LECTURE NOTE ON MATHEMATICS FOR SOCIAL SCIENTISTS II (MTS 108)

ADEOSUN S.A www.crescent-university.edu.ng

1 | P a g e

LECTURE NOTE

ON

MATHEMATICS FOR SOCIAL SCIENTISTS II

(MTS 108)

BY

ADEOSUN SAKIRU ABIODUN

E-mail: [email protected]

mailto:[email protected]


2 | P a g e

Course Outline

Functions and equations: Definition and types of functions, definition and types of

equations. Matrices. Introduction to statistical distribution and density functions,

especially the binomial, Poisson and normal. Introduction to calculus functions of the

variable and their continuity. Techniques of differentiation; logarithmic, trigonometric

and exponential functions. Integral calculus; Optimization of functions, maximal,

minimal and in flexional points, and the application of these concepts in Business

and Economics.


3 | P a g e

§ 1.0 FUNCTIONS

Mathematical modeling is an attempt to describe some part of the real world

in mathematical terms. Our models will be functions that show the relationship

between two or more variables. These variables will represent quantities that we

wish to understand or describe. We will begin by reviewing the basic concept of

functions. In short, we call any rule that assigns or corresponds to each element in

one set precisely one element in another set a function.

Definition of a function: A function 𝑓 from 𝐷 to 𝑅 is a rule that assigns to each

element 𝑥 in 𝐷 one and only one (unique) element 𝑦 = 𝑓(𝑥) in 𝑅.

The set 𝐷 in the definition is called the domain of f. We might think of the

domain as the set of inputs. We then think of the values 𝑓(𝑥) as outputs. The set of

all the possible outputs (set of functional), 𝑅 is called the range of f. The letter

representing elements in the domain is called the independent variable and the letter

representing elements in the range is called the dependent variable. Thus, if

𝑦 = 𝑓(𝑥), 𝑥 is the independent variable while 𝑦 is the dependent variable. Note that

𝑦 = 𝑓(𝑥) is read as “𝑦 is a function of 𝑥”.

Example 1: A restaurant serves a steak special for N500. Write a function that

models the amount of revenue made from selling these specials. How much revenue

will 15 steak specials earn?

Solution: We first need to decide if the independent variable is the price of the

steak specials, the number of specials sold, or the amount of revenue earned. Since

the price is fixed at N500 per special and revenue depends on the number of special

sold, we choose the independent variable, 𝑥, to be the number of specials sold and

the dependent variable, 𝑅 = 𝑓(𝑥) to be the amount of revenue.

∴ 𝑅 = 𝑓 𝑥 = 500𝑥. We note that 𝑥 must be a whole number, so the domain is

= 0,1,2, … . Hence, when selling 15 steak specials, 𝑅 = 𝑓 15 = 500 15 = 7500. So

the revenue is N7500.

Example 2: If 𝑓 𝑥 = 𝑥3 − 3𝑥 + 4, find the values of : 𝑓(0), 𝑓(−1), 𝑓(𝑎), 𝑓(3𝑥).

Solution: 𝑓 0 = 03 − 3 0 + 4 = 4


4 | P a g e

𝑓 −1 = −1 3 − 3 −1 + 4 = 6

𝑓 𝑎 = 𝑎3 − 3𝑎 + 4

𝑓 3𝑥 = 3𝑥 3 − 3 3𝑥 + 4 = 27𝑥3 − 9𝑥 + 4

Example 3: If 𝑓 𝑥 =𝑥+1

𝑥2+𝑥−2, determine

(a) the image of −1

(b) 𝑓(𝑥 − 1)

(c) What values of 𝑥 will have no images for this function?

Solution: (a) The image of 𝑥 = −1 is 𝑓(−1):

𝑓 −1 =−1+1

−1 2+ −1 −2=

0

−2= 0. (Note 𝑥 = −1 is the zero for this function)

(b) 𝑓 𝑥 − 1 =𝑥−1+1

𝑥−1 2+ 𝑥−1 −2=

𝑥

𝑥2−𝑥−2

(c) 𝑓 𝑥 =𝑥+1

𝑥2+𝑥−2=

𝑥+1

𝑥+2 𝑥−1 .

When 𝑥 = −2, 𝑓 −2 =−1

0 and when 𝑥 = 1, 𝑓 1 =

2

0. And since division by 0 is

meaningless (undefined), therefore 𝑓(−2) and 𝑓(1) do not exist. Hence the values

𝑥 = −2 and 𝑥 = 1 have no images for this function.

Example 4: Given the functions 𝑓 𝑥 = 𝑥2 − 1, 𝑔 𝑥 = 2𝑥 + 3 and 𝑕 𝑥 =1

𝑥, find

(i) the domain and range of 𝑓(𝑥)

(ii) the range of 𝑕(𝑥)

(iii) 𝑓 ⃘ 𝑔 𝑥 (called the composition function)

Solution: (i) The domain and range of 𝑓(𝑥) is the set of all real numbers i.e ℝ.

(ii) Note that one way of determining the range is to express 𝑥 as a function of

𝑦; 𝑥 = 𝑝(𝑦), then the range is the set of all real values of 𝑦 for which the function 𝑝

is defined.

Solve for 𝑥:

𝑥 =1

𝑦 define for all real numbers except when 𝑦 = 0. Therefore the range of 𝑕(𝑥)

is the set of real numbers when zero is deleted i.e. Rng h = ℝ\ 0 .

(iii) 𝑓 ⃘ 𝑔 𝑥 = 𝑓 𝑔(𝑥) = 2𝑥 + 3 2 − 1 = 4𝑥2 + 12𝑥 + 8 .


5 | P a g e

Different between a Function and an Equation

Function Equation

Let 𝑦 = 𝑓 𝑥 = 𝑥2 − 3𝑥 − 10 = 𝑥 − 5 𝑥 + 2 .

This states the function. A function cannot be

solved but is used to find the images of 𝑥. The

values of 𝑥 which make 𝑓 𝑥 = 0 are called the

zeros of 𝑓. The zeros of this function are

𝑥 = 5 and 𝑥 = −2.

For example 𝑥2 − 3𝑥 − 10 = 0. An

equation is solved to find its roots

which are the values of 𝑥 which

satisfy the equation. The roots of

this equation are 𝑥 = 5 or 𝑥 = −2.

Some important types of functions:

1. Linear function 𝑦 = 2𝑥 − 4

2. Quadratic function 𝑦 = 𝑥2 − 𝑥 + 12

3. Polynomial function 𝑦 = 𝑎𝑛𝑥𝑛 + 𝑎𝑛−1𝑥𝑛−1 + ⋯ + 𝑎1𝑥 + 𝑎0

4. Trigonometric function 𝑦 = sin 𝑥

5. Logarithmic function 𝑦 = log 𝑥

6. Exponential function 𝑦 = 3𝑥 or 𝑒𝑥

Many other functions will be defined by formula. For example, area, A, of a circle is a

function of radius r, as 𝐴 = 𝜋𝑟2, also Total value (T) = Price(P) × Quantity (Q) and

so on.

Exercise 1

1. Find the values stated for these functions:

(a) 𝑓 𝑥 = 3𝑥 − 1 ; 𝑓(0), 𝑓(−1), 𝑓(1)

(b) 𝑓 𝑥 = 𝑥2 + 𝑥 + 1 ; 𝑓(−1), 𝑓(0), 𝑓(𝑥 + 1)

(c) 𝑔 𝑥 = 𝑥2 − 𝑥 − 2 ; 𝑔 −1 , 𝑔 −1

2 , 𝑔 0 , 𝑔 1

2

(d) 𝑕 𝑥 =𝑥

𝑥+1; 𝑕 0 , 𝑕 1 , 𝑕(𝑥 − 1)

2. Find the zeros of the following functions:

(a) 𝑓 𝑥 = 𝑥 − 2 (b) 𝑓 𝑥 = 𝑥 + 3 2

(c) 𝑔 𝑥 =𝑥−1

𝑥+6 (d) 𝑓 𝑥 =

𝑥2+2𝑥−15

𝑥2−1

3. If 𝑓 𝑥 = 3𝑥 + 2, what is the value of 𝑥 whose image is 5? If also given that

𝑔 𝑥 = 2𝑥 − 3 and 𝑓 𝑥 + 𝑔 𝑥 = 4, find the value of 𝑥.


6 | P a g e

4. Given the function 𝑓 𝑥 = 2𝑥2 − 𝑥 + 1, express 𝑓 𝑥+𝑕 −𝑓(𝑥)

𝑕 in its simplest form.

§ 2.0 MATRICES AND DETERMINANTS

2.1 Basic Concepts

A matrix is an array (enclosed between brackets) of real numbers arranged in

𝑚 rows and 𝑛 columns. The matrix is then referred to as an 𝑚 by 𝑛 matrix or

(𝑚 × 𝑛) matrix.

For example,

(a) 𝐴 = 1 2 3 is a 1 by 3 matrix.

(b) 𝐵 = 2 4 12 0 3

is a 2 by 3 matrix

(c) 𝐶 = 8 24 50 1

is a 3 by 2 matrix

(d) 𝐷 = 1 2 34 5 67 8 9

is a 3 by 3 matrix.

Any 𝑛 by 𝑛 matrix is called a square matrix.

2.2 Types of Matrices

(I) Equal matrices: Two matrices 𝐴 and 𝐵 are said to be equal when

(i) The number of rows of 𝐴 = the number of rows of 𝐵 and

(ii) The number of columns of 𝐴 = the number of columns of B and

(iii) The entries in corresponding position are the same in both 𝐴 and 𝐵.

For example:

1. If 𝐴 = 3 12 2

and 𝐵 = 3 1 22 2 0

, then 𝐴 ≠ 𝐵.

2. 𝑋 = 2 0 33 1 00 0 6

, 𝑌 =1

2 4 0 66 2 00 0 12

, then 𝑋 = 𝑌 since entries in

corresponding positions in both matrices are equal

(II) .Identity Matrix: A square matrix 𝐴 is said to be an identity matrix when each

entry on the leading diagonal of 𝐴 is unity and other entries off the leading

diagonal are zeros. For example, the following are identity matrices:


7 | P a g e

I2 = 1 00 1

, 𝐼3 = 1 0 00 1 00 0 1

, 𝐼4 =

1000

0100

0010

0001

(III)Diagonal Matrix: A square matrix with zero entries off the leading diagonal is a

diagonal matrix provided the entries on the leading diagonal are non-zero. For

example,

𝐴 = 3 0 00 6 00 0 9

, 𝐵 =

4000

0500

0090

0002

are diagonal matrices.

(IV) Triangular Matrix

A matrix whose entries above or below the leading diagonal are zeros is

referred to as a triangular matrix. It is said to be upper triangular where all

entries below the leading diagonal are zeros. It is lower triangular when all

entries above the leading diagonal are zeros. Following are the two examples:

𝑋 = 1 4 30 7 50 0 8

, 𝑌 = 1 0 03 6 04 5 7

.

(V) The Transpose of a Matrix

A matrix 𝐵 is the transpose of 𝐴 if when the rows of 𝐵 are turned into columns,

the resulting matrix is the same as 𝐴. For example, let 𝐴 = 1 3 59 6 32 1 4

, if the

rows of 𝐴 are now put into the columns of another matrix 𝐵, we have

𝐵 = 1 3 59 6 32 1 4

. Then 𝐵 is the transpose of 𝐴. This is written as 𝐵 = 𝐴𝑇 .

(VI) Symmetric Matrix

A matrix 𝑋 is said to be symmetric if 𝑋 = 𝑋𝑇. In other words, a matrix is

symmetric if the matrix and its transpose are the same. For example,

𝑋 = 2 8 18 9 21 2 3

is such that 𝑋 = 𝑋𝑇. Hence 𝑋 is a symmetric matrix.


8 | P a g e

(VII) Skew Symmetric Matrix

A matrix 𝑋 is said to be skew symmetric if 𝑋𝑇 = −𝑋. We noted that this could

only be possible if all the leading diagonal entries of 𝑋 are zeros. For

example, 𝑋 = 0 1 −4

−1 0 64 −6 0

, 𝑋𝑇 = 0 −1 41 0 −6

−4 6 0 , −𝑋 =

0 −1 41 0 −6

−4 6 0 .

Observed that 𝑋𝑇 = −𝑋. Therefore, 𝑋 is a skew symmetric matrix.

(VIII) Zero Matrix or Null Matrix

A null matrix is a matrix with all its entries zero. For example, 𝑋 = 0 0 00 0 00 0 0

.

2.3 Basic Operations on Matrices

(a) Multiplication of a Matrix by a Constant (Scalar)

Let 𝑐 be a constant and 𝑋 a matrix. The matrix 𝑌 = 𝑐𝑋, is obtained from 𝑋 by

multiplying each entry of 𝑋 by 𝑐.

Example: If 𝐴 = 3 2 11 4 5

, then 𝑐𝐴 = 3𝑐 2𝑐 𝑐𝑐 4𝑐 5𝑐

= 12 8 44 16 20

when

𝑐 = 4.

(b) Addition of Two Matrices

Let 𝐴 and 𝐵 be two matrices. The sum 𝐴 + 𝐵 exists only when the number

of rows of both matrices are the same, and the number of columns of both

matrices are also the same. That is, both matrices have the same order.

When 𝐴 + 𝐵 exists, then the sum is obtained by adding corresponding entries

of 𝐴 and 𝐵 and putting the sum in the corresponding positions of a new matrix

𝐴 + 𝐵.

Example: Let 𝐴 = 2 14 79 1

, 𝐵 = 5 44 12 3

. Then

𝐴 + 𝐵 = 2 + 5 1 + 44 + 4 7 + 19 + 2 1 + 3

= 7 58 8

11 4 .


9 | P a g e

(c) Difference of Two Matrices

The difference of two matrices 𝐴 and 𝐵 which are of the same order is defined

as the sum of the two matrices 𝐴 and (−𝐵) where –𝐵 = −1 ∙ 𝐵. Thus if

𝐴 =

𝑎11 ⋯ 𝑎1𝑛

⋮ ⋱ ⋮𝑎𝑚1 ⋯ 𝑎𝑚𝑛

and 𝐵 = 𝑏11 ⋯ 𝑏1𝑛

⋮ ⋱ ⋮𝑏𝑚1 ⋯ 𝑏𝑚𝑛

then

𝐴 − 𝐵 = 𝑎11 − 𝑏11 ⋯ 𝑎1𝑛 − 𝑏1𝑛

⋮ ⋱ ⋮𝑎𝑚1 − 𝑏𝑚1 ⋯ 𝑎𝑚𝑛 − 𝑏𝑚𝑛

.

Example: If 𝐴 = 5 2 13 8 7

and 𝐵 = 6 1 −39 7 5

, evaluate (i) 𝐴 − 𝐵 (ii)

4𝐴 − 3𝐵.

Solution:

𝐴 − 𝐵 = 5 − 6 2 − 1 1 − (−3)3 − 9 8 − 7 7 − 5

= −1 1 4−6 1 2

.

4𝐴 − 3𝐵 = 20 − 18 8 − 3 4 + 912 − 27 32 − 21 28 − 15

= 2 5 13

−15 11 13 .

Note: 𝐴 + 𝐵 = 𝐵 + 𝐴, and 𝐵 − 𝐴 = − 𝐴 − 𝐵 ≠ 𝐴 − 𝐵.

(d) Product of Two Matrices

Let 𝐴 and 𝐵 be two matrices. The condition that the product 𝐴𝐵 (in that order)

exists is that the number of columns of 𝐴 must be equal to the number of

rows of 𝐵. When the product 𝐴𝐵 exists, then the entry in row 𝑟 and column 𝑠

of 𝐴𝐵 is obtained as the scalar product of the entries in row 𝑟 of 𝐴 and those

in the column 𝑠 of 𝐵. For instance, if 𝐴 = 𝑎11 𝑎12

𝑎21 𝑎22 , 𝐵 =

𝑏11 𝑏12 𝑏21 𝑏22

𝑏13

𝑏23 .

Let 𝐴𝐵 = 𝐶 = 𝑐11 𝑐12 𝑐13

𝑐21 𝑐22 𝑐23 then the matrices 𝐴 and 𝐵 are comfortable for

product where 𝑐11 = 𝑎11 𝑎12 𝑏11

𝑏21 = 𝑎11𝑏11 + 𝑎12𝑏21

𝑐12 = 𝑎11 𝑎12 𝑏12

𝑏22 = 𝑎11𝑏12 + 𝑎12𝑏22

𝑐13 = 𝑎11 𝑎12 𝑏13

𝑏23 = 𝑎11𝑏13 + 𝑎12𝑏23

𝑐21 = 𝑎21 𝑎22 𝑏11

𝑏21 = 𝑎21𝑏11 + 𝑎22𝑏21

𝑐22 = 𝑎21 𝑎22 𝑏12

𝑏22 = 𝑎21𝑏12 + 𝑎22𝑏22


10 | P a g e

𝑐23 = 𝑎21 𝑎22 𝑏13

𝑏23 = 𝑎21𝑏13 + 𝑎22𝑏23.

Note: If 𝐴 is an 𝑚 × 𝑟 matrix and 𝐵 is an 𝑟 × 𝑛 matrix, then the product 𝐴𝐵 is

an 𝑚 × 𝑛 matrix.

Examples

1. If 𝐴 = 6 1 45 3 27 0 5

and 𝐵 = 2 48 61 7

, evaluate 𝐴𝐵.

Solution

𝐴𝐵 =

6 2 + 1 8 + 4 (1) 6 4 + 1 6 + 4 (7) 5 2 + 3 8 + 2 (1) 5 4 + 3 6 + 2 (7) 7 2 + 0 8 + 5 (1) 7 4 + 0 6 + 5 (7)

= 24 5836 5219 63

.

2. Find the value of 𝑎 and 𝑏 so that 2 7

−3 0

𝑎 𝑏1 5

= 11 29−6 9

.

Solution

2 7

−3 0

𝑎 𝑏1 5

= 2𝑎 + 7 2𝑏 + 35−3𝑎 −3𝑏

= 11 29−6 9

⇒ 2𝑎 + 7 = 11, 2𝑏 + 35 = 29, − 3𝑎 = −6, − 3𝑏 = 9

⇒ 𝑎 = 2 and 𝑏 = −3.

2.4 Determinants

The determinant of a square matrix 𝑚 × 𝑛 is a specific number associated

with the matrix, and is usually denoted by 𝐴 or 𝑑𝑒𝑡 𝐴 . The determinant may be

positive, negative or zero. A square matrix 𝐴 is called singular matrix if its

determinant 𝐴 = 0.

Given matrix 𝐴 = 𝑎11 𝑎12

𝑎21 𝑎22 , 𝐴 can be obtained as

𝐴 = 𝑎11 𝑎12

𝑎21 𝑎22 = 𝑎11𝑎22 − 𝑎12𝑎21. Also given a 3 × 3 matrix 𝐴 =

𝑎11 𝑎12 𝑎13

𝑎21 𝑎22 𝑎23

𝑎31 𝑎32 𝑎33

.

The determinant of 𝐴 is given by

𝑑𝑒𝑡 𝐴 =

𝑎11 𝑎12 𝑎13

𝑎21 𝑎22 𝑎23

𝑎31 𝑎32 𝑎33


11 | P a g e

= 𝑎11 𝑎22 𝑎23

𝑎32 𝑎33 − 𝑎12

𝑎21 𝑎23

𝑎31 𝑎33 + 𝑎13

𝑎21 𝑎22

𝑎31 𝑎32

= 𝑎11𝑎22𝑎33 + 𝑎12𝑎23𝑎31 + 𝑎13𝑎21𝑎32 − 𝑎11𝑎23𝑎32 − 𝑎12𝑎21𝑎33 − 𝑎13𝑎22𝑎31.

Examples

1. If 𝐴 = 2 −35 4

then 𝐴 = 2 4 − −3 5 = 23.

2. If 𝑋 = −2 28 4

then 𝑋 = −2 4 − 2 8 = −24.

3. Given that 𝑌 = 1 0 −24 6 12 −3 1

, find 𝑌 .

Solution 𝑌 = +1 6 1

−3 1 − 0

4 12 1

+ −2 4 62 −3

= 6 − −3 − 0 − 2 −12 − 12

= 9 + 48 = 57

4. Given that 𝑥 + 1 1−1 𝑥 − 1

= 4, find the value of 𝑥.

Solution

𝑥 + 1 𝑥 − 1 − −1 1 = 4

⇒ 𝑥2 − 𝑥 + 𝑥 − 1 + 1 = 4

⇒ 𝑥2 = 4

⇒ 𝑥 = ±2

2.5 Minor, Cofactor and Adjoint of a Matrix

I. A minor of a matrix 𝐴 is any square submatrix of 𝐴. For example, if

𝐴 = 1 2 34 5 67 8 9

, then 1 24 5

, 1 34 6

, 5 68 9

, 4 57 8

are submatrices of 𝐴.

Each of them is therefore a minor of 𝐴. Some particular types of minors of a

matrix 𝐴 are obtained by deleting from 𝐴 its 𝑟th row and the 𝑠th column. The

determinant of such a minor is denoted by 𝐴𝑟𝑠 .

Example 1: Given the matrix 𝐴 = 1 2 34 5 67 8 9

, find the determinant of all 2 × 2

minors of 𝐴.

Solution: The (1, 1) minor is obtained by deleting from 𝐴 the first row and the

first column. The determinants are as follows:


12 | P a g e

𝐴11 = 5 68 9

= 45 − 48 = −3

𝐴12 = 4 67 9

= 36 − 42 = −6

𝐴13 = 4 57 8

= 32 − 35 = −3

𝐴21 = 2 38 9

= 18 − 24 = −6

𝐴22 = 1 37 9

= 9 − 21 = −12

𝐴23 = 1 27 8

= 8 − 14 = −6

𝐴31 = 2 35 6

= 12 − 15 = −3

𝐴32 = 1 34 6

= 6 − 12 = −6

𝐴33 = 1 24 5

= 5 − 8 = −3.

II. Cofactor of a Matrix

The (𝑟, 𝑠) cofactor of a matrix is the determinant of the matrix obtained by

deleting the 𝑟th and the 𝑠th column of 𝐴 and then multiplying the result by

−1 𝑟+𝑠. Let 𝐶𝑟𝑠 denote the (𝑟, 𝑠) cofactor. Then 𝐶𝑟𝑠 = −1 𝑟+𝑠𝐴𝑟𝑠 .

Example 2: Find the cofactor of 𝐴 in the example 1 above.

Solution: Using 𝐶𝑟𝑠 = −1 𝑟+𝑠𝐴𝑟𝑠 , we have the following.

𝐶11 = −1 1+1𝐴11 = −3

𝐶12 = −1 1+2𝐴12 = 6

𝐶13 = −1 1+3𝐴13 = −3

𝐶21 = −1 2+1𝐴21 = 6

𝐶22 = −1 2+2𝐴22 = −12

𝐶23 = −1 2+3𝐴22 = 6

𝐶31 = −1 3+1𝐴31 = −3

𝐶32 = −1 3+2𝐴32 = 6

𝐶33 = −1 3+3𝐴33 = −3.

Therefore, the cofactor of 𝐴 is given by


13 | P a g e

𝐶 = −3 6 −36 −12 6

−3 6 −3 .

III. The Adjoint of a Matrix

The Adjoint of a matrix 𝐴 is the transpose of its matrix of cofactors. Let 𝐴 be a

matrix whose matrix of cofactor is 𝐶. Then 𝐴𝑑𝑗 𝐴 = 𝐶𝑇. Thus in the example

two above,

𝐴𝑑𝑗 𝐴 = −3 6 −36 −12 6

−3 6 −3

𝑇

= −3 6 −36 −12 6

−3 6 −3 .

2.6 The Inverse of a Matrix 𝑨 𝑨−𝟏

Let 𝐴 be an 𝑛 × 𝑛. Let 𝐼𝑛 be the 𝑛 × 𝑛 identity matrix. Then 𝐴𝐼 = 𝐼𝐴 = 𝐴.

Furthermore, 𝐴−1𝐴 = 𝐴𝐴−1 = 𝐼. Note that we do not write 𝐴−1 as 1

𝐴 since

1

𝐴 has no

meaning in the theory of matrices.

Let 𝐴 be an 𝑚 × 𝑛 matrix such that 𝐴−1 exist. Then, the expression for the

inverse of matrix 𝐴 is given as:

𝐴−1 =𝐴𝑑𝑗 𝐴

𝐴 provided 𝐴 ≠ 0.

Matrix 𝐴 is said to be singular when 𝐴 = 0 and so inverse not exist. While matrix 𝐴

is non singular when 𝐴 ≠ 0.

Examples

1. Find 𝑋−1 if 𝑋 = 8 43 1

.

Solution:

𝑋 = 8 43 1

= 8 − 12 = −4.

Cofactors are:

𝐶11 = −1 1+1𝐴11 = 1 𝐶12 = −1 1+2𝐴12 = −3

𝐶21 = −1 2+1𝐴21 = −4 𝐶22 = −1 2+2𝐴22 = 8.

Then 𝐶 = 1 −3

−4 8

𝐴𝑑𝑗 𝑋 = 1 −3

−4 8

𝑇

= 1 −4

−3 8


14 | P a g e

∴ 𝑋−1 =𝐴𝑑𝑗 𝑋

𝑋 =

1 −4

−3 8

−4=

−1

41

3

4−2

.

Check: 𝑿𝑿−𝟏 = 𝑰𝟐. As exercise.

2. If 𝐴 = 2 1 01 3 23 1 −1

, find 𝐴−1.

Solution:

𝐴 = 2 3 21 −1

− 1 1 23 −1

+ 0 1 33 1

= 2 −3 − 2 — −1 − 6 + 0 = −3

Cofactor of matrix 𝐴:

𝐶 =

+

3 21 −1

− 1 23 −1

+ 1 33 1

− 1 01 −1

+ 2 03 −1

− 2 13 1

+ 1 03 2

− 2 01 −1

+ 2 11 3

= −5 7 −81 −2 12 −4 5

𝐴𝑑𝑗 𝐴 = 𝐶𝑇 = −5 1 27 −2 −4

−8 1 5

∴ 𝐴−1 =𝐴𝑑𝑗 𝐴

𝐴 =

−5 1 27 −2 −4

−8 1 5

−3=

1

3

5 −1 −2−7 2 48 −1 −5

.

Check that 𝐴𝐴−1 = 𝐼3

2.7 SYSTEMS OF LINEAR EQUATIONS

A system of linear equations is of the form:

𝑎11𝑥1 + 𝑎12𝑥2 + ⋯ + 𝑎1𝑛𝑥𝑛 = 𝑏1

𝑎21𝑥1 + 𝑎22𝑥2 + ⋯ + 𝑎2𝑛𝑥𝑛 = 𝑏2

⋮

𝑎𝑚1𝑥1 + 𝑎𝑚2𝑥2 + ⋯ + 𝑎𝑚𝑛𝑥𝑛 = 𝑏𝑚

The method of solution could be

(a) Solution by direct algebraic method (substitution method or elimination

method)

(b) Solution using the matrix inverse

(c) Solution using Cramer’s rule.


15 | P a g e

For solution by direct algebraic method, it has been discussed in MTS 107 and so

we omit it here.

(b) Solution using Inverse Method

In matrix form:

𝑎11 ⋯ 𝑎1𝑛

⋮ ⋱ ⋮𝑎𝑚1 ⋯ 𝑎𝑚𝑛

𝑥1

⋮𝑥𝑛

= 𝑏1

⋮𝑏𝑚

𝐴𝑋 = 𝐵

𝐴−1𝐴𝑋 = 𝐴−1𝐵

𝐼𝑋 = 𝐴−1𝐵

⇒ 𝑋 = 𝐴−1𝐵 provided 𝐴 ≠ 0

Example: Solve the following equations by inverse method:

1. 𝑥1 + 4𝑥2 = 22

𝑥1 + 𝑥2 = 7

2. 3𝑥1 − 2𝑥2 + 𝑥3 = 2

𝑥1 + 3𝑥2 + 4𝑥3 = 19

5𝑥1 + 4𝑥2 − 3𝑥3 = 4.

Solution:

1. In matrix form:

1 41 1

𝑥1

𝑥2 =

227

Where 𝐴 = 1 41 1

, 𝑋 = 𝑥1

𝑥2 and 𝐵 =

227

.

The solution of which is

𝑥1

𝑥2 =

1 41 1

−1

227

= −1

3

1 −4−1 1

227

= −1

3

−6−15

= 25

That is, 𝑥1 = 2 and 𝑥2 = 5.

2. This can be written in the form

3 −2 11 3 45 4 −3

𝑥1

𝑥2

𝑥3

= 2

194

Or 𝐴𝑋 = 𝐵


16 | P a g e

Hence, 𝑋 = 𝐴−1𝐵 or

𝑥1

𝑥2

𝑥3

= 3 −2 11 3 45 4 −3

−1

2

194

thus we need 𝐴−1.

𝐴 = 3𝐴11 − −2 𝐴12 + (1)𝐴13

= 3 −25 + 2 −23 + (−11)

= −132.

Cof 𝐴 =

+

3 44 −3

− 1 45 −3

+ 1 35 4

− −2 14 −3

+ 3 15 −3

− 3 −25 4

+ −2 13 4

− 3 11 4

+ 3 −21 3

= −25 23 −11−2 −14 −22−11 −11 11

𝐴𝑑𝑗 𝐴 = 𝐶𝑇 = −25 −2 −1123 −14 −11

−11 −22 11

𝐴−1 =𝐴𝑑𝑗 𝐴

𝐴 =

−25 −2 −1123 −14 −11

−11 −22 11

−132=

1

132

25 2 11−23 14 1111 22 −11

Therefore,

𝑥1

𝑥2

𝑥3

= 𝐴−1 2

194

=1

132

25 2 11−23 14 1111 22 −11

2

194

=1

132

50 + 38 + 44−46 + 266 + 4422 + 418 − 44

=1

132 132264392

= 123 .

(c) Solution by Cramer’s rule

1. The 2 × 2 matrix case:

𝑥1 + 4𝑥2 = 22

𝑥1 + 𝑥2 = 7 Or 1 41 1

𝑥1

𝑥2 =

227

i.e 𝐴𝑋 = 𝐵.


17 | P a g e

By Cramer’s rule,

𝑥1 = 22 47 1

1 41 1

, 𝑥2 =

1 221 7

1 41 1

𝑥1 =∆1

∆ , 𝑥2 =

∆2

∆

Where ∆≝ 𝐴 , ∆1 is the determinant obtained by replacing the first column of 𝐴 by

the matrix 227

and ∆2 is the determinant obtained by replacing the second column

of 𝐴 by the matrix 227

. Thus,

∆= 1 41 1

= 1 − 4 = −3

∆1= 22 47 1

= 22 − 28 = −6

∆2= 1 221 7

= 7 − 22 = −15

Hence,

𝑥1 =∆1

∆=

−6

−3= 2 and 𝑥2 =

∆2

∆=

−15

−3= 5 .

Therefore, 𝑥1

𝑥2 =

25 .

2. The (3 × 3) matrix case:

3 −2 11 3 45 4 −3

𝑥1

𝑥2

𝑥3

= 2

194

∆= 3 −2 11 3 45 4 −3

= 132

∆1= 2 −2 1

19 3 44 4 −3

= −132

∆2= 3 2 11 19 45 4 −3

= −264

∆3= 3 −2 21 3 195 4 4

= −396

Hence,

𝑥1 =∆1

∆=

−132

−132= 1

𝑥2 =∆2

∆=

−264

−132= 2


18 | P a g e

𝑥3 =∆3

∆=

−396

−132= 3.

2.8 BUSINESS APPLICATION

Example: There are 30 secondary schools and 60 primary schools in a local

government area in a certain state. Each of the secondary schools and the primary

schools has 1 Messenger (ME), 5 Clerks (CL) and 1 Cashier (CA). Each secondary

school in addition has 1 Accountant (AC) and 1 Head Clerk (HC). The monthly salary

(N’000) of each of them is as follows:

Messenger – N20; Clerk – N40, Cashier – N35, Accountant – N50, and Head clerk –

N60. Use matrices to find

(i) The total number of posts of each kind of primary and secondary schools

taken together.

(ii) The total monthly salary of each primary school and each secondary

school separately.

(iii) The total monthly salary bill of all secondary and primary schools.

Solution: Consider the row matrix of the two types of schools:

𝑋 = 30 60

Representing the number of secondary schools and primary schools respectively,

we have

Let ME CL CA AC HC

𝑌 = 1 5 11 5 1

10

10

This is the matrix of the number of person in each position for each type of

school. Consider the product matrix:

𝑋𝑌 = 30 60 1 5 11 5 1

10

10

= 90 450 90 30 30 .

Then, there are in the two types of schools, 90 Messengers, 450 Clerks, 90

Cashiers, 30 Accountants and 30 Head Clerks.


19 | P a g e

(ii) The monthly salaries (N’000) can be put as a column matrix 𝑍 =

2040355060

.

Total monthly bill of each kind of school is

𝑌𝑍 = 1 5 11 5 1

10

10

2040355060

= 20 + 200 + 35 + 50 + 60

20 + 200 + 35 + 0 + 0

= 365255

Total monthly salary bill for secondary schools is N365,000 while total monthly for

primary schools is N255,000.

(iii) 𝑋 𝑌𝑍 gives the total monthly salary bill for all secondary and primary schools.

𝑋 𝑌𝑍 = 30 60 365255

= 𝑁26,250 (’000)

= 𝑁26,250,000.

Exercise 2

1. Use the following information to answer question (i) – (iv):

𝐴 = 2 4 13 −5 0

and 𝐵 = 6 7 10 0 2

.

(i) Find 𝐴 + 𝐵

(ii) Find 𝐴𝐵

(iii) 𝐴𝑇𝐵

(iv) 𝐴𝑇𝐵 .

2. Evaluate 3 5 18 0 24 0 6

, −10 15

2 2 ,

1 −2 3−4 5 67 8 −9

.

3. Use the following information to answer question (a) – (c):

2 34 5

𝑥𝑦 =

2543

is written as 𝐴𝑋 = 𝐵.

(a) Write down 𝐴 and 𝐵.


20 | P a g e

(b) Find 𝐴−1

(c) Find the matrix 𝑋.

(d) Find 𝐴 − 𝜆𝐼

4. A woman invested different amounts at 8%, 83

4% and 9%, all at simple

interest. Altogether she invested N40,000 and earns N3,455 per year. How

much does she have invested at each rate if she has N4,000 more invested at

9% than 8%? Solve by using matrices.

5. A salesman has the following record of sales during three months for three

items 𝐴, 𝐵 and 𝐶 which have different rates of commission.

Months Sales units Total commission drawn

(in £) A B C

May 90 100 20 800

June 130 50 40 900

July 60 100 30 850

Find out the rates of commission on the items A, B and C. Solve by Cramer’s

rule (determination method)

§ 3.0 STATISTICAL DISTRIBUTION

3.1 BINOMIAL DISTRIBUTION

An experiment consisting of 𝑛 repeated trials such that

a) the trials are independent and identical

b) each trial result in only one or two possible outcomes

c) the probability of success 𝑝 remains constant

d) the random variable of interest is the total number of success.

The binomial distribution is one of the widely used in statistics and it used to

find the probability that an outcome would occur 𝑥 times in 𝑛 performances of an

experiment and its probability density function is given by

𝑓 𝑥; 𝑛, 𝑝 = 𝑛𝑥 𝑝𝑥 1 − 𝑝 𝑛−𝑥 , 𝑥 = 0,1, … , 𝑛

Where 𝑛 = Total number of trials

𝑝 = Probability of success

1 − 𝑝 = Probability of failure


21 | P a g e

𝑛 − 𝑥 = Number of failure in 𝑛 – trials

To find the probability of 𝑥 success in 𝑛 trials, the only values we need are that of 𝑛

and 𝑝.

Properties of Binomial Distribution

Mean 𝜇 = 𝑛𝑝

Variance 𝜎2 = 𝑛𝑝𝑞 ; 𝑞 = 1 − 𝑝

Standard deviation 𝜎 = 𝑛𝑝𝑞

Note: 𝑋~𝐵(𝑛, 𝑝) reads as 𝑋 follows binomial distribution.

Examples:

1. Observation over a long period of time has shown that a particular sales man

can make a sale on a single contact with the probability of 20%. Suppose the

same person contact four prospects,

(a) What is the probability that exactly 2 prospects purchase the product?

(b) What is the probability that at least 2 prospects purchase the product?

(c) What is the expected value (mean) of the prospects that would purchase the

product?

Solution: Let 𝑋 denote the number of prospect.

Let 𝑝 denote the probability of (success) purchase = 0.2

Then 𝑋~𝐵(4,0.2).

𝑓 𝑥 = 𝑛𝑥 𝑝𝑥𝑞𝑛−𝑥 𝑥 = 0,1,2,3,4

= 0, 𝑒𝑙𝑠𝑒𝑤𝑕𝑒𝑟𝑒

𝑓 𝑥 = 4

𝑥 0.2 𝑥 0.8 4−𝑥

(a) 𝑓 𝑥 = 2 = 42 0.2 2 0.8 4−2 = 0.1536

(b) 𝑓 𝑥 ≥ 2 = 𝑓 2 + 𝑓 3 + 𝑓(4) ( OR 1 − 𝑓 𝑥 ≤ 1 = 1 − {𝑓 0 + 𝑓(1)})

= 0.1536 + 43 0.2 3 0.8 4−3 + 4

4 0.2 4 0.8 4−4

= 0.1536 + 0.0256 + 0.0016

= 0.1808

(c) Expected value = 𝑛𝑝

= 4 × 0.2

= 0.8


22 | P a g e

2. The probability that a student is accepted to a prestigious college is 0.3. If 5

students from the same school apply, what is the probability that at most 2 are

accepted?

Solution:

𝑓 𝑥 ≤ 2 = 𝑓 0 + 𝑓 1 + 𝑓(2)

= 5C0 0.3 0 0.7 5−0 + 5C1 0.3 1 0.7 5−1 + 5C2 0.3 2 0.7 5−2

= 0.1681 + 0.3602 + 0.3087

= 0.837

3.2 POISSON DISTRIBUTION

When the size of the sample (𝑛) is very large and the probability of obtaining

success in any one trial very small, then Poisson distribution is adopted. Given an

interval of real numbers, assumed counts of occur at random throughout interval, if

the interval can be partition into sub interval of small enough length such that

a) The probability of more than one count sub interval is 0

b) The probability of one count in a sub interval is the same for all sub intervals

and proportional to the length of the sub interval

c) The count in each of the sub interval is independent of all other sub intervals.

A random experiment of this type is called a Poisson Process. If the mean

number of count in an interval is 𝜆 > 0, the random variable 𝑥 that equals the number

of count in an interval has a Poisson distribution with parameter 𝜆 and the probability

density function is given by

𝑓 𝑥; 𝜆 =𝜆𝑥𝑒−𝜆

𝑥 !; 𝑥 = 0,1, … , 𝑛.

For a Poisson distribution, the mean and the variance is given by 𝐸 𝑥 = 𝜆, 𝑉 𝑥 = 𝜆.

Poisson distribution probability density function can also be used to approximate

binomial distribution when 𝑛 is large (𝑛 ≥ 30) and 𝑛𝑝 < 5.

Examples

1. Attendance in a factory shows 7 absences. What is the probability that on a

given day there will be more than 8 people absent?

Solution

𝑃 𝑋 > 8 = 1 − 𝑃 𝑋 ≤ 8


23 | P a g e

= 1 − 𝑃 1 + 𝑃 2 + 𝑃 3 + 𝑃 4 + 𝑃 5 + 𝑃 6 + 𝑃 7 + 𝑃(8)

= 1 − 71𝑒−7

1!+

72𝑒−7

2!+

73𝑒−7

3!+

74𝑒−7

4!+

75𝑒−7

5!+

76𝑒−7

6!+

77𝑒−7

7!+

78𝑒−7

8!

= 1 − 0.0064 + 0.0223 + 0.0521 + 0.0912 + 0.1277 + 0.149

+0.149 + 0.1304

= 0.2709

2. An automatic production line breaks down every 2 hours. Special production

requires uninterrupted operation for 8 hours. What is the probability that this

can be achieved?

Solution 𝜆 =8

2= 4, 𝑥 = 0 (no breaks down)

𝑃 𝑥 = 0 =40𝑒−4

0! = 0.0183 = 1.83%

3.3 NORMAL DISTRIBUTION

The normal distribution is the most important and the most widely used

among all continuous distribution in the statistics. The graph of a normal distribution

is a bell – shaped curved that extends indefinitely in both direction.

Normal curve

1 or 100%

𝜇

3.3.1 Features (Properties) of Normal Curve

1. The curve is symmetrical about the vertical axis through the mean 𝜇.

2. The mode is the highest point on the horizontal axis where the curve is

maximum and occurs where 𝑥 = 𝜇.

3. The normal curve approaches the horizontal axis asymptotically.

4. The total area under the curve is one (1) or 100%.

It is clear from these properties that a knowledge of the population means and

standard deviation gives a complete picture of the distribution of all the values.


24 | P a g e

Notation: Instead of saying that the values of a variable 𝑥 are normally distributed

with mean 𝜇 and standard deviation 𝜎, we simply say that 𝑥 has an 𝑁 𝜇, 𝜎2 or 𝑥 is

𝑁 𝜇, 𝜎2 or 𝑋~𝑁 𝜇, 𝜎2 .

A random variable 𝑥 is said to have a normal distribution if its probability

density function is given by

𝑓 𝑥; 𝜇, 𝜎 =1

2𝜋𝜎 𝑒−1

2 𝑥−𝜇

𝜎

2

, −∞ < 𝑥 < ∞

Where 𝜎 and 𝜇 are the parameters of the distribution (standard deviation and mean

respectively).

3.3.2 The Standard Normal Curve

The standard normal distribution is a special case. If 𝑋~𝑁 0,1 , then 𝑥 is

called the standard normal variable with probability density function 𝑓 𝑧 =

1

2𝜋𝑒−𝑧2

2 , −∞ < 𝑧 < ∞.

The table usually used to determine the probability that a random variable 𝑥

drawn from a normal population with no mean and standard deviation 1 is the

standard normal distribution table (or 𝑧- scores table).

We shall note the following point when using the table:

1. 𝑃 𝑧 ≤ 0 = 0.5

2. 𝑃 𝑧 ≤ −𝑧 = 1 − 𝑃 𝑧 ≤ 𝑧

= 𝑃(𝑧 ≥ 𝑧)

3. 𝑃 −𝑧1 < 𝑧 < 𝑧2 = 𝑃 𝑧 < 𝑧2 − 𝑃 𝑧 > −𝑧1

= 𝑃 𝑧 < 𝑧2 − [1 − 𝑃 𝑧 < 𝑧1 ]

= 𝑃 𝑧 < 𝑧2 + 𝑃 𝑧 < 𝑧1 − 1

Example 1: Find the probability that a random variable having the standard normal

distribution will take a value

(a) Less than 1.53 (b) Less than –0.82

(c) Between 1.25 and 2.34 (d) between –0.35 and 1.26

Solution:

(a) 𝑃 𝑧 < 1.53 = 0.9370 (using the table)

(b) 𝑃 𝑧 < −0.82 = 0.2061


25 | P a g e

(c) 𝑃 1.25 ≤ 𝑧 ≤ 2.34 = 𝑃 𝑧 ≤ 2.34 − 𝑃 𝑧 ≥ 1.25

= 𝑃 𝑧 ≤ 2.34 − 1 − 𝑃(𝑧 < 1.25)

= 0.9904 − 1 + 0.8944

= 0.8848

(d) 𝑃 −0.35 ≤ 𝑧 ≤ 1.26 = 𝑃 𝑧 ≤ 1.26 − 𝑃 𝑧 ≥ −0.35

= 𝑃 𝑧 ≤ 1.26 − 1 − 𝑃(𝑧 < 0.35)

= 𝑃 𝑧 ≤ 1.26 + 𝑃 𝑧 < 0.35 − 1

= 0.8962 + 0.6368 − 1

= 0.533

3.3.3 STANDARDIZED NORMAL VARIABLE

In real world application, the given continuous random variable may have a

normal distribution in value of a mean and standard deviation different from 0 and 1.

To overcome this difficulty, we obtain a new variable denoted by 𝑧 and this is given

by

𝑧 =Value 𝑥−Mean

Standard deviation =

𝑋−𝜇

𝜎 (for 𝜇 ≠ 0 and 𝜎 ≠ 1)

Example 2: If 𝑋~𝑁(510𝑔, 6.25𝑔), calculate 𝑃(𝑋 = 507.5𝑔)

Solution: 𝑧 =𝑋−𝜇

𝜎

=507.5−510

2.5

= −1

∴ 𝑃 𝑋 = 507.5𝑔 = 𝑃 𝑧 ≤ −1

= 0.1587

Example 3: It is known from the previous examination results that the marks of

candidates have a normal distribution with mean 55 and standard deviation 10. If the

pass mark in a new examination is set at 40, what percentages of the candidates will

be expected to fail?

Solution: We have 𝑥~𝑁(55,100) and the required proportion of 𝑥 −values that are

below 40 is 𝑃(𝑥 < 40).

𝑃 𝑥 < 40 = 𝑃(𝑧 <𝑋−𝜇

𝜎)


26 | P a g e

= 𝑃 𝑧 <40−55

10

= 𝑃(𝑧 ≤ −1.5)

= 0.0668

∴ 6.68% of the candidates are expected to fail.

Exercise 3

1. A surgery is successful for 75% patients. What is the probability of its success

in at least 7 cases out of randomly selected 9 patients?

2. If the 5% of the electric bulbs manufactured by a company are defective. Find

the probability that in the sample of 130 bulbs, at most 3 doors are defective.

a) Use binomial to solve the problem

b) Use Poisson distribution and compare your results.

3. It is known that the marks in a University direct entry examination are normally

distributed with mean 70 and standard deviation 8. Given that your score is

69, what percentage of all the candidates will be expected to score more than

you?

§ 4.0 LIMIT AND CONTINUITYOF FUNCTIONS

4.1 Definition

A given function 𝑓(𝑥) is said to have a limit 𝐿 as 𝑥 approaches 𝑐 (in symbol

lim𝑥→𝑐 𝑓(𝑥) = 𝐿) if and only if, 𝑓(𝑥) is as near to 𝐿 as we please for all values of 𝑥 ≠ 𝑐

but sufficiently near to 𝑐.

We say that 𝑓(𝑥) approaches the limit in real number as 𝑥 approaches a fixed

number 𝑐, if and only if, lim𝑥→𝑐− 𝑓(𝑥) = 𝐿 = lim𝑥→𝑐+ 𝑓(𝑥). When this is so, we

write lim𝑥→𝑐 𝑓(𝑥) = 𝐿. Alternatively, this simply means that 𝑓(𝑥) is arbitrary close to 𝐿

whenever the variable 𝑥 is sufficiently near to the number 𝑐.

Remark: A function 𝑓(𝑥) cannot have more than one limit.

4.2 Rules for evaluating limits

If lim𝑥→𝑐 𝑓(𝑥) and lim𝑥→𝑐 𝑔(𝑥) exist, then the following rules hold.

(i) lim𝑥→𝑐 𝑘 = 𝑘 where 𝑘 is a constant.


27 | P a g e

(ii) lim𝑥→𝑐 𝜆𝑓 𝑥 = 𝜆 lim𝑥→𝑐 𝑓(𝑥) for any 𝜆 real number

(iii) lim𝑥→𝑐 𝑓(𝑥) ± 𝑔(𝑥) = lim𝑥→𝑐 𝑓(𝑥) ± lim𝑥→𝑐 𝑔(𝑥)

(iv) lim𝑥→𝑐 𝑓(𝑥)

𝑔(𝑥) =

lim 𝑥→𝑐 𝑓(𝑥)

lim 𝑥→𝑐 𝑔(𝑥) , provided lim𝑥→𝑐 𝑔(𝑥) ≠ 0.

Example: Evaluate the following limits:

(1) lim𝑥→0(𝑥3 − 5𝑥2 + 2) (2) lim𝑥→0 9+𝑥−3

𝑥 (3) lim𝑥→2

𝑥2−𝑥−2

𝑥2−4 (4) lim𝑥→∞

4𝑥3

𝑥3+3 .

Solution

(1) lim𝑥→0(𝑥3 − 5𝑥2 + 2) = lim𝑥→0

𝑥3 − lim𝑥→0

5𝑥2 + lim𝑥→0

2

= 0 2 − 5 02 + 2

= 2

(2) lim𝑥→0 9+𝑥−3

𝑥= lim

𝑥→0

9+𝑥−3

𝑥×

9+𝑥+3

9+𝑥+3

= lim𝑥→0

𝑥

𝑥 9+𝑥+3

= lim𝑥→0

1

9+𝑥+3

=1

9+3

=1

6

(3) lim𝑥→2𝑥2−𝑥−2

𝑥2−4= lim𝑥→2

𝑥+1 𝑥−2

𝑥+2 𝑥−2

= lim𝑥→2

𝑥+1

𝑥+2

=lim𝑥→2

(𝑥+1)

lim𝑥→2

(𝑥+2)

=3

4

(4) lim𝑥→∞4𝑥3

𝑥3+3= lim

𝑥→∞

4𝑥3

𝑥3

𝑥3

𝑥3+3

𝑥3

= lim𝑥→∞

4

1+3

𝑥3

=lim𝑥→∞

4

lim𝑥→∞

1+lim3

𝑥3𝑥→∞

= 4


28 | P a g e

4.3 CONTINUITY OF FUNCTION

We say that the function 𝑓 is continuous at 𝑥 = 𝑐 if lim𝑥→𝑐 𝑓(𝑥) = 𝑓(𝑐). We say

that 𝑓 is continuous if it is continuous at each point of its domain.

Remark:

(1) The continuity definition requires that the following conditions be met if 𝑓 is to

be continuous at 𝑐 (a point): (a) 𝑓(𝑐) is defined as a finite real number, (b)

lim𝑥→𝑐− 𝑓(𝑥) exists and equals 𝑓(𝑐), (c) lim𝑥→𝑐− 𝑓(𝑥) = f(c) = lim𝑥→𝑐+ 𝑓(𝑥) .

When a function 𝑓 is not continuous at 𝑐, one or more, of these conditions are

not met.

(2) All polynomials, sin 𝑥, cos 𝑥 , 𝑒𝑥 , sinh 𝑥, cosh 𝑥, 𝑏𝑥 , 𝑏 ≠ 1 are continuous for all

real values of 𝑥. All logarithmic functions, log𝑏 𝑥 , 𝑏 > 0, 𝑏 ≠ 1 are continuous

for all 𝑥 > 0. Each rational function, 𝑝(𝑥)/𝑞(𝑥), is continuous where 𝑞(𝑥) ≠ 0.

Examples

1. Determine whether the function 𝑓 𝑥 = 3𝑥2 + 4𝑥 − 10 is continuous at 𝑥 = 3

or not.

Solution:

𝑓 3 = 3 32 + 4 3 − 10 = 29.

lim𝑥→3 𝑓 𝑥 = lim𝑥→3 3𝑥2 + lim𝑥→3 4𝑥 − lim𝑥→3 10

= 27 + 12 − 10

= 29

∴ lim𝑥→3

𝑓 𝑥 = 𝑓 3 = 29

𝑓 𝑥 is continuous at 𝑥 = 3.

2. Verify the continuity of the function 𝑓 𝑥 =𝑥−2

𝑥2−5𝑥+6 at 𝑥 = 2.

Solution:

𝑓 𝑥 =𝑥−2

𝑥2−5𝑥+6=

𝑥−2

𝑥−2 𝑥−3 =

1

𝑥−3

𝑓 2 =1

2−3= −1

lim𝑥→2 𝑓 𝑥 = lim𝑥→31

𝑥−3

=lim 𝑥→2 1

lim 𝑥→2 𝑥−3


29 | P a g e

=1

2−3= −1

∴ lim𝑥→2

𝑓 𝑥 = 𝑓 2 = −1

𝑓 𝑥 is continuous at 𝑥 = 2.

3. The function 𝑓 𝑥 =𝑥2−𝑥−6

𝑥−3 is not continuous at 𝑥 = 3 since 𝑓 3 is not

defined.

Exercise 4

1. Evaluate each of the following limits.

(a) lim𝑥→1𝑥2−1

𝑥3−1

(b) lim𝑥→∞𝑥2−𝑥

𝑥3+2𝑥

(c) lim𝑥→4 𝑥−2

𝑥−4

2. Is function 𝑓 𝑥 =2𝑥2+3

2𝑥+1 continuous at the point 𝑥 = −

1

2?

§ 5.0 BASIC CONCEPT OF DIFFERENTIATION

Differentiation is a mathematical concept dealing with the rate of change. This

rate of change is generally termed the derivative, slope, gradient or marginal

measure in various areas of usage. That is, the process of finding the derivative of a

function is known as differentiation. Formally, differentiation (the differential calculus)

is the process of finding the derivative of a function. And it has a lot of applications in

many areas of human endeavours such as the field of engineering, science,

economics, business, and so on.

5.1 Determination of Derivative from First Principle

There is need to apply the limits theory to the incremental principle of the

variables involved in the function. For instance, let us consider the function 𝑦 = 𝑓(𝑥).

A small increment in 𝑥 (denoted by ∆𝑥) will cause a small increment in 𝑦 (also

denoted by ∆𝑦). Then for

𝑦 = 𝑓(𝑥) ……………………. (i)

We have 𝑦 + ∆𝑦 = 𝑓 𝑥 + ∆𝑥 ………… (ii)


30 | P a g e

Subtract equation (i) from equation (ii), gives

𝑦 + ∆𝑦 − 𝑦 = 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥)

∆𝑦 = 𝑓 𝑥 + ∆𝑥 − 𝑓(𝑥)

Divide all through by ∆𝑥, we have

Δ𝑦

Δ𝑥=

𝑓 𝑥+∆𝑥 −𝑓(𝑥)

∆𝑥 ……………. (iii)

As ∆𝑥 → 0, we have

lim∆𝑥→0Δ𝑦

Δ𝑥= lim∆𝑥→0


∆𝑥 …. (iv)

By the limit’s theory, we have

𝑑𝑦

𝑑𝑥= lim∆𝑥→0


∆𝑥 ……….. (v) (i.e.

𝑑𝑦

𝑑𝑥≡ lim∆𝑥→0

Δ𝑦

Δ𝑥)

Note that 𝑑𝑦

𝑑𝑥 is also represented by 𝑓 ′ (𝑥) which is called the derivative of 𝑓(𝑥).

Others are 𝑑

𝑑𝑥 𝑓 ,

𝑑𝑓

𝑑𝑥, 𝑦′ .

If the left hand side of the equation (v) exists, then 𝑑𝑦

𝑑𝑥 is called the differential

coefficient of 𝑦 with respect to 𝑥.

Example: Find the derivative of (i) 𝑦 = 𝑥3 + 2 (ii) 𝑦 = 𝑎𝑥𝑛 from the first principle.

Solution:

(i) 𝑦 = 𝑥3 + 2 ------------- (*)

𝑦 + ∆𝑦 = 𝑥 + ∆𝑥 3 + 2 -------------- (**)

Then subtracting equation (*) from equation (**), we have

𝑦 + ∆𝑦 − 𝑦 = 𝑥 + ∆𝑥 3 + 2 − 𝑥3 + 2

∆𝑦 = 𝑥3 + 3𝑥2 ∆𝑥 + 3𝑥 ∆𝑥 2 + ∆𝑥 3 + 2 − 𝑥3 − 2

∆𝑦 = 3𝑥2 ∆𝑥 + 3𝑥 ∆𝑥 2 + ∆𝑥 3

Divide through by ∆𝑥, we have

Δ𝑦

Δ𝑥= 3𝑥2 + 3𝑥 ∆𝑥 + ∆𝑥 2

limΔ𝑥→0

Δ𝑦

Δ𝑥= lim

Δ𝑥→0 3𝑥2 + 3𝑥 ∆𝑥 + ∆𝑥 2

∴𝑑𝑦

𝑑𝑥= 3𝑥2 + 0 + 0

= 3𝑥2


31 | P a g e

(ii) Given 𝑦 = 𝑎𝑥𝑛 .

Let 𝑦 + ∆𝑦 = 𝑎 𝑥 + ∆𝑥 𝑛

Then

𝑦 + ∆𝑦 − 𝑦 = 𝑎 𝑥 + ∆𝑥 𝑛 − 𝑎𝑥𝑛

∆𝑦 = 𝑎𝑥𝑛 1 +∆𝑥

𝑥

𝑛

− 1 Using 1 + 𝑥 𝑞 = 1 + 𝑞 𝑞−1 𝑞−2 …(𝑞−𝑟+1)

𝑟 !∞𝑟=1 𝑥𝑟

∆𝑦 = 𝑎𝑥𝑛 1 + 𝑛 ∆𝑥

𝑥 +

𝑛(𝑛 − 1)

2! ∆𝑥

𝑥

2

+ ⋯ − 1

= 𝑎 𝑛𝑥𝑛−1 ∆𝑥 +𝑛(𝑛−1)

2!𝑥𝑛−2 ∆𝑥 2 + ⋯

∴Δ𝑦

Δ𝑥= 𝑎 𝑛𝑥𝑛−1 + terms containing higher power of ∆𝑥

limΔ𝑥→0

Δ𝑦

Δ𝑥= lim

∆𝑥→0 𝑎𝑛𝑥𝑛−1 + terms containing higher power of ∆𝑥

∴𝑑𝑦

𝑑𝑥= 𝑎 𝑛𝑥𝑛−1 ……………………. (vi)

We note that equation (vi) is the derivative of 𝒚 = 𝒂𝒙𝒏 and it shall be applied as

a rule when the derivative from first principle is not required.

Example: (a) Obtain the derivative for the following. (i) 𝑦 = 𝑥4 (ii) 𝑦 = 4𝑥5.

(b) Obtain the value of the derivative for the following functions at 𝑥 = 2: (i) 𝑦 = 2𝑥3

(ii) 𝑦 = 3𝑥6.

Solution:

(a) (i) 𝑑𝑦

𝑑𝑥= 4 𝑥4−1 = 4𝑥3

(ii) 𝑑𝑦

𝑑𝑥= 4 5 𝑥5−1 = 20𝑥4

(b) (i) 𝑑𝑦

𝑑𝑥= 6𝑥2

At 𝑥 = 2, 𝑑𝑦

𝑑𝑥= 6 22 = 24

(ii) 𝑑𝑦

𝑑𝑥= 18𝑥5

At 𝑥 = 2,𝑑𝑦

𝑑𝑥= 18 25 = 576


32 | P a g e

5.2 Rules of Differentiation

The following rules shall be applied to various functions:

Rule 1: If a polynomial 𝑦 = 𝑢 ± 𝑣 ± 𝑤 ± ⋯ where 𝑢, 𝑣, 𝑤 are functions of 𝑥, then

𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥 𝑢 ± 𝑣 ± 𝑤 ± ⋯

=𝑑𝑢

𝑑𝑥±

𝑑𝑣

𝑑𝑥±

𝑑𝑤

𝑑𝑥± ⋯

Hence, the derivative of a sum is the of the derivatives and the derivative of a

difference is the difference of the derivatives.

Example: Find the derivative of each of the following functions.

(i) 𝑦 = 2𝑥5 − 3𝑥6 + 6𝑥2; (ii) 𝑓 𝑥 = 2𝑥3 + 3𝑥4 −2

𝑥 .

Solution:

(i) 𝑑𝑦

𝑑𝑥= 2 5 𝑥5−1 − 3 6 𝑥6−1 + 6(2) 𝑥2−1

= 10𝑥4 − 18𝑥5 + 12𝑥

(ii) 𝑑𝑓

𝑑𝑥= 2 3 𝑥3−1 + 3 4 𝑥4−1 − 2 −1 𝑥−1−1

= 6𝑥2 + 12𝑥3 +2

𝑥2 .

Rule 2: If 𝑦 = 𝐶, where 𝐶 is a constant, then 𝑑𝑦

𝑑𝑥= 0. E.g. (a) 𝑦 = 2,

𝑑𝑦

𝑑𝑥= 0.

(b) 𝑦 = 𝑎2 , 𝑑𝑦

𝑑𝑥= 0 since 𝑎 is a constant and has no degree of 𝑥 from 1 and above.

Rule 3: If a product 𝑦 = 𝑢𝑣, where 𝑢 and 𝑣 are functions of 𝑥, then

𝑑𝑦

𝑑𝑥= 𝑣

𝑑𝑢

𝑑𝑥+ 𝑢

𝑑𝑣

𝑑𝑥 .

By extension, if 𝑦 = 𝑢𝑣𝑤, where 𝑢, 𝑣 and 𝑤 are functions of 𝑥, then

𝑑𝑦

𝑑𝑥= 𝑣𝑤

𝑑𝑢

𝑑𝑥+ 𝑢𝑤

𝑑𝑣

𝑑𝑥+ 𝑣𝑢

𝑑𝑤

𝑑𝑥 .

Example: Find the derivative of the following functions:

(a) 𝑦 = 2𝑥 + 2 3𝑥2 + 2𝑥 (b) 𝑦 = 2𝑥 + 4 𝑥2 + 3𝑥 2𝑥3 + 5𝑥 .


33 | P a g e

Solution:

(a) Let 𝑢 = 2𝑥 + 2, 𝑑𝑢

𝑑𝑥= 2

𝑣 = 3𝑥2 + 2𝑥, 𝑑𝑣

𝑑𝑥= 6𝑥 + 2 .

By product rule,

𝑑𝑦

𝑑𝑥= 𝑣

𝑑𝑢

𝑑𝑥+ 𝑢

𝑑𝑣

𝑑𝑥

= 3𝑥2 + 2𝑥 2 + 2𝑥 + 2 6𝑥 + 2

= 6𝑥2 + 4𝑥 + 12𝑥2 + 4𝑥 + 12𝑥 + 4

= 18𝑥2 + 20𝑥 + 4 .

(b) Let 𝑢 = 2𝑥 + 4, 𝑑𝑢

𝑑𝑥= 2

𝑣 = 𝑥2 + 3𝑥, 𝑑𝑣

𝑑𝑥= 2𝑥 + 3

𝑤 = 2𝑥3 + 5𝑥, 𝑑𝑤

𝑑𝑥= 6𝑥2 + 5

But 𝑑𝑦

𝑑𝑥= 𝑣𝑤

𝑑𝑢

𝑑𝑥+ 𝑢𝑤

𝑑𝑣

𝑑𝑥+ 𝑣𝑢

𝑑𝑤

𝑑𝑥

= 𝑥2 + 3𝑥 2𝑥3 + 5𝑥 2 + 2𝑥 + 4 2𝑥3 + 5𝑥 2𝑥 + 3

+ 𝑥2 + 3𝑥 2𝑥 + 4 6𝑥2 + 5

Rule 4: If quotient 𝑦 =𝑢

𝑣, where 𝑢 and 𝑣 are functions of 𝑥, then

𝑑𝑦

𝑑𝑥=

𝑣 𝑑𝑢

𝑑𝑥 − 𝑢

𝑑𝑣

𝑑𝑥

𝑣2 .

Example: Differentiate the following with respect to 𝑥.

(i) 𝑦 =1+𝑥2

3𝑥 (ii) 𝑦 =

2+3𝑥

𝑥2+3𝑥+5

Solution:

(i) Let 𝑢 = 1 + 𝑥2 , 𝑑𝑢

𝑑𝑥= 2𝑥

𝑣 = 3𝑥, 𝑑𝑣

𝑑𝑥= 3

By Quotient Rule:

𝑑𝑦

𝑑𝑥=

𝑣 𝑑𝑢

𝑑𝑥 − 𝑢

𝑑𝑣

𝑑𝑥

𝑣2

=3𝑥 2𝑥 − 1+𝑥2 (3)

3𝑥 2


34 | P a g e

=6𝑥2−3−3𝑥2

9𝑥2

=𝑥2−1

3𝑥2 .

(ii) Let 𝑢 = 2 + 3𝑥, 𝑑𝑢

𝑑𝑥= 3

𝑣 = 𝑥2 + 3𝑥 + 5, 𝑑𝑣

𝑑𝑥= 2𝑥 + 3

But

𝑑𝑦

𝑑𝑥=

𝑣 𝑑𝑢

𝑑𝑥 − 𝑢

𝑑𝑣

𝑑𝑥

𝑣2

= 𝑥2+3𝑥+5 (3)− 2+3𝑥 (2𝑥+3)

𝑥2+3𝑥+5 2

=3𝑥2+9𝑥+15−4𝑥−6−6𝑥2−9𝑥

𝑥2+3𝑥+5 2

=−3𝑥2−4𝑥+9

𝑥2+3𝑥+5 2 .

Rule 5: If a function of function 𝑦 = 𝑎𝑥 + 𝑏 𝑛 where 𝑎 and 𝑏 are constants and 𝑛 is

the index, then 𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥 𝑎𝑥 + 𝑏 𝑛 = 𝑛 𝑎𝑥 + 𝑏 𝑛−1

𝑑

𝑑𝑥(𝑎𝑥 + 𝑏) .

This rule is also called the composite or chain rule. In another way, this can

be written as: 𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢

𝑑𝑢

𝑑𝑥

Where 𝑦 is a function of 𝑢 and 𝑢 is a function of 𝑥. It is called function of a

function.

Example: Differentiate the following: (i) 𝑦 = 𝑥 + 3 4 (ii) 𝑦 = 3𝑥2 − 5 6 .

Solution:

(i) Let 𝑢 = 𝑥 + 3; 𝑑𝑢

𝑑𝑥= 1 then 𝑦 = 𝑢4 ,

𝑑𝑦

𝑑𝑢= 4𝑢3 .

∴𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢×

𝑑𝑢

𝑑𝑥

= 4𝑢3(1)

= 4𝑢3

= 4 𝑥 + 3 3 .

(ii) Let 𝑢 = 3𝑥2 − 5; 𝑑𝑢

𝑑𝑥= 6𝑥 then 𝑦 = 𝑢6 ,

𝑑𝑦

𝑑𝑢= 6𝑢5 .

∴𝑑𝑦

𝑑𝑥=

𝑑𝑦

𝑑𝑢×

𝑑𝑢

𝑑𝑥

= 6𝑢5 (6𝑥)


35 | P a g e

= 36𝑥𝑢5

= 36𝑥 3𝑥2 − 5 5 .

Rule 6: For the implicit function, the relationship between 𝑦 and 𝑥 variables may not

be expressed explicitly (e.g. 𝑥𝑦2, 𝑥𝑦 + 𝑥2 = 3𝑥2𝑦5). In this case, treat 𝑦 as if

it is a function of 𝑥 and the rules of differentiation discussed above are

applied as appropriate.

Example: Differentiate the following implicitly:

(i) 𝑥3 + 𝑦4 = 7 (ii) 6𝑦2𝑥 − 4𝑥2𝑦3 + 2𝑦 = 0

Solution:

(i) Differentiate the function term by term with respect to 𝑥

3𝑥2 𝑑𝑥

𝑑𝑥 + 4𝑦3

𝑑𝑦

𝑑𝑥= 0

3𝑥2 + 4𝑦3 𝑑𝑦

𝑑𝑥= 0

⇒ 4𝑦3 𝑑𝑦

𝑑𝑥= −3𝑥2

∴𝑑𝑦

𝑑𝑥=

−3𝑥2

4𝑦3 .

(ii) 6 𝑦2 𝑑𝑥

𝑑𝑥+ 2𝑦𝑥

𝑑𝑦

𝑑𝑥 − 4 2𝑥

𝑑𝑥

𝑑𝑥𝑦3 + 3𝑥2𝑦2 𝑑𝑦

𝑑𝑥 + 2

𝑑𝑦

𝑑𝑥= 0

6𝑦2 + 12𝑥𝑦𝑑𝑦

𝑑𝑥− 8𝑥𝑦3 − 12𝑥2𝑦2

𝑑𝑦

𝑑𝑥+ 2

𝑑𝑦

𝑑𝑥= 0

⇒ 12𝑥𝑦 − 12𝑥2𝑦2 + 2 𝑑𝑦

𝑑𝑥= 8𝑥𝑦3 − 6𝑦2

∴𝑑𝑦

𝑑𝑥=

8𝑥𝑦3 − 6𝑦2

12𝑥𝑦 − 12𝑥2𝑦2 + 2

=4𝑥𝑦3−3𝑦2

6𝑥𝑦−6𝑥2𝑦2+1 .

Rule 7: If the exponential function 𝑦 = 𝑒𝑎𝑥 +𝑏 , where 𝑎 and 𝑏 are constants, then

𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥 𝑒𝑎𝑥+𝑏

=𝑑

𝑑𝑥 𝑎𝑥 + 𝑏 𝑒𝑎𝑥 +𝑏

= 𝑎𝑒𝑎𝑥 +𝑏


36 | P a g e

Example: Differentiate the following with respect to 𝑥. (i) 𝑦 = 6𝑒2𝑥2+𝑥 (ii) 𝑦 = 𝑥3𝑒𝑥3.

Solution:

(i) 𝑑𝑦

𝑑𝑥=

𝑑

𝑑𝑥 2𝑥2 + 𝑥 ∙ 6𝑒2𝑥2+𝑥

= 4𝑥 + 1 ∙ 6𝑒2𝑥2+𝑥

= 6 4𝑥 + 1 𝑒2𝑥2+𝑥 .

(ii) Applying product rule:

Let 𝑢 = 𝑥3; 𝑑𝑢

𝑑𝑥= 3𝑥2

𝑣 = 𝑒𝑥3;

𝑑𝑣

𝑑𝑥= 3𝑥2𝑒𝑥3

But 𝑑𝑦

𝑑𝑥= 𝑣

𝑑𝑢

𝑑𝑥+ 𝑢

𝑑𝑣

𝑑𝑥

= 𝑒𝑥3 3𝑥2 + 𝑥3 3𝑥2𝑒𝑥3

= 3𝑥2𝑒𝑥3+ 3𝑥5𝑒𝑥3

= 𝑒𝑥3 3𝑥2 + 3𝑥5 .

Rule 8: If the logarithmic function 𝑦 = log𝑒 𝑎𝑥 + 𝑏 = ln 𝑎𝑥 + 𝑏 where 𝑎 and 𝑏 are

constants, then 𝑑𝑦

𝑑𝑥=

1

𝑎𝑥 +𝑏∙

𝑑

𝑑𝑥 𝑎𝑥 + 𝑏 =

𝑎

𝑎𝑥 +𝑏 .

Example: Differentiate the following with respect to 𝑥.

(i) 𝑦 = ln 𝑥 (ii) 𝑦 = log𝑒 5𝑥 + 2 (iii) 𝑦 = ln 6𝑥 − 1 2 .

Solution:

(i) 𝑑𝑦

𝑑𝑥=

1

𝑥

(ii) 𝑑𝑦

𝑑𝑥=

1

5𝑥+2∙

𝑑

𝑑𝑥 5𝑥 + 2

=5

5𝑥+2

(iii) 𝑑𝑦

𝑑𝑥=

1

6𝑥−1 2 ∙𝑑

𝑑𝑥 6𝑥 − 1 2

=1

6𝑥−1 2 6 2 (6𝑥 − 1)

=12

6𝑥−1 .


37 | P a g e

Rule 9: Differentiation of Trigonometric functions:

(a) If 𝑦 = sin 𝑥 then 𝑑𝑦

𝑑𝑥= cos 𝑥

(b) If 𝑦 = cos 𝑥 then 𝑑𝑦

𝑑𝑥= − sin 𝑥

(c) If 𝑦 = tan 𝑥 then 𝑑𝑦

𝑑𝑥= sec2 𝑥

Examples:

1. 𝑦 = cos 7𝑥 , 𝑑𝑦

𝑑𝑥= −7 sin 7𝑥

2. 𝑦 = sin 3𝑥 − 4 , 𝑑𝑦

𝑑𝑥= 3 cos 3𝑥 − 4

3. 𝑦 = cos 𝜋

4− 2𝑥 ,

𝑑𝑦

𝑑𝑥= 2 sin

𝜋

4− 2𝑥

5.3 SECOND ORDER DERIVATIVE

Second order derivative is one of higher derivatives where successive

differentiations are carried out. The process of differentiating a function more that

one is called successive differentiation.

If 𝑦 = 𝑓 𝑥 ,𝑑𝑦

𝑑𝑥 is also a function of 𝑥. The derivative of

𝑑𝑦

𝑑𝑥 with respect to 𝑥

𝑑

𝑑𝑥

𝑑𝑦

𝑑𝑥 . The expression

𝑑

𝑑𝑥

𝑑𝑦

𝑑𝑥 is called the second order derivative of 𝑦 with respect

to 𝑥 and it is denoted by 𝑑2𝑦

𝑑𝑥2 or 𝑦′′ or 𝑓 ′′ (𝑥). It is useful when determining the turning

point of a function.

Example: Find the first, second and third derivatives of the following.

(i) 𝑦 = 4𝑥5 (ii) 𝑦 = 3𝑥6 − 2𝑥5 + 𝑒𝑥 + 3𝑥2 − 8 .

Solution:

(i) 𝑑𝑦

𝑑𝑥= 20𝑥4

𝑑2𝑦

𝑑𝑥2= 80𝑥3

𝑑3𝑦

𝑑𝑥3 = 240𝑥2 .

(ii) 𝑑𝑦

𝑑𝑥= 18𝑥5 − 10𝑥4 + 𝑒𝑥 + 6𝑥


38 | P a g e

𝑑2𝑦

𝑑𝑥2= 90𝑥4 − 40𝑥3 + 𝑒𝑥 + 6

𝑑3𝑦

𝑑𝑥3 = 360𝑥3 − 120𝑥2 + 𝑒𝑥 .

5.4 APPLICATION OF DIFFERENTIATION

5.4.1 Application to maximum and minimum values

The concept of maximum and minimum values is termed the zero slope

analysis. The focus here is to find a point where the slope of the function to be

optimized is zero. It should be noted that the concept is not restricted to plotting of

curve alone, but more importantly to determine the maximum and minimum values of

a function such as maximizing profit or minimizing cost, etc.

We have the following procedures for the determination of maximum and

minimum values of a function 𝑦 = 𝑓(𝑥) if it exists:

Step 1: Obtain the first derivative and set it equal to zero i.e. 𝑑𝑦

𝑑𝑥= 0. This occur at

turning point.

Step 2: From the 𝑑𝑦

𝑑𝑥= 0, determine the stationary point.

Step 3: Compute 𝑑2𝑦

𝑑𝑥2 at these stationary points.

If 𝑑2𝑦

𝑑𝑥2 < 0, the stationary point is maximum.

If 𝑑2𝑦

𝑑𝑥2 > 0, the stationary point is minimum.

If 𝑑2𝑦

𝑑𝑥2 = 0, (that is point of inflection) higher derivative (than second) can be

used to decide.

Example 1: ADAS company planning to have a new product in the market came up

with a total sales function 𝑆 = −1000𝑝2 + 10000𝑝 and the total cost function

𝐶 = −2000𝑝 + 2500, where 𝑝 and 𝐶 are respectively the price and cost (in US dollar)

of the new product. Find the optimal price for the new product and the maximum

profit expected for the company.

Solution: The profit (𝜑) is obtained by the difference in total sales and total cost.

∴ 𝜑 = 𝑆 − 𝐶

= −1000𝑝2 + 10000𝑝 − −2000𝑝 + 2500

= −1000𝑝2 + 10000𝑝 + 2000𝑝 − 2500


39 | P a g e

= −1000𝑝2 + 12000𝑝 − 2500

𝑑𝜑

𝑑𝑝= −2000𝑝 + 12000

But at turning point, 𝑑𝜑

𝑑𝑝= 0, then

0 = −2000𝑝 + 12000

⇒ 𝑝 =12000

2000= 6.

But 𝑑2𝜑

𝑑𝑝2= −2000 (which is less than zero, indicating maximum)

Hence, 𝑝 = 6, gives a maximum point. Therefore, the optimal price is $6. And the

maximum profit 𝜑 = −1000 6 2 + 12000(6) − 2500

= $33500 .

Example 2: 𝑛 articles are produced at a total cost of N 2𝑛2 + 30𝑛 + 20 , and each

one is sold for N 𝑛

3+ 100 . Determine the value of 𝑛 which gives the greatest profit,

and find this profit.

Solution: Total cost of articles = N 2𝑛2 + 30𝑛 + 20

Total money received for 𝑛 articles sold = N𝑛 𝑛

3+ 100 .

Then the profit N𝑃 = N 𝑛 𝑛

3+ 100 − 2𝑛2 + 30𝑛 + 20 .

So, 𝑃 = 70𝑛 −5

3𝑛2 − 20.

The minimum value of 𝑃 will be given by 𝑑𝑃

𝑑𝑛= 0

𝑑𝑃

𝑑𝑛= 70 −

10

3𝑛 = 0 ⇒ 𝑛 = 21.

𝑑2𝑃

𝑑𝑛2= −

10

3< 0.

Therefore, the maximum profit = N 70 × 21 −5

3 21 2

= N715 .

5.4.2 Application to Marginal Cost and Revenue

For the cost analysis (that is, relationship between average and marginal cost)

Let 𝑇 = Total cost

𝑞 = Quantity demand

Then, the Average Cost 𝐴𝐶 =𝑇

𝑞


40 | P a g e

And the Marginal Cost 𝑀𝐶 =𝑑𝑇

𝑑𝑞, i.e.

𝑑

𝑑𝑞 𝑇

Also the slope of average cost is obtained by:

𝑑

𝑑𝑞

𝑇

𝑞 =

𝑞 𝑑𝑇

𝑑𝑞 –𝑇

𝑑𝑞

𝑑𝑞

𝑞2 (Quotient Rule)

=

𝑑𝑇

𝑑𝑞−

𝑇

𝑞

𝑞

=1

𝑞

𝑑𝑇

𝑑𝑞−

𝑇

𝑞

=1

𝑞𝑀𝐶 −

1

𝑞𝐴𝐶

It is established that if the cost curve is U – shape and the 𝐴𝐶 has a sloping

downward curve, then 𝑑

𝑑𝑞

𝑇

𝑞 < 0 which means that 𝑀𝐶 < 𝐴𝐶. When 𝐴𝐶 decreases

𝑀𝐶 < 𝐴𝐶 and on the lowest point of 𝐴𝐶 curve, the tangent will be horizontal.

Therefore 𝑑

𝑑𝑞

𝑇

𝑞 = 0 i.e. 𝑀𝐶 = 𝐴𝐶. Also when 𝐴𝐶 increasing, 𝑀𝐶 > 𝐴𝐶 and

𝑑

𝑑𝑞

𝑇

𝑞 > 0.

Relationship between Average and Marginal revenue

Let Total revenue = 𝑅 and we know that 𝑅 = 𝑃𝑟𝑖𝑐𝑒 × 𝑞𝑢𝑎𝑛𝑡𝑖𝑡𝑦 i.e. 𝑅 = 𝑃𝑞.

Then, Average Revenue 𝐴𝑅 =𝑅

𝑞

And, the Marginal Revenue 𝑀𝑅 =𝑑

𝑑𝑞 𝑅 = 𝑃 + 𝑞

𝑑𝑝

𝑑𝑞. (Product Rule)

Example 1: Determine the minimum average cost if the cost function is given by:

𝑇 = 36𝑞 − 20𝑞2 + 4𝑞3. Also obtain the marginal cost at the point of minimum

average cost.

Solution:

Given 𝑇 = 36𝑞 − 10𝑞2 + 2𝑞3

∴ 𝐴𝐶 =𝑇

𝑞= 36 − 10𝑞 + 2𝑞2

For maximum or minimum, 𝑑

𝑑𝑞

𝑇

𝑞 = 0

i.e. −10 + 4𝑞 = 0

⇒ 𝑞 =5

2.


41 | P a g e

More so, 𝑑2

𝑑𝑞2 𝑇

𝑞 at 𝑞 = 5

2 = 4, which is greater than zero.

Hence, 𝑞 =5

2 is at minimum point.

∴ The Minimum Average Cost = 36 − 10 5

2 + 2

5

2

2

= 23.5

For the Marginal Cost (𝑀𝐶), we have

𝑀𝐶 =𝑑𝑇

𝑑𝑞= 36 − 20𝑞 + 6𝑞2

𝑀𝐶 at 𝑞 =5

2 is equal to 36 − 20

5

2 + 6

5

2

2

= 23.5 .

Example 2: Find the maximum profit of a company with revenue function

𝑅 = 200𝑞 − 2𝑞2 and the cost function 𝑇 = 2𝑞3 − 57𝑞2.

Solution:

Profit 𝜑 = 𝑅 − 𝑇

= 200𝑞 + 55𝑞2 − 2𝑞3

At turning point, 𝑑𝜑

𝑑𝑞= 0

𝑑𝜑

𝑑𝑞= 200 + 110𝑞 − 6𝑞2 = 0

⇒ 100 + 55𝑞 − 3𝑞2 = 0

3𝑞2 − 55𝑞 − 100 = 0

3𝑞2 − 60𝑞 + 5𝑞 − 100 = 0

3𝑞 𝑞 − 20 + 5 𝑞 − 20 = 0

3𝑞 + 5 𝑞 − 20 = 0

⇒ 𝑞 = −5

3 or 20

But 𝑑2𝜑

𝑑𝑞2= 110 − 12𝑞

At 𝑞 = −5

3,

𝑑2𝜑

𝑑𝑞2 = 110 − 12 5

3 = 90 > 0 indicating minimum point.

At 𝑞 = 20, 𝑑2𝜑

𝑑𝑞2 = 110 − 12 20 = −130 < 0 indicating maximum point.

Hence, at 𝑞 = 20, we have the maximum profit.

∴ The maximum profit = 200 20 + 55 20 2 − 2 20 3

= 10,000


42 | P a g e

5.4.3 Application of differentiation to the elasticity of demand

The price elasticity of demand is the rate of change in response to quantity

demanded to the change in price. Let 𝑞 = 𝑓(𝑝) be the demand function, where 𝑞 is

the demand and 𝑝 is the price. Then the elasticity of demand is

𝑒𝑑 = −𝑃

𝑞∙

𝑑𝑞

𝑑𝑃 When 𝑒𝑑 > 1, the demand is elastic and, when 𝑒𝑑 < 1, the demand is

inelastic.

For elasticity of supply, Let 𝑥 = 𝑓 𝑝 be the supply function, where 𝑥 is the

supply and 𝑝 is the price. The elasticity of supply is defined as 𝑒𝑠 =𝑝

𝑥∙

𝑑𝑥

𝑑𝑝 .

Example: Given a demand function 𝐷 = 60 − 𝑝 − 𝑝2 , determine its elasticity of

demand when 𝑝 = 4

Solution: 𝑑𝐷

𝑑𝑝= −1 − 2𝑝

But 𝑒𝑑 = −𝑝

𝐷

𝑑𝐷

𝑑𝑝

= −𝑝

60−𝑝−𝑝2 −1 − 2𝑝

=2𝑝2+𝑝

60−𝑝−𝑝2 .

When 𝑝 = 4 we get

𝑒𝑑 = 0.9

Exercise 5

1. A firm produces 𝑥 tonnes of output at a total cost 𝐶 =1

10𝑥3 − 5𝑥2 + 10𝑥 − 32.

At what level of output will the marginal cost and the average variable cost

attain their respective minimum?

2. A certain manufacturing company has total cost function

𝐶 = 15 + 9𝑥 − 6𝑥2 + 𝑥3 . Find 𝑥, when the total cost is minimum.

3. The relationship between profit 𝑃 and advertising cost 𝑥 is given by

𝑃 =4000𝑥

500+𝑥− 𝑥. Find 𝑥 which maximizes 𝑃.

4. The total cost and total revenue of a firm are given by

𝐶 = 𝑥3 − 12𝑥2 + 48𝑥 + 11 and 𝑅 = 83𝑥 − 4𝑥2 − 21.

Find the output (i) when the revenue is maximum (ii) when profit is maximum.


43 | P a g e

5. A GSM company has a profit of N2 per SIM when the number of SIM in the

exchanges is not over 10,000. The profit per SIM decreases by 0.01 kobo for

each SIM over 10,000. What is the maximum profit?

6. The total cost function of a firm is 𝐶 =1

3𝑥3 − 5𝑥2 + 28𝑥 + 10 where 𝑥, is the

output. A tax at N2 per unit of output is imposed and the producer adds it to

his cost. If the market demand function is given by 𝑝 = 2530 − 5𝑥, where N𝑝

is the price per unit of output. Determine the profit maximizing output and

price.

7. Investigate the maxima and minima of the function 2𝑥3 − 3𝑥2 − 36𝑥 + 10 .

8. For the cost function 𝐶 = 2000 + 1800𝑥 − 75𝑥2 + 𝑥3 find when the total cost

(𝐶) is increasing and when it is decreasing.

9. Find the equation of the tangent and normal to the demand curve

𝑦 = 10 − 3𝑥2 at (1, 7).

10. ADAS produces 𝑥 tonnes of output at a total cost

𝐶 𝑥 = 1

10𝑥3 − 4𝑥2 + 20𝑥 + 5 . Find (i) Average Cost (ii) Average Variable

Cost (iii) Average Fixed Cost (iv) Marginal Cost and (v) Marginal Average

Cost.

11. The total cost 𝑌 of making 𝑛 units of product is

𝑌 = 0.00005𝑛3 − 0.06𝑛2 + 10𝑛 + 79020. Find the marginal cost at 1000 units

of output.

12. Find the elasticity of the following functions.

(a) 𝑥 = 2𝑝2 + 8𝑝 + 10

(b) 𝑦 = 4𝑥 − 8 when 𝑥 = 5.


44 | P a g e

§ 6.0 INTEGRAL CALCULUS

Integration is the reverse of differentiation. If 𝐹(𝑥) is a function whose

derivative 𝐹′ 𝑥 = 𝑓(𝑥), then 𝐹(𝑥) is called an integral of 𝑓(𝑥). For example,

𝐹 𝑥 = 𝑥3 is an integral of 𝑓 𝑥 = 3𝑥2 since 𝐹′ 𝑥 = 3𝑥2 = 𝑓(𝑥).

6.1 Indefinite Integral

Let 𝑦 = 𝑓(𝑥) be a function of 𝑥. Suppose that 𝑑𝑦

𝑑𝑥= 𝜑(𝑥) where 𝜑(𝑥) is a

known function of 𝑥. Then 𝑦 = 𝜑(𝑥) 𝑑𝑥 is called an indefinite integral of 𝜑(𝑥) with

respect to 𝑥.

Remark

1. Let 𝑐 be an arbitrary constant. Since 𝑑

𝑑𝑥 𝑐 = 0, it follows that 𝑐 = 0 𝑑𝑥 .

2. If 𝑓(𝑥) is an indefinite integral of a function 𝜑(𝑥) and 𝑐 is an arbitrary constant,

then all positive indefinite integrals of 𝜑(𝑥) are of the form

𝜑(𝑥) 𝑑𝑥 = 𝑓 𝑥 + 𝑐. In this case, 𝑐 is called a constant of integration.

For all rational values of 𝑛 except when 𝑛 = −1, 𝑑

𝑑𝑥 𝑥𝑛+1

𝑛+1 =

1

𝑛+1∙ 𝑛 + 1 𝑥𝑛 .

Therefore, if 𝑛 ≠ −1, the indefinite integral of 𝑥𝑛 is 𝑥𝑛 𝑑𝑥 =𝑥𝑛+1

𝑛+1+ 𝑐 . Note also that

constant different from zero, the integral is obtained by simply multiply the constant

by the independent variable and add the constant of integration. E.g. 2 𝑑𝑥 = 2𝑥 + 𝑐.

Example

1. If 𝑑𝑦

𝑑𝑥= 4𝑥2, find 𝑦.

Solution

𝑑𝑦 = 4𝑥2𝑑𝑥

𝑑𝑦 = 4𝑥2 𝑑𝑥

⇒ 𝑦 =4𝑥2+1

2+1+ 𝑐 =

4

3𝑥3 + 𝑐

2. Integrate 6𝑡3 + 6 with respect to 𝑡.

Solution

6𝑡3 + 6 𝑑𝑡 =6

4𝑡4 + 6𝑡 + 𝑐

=3

2𝑡4 + 6𝑡 + 𝑐 .


45 | P a g e

6.2 Some Standard form of Integration

(i) 𝑒𝑥 𝑑𝑥 = 𝑒𝑥 + 𝑐

(ii) 𝑑𝑥

𝑥= log𝑒 𝑥 + 𝑐

(iii) sin 𝑥 𝑑𝑥 = − cos 𝑥 + 𝑐

(iv) cos 𝑥 𝑑𝑥 = sin 𝑥 + 𝑐

(v) 𝑑𝑥

𝑥2+𝑎2 =1

𝑎tan−1 𝑥

𝑎+ 𝑐

(vi) 𝑑𝑥

𝑎2−𝑥2= sin−1 𝑥

𝑎+ 𝑐

Example: Integrate the following with respect to 𝑥.

(a) 𝑑𝑥

16−𝑥2=

𝑑𝑥

42−𝑥2= sin−1 𝑥

4+ 𝑐

(b) 𝑑𝑥

𝑥2+25=

1

5tan−1 𝑥

5+ 𝑐

6.3 Rules of Integration

Rule 1: The integration of a sum (difference) of a finite number of functions is the

sum (difference) of their separate integrals. E.g.

𝑥2 − 𝑥 + 5 𝑑𝑥 = 𝑥2 𝑑𝑥 − 𝑥 𝑑𝑥 + 5 𝑑𝑥

=𝑥3

3−

𝑥2

2+ 5𝑥 + 𝑐 .

Rule 2: A constant factor may be brought outside the integral sign. E.g.

18𝑥3 𝑑𝑥 = 18 𝑥3 𝑑𝑥 = 18 𝑥4

4 + 𝑐 =

9

2𝑥4 + 𝑐 .

Rule 3: The addition of a constant to the variable makes no difference to the form of

the result. E.g.

(i) 𝑑𝑥

𝑥−3 2+32=

1

3tan−1 𝑥−3

3+ 𝑐

(ii) 𝑑𝑥

𝑥+7= log𝑒 𝑥 + 6 + 𝑐

Rule 4: Multiplying the variables by the constant also makes no difference to the

form of the result, but we have to divide by the constant. (Note integration by

substitution can be used to verify this claim)


46 | P a g e

Examples

1. cos 3𝑥 𝑑𝑥 =1

3sin 3𝑥 + 𝑐

2. 𝑒4𝑥 𝑑𝑥 =1

4𝑒4𝑥 + 𝑐

3. 𝑒2𝑥−5 𝑑𝑥 =1

2𝑒2𝑥−5 + 𝑐

4. 2𝑥 + 3 12 𝑑𝑥 =

1

2 2𝑥+3

32

32

+ 𝑐 =1

3 2𝑥 + 3

32 + 𝑐

Rule 5: The integration of a fraction whose numerator is derivative of the

denominator is obtained by the logarithm of the denominator. i.e.

𝑓 ′ 𝑥

𝑓 𝑥 𝑑𝑥 = log𝑒 𝑓 𝑥 + 𝑐

Examples

(a) 2𝑥

𝑥2+12𝑑𝑥 = ln 𝑥2 + 12 + 𝑐

(b) 𝑥2

𝑥3+2𝑑𝑥 =

1

3

3𝑥2

𝑥3+2𝑑𝑥 =

1

3ln 𝑥3 + 2 + 𝑐

(c) 𝑥𝑒𝑥2

𝑒𝑥2+3

𝑑𝑥 =1

2

2𝑥𝑒𝑥2

𝑒𝑥2+3

𝑑𝑥 =1

2ln 𝑒𝑥2

+ 3 + 𝑐

(d) 𝑑𝑥

𝑥 log 𝑒 𝑥=

1

𝑥𝑑𝑥

log 𝑒 𝑥= log𝑒 log𝑒 𝑥 + 𝑐

Rule 6: (Integration by Substitution)

Consider the indefinite integral 𝑓(𝑥) 𝑑𝑥 of an arbitrary function 𝑦 = 𝑓(𝑥). Suppose

that 𝑓(𝑥) can be written as 𝑓 𝑥 = 𝑔(𝑢)𝑑𝑢

𝑑𝑥 for some function 𝑔(𝑢) of suitably chosen

variable 𝑢 = 𝑢(𝑥), then 𝑓(𝑥) 𝑑𝑥 = 𝑔(𝑢) 𝑑𝑢. That is, we get 𝑓(𝑥) 𝑑𝑥 by

evaluating 𝑔(𝑢) 𝑑𝑢 as a function of 𝑢 and then substituting 𝑢 = 𝑢(𝑥) to get a

function of the original variable 𝑥.

Example 1: Evaluate 𝑥

1−2𝑥2𝑑𝑥.

Solution: Set 𝑢 = 1 − 2𝑥2; 𝑑𝑢

𝑑𝑥= −4𝑥 ⇒ 𝑑𝑥 =

1

−4𝑥𝑑𝑢.

Therefore, 𝑥

1−2𝑥2𝑑𝑥 =

𝑥

𝑢

1

−4𝑥𝑑𝑢

= −1

4

1

𝑢𝑑𝑢

= −1

4 𝑢−1

2 𝑑𝑢


47 | P a g e

= −1

4 2 𝑢

12 + 𝑐

= −1

2 1 − 2𝑥2 + 𝑐

Example 2: Evaluate 𝑒2−5𝑥 𝑑𝑥

Solution: Let 𝑢 = 2 − 5𝑥; 𝑑𝑢

𝑑𝑥= −5 ⇒ 𝑑𝑥 =

𝑑𝑢

−5.

Hence, 𝑒2−5𝑥 𝑑𝑥 = 𝑒𝑢 𝑑𝑢

−5

= −1

5 𝑒𝑢 𝑑𝑢

= −1

5𝑒𝑢 + 𝑐

= −1

5𝑒2−5𝑥 + 𝑐

Rule 7: (Integration by Parts)

If 𝑓 and 𝑔 are two differentiable functions of 𝑥. Then 𝑑

𝑑𝑥 𝑓𝑔 = 𝑓 𝑥

𝑑𝑔

𝑑𝑥+ 𝑔(𝑥)

𝑑𝑓

𝑑𝑥

which implies that 𝑓 𝑥 𝑑𝑔

𝑑𝑥=

𝑑

𝑑𝑥 𝑓𝑔 − 𝑔(𝑥)

𝑑𝑓

𝑑𝑥 . Using the formula for the integral of a

sum which is sum of the integrals, we obtain

𝑓 𝑥 𝑑𝑔

𝑑𝑥𝑑𝑥 = 𝑓 𝑥 𝑔 𝑥 − 𝑔(𝑥)

𝑑𝑓

𝑑𝑥𝑑𝑥

This is called the formula for integration by parts. If 𝑢 = 𝑓(𝑥) and 𝑣 = 𝑔(𝑥), the

formula can be written as 𝑢 𝑑𝑣 = 𝑢𝑣 − 𝑣 𝑑𝑢 .

Example: Evaluate the following integrals:

1. 2𝑥 sin 𝑥 𝑑𝑥

Solution: Let 𝑢 = 2𝑥, 𝑑𝑢

𝑑𝑥= 2 ⇒ 𝑑𝑢 = 2𝑑𝑥

𝑑𝑣 = sin 𝑥 𝑑𝑥 ⇒ 𝑣 = cos 𝑥

Hence, by integration by parts, we have

2𝑥 sin 𝑥 𝑑𝑥 = 2𝑥 cos 𝑥 − cos 𝑥 2𝑑𝑥

= 2𝑥 cos 𝑥 − 2 cos 𝑥 𝑑𝑥

= 2𝑥 cos 𝑥 − 2 sin 𝑥 + 𝑐

2. 𝑥𝑒𝑥 𝑑𝑥 = 𝑥𝑒𝑥 − 𝑒𝑥 𝑑𝑥 = 𝑥𝑒𝑥 − 𝑒𝑥 + 𝑐


48 | P a g e

3. 𝑥 log𝑒 𝑥 𝑑𝑥

Solution: Let 𝑢 = log𝑒 𝑥 , 𝑑𝑢

𝑑𝑥=

1

𝑥 ⇒ 𝑑𝑢 =

1

𝑥𝑑𝑥

𝑑𝑣 = 𝑥 𝑑𝑥 ⇒ 𝑣 =𝑥2

2

Hence, by integration by parts, we have

𝑥 log𝑒 𝑥 𝑑𝑥 =𝑥2

2log𝑒 𝑥 −

𝑥2

2

1

𝑥𝑑𝑥

=𝑥2

2log𝑒 𝑥 −

1

2 𝑥 𝑑𝑥

=𝑥2

2log𝑒 𝑥 −

1

4𝑥2 + 𝑐

PP: (a) 𝑥2𝑒𝑥 𝑑𝑥 (b) 𝑒𝑥 sin 𝑥 𝑑𝑥 .

6.4 Definite Integral

Let 𝑓 be a function of 𝑥 and 𝑎, 𝑏 be an interval, 𝑓(𝑥)𝑏

𝑎𝑑𝑥 is called definite

integral with respect to 𝑥 within the limit 𝑎 and 𝑏.

THEOREM (Fundamental Theorem of Calculus): Suppose that 𝑓 is a strong

continuous function over a closed interval 𝑎, 𝑏 and that 𝐹 is anti – derivative of 𝑓.

Then 𝑓(𝑥)𝑏

𝑎𝑑𝑥 = 𝐹(𝑥)

𝑏𝑎

= 𝐹 𝑏 − 𝐹(𝑎).

Examples

1. 𝑥23

1𝑑𝑥 =

𝑥3

3

31

= 33

3 −

13

3 = 8

2

3

2. 4𝑥5

0𝑑𝑥 = 2𝑥2 5

0= 50

PP: (a) 𝑒2−5𝑥1

0𝑑𝑥 (b) 𝑥 ln 𝑥

1

0𝑑𝑥

6.5 APPLICATIONS OF INTEGRATION TO ECONOMICS AND COMMERCE

We learnt already that the marginal function is obtained by differentiating the

total function. We were given the total cost, total revenue or demand function and

we obtained the marginal cost, marginal revenue or elasticity of demand. Now we

shall obtain the total function when marginal function is given.


49 | P a g e

6.5.1 The cost function and average cost function from marginal cost

function:

If 𝐶 is the cost of producing an output 𝑥, then marginal cost function,

𝑀𝐶 =𝑑𝐶

𝑑𝑥. Using integration as reverse process of differentiation, we obtain,

Cost function, 𝐶 = 𝑀𝐶 𝑑𝑥 + 𝑘

where 𝑘 is the constant of integration which can be evaluated if the fixed cost is

known. If the fixed cost is not known, then 𝑘 = 0.

Average cost function, 𝐴𝐶 =𝐶

𝑥, 𝑥 ≠ 0

Example 1: The marginal cost function of manufacturing 𝑥 units of a commodity is

8 + 16𝑥 − 9𝑥2. Find the total cost and average cost, given that the total cost of

producing 1 unit is 20.

Solution: Given that,

𝑀𝐶 = 8 + 16𝑥 − 9𝑥2

𝐶 = 𝑀𝐶 𝑑𝑥 + 𝑘

= 8 + 16𝑥 − 9𝑥2 𝑑𝑥 + 𝑘

= 8𝑥 + 8𝑥2 − 3𝑥3 + 𝑘 ------------------ (*)

Given, when 𝑥 = 1, 𝐶 = 20

∴ ∗ ⇒ 20 = 8 1 + 8 12 − 3 13 + 𝑘 ⇒ 𝑘 = 13

∴ Total Cost function, 𝐶 = 8𝑥 + 8𝑥2 − 3𝑥3 + 13

Average Cost function, 𝐴𝐶 =𝐶

𝑥, 𝑥 ≠ 0

= 8 + 8𝑥 − 23 +13

𝑥 .

Example 2: The marginal cost function of manufacturing 𝑥 units of a commodity is

3𝑥2 − 2𝑥 + 32. If there is no fixed cost, find the total cost and average cost

functions. Solution: Given that,

𝑀𝐶 = 3𝑥2 − 2𝑥 + 32

𝐶 = 𝑀𝐶 𝑑𝑥 + 𝑘

= 3𝑥2 − 2𝑥 + 32 𝑑𝑥 + 𝑘

= 𝑥3 − 𝑥2 + 32𝑥 + 𝑘

No fixed cost ⇒ 𝑘 = 0


50 | P a g e

∴ Total cost, 𝐶 = 𝑥3 − 𝑥2 + 32𝑥

Average Cost, 𝐴𝐶 =𝐶

𝑥, 𝑥 ≠ 0

= 𝑥2 − 𝑥 + 32 .

6.5.2 The revenue function and demand function from marginal revenue

function:

If 𝑅 is the total revenue function when the output is 𝑥, then marginal revenue

𝑀𝑅 =𝑑𝑅

𝑑𝑥.

Integrating this with respect to 𝑥 we have

Revenue function, 𝑅 = 𝑀𝑅 𝑑𝑥 + 𝑘

where 𝑘 is the constant of integration which can be evaluated under given

conditions.

If the total revenue 𝑅 = 0, when 𝑥 = 0,

Demand function, 𝑝 =𝑅

𝑥, 𝑥 ≠ 0

Example 1: If the marginal revenue for a commodity is 𝑀𝑅 = 10 − 6𝑥2 + 2𝑥, find

the total revenue and demand function.


𝑀𝑅 = 10 − 6𝑥2 + 2𝑥

𝑅 = 𝑀𝑅 𝑑𝑥 + 𝑘

= 10 − 6𝑥2 + 2𝑥 𝑑𝑥 + 𝑘

= 10𝑥 − 2𝑥3 + 𝑥2 + 𝑘

Since 𝑅 = 0 when 𝑥 = 0, then 𝑘 = 0.

∴ 𝑅 = 10𝑥 − 2𝑥3 + 𝑥2

𝑝 =𝑅

𝑥, 𝑥 ≠ 0

⇒ 𝑝 = 10 − 2𝑥2 + 𝑥.

Example 2: For the marginal revenue function 𝑀𝑅 = 5 + sin 4𝑥 − 2𝑥 − 𝑥2, find the

revenue function and demand function.


𝑀𝑅 = 5 + sin 4𝑥 − 2𝑥 − 𝑥2


51 | P a g e

𝑅 = 𝑀𝑅 𝑑𝑥 + 𝑘

= 5 + sin 4𝑥 − 2𝑥 − 𝑥2 𝑑𝑥 + 𝑘

= 5𝑥 −1

4coss 4𝑥 + 𝑥2 +

𝑥3

3+ 𝑘

Since 𝑅 = 0 when 𝑥 = 0, then 𝑘 = −1

4.

∴ 𝑅 = 5𝑥 −1

4coss 4𝑥 + 𝑥2 +

𝑥3

3−

1

4

𝑝 =𝑅

𝑥, 𝑥 ≠ 0

⇒ 𝑝 = 5 −1

4𝑥coss 4𝑥 + 𝑥2 +

𝑥3

3−

1

4𝑥.

6.5.3 The demand function when the elasticity of demand is given

We know that,

Elasticity of demand 𝑒𝑑 = −𝑝

𝑥∙

𝑑𝑥

𝑑𝑝

⇒ −𝑑𝑝

𝑝=

𝑑𝑥

𝑥

1

𝑒𝑑

Integrating both sides

− 𝑑𝑝

𝑝=

1

𝑒𝑑

𝑑𝑥

𝑥

This equation yields the demand function 𝑝 as a function of 𝑥.

The revenue function can be found out by using the relation, 𝑅 = 𝑝𝑥.

Example 1: The elasticity of demand with respect to price 𝑝 for a commodity is

𝑥−6

𝑥, 𝑥 > 6 when the demand is 𝑥. Find the demand function if the price is 2 when

demand is 8. Also find the revenue function.


Elasticity of demand, 𝑒𝑑 = −𝑝

𝑥∙

𝑑𝑥

𝑑𝑝=

𝑥−6

𝑥

⇒ 𝑑𝑥

𝑥−6= −

𝑑𝑝

𝑝

Integrating both sides,

𝑑𝑥

𝑥−6= −

𝑑𝑝

𝑝+ 𝑐

⇒ log𝑒 𝑥 − 6 = − log𝑒 𝑝 + log𝑒 𝑘 (since c is a constant, 𝑐 ≡ log𝑒 𝑘)

⇒ log𝑒 𝑥 − 6 + log𝑒 𝑝 = log𝑒 𝑘


52 | P a g e

⇒ log𝑒𝑝 𝑥 − 6 = log𝑒 𝑘

⇒ 𝑝 𝑥 − 6 = 𝑘 -------------- (1)

When 𝑝 = 2, 𝑥 = 8, from (1) we get

𝑘 = 4.

∴ The demand function is,

𝑝 =4

𝑥−6, 𝑥 > 6.

Revenue, 𝑅 = 𝑝𝑥 or 4𝑥

𝑥−6, 𝑥 > 6.

Example 2: The elasticity of demand with respect to price for a commodity is a

constant and is equal to 2. Find the demand function and hence the total revenue

function, given that when the price is 1, the demand is 4.


Elasticity of demand, 𝑒𝑑 = 2

⇒ −𝑝

𝑥∙

𝑑𝑥

𝑑𝑝= 2

⇒ 𝑑𝑥

𝑥= −2

𝑑𝑝

𝑝

Integrating both sides,

⇒ 𝑑𝑥

𝑥= −2

𝑑𝑝

𝑝

ln 𝑥 = −2 ln 𝑝 + ln 𝑘

ln 𝑥 + ln 𝑝2 = ln 𝑘

ln 𝑥𝑝2 = ln 𝑘

𝑥𝑝2 = 𝑘 --------------- (1)

Given, when 𝑥 = 4, 𝑝 = 1

From (1) we get 𝑘 = 4

∴ (1) ⇒ 𝑥𝑝2 = 4 or 𝑝 =2

𝑥, 𝑥 > 0

Demand function 𝑝 =2

𝑥, 𝑥 > 0 ;

Revenue 𝑅 = 𝑝𝑥 =2𝑥

𝑥= 2 𝑥, 𝑥 > 0.


53 | P a g e

Exercise 6

1. The marginal cost function of manufacturing 𝑥 units of a commodity is

3 − 2𝑥 − 𝑥2. If the fixed cost is 200, find the total cost and average cost

function.

2. If the marginal revenue for a commodity is 𝑀𝑅 =𝑒2𝑥

100+ 𝑥 + 𝑥2, find the

revenue function.

3. The marginal cost and marginal revenue with respect to a commodity of a firm

are given by 𝑀𝐶 = 4 + 0.8𝑥 and 𝑀𝑅 = 12. Find the total profit, given that the

total cost at zero output is zero. Hence, determine the profit when the output

is 50.

4. The marginal revenue function (in thousands of naira) of a commodity is

7 + 𝑒−0.05𝑥 where 𝑥 is the number of units sold. Find the total revenue from

the sale of 100 units 𝑒−5 = 0.0067

5. The marginal cost and marginal revenue are given by 𝑀𝐶 = 20 +𝑥

20 and

𝑀𝑅 = 30. The fixed cost is N200. Find the maximum profit.

6. ADAS company determines that the marginal cost of producing 𝑥 units is

𝑀𝐶 = 10.6𝑥. the fixed cost is N50. The selling price per unit is N5. Find

(i) Total Cost Function

(ii) Total Revenue Function

(iii) Profit Function

7. Find the cost of producing 3000 units of commodity if the marginal cost in

naira per unit is 𝐶′ 𝑥 =𝑥

3000+ 2.50.

8. The marginal cost of a production level of 𝑥 units is given by 𝐶′ 𝑥 = 85 +375

𝑥2 .

Find the cost of producing 10 incremental units after 15 units have been

produced.

Education

LECTURE NOTE ON MATHEMATICS FOR SOCIAL SCIENTISTS II (MTS 108)