 5 − 3x. Therefore x = 5−ln
.
3
3. STEP 1: The original graph is the graph of f0 (x) = ln x:
y
x
1
f0
STEP 2: Shift the graph of f0 2 units to the right to get f1 (x) = ln(x − 2):
47
3.17. LOGARITHMIC FUNCTIONS
By Dr. M. Nxumalo
y
x
3
f1
STEP 3: Stretch the graph of f1 vertically by a factor of 2 to get f2 (x) = 2 ln(x − 2):
y
x
3
f2
STEP 4: Shift the graph of f2 1 unit down to get f (x) = 2 ln(x − 2) − 1:
y
x
f
48
3.18. INVERSE TRIGONOMETRIC FUNCTIONS
3.18
By Dr. M. Nxumalo
Inverse trigonometric functions
The graphs of trigonometric functions in Section 3.11 are not injective. This means that they
do not have inverses. With restrictions on their domain, we can get inverses. This is the aim
of this section.
The function f (x) = sin x, − π2 ≤ x ≤ π2 is bijective, and hence it has an inverse. Its inverse
is called the acrsine function. For f (x) = cos x, 0 ≤ x ≤ π and g(x) = tan x, − π2 < x < π2 ,
we have the following cancellation equations:
1. sin−1 (sin x) = x for − π2 ≤ x ≤ π2
2. sin(sin−1 (x)) = x for −1 ≤ x ≤ 1
3. cos−1 (cos x) = x for 0 ≤ x ≤ π
4. cos(cos−1 (x)) = x for −1 ≤ x ≤ 1
Example 3.18.1. Simplify the expression cos(tan−1 x).
Solution:
Let y = tan−1 (x). Then tan y = x and − π2 < y < π2 . Observe that
sec2 y = 1 + tan2 y = 1 + x2 .
So, sec y =
√
1 + x2 .
This makes,
cos(tan−1 x) = cos y =
49
1
1
=√
.
sec y
1 + x2
Chapter 4
Limits
4.1
The tangent problem
The word tangent is derived from the Latin word tangens, which means “touching”. Thus,
a tangent to a curve is a line that touches the curve. Therefore, a tangent line should have
the same direction as the curve at the point of contact. In this section, we plan to make this
precise.
Let us try to find a tangent line t to the parabola y = x2 :
y
t
y = x2
Q(x, x2 )
P (1, 1)
x
0
The line P Q is called a secant line which is defined as a line between two points of a curve.
We need to find the slope m of t. Since we are only given one point, we compute an
approximation to m by choosing a nearby point Q(x, x2 ) on the parabola y = x2 .
50
4.2. THE VELOCITY PROBLEM
By Dr. M. Nxumalo
We choose x 6= 1 so that P 6= Q. Therefore
x2 − 1
x−1
(x + 1)(x − 1)
=
x−1
= x + 1.
mP Q =
For several values of x closer to 1, mP Q = 2. In fact, the closer Q is to P , the closer x is
to 1 and, hence the closer mP Q is to 2. This suggests that the slope of the line t should be 2.
We say that the slope of the tangent line is the limit of the slopes of the secant lines. We
express this symbolically by writing
lim mP Q = m
Q→P
4.2
and
lim x + 1 = 2.
x→1
The velocity problem
When driving a car, you will notice that the speedometer’s needle does not stay still for very
long. This means that the speed (or velocity) of the car is not constant.
We wish to find the instantaneous velocity which is the velocity of an object under
motion at a specific point of time.
Consider the following example:
Example 4.2.1. Suppose that a ball is dropped from a building, 450m above the ground.
Find the velocity of the ball after 5 seconds.
Solution: Denote by s(t) the distance fallen after t seconds. Then
s(t) = 4.9t2
Simple calculations will show that the slope of the line tangent that intersects s(t) at t = 5
seconds is given by m = 49 m.s−1 . (Use the two points (t, 4.9t2 ) and (5, 4.9 × 52 )).
51
4.3. THE LIMIT OF A FUNCTION
4.3
By Dr. M. Nxumalo
The limit of a function
We have noticed how limits arise when we want to find the tangent to a curve. We now turn
our attention to limits in general.
Let us investigate the behavior of the function f defined by f (x) = x2 − x + 2 for values of
x near 2.
y
y = x2 − x + 2
4
0
2
x
From the above graph, we see that when x approaches 2 (on either side of 2), f (y) gets 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 notion for this is
lim (x2 − x + 2) = 4
x→2
Formally, we write
lim f (x) = L
x→a
to mean that “the limit of f (x), as x approaches a, equals L.”
This says that the values of f (x) get closer and closer to L as x gets closer and closer to a
(from either side of a) but x 6= a.
52
4.4. ONE-SIDED LIMITS
By Dr. M. Nxumalo
The notion x 6= a means that f (x) need not be defined when x = a. We only care about
how f is defined near a.
4.4
One-sided limits
Consider the following function:
(
x+2
f (x) =
−x
if x ≥ −2
if x < −2.
We get the following graph:
y
f
2
x
−2 0
We can see that f (x) approaches 2 as x approaches −2 from the left, and f (x) approaches 0
as x approaches −2 from the right. We indicate this situation symbolically by writing
lim f (x) = 2 and
x→−2−
lim = 0
x→−2+
The symbol “x → −2− ” indicates that we consider only values of x that are less than −2.
Likewise, “x → −2+ ” indicates we consider only values of x that are greater than −2.
Formally, we write
lim f (x) = L
x→a−
to mean that “the left-hand limit of f (x), as x approaches a (or the limit of f (x) as x approaches
x from the left) equals L”. We also write
lim f (x) = L
x→a+
53
4.5. INFINITE LIMITS
By Dr. M. Nxumalo
to mean that “the right-hand limit of f (x), as x approaches a (or the limit of f (x) as x
approaches x from the right) equals L”.
It is clear that
lim f (x) = L if and only if lim− f (x) = L and lim+ f (x) = L
x→a
x→a
x→a
The above statement tells us that there is no value of f (x) as x approaches a if the sided
limits do not have the same value. In other words, lim f (x) exists only if
x→a
lim f (x) = lim+ f (x).
x→a−
4.5
x→a
Infinite limits
Consider the function f (x) =
1
whose graph is given by:
(x − 1)2
y
f
x
0
1
1
becomes
(x − 1)2
very large. From the graph, we can see that f (x) does not approach a number whenever x
1
approaches 1. So, lim
does not exist.
x→1 (x − 1)2
As x approaches 1 (from both sides), (x − 1)2 becomes close to 0 and
We use the notation
1
=∞
x→a (x − 1)2
lim
to indicate the kind of behavior exhibited above.
This does not mean that we are regarding ∞ as a number, nor does it mean that the limit
exists. It simply expresses the particular way in which the limit does not exist.
54
4.5. INFINITE LIMITS
By Dr. M. Nxumalo
In general, we write
lim f (x) = ∞
x→a
to indicate that the values of f (x) become larger and larger (or “increase without bound”) as
x becomes closer and closer to a. We also write
lim f (x) = −∞
x→a
to indicate that the values of f (x) become smaller and smaller (or “decrease without bound”)
as x becomes closer and closer to a.
Formally: For a function f defined on both sides of a, except possibly at a itself, we write
lim f (x) = ∞
x→a
to mean that the values of f (x) can be made arbitrarily large (as large as we please) by taking
x sufficiently close to a but not equal to a.
We write
lim f (x) = −∞
x→a
to mean that the values of f (x) can be made arbitrarily large negative by taking x sufficiently
close to a but not equal to a.
The symbol lim f (x) = ∞ can be read as “the limit of f (x) as x approaches a, is infinity”
x→a
and the symbol lim f (x) = −∞ can be read as “the limit of f (x) as x approaches a, is negative
x→a
infinity”.
Definition 4.5.1. The line x = a is called a vertical asymptote of the curve y = f (x) if at
least one of the following statements is true:
1. lim f (x) = ∞
x→a
2. lim f (x) = −∞
x→a
3. lim− f (x) = ∞
x→a
4. lim f (x) = −∞
x→a
5. lim+ f (x) = ∞
x→a
6. lim+ f (x) = −∞
x→a
55
4.6. CALCULATING LIMITS USING THE LIMIT LAWS
By Dr. M. Nxumalo
4.6
Calculating limits using the limit laws
In this chapter, we try to avoid using calculators and graphs to guess the values of limits. We
use the following properties of limits.
LIMIT LAWS: Suppose that c is a constant and the limits
lim f (x) and lim g(x)
x→a
x→a
exist. Then
1. SUM RULE: lim [f (x) + g(x)] = lim f (x) + lim g(x)
x→a
x→a
x→a
2. DIFFERENCE RULE: lim [f (x) − g(x)] = lim f (x) − lim g(x)
x→a
x→a
x→a
3. PRODUCT RULE: lim [f (x)g(x)] = lim f (x) · lim g(x)
x→a
x→a
x→a
lim f (x)
f (x)
4. QUOTIENT RULE: lim
= x→a
if lim g(x) 6= 0
x→a g(x)
lim g(x) x→a
x→a
Example 4.6.1. Use the Limit Laws and the graphs of f and g
5
4
y
g
x
−2
f
0 1
−4
to evaluate the following limits, if they exist:
1.
lim [f (x) + 5g(x)]
x→−2−
2. lim [f (x)g(x)]
x→1
3.
lim
x→−2+
f (x)
g(x)
56
3
4.6. CALCULATING LIMITS USING THE LIMIT LAWS
By Dr. M. Nxumalo
Solution:
1.
lim [f (x) + 5g(x)] = lim − f (x) + lim − 5g(x)
x→−2−
x→−2
x→−2
= lim − f (x) + 5 lim − g(x)
x→−2
x→−2
= −4 + 5(5)
= 21.
2.
lim [f (x)g(x)] = lim f (x) · lim g(x)
x→1
x→1
x→1
= −4 · −4
= 16.
3.
lim
x→−2+
lim f (x)
f (x)
x→−2+
=
g(x)
lim + g(x)
x→−2
0
5
= 0.
=
We further have the following properties of limits which result from the Limit laws:
h
i
6 POWER RULE: lim [f (x)]n = lim f (x) where n ∈ N
x→a
x→a
7 CONSTANT RULE: lim c = c
x→a
8 lim [cf (x)] = c lim f (x) (This is a result of a product rule)
x→a
x→a
9 lim x = a
x→a
10 lim xn = an where n ∈ N (This is a result of a power rule)
x→a
√
√
11 lim n x = n a where n ∈ N. (If n is even, we assume a > 0) (This is a result of a
x→a
power rule)
57
4.6. CALCULATING LIMITS USING THE LIMIT LAWS
By Dr. M. Nxumalo
q
p
12 lim n f (x) = n lim f (x) where n ∈ N. (If n is even, we assume that lim f (x) > 0)
x→a
x→a
x→a
(This is a result of a power rule)
13 If f is a polynomial or a rational function and a is in the domain of f , then lim f (x) =
x→a
f (a)
Example 4.6.2.
1. Evaluate:
(a) lim [2x2 − 3x + 4]
x→5
x3 + 2x2 − 1
x→−2
5 − 3x
(b) lim
(c) lim g(x) where
x→1
(
x+1
g(x) =
π
√
(d) lim
x→0
if x 6= 1
if x = 1.
x2 + 9 − 3
x2
2. Show that
(a) lim |x| = 0
x→0
|x|
does not exist.
x→0 x
(b) lim
3. The greatest integer function is defined by bxc = the largest integer that is less or
equal to x. (For instance, b1c = 1, b8.72c = 8). Show that lim bxc does not exist.
x→3
Solution:
1. (a)
lim [2x2 − 3x + 4] = 2 lim x2 − 3 lim x + lim 4 = 2(52 ) − 3(5) + 4 = 39.
x→5
x→5
58
x→5
x→5
4.6. CALCULATING LIMITS USING THE LIMIT LAWS
By Dr. M. Nxumalo
(b)
lim (x3 + 2x2 − 1)
x3 + 2x2 − 1
x→−2
lim
=
x→−2
5 − 3x
lim (5 − 3x)
x→−2
lim x3 + 2 lim (x2 ) − lim 1
=
x→−2
x→−2
x→−2
lim 5 − 3 lim x
x→−2
3
x→−2
2
(−2) + 2(−2) − 1
5 − 3(−2)
1
=− .
11
=
(c) Does not exist.
(d) We use a method of rationalizing expressions:
√
lim
x→0
√
x2 + 9 − 3
x2 + 9 + 3
√
·
x2
x2 + 9 + 3
x2 + 9 − 9
√
= lim
x→0 x2 ( x2 + 9 + 3)
1
= lim √
x→0
x2 + 9 + 3
1
=q
lim x2 + 9 + 3
x2 + 9 − 3
= lim
x→0
x2
√
x→0
1
= .
6
2. (a) We have
(
x
|x| =
−x
if x ≥ 0
if x < 0
For x ≥ 0, lim+ |x| = lim+ x = 0.
x→0
x→0
For x < 0, lim− |x| = lim− (−x) = 0.
x→0
x→0
It follows that lim |x| = 0.
x→0
|x|
|x|
= lim+ 1 = 1 and lim−
= lim− (−1) = −1.
x→0
x→0
x→0
x→0
x
x
|x|
|x|
Therefore lim+
6= lim−
, so the limit does not exist.
x→0
x→0
x
x
(b) lim+
59
4.6. CALCULATING LIMITS USING THE LIMIT LAWS
By Dr. M. Nxumalo
3. We have
(
3
bxc =
2
if x ≥ 3
if 2 ≤ x < 3
So
lim bxc = 3 6= lim− bxc = 2.
x→3+
x→3
Therefore, the limit does not exist.
We give the following two theorems:
Theorem 4.6.3. If f (x) ≤ g(x) when x is near a (except possibly at a) and the limits of f
and g both exist as x approaches a, then
lim f (x) ≤ lim g(x)
x→a
x→a
Theorem 4.6.4 (The Squeeze Theorem). If f (x) ≤ g(x) ≤ h(x) when x is near a (except
possibly at a) and
lim f (x) = lim h(x) = L
x→a
x→a
then
lim g(x) = L.
x→a
1
Example 4.6.5. Show that lim x2 sin
= 0.
x→0
x
Solution:
We have
1
≤ 1,
−1 ≤ sin
x
so
1
≤ x2 .
−x ≤ x sin
x
2
2
1
Now, lim (−x ) = 0 and lim (x ) = 0. By the Squeeze Theorem, lim x sin
= 0.
x→0
x→0
x→0
x
2
2
2
60
4.7. THE PRECISE DEFINITION OF A LIMIT
4.7
By Dr. M. Nxumalo
The precise definition of a limit
The phrases “f (x) is closer to L” and “x is closer and closer to a” are vague. Here, we give a
precise definition of a limit.
Consider the function
(
2x − 1
f (x) =
6
if x 6= 3
if x = 3
with the following graph:
f
y
5+
5
5−
f
x
3
3−δ
3+δ
Intuitively, it is clear that when x is close to 3 but x 6= 3, then f (x) is close to 5, and
lim f (x) = 5.
x→3
We tackle the following question:
How close to 3 does x have to be so that f (x) differs from 5 by 0.1?
The distance from x to 3 is |x − 3| and the distance from f (x) to 5 is |f (x) − 5|. The
problem is to find a real number δ such that
If
|x − 3| < δ
but
x 6= 3,
then
|f (x) − 5| < 0.1.
If |x − 3| > 0, then x 6= 0, so an equivalent formulation of the problem is to find a number
δ such that
If
0 < |x − 3| < δ
but
x 6= 3,
then
If
0 < |x − 3| <
then
61
0.1
= 0.05,
2
|f (x) − 5| < 0.1.
4.7. THE PRECISE DEFINITION OF A LIMIT
By Dr. M. Nxumalo
|f (x) − 5| = |(2x − 1) − 5| = |2x − 6| = 2|x − 3| < 0.1.
That is
If
0 < |x − 3| < 0.05
x 6= 3,
but
then
|f (x) − 5| < 0.1.
The answer to the problem is δ = 0.05.
We give a precise definition of a limit.
Definition 4.7.1. Let f be a function defined on some open interval that contains the number
a, except possibly at a itself. Then we say that the limit of f (x) as x approaches a in L,
and we write
lim f (x) = L
x→a
if for every real number > 0 there is a real number δ > 0 such that
0 < |x − a| < δ
implies
|f (x) − L| < In words:
lim f (x) = L means that the distance between f (x) and L can be made arbitrary small by
x→a
taking the distance from x to a sufficiently small (but not 0).
Alternatively,
lim f (x) = L means that for every postive real number (no matter how small it is), we
x→a
can find δ > 0 such that if x lies in the open interval (a − δ, a + δ) and x 6= a, then f (x) lies
in the open interval (L − , L + ).
The definition of limit says that if any small interval (L − , L + ) is given around L, then
we can find an interval (a − δ, a + δ) around a such that f maps all the points in (a − δ, a + δ)
(except possibly a) into the interval (L − , L + ):
f
f (x)
x
a−δ
a
L−
a+δ
62
L
L+
4.8. LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES
By Dr. M. Nxumalo
Example 4.7.2. Prove that lim (4x − 5) = 7.
x→3
Solution:
We pick a random > 0. (We must find δ > 0 and verify that it satisfies the condition
that “If 0 < |x − 3| < δ, then |f (x) − 7| < ”).
Let us guess the value of δ. We work backwards:
|f (x) − 7| < ⇐⇒ |4x − 5 − 7| < ⇐⇒ |4x − 12| < ⇐⇒ |4(x − 3)| < ⇐⇒ 4|x − 3| < ⇐⇒ |x − 3| < .
4
Set δ = 4 . Then δ > 0.
(We verify that our δ works).
Now, if |x − 3| < δ, then
|x − 3| <
=⇒ 4|x − 3| < 4
=⇒ |4x − 12| < =⇒ |4x − 5 − 7| < =⇒ |f (x) − 7| < .
Thus lim (4x − 5) = 7.
x→3
4.8
Limits at infinity: horizontal asymptotes
In Section 4.5, we investigated infinite limits and vertical asymptotes. There we let x approach
a number and the result was that the values of y became arbitrarily large (positive or negative).
In this section we let x become arbitrarily large (positive or negative) and see what happens
to y.
63
4.8. LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES
By Dr. M. Nxumalo
Let us investigate the behaviour of the function f defined by
x2 − 1
f (x) = 2
x +1
as x becomes large:
y
y=1
y = f (x)
0
x
1
As x grows larger and larger, the values of f (x) get closer and closer to 1. The situation is
expressed symbolically by
x2 − 1
=1
x→∞ x2 + 1
lim
In general, we use the notation
lim f (x) = L
x→∞
to indicate that the values of f (x) becomes closer and closer to L as x becomes larger and
larger.
Definition 4.8.1. Let f be a function defined on some interval (a, ∞). Then
lim f (x) = L
x→∞
means that the values of f (x) can be made arbitrarily close to L by taking x sufficiently large.
We read lim f (x) = L as “the limit of f (x), as x approaches infinity, is L” or “the limit
x→∞
of f (x), as x becomes infinite, is L”. The symbol ∞ does not represent a number.
From the above graph, we see that as x grows large negative, the values of f (x) get closer
and closer to 1. The situation is expressed symbolically by
x2 − 1
=1
x→−∞ x2 + 1
lim
Definition 4.8.2. Let f be a function defined on some interval (−∞, a). Then
lim f (x) = L
x→−∞
means that the values of f (x) can be made arbitrarily close to L by taking x sufficiently large
negative.
64
4.8. LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES
By Dr. M. Nxumalo
Definition 4.8.3. The line y = L is called a horizontal asymptote of the curve y = f (x)
if either
lim f (x) = L
x→∞
Example 4.8.4.
or
lim f (x) = L
x→−∞
1. For the function f (x) = 1 +
y
2x
with the following graph:
3x2 + 1
f (x) = 1 +
y=1
2x
3x2 + 1
x
0
the number 1 is its horizontal asymptote.
2. The function f (x) = tan−1 x with the following graph:
y
y = π2
x
0
y = − π2
has two horizontal asymptotes: y = − π2 and y = π2 .
f (x) = arctan x
3. Find the infinite limits, limits at infinity, and asymptotes for the function f whose graph
is shown below:
y
2
1
−1.5
0
x
1.5
Solution:
(a) lim f (x) = 1 (horizontal asymptote: y = 1).
x→∞
(b) lim f (x) = 2 (horizontal asymptote: y = 2).
x→−∞
(c)
(d)
lim
f (x) = −∞ (vertical asymptote: x = −1.5).
lim
f (x) = −∞ (vertical asymptote: x = −1.5).
x→−1.5−
x→−1.5+
(e) lim − f (x) = ∞ (vertical asymptote: x = 1.5).
x→1.5
(f) lim + f (x) = −∞ (vertical asymptote: x = 1.5).
x→1.5
65
4.8. LIMITS AT INFINITY: HORIZONTAL ASYMPTOTES
By Dr. M. Nxumalo
Theorem 4.8.5. If r > 0 is a rational number, then
1
=0
x→∞ xr
lim
If r > 0 is a rational number such that xr is defined for all x, then
1
=0
x→−∞ xr
lim
To evaluate the limit at infinity of any rational function, we must first divide both the
numerator and denominator by the highest power of x that occurs in the denominator.
Example 4.8.6.
1. Evaluate:
3x2 − x − 2
.
x→∞ 5x2 + 4x + 1
lim
2. Compute lim
√
x→∞
x2 + 1 − x .
3. Evaluate lim− sin x.
x→0
Solution:
1.
2
3x
− xx2 − x22
3x2 − x − 2
x2
=
lim
lim
2
x→∞ 5x + 4x + 1
x→∞ 5x2 + 4x + 1
x2
x2
x2
3 − x1 − x22
x→∞ 5 + 4 + 12
x
x
= lim
3 − x1 − x22
x→∞ 5 + 4 + 12
x
x
3
= .
5
= lim
2.
lim
x→∞
√
√
x2 + 1 − x √ 2
x2 + 1 − x = lim √
· x +1+x
x→∞
x2 + 1 + x
1
= lim √
x→∞
x2 + 1 + x
= lim q
x→∞
=0
3. lim− sin x = sin(0) = 0.
x→0
66
1
x
x2
+ x12 + xx
x2
4.9. INFINITE LIMITS AT INFINITY
4.9
By Dr. M. Nxumalo
Infinite limits at infinity
The notation
lim f (x) = ∞
x→∞
is used to indicate that the values of f (x) become large as x becomes larger. Similar meanings
are attached to the following symbols:
lim f (x) = ∞
x→−∞
Example 4.9.1.
lim f (x) = −∞
x→∞
lim f (x) = −∞
x→−∞
1. Evaluate:
(a) lim x3
x→∞
(b) lim x3
x→−∞
Solution:
1. (a) lim x3 = ∞.
x→∞
(b) lim x3 = −∞.
x→−∞
4.10
Tangents
We return to tangent problems with the ability to calculate slopes of tangents using the
discussed limit laws.
Consider the following diagram:
y
Q(x, f (x))
P (a, f (a))
f (x) − f (a)
x
a
x
x−a
For a curve C with the equation y = f (x), we want to find the tangent line to C at the
point P (a, f (a)), so we consider a nearby point Q(x, f (x)), where x 6= a, and compute the
slope of the secant line P Q:
67
4.10. TANGENTS
By Dr. M. Nxumalo
mP Q =
f (x) − f (a)
x−a
Then we let Q approach P along the curve C by letting x approach a. If mP Q approaches a
number m, then we define the tangent t to be the line through P with slope m. This amounts
to saying that the tangent line is the limiting position of the secant line P Q as Q approaches
P.
Definition 4.10.1. The tangent line to the curve y = f (x) at the point P (a, f (a)) is the
line through P with slope
f (x) − f (a)
x→a
x−a
m = lim
provided that this limit exists.
Example 4.10.2. Find the equation of the tangent line to the parabola f (x) = x2 + 3 at the
point P (1, 1).
Solution: The slope of the tangent line in question f is given by
f (x) − f (1)
x→1
x−1
2
x + 3 − ((1)2 + 3)
= lim
x→1
x−1
x2 − 1
= lim
x→1 x − 1
(x − 1)(x + 1)
= lim
x→1
x−1
= lim (x + 1)
m = lim
x→1
= 2.
For the y-intercept c:
f (1) = (1)2 + 3 = 4.
Therefore y = 2x + 4.
Alternatively, if we let h = x − a, we get that x = a + h. So, the slope of the secant line
P Q is
mP Q =
From the diagram:
68
f (a + h) − f (a)
h
4.11. VELOCITIES
By Dr. M. Nxumalo
y
Q(a + h, f (a + h))
P (a, f (a))
f (a + h) − f (a)
x
a a+h
h
notice that as x approaches a, h approaches 0 (because h = x − a) and so the expression
for the slope of the tangent line in Definition 4.10.1 becomes
m = lim
h→0
f (a + h) − f (a)
h
Example 4.10.3. Find the equation of the tangent line to the hyperbola f (x) = 3/x at the
point (3, 1).
Solution: The slope of the tangent line in question is given by
f (3 + h) − f (3)
h→0
h
3
−1
3
+
h
= lim
h→0
h
−h
= lim
h→0 h(3 + h)
f
−1
= lim
h→0 3 + h
1
=− .
3
m = lim
1
For the y-intercept c: Using the point (3, 1), y − y1 = m(x − x1 ) gives y − 1 = − (x − 3).
3
1
y = − x + 2.
3
4.11
Velocities
Suppose an object moves along a straight line according to the equation of motion s = f (t),
where s is the displacement (directed distance) of the object from the origin at time t. The
function f that describes the motion is called the position function of the object. In the
69
4.11. VELOCITIES
By Dr. M. Nxumalo
time interval from t = a to t = a + h the change in position is f (a + h) − f (a). The average
velocity over this time is
average velocity =
displacement
f (a + h) − f (a)
=
time
h
which is the same as the slope of the secant line P Q in the following diagram:
y
Q(a + h, f (a + h))
P (a, f (a))
f (a + h) − f (a)
x
a a+h
h
Now, suppose we compute average velocities over shorter and shorter time intervals [a, a + h].
In other words, we let h approach 0. We define the velocity (or instantaneous velocity)
v(a) at time t = a to be the limit of these average velocities:
f (a + h) − f (a)
h→0
h
This means that the velocity at time t = a is equal to the slope of the tangent line at P .
v(a) = lim
Example 4.11.1. Suppose that a ball is dropped from the top of a building, 450m above the
ground.
1. What is the velocity of the ball after 5 seconds?
2. How fast is the ball travelling when it hits the ground?
Solution: We have the following equation of motion: f (t) f= 4.9t2 . Therefore
f (t + h) − f (t)
h→0
h
4.9(t + h)2 − 4.9t2
= lim
h→0
h
2
4.9(t + 2th + h2 − t2 )
= lim
h→0
h
4.9h(2t + h)
= lim
h→0
h
= 9.8t.
v(a) = lim
70
4.12. OTHER RATES OF CHANGE
By Dr. M. Nxumalo
1. v(5) = 9.8(5) = 49 m/s.
2. The ball will hit the ground when f (t) = 450. Therefore 4.9t2 = 450, making
r
450
t=
= 9.6s.
4.9
Therefore, the velocity of the ball as it hits the ground is therefore
v(t) = 9.8t = 9.8(9.6) = 94 m/s.
4.12
Other rates of change
Suppose y is a quantity that depends on another quantity x. Thus, y is a function of x and we
write y = f (x). If x changes from x1 to x2 , then the change in x (also called the increment
of x) is
∆x = x2 − x1
and the corresponding change in y is
∆y = f (x2 ) − f (x1 )
The difference quotient
f (x2 ) − f (x1 )
∆y
=
∆x
x2 − x1
is called the average rate of change of y with respect to x over the interval [x1 , x2 ].
By analogy with velocity, we consider the average rate of change over smaller and smaller
intervals by letting x2 approach x1 and therefore letting ∆x approach 0. The limit of these
average rates of change is called the (instantanenous) rate of change of y with respect
to x at x = x1 :
∆y
f (x2 ) − f (x1 )
= lim
x2 →x1
∆x→0 ∆x
x2 − x1
instantaneous rate of change = lim
Example 4.12.1. Temperature readings T (in degrees Celsius) were recorded every hour
starting at midnight on a day in April in Grahamstown, South Africa. The time x is measured
in hours from midnight. The data are given in the table below:
71
4.12. OTHER RATES OF CHANGE
x(h)
0
1
2
3
4
T (◦ C)
6.5
6.1
5.6
4.9
4.2
x(h)
5
6
7
8
9
T (◦ C)
4.0
4.0
4.8
6.1
8.3
By Dr. M. Nxumalo
x(h)
10
11
12
13
14
T (◦ C)
10.0
12.1
14.3
16.0
17.3
x(h)
15
16
17
18
19
T (◦ C)
18.2
18.8
17.6
16.0
14.1
x(h)
20
21
22
23
24
T (◦ C)
11.5
10.2
9.0
7.9
7.0
1. Find the average rate of change of temperature with respect to time
(a) from noon to 3 P.M.
(b) from noon to 2 P.M.
(c) from noon to 1 P.M.
2. Estimate the instantaneous rate of change at noon.
Solution: 1. (a) From noon to 3 P.M., the temperature changes from 14.3◦ C to 18.2◦ C.
Therefore,
∆T = T (15) − T (12) = 18.2 − 14.3 = 3.9◦ C
while the rate of change in time is ∆x = 3 h. Therefore, the average rate of change of
temperature with respect to time is
3.9
∆T
=
= 1.3◦ C/h.
∆x
3
(b)
T (14) − T (12)
17.3 − 14.3
3.9
∆T
=
=
=
= 1.5◦ C/h.
∆x
2
2
3
(c)
∆T
T (13) − T (12)
16.0 − 14.3
3.9
=
=
=
= 1.7◦ C/h.
∆x
1
1
3
2. Given the points P (12, 14.3) and Q(17, 17.6) from the table, we find that the slope of
the secant P R is
mP R =
17.6 − 14.3
= 0.66.
17 − 12
If we repeat this process for all the points R from the table, with P (12, 14.3), we get
72
4.12. OTHER RATES OF CHANGE
R
(0, 6.5)
(1, 6.1)
(2, 5.6)
(3, 4.9)
(4, 4.2)
(5, 4.0)
(6, 4.0)
(7, 4.8)
(8, 6.1)
mP R
0.65
0.75
0.87
1.04
1.26
1.47
1.72
1.9
2.05
R
(9, 8.3)
(10, 10.0)
(11, 12.1)
(13, 16.0)
(14, 17.3)
(15, 18.2)
(16, 18.8)
(17, 17.6)
(18, 16.0)
mP R
2
2.15
2.2
1.7
1.5
1.3
1.13
0.66
0.28
By Dr. M. Nxumalo
R
(19, 14.1)
(20, 11.5)
(21, 10.2)
(22, 9.0)
(23, 7.9)
(24, 7.0)
mP R
−0.03
−0.35
−0.46
−0.53
−0.58
−0.61
From the above table, we expect the slope of the tangent line at x = 12 to lie somewhere
between 1.7 and 2.3. The average of the slopes of the two closest secant lines is 2 (1.95 from a
calculator). So, the instantaneous rate of change of temperature with respect to time at noon
is about 2◦ C.
73
Chapter 5
Continuity
In every day language, a continuous process is one that takes place gradually, without interruption or abrupt change. In this section, we explore the concept of being continuous in the
language of mathematics.
Informally, we expect to deal with functions that do not have gaps.
Definition 5.0.1. A function f is continuous at a number a if
1. f (a) is defined (that is a is in the domain of f )
2. lim f (x) exists (that is lim− f (x) = lim+ f (x))
x→a
x→a
x→a
3. lim f (x) = f (a).
x→a
A noncontinuous function is said to be discontinuous. Clearly, a function is discontinuous
if at least one of the conditions of Definition 5.0.1 does not hold.
From Definition 5.0.1, we can see that it is necessary to show that f (a) is defined and
lim f (x) = f (a)
x→a
for a function f to be continuous at a.
Example 5.0.2. Where are each of the following functions discontinuous?
1. f (x) =
x2 − x − 2
x−2
74
By Dr. M. Nxumalo
(
2. f (x) =
3. f (x) =
1
x2
if x 6= 0
if x = 0
1
( 2
x −x−2
x2
x−2
if x 6= 2
if x = 2
Solution: 1. The domain of f is given by Domf = {x ∈ R : x 6= 2}. So, f is discontinuous
at x = 2. This is because f (2) is not defined in R (or 2 is not in the domain of f ).
2. We have that lim f (x) = ∞. Therefore lim f (x) 6= f (0). So, f is discontinuous at 0.
x→0
x→0
3. We have that
x2 − x − 2
x→2
x2
(x − 2)(x + 1)
= lim
x→2
x−2
= lim (x + 1)
lim f (x) = lim
x→2
x→2
= 3.
But f (2) = 1 6= lim f (x), so f is discountinuous at 2.
x→2
Definition 5.0.3. A function f is continuous from the right at a number a if
lim f (x) = f (a)
x→a+
and continuous from the left at a number a if
lim f (x) = f (a)
x→a−
Example 5.0.4. At each integer n, the function f (x) = bxc is continuous from the right but
discontinuous from the left.
Solution: Follows since
lim bxc = a = bac
x→a+
but
lim bxc =
6 bac
x→a−
for all a ∈ N.
75
By Dr. M. Nxumalo
Definition 5.0.5. A function f is continuous on an interval if it is continuous at every
number in the interval. (If f is defined only on one side of an endpoint of the interval, we
understand continuous at the endpoint to mean continuous from the right or continuous from
the left.)
Example 5.0.6. Show that the function f (x) = 1 −
√
1 − x2 is continuous on the interval
[−1, 1].
Solution: The domain of f is given by
Domf = {x ∈ R : 1 − x2 ≥ 0}
= {x ∈ R : −1 ≤ x ≤ 1}
= [−1, 1]
So, f is defined on each x ∈ [−1, 1].
For each a ∈ [−1, 1], it is clear that
lim f (x) = f (a).
x→a
Theorem 5.0.7. If f and g are continuous at a and c is a constant, then the following
functions are also continuous at a:
1. f + g
2. f − g
3. cf
4. f g
5. fg if g(a) 6= 0
We only prove 1.
Given:
1. f is continuous at a: f (a) is defined and lim f (x) = f (a).
x→a
76
By Dr. M. Nxumalo
2. g is continuous at a: g(a) is defined and lim g(x) = g(a).
x→a
To show: (f + g)(a) is defined and lim (f + g)(x) = (f + g)(a).
x→a
Proof. Since f and g are continuous at a, f (a) and g(a) are defined. So, (f +g)(a) = f (a)+g(a)
is defined.
Furthermore, we have that lim f (x) = f (a) and lim g(x) = g(a).
x→a
x→a
Now,
lim (f + g)(x) = lim (f (x) + g(x)),
x→a
x→a
since (f + g)(x) = f (x) + g(x)
= lim f (x) + lim g(x)
x→a
x→a
= f (a) + g(a),
by hypothesis
= (f + g)(a).
Theorem 5.0.8.
1. Any polynomial is continuous everywhere; that is, it is continuous on
R.
2. Any rational function is continuous wherever it is defined; that is, it is continuous on its
domain.
We start by proving 1.
Given: Assume P is a polynomial function: P (x) = cn xn + cn−1 xn−1 + ... + c2 x2 + c1 x + c0 .
To show: P is continuous at a ∈ R: For any a ∈ R, lim P (x) = P (a). That is,
x→a
lim cn xn + cn−1 xn−1 + ... + c2 x2 + c1 x + c0 = cn an + cn−1 an−1 + ... + c2 a2 + c1 a + c0 .
x→a
Proof. Assume P is a polynomial function. Then
P (x) = cn xn + cn−1 xn−1 + ... + c2 x2 + c1 x + c0
where c0 , c1 , c2 , ..., cn are constants.
77
By Dr. M. Nxumalo
Now, for each a ∈ R, it is clear that P (a) is defined and
lim P (x) = lim [cn xn + cn−1 xn−1 + ... + c2 x2 + c1 x + c0 ]
x→a
x→a
= lim [cn xn ] + lim [cn−1 xn−1 ] + ... + lim [c2 x2 ] + lim [c1 x] + lim c0
by Limit Law 1
= cn lim [xn ] + cn−1 lim [xn−1 ] + ... + c2 lim [x2 ] + c1 lim [x] + lim c0
by Limit Law 3
x→a
x→a
x→a
x→a
x→a
x→a
x→a
x→a
= cn an + cn−1 an−1 + ... + c2 a2 + c1 a + lim c0
x→a
= cn an + cn−1 an−1 + ... + c2 a2 + c1 a + c0
x→a
x→a
by Limit Law 9
by Limit Law 7
= P (a).
We prove 2.
Given: Assume f is a rational function:
f (x) =
P (x)
Q(x)
where P and Q are polynomials.
To show: f is continuous at every a ∈ Domf .
Proof. Assume f is a rational function. Then
f (x) =
P (x)
Q(x)
where P and Q are polynomials.
The domain of f is given by Domf = {x ∈ R : Q(x) 6= 0}. This means Q(a) 6= 0. By
P
is continuous at a. Thus f is continuous wherever it is defined.
Theorem 5.0.7(5), Q
Theorem 5.0.9. The following types of functions are continuous at every number in their
domain: (1) Polynomials; (2) Rational functions; (3) Root functions; (4) Trigonometric functions; (5) Inverse trigonometric functions; (6) Exponential functions; (7) Logarithmic functions.
Theorem 5.0.10. If f is continuous at b and lim g(x) = b, then lim f (g(x)) = f (b). In other
x→a
x→a
words,
lim f (g(x)) = f (lim g(x))
x→a
78
x→a
By Dr. M. Nxumalo
Theorem 5.0.11. If g is continuous at a and f is continuous at g(a), then the composite
function f ◦ g, given by (f ◦ g)(x) = f (g(x)), is continuous at a.
Example 5.0.12.
x3 + 2x2 − 1
x→−2
5 − 3x
1. Find lim
ln x + tan−1 x
continuous?
x2 − 1
√ 1− x
.
3. Evaluate lim arcsin
x→1
1−x
2. Where is the function f (x) =
4. Where is the function f (x) = ln(1 + cos x) continuous?
Solution: 1. This is a rational function which, by Theorem 5.0.8(2), it is continuous
everywhere in its domain. The domain of f is given by
5
Domf = {x ∈ R : 5 − 3x 6= 0} = x ∈ R : x 6=
.
3
Since −2 6= 35 , we have that −2 ∈ Domf so that f is continuous at −2.
Therefore
(−2)3 + 2(−2)2 − 1
1
x3 + 2x2 − 1
=
=− .
5 − 3x
5 − 3(−2)
10
2. This is a rational function, so it is continuous everywhere in its domain. We must
find the domain of f : The domain for f1 (x) = ln x is given by (0, ∞) and the domain for
f2 (x) = tan−1 x is given by R. According to combinations of functions, the domain for f1 + f2
is given by
Domf1 +f2 = (0, ∞) ∩ R = (0, ∞).
The domain for f3 (x) = x2 − 1 is given by Domf3 = R.
So,
Dom(f1 +f2 )/f3 = {x ∈ R : x ∈ Domf1 +f2 ∩ Domf3 , f3 (x) 6= 0}
= {x ∈ R : x ∈ (0, ∞), x2 − 1 6= 0}
= {x ∈ R : x ∈ (0, ∞), x ∈
/ {−1, 1}}
= {x ∈ R : x ∈ (0, 1) ∪ (1, ∞)}.
Therefore, f is continuous at all x lying in (0, 1) ∪ (1, ∞).
79
By Dr. M. Nxumalo
√
1− x
3. This is the composition of the functions f (x) = arcsin x and g(x) =
. We shall
1−x
follow Theorem 5.0.10 in our argument.
The domain of f is R. So, f is continuous everywhere in R.
We search for b:
√
1− x
lim g(x) = lim
x→1
x→1 1 − x
√
√
(1 − x)(1 + x)
√
= lim
x→1 (1 − x)(1 +
x)
1
√
= lim
x→1 1 +
x
1
= .
2
1
is our b.
2
But f is continuous everywhere, so f is continuous at b = 12 . It follows from Theorem
5.0.10 that
lim arcsin
x→1
√ 1− x
1
1
π
=f
= arcsin
= .
1−x
2
2
6
4. This is the composition of the functions t(x) = ln x and g(x) = 1 + cos x.
The domain of f (x) is given by Domf = {x ∈ R : x ∈
/ {±π, ±3π, ±5π, ...}}.
Therefore, f is continuous everywhere except at points in {±π, ±3π, ±5π, ...}.
Theorem 5.0.13. The Intermediate Value Theorem: Suppose that f is continuous on
the closed interval [a, b] and let N be any number between f (a) and f (b), where f (a) 6= f (b).
Then there exists a number c in (a, b) such that f (c) = N .
The Intermediate Value Theorem (IVT) states that a continuous function takes on every
intermediate value between the function values f (a) and f (b):
y
f (b)
y = f (x)
N
f (a)
x
a cb
80
By Dr. M. Nxumalo
The use of the IVT is in locating roots of equations.
Example 5.0.14. Show that there is a root of the equation
4x3 − 6x2 + 3x − 2 = 0
between 1 and 2.
Solution: Set f (x) = 4x3 − 6x2 + 3x − 2.
Since f is polynomial, it is continuous on [1, 2].
We have that f (1) = 4 − 6 + 3 − 2 = −1 < 0 and f (2) = 32 − 24 + 6 − 2 = 12 > 0.
So, choose N = 0.
Therefore f (1) < N < f (2), meaning N is a number between f (1) and f (2). By the IVT,
there is a number c in (1, 2) such that f (c) = 0.
In other words, the equation 4x3 − 6x2 + 3x − 2 = 0 has at least one root c in the interval
(1, 2).
81
Chapter 6
Derivatives
Limits of the form
f (a + h) − f (a)
h→0
h
lim
arise whenever we calculate a rate of change in any of the sciences or engineering, such as the
rate of reaction in chemistry or a marginal cost in economics. Since this limit occurs so widely,
it is given a special name and notation.
Definition 6.0.1. The derivative of a function f at a number a, denoted by f 0 (a), is
f (a + h) − f (a)
h→0
h
f 0 (a) = lim
if this limit exists.
f 0 (a) is read as f prime of a.
If we write x = a + h, then h = x − a and h approaches 0 if and only if x approaches a.
Therefore
f (x) − f (a)
x→a
x−a
f 0 (a) = lim
Given any number x for which the limit
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
exists, we assign x to the number f 0 (x). So, we have f 0 as a new function, called the derivative
of f . The value of f 0 (x) can be interpreted geometrically as the slope of the tangent line to
the graph of f at the point (x, f (x)).
82
By Dr. M. Nxumalo
The function f 0 is called the derivative of f because it has been “derived” from f by the
limiting operation given in the above equation.
The domain of f 0 is the set {x : f 0 (x) exists} and may be smaller than the domain of f .
Example 6.0.2.
2. If f (x) =
1. Find the derivative of the function f (x) = x2 − 8x + 9 at a number c.
√
x − 1, find the derivative of f . State the domain of f 0 .
Solution: 1.
f (c + h) − f (c)
h→0
h
(c + h)2 − 8(c + h) + 9 − (c2 − 8c + 9)
= lim
h→0
h
2
2
c + 2ch + h − 8c − 8h + 9 − c2 + 8c − 9
= lim
h→0
h
= lim (2c + h − 8)
f 0 (c) = lim
h→0
= 2c − 8.
2.
f (x + h) − f (x)
h→0
h
√
√
x+h−1− x
= lim
h→0
√ h
√
√
√
x+h−1− x
√
= lim √
· x+h−1+ x−1
h→0 h x + h − 1 +
x−1
1
√
= lim √
h→0 h x + h − 1 +
x−1
1
√
= √
h x−1+ x−1
1
= √
.
2 x−1
f 0 (x) = lim
The domain of f 0 is given by
√
Domf 0 = {x ∈ R : 2 x − 1 6= 0, x − 1 ≥ 0}
= {x ∈ R : x − 1 6= 0, x ≥ 1}
= {x ∈ R : x 6= 1, x ≥ 1}
= {x ∈ R : x > 1}.
83
By Dr. M. Nxumalo
For y = f (x), we have the following common alternative notations for the derivative:
f 0 (x) = y 0 =
dy
df
d
=
=
f (x) = Df (x) = Dx f (x).
dx
dx
dx
d
are called differentiation operators because they indicate the
dx
operation of differentiation, which is the process of calculating a derivative.
The symbols D and
We use the following notation to indicate the value of a derivative
dy
at a specific number
dx
a:
dy
dx x=a
or
dy
dx x=a
Definition 6.0.3. A function f is differentiable at a if f 0 (a) exists. It is differentiable on
an open interval (a, b) (or (0, ∞) or (−∞, ∞)) if it is differentiable at every number in the
interval.
Example 6.0.4. Where is the function f (x) = |x| differentiable?
Solution: For x > 0, we have that
f (x + h) − f (x)
|h + x| − |x|
h+x−x
= lim
= lim
=1
h→0
h→0
h→0
h
h
h
lim
and for x < 0,
f (x + h) − f (x)
|h + x| − |x|
−(h + x) + x
−x − h + x
= lim
= lim
= lim
= −1.
h→0
h→0
h→0
h→0
h
h
h
h
lim
For x = 0,
lim+
h→0
|h + 0| − |0|
h
= lim = 1
h→0 h
h
and
lim−
h→0
|h + 0| − |0|
−(h + 0) + 0
−h
= lim
= lim
= −1.
h→0
h→0 h
h
h
Therefore the limit lim f (x) does not exist for x = 0. Hence f is not differentiable at x = 0.
h→0
Theorem 6.0.5. If a function f is differentiable at a, then f is continuous at a.
Three ways for f not to be differentiable at a:
1. If the graph of a function f has a “corner” or “kink” in it, then the graph of f has no
tangent at this point and f is not differentiable there.
84
By Dr. M. Nxumalo
2. If f is not continuous at a, then it is not differentiable at a.
3. If f has a vertical tangent line when x = a (i.e., f is continuous at a and
lim |f 0 (x)| = ∞)
x→0
then f is not differentiable at a.
85
Chapter 7
PART 2: Differentiation Rules
Rule 1 (Derivative of a Constant Function): For every c ∈ R,
d
(c) = 0.
dx
Proof. Let c ∈ R. For f (x) = c, we have
d
f (x + h) − f (x)
(f (x)) = f 0 (x) = lim
h→0
dx
h
c−c
= lim
h→0
h
=0
which completes the proof.
————————————————————————————————————
Rule 2 (Derivative of a Power Function): If n is a positive integer, then
d n
(x ) = nxn−1 .
dx
In order to prove this result, we recall the following:
Theorem 7.0.1 (The Binomial Theorem). For any positive integer n,
n(n − 1) n−2 2 n(n − 1)(n − 2) n−3 3
a b +
a b
1×2
1×2×3
n(n − 1)(n − 2)(n − 3) n−4 4
+
a b + ... + nabn−1 + bn .
1×2×3×4
(a + b)n = an + nan−1 b +
Now, let us prove Rule 2. Our a = x and b = h.
86
By Dr. M. Nxumalo
Proof. Let n ∈ N and set f (x) = xn . Then
d
f (x + h) − f (x)
(f (x)) = f 0 (x) = lim
h→0
dx
h
n
(x + h) − xn
= lim
h→0
h
n(n − 1) n−2 2
n−1
n
x h + ... + nxh
+ h − xn
x + nx h +
2
= lim
h→0
h
n(n − 1) n−2 2
n−1
n−1
n
nx h +
x h + ... + nxh
+h
2
= lim
h→0
h
n(n
−
1)
n−2
n−2
n−1
n−1
x h + ... + nxh
+h
h nx
+
2
= lim
h→0
h
n(n − 1) n−2
n−2
n−1
n−1
= lim nx
x h + ... + nxh
+h
+
h→0
2
n(n − 1) n−2
n−1
x h + ... + lim (nxhn−2 ) + lim hn−1
= lim nx
+ lim
h→0
h→0
h→0
h→0
2
n−1
= nx .
n
n−1
Rule 2 can also be extended to all real numbers, instead of just natural numbers.
Example 7.0.2. Differentiate y =
Solution:
√
3
x2 .
√ 3
d
x2
2
d x3
2 2
dy
2 1
=
=
= x 3 −1 = x− 3 .
dx
dx
dx
3
3
————————————————————————————————————
Rule 3 (Derivative of a Constant times a Function): If c ∈ R and f is differentiable,
then
d
d
(cf (x)) = c f (x).
dx
dx
87
By Dr. M. Nxumalo
Proof. Let c ∈ R and f be a differentiable function. Then
d
cf (x + h) − cf (x)
(cf (x)) = (cf )0 (x) = lim
h→0
dx
h
c[f (x + h) − f (x)]
= lim
h→0
h
(f (x + h) − f (x))
= lim c ·
h→0
h
(f (x + h) − f (x))
= lim c · lim
h→0
h→0
h
0
= cf (x)
d
= c f (x).
dx
————————————————————————————————————
Rule 4 (Derivative of the Sum of Two Functions): If f and g are differentiable
functions, then
d
d
d
[f (x) + g(x)] =
f (x) + g(x).
dx
dx
dx
Proof. Let f and g be differentiable functions. Then
d
(f + g)(x + h) − (f + g)(x)
d
[f (x) + g(x)] =
[(f + g)(x)] = lim
h→0
dx
dx
h
f (x + h) + g(x + h) − [f (x) + g(x)]
= lim
h→0
h
f (x + h) − f (x) + g(x + h) − g(x)
= lim
h→0
h
f (x + h) − f (x) g(x + h) − g(x)
= lim
+
h→0
h
h
f (x + h) − f (x)
g(x + h) − g(x)
= lim
+ lim
h→0
h→0
h
h
0
0
= f (x) + g (x)
d
d
=
f (x) + g(x).
dx
dx
————————————————————————————————————
88
By Dr. M. Nxumalo
Rule 5 (Derivative of the Difference of Two Functions): If f and g are differentiable
functions, then
d
d
d
[f (x) − g(x)] =
f (x) − g(x).
dx
dx
dx
Proof. Let f and g be differentiable functions. Then
d
d
d
[f (x) − g(x)] =
[(f − g)(x)] =
[(f + (−1 · g)(x)]
dx
dx
dx
d
d
=
f (x) + (−1 · g)(x) by Rule 4
dx
dx
d
d
f (x) + (−1) g(x) by Rule 3
=
dx
dx
d
d
=
f (x) − g(x).
dx
dx
In the following example, we apply Rule 1-5.
Example 7.0.3. Find an equation of the tangent line to the curve
f (x) = x8 + 12x5 − 4x4 + 10x3 − 6x + 5
at the point (1, 18).
Solution: We have
df
d
=
x8 + 12x5 − 4x4 + 10x3 − 6x + 5
dx
dx
d 8
d
d
d
d
d
=
x + 12x5 − 4x4 + 10x3 − 6x + 5
dx
dx
dx
dx
dx
dx
7
4
3
2
= 8x + 12(5x ) − 4(4x ) − 3(10x ) − 6(1) + 0
= 8x7 + 60x4 − 16x3 − 30x2 − 6.
So, the slope of the tangent line at (1, 18) is
f 0 (1) = 8(1)7 + 60(1)4 − 16(1)3 − 30(1)2 − 6 = 16.
Therefore
y − 18 = 16(x − 1) =⇒ y = 16x + 2.
89
By Dr. M. Nxumalo
————————————————————————————————————
Rule 6 (Derivative of the Product of Two Functions): If f and g are differentiable
functions, then
d
d
d
[f (x) · g(x)] = f (x) g(x) + g(x) f (x).
dx
dx
dx
Proof. Let f and g be differentiable functions. Then
d
d
[f (x) · g(x)] =
[(f g)(x)]
dx
dx
f g(x + h) − f g(x)
= lim
h→0
h
f (x + h)g(x + h) − f (x)g(x)
= lim
h→0
h
f (x + h)g(x + h) + f (x)g(x + h) − f (x)g(x + h) − f (x)g(x)
= lim
h→0
h
f (x + h)g(x + h) − f (x)g(x + h) f (x)g(x + h) − f (x)g(x)
= lim
+
h→0
h
h
f (x + h)g(x + h) − f (x)g(x + h)
f (x)g(x + h) − f (x)g(x)
= lim
+ lim
h→0
h→0
h
h
g(x + h)[f (x + h) − f (x)]
f (x)[g(x + h) − g(x)]
= lim
+ lim
h→0
h→0
h
h
g(x + h) − g(x)
f (x + h) − f (x)
+ lim f (x) lim
= lim g(x + h) lim
h→0
h→0
h→0
h→0
h
h
0
0
= g(x)f (x) + f (x)g (x)
d
d
= g(x) f (x) + f (x) g(x).
dx
dx
Example 7.0.4. If y =
√
x(1 − x2 ), then
dy
d √
x(1 − x2 )
=
dx
dx
√ d
d√
= (1 − x2 )
x + x (1 − x2 )
dx
dx
√
1
1
x− 2 + x(−2x)
= (1 − x2 )
2
√
1 1
= x− 2 − 2x x.
2
————————————————————————————————————
90
By Dr. M. Nxumalo
Rule 7 (Derivative of the Quotient of Two Functions): If f and g are differentiable
functions with g(x) 6= 0, then
d f (x)
=
dx g(x)
g(x)
d
d
[f (x)] − f (x) [g(x)]
dx
dx
.
[g(x)]2
Proof. If f and g are differentiable functions with g(x) 6= 0. Then
d f (x)
d f
=
(x)
dx g(x)
dx g
f
f
(x + h) − (x)
g
g
= lim
h→0
h
f (x + h) f (x)
−
g(x + h)
g(x)
= lim
h→0
h
g(x)f (x + h) − f (x)g(x + h)
g(x + h)g(x)
= lim
h→0
h
g(x)f (x + h) − f (x)g(x + h)
h
= lim
h→0
g(x + h)g(x)
g(x)f (x + h) + f (x)g(x) − f (x)g(x) − f (x)g(x + h)
h
= lim
h→0
g(x + h)g(x)
g(x)(f (x + h) − f (x)) f (x)(g(x + h) − g(x))
−
h
h
= lim
h→0
g(x + h)g(x)
g(x)(f (x + h) − f (x))
f (x)(g(x + h) − g(x))
lim
− lim
h→0
h
h
= h→0
lim g(x + h)g(x)
h→0
f (x + h) − f (x)
g(x + h) − g(x)
− f (x) lim
h→0
h→0
h
h
=
g(x)g(x)
g(x)f 0 (x) − f (x)g 0 (x)
=
[g(x)]2
d
d
g(x) [f (x)] − f (x) [g(x)]
dx
dx
=
.
2
[g(x)]
g(x) lim
91
By Dr. M. Nxumalo
Example 7.0.5. Let y =
x2 + x − 2
. Then
x3 + 6
d
d
(x3 + 6) dx
(x2 + x − 2) − (x2 + x − 2) dx
(x3 + 6)
y =
(x3 + 6)2
(x3 + 6)(2x + 1) − (x2 + x − 2)(3x2 )
=
(x3 + 6)2
−x4 − 2x3 + 6x2 + 12x + 6
=
(x3 + 6)2
————————————————————————————————————
0
Rule 8 (Derivative of an Exponential Function): If a is a positive real number, then
d x
a = lna · ax .
dx
To prove the Rule 8, we give the following result whose proof will be given later.
Proposition 7.0.6. . Let a be a positive real number. Then
ah − 1
= lna.
h→0
h
lim
Let us prove Rule 8.
Proof. Let a be a positive real number and f be a differentiable function. Then
ax+h − ax
d x
a = lim
h→0
dx
h
ax ah − ax
= lim
h→0
h
x h
a (a − 1)
= lim
h→0
h
h
a
−1
= ax lim
h→0
h
x
= a lna by Proposition 7.0.6.
From Rule 8, we get that
d x
e = ex lne = ex .
dx
————————————————————————————————————
We want to consider derivatives of some trigonometric functions. Recall that:
92
By Dr. M. Nxumalo
1. tanx =
sinx
cosx
2. secx =
1
cosx
1
sinx
cosx
4. cotx =
sinx
3. cscx =
Rule 9 (Derivative of sinx):
d
sinx = cosx
dx
We shall freely use the facts:
sin(x + h) = sinx cosh + cosx sinh
and
sinh
= 1.
h→0 h
lim
We verify that
cosh − 1
= 0.
h→0
h
lim
Indeed,
cosh − 1
cosh − 1 cosh + 1
= lim
·
lim
h→0
h→0
h
h
cosh + 1
2
cos h − 1
= lim
h→0 h(cosh + 1)
− sin2 h
= lim
h→0 h(cosh + 1)
sinh
sinh
= − lim
·
h→0
h
cosh + 1
sinh
sinh
= − lim
· lim
h→0 h
h→0 cosh + 1
0
= −1 ·
= 0.
0+1
Let us prove Rule 9.
93
By Dr. M. Nxumalo
Proof.
d
sin(x + h) − sinx
sinx = lim
h→0
dx
h
sinx cosh + cosx sinh − sinx
= lim
h→0
h
sinx(cosh − 1) + cosx sinh
= lim
h→0
h
sinx(cosh − 1)
cosx sinh
= lim
+ lim
h→0
h→0
h
h
sinh
cosh − 1
+ cosx lim
= sinx lim
h→0 h
h→0
h
= sinx(0) + cosx(1)
= cosx.
————————————————————————————————————
Rule 10 (Derivative of cosx):
d
cosx = − sinx.
dx
Proof.
d
cos(x + h) − cosx
cosx = lim
h→0
dx
h
cosx cosh − sinx sinh − cosx
= lim
h→0
h
cosx(cosh − 1) + sinx sinh
= lim
h→0
h
cosx(cosh − 1)
sinx sinh
= lim
− lim
h→0
h→0
h
h
cosh − 1
sinh
= cosx lim
− sinx lim
h→0
h→0 h
h
= cosx(0) − sinx(1)
= − sinx.
————————————————————————————————————
94
By Dr. M. Nxumalo
The Chain Rule: If f and g are differentiable and F = f ◦ g is the composite function
defined by F (x) = f (g(x)), then F is differentiable and F 0 is given by the product
F 0 (x) = f 0 (g(x))g 0 (x).
This rule tells us that we find f 0 (x), then f 0 (g(x)) and then g 0 (x) and finally the product
f 0 (g(x))g 0 (x).
Let us use the Chain Rule to prove Rule 8: For any positive real number a,
d x
a = lna · ax .
dx
Before we start, let us recall that a = elna .
Now, we have that
ax = (elna )x .
Therefore
ax = e(lna)x .
Let f (x) = ex and g(x) = x lna. Then f 0 (x) = ex , g 0 (x) = lna and f 0 (g(x)) = ex lna .
Therefore
d x
a = e(lna)x · lna = ax lna.
dx
ex
Example 7.0.7. Differentiate y = cos 4 +
.
1 + 3x2
Solution:
d
ex
ex
×
4+
y = − sin 4 +
1 + 3x2
dx
1 + 3x2
ex
(1 + 3x2 )(ex ) − ex (6x)
= − sin 4 +
× 0+
1 + 3x2
(1 + 3x2 )2
ex
(1 + 3x2 )(ex ) − 6xex
= − sin 4 +
1 + 3x2
(1 + 3x2 )2
0
————————————————————————————————————
Implicit Differentiation
95
By Dr. M. Nxumalo
We sometimes encounter equations involving two variables where it is not easy to make
d
one of the variables the subject of the formula to be able to apply
as we did in the previous
dx
rules. For instance, in the equation
sin(xy) + x2 y 3 = 2,
it won’t be easy to make either x or y the subject of the formula. In such cases, we apply what
we call implicit differentiation which involves differentiating both sides of the equation with
respect to x and then solving the resulting equation for y 0 .
Example 7.0.8.
1. If sin(xy) + x2 y 3 = 2, find
dy
.
dx
Solution:
d
d
d
d
(sin(xy) + x2 y 3 ) =
2 =⇒
sin(xy) + x2 y 3 = 0
dx
dx
dx
dx
=⇒ [y + xy 0 ] cos(xy) + 2xy 3 + x2 3y 2 y 0 = 0
=⇒ y cos(xy) + x cos(xy)y 0 + 2xy 3 + 3x2 y 2 y 0 = 0
=⇒ y 0 (x cos(xy) + 2xy 3 + 3x2 y 2 ) = −y cos(xy)
y cos(xy)
=⇒ y 0 = −
x cos(xy) + 2xy 3 + 3x2 y 2
————————————————————————————————————
Rule 11 (Derivative of sin−1 x):
d
1
(sin−1 x) = √
.
dx
1 − x2
Proof. Let y = sin−1 x. Then
x = siny
and
−
π
π
≤y≤
2
2
Therefore
d
d
x=
siny
dx
dx
=⇒ 1 = cosyy 0
=⇒ y 0 =
p
Since cos2 y = 1 − sin2 y, we have cosy = ± 1 − sin2 x.
Because −
π
π
≤ y ≤ , so cosy ≥ 0. Therefore
2
2
p
√
cosy = 1 − sin2 x = 1 − x2 .
96
1
.
cosy
By Dr. M. Nxumalo
Thus
y0 = √
1
.
1 − x2
We give the following derivatives of inverses of trigonometric functions without proofs:
1.
d
1
.
(cos−1 x) = − √
dx
1 − x2
2.
d
1
(tan−1 x) =
.
dx
1 + x2
3.
d
1
(csc−1 x) = − √
.
dx
x x2 − 1
4.
d
1
(sec−1 x) = √
.
dx
x x2 − 1
5.
d
1
(cot−1 x) = −
.
dx
1 + x2
————————————————————————————————————
Rule 12 (Derivative of Logarithmic Functions):
d
1
(loga x) =
.
dx
x lna
Proof. Let y = loga x. Then
ay = x.
Therefore
d y
d
a =
x
dx
dx
=⇒ ay · lnay 0 = 1
97
=⇒ y 0 =
1
ay lna
=
1
.
x lna
By Dr. M. Nxumalo
Since loge x = lnx, we get that
d
d
1
1
(lnx) =
(loge x) =
= .
dx
dx
x lne
x
————————————————————————————————————
We want to consider derivatives of hyperbolic functions. Firstly, let us start by discussing
what are hyperbolic functions. Hyperbolic functions are used in problems such as computing
tension in a cable suspended by its two ends, as in an electric transmission line.
Recall that an even function satisfies f (−x) = f (x), while an odd function satisfies f (−x) =
−f (x). Every function f that is defined on an interval centered at the origin can be written
in a unique way as the sum of one even function and one odd function:
f (x) =
f (x) + f (−x) f (x) − f (−x)
+
2
2
where the first part is the even part of f (x) and the second part is the odd part of f (x).
If we write ex this way, we get
ex =
ex + e−x ex − e−x
+
2
2
The even part of ex is called the hyperbolic cosine of x and the odd part of ex is called
the hyperbolic sine of x. They describe the motions of waves in elastic solids and the
temperature distributions in metal cooling fins.
We write
coshx =
ex + e−x
2
sinhx =
ex − e−x
2
for the hyperbolic cosine of x and
for the hyperbolic sine of x.
Rule 13 (Derivative of sinhx):
d
(sinhx) = coshx
dx
98
By Dr. M. Nxumalo
Proof.
d
d ex − e−x
(sinhx) =
dx
dx
2
x
e−x
d e
−
=
dx 2
2
x
−x
e
e
=
+
= coshx.
2
2
Rule 14 (Derivative of coshx):
d
(coshx) = sinhx
dx
Rule 15 (Derivative of tanhx):
d
(tanhx) = sech2 x
dx
————————————————————————————————————
Higher Derivatives
If f is a differentiable function, then the derivative f 0 is also a function, so f 0 may have a
derivative of its own, denoted by (f 0 )0 = f 00 . This new function is called the second derivative
d2 y
of f . Another notation for the second derivative of f is
or D2 f (x), where y = f (x).
dx
d(n) y
or D(n) f (x), where y = f (x).
For n ∈ N, the n-th derivative of f is given by f (n) (x) or
dx
Example 7.0.9. Find y 0 , y 00 , y 000 and y (4) for
y = x3 − 6x2 − 5x + 3.
Solution:
0
00
y = 3x2 − 12x − 5
y = 6x − 12
000
y =6
000
y =0
Example 7.0.10. Find D27 cos x.
Solution:
D1 cos x = − sin x
D2 cos x = − cos x
99
D3 cos x = sin x
D4 cos x = cos x
By Dr. M. Nxumalo
D5 cos x = − sin x
D6 cos x = − cos x
D7 cos x = sin x
D8 cos x = cos x
D9 cos x = − sin x
D10 cos x = − cos x
D11 cos x = sin x
D12 cos x = cos x
From here, we can see that D(4n+3) cos x = sin x where n ∈ N ∪ {0}. Since 27 = 4(3) + 3, it
follows that D27 cos x = sin x.
————————————————————————————————————
We now consider derivatives of inverses of hyperbolic functions. We start by considering
inverses of hyperbolic functions and then consider their derivatives.
We have that
−1
sinh x = ln x +
√
x2 + 1
:
Let y = sinh−1 x. Then
x = sinhy =
ey − e−y
2
Therefore
ey − e−y = 2x =⇒ ey − 2x − e−y = 0.
Multiplying both sides by ey gives
e2y − 2xey − 1 = 0 =⇒ (ey )2 − 2x(ey ) − 1 = 0
√
2x ± 4x2 + 4
y
=⇒ e =
2
√
y
=⇒ e = x ± x2 + 1.
Since x <
√
x2 + 1, we must have that x −
√
x2 + 1 < 0. This cannot be the choice of ey since
ey > 0.
Therefore
ey = x +
√
x2 + 1
This gives
y = ln x +
√
x2 + 1
.
Thus
√
sinh−1 x = ln x + x2 + 1 .
Similarly:
100
By Dr. M. Nxumalo
√
1. cosh−1 x = ln x + x2 − 1 .
1+x
−1
1
2. tanh x = 2 ln
.
1−x
Rule 16 (Derivative of sinh−1 x):
d
1
(sinh−1 x) = √
dx
1 + x2
Proof.
√
√
d d 1
d
√
·
(sinh−1 x) =
ln x + x2 + 1 =
x + x2 + 1
dx
dx
x + x2 + 1 dx
1
x
√
=
· 1+ √
x + x2 + 1
x2 + 1
1
=√
x2 + 1
Rule 17 (Derivative of cosh−1 x):
d
1
(cosh−1 x) = √
2
dx
x −1
Rule 18 (Derivative of tanh−1 x):
d
1
(tanh−1 x) =
dx
1 − x2
————————————————————————————————————
Related Rates
We are now focusing on using differentiation rules to solve real-world problems.
Here are recommended steps to use when solving these problems:
1. Read the problem carefully.
2. Draw a diagram if possible.
3. Introduce notation. Assign symbols to all quantities that are functions of time.
4. Express the given information and the required rate in terms of derivatives.
101
By Dr. M. Nxumalo
5. Write an equation that relates the various quantities of the problem. If necessary, use
the geometry of the situation to eliminate one of the variables by substitution.
6. Use the Chain Rule to differentiate both sides of the equation with respect to y.
7. Substitute the given information into the resulting equation and solve for the unknown
result.
Example 7.0.11. Air is being pumped into a spherical balloon so that its volume increases at
a rate of 100cm3 /s. How fast is the radius of the balloon increasing when the diameter is 50cm?
Solution:
Set
V = volume of the balloon
and
r = radius of the balloon
Both the volume and radius of the balloon are functions of the time t.
dV
The rate of increase of the volume with respect to time is given by
and the rate of
dt
dr
increase of the radius with respect to time is given by
.
dt
Therefore
dV
= 100cm3 /s.
dt
Since this is a “sphere” problem, let us relate V and r by the formula for the volume of
a sphere:
4
V = πr3 .
3
Using the Chain Rule,
dV
dV
=
dt
dt
4 3
πr
3
= 4πr2
dr
.
dt
We get that
dr
1 dV
=
.
dt
4πr2 dt
Therefore,
dr
1
1
=
(100) =
.
2
dt
4π(25)
25π
The radius of the balloon is increasing at the rate of
1
cm3 /s.
25π
Example 7.0.12. A ladder 5m long rests against a vertical wall. If the bottom of the ladder
102
By Dr. M. Nxumalo
slides away from the wall at a rate of 1m/s, how fast is the top of the ladder sliding down the
wall when the bottom of the ladder is 4m from the wall.
Solution:
Set
r = length of the ladder
y = distance from the top of the ladder to the ground
and
x = distance from the bottom of the ladder to the wall
Both x and y are functions of the time t.
The rate of sliding down the wall of the top of the ladder with respect to time is given
dy
and the rate of sliding of the bottom of the ladder from the wall with respect to
by
dt
dx
time is given by
.
dt
Therefore
dx
= 1m/s.
dt
Since this is a “Pythagorean Theorem” problem, let us relate x, y and r by the
Pythagorean Theorem:
x2 + y 2 = 5.
Using the Chain Rule,
2x
dy
dx
+ 2y
= 0.
dt
dt
We get that
dy
x dx
=−
.
dt
y dt
√
When x = 4, the Pythagorean Theorem gives y = 11.
Therefore,
dy
4
4
= − √ (1) = − √ .
dt
11
11
4
The top of the ladder slides down the wall at the rate of √ m/s.
11
103
Chapter 8
Applications of Differentiation
8.1
Maximum and Minimum Values
Some of the most important applications of differential calculus are optimization problems, in
which we are required to find the optimal (best) way of doing something. These applications
often involve finding the maximum or minimum values of a function.
Definition 8.1.1. A function f has an absolute maximum at c if
f (c) ≥ f (x) for all x ∈ D
where D is the domain of f . The number f (c) is called the maximum value of f on D.
Definition 8.1.2. A function f has an absolute minimum at c if
f (c) ≤ f (x) for all x ∈ D
where D is the domain of f . The number f (c) is called the minimum value of f on D.
The maximum and minimum values of f are called the extreme values of f .
Example 8.1.3. Consider the following graph of f :
104
8.1. MAXIMUM AND MINIMUM VALUES
5
4
By Dr. M. Nxumalo
y
f
x
−4.5
0 1
−2
3
6 7
−4
f has an absolute minimum at 1 and absolute maximum at −2. Both −4 and 5 are extreme
values of f .
Sometimes we restrict values of x to a small regions and obtain new types of minimum and
maximum values.
For instance, if in Example 8.1.3 we only consider the region (2, 5), we get that 4 is the
maximum value of f in that region. Also, 0 is the minimum value of f in the region (5, 7).
We have names for these types of extreme values.
Definition 8.1.4. A function f has a local minimum at c if
f (c) ≤ f (x) for all x in some open interval containing c.
Definition 8.1.5. A function f has a local maximum at c if
f (c) ≥ f (x) for all x in some open interval containing c.
Example 8.1.6. Consider the following graph of f :
y
4
f
x
−4.5
−2
−4
−5
105
0 1
3
6 7
8.1. MAXIMUM AND MINIMUM VALUES
By Dr. M. Nxumalo
1. f has a local minimum at the following points:
−2, 1 and 6
2. f has a local maximum at the following points:
1 and 3
3. f does not have local maxima at −4.5 and 7. This is because, for instance, at −4.5 there
is no open interval we can find around −4.5.
We have the following result telling us when a continous function has absolute minima and
absolute maxima.
Theorem 8.1.7 (The Extreme Value Theorem). If f is continuous on a closed interval
[a, b], then f attains an absolute maximum value f (c) and an absolute minimum value f (d) at
some numbers c and d in [a, b].
Sometimes we only need to investigate only a few values to find a function’s extrema.
Theorem 8.1.8 (Fermat’s Theorem). If f has a local maximum or minimum value at c,
and if f 0 is defined at c, then f 0 (c) = 0.
Definition 8.1.9. A critical point of a function f is a number c in the domain of f such
that either f 0 (c) = 0 or f 0 (c) does not exist.
Result 8.1.10. If f has a local minimum or maximum at c, then c is a critical point of f .
The only domain points where a function can assume extreme values are critical points
and end points.
In summary, to find the absolute maximum and absolute minimum values of a continuous
function f on a closed interval [a, b]:
1. Find the values of f at the critical numbers of f in [a, b].
2. Find the values of f at the endpoints of the interval.
106
8.2. THE MEAN VALUE THEOREM
By Dr. M. Nxumalo
3. The largest of these values from 1 to 2 above is the absolute maximum value; the smallest
of these values is the absolute minimum value.
Example 8.1.11. Find the absolute maximum and minimum values of the function
f (x) = x3 − 3x2 + 1
−
1
≤ x ≤ 4.
2
Solution:
Values of f at critical numbers of f in [− 12 , 4]: f 0 (x) = 0 gives 3x2 − 6x = 0. Therefore,
x = 0 or x = 2. We get that
f (0) = 1
and
f (2) = −3.
Values of f at the endpoints of the interval [− 12 , 4]:
1
1
f −
and f (4) = 17.
=
2
8
The absolute minimum value of f is given by −3 and the absolute maximum value is given
by 17.
8.2
The Mean Value Theorem
Theorem 8.2.1 (Rolle’s Theorem). Let f be a function that satisfies the following three
properties:
1. f is continuous on the closed interval [a, b].
2. f is differentiable on the open interval (a, b).
3. f (a) = f (b).
Then there is a number c in (a, b) such that f 0 (c) = 0.
Theorem 8.2.2 (The Mean Value Theorem). Let f be a function that satisfies the following hypothesis:
107
8.2. THE MEAN VALUE THEOREM
By Dr. M. Nxumalo
1. f is continuous on the closed interval [a, b].
2. f is differentiable on the open interval (a, b).
Then there is a number c in (a, b) such that
f 0 (c) =
f (b) − f (a)
.
b−a
Proof. We picture f as the following curve:
y
B(b, f (b))
f
A(a, f (a))
x
a
b
g
The equation of the line g is given by
g(x) − f (a) =
f (b) − f (a)
(x − a)
b−a
g(x) = f (a) +
f (b) − f (a)
(x − a).
b−a
which makes
The vertical difference between the graph of f and g is given by
h(x) = f (x) − g(x)
= f (x) − f (a) −
f (b) − f (a)
(x − a).
b−a
The function h satisfies the hypotheses of the Rolle’s Theorem:
1. Both f and g are continuous on [a, b], so h is also continuous there.
2. Both f and g are differentiable on (a, b), so h is also continuous there.
108
8.2. THE MEAN VALUE THEOREM
By Dr. M. Nxumalo
3.
h(a) = f (a) − f (a) −
f (b) − f (a)
(a − a)
b−a
=0
and
f (b) − f (a)
(b − a)
b−a
= f (b) − f (a) − f (b) + f (a)
h(b) = f (b) − f (a) −
= 0.
Therefore h(a) = h(b).
By the Rolle’s Theorem, there is a number c in (a, b) such that h0 (c) = 0. Since
h0 (c) = f 0 (c) −
f (b) − f (a)
b−a
we have that
f 0 (c) =
f (b) − f (a)
b−a
as required.
Example 8.2.3. Verify that the function
f (x) = 3x2 + 2x + 5
satisfies the hypothesis of the Mean Value Theorem on the interval [−1, 1]. Then find all
numbers c that satisfy the conclusion of the Mean Value Theorem.
Solution:
The function f is a polynomial. It follows that it is both continuous on [−1, 1] and also
differentiable on (−1, 1).
By the Mean Value Theorem, there is c ∈ [−1, 1] such that
f 0 (c) =
10 − 6
f (1) − f (−1)
=
= 2.
1 − (−1)
2
Therefore
2 = 6c + 2 =⇒ c = 0.
109
8.3. HOW DERIVATIVES AFFECT THE SHAPE OFBy
A Dr.
GRAPH
M. Nxumalo
8.3
How Derivatives Affect the Shape of a Graph
Here, we learn how how we can deduce facts about a function f from its derivatives.
What does f 0 say about f ?
8.3.1
Firstly, f 0 gives us information about increasing/decreasing of the graph f .
Consider the following graph:
y
D
B
C
A
x
0
Between A and B and between C and D, the tangent lines have positive slopes and so
f 0 (x) > 0. Between B and C, the tangent lines have negative slopes and so f 0 (x) < 0.
We have the following result.
Theorem 8.3.1 (Increasing/Decreasing Test). Let f be a function.
1. If f 0 (x) > 0 on an interval, then f is increasing on that interval.
2. If f 0 (x) < 0 on an interval, then f is decreasing on that interval.
Example 8.3.2. Find where the function f (x) = 3x4 − 4x3 − 12x2 + 5 is increasing and where
it is decreasing.
Solution:
110
8.3. HOW DERIVATIVES AFFECT THE SHAPE OFBy
A Dr.
GRAPH
M. Nxumalo
We have that
f 0 (x) = 12x3 − 12x2 − 24x = 12x(x2 − x − 2) = (12x)(x − 2)(x + 1).
Our critical numbers are −1, 0 and 2.
We have the following table of symbols:
12x
x−2
x+1
f 0 (x)
-
-1
0
0
+
+
0
0
+
0
+
+
-
2
+
0
+
0
+
+
+
+
Increasing (f 0 (x) > 0): (−1, 0) and (2, ∞).
Decreasing (f 0 (x) < 0): (−∞, −1) and (0, 2).
Secondly, f 0 helps us determine whether or not a function f has a local maximum or
minimum at a critical number.
Theorem 8.3.3 (The First Derivative Test). Let f be a continuous function and let c be
a ctitical point of f .
1. If f 0 changes from positive to negative at c, then f has a local maximum at c.
2. If f 0 changes from negative to positive at c, then f has a local minimum at c.
3. If f 0 does not change sign at c (for example, if f 0 is positive on both sides of c or negative
on both sides), then f has no local maximum or minimum at c.
Example 8.3.4. Find the local maximum and minimum values of the function
g(x) = x + 2 sin x 0 ≤ x ≤ 2π.
Solution: We have that
g 0 (x) = 1 + 2 cos x.
Our critical values are: 2π
and 4π
.
3
3
We have the following table of symbols:
111
8.3. HOW DERIVATIVES AFFECT THE SHAPE OFBy
A Dr.
GRAPH
M. Nxumalo
x − 2π
3
x − 4π
3
g 0 (x)
2π
3
+
0
0
4π
3
+
-
+
0
0
+
+
+
There is a local minimum at 4π
. Therefore
3
4π
4π
4π
4π √
g
=
+ 2 sin
=
− 3
3
3
3
3
is a local minimum value.
. Therefore
There is a local maximum at 2π
3
2π
2π
2π
2π √
g
=
+ 2 sin
=
+ 3
3
3
3
3
is a local maximum value.
What does f 00 say about f ?
8.3.2
Firstly, f 00 gives us information about concavity of the graph f .
Consider the following graph:
y
E
B
C
D
A
x
0
Here, the tangent lines in the regions A − B and D − E have positive slopes but they do
not behave the same way. For instance, in the region A − B, the tangent lines are above values
of f (x) and in the region D − E, the tangent lines are below all values of f (x).
Consider the following definition which captures the above paragraph.
112
8.3. HOW DERIVATIVES AFFECT THE SHAPE OFBy
A Dr.
GRAPH
M. Nxumalo
Definition 8.3.5. If the graph of f lies above all of its tangents on an interval I, then it is
called concave upward on I. If it lies below all of its tangents on an interval I, then it is
called concave downward on I.
In the above graph, f is concave downward in the regions: A − B and B − C. It is concave
upward in the regions C − D and D − E.
Let us see how the second derivative helps determine the intervals of concavity.
In the region A − B, the slope of f is decreasing, making f 0 a decreasing function, hence
f 00 (x) < 0. Similarly, in the region C −D, the slope of f is increasing, so that f 0 is an increasing
function; therefore f 00 (x) > 0.
Theorem 8.3.6 (Concavity Test). Let f be a function.
1. If f 00 (x) > 0 for all x ∈ I, then the graph of f is concave upward on I.
2. If f 00 (x) < 0 for all x ∈ I, then the graph of f is concave downward on I.
Example 8.3.7. Sketch a possible graph of a function f that satisfies the following conditions:
1. f 0 (x) > 0 on (−∞, 1), f 0 (x) < 0 on (1, ∞).
2. f 00 (x) > 0 on (−∞, −2) and (2, ∞), f 00 (x) < 0 on (−2, 2)
3. lim f (x) = −2, lim f (x) = 0.
x→−∞
x→∞
Solution:
y
−2
0
1
113
2
x
8.3. HOW DERIVATIVES AFFECT THE SHAPE OFBy
A Dr.
GRAPH
M. Nxumalo
We define an inflection point.
Definition 8.3.8. A point P on a curve f is called an inflection point if f is continuous there
and the curve changes from concave upward to concave downward or from concave downward
to concave upward at P .
Secondly, f 00 gives us information about maximum and minimum values of f .
Theorem 8.3.9 (The Second Derivative Test). Let f be a function such that f 00 is continuous near c.
1. If f 0 (c) = 0 and f 00 (c) > 0, then f has a local minimum at c.
2. If f 0 (c) = 0 and f 00 (c) < 0, then f has a local maximum at c.
Example 8.3.10. Discuss the curve y = x4 −4x3 with respect to increase/ decrease, concavity,
points of inflection, and local maxima and minima. Use this information to sketch the curve.
Solution:
0. Domain: Domy = R.
1. Increase/ Decrease:
We have that
y 0 = 4x3 − 12x2 = 4x2 (x − 3) = (2x)(2x)(x − 3)).
Critical values of y are: 0 and 3.
We have the following table of signs:
2x
2x
x−3
y0
-
0
0
0
0
+
+
-
3
+
+
0
0
+
+
+
+
Increase: y increases on (3, ∞).
Decrease: y decreases on (−∞, 3).
2. Concavity:
y 0 = 4x3 − 12x2 =⇒ y 00 = 12x2 − 24x = 12x(x − 2).
114
8.3. HOW DERIVATIVES AFFECT THE SHAPE OFBy
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GRAPH
M. Nxumalo
Critical values of y 0 are: 0 and 2.
We have the following table of signs:
12x
x−1
y 00
+
0
0
0
+
-
2
+
0
0
+
+
+
The curve is concave downward on (0, 2).
It is concave upward on (−∞, 0) and (2, ∞).
3. Points of inflection: (0, 0) (y changes from concave upward to concave downward) and
(2, −16) (y changes from concave downward to concave upward).
4. Local maxima and minima:
From 1. above, there is a local minimum at 3. Therefore
y(3) = (3)4 − 4(3)3 = −27
is a local minimum value.
The First Derivative Test tells us that y does not have a local minimum or maximum at 0.
5. Sketch:
y
0
2
x
3
1
Example 8.3.11. Use the first and second derivatives of f (x) = e x , together with asymptotes,
to sketch its graph.
115
8.3. HOW DERIVATIVES AFFECT THE SHAPE OFBy
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GRAPH
M. Nxumalo
Solution:
0. Domain:
Domf = {x ∈ R : x 6= 0}.
1. Asymptotes: Let t = x1 . Then
1
lim+ e x = lim et = ∞.
t→∞
x→0
Therefore, x = 0 is a vertical asymptote of f .
Then
1
lim− e x = lim et = 0.
t→−∞
x→0
We also have that
1
lim e x = e0 = 1.
x→±∞
Therefore y = 1 is a horizontal asymptote of f .
2. Increase/Decrease:
We have that
1
f 0 (x) = −
ex
.
x2
We have the following table of signs:
1
x
e
x
x
0
f (x)
+
+
0
Undefined
0
0
Undefined
+
+
+
Increase: f increases on (−∞, 0) and (0, ∞).
3. Concavity:
1
e x (2x + 1)
f (x) = −
.
x4
00
We have the following table of signs for the values 0 and − 12 :
1
x
e
(2x + 1)
x4
f 00 (x)
+
+
-
− 12
+
0
+
0
+
+
+
+
0
+
0
0
Undefined
116
+
+
+
+
8.4. INDETERMINATE FORMS AND L’HOSPITAL’S
ByRULE
Dr. M. Nxumalo
The curve is concave downward on (−∞, − 12 ).
It is concave upward on (− 12 , 0) and (0, ∞).
4. Points of inflection: (− 12 , e−2 ) (f changes from concave downward to concave upward).
5. Local maxima and minima:
From 2. above, The First Derivative Test tells us that f does not have a local maximum
or minimum.
6. Sketch:
y
y=1
x
0
8.4
Indeterminate Forms and L’Hospital’s Rule
Sometimes we want to evaluate limits of functions such as
ex − 1
x→0 2x − 1
lim
Attempting this using the quotient rule, we get
lim (ex − 1)
ex − 1
x→0
lim
=
x→0 2x − 1
lim (2x − 1)
x→0
0
=
0
whose value is not obvious since
0
is not a real number.
0
Such a limit is called an indeterminate form of type
117
0
.
0
8.4. INDETERMINATE FORMS AND L’HOSPITAL’S
ByRULE
Dr. M. Nxumalo
We have the following list of indeterminate forms:
0
0
∞
2.
∞
1.
−∞
−∞
∞
4.
−∞
3.
5.
−∞
∞
6. ∞ − ∞
7. 00
8. ∞0
9. 1∞
To evaluate limits of the above inderterminate forms, we introduce a systematic method
known as l’Hospital’s Rule.
Theorem 8.4.1. Suppose f and g are differentiable and g 0 (x) 6= 0 near a (except possibly at
a). If
lim f (x) = 0
and
lim f (x) = ±∞
and
x→a
lim g(x) = 0
x→a
or if
x→a
lim g(x) = ±∞,
x→a
then
f 0 (x)
f (x)
= lim 0
x→a g(x)
x→a g (x)
lim
provided that the limit on the right side exists (or is ∞ or −∞).
The above can be extended to indeterminate differences, indeterminate products and indeterminate powers to retrive the indeterminate forms listed above.
118
8.5. SUMMARY OF CURVE SKETCHING
By Dr. M. Nxumalo
tan x − x
.
x→0
x3
Example 8.4.2. Find lim
Solution:
This limit is indeterminate with a form of type 00 . By l’Hospital’s rule,
tan x − x
sec2 x − 1
= lim
lim
x→0
x→0
x3
3x2
This limit is indeterminate with a form of type 00 . By l’Hospital’s rule again,
tan x − x
sec2 x − 1
2 sec2 x tan x
=
lim
=
lim
x→0
x→0
x→0
x3
3x2
6x
lim
This limit is indeterminate with a form of type 00 . By l’Hospital’s rule again,
tan x − x
sec2 x − 1
2 sec2 x tan x
=
lim
=
lim
x→0
x→0
x→0
x3
3x2
6x
2 sec2 x
= lim
x→0
6
1
= .
3
lim
Example 8.4.3. Calculate lim+ (1 + sin 4x)cot x .
x→0
Solution:
This limit is indeterminate with the form of type 0∞ .
Let y = (1 + sin 4x)cot x . Then
ln y = ln (1 + sin 4x)cot x = cot x(ln(1 + sin 4x)).
By l’Hospital’s rule,
4 cos 4x
4x = 4.
lim (ln y) = lim+ cot x(ln(1 + sin 4x)) = lim+ 1 + sin
x→0
x→0
x→0+
sec2 x
Because y = eln y , we have that
lim (1 + sin 4x)cot x = lim+ y = lim+ eln y = e4 .
x→0+
8.5
x→0
x→0
Summary of Curve Sketching
Here is a checklist intended as a guide to sketching a curve by hand.
119
8.5. SUMMARY OF CURVE SKETCHING
By Dr. M. Nxumalo
1. Domain
2. Intercepts
3. Symmetry
4. Asymptotes
5. Intervals of Increase or Decrease
6. Local Maximum and Minimum Values
7. Concavity and Points of Inflection
8. Sketch the Curve
Example 8.5.1. Use the guidelines to sketch the curve f (x) = √
x2
.
x+1
Solution:
1. Domain:
Domf = {x ∈ R : x + 1 ≥ 0, x + 1 6= 0}
= {x ∈ R : x + 1 > 0}
= {x ∈ R : x > −1}
= (−1, ∞).
2. Intercepts:
x-intercepts: Let f (x) = 0. Then
0 = x2 =⇒ x = 0.
Therefore, the curve f cuts the x-axis at (0, 0).
y-intercepts: Let x = 0. Then f (x) = 0. Therefore, the curve f cuts the y-axis at (0, 0).
3. Symmetry:
Since
f (−x) = √
120
x2
6= f (x)
1−x
8.5. SUMMARY OF CURVE SKETCHING
By Dr. M. Nxumalo
and
f (−x) = √
x2
x2
6= −f (x) = √
1−x
− x+1
we have that f is neither even nor odd. So, there is no symmetry.
4. Asymptotes:
Horizontal Asymptotes:
2
x
x2
1
2
lim √
= lim q x
= lim q
= ∞.
x→∞
x→∞
x
1
1
1
x + 1 x→∞
+
+
x2
x2
x2
x
So, there is no horizontal asymptote.
Vertical Asymptotes:
lim + √
x→−1
x2
= ∞.
x+1
Therefore, x = −1 is a vertical asymptote of f .
5. Intervals of increase or decrease:
We have that
f 0 (x) =
x(3x + 4)
2(x + 1)
3
2
=
x(3x + 4)
1
2((x + 1) 2 )3
.
The only critical value we have is x = 0 which is obtained when f 0 (x) = 0.
We have the following table of signs for values − 43 , −1 and 0:
3x + 4
x
3
(x + 1) 2
f 0 (x)
Undefined
Undefined
− 43
0
Undefined
Undefined
+
+
Undefined
Undefined
−1
+
0
Undefined
+
+
-
0
+
0
+
0
+
+
+
+
Increase: f increases on (0, ∞).
Decrease: f decreases on (−1, 0).
One could also omit the part of − 43 since it does not feature in the domain of f .
6. Local maxima and minima:
From 5. above, we get that f has a local minimum at 0. It does not have a local maximum
(by The First Derivative Test).
121
8.5. SUMMARY OF CURVE SKETCHING
By Dr. M. Nxumalo
7. Concavity and Points of inflection:
We have
f 00 (x) =
3x2 + 8x + 8
5
4(x + 1) 2
.
5
In the region (−1, ∞), it is clear that both 3x2 + 8x + 8 and 4(x + 1) 2 are always positive.
So, f 00 (x) > 0 for all points in Domf . Therefore f is concave upward on (−1, ∞). There is no
point of inflection.
8. Sketch:
y
x
0
x = −1
8.5.1
Slant Asymptotes
Some curves have asymptotes that are oblique, that is, neither horizontal nor vertical. If
lim [f (x) − (mx + b)] = 0
x→∞
then the line y = mx + b is called the slant asymptote because the vertical distance between
the curve y = f (x) and the line y = mx + b approaches 0.
For rational functions, slant asymptotes occur when the degree of the numerator is one
more than the degree of the denominator.
Example 8.5.2. Sketch the graph of f (x) =
122
x3
.
x2 + 1
8.5. SUMMARY OF CURVE SKETCHING
By Dr. M. Nxumalo
1. Domain:
Domf = {x ∈ R : x2 + 1 6= 0}
= R.
2. Intercepts:
x-intercepts: Let f (x) = 0. Then
0 = x3 =⇒ x = 0.
Therefore, the curve f cuts the x-axis at (0, 0).
y-intercepts: Let x = 0. Then f (x) = 0. Therefore, the curve f cuts the y-axis at (0, 0).
3. Symmetry:
Since
x3
= −f (x)
f (−x) = − 2
x +1
we have that f is odd. So, it is symmetric about the origin.
4. Asymptotes:
Horizontal Asymptotes:
x3
= ∞.
x→∞ x2 + 1
lim
So, there is no horizontal asymptote.
Vertical Asymptotes: No vertical asymptote since f is well-defined on R.
Slant Asymptotes: We have that
2
x3
x2
x +1−1
x
f (x) = 2
=x
=x
=x− 2
.
2
2
x +1
x +1
x +1
x +1
Therefore,
lim (f (x) − x) = 0.
x→∞
Thus y = x is a slant asymptote.
5. Intervals of increase or decrease:
123
8.5. SUMMARY OF CURVE SKETCHING
By Dr. M. Nxumalo
We have that
f 0 (x) =
x2 (x2 + 3)
.
(x2 + 1)2
The only critical value we have is x = 0 which is obtained when f 0 (x) = 0.
It is clear that f 0 (x) > 0 for all x ∈ R r {0}.
Increase: f increases on (−∞, 0) and (0, ∞).
6. Local maxima and minima:
From 5. above, we get that f does not have a local minimum or local maximum.
7. Concavity and Points of inflection:
We have
2x(3 − x2 )
.
(x2 + 1)3
√
We get f 00 (x) = 0 whenever x = 0 or x = ± 3.
f 00 (x) =
We have the following table of signs:
√
√
− 3
0
3
2x
- 0 + +
2
3−x
0
+ + + 0
(x2 + 1)3 +
+
+ + + +
00
f (x)
+
0
- 0 + 0
+
+
-
√
√
Concave upward: f is concave upward in the regions (−∞, − 3) and (0, 3).
√
√
Concave downward: f is concave downward in the regions (− 3, 0) and ( 3, ∞).
√
√ √
√
Points of inflection: The points of inflection are (− 3, −3 3/4), (0, 0) and ( 3, 3 3/4).
8. Sketch:
124
8.6. OPTIMIZATION PROBLEMS
By Dr. M. Nxumalo
y
y=x
√
− 3
√
0
x
3
inflection points
8.6
Optimization Problems
What we learned that finding extreme values have practical applications in many areas of
life: A business person wants to minimize costs and maximize profits; A traveller wants to
minimize transportation time. In solving such practical problems the greatest challenge is
often to convert the world problem into a mathematical optimization problem by setting up
the function that is to be maximized or minimized.
Steps in solving optimization problems:
1. Understand the problem.
2. Draw a diagram.
3. Introduce notation: Let’s use Q for the quantity to be maximized or minimized.
125
8.6. OPTIMIZATION PROBLEMS
By Dr. M. Nxumalo
4. Express Q in terms of other given symbols.
5. If Q has been expressed as a function of more than one variable, attempt to express it
as a function of one variable and write its domain.
6. Find absolute maximum or minimum as required.
Example 8.6.1. A farmer has 240m of fencing and wants to fence off a rectangular field that
borders a straight river. He needs no fence along the river. What are the dimensions of the
field that has the largest area?
Solution:
Let us try to understand the given problem.
1. The field is a rectangle.
2. The field is a rectangle whose bottom does not have a fence.
3. The perimeter of the fencing is 240m.
4. We must find a combination of length and width which will ensure that the fence cover
the largest area.
Let A be the area of the rectangle, P its perimeter, x its width and y its length.
Then
A = xy
and
P = 2x + y.
Therefore 2x + y = 240, making y = 240 − 2x.
Therefore
A(x) = x(240 − 2x) = 240x − 2x2 .
We maximize A(x). Firstly, we note that DomA = [0, 120].
Next, we have that
A0 (x) = 240 − 4x.
60 is the only critical point of A.
126
8.6. OPTIMIZATION PROBLEMS
By Dr. M. Nxumalo
We have the following table of signs:
240 − 4x
A0 (x)
+
+
60
0
0
-
Clearly, A has the absolute maximum at x = 60.
Therefore y = 240 − 2(60) = 120.
This means that the rectangular field should have a width of 60m and length of 120m.
Example 8.6.2. A cylindrical can is to be made to hold 1L of oil. Find the dimension that
will minimize the cost of the metal to manufacture the can.
Solution:
Let us try to understand the given problem.
1. The can is a cylinder.
2. The volume of the can is 1L.
3. We must find a combination of radius and height which will ensure that the cost of the
cylinder is minimal. To minimize the cost of a cylinder, we must minimize the total
surface area of the cylinder.
Let A be the surface area of the can, V its volume, r its radius and h its height.
Then
A = 2πr2 + 2πrh
and
V = πr2 h.
We know that 1L = 1000cm3 , so πr2 h = 1000, making h =
1000
.
πr2
Therefore
2
A(r) = 2πr + 2πr
1000
πr2
= 2πr2 +
2000
.
r
We minimize A(r). Firstly, we note that DomA = (0, ∞]. Reason for this is because, the
radius cannot be nonpositive.
Next, we have that
2000
4(πr3 − 500)
.
A (r) = 4πr − 2 =
r
r2
0
127
8.6. OPTIMIZATION PROBLEMS
q
3
By Dr. M. Nxumalo
500
is the only critical point of A.
π
We have the following table of signs:
q
3
4(πr3 − 500)
A0 (r)
-
500
π
0
0
+
+
q
Clearly, A has the absolute minimum at r = 3 500
.
π
Therefore
h=
1000
=
πr2
π
1000
q 2 .
3
500
π
q
This means that the can should have a radius of 3 500
cm and height of
π
π
1000
q 2 cm.
3
500
π
Example 8.6.3. Find the area of the largest rectangle that can be inscribed in a semicircle
of radius r.
Solution:
Let us try to understand the given problem.
1. We have a semicircle and a rectangle.
2. We must find the area of the largest inscribed rectangle.
We take the semicircle to be the upper half of the circle x2 + y 2 = r2 with center at the
origin. Then the word inscribed means that the rectangle has two vertices on the semicircle
and two vertices on the x-axis.
Let (x, y) be the vertex that lies in the first quadrant. Then the rectangle has sides of
lengths 2x and y. So, its area is
A = 2xy.
Using y =
√
r2 − x2 , we get
√
A(x) = 2x r2 − x2
128
8.7. NEWTON’S METHOD
By Dr. M. Nxumalo
whose domain is given by [0, r]. We maximize A(x).
Observe that
2(r2 − 2x2 )
A0 (x) = √
.
r 2 − x2
Clearly, √r2 is the critical value of A.
√r
2
2(r2 − 2x2 )
A0 (x)
+
+
0
0
-
We get that A has the absolute maximum at x = √r2 .
Therefore
s
y=
r2 −
r
√
2
r
2
=
r2 −
r2
.
2
The area of the largest inscribed rectangle is given by
r
r
r2
A=2 √
r2 −
= r2 .
2
2
8.7
Newton’s Method
For polynomials of degree 0 to 4, there are formulas to find the roots. However, for polynomials
of degree 5 or higher, there is no such formula. For equations where the process of finding
roots is complicated, we resort to finding approximate solutions.
Consider the following graph:
129
8.7. NEWTON’S METHOD
By Dr. M. Nxumalo
y
(x1 , f (x1 ))
(x2 , f (x2 ))
0
r
x3
x2
x1
x
In the above graph, (r, 0) is a root of the function f . To approximate the value of r:
1. Guess a value x1 closer to r.
2. Draw a tangent line touching f at (x1 , f (x1 )). The x-intercept of this line will be x2 . To
get the value of x2 , we use the equation
y − f (x1 ) = f 0 (x1 )(x − x1 )
which is fitting since the line passes through (x1 , f (x1 )). Therefore for f 0 (x1 ) 6= 0,
x2 = x1 −
f (x1 )
.
f 0 (x1 )
3. Draw another tangent line touching f at (x2 , f (x2 )). The x-intercept of this line will be
x3 . To get x3 , we use the equation
y − f (x2 ) = f 0 (x2 )(x − x2 )
which is fitting since the line passes through (x2 , f (x2 )). Therefore for f 0 (x2 ) 6= 0,
x3 = x2 −
130
f (x2 )
.
f 0 (x2 )
8.7. NEWTON’S METHOD
By Dr. M. Nxumalo
In general, if the nth approximation is xn and f 0 (xn ) 6= 0, then the next approximation is
given by
xn+1 = xn −
f (xn )
.
f 0 (xn )
The above process finding approximate solutions is called the Newton’s method.
Example 8.7.1. Starting with x1 = 2, find the third approximation x3 to the root of the
equation x3 − 2x − 5 = 0.
Solution:
We use Newtons method:
We have that f 0 (x) = 3x2 − 2.
Therefore
xn+1 = xn −
x3n − 2xn − 5
.
3x2n − 2
With x1 = 2, we get, for n = 1,
x2 = x1 −
x31 − 2x1 − 5
23 − 2(2) − 5
=
x
−
= 2.1.
1
3x21 − 2
3(2)2 − 2
For n = 2,
x 3 = x2 −
x32 − 2x1 − 5
(2.1)3 − 2(2.1) − 5
=
x
−
= 2.0946.
1
3x22 − 2
3(2.1)2 − 2
We sometimes want to use Newton’s method to achieve a given accuracy, say to eight
decimal places. Here, we stop when successive approximations xn and xn+1 agree to eight
decimal places.
√
Example 8.7.2. Use Newton’s method to find 6 2 correct to eight decimal places.
Solution:
√
Finding 6 2 is equivalent to finding the positive root of the equation
x6 − 2 = 0.
Set f (x) = x6 − 2. Then f 0 (x) = 6x5 . From Newton’s method,
xn+1 = xn −
131
x6n − 2
.
6x5n
8.8. ANTIDERIVATIVES
By Dr. M. Nxumalo
If we choose x1 = 1, then we obtain
x2 ≈ 1.16666667
x3 ≈ 1.12644368
x4 ≈ 1.12246205
x5 ≈ 1.12246205
Since x5 and x6 agree to eight decimal places, we conclude that
√
6
2 ≈ 1.12246205
to eight decimal places
8.8
Antiderivatives
A physicist who knows how the velocity of a particle might wish to know its position at a given
time. An engineer who can measure the variable at which water is leaking from a tank wants to
know the amount leaked over a certain time period. A biologist who knows the rate at which
a bacteria population is increasing might want to deduce what the size of the population will
be at some future time. In each case, the problem is to find a function F whose derivative is
a known function f . If such a function F exists, it is called an antiderivative of f .
Definition 8.8.1. A function F is called an antiderivative of f on an interval I if F 0 (x) =
f (x) for all x ∈ I.
Consider the function f (x) = x2 . We have that F (x) = 13 x3 because F 0 (x) = f (x).
But F1 (x) = 13 x3 + 50 is still another antiderivative of f . In fact, any function of the form
H(x) = 13 x3 + C, where C is a constant, is an antiderivative of f .
Theorem 8.8.2. If F is an antiderivative of f on an interval I, then the most general antiderivative of f on I is
F (x) + C
where C is an arbitrary constant.
132
8.8. ANTIDERIVATIVES
By Dr. M. Nxumalo
Example 8.8.3. Find the most general antiderivative of the following functions:
1. f (x) = sin x
2. f (x) = x1
2x5 −
3. f (x) = 4 sin x +
x
√
x
Solution:
1. F (x) = − cos x + C where C is an arbitrary constant. This is because,
F 0 (x) = sin x.
2.
(
ln x + C1
F (x) =
− ln(−x) + C2
if x > 0
if x < 0
where C1 and C2 are arbitrary constants.
3. Observe that
1
f (x) = 4 sin x + 2x4 − √ .
x
So,
√
2
F (x) = −4 cos x + x5 − 2 x + C
5
where C is an arbitrary constant.
Example 8.8.4. Find f if f 0 (x) = ex + 20(1 + x2 )−1 and f (0) = −2.
Solution:
We have that f (x) = ex + 20 tan−1 x + C where C is an arbitrary constant. To determine
C, we use the fact that f (0) = −2:
f (0) = e0 + 20 tan−1 (0) + C = −2
=⇒
Therefore
f (x) = ex + 20 tan−1 x − 3.
133
C = −3.
8.8. ANTIDERIVATIVES
By Dr. M. Nxumalo
Example 8.8.5. Find f if f 00 (x) = 12x2 + 6x − 4, f (0) = 4, and f (1) = 1.
Solution:
We have that
f 0 (x) = 4x3 + 3x2 − 4x + C.
Using the antiderivative rules once more, we find that
f (x) = x4 + x3 − 2x2 + Cx + D.
To determine C and D, we use the facts that f (0) = 4 and f (1) = 1:
Since f (0) = 0 + D = 4, we have that D = 4.
Also, since f (1) = 1 + 1 − 2 + C + 4 = 1, we get that C = −3.
Therefore
f (x) = x4 + x3 − 2x2 − 3x + 4.
Example 8.8.6. The graph of a function f is given below:
y
0
1
2
3
x
4
Make a rough sketch of an antiderivative F , given that F (0) = 2.
Solution:
Here, we have that f (x) = F 0 (x). We must find F . We use our knowledge of a relationship
between a function and its derivative:
1. F (0) = 2.
2. Since F 0 (x) < 0 for all x in the regions (0, 1) and (3, ∞), we have that F is decreasing
on these intervals.
134
8.8. ANTIDERIVATIVES
By Dr. M. Nxumalo
3. Since F 0 (x) > 0 for all x in the region (1, 3), we have that F is increasing on the interval
(1, 3).
4. Since F 0 (x) = 0 at x = 1 and x = 3, we have that both 1 and 3 are critical values of F .
5. Since the graph changes of F 0 changes from negative to positive at x = 1, we have that
F 0 has a local minimum at x = 1.
6. Since the graph of F 0 changes from positive to negative at x = 3, we have that F 0 has a
local maximum at x = 3.
7. Since the graph of F 0 approaches 0 as x approaches infinity, we have that F becomes
flatter as x → ∞.
8. Since F 00 (x) = f 0 (x) changes from positive to negative at x = 2, we have that F has an
inflection point at x = 2.
9. Since F 00 (x) = f 0 (x) changes from negative to positive at x = 4, we have that F has an
inflection point at x = 4.
The sketch:
y
4
0
1
2
3
x
4
135
Chapter 9
Introduction to Integration
9.1
Areas
In Mathematics, we sometimes wish to know the total space taken by a flat 2D shape of an
object. This is what we call the area of an object. It is easy to find the area for regions with
straight sides. For a rectangle, the area is the product of its length and the width. The area of
a triangle is half the base times the height. The area of a polygon is found by dividing it into
triangles and adding the areas of the triangles. What about the area for regions with curved
sides?:
y
y = f (x)
S
a
x
b
It is not easy to find it. In this section, we attempt to find it.
In defining a tangent, we first approximated the slope of the tangent line by slopes of secant
lines and then we took the limit of these approximations. We pursue a similar idea for areas.
136
9.1. AREAS
By Dr. M. Nxumalo
We first approximate the region S by rectangles and then we take the limit of the areas of
these rectangles as we increase the number of rectangles.
Consider the following example.
Example 9.1.1. Use rectangles to estimate the area under the parabola y = x2 from 0 to 1:
y
y = x2
(1, 1)
S
x
0
1
Solution:
Let us divide S into eight strips as follows:
y
y = x2
(1, 1)
x
0
1 1 3 1 5 3 7 1
8 4 8 2 8 4 8
If we draw rectangles in each strip, whose base is the same as the strip and whose height is
the same as the right edge of the strip, as below:
137
9.1. AREAS
By Dr. M. Nxumalo
y
y = x2
(1, 1)
x
0
1 1 3 1 5 3 7 1
8 4 8 2 8 4 8
we get that the heights of these rectangles are the values of the function f (x) = x2 at
the right end points of the subintervals [0, 81 ], 81 , 14 , 14 , 83 , 38 , 21 , 12 , 58 , 85 , 43 , 43 , 78 , 78 , 1 .
Each rectangle has width 18 and the heights are ( 18 )2 , ( 41 )2 , ( 38 )2 , ( 21 )2 , ( 85 )2 , ( 43 )2 , ( 78 )2 and 12 .
Let R8 be the sum of the areas of these rectangles. We get that
2
2
2
2
2
2
2
1 1
1 1
1 3
1 1
1 5
1 3
1 7
1
R8 = ·
+ ·
+ ·
+ ·
+ ·
+ ·
+ ·
+ ·(1)2 = 0.3984375.
8 8
8 4
8 8
8 2
8 8
8 4
8 8
8
It is clear that the area A of S is lesser than R8 :
A < 0.3984375.
If we only consider rectangles whose heights are the values of f at the left end points of
the subintervals:
y
y = x2
(1, 1)
x
0
1 1 3 1 5 3 7 1
8 4 8 2 8 4 8
we get that the sum of the areas of these approximating rectangles is
2
2
2
2
2
2
2
1
1
1 1
1 3
1 1
1 5
1 3
1 7
2 1
R8 = ·(0) + ·
+ ·
+ ·
+ ·
+ ·
+ ·
+ ·
= 0.2734377.
8
8 8
8 4
8 8
8 2
8 8
8 4
8 8
138
9.1. AREAS
By Dr. M. Nxumalo
Clearly
0.2734377 < A < 0.46875.
So one possible answer to the question is to say that the true area of S lies somewhere
between 0.2734375 and 0.3984375.
We could obtain better estimates by increasing the number of strips.
Example 9.1.2. For region S in Example 9.1.1, show that the sum of the areas of the upper
approximating rectangles approaches 13 , that is,
1
lim Rn = .
n→∞
3
Sulution:
Rn is the sum of the areas of n rectangles. Each rectangle has n1 width and the heights
are the values of the function f (x) = x2 at the points n1 , n2 , n3 , ..., nn ; that is, the heights are
( n1 )2 , ( n2 )2 , ( n3 )2 , ..., ( nn )2 .
Thus
!
2
2
2
1
1
1 n 2
1
2
3
+ ·
+ ·
+ ... + ·
lim (Rn ) = lim
n→∞
n→∞
n
n
n
n
n
n
n
1 1 2
· (1 + 22 + 32 + ... + n2 )
= lim
n→∞
n n2
1 2
2
2
2
= lim
(1 + 2 + 3 + ... + n )
n→∞
n3
1 n(n + 1)(2n + 1)
= lim
n→∞
n3
6
(n + 1)(2n + 1)
= lim
n→∞
6n2
1
= .
3
1
·
n
It can also be shown that
1
lim Ln = .
3
n→∞
Define the area A of a region S to be the limit of the sums of the areas of the approximating
rectangles, that is,
A = lim Rn = lim Ln .
n→∞
139
n→∞
9.1. AREAS
By Dr. M. Nxumalo
Let us consider a more general region S:
y
y = f (x)
S
a
x
b
We start by dividing S into n strips S1 , S2 , S3 , ..., Sn of equal width:
y
y = f (x)
S1 S2 S3
Si
a
Sn
x
b
The width of the interval [a, b] is b − a, so the width of each of the n strips is
∆x =
b−a
.
n
These strips divide the interval [a, b] into n subintervals
[x0 , x1 ], [x1 , x2 ], [x2 , x3 ], ..., [xn−1 , xn ]
where x0 = a and xn = b. The right end points of the subintervals are
x1 = a + ∆x,
x2 = a + 2∆x,
140
x3 = a + 3∆x, ...
9.1. AREAS
By Dr. M. Nxumalo
If we approximate the ith strip Si by a rectangle with width ∆x and height f (xi ), which is
the value of f at the endpoint, then the area of the ith rectangle is given by f (xi )∆x. More
generally, we have the following definition of area.
Definition 9.1.3. The area A of the region S that lies under the graph of the continuous
function f is the limit of the sum of the areas of approximating rectangles:
A = lim Rn = lim [f (x1 )∆x + f (x2 )∆x + ... + f (xn )∆x] .
n→∞
n→∞
In sigma notation, we write
A = lim Rn = lim
n→∞
n
X
n→∞
f (xi )∆x.
i=1
Using left end points, we have
A = lim Ln = lim [f (x0 )∆x + f (x1 )∆x + ... + f (xn−1 )∆x] .
n→∞
n→∞
In sigma notation, we write
A = lim Ln = lim
n→∞
n→∞
n
X
f (xi−1 )∆x.
i=1
Instead of using left endpoints or right endpoints, we could take the height of the ith
rectangle to be the value of f at any number x∗i in the ith subinterval [xi−1 , xi ]. We call the
numbers x∗1 , x∗2 , ..., x∗n the sample points. So, a more general expression for the area of S is
A = lim [f (x∗1 )∆x + f (x∗2 )∆x + ... + f (x∗n )∆x] .
n→∞
Example 9.1.4. Let A be the area of the region that lies under the graph of f (x) = e−x
between x = 0 and x = 2. Using right endpoints, find an expression for A as a limit. Do not
evaluate the limit.
Solution:
Set a = 0 and b = 2. Then
∆x =
Therefore x1 = 0 + n2 = n2 , x1 = 0 + 2
2
n
141
2−0
2
= .
n
n
= n4 , x3 = n6 , ...,xn = 2n
.
n
9.2. THE DEFINITE INTEGRAL
By Dr. M. Nxumalo
The area of the approximating rectangles is
lim Rn = lim (f (x1 )∆x + f (x2 )∆x + f (x3 )∆x + ... + f (xn )∆x)
n→∞
= lim e−x1 ∆x + e−x2 ∆x + e−x3 ∆x + ... + e−xn ∆x
n→∞
2
2
2
2
−x1
−x2
−x3
−xn
= lim e
+e
+e
+ ... + e
n→∞
n
n
n
n
2
4
6
2n
2
2
2
2
= lim e− n
+ e− n
+ e− n
+ ... + e− n
n→∞
n
n
n
n
4
6
2n
2 −2
e n + e− n + e− n + ... + e− n
= lim
n→∞
n
!!
n
X
2i
2
= lim
e− n
.
n→∞
n i=1
n→∞
9.2
The Definite Integral
Definition 9.2.1 (Definition of a Definite Integral). If f is a continuous function defined
for a ≤ x ≤ b, we divide the interval [a, b] into n subintervals of equal width ∆x = b−a
. We
n
let x0 (= a), x1 , x2 , ..., xn (= b) be the endpoints of these subintervals and we let x∗1 , x∗2 , ..., x∗n be
any sample points in these subintervals, so x∗i lies in the ith subinterval [xi−1 , xi ]. Then the
definite integral of f from a to b is
Z b
n
X
f (x)dx = lim
f (x∗i )∆x.
n→∞
a
i=1
Because f is continuous, the limit in Definition 9.2.1 always exists and gives the same
value no matter how we choose the sample points x∗i . If we take the sample points to be right
endpoints, then x∗i = xi , and the definition of an integral becomes
Z b
n
X
f (x)dx = lim
f (xi )∆x.
n→∞
a
i=1
If we take the sample points to be left endpoints, then x∗i = xi−1 , and the definition of an
integral becomes
Z b
f (x)dx = lim
n→∞
a
The symbol
R
n
X
f (xi−1 )∆x.
i=1
was introduced by Leibniz and is called the integral sign. For
Rb
a
f (x)dx,
we say f (x) is the integrand and a and b are the limits of integration; a is the lower limit
142
9.2. THE DEFINITE INTEGRAL
By Dr. M. Nxumalo
and b is the upper limit. The symbol dx has no official meaning by itself. The procedure of
calculating an integral is called integration.
The sum
n
X
f (x∗i )∆x
i=1
is called a Riemann sum.
The definite integral
Rb
a
f (x)dx can be interpreted as the area under the curve y = f (x)
from a to b.
If f happens to be positive, then the Riemann sum can be interpreted as a sum of areas
approximating rectangles. If f takes on both positive and negative values, then the Riemann
sum is the sum of the areas of the rectangles that lie above the x-axis and the negative of the
areas of the rectangles that lie below the x-axis.
Let us recall the following formulas for sums.
1.
Pn
i=1 i =
n(n + 1)
2
n(n + 1)(2n + 1)
6
2
Pn 3
n(n + 1)
3.
i=1 i =
2
Pn
4.
i=1 c = nc
2.
Pn
5.
Pn
6.
Pn
Pn
Pn
7.
Pn
Pn
Pn
i=1 i
2
=
i=1 cai = c
Pn
i=1 ai
i=1 (ai + bi ) =
i=1 (ai − bi ) =
i=1 ai +
i=1 ai −
i=1 bi
i=1 bi
Example 9.2.2. Express
lim
n→∞
n
X
i=1
as an integral on the interval [0, π].
Solution:
143
(x3i + xi sin xi )∆x
9.2. THE DEFINITE INTEGRAL
We have
By Dr. M. Nxumalo
Z π
n
X
3
(x3 + x sin x)dx.
lim
(xi + xi sin xi )∆x =
n→∞
0
i=1
Example 9.2.3. Evaluate the Riemann sum for f (x) = x3 − 6x taking the samples to be right
endpoints and a = 0, b = 3 and n = 6.
Solution:
With n = 6, we have
∆x =
b−a
3−0
1
=
=
n
6
2
and the right end points are x1 = 0.5, x2 = 1.0, x3 = 1.5, x4 = 2.0, x5 = 2.5, and x6 = 3.0.
So, the Riemann sum is
R6 =
6
X
f (xi )∆x
i=1
= f (0.5)∆x + f (1.0)∆x + f (1.5)∆x + f (2.0)∆x + f (2.5)∆x + f (3.0)∆x
1
= (−2.875 − 5 − 5.625 − 4 + 0.625 + 9)
2
= −3.9375.
f is not a positive function, so the Riemann sum does not represent a sum of areas of rectangles.
But it does represent the sum of the areas of the rectangles that lie above the x-axis and the
negative of the areas of the rectangles that lie below the x-axis.
Example 9.2.4. Evaluate
R3
0
(x3 − 6x)dx.
Solution:
With n subintervals, we have
∆x =
b−a
3
= .
n
n
Therefore x0 = 0, x1 = n3 , x2 = n6 , x3 = n9 , and in general, xi = 3i
.
n
144
9.2. THE DEFINITE INTEGRAL
By Dr. M. Nxumalo
Since we are using right endpoints, we have
Z 3
n
n
X
X
3
3i
3
(x − 6x)dx = lim
f (xi )∆x = lim
f
n→∞
n→∞
n
n
0
i=1
i=1
!
3
n
3X
3i
3i
= lim
−6
n→∞ n
n
n
i=1
n 3 X 27 3 18
i − i
= lim
n→∞ n
n3
n
i=1
!
n
n
81 X 3 54 X
= lim
i − 2
i
n→∞
n4 i=1
n i=1
!
2
54 n(n + 1)
81 n(n + 1)
− 2
= lim
n→∞
n4
2
n
2
2
!
1
1
81
− 27 1 +
1+
= lim
n→∞
4
n
n
=
81
27
− 27 = − = −6.75.
4
4
This integral cannot be interpreted as an area because f takes on both positive and negative
values. But it can be interpreted as the sum of the area of the part of the curve that lies above
the x-axis and the negative of the area of the part of the curve that lies below the x-axis.
R3
Example 9.2.5. Set up an expression for 1 ex dx.
Solution:
We have f (x) = ex , a = 1, b = 3, and
∆x =
3−1
2
b−a
=
= .
n
n
n
So, x0 = 1, x1 = 1 + n2 , x2 = 1 + n4 , x3 = 1 + n6 , and
xi = 1 +
2i
.
n
Therefore,
Z 3
x
e dx = lim
n→∞
1
= lim
n→∞
n
X
i=1
n
X
f (xi )∆x
f
i=1
n
2i
1+
n
2 X 1+ 2i
= lim
e n.
n→∞ n
i=1
145
2
n
9.2. THE DEFINITE INTEGRAL
By Dr. M. Nxumalo
We often choose the sample point x∗i to be the right endpoint of the ith subinterval because
it is convenient for computing the limit. But if the purpose is to find an approximation to an
integral, it is usually better to choose x∗i to be the midpoint of the interval, which we denote
by xi . Any Riemann sum is an approximation to an integral, but if we use midpoints we get
the following approximation.
Midpoint Rule:
Z b
n
X
f (x)dx ≈
f (xi )∆x = ∆x (f (x1 ) + f (x2 ) + ... + f (xn ))
a
i=1
where
∆x =
b−a
n
and
1
xi = (xi−1 + xi ) = midpoint of [xi−1 , xi ].
2
Example 9.2.6. Use the Midpoint Rule with n = 5 to approximate
R2 1
1 x
dx.
Solution:
The endpoints of the five subintervals are 1, 1.2, 1.4, 1.6, 1.8, and 2.0. So, the midpoints
are 1.1, 1.3, 1.5, 1.7, and 1.9.
The width of the subintervals is given by
∆x =
2−1
1
= .
5
5
The Midpoint Rule gives
Z 2
1
dx ≈ ∆x (f (1.1) + f (1.3) + f (1.5) + f (1.7) + f (1.9))
1 x
1 1
1
1
1
1
=
+
+
+
+
5 1.1 1.3 1.5 1.7 1.9
≈ 0.691908.
Properties of the integral:
1.
Rb
2.
Rb
a
a
cdx = c(b − a) where c is any constant.
(f (x) + g(x))dx =
Rb
a
Rb
f (x)dx + a g(x)dx.
146
9.2. THE DEFINITE INTEGRAL
3.
Rb
4.
Rb
5.
Rb
6.
Rb
a
By Dr. M. Nxumalo
Rb
cf (x)dx = c a f (x)dx where c is any constant.
Ra
(f
(x))dx
=
−
(f (x))dx.
a
b
a
(f (x) − g(x))dx =
f (x)dx =
a
Rb
a
Rb
f (x)dx − a g(x)dx.
Rb
f
(x)dx
+
f (x)dx.
c
a
Rc
Example 9.2.7. Use the properties of integrals to evaluate
R1
0
(4 + 3x2 )dx.
Solution:
Z 1
Z 1
2
(4 + 3x )dx =
0
Z 1
4dx +
Z0 1
0
Z 1
4dx + 3
=
0
We have that
For
R1
0
R1
0
3x2 dx
x2 dx
0
4dx = 4(1 − 0) = 4.
x2 :
With n subintervals, we have
1
b−a
= .
n
n
1
i
Therefore x0 = 0, x1 = n , and in general, xi = n .
∆x =
Since we are using right endpoints, we have
Z 1
n
n
X
X
i
1
2
x dx = lim
f (xi )∆x = lim
f
n→∞
n→∞
n
n
0
i=1
i=1
n 2
1X i
= lim
n→∞ n
n
i=1
n
1X 1 2
i
= lim
n→∞ n
n2
i=1
n
1 X 2
i
= lim 3
n→∞ n
i=1
1 n(n + 1)(2n + 1)
= lim 3
n→∞ n
6
1
1
1
= lim
+
+
n→∞
3 2n 6n2
1
= .
3
147
9.2. THE DEFINITE INTEGRAL
By Dr. M. Nxumalo
Therefore
Z 1
1
(4 + 3x )dx = 4 + 3
= 5.
3
0
R 10
R8
R 10
Example 9.2.8. If it is known that 0 f (x)dx = 17 and 0 f (x)dx = 12, find 8 f (x)dx.
2
Solution:
We have
Z 8
f (x)dx
f (x)dx =
f (x)dx +
=⇒
f (x)dx = 17
12 +
8
0
8
0
Z 10
Z 10
Z 10
So,
Z 10
f (x)dx = 5.
8
Comparison Properties of the integral:
1. If f (x) ≥ 0 for a ≤ x ≤ b, then
Rb
a
2. If f (x) ≥ g(x) for a ≤ x ≤ b, then
f (x)dx ≥ 0.
Rb
a
f (x)dx ≥
Rb
a
g(x)dx.
3. If m ≤ f (x) ≤ M for a ≤ x ≤ b, then
Z b
m(b − a) ≤
f (x)dx ≤ M (b − a).
a
Example 9.2.9. Estimate
R 1 −x2
e dx.
0
Solution:
2
f (x) = e−x is a decreasing function on [0, 1]. It has the absolute maximum of M = f (0) =
1 and the absolute minimum of m = f (1) = e−1 . That is e−1 ≤ f (x) ≤ 1 for 0 ≤ x ≤ 1.
It follows that
Z 1
−1
e (1 − 0) ≤
2
e−x dx ≤ 1(1 − 0).
0
Therefore
e
−1
Z 1
≤
2
e−x dx ≤ 1.
0
THE END!!
148