Chapter Three

Functions

3.1. Review of relations and functions

Definition 3.1: Let and be sets. The set is called the Cartesian product of the sets and. In other
words, is the set of all ordered pairs, where and.

Definition 3.2: (Equality of ordered Pairs): Two ordered pairs ( a, b) and ( c, d ) are equal if
and only if a  c and b  d .
Definition 3.3: Let and be sets. relation R from to (or between A and B) is any subset of
the Cartesian product.

A relation on a set is any subset of the Cartesian product. If is a relation from to , then

Domain of
Range of

Remark: If R is a relation from the set A to the set B , then the set B is called the codomain of
the relation R . The range of relation is always a subset of the codomain.
Example 3.1. Let and . Since A is a finite set we can list the elements of the relation.

We can represent this relation diagrammatically as in figure 3.1 by putting its elements on the
arrow diagram of the Cartesian product (as )

Figure3.1
Example 3.2: Draw the graph of and give domain and range of R.
Solution: First find the intersection between the graph of and. Hence we have, and .

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 Figure1.2
Form the graph, Domain of and Range of R is .
Example 3.3: Sketch the graph of the relation and give the domain and range of R.
Solution: Here we find the intersection of the circle and the parabola . Thus, we get and . That is,
Or . But we choose because from the given information is a positive real number. When we
solve for , we get.

Figure1.3
From the graph, Domain of , Range of
Dentition 3.4: Let R be a relation on set A. We say that R is:
i. Reflexive if and only if for every
ii. Symmetric if and only if implies for every
iii. Transitive if and only if and implies for every.
Examples 3.3:
1. Let be the relation on the set of real numbers defined by if and only if . Then
(i) R is reflexive because for every .
(ii) R is not symmetric, for example, but , so does not imply
(iii)R is transitive because, if and , then for all
2. Let ( means set of positive real numbers) and let be a relation on defined by if and only if.
Show that R is reflexive, symmetric and transitive.
Solution:
i. For all positive real numbers and, because addition under positive real numbers satisfies
commutative property, so for every.
Therefore, R is reflexive.
ii. R is symmetric since if , then which implies that

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 , hence (c, d) R (a, b). Therefore, R is symmetric.
iii.To show that R is transitive, suppose (a, b) R (c, d) and (c, d) R (e, f ). That is,
and
Adding these equations give
Therefore,. That is, R is transitive.
Dentition 3.5: If a relation R on a set A is reflexive, symmetric and transitive, then it is called
an equivalence relation.
 Inverse of a Relation.
Definition 3.6: If is a relation from A to B, then the inverse relation of denoted by is a
relation from B to A defined by.
The domain of is the range of and the range of is the domain of .
That is, Domain ofRange of and Range ofDomain of.
Note: The inverse of a relation is formed by interchanging the coordinates of each of the ordered
pairs in the given relation.
Example 3.4:
1. Let . Find the inverse of.
Solution: If we interchange the coordinates of each of the ordered pair, we get . Therefore, .
2. Find the inverse of the relation .
Solution: (interchange variables and )
(solve for ). Therefore,
3. Find the inverse of the relation .
Solution: (given)
(interchange variables and )
(solving for ). Hence,.
4. Find the inverse of the relation .
Solution: (given)
(Interchange variables and )
(Solving for )

 Graphs of Inverse relation

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The graphs of relations and their inverses are related in an interesting way. First, note that the
graphs of the ordered pairs and always have symmetry about the line .

Figure1.12
Example 1.5:
1. Sketch the graph of the relation and its inverse.
Solution: As in the example 1.4 above the inverse of the relation is . Thus, the graph of and are
given as below.

Figure1.13
2. Sketch the graph of the relation and its inverse.
Solution: (By definition)
(Interchange variables and )
(Solving for )

Figure1.14
3. Sketch the graph of the relation and its inverse.
Solution: (given)
(Interchange variables and )
(Solving for )

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 Figure1.15
3.2. Functions
Definition 3.7: Let A and B be two sets, a function from A to B, written as , is a rule which
assigns each (input) to a unique element (output).
From definition 3.7, the domain of is and the range of is the set of all possible values of as
varies throughout the domain. Symbolically, domain of and range of
. The set is called the co domain.
Remark: We should not confuse about the range and co domain of a function. That is, if is a
function , it is not generally true that is the range of rather the range of is the sub set of co
domain .
Note:
1. In definition 3.7, the term unique means ‘exactly one’. Thus, a function cannot assign two
different outputs to the same input.
2. A function is a special type of relation.
Example 1.6:
1. Which one of the following are functions?
a. d.
b. e.
c. f.
Solution:(b) and (c) are functions because every input has exactly on output. But (a), (d), (e) and
(f) are not functions because some inputs have more than one output.
2. Let . Answer the following questions.
a. For what value of is ?
b. For what values of is ?
Solution:
a.

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 or . Therefore, for .
b. . Using sign chart;

Therefore, , for .
3. The domain and range of some of the following functions are given below.

Function Domain Range

Note: The vertical line test. A curve in the -plane is the graph of some function if and only if no
vertical line intersects the curve more than once.

Figure1.16
Example 3.7: The graph of the equation is a circle of radius , centered at the origin, and hence
there are vertical lines that cut the graph more than once. Solving for in terms of ; . This
equation does not define as a function of .

Figure1.17
But and define as a function of .

(b)
(a)

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 Figure1.18
 Absolute Value (Modulus) Function
Definition 1.8: The absolute value (modulus) of any real number ‘’ denoted by is defined as
The absolute value of a number is defined as its distance from the origin.
Properties of Absolute value
If and are real numbers, then
i. iv. ,
ii. v. (Triangular inequality)
iii. vi.

Definition 1.9: The absolute value function is defined by

Domain of and range of .

 Signum Function
Definition 1.10: The Signum function, read as signum , usually written as sgn is a function
given by
Since,
A signum function is an example of a piecewise-defined function with different properties.
Some Properties of Signum Function
 For all , 
 For all ,  If ,then

 If , then
 If , then
  If , then

Example 1.10:
1. Simplify the following.
a. b.
Solution:
a.
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 b.
 Greatest Integer (Stair Case) Function
Here we discuss on an interesting and useful function called the greatest integer function.
Deintion1.11: The greatest integer of a real number, denoted by , is an integer such that . That
is, is the largest integer less than or equal to ( ).
The greatest integer function is a function defined by the equation
The domain of is the set of real numbers, and its range is .

Example 1.11:
1. Consider the following.
a. The real number is between the integers and . Thus, the greatest integer less than or equal to
is . Therefore, .
b. ( that is, the largest integer less than or equal to )
c. (that is, the largest integer less than or equal to )
d. (that is, the largest integer less than or equal to )
e.
f.
g.
Note: For any integer, (the greatest integer of an integer is the integer itself).
 Operations on Functions
 Combination of functions
Two functions, and , can be added, subtracted, multiplied, and divided in a natural way to form
new functions and.
Definition 1.18: Given two functions and , we define the following.
a. c.
b. d. ,
For the functions and, the domain is the intersection of the domains of and and the domain of
the function is the intersection of the domains and except with the points where (this is to
avoid division by zero).
Example 1.22: Let .

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Find ; state their domains.
Solution:

Domain of , Domain of and [domain of ] [domain of].

Therefore, by definition 1.18, domain of domain of ,
domain of domain of domain of and
domain of because at .
Remark: There are situations in which the natural domain associated with the formula resulting
from an operation on two functions is not the correct domain for the new function.
For example, if and , then according to definition 1.18, the domain of should be . However, ,
which has a natural domain of which is incorrect. Thus, to be precise in describing the formula
for , we must write
.
 Composition of functions
Definition 1.19: Given functions and, the composition of with, denoted by , is a function
defined by . The domain of is defined to consist of all in the domain of for which is in the
domain of .
Example 1.23: Let and. Find
a. b.
Solution:
a. The formula for . Since the domain of is and the domain of is , the domain of consists of
all in such that lies in . Thus, the domain of is .
Therefore, .
b. The formula for is . Since the domain of is and the domain of is , the domain of consists of
all in such that
lies in . Thus, the domain of is . Therefore, .
Here, no need to indicate that the domain is , since this is the natural domain of .

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Remark: The functions and in the preceding example are not the same. Thus, the order in
which functions are composed can (and usually will) make a difference in the end result.
Example 1.24: Express as a composition of two functions.
Solution: To evaluate for a given value of , we would first compute and then raise the result to
the fifth power. Therefore, the inside function (first operation) is and the outside function
(second function) is . So, .
As a check, .
Types of functions
A function is classified in three as follow.
a. One -to-One functions
Definition 1.15: A function is a one-to-one (an injective) if and only if for any ,
[equivalently, ]. That is, a function is a one-to-one (an injective) if and only if every element
of the range has exactly one and only one pre-image in the domain.
Note: The statement is the contra positive of the statement
and are equivalent statements.
Example 1.17: Show that defined by is a one-to-one function.
Solution: For any, let , then
. Or for any , let , then
. Hence by definition 1.15, the given function is a one-to-one function.
Example 1.18: Show that the function given by is not one-to-one.
Solution: Let’s take and, then clearly .
But . This implies that there are numbers such that but . Therefore, is not a one-to-one function.
Note: A function is one-to-one if and only if any horizontal line crosses the graph of at most
once.
b. Onto functions
Definition 1.16: A function is said to be an onto (a surjective) if and only if every element of
has a pre image in . That is, a function is onto if and only if range of co domain (Or range ).
Example 1.19: Let defined by. Show that is an onto function.
Solution: To show that is an onto, we have to show that, for every , there is (in the domain of )
such that and this is done by solving for from .

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Let . Therefore, for every , there exists such that , that is, . Which implies that is an onto
function. Or simply the range of and co domain of range of co domain of and hence it is an
onto function.
Example 1.20: The function given by is not an onto function because if we take , there is no
such that . Or in other words range of (co domain)
since range of .

c. One-to-one correspondence
Definition 1.17: A function is a one-to-one correspondence (a bijective) if and only if is both
one-to-one and onto.
Example 1.21: Show that the function given by is a one-to-one
correspondence.
Solution: To show that the given function is a one-to-one correspondence, we have to show that,
it is both one-to-one and onto. For any , let
and thus is a one-to-one function.
Clearly, the range of is . That is, range of which implies that is an onto function.

Even and odd functions

a. A function with the property that is called an even function.
For example, the functions and are even functions.
b. A function with the property that is called an odd function.
For example, the functions and are odd functions.
Example 1.25: The function is an odd function because
.
The function is an even function, since .
 Inverse functions and their graphs
Definition 1.20: Suppose be a function. A function is called the inverse function of if
(identity function) and. If is the inverse function of , then we often rename as .
Note: is function if and only if is a one-to-one correspondence function.
Steps for finding inverse of a function :
i. First interchange and in the formula of
ii. Solve for in terms of

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 iii. Write
Example 1.26: Let be the function defined by , and let be the function defined by . Show that
is the inverse of .
Solution: and

Hence, by the definition is the inverse of .

Example 1.27: Find the inverse of the function .
Solution: (Original function)
(Interchange and in the formula of )
(Solve for in terms of )
(Write )
Example 1.28: Let . Show that is not a one-to-one on .
Solution: Take , then clearly but .
Theorem 1.1 (The Horizontal line test): A function has an inverse if and only if its graph is
cut at most once by a horizontal line.
Example 1.29:
1. The function does not have an inverse and this is not confirmed by the horizontal line
test. Since the graph of is cut more than once by certain horizontal lines.

Figure1.41
2. The function has an inverse since its graph passes the horizontal line test.

Figure1.42
 Graphs of Inverse Functions
Next we explore the relationship between the graphs of and. For this , it is desirable to use as the
independent variable for both functions, which means that we will compare the graphs of

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and.
If is a point on the graph of , then . This is equivalent to the statement that , which means that is
a point on the graph of . In short, reversing the coordinates of a point on the graph of produces a
point on the graph of . Similarly, reversing the coordinates of a point on the graph of produces a
point on the graph of . Therefore, the graphs of and are reflections of one another about the line .
Theorem 1.2: If has an inverse, then the graphs of are reflections of one another about the
line ; that is, each is the mirror image of the other with respect to .

Figure1.43
Example 1.30: The following graphs show graphs of the functions and
with their respective inverses.

a. b.
c.
Figure1.44

Polynomials, zeros of polynomials, rational functions and their graphs

 Polynomial functions
Definition 3.13: A polynomial function is a function of the form

y  a n x n  a n 1 x n 1   a1 x  a 0 , a n  0.
n
Each ai is assumed to be a real number, and is a non-negative integer, a n is called the
leading coefficient. Such a polynomial is said to be of degree n.
Remark:
1. The domain of a polynomial function is always the set of real numbers.

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 2. (Types of polynomials)
- A polynomial of degree 1 is called a linear function.
- A polynomial of degree 2 is called quadratic function.
- A polynomial of degree 3 is called a cubic function.
i.e p ( x)  a 3 x  a 2 x  a1 x  a 0 , a 3  0.
3 2

Example 3.18: p( x )  2 x  1 , q( x )  3 x  2 x   f ( x )  2 x 3 are examples of

2 4
and
polynomial functions.
 Properties of polynomial functions
1. The graph of a polynomial is a smooth unbroken curve.
2. If p is a polynomial of degree n , then it has at most n zeros. Thus, a quadratic
polynomial has at most 2 zeros.
3. The graph of a polynomial function of degree n can have at most n  1 turning points.
Thus, the graph of a polynomial of degree 5 can have at most 4 turning points.
4. The graph of a polynomial always exhibits the characteristic that as x gets very large,
y gets very large.

 Zeros of a polynomial
A number a is said to be a zero of a polynomial function p if p( a )  0 .

Division Algorithm

Let p(x ) and d (x ) be polynomials with d ( x )  0 , and with the degree of d (x ) less than or
equal to the degree of p(x ) . Then there are polynomials q(x ) and R(x ) such that
p ( x)  d ( x). q ( x)  R( x) R( x )  0 R(x )
dividend divisor quotient remainder , where either or the degree of is less than degree
of d (x ) .
Example 3.19: Divide .
Solution: Using long division we have

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 x2  2x  4
x 2  2x x 4  0x 3  0x 2  0x  1
 (x4  2x3 )
 2x 3  0x 2
 ( 2 x 3  4 x 2 )
4x 2  0x
 (4 x 2  8 x )
 8x  1

x 4  1  ( x 2  2 x ). ( x 2  2 x  4)  ( 8 x  1)
This long division means dividend divisor quotient remainder .

In long division algorithm if the divisor d (x ) is linear (that is of the form x  r ), then

p ( x )  ( x  r ) q( x )  R .

Note that since the divisor is of the first degree, the remainder R , must be a constant and
substitute x  r , into this equation, we get
P ( r )  ( r  r ) q( r )  R  0  q( r )  R

Therefore, p( r )  R .

The Remainder Theorem

When a polynomial p(x ) of degree at least 1 is divided by x  r , then the remainder is p(r ) .

Example 3.20: The remainder when P( x )  x  x  3x  1 is divided by x  2 is p( 2)  9 .

3 2

NB: From the remainder theorem, if x  r is a factor of p(x ) , then the remainder is 0.
Conversely, if the remainder is 0, then x  r , is a factor of p(x ) . This is known as the Factor
Theorem.

The Factor Theorem

x  r is a factor of p(x ) if and only if p( r )  0 .

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 Location theorem

Let f be a polynomial function and a and b such that a  b . If f ( a ) f (b)  0 , then there
is at least one zero of f between a and b .

Fundamental Theorem of Algebra

If p(x ) is a polynomial of degree n  0 whose coefficients are complex numbers, then p(x )
has at least one zero in the complex number system.

The linear Factorization Theorem

n 1
If p( x )  a n x  a n 1 x 
n
 a1 x  a 0 , where n  1 and an  0 , then

p ( x)  a n ( x  r1 ) ( x  r2 ) ( x  rn ) , where the ri are complex numbers (possible real and

not necessarily distinct).
NB: From the linear factorization theorem, every polynomial of degree n  1 has exactly n
zeros in the complex number system, where a root of multiplicity k counted k times.
Example 3.21: Express each of the polynomials in the form described by the Linear
Factorization Theorem. List each zero and its multiplicity.

a) p( x )  x 3  6 x 2  16 x
b) q( x )  3x  10 x  8
2

c) f ( x )  2 x  8 x  10 x
4 3 2

Solution:

a) We may factorize p(x ) as follows:

p( x )  x 3  6 x 2  16 x  x ( x 2  6 x  16)
 x ( x  8)( x  2)
 x ( x  8)( x  ( 2))

The zeros of p(x ) are 0, 8, and – 2 each of multiplicity one.

b) We may factorize q(x ) as follows:

q( x )  3x 2  10 x  8  (3x  4) ( x  2)
4
 3( x  )( x  2)
3

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 q(x ) 4
Thus, the zeros of are 3 and 2, each of multiplicity one.

c) We may factorize f (x ) as follows:

f ( x )  2 x 4  8 x 3  10 x 2  2 x 2 ( x 2  4 x  5)
 2 x 2 ( x  ( 2  i ))( x  ( 2  i ))

The zeros of f(x) are 0 with multiplicity two and  2  i and  2  i each with multiplicity one.
Example 3.22:

1. Find a polynomial p(x ) with exactly the following zeros and multiplicity.
zeros multiplicity
1 3
2 4
5 2
Are there any other polynomials that give the same roots and multiplicity?
2. Find a polynomial f (x) having the zeros described in part (1) such that f(1) = 32.

Solution:
1. Based on the Factor Theorem, we may write the polynomial as:
p ( x)  ( x  (1)) 3 ( x  2) 4 ( x  5) 2  ( x  1) 3 ( x  2) 4 ( x  5) 2

which gives the required roots and multiplicities.

Any polynomial of the form kp (x ) , where k is a non-zero constant will give the same
roots and multiplicities.

2. Based on part (1), we know that f ( x)  k ( x  1) ( x  2) ( x  5) . Since we want

3 4 2

f ( x )  32 , we have
f (1)  k (1  1) 3 (1  2) 4 (1  5) 2
32  k (8)(1)(16)  k  1
4

Thus, f ( x )  4 ( x  1) ( x  2) ( x  5) .
1 3 4 2

Conjugate Roots Theorem

Let p(x ) be a polynomial with real coefficients. If a complex number a  bi (where ) is a

zero of p(x ) , then so is its conjugate a  bi .

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Example 3.23: Let r ( x)  x  2 x  9 x  26 x  20. Given that 1  3 i is a zero, find the other
4 3 2

zero of r (x ) .

Solution: By the Conjugate Roots Theorem, if 1  3 i is a zero, then its conjugate, 1  3 i

must also be a zero. Therefore, x  (1  3 i ) and x  (1  3 i ) are both factors of r (x ) , and so
their product must be a factor of r (x ) . That is, [ x  (1  3 i )] [ x  (1  3 i )]  x  2 x  4 is a
2

factor of r (x ) . Dividing r (x ) by x 2  2 x  4 , we obtain

r ( x )  ( x 2  2 x  4)( x 2  4 x  5)  ( x 2  2 x  4) ( x  5) ( x  1).

Thus, the zeros of r (x ) are 1  3 i , 1  3 i ,  5 and 1.

The Rational Root Theorem

Suppose that  a1 x  a 0 , where n  1, a n  0 is an n th degree

f ( x )  a n x n  a n 1 x n 1 
p f ( x)  0 p q
polynomial with integer coefficients. If q is a rational root of , where and
p q
have no common factor other than  1 , then is a factor of a 0 and is a factor of a n .
3
To get a feeling as to why this theorem is true, suppose 2 is a root of
a 3 x 3  a 2 x 2  a1 x  a 0  0 .
3 2
 3  3  3
a 3    a 2    a1    a 0  0
Then,  2  2 2 which implies that
27 a 3 9a 2 3a1
   a0  0
8 4 2 multiplying both sides by 8
27 a 3  18a 2  12a1  8a 0 ...................................................(1)

27 a 3  18a 2  12a1  8a 0 ...................................................(2)

If we look at equation (1), the left hand side is divisible by 3, and therefore the right hand side
must also be divisible by 3. Since 8 is not divisible by 3, a 0 must be divisible by 3. From
equation (2), a 3 must be divisible by 2.

Example 3.24: Find all the zeros of the function p( x )  2 x  3x  23x  12.
3 2

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 p
Solution: According to the Rational Root Theorem, if q is a rational root of the given equation,
then p must be a factor of  12 and q must be a factor of 2. Thus, we have

possible values of p :  1,  2,  3,  4,  6,  12

possible values of q :  1,  2
p 1 3
 1,  ,  2,  3,  ,  4,  6,  12
possible rational roots q : 2 2

We may check these possible roots by substituting the value in p(x ) . Now p(1)  30 and
p( 1)  12 . Since p(1) is negative and p(1) is positive, by location theorem, p(x ) has a zero
between  1 and 1. Since P  2   0 , then x  2  is a factor of p(x ) . Using long division, we
1 1

obtain

p( x )  2 x 3  3x 2  23x  12  ( x  12 )( 2 x 2  2 x  24)
 2( x  12 )( x  4)( x  3)

Therefore, the zeros of p(x) are  2 ,  4 and 3.

1

 Rational Functions and Their Graphs

Definition 2.5: A function is called a rational function if it can be written in the form , where
and are polynomial functions and .
Note:
i. The domain of the rational function is a set of all real numbers except the points for which the
denominator gets zero.
ii. If the functions and have no common factor other than, then the
rational function is said to be in its simplified form.
Example 2.15:
1. The following are examples of rational functions.
, , ,
2. Given , then

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 d. (undefined)

 Asymptotes
A rational function often has asymptotes: vertical and/or horizontal/or oblique. Informally
speaking, an asymptote is a straight line (vertical, horizontal or slanted) toward which the graph
comes near.

(a) Vertical and horizontal asymptotes (b) Vertical and oblique asymptotes
Figure1.6
Before defining asymptotes, note the following.
Symbol Meaning
approaches to from the left
approaches to from the right
goes to negative infinity; that is, decreases with out bound
goes to positive infinity; that is, increases with out bound
i. The line is called a vertical asymptote of the function if approaches
as approaches from the right or left.

(a) as (b) as

(c) as (d) as
Figure2.7
ii. The line is called a horizontal asymptote of the function if approaches as
approaches .

(b) as

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 (a) as
Figure2.8
Asymptotes of a rational functions
Let be a rational function where are leading
coefficient and degree of and are leading coefficient and degree of respectively.
i. The vertical asymptotes are the lines . i.e., at the point where and .
If is a common factor of and then the function has a hole at or it requires farther
simplification.
ii. If , then the function has horizontal asymptote .
iii. If , then the function has horizontal asymptote .
iv. If , then the function has no horizontal asymptote.
v. If , then the function has oblique (slant) asymptote.
vi. If , then the function has neither horizontal nor oblique asymptotes but it has an asymptote
called a curve linear asymptote.
Example 2.16: Identify the asymptotes of the rational function .
Solution: First factor the denominator of the rational function, and is not the factor of the
numerator as . The degree of the numerator and denominator are equal and.
Therefore, vertical asymptote is the line and horizontal asymptote is the line .
Note:
1. A rational function can have only one horizontal/oblique asymptote, but it may have many
vertical asymptotes.
2. If a rational function has a horizontal asymptote, then it does not have an oblique one.
3. The graph of a rational function can cross its horizontal/oblique asymptote, but does not
cross its vertical asymptote.
4. Horizontal/oblique asymptotes describe the behavior of function for with large absolute
value; vertical asymptotes describe the behavior of function near a point.
Steps for Sketching Graph of a Rational Function
i. Find the domain: (a) solve (b) Domain of .
ii. Find -and -intercepts: -intercept: and -intercepts: at numerator .
iii. Find vertical asymptotes, if any.
iv. Find the horizontal/oblique asymptote, if any.

21 | P a g e
v. Find the points where the graph crosses the horizontal/oblique asymptote by solving the
equation .
vi. Check for symmetries
a. If then the graph is symmetric about - axis;
b. If , then the graph is symmetric about the origin.
Remark: If the graph is symmetric then only sketch graph the function for and use symmetry
to sketch the corresponding part for.
vii. Use the sign chart for the “reduced”, if , then the graph of is above the
-axis and if , then the graph is below the -axis.
viii. Sketch the graph of using steps (i) – (vii).
Example 2.17: Sketch the graph of the rational function.
Solution:
i. Domain: .
ii. Intercepts: -intercepts are and -intercept is .
iii. Vertical asymptote are the lines and .
iv. Degree of numerator degree of the denominator, then is the horizontal asymptote.
v. Intersection with asymptote: Here solve. That is
or .
The graph crosses its horizontal asymptote at . That is, is the intersection point.
vi. Symmetries:
and which implies .
Hence, is neither symmetric about -axis nor the origin.
vii. Sign chart for

The graph is above the -axis on and is below the -axis on

.
viii. Graph is

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 Figure2.9
Examples:
1. Find the asymptotes of the function .
Solution: First factor both denominator and numerator of the function and then simplify if
possible. Thus we have .
Vertical asymptote: . But the line is not the vertical asymptote though the denominator is zero at
rather it has a hole at .
Horizontal asymptote: (that is, the line ).
2. Find the asymptotes of the rational function .
Solution:
Vertical asymptote is the line .
Has no horizontal asymptote because degree of numerator is greater than the degree of
denominator.
Oblique (slant) asymptote: Since degree of the numerator degree of the denominator, then by
long division, the line is the oblique asymptote.
3. Sketch .
Solution: In similar fashion to example 2.17 above we have the following.
i. Domain : .
ii. Intercepts: intercept is and -intercept is .
iii.Vertical asymptote is the line .
iv.Has no Horizontal asymptote (since degree of numerator is greater than the degree of
denominator). Since degree of numerator degree of denominator, then using long the line is
the oblique asymptote.
v. is neither odd nor even and its graph is given below.

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 Figure2.10
4. Sketch . At what point does the graph intersect its horizontal asymptote?

Solution:
i. Domain: .
ii. Intercepts: -intercept are and -intercept is .
iii. Vertical asymptote is the line .
iv. Horizontal asymptote is the line .
v. No oblique asymptote (since it has a horizontal asymptote).
vi. Intersection with horizontal asymptote: solve the equation , we get
. Thus, intersects its horizontal asymptote
at.Therefore, the graph is

Figure2.11
5. Sketch
Solution: (using long division)
i. Domain:
ii. Intercepts: intercept is and -intercept is
iii. Vertical asymptotes are the lines and
iv. Has no Horizontal asymptote (Since degree of numerator degree of denominator)
v. As degree of numerator degree of denominator, the line is the Oblique asymptote
vi. The graph intersects its oblique asymptote at which is evaluated by solving
Therefore the graph is

24 | P a g e
 Figure2.12
6. Identify the asymptote and sketch .
Solution:
i. Domain:
ii. Intercepts: -intercept is and it has no -intercept since is not in the domain.
iii. Vertical asymptote is the line or -axis.
iv. No Horizontal asymptote (since degree of numerator degree of denominator).
v. No oblique asymptote (since degree of numerator degree of denominator).
vi. Using long division, . Thus, function has a curve linear
asymptote and its graph is as below.

Figure2.13
Note: The graph of a function can intersect its curve linear asymptote.
7. At what point does the graph of the function intersect its oblique asymptote?
Solution: By using long division, . Hence, the line is the oblique asymptote of the graph of the
given function. Here, to find the intersection point of the graph with its oblique asymptote, solve
the equation.
Next, solve or or or
and . Therefore, the function intersects its oblique asymptote at the point and the graph of the
function is

Figure2.14
Its domain is and the line or simply -axis is its vertical asymptote.

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 Exponential and Logarithmic Functions
 Exponential functions
Definition 1.12: A function of the form, where, and , is called an exponential function with
base .
The domain of the exponential function is .
The range of this exponential function is .
Note: An exponential function has a constant base and a variable exponent. Hence, functions of
the form , are not classified as exponential functions, because they have a variable base and a
constant exponent.

The Exponential function y  f ( x )  b

x

1. The graph of y  b exhibits exponential growth if b  1 or exponential decay if

x

0  b  1.
2. The y  intercept is 1.
3. The x  intercept is a horizontal asymptote
4. The exponential function is 1 – 1. Algebraically if b x  b y , then x  y
Example 1.12: The functions, , , , , and are exponential functions.
The graph of an exponential function has one of the basic two shapes depending on whether or .

Figure1.31
Example 1.13: Graphs of some specific exponential functions are given below.

Figure1.32
Note: If , then the function becomes a constant, since . That’s why we exclude it from the
families of exponential functions. An exponential function with fixed base
(that is, ) is called a natural exponential function.
Laws of Exponential Expressions
For all real numbers and for , the following laws hold.
i. iii.
ii. iv.

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Note: An exponential function obeys the following properties.
d. f.
e. g.
Example : Solve the following equations. (a) (b)
Solution:
a. b. ()
Example: Sketch the graph of
Solution: First reflect the graph of about the -axis to get the graph of . Then shift the graph of
upward units to obtain the graph of as in the figure below. The domain is and range is .

(a) (b) (c)

Exercise: Sketch the graph of the function
Answer: Similarly, reflect the graph of about the -axis to obtain the graph of . Then compress the graph
vertically by a factor of to get the graph of . Finally, shift the graph downward one unit to get the of . The
domain is and range is . Then the graph of the function is;

 Logarithmic Functions
Definition 1.13: If and , then the function given by for , is called the logarithmic function with base .
Domain of the function is and Range of the function is
Note: The graphs exponential function and logarithmic function are reflections of one another
about the line and hence they are inverse functions of each other.
If we take and , and if we keep in mind that the domain of is the same as the range of , then we
get the following.
for all and for all

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A logarithmic function with fixed base is called a natural logarithm and is given by .

The Logarithmic Function y  log b x have the following properties

1. Its graph exhibits logarithmic growth if b  1 and logarithmic decay if 0  b  1 .
2. The x  intercept is 1. There is no y  intercept.
3. The y  axis is a vertical asymptote.
Laws of Logarithmic Expressions
If and are positive numbers and then:
i. v. (change of base)
ii. vi.
iii. ; for any real number vii.
iv.

Example:
1. Solve the following equations.
a. b.
Solution:
a.
b.
2. Sketch the graph of the function.
Solution: To sketch this graph, shift two units to the right to get the graph of and then shift one unit
downward to get the graph of .

(a) (b) (c)

Solution: To sketch the graph of , shift the graph of by units to the left to get the graph of
and then shift the graph of by 1 unit upward to obtain the graph of .

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Exercise: Sketch he graph of the function

Answer: To sketch the graph of the function, , first sketch the graph of the function , and then
reflect this graph about the -axis to get the graph of the function

More Examples:
1. Solve the equation
Solution:
Original equation
Letting
Factorizing
or Solving for
or Since we let
has no solution Because for all

Solving for

2. Solve
Solution:

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 . Since , then the solution is
 Trigonometric functions and their graphs

Trigonometric functions are used to model the regular periodic patterns that play an important
role in the social, biological, and physical sciences: business cycles, agricultural seasons, heart
rhythms, and hormone level fluctuations, and tides and planetary motions.
Angle Measurement
An angle is formed by the union of two rays with common edges/end point an in fig below. The
common end point is called the vertex of the angle and rays are the sides of the angle.

In forming the angle, one side remains fixed while the other side rotates. The fixed side is called
the initial side and the side that rotates is called the terminal side. If the terminal side rotates in a
counter clockwise direction, the angle positive angle, and if the terminal side rotates in a
clockwise direction, the angle negative angle.

B B

The degree and radian measures are related as

Example 3.39:
1. Convert each of the following radian measures to degrees.
 3
a) 6 b) 5

Solution: a)
b)
2. Convert to radian measures
a) 90 b) 270

Solution: a) .
b) .

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Angle position: an angle is said to be in its standard position if the vertex of  is placed at the
origin and initial side of  is placed along the positive x  axis . Thus the location of the
terminal side of  will, of course, depend on the size of  .

Y
Y
P(x,y)
ϴ
r
X
X

When we locate a point (other than the origin) on the terminal side of  , we identify its
coordinates ( x, y ) and its distance to the origin, denoted by r . Then, r is positive.
Let an angle  be in its standard position, we define the six trigonometric functions of  as
follows:
Definition 3.23
Name of function Abbreviation Definition
 sin  y
sin  
Sine r
 cos x
cos 
Cosine r
 tan  y
tan  
Tangent x
 csc  r
csc  
Cosecant y
 sec  r
sec  
Secant x
 cot  x
cot  
Cotangent y
Note: The domain of and is the set of real numbers and their range is .

 The graph of y  sin 

We can plot using tabular method or unit circle in the figure below.

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 (0,1)
(x,y)

θ
(-1,0) (1,0)

(0,-1)

As the terminal side of  moves through the first

y 

quadrant, increases from 0 (when   0 ) to 1(when   2 ). Thus, as


increases from 0 to

2 ,
y  sin  steadily increases from 0 to 1.


As  increases from 2 to , y  sin  decreases form 1 to 0. A similar analysis reveals that as


 increases from  to 32 , sin  decreases from 0 to – 1; and as  increases from 32 to 2 ,
sin  increases from – 1 to 0.

Based on this analysis, we have the graph of f ( )  sin  in the interval [0,2 ] as show below.

y = sin x

Since the values of f ( )  sin  depend only on the position of the terminal side, adding or
subtracting multiples of 2 to  will leave the value of f ( )  sin  unchanged. Thus, the
values of f ( )  sin  will repeat every 2 units. The complete graph of f ( )  sin  appears
below.

The graph of y  sin x , which is called the basic sine curve.

 The graph of y  cos 

Similarly, the graph of is given below.

32 | P a g e
  The graph of y  tan 

y x  0 tan 
tan  
Since x is undefined whenever , is undefined whenever the terminal side of
the angle corresponding to  falls on the y  axis . This happens for   2 , to which we can


add or subtract any multiple of  that will again bring the terminal side back to the y  axis .
n
Thus, domain of tan  is { :   2  n } , where is an integer and the graph is given below.


Note the following.

1. f ( x )  sin x Domain = All real numbers

2. f ( x )  cos x Domain = All real numbers
Domain = {x : x  2  n }

3. f ( x )  tan x
f ( x )  csc x Domain = { x : x  n }
4.
Domain = {x : x  2  n }

5. f ( x )  sec x
6. f ( x )  cot x Domain = {x : x  n }
where n is an integer
Definition (Periodic Function)
A function y  f (x ) is called periodic if there exists a number p such that
f ( x  p )  f ( x ) for all x in the domain of f . The smallest such number p is called the
period of the function
A periodic function keeps repeating the same set of y  values over and over again. The graph of
a periodic function shows the same basic segment of its graph being repeated. In the case of sine
and cosine functions, the period is 2 . The period of the tangent function is  .
Definition (Amplitude of a periodic function)

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 The amplitude of a periodic function f (x ) is
1 f (x )  f (x )]
A [
2 maximum value of minimum value of
Amplitude is the height from the center line to the peak (or to the trough).

Thus, the amplitude of the basic sine and cosine function is 1.

Definition (Frequency of a periodic function)

The number of complete cycles a sine or cosine graph makes on an interval of length equal to
2 is called its frequency.

Graphs of Transformed Sine and Cosine Functions

The transformed sine and cosine function are given by and respectively, where and are
constants with

 is the amplitude (or the maximum value of the function)

 is the period of the function
 is the frequency
 is the phase shift (or horizontal shift) and
 is the vertical shift.

Example: Sketch the graph of . Find the amplitude, the period, and the phase shift.
Solution: Note that . Then we have the following:
Amplitude , Period and the phase shift
To sketch the final graph, begin with the basic sine curve, . Then sketch graphs of each of the
following equations in sequence.

34 | P a g e
Example: Graph . Find the amplitude, and the period.
Solution: First we evaluate the following;
Amplitude
The period of is and the period of is . The period of the sum is the least common multiple of
and which is . That is, period . To sketch the graph of , graph and using the same coordinate
system.

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Now graphically add some coordiates, or ordinates, to obtain points on the graph that we seek.
At , we transfer the distance , which is the value of , up to add it to the value of , point is on the
graph that we seek. At , use a similar procedure, but this time both ordinates are negative, point
is on the graph. At , add the negative ordinate of to the positive ordinate of , point is also on the
graph. Continue to plot points in this fashion and then connect them to get the desired graph, as
shown below. This method is called addition of ordinates, because we add the values (ordinates)
of .

Example: Find the maximum value of the function

Solution: The maximum value is .
Exercise: Sketch the graph of the following and find the amplitude, the period, and the phase
shift.
a.
b.

Some Trigonometric Identities

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The reciprocal Identities 1.

2.

3.

The quotient Identities 4.

5.
The Pythagorean Identities 6. sin 2 x  cos 2 x  1
7. tan 2 x  1  sce 2 x
8. 1  cot 2 x  csc 2 x

The addition formula 9. (a) sin( x  y )  sin x cos y  cos x sin y

(b) sin( x  y )  sin x cos y  cos x sin y
10. (a) cos( x  y )  cos x cos y  sin x sin y
(b) cos( x  y )  cos x cos y  sin x sin y
tan x  tan y
tan( x  y ) 
11. (a) 1  tan x tan y
tan x  tan y
tan( x  y ) 
(b) 1  tan x tan y

The double angle formula 12. sin 2 x  2 sin x cos x

13. cos 2 x  cos 2 x  sin 2 x  1  2 sin 2 x  2 cos 2 x  1
2 tan x
tan 2 x 
14. 1  tan 2 x

x 1  cos x
sin 
The half-angle formula 15. 2 2
x 1  cos x
cos 
16. 2 2
x 1  cos x
tan 
17. 2 1  cos x

 Hyperbolic functions and their graphs

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The hyperbolic functions are certain combinations of exponential functions, that occur in various
applications, with properties similar to those of the trigonometric functions. The two basic
hyperbolic functions are the hyperbolic sine and hyperbolic cosine functions. They are defined as
follows:

Definition
1. The hyperbolic sine function is 2. The hyperbolic cosine function is
defined by: defined by:
e x  ex e x  ex
sinh x  cosh x 
2 2
The domain of sinh x is  . The domain of cosh x is also .

Remark:

1. Since e  0 for all x   , we see that cosh x  sinh x for every x   .

x

e x  ex ex  e x e x  ex cosh x

f ( x)  f ( x)    f ( x)
2. If 2 , then 2 2 . Thus, is an even
function.
3. Sinhx is an odd function.
4. In contrast to sine and cosine, the hyperbolic functions are not periodic.

Example 3.28: Using the above definitions, show that

1. cosh x  sinh x  1
2 2

2. sinh( x  y )  cosh x sinh y  sinh x cosh y

3. sinh( x  y )  sinh x cosh y  cosh x sinh y

Solution:
1. We have
2 2
 e x  ex   e x  ex   e 2 x  2  e 2 x   e 2 x  2  e 2 x 
            1
 2   2   4   4 
e x  y  e  x  y e x e y  e  x e  y 2e x e y  2e  x e  y
sinh( x  y )   
2. 2 2 4

e x e y  e x e  y  e  x e y  e  x e  y e x e y  e x e  y  e  x e y  e  x y
 
4 4

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  e x  ex  e y  e  y   e x  ex  e y  e  y 
       
 2  2   2  2 

 cosh x sinh y  sinh x cosh y

3. Left as an exercise.

 The graph of y  cosh x

Since cosh x is an even function, its graph is symmetric about the y  axis . Its y  intercept is
(0,1) cosh(0)  1 x e x ex
cosh x  
, because . As tends to infinity, 2 2 tends to infinity because
e x
e x x cosh x
2 goes to infinity and 2 approaches to 0. When is a large negative number acts
e x
e x y  cosh x
like 2 , because 2 gets close to 0. Thus the graph of looks like:

x
This graph can also be obtained by geometrically adding the two curves y  e and y  e ,
x

and taking half of each resulting y  value . Observe that range of cosh x is [1, ) .

 The graph of y  sinh x

Since sinh x is an odd function, its graph is symmetric about the origin. The graph passes
sinh(0)  0 x sinh x ex x
through the origin because . As gets large acts like 2 and when is a
sinh x e x y  sinh x

large negative number, acts like 2 . Thus, the graph of looks like:

39 | P a g e
 The remaining four hyperbolic functions are defined in terms of cosh x and sinh x by
analogy with trigonometry.

sinh x e x  e  x tanh x )
tanh x  
cosh x e x  e  x (The domain of is .
cosh x e x  e  x coth x  \ {0}
coth x  
sinh x e x  e  x (The domain of is )
1 2 sec hx 
sec h x   x
cosh x e  e  x (The domain of is )
1 2 csc h x  \ {0}
csc h x   x
sinh x e  e  x (The domain of is )

Some identities of hyperbolic functions

cosh 2 x  sinh 2 x  1 (1)

1  tanh 2 x  sec h 2 x (2)
coth 2 x  1  csc h 2 x (3)
sinh( x  y )  sinh x cosh y  cosh x sinh y (4)
cosh( x  y )  cosh x cosh y  sinh x sinh y (5)

The trigonometric functions are sometimes called circular functions because the point
(cos , sin  ) lies on the circle x 2  y 2  1 for all  . Similarly, identity (1) tells us that the point
(cosh  , sinh  ) lies on the hyperbola x 2  y 2  1 , and this is the reason for the name hyperbolic
functions.

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