Chapter Two

Functions
Our everyday lives are filled with situations in which we encounter relationships between two
sets. For example,
 To each automobile, there corresponds a license plate number
 To each circle, there corresponds a circumference
 To each number, there corresponds its square

In order to apply mathematics to a variety of disciplines, we must make the idea of a

“relationship” between two sets mathematically precise.

On completion of this chapter students will be able to:

 understand the concept of real numbers
 use properties of real numbers to solve problems
 determine whether a given real number is rational number or not
 solve linear equations and inequalities
 solve quadratic equations and inequalities
 understand the notion of relation and function
 determine the domain and range of relations and functions
 find the inverse of a relation
 define polynomial and rational functions
 perform the fundamental operations on polynomials
 find the inverse of an invertible function
 apply the theorems on polynomials to find the zeros of polynomial functions
 apply theorems on polynomials to solve related problems
 sketch and analyze the graphs of rational functions
 define exponential, logarithmic, and trigonometric functions
 sketch the graph of exponential, logarithmic, and trigonometric functions
 use basic properties of logarithmic, exponential and trigonometric functions to solve
problems

In this chapter, before discussing the idea of relations and functions we first review the system of
real numbers, linear and quadratic equations and inequalities.

39
 1.1The real number systems

At the end of this section, students will be able to:

 understand the concept of real numbers
 use properties of real numbers to solve problems
 determine whether a given real number is rational number or not
In this section we will define what the real numbers are and what are their properties? To answer,
we start with some simpler number systems.

 The integers and the rational numbers

The simplest numbers of all are the natural numbers,

1, 2, 3, 4, 5, 6, ⋯

With them we can count: our books, our friends, and our money. If we adjoin their negatives
and zero, we obtain the integers;
⋯, −3, −2, −1, 0, 1, 2, 3,⋯
When we try to measure length, weight, or voltage, the integers are inadequate. They are spaced
too far apart to give sufficient precision. Thus, we are led to consider quotients (ratios) of
integers, numbers such as:
3 −7 21 19 16 −17
, , , ,
4 8 5 −2 2 and 1
16 −17
Note that we included 2 and 1 , though we would normally write them as 8 and – 17,
5
since they are equal to the latter by the ordinary meaning of division. We did not include 0 or
−9
0 , since it is impossible to make sense out of these symbols. In fact, let us agree once and for
m
all to banish division by zero from this section. Numbers which can be written in the form n ,

where m and n are integers with n≠0 , are called rational numbers.

Do the rational numbers serve to measure all lengths? No. This surprising fact was discovered by

the ancient Greeks long ago. They showed that while √2 measures the hypotenuse of a right
triangle with sides of length 1, it cannot be written as a quotient of two integers(see exercise…).

40
 3
Thus, Thus, √2 is an irrational (not rational) number. So are √ 3 , √5 , √ 7 , π and a host of
other numbers.
 The real numbers

Consider the set of all numbers (rational and irrational) that can measure lengths, together with
their negatives and zero. We call these numbers the real numbers.
The set of real numbers denoted by ℜ can be described as the union of the set of rational and

irrational numbers. i.e ℜ = {x : x is a rational number or an irrational number}.

The real numbers may be viewed as labels for points along a horizontal line. There they measure
the distance to the right or left (the directed distance) from a fixed point called the origin and
labeled 0. Each point on the number line corresponds a unique real number and vice-versa.

Most students will remember that the number system can be enlarged still more to the so-called

complex numbers. These are numbers of the form a+b √−1 , where a and b are real
numbers.
 The four arithmetic operations

Give two real numbers x and y , we may add or multiply them to obtain two new real

numbers x+ y and x⋅y (also written simply as xy ). The real numbers along with the

operations of addition (+) and multiplication ( ¿) , obey the 11 properties listed below. Most of
these properties are straightforward and may seem trivial. Nevertheless, we shall see that these
11 basic properties are quite powerful in that they are the basis for simplifying algebraic
expressions.

The commutative Properties

1. For addition: a+b=b+ a
2. For multiplication: ab=ba

The associative properties

41
 3. For addition: a+(b+c )=(a+ b )+ c
4. For multiplication: a(bc )=(ab )c

The distributive property

5. a(b+ c )=ab+ac or (b+c )a=ba+ca

Identities
6. For addition: There is a unique number called the additive identity, represented by 0,
which has the property that a+0=a=0+ a for all real numbers a .
7. For multiplication: There is a unique real number called the multiplicative identity,
represented by 1, which has the property that a⋅1=a=1⋅a for all real numbers
a .

Inverses
8. For addition: Each real number a has a unique additive inverse, represented by
−a , which has the property that a+(−a)=0=(−a)+a
9. For multiplication: Each real number a , except 0, has a unique multiplicative inverse,
1 1 1
represented by a , which has the property that a⋅( a )=1=( a ) a .

Closure properties
10. For addition: The sum of two real numbers is a real number.
11. For multiplication: The product of two real numbers is a real number.

Subtraction and division are defined by:

1
x− y=x +(− y ) and x÷ y=x⋅ y , where y≠0 .

In the product ab , a and b are called factors, in the sum a+b , a and b are
called terms.

Example 2.1: The set of irrational numbers is not closed under addition and multiplication,

because √ 2+(−√2 )=0 and √ 2 √ 8=√16=4 , which are rational numbers.

 The order relation on the set of real numbers

42
The nonzero real numbers separate nicely into two disjoint sets – the positive real numbers and
the negative real numbers. This fact allows us to introduce the order relation < (read “is less
than”) by
x< y ⇔ y −x is positive

We agree that x< y and y > x will mean the same thing. The order relation ¿ (read”is
less than or equal to”) is a first cousin of <. It is defined by
x≤ y ⇔ y−x is positive or
zero

The order relation < has the following properties:

The order property

1. Trichotomy: If x and y are numbers, exactly one of the following
holds:
x< y or x= y or x> y
2. Transitivity: x< y and y < z ⇒ x< z

3. Addition: ¿ y ⇔ x + z < y + z
4. Multiplication: When z is positive, x< y ⇔ xz < yz ,
When z is negative, x< y ⇔ xz > yz

 Intervals

Let a and b be two real numbers such that a ¿ b, then the intervals which are subsets of R with
end points a and b are denoted and defined as below:
i) (a , b ) ={ x : a<x<b } open interval from a to b.
ii) [ a , b ] ={ x : a≤x≤b } closed interval from a to b.
iii) (a , b ] ={ x : a<x≤b } open-closed interval from a to b.
iv) [ a , b ) ={ x: a≤x<b } closed-open interval from a to b.

Exercise 2.1
1. Simplify as much as possible:
5 1 2
a) 4−3( 8−12 )−6 c) 6
−( 4 + 3 )
1 3 7
2
−4+8
1
+ 34 − 78
b) 2[3−2(4−8)] d) 2

2. Which of the following statements are true and which of them are false?

43
 a) The sum of any two rational numbers is rational.
b) The sum of any two irrational numbers is irrational.
c) The product of any two rational numbers is rational.
d) The product of any two irrational numbers is irrational.
3. Find the value of each of the following, if undefined, say so.
0 0
a) 0⋅0 c) 0 e) 8
8 0 8
b) 0 d) 8 f) 0
a
4. Show that division by 0 is meaningless as follows: Suppose a≠0 . If 0
=b , then
0
a=0⋅b=0 , which is a contradiction. Now find a reason why 0 is also
meaningless.
5. Prove each if a>0 , b>0
2 2 1 1
a) a<b ⇔ a < b b) a<b ⇔ a > b

6. Which of the following are always correct if a≤b ?

2 2 2
a) a−4≤b−4 b) −a≤−b c) a ≤ab d) a ≤a b
2.2 Equations and Inequalities: Linear and Quadratic

At the end of this section, students will be able to:

 solve linear equations and inequalities
 solve quadratic equations and inequalities identify the notions of the common sets of
numbers

 Linear Equations and inequalities

An equation is a symbolic statement of equality. That is, rather than writing “twice a number is
four less than the number,” we write 2 x =x−4 . Our goal is to find the solution to a given
equation. By solution we mean the value or values of the variable that make the algebraic
statement true.

Definition 2.1: (Linear Equation)

A linear equation in one variable is an equation that can be put in the form ax +b=0 ,
where a and b are constants, and a≠0 .

Equations that have the same solutions are called equivalent equations. For example, 3x  1  5
and 3x  6 are equivalent equations because the solution set of both equations is {2}. Our goal
here is to take an equation and with the help of a few properties, gradually, change the given

44
equation into an equivalent equation of the form x=a , where x is the variable for which
we are solving. These properties are:

1. The addition property

If a=b , then a+ c=b+c . That is, adding the same quantity to both sides of an
equation will produce an equivalent equation.

2. The multiplication property

If a=b , then ac=bc . That is, multiplying both sides of an equation by the same
nonzero quantity will produce an equivalent equation.

Example 2.2:
1. Solve for x
a) 820 x=10 x+30 (50−x ) b) 3(2 x  1)  2(1  5 x)  6 x  11

Solution:
a) 820 x=10 x+30 (50−x ) Simplify the right hand side
820 x=10 x+1500−30 x
820 x=1500−20 x Applying the addition property (add 20 x to both sides)
840 x=1500
1500 25
x= =
Thus, 840 14 .
25
Remember to check by substituting 14 for x in the original equation.

b) 3(2 x  1)  2(1  5 x)  6 x  11 (The given equation)

6 x  3  2  10 x  6 x  11 (Removing parentheses by distribution)
6 x  10 x  6 x  2  11  3 (Collecting like terms: ‘variables to the left and
numbers to the right’ )
10 x  10
x 1 (Dividing both sides by 10)
Therefore, the solution set (S.S) is {1}.

8x  3 5
 5( x  2)  3( x  )
2. Find the solution set of 2 6

8x  3 5
 5( x  2)  3( x  )
Solution: 2 6 (The given equation)
This gives us:

45
 3 5
4x   5 x  10  3 x 
2 2
5 3
4 x  5 x  3x     10
2 2 Using addition property
2x  6
Hence, x  3 . That is, the solution set is {3}.

3. A computer discount store held an end of summer sale on two types of computers. They
collected Birr 41,800 on the sale of 58 computers. If one type sold for Birr 600 and the
other type sold for Birr 850, how many of each type were sold?

Solution: If we let x to be the number of Birr 600 computers sold, then 58−x = the
number of computers that are sold for Birr 850 (since 58 were sold all together).
Our equation involves the amount of money collected on the sale of each type of computer that
is, the value of computers sold). Thus we have:
600 x+850(58−x )=41 , 800 , which yields
x=30
Hence, there were 30 computers sold at Birr 600 and 28 computers sold at 850.

Remark: The solution set of some equation can be the set of all rational numbers. This is the
case when the equation is satisfied by every rational number.

Example 2.3: Find the solution set of 5 x  2( x  1)  4  3( x  2)

Solution: 5 x  2( x  1)  4  3( x  2) (The given equation)

5 x  2 x  2  4  3x  6 (Removing parentheses by distribution)
3x  6  3 x  6 (Combining like terms)

This is always true whatever the value of x is. In fact, subtracting 3x from both sides of the last
equation we get 6=6 which is always true. This means the given equation is satisfied if you take
any number for x as you wish. Thus, S.S = ℜ .
Remark: There are also some equations which cannot be satisfied by any number. For example,
the equation x+10 = x says ‘If you increase a number x by 10, the result is x itself (unchanged)’.
Obviously, there is no such a number. The solution set of such equation is empty set. If you try
to solve such equation, you end up with a false statement (false equality). For example, an
attempt to solve x+10 = x leads to the following:
10+x  x = x  x (Subtracting x from both sides of the equation)
10 = 0, which is false.

46
Hence, the solution set of x+10 = x is  (empty set).

Example 2.4: Find the solution set of 6  3(1  x)  2(1  5 x )  7 x

Solution: 6  3(1  x)  2(1  5 x)  7 x (The given equation)
6  3  3 x  2  10 x  7 x (Removing parentheses by distribution)
9  3x  2  3x (Combining like terms)
9  3 x  3 x  2  3 x  3x (Adding 3x to both sides)
9 = 2, which is false.
This means the solution set of the given equation is empty, .

Example 2.5: A man has a daughter and a son. The man is five times older than his daughter.
Moreover, his age is twice of the sum of the ages of his daughter and son. His daughter is 3 years
younger than his son. How old is the man and his children?

Solution: The unknowns in the problem are age of the man, age of his daughter, and age of his
son. So, let m = Age of the man; d = Age of the daughter; and s = Age of the son. Then, ‘The
man is 5 times older than his daughter’ means m=5d . Moreover, ‘Age of the man is twice the
sum of the ages of his daughter and son’ means m=2(d+s) . ‘His daughter is 3 years younger
than his son’ means d = s 3.
Now, from the last (3rd ) equation you can get s = d +3. Substitute this in the 2nd equation to get
m=2(d +d+3) = 2(2d+3). Thais is, m=4d+6. Next substitute this in the 1st equation to get
4d+6 = 5d or 6 = 5d4d=d. Hence, d= 6. From this, s = d +3 = 6+3 = 9, and m=5d =56= 30.
Therefore, the age of the man is 30, age of his daughter is 6 and age of his son is 9.

Definition 2.2: (Linear Inequalities)

A linear inequality is an inequality that can be put in the form ax +b<0 , where a and
b are constants with a≠0 . (The ¿ symbol can be replaced with ¿, ≤ or ¿ )

To solve inequalities, we will need the following properties of inequalities.

For a, b, c∈ ℜ , if a<b , then

1) a+ c<b +c 2) ac <bc , when c >0 3) ac >bc , when
c <0

Thus, to produce an equivalent inequality, we may add (subtract) the same quantity to (from)
both sides of an inequality, or multiply (divide) both sides by the same positive quantity. On the

47
other hand, we must reverse the inequality symbol to produce an equivalent inequality if we
multiply (divide) both sides by the same negative quantity.

Example 2.6:
1. Solve the linear inequality 5 x+8 (20−x )≥2( x−5) .

Solution: 5 x+8 (20−x )≥2( x−5) Simplify each side

5 x+160−8 x≥2 x−10
160−3 x≥2 x−10 Now apply the inequality property
−5 x≥−170 Divide both sides by – 5
x≤34 Note that the inequality symbol is reversed
Thus, the solution set is {x ∈ ℜ: x≤34 }=(−∞ , 34 ] .

Example 2.7: Find the solution set of the inequality 3x 5(x+2)  0.

Solution: 3x 5(x 2)  0 (The given inequality)
3x 5x + 10  0 (Removing the parentheses by distribution)
2x + 10  0 (Combining like terms)
2x  10 (Subtracting 10 from both sides)
10
x  2 (Dividing both sides by 2 reverse the inequality)
That is, x  5. Therefore, S.S = {x: x  5}, the set of all real numbers less 5.
The solution of an inequality is sometimes required to be only in a given domain (set). If so, a
solution set should contain only those solutions that belong to the specified domain.

Example 2.8: Find the solution set of x  4( x  1)  13  ( x  2) in the set of natural numbers, ℕ.
Solution: x  4( x  1)  13  ( x  2) (The given inequality)
x  4 x  4  13  x  2 (Removing parentheses by distribution)
3 x  4  11  x (Combining like terms; i.e., x  4 x  3x and 13+2= 11)
3x  x  11  4 (Collecting like terms)
2 x  7 (Next, division of both sides of this by 2 reverses the
inequality)
7
x
2; i.e., x  3.5
Thus, the solution of the given inequality in ℕ is {1, 2, 3}. (Recall: ℕ = {1, 2, 3, … })
Some inequalities may have no solution in the specified domain as in the following example.

Example 2.9: Find the solution set of 7 x  6  3 x  2 in the set of whole numbers, W.
Solution: 7 x  2  3x  6 (The given inequality)

48
 7 x  3 x  6  2 (Collecting like terms)
4 x  4
4 x 4

4 4 or x  1
However, there is no negative whole number. Therefore, the solution set of the given inequality
in W is , empty set. (Recall: W = {0, 1, 2, 3, … } )
1 1 3 3
( x  3)  x   ( x  1)
Example 2.10: Find the solution set of the inequality 6 2 2 2 in ℚ.
Solution: The inequality involves fractional numbers. Thus, like for the case of linear equations,
clear the denominators by multiplying both sides of the inequality by the LCM of the
denominators. The denominators in this equation are 6 and 2; and their LCM is 6. Thus,
multiply every term in both sides of the given inequality by 6. That is,
1  1  3 3 
6  ( x  3)   6  x   6    6  ( x  1) 
6  2  2 2  (The inequality is not reversed because 60)
x  3  3x  9  9( x  1) (Simplifying/clear denominators)
4x  6  9 x  9
4x  9 x  9  6 (Collecting like terms)
5 x  15 (Next, division of both sides by 5)
15
x
5 or x  3 .
Therefore, S.S = { x ℚ | x  3 }.

 Quadratic Equations and Inequalities

A quadratic equation is a polynomial equation in which the highest degree of the variable is 2.
2
We define the standard form of a quadratic a quadratic equation as Ax + Bx+ c=0 , where
A≠0 .
As with linear equations, the solutions of quadratic equations are values of the variable that make
2
the equation a true statement. The solutions of Ax + Bx+C=0 are also called the roots of
2
the polynomial equation Ax + Bx+C=0 .

2 2
In solving the equation Ax + Bx+C=0 , if the polynomial Ax +Bx+C can be factored,
the we can use the zero product rule (which is stated below) to reduce the problem to that of

49
 2
x + x−6=0 , we van factor
solving two linear equations. For example, to solve the equation
the left hand side to get ( x−2)( x +3 )=0 . Hence, we can conclude that x−2=0 or
x+ 3=0 , which yields x=2 or x=−3 .

The Zero-Product Rule: If a⋅b=0 , then a=0 or

b=0

Another method is to apply the Square Root Theorem.

2
The Square Root Theorem: If x =d , then
x=±√ d .

Example 2.11: Solve the following

2 2 2
a) 4 x + 10 x=6 b) 5 x −6=8 c) ( x−2) =6
2
Solution: a) 4 x + 10 x=6 Put into standard form
2
4 x + 10 x−6=0 Factor the left hand side
2(2 x−1)( x+3)=0 Hence we have
2 x −1=0 or x+ 3=0 Solving each linear equation, we get
1
x= 2
or x=−3
b) We note that there is no first-degree term, so our approach will be to apply the Square
Root Theorem.
2 2
5 x −6=8 Isolate x on the left-hand side before applying
the
square root theorem
2
5 x =14
14
x 2= 5 Applying the square root theorem we get
14
x=± √ 5
c) Since it is in the form of a squared quantity equal to a number, we will apply the
Square Root Theorem to get x=2± √6 .

Part (c) of the above example illustrates that if we can construct a perfect square binomial from a
quadratic equation (i.e., get the equation in the form ( x+ p )2 =d ) , then we can apply the

Square Root Theorem and solve for x to get x=− p± √ d .

50
The method of constructing a perfect square is called completing the square. It is based on the
2
fact that in multiplying out the perfect square ( x+ p ) , with p a constant, we get
2 2 2
( x+ p ) =x +2 px + p
2
Notice the relationship between the constant term, p , and the coefficient of the middle term,
2 p : The constant term is the square of half the coefficient of the middle term.

2
Example 2.12: Solve by completing the square: 2 x −8 x +4=6 .
2 2
Solution: 2 x −8 x +4=6 Divide both sides by 2, the coefficient of x
2
x −4 x +2=3 Isolate the constant term on the right-hand side
2
x −4 x=1 Take half the middle term coefficient, square it
2
( 12 (−4 )) =4 , we add 4 to both sides of the
equation
2
x −4 x +4=1+ 4 Factor the left hand side
2
( x−2) =5 Solve for x using the Square Root Theorem
x=2± √ 5 .

Unlike the factoring method, all quadratic equations can be solved by completing the square. If
2
we were to complete the square for the general quadratic equation Ax +Bx+C=0 , A≠0 ,
we would arrive at the formula given below.
2
The Quadratic Formula: If Ax + Bx+C=0 and A≠0 , then
−B±√ B 2−4 AC
x=
2A
2
Example 2.13: Solve the following using the quadratic formula: x −8=−6 x .
2
Solution: Writing the equation in standard form we get, x +6 x−8=0 . By the quadratic
formula we have:
−6± √ 6 2−4(1 )(−8) −6±√68 −6±2 √ 17
x= = = =−3± √ 17
2(1) 2 2
Thus, the solution set is {−3−√17 , −3+ √17 } .
2
A quadratic inequality is in standard form if it is in the form Ax + Bx+C <0 . (We can
replace ¿ with ¿, ≤, or ¿ .)

51
If we keep in mind that u>0 means u is positive, then solving an inequality such as
2 2
2 x + 5 x−3>0 means we are interested in finding the values of x that will make 2x +
2
5x −3 positive. Or, since 2 x +5 x−3=(2 x−1 )(x +3 ) , we are looking for values of
x that make (2 x−1 )( x+3 ) positive. For (2 x−1 )( x+3 ) to be positive, the factors must
be either both positive or both negative. To determine when this happens, we first find the values
of x for which (2 x−1 )( x+3 ) is equal to 0; we call these the cut points of
1
(2 x−1 )( x+3 ) . The cut points are 2 and −3 .

Thus, our approach in solving quadratic inequalities will be primarily algebraic. After putting the
inequality in standard form, we will determine the sign of each factor of the expression for
various values of x . Then, we determine the solution by examining the sign of the product.
This process is called a sign analysis.

2
Returning to the problem 2 x + 5 x−3>0 , we draw a number line and examine the sign of
each factor as x takes on various values on the number line, especially around the cut points.

Sign of x+ 3 − − − − − − ++ + + + + + + + + + + +
Sign of 2 x −1 − − − − − − − − − − + + + + + + +

−6 −5 −4 −3 −2 −1 0 1 2 3 4 5 6

The above figure illustrates that the factor x+ 3 is negative when x<−3 and positive when
1 1
x>−3 . It is also shown that 2 x −1 is negative when x< 2 and positive when x> 2 .
1
Thus the product of the two factors is positive when x<−3 and x> 2 . Therefore, the
1
solution set is (−∞ ,−3 )∪( ,∞ ) .
2

Remark: 1. The cut points of the inequalities will break up the number line into intervals.
2. The sign of the product does not change within an interval, i.e., if the expression is
positive (or negative) for one value within the interval, it is positive (or negative) for all
values within the interval.
2
Example 2.14: Solve the quadratic inequality x −2 x−2<0 .
2
Solution: Since we cannot factor x −2 x−2 , we use the quadratic formula to find that its
2
roots are 1±√ 3 . This gives the cut points for the polynomial x −2 x−2 . We use the sign

analysis (see the figure below) with the test points given. Note: 1+ √ 3≈2 .7∧1 √ 3≈−0. 7 .
52
 2
Sign of x −2 x−2 + 0 – 0 +
x=−10 1−√ 3 x=1 1+ √ 3 x=100
2
Substituting the test values – 10, 1, and 100 for x in the expression x −2 x−2 , we find

that
2
x −2 x−2 is negative only when x is in the interval (1−√ 3 ,1+ √3) .

Exercise 2.2
1. Solve the linear equations
2 5−x
+4=
a) 2−3( x−4 )=2( x−1 ) d) x +3 x +3
6 12 1
= +
b) 3 x−[ 2+3(2−x )]=5−(3−x ) e) x −3 x x x−3
2

3 2
c) 4
( 2 x −3)= 3
x+ 5
2. Solve the linear inequalities
5 x−2 x+3
2 1 ≥
a) 4 x + 3 ≤2 x−( 3 x +1) b) 5 x−2>3 x−( x− 5 ) c) 3 4
3. A truck carries a load of 50 boxes; some are 20 kg boxes and the rest are 25 kg boxes. If
the total weight of all boxes is 1175 kg, how many of each type are there?
9
4. The product of two numbers is 5. If their sum 2 , find the numbers.
5. Solve
2 2 2
a) 2 x −7 x=15 c) x +2 x−4=0 e) 3 x −6 x +5=0
1 1 3
x−3= + =4
b) x +3 d) x−5 x+ 2
6. Solve the quadratic inequalities
2 2
a) x +2 x−24> 0 d) 2 x −x−2≥0
2 2
b) x −5 x≤24 e) x ≤16
2
c) x −3 x−3<0
3
>4
7. A student was given the inequality: x−2 . The first step the student took in
solving this inequality was to transform it into 3>4( x−2 ) . Explain what the student
did wrong.

53
 2.3. Review of relations and functions

After completing this section, the student should be able to:

 define Cartesian product of two sets
 understand the notion of relation and function
 know the difference between relation and function
 determine the domain and range of relations and functions
 find the inverse of a relation

The student is familiar with the phrase ordered pair. In the ordered pair (2,3),(−2,4) and
(a,b) ; 2, −2 and a are the first coordinates while 3, 4 and b are the second
coordinates.

 Cartesian Product

Given sets A={3, 4} and B={ 4 , 5 , 9} . Then, the set {(3,4 ),(3,5 ),(3,9),(4,4 ),(4,5),(4,9 )}
is the Cartesian product of A and B , and it is denoted by A×B .

Definition 2.3: Suppose A and B are sets. The Cartesian product of A and B ,
denoted by A×B , is the set which contains every ordered pair whose first coordinate is an
element of A and second coordinate is an element of B , i.e.
A × B =¿ ¿ and b∈B¿¿ .

Example 2.15: For A={2, 4} and A={−1, 3} , we have

a) A×B={(2,−1),(2,3),( 4 ,−1),(4,3 )} , and
b) B× A={(−1,2 ),(−1,4 ),(3,2),(3,4 )} .

From this example, we can see that A×B and B× A are not equal. Recall that two sets are
equal if one is a subset of the other and vice versa. To check equality of Cartesian products we
need to define equality of ordered pairs.

Definition 2.4: (Equality of ordered Pairs)

Two ordered pairs (a,b) and (c ,d ) are equal if and only if a=c and b=d .

Example 2.16: Let A={1,2,3} and B={a , b , c} . Then,

A×B={(1,a),(1,b),(1,c ),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c)} .

54
 Definition 2.5: (Relation)
If A and B are sets, any subset of A×B is called a relation from A into B.

Suppose R is a relation from a set A to a set B. Then, R A×B and hence for each
(a ,b )∈ A×B , we have either (a, b )∈ R (a, b )∉ R . If (a, b )∈ R , we say “ is R-
or a

related (or simply related) to b”, and write aRb . If (a, b )∉ R , we say that “a is not related
to b”.
In particular if R is a relation from a set A to itself, then we say that R is a relation on A.

Example 2.17:
1. Let A={1,3,5,7} and B={6,8} . Let R be the relation “less than” from A to
B . Then, R={(1,6 ),(1,8),((3,6 ),(3,8),(5,6), (5,8 ),(7,8)} .

2. Let A={1,2,3,4,5} and B={a , b , c} .

a) The following are relations from A into B ;
i) R1 ={(1, a )}

ii) R2 ={(2 ,b ),(3 , b),( 4, c),(5 ,a)}

iii) R3 ={(1 ,a ),((2 , b),(3 , c)}

b) The following are relations from B to A ;
i) R4 ={( a ,3 ),(b ,1 )}

ii) R5 ={(b,2),(c ,4),(a,2),(b,3)}

iii) R6 ={(b , 5)}

Definition 2.6: Let R be a relation from A into B . Then,

a) the domain of R , denoted by Dom( R) , is the set of first coordinates of the
elements of R , i.e
Dom( R)={a∈ A :(a , b )∈ R}
b) the range of R , denoted by Range( R) , is the set of second coordinates of
elements of R , i.e
Range( R)={b ∈ B :(a , b)∈ R}

Remark: If R is a relation form 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 2.18:
55
 1. The set R={( 4,7),(5,8),(6 ,10)} is a relation from the set A={1,2,3,4,5,6} to the set
B =¿ ¿ . The domain of R is {4,5,6} , the range of R is {7,8,10} and
the codomain of R is {6,7,8,9 ,10} .
2. The set of ordered pairs R={(8,2 ),(6,−3),(5,7),(5 ,−3)} is a relation between the sets
{5,6,8} and {2,−3,7} , where {5,6,7} is the domain and {2,−3,7} is the range.

Remark:
1. If (a, b )∈ R for a relation R , we say is related to (or paired with) b . Note
a

that a may also be paired with an element different from b . In any case, b is
called the image of a while a is called the pre-image of b .
2. If the domain and/or range of a relation is infinite, we cannot list each element
assignment, so instead we use set builder notation to describe the relation. The situation
we will encounter most frequently is that of a relation defined by an equation or formula.
For example,
R={( x , y ): y=2 x −3 , x , y ∈ IR}
is a relation for which the range value is 3 less than twice the domain value. Hence,
(0, −3),(0.5, −2) and (−2,−7) are examples of ordered pairs that are of the
assignment.

Example 2.19:
1. Let A={1, 2, 3, 4, 6}
Let R be the relation on A defined by R=¿ ¿ is a factor of b¿¿ .
Find the domain and range of R .

Solution: We have
R={(1,1),(1,2),(1,3),(1,4),(1,6 ),(2,2),(2,4 ),(2,6),(3,3 ),(3,6),(4,4 ),(6,6 )} .
Then, Dom( R)={1,2,3,4,6} and Range( R)={1,2,3,4,6} .

2. Let A={1,2,3,4,5} and B={1, 2, 3, ⋯, 67} .

Let R=¿ ¿ is cube root of y } . Find a) R b) Dom( R) c)
Range( R)

56
 3, 3 3 3 3
Solution: We have 1= √1 , 2=√ 8 , 3= √27 , 4=√ 64 , 5= √125 and 1,8,27 and 64 are in
B whereas 125 is not in B . Thus, R={(1,1 ),(2,8),(3 ,27 ),(4 ,64 )} , Dom( R)={1,2,3,4 }
and R={1,8 ,27 ,64} .

Remark:
1. A relation R on a set A is called
i) a universal relation if R= A× A
ii) identity relation if R={( a ,a ):a ∈ A}
iii) void or empty relation if R=φ
2. If R is a relation from A to B , then the inverse relation of R , denoted by
R−1 , is a relation from B to A and is defined as:
−1
R ={( y , x):( x , y )∈ R} .
Observe that Dom( R)=Range( R−1 ) and Range( R)=Dom( R−1 ) . For instance, if
R={(1,4),(9 , 15),(10 , 2)} is a relation on a set A={1,2,3 ,⋯,20} , then
R−1 ={(4,1 ),(15,9 ),(2 ,10)}

Example 2.20: Let R be a relation defined on IN by R={( a ,b ):a , b ∈ IN , a+2 b=11} .

b) Dom( R) Range( R) −1
Find a) R c) d) R

Solution: The smallest natural number is 1.

b=1 ⇒ a+2(1 )=11 ⇒a=9
b=2 ⇒ a+2( 2 )=11 ⇒ a=7
b=3 ⇒ a+ 2(3 )=11 ⇒ a=5
b=4 ⇒ a+2( 4 )=11 ⇒ a=3
b=5 ⇒ a+ 2(5 )=11 ⇒ a=1
b=6 ⇒ a+2( 6 )=11 ⇒a=−1 ∉ IN
Therefore, R={(9,1 ),(7,2),(5,3),(3,4 ),(1,5)} , Dom( R)={1,3,5,7,9} ,
−1
Range( R)={1,2,3,4,5} and R ={(1,9),(2,7),(3,5 ),(4,3 ),(5,1 )} .

 Functions

Mathematically, it is important for us to distinguish among the relations that assign a unique
range element to each domain element and those that do not.

57
 Definition 2.7: (Function)
A function is a relation in which each element of the domain corresponds to exactly one
element of the range.

Example 2.21: Determine whether the following relations are functions.

a) R={(5 ,−2),(3,5),(3,7)} b) {(2,4),(3,4),(6,-4)}

Solution:
a) Since the domain element 3 is assigned to two different values in the range, 5 and 7, it is
not a function.
b) Each element in the domain,
{2,3,6 } , is assigned no more than one value in the range,
2 is assigned only 4, 3 is assigned only 4, and 6 is assigned only – 4. Therefore, it is a
function.

Remark: Map or mapping, transformation and correspondence are synonyms for the word

function. If f is a function and ( x , y)∈ f , we say x is mapped to y.

Definition 2.8: A relation f from A into B is called a function from A into B, denoted by
f : A→ B or A ⃗f B
if and only if
(i) Dom(f )= A

(ii) No element of A is mapped by f to more than one element in B, i.e. if

( x , y)∈ f and
( x, z)∈ f , then y=z .

Remark: 1. If to the element x of A corresponds y (∈ B ) under the function f , then we

write f (x )= y and y is called the image of x under y and x is called a pre-image of

y under f .
2. The symbol f (x ) is read as “ f of x” but not “ f times x”.
3. In order to show that a relation f from A into B is a function, we first show that the

domain of f is A and next we show that f well defined or single-valued, i.e. if

x= y in A, then f (x )=f ( y ) in B for all x, y ∈ A .

58
Example 2.22:
1. Let
A={1,2,3,4} and B={1,6,8,11,15} . Which of the following are functions from
A to B .

a) f defined by f (1 )=1 , f (2 )=6 , f (3)=8 , f ( 4 )=8

b) f defined by f (1 )=1 , f (2 )=6 , f (3)=15
c) f defined by f (1 )=6 , f (2)=6 , f (3 )=6 , f (4 )=6
d) f defined by f (1 )=1 , f (2 )=6 , f (2)=8 , f (3 )=8 , f (4 )=11
e) f defined by f (1 )=1 , f (2 )=8 , f (3)=11, f ( 4 )=15

Solution:
a) f is a function because to each element of A there corresponds exactly one element
of B .

b) f is not a function because there is no element of B which correspond to 4( ∈

A).
c) f is a function because to each element of A there corresponds exactly one element
of B. In the given function, the images of all element of A are the same.

d) f is not a function because there are two elements of B which are corresponding
to 2.
In other words, the image of 2 is not unique.
e) f is a function because to each element of A there corresponds exactly one
element
of B.

As with relations, we can describe a function with an equation. For example, y=2x+1 is a
function, since each x will produce only one y .

2
2. Let f ={( x , y ): y=x } . Then, f maps:

1 to 1 -1 to 1
2 to 4 -2 to 4
3 to 9 -3 to 9

59
More generally any real number x is mapped to its square. As the square of a number is unique,
f maps every real number to a unique number. Thus, f is a function from ℜ into ℜ .
We will find it useful to use the following vocabulary: The independent variable refers to the
variable representing possible values in the domain, and the dependent variable refers to the
variable representing possible values in the range. Thus, in our usual ordered pair notation
( x, y) , x is the independent variable and y is the dependent variable.

 Domain, Codomain and range of a function

For the function f : A→ B

(i) The set A is called the domain of f
f
(ii) The set B is called the codomain of
(iii) The set {f ( x ): x ∈ A } of all image of elements of A is called the range of
f
Example 2.23:
1. Let
A={1,2,3} and
B={1,2,3 ,⋯,10} . Let f : A→ B be the correspondence
which assigns to each element in A , its square. Thus, we have
f (1 )=1 , f (2 )=4 , f (3 )=9 . Therefore, f is a function and Dom(f )={1,2,3} ,
Range(f )={1,4,9} and codomain of f is {1,2,3,⋯,10} .

2. Let
A={2,4,6,7,9}, B=IN . Let x and y represent the elements in the sets A

and B , respectively. Let f : A→ B be a function defined by

f (x )=15 x+17 , x ∈ A .

The variable x can take values 2, 4, 6, 7, 9. Thus, we have

f (2 )=15(2 )+17=47 , f ( 4 )=77 , f (6 )=107 , f (7 )=122 , f (9 )=152 .

This implies that Dom(f )={2,4,6,7,9}, Range( f )={47 ,77 ,107 , 122 ,152} and
codomain
of f is IN .

60
 p

3. Let f be the subset of Q×Z defined by

f=
{( q )
, p : p , q ∈ Z , q≠ 0
} . Is f a
function?
Solution: First we note that Dom(f )=Q . Then, f satisfies condition (i) in the
2 4 2 4
( 3
, 2)∈ f ( 6
, 4) ∈ f =6
definition of a function. Now, , and 3
but
f ( 23 )= 2≠4=f ( 46 ) . Thus f is not well defined. Hence, f is not a function from
Q to Z .

4. Let f be the subset of Z ×Z defined by f ={(mn, m+n ):m , n ∈ Z} . Is f a

function?
Solution: First we show that f satisfies condition (i) in the definition. Let x be any

element of Z . Then, x=x⋅1 . Hence, ( x , x+1)=( x⋅1, x +1)∈ f . This implies

that x∈ Dom (f ) . Thus, Z ⊆ Dom(f ) . However, Dom(f )⊆ Z and so

Dom(f )=Z . Now, 4  Z and 4=4⋅1=2⋅2 . Thus, (4⋅1,4+1) and
(2⋅2,2+2) are in f . Hence we find that 4⋅1=2⋅2 and
f (4⋅1)=5≠4=f (2⋅2 ) . This implies that f is not well defined, i.e, f does not

satisfy condition (ii). Hence, f is not a function from Z to Z .

5. Determine whether the following equations determine y as a function of x , if so,

find the domain.

2x
y= 2
a) y=−3 x+5 b) 3 x−5 c) y =x

Solution:
a) To determine whether y=−3 x+5 gives y as a function of x , we need to

know whether each x-value uniquely determines a y-value. Looking at the equation
y=−3 x+5 , we can see that once x is chosen we multiply it by – 3 and then add

5. Thus, for each x there is a unique y . Therefore, y=−3 x+5 is a function.

61
 2x
y=
b) Looking at the equation 3 x−5 carefully, we can see that each x-value
uniquely determines a y-value (one x-value can not produce two different y-values).
2x
y=
Therefore, 3 x−5 is a function.

As for its domain, we ask ourselves. Are there any values of x that must be
2x
y=
excluded? Since 3 x−5 is a fractional expression, we must exclude any value
of x that makes the denominator equal to zero. We must have
5
3 x−5≠0 ⇔ x≠
3
5
Therefore, the domain consists of all real numbers except for 3 . Thus,
5
{x : x≠ }
Dom(f )= 3 .
2 2
c) For the equation y =x , if we choose x=9 we get y =9 , which gives
y=±3 . In other words, there are two y− values associated with x=9 .
2
Therefore, y =x is not a function.

6. Find the domain of the function

y=√ 3 x−x 2 .

Solution: Since y is defined and real when the expression under the radical is non-
negative, we need x to satisfy the inequality
3 x−x 2≥0 ⇔ x (3−x )≥0
This is a quadratic inequality, which can be solved by analyzing signs:

⃗
− − − | + + +| − − −

Sign of 3 x−x
2
0 3
2
Since we want 3 x−x =x (3−x ) to be non-negative, the sign analysis shows us that

the domain is
{x :0≤x ≤3} or [0,3] .

62
Exercise 2.3
1. Let R be a relation on the set
A={1,2,3,4,5,6} defined by R={( a ,b ):a+b≤9} .
i) List the elements of R
ii) Is R=R−1

2. Let R be a relation on the set

A={1,2,3,4,5,6,7} defined by R=¿ ¿ divides
a−b } .
i) List the elements of R
ii) Find Dom( R) ∧ Range( R )
−1
iii) Find the elements of R

iv) Find Dom( R−1 ) ∧ Range (R−1 )

3. Let
A={1,2,3,4,5,6} . Define a relation on A by R={( x , y ): y=x+1} . Write
−1
down the domain, codomain and range of R . Find R .

4. Find the domain and range of the relation

{( x, y):|x|+ y≥2} .
5. Let
A={1,2,3} and B={3,5,6,8} . Which of the following are functions from A to
B ?

a) f ={(1,3),(2,3 ),(3,3 )} c) f ={(1,8),(2,5 )}

b) f ={(1,3),(2,5 ),(1,6)} d) f ={(1,6),(2,5 ),(3,3)}
6. Determine the domain and range of the given relation. Is the relation a function?
1 1 1 1
a) {(−4 ,−3),(2,−5),(4,6),(2,0)} d)
{(− 2 , 6 ),(−1,1 ),( 3 , 8 )}
3
{(8 ,−2 ),(6 ,− 2 ),(−1,5)}
b) e) {(0,5),(1,5),(2,5),(3,5),(4,5 ),(5,5)}

c) {(− √3,3),(−1,1),(0,0),(1,1),( √3,3)} f) {(5,0),(5,1),(5,2),(5,3),(5,4),(5,5)}

7. Find the domain and range of the following functions.

a) f (x )=1+8 x−2 x2 c)
f (x )=√ x2 −6 x +8

b)
f ( x )= 2
1
x −5 x+ 6 d)
f(x)=¿ {3 x+4,−1≤x<2¿¿¿¿
63
 8. Given
f(x)=¿ {3 x−5, x<1¿¿¿¿ .
Find a) f (−3 ) b) f (1) c) f (6)

2.4 Real Valued functions and their properties

After completing this section, the student should be able to:

 perform the four fundamental operations on polynomials

 compose functions to get a new function
 determine the domain of the sum, difference, product and quotient of two functions
 define equality of two functions

Let f be a function from set A to set B . If B is a subset of real number system ℜ ,

then f is called a real valued function, and in particular if A is also a subset of ℜ , then
f : A→ B is called a real function.

Example 2.24: 1. The function f : ℜ→ℜ defined by f (x )=x 2 +3 x+7 , x∈ ℜ is a real

function.

2. The function f : ℜ→ℜ defined as f (x )=|x| is also a real valued function.

 Operations on functions

Functions are not numbers. But just as two numbers a and b can be added to produce a
new number a+b , so two functions f and g can be added to produce a new function
f +g . This is just one of the several operations on functions that we will describe in this
section.
x −3
f (x )=
Consider functions f and g with formulas 2 , g( x)= √ x . We can
x−3
+√x
make a new function f +g by having it assign to x the value 2 , that is,
x−3
(f + g )( x )=f ( x )+g( x )= +√ x
2 .

64
 Definition 2.9: Sum, Difference, Product and Quotient of two functions

Let f (x ) and g( x) be two functions. We define the following four functions:

1. (f + g )( x )=f ( x )+g( x ) The sum of the two functions

2. (f −g)( x )=f ( x )−g( x ) The difference of the two functions
3. (f⋅g )( x )=f ( x ) g( x ) The product of the two functions
f f ( x)
4.
()g
( x )=
g( x ) The quotient of the two functions (provided
g( x )≠0)

Since an x− value must be an inout into both f and g , the domain of (f +g )( x ) is

the set of all x common to the domain of f and g . This is usually written as
Dom(f + g )=Dom (f )∩Dom( g) . Similar statements hold for the domains of the difference
and product of two functions. In the case of the quotient, we must impose the additional
restriction that all elements in the domain of g for which g( x)=0 are excluded.

Example 2.25:
2
1. Let f (x )=3 x +2 and g( x)=5 x−4 . Find each of the following and its domain
f
a) (f +g )( x ) b) (f −g)( x) c) (f . g )( x) d) g
(x) ()
Solution:
a) (f +g )( x )=f ( x )+g( x )=(3 x2 +2)+(5 x−4 )= 2
3 x +5 x−2
b) (f −g)( x )=f ( x )−g( x )=(3 x 2 +2)−(5 x −4 )= 2
3 x −5 x +6
2 3 2
c) (f⋅g )( x )=(3 x +2)(5 x−4 )= 15 x −12 x +10 x−8
f f ( x) 3 x 2 +2
d)
()
g
( x )=
g( x )
=
5 x−4

We have
Dom(f +g )=Dom (f −g )=Dom (fg)=Dom(f )∩Dom( g )=ℜ∩ℜ=ℜ
f 5
g ()
Dom = [ Dom (f )∩Dom( g) ] {x :g ( x )=0 ¿=ℜ¿ { ¿ }¿
4

65
 2. Let
4
f (x )=√ x+1 and g( x)= √9−x 2 , with respective domains [−1,∞) and
f
f +g , f −g , f⋅g , 3
[−3,3] . Find formulas for g and f and give their domains.

Solution:
Formula Domain
4 2
(f +g )( x)=f ( x )+g( x)= √ x +1+ √ 9−x [−1, 3]
4
(f −g)( x)=f ( x)−g( x)= √ x +1−√ 9−x 2 [−1, 3]
4
(f⋅g )( x)=f ( x)⋅g( x )=√ x+1⋅√9−x 2 [−1, 3]
4

( gf )( x )= fg(( xx)) = √√9−x

x +1
2
[−1, 3)
3
4 3
f 3 ( x )=( f ( x ) )3 =( √ x +1 ) =( x +1 ) 4
[−1,∞)

There is yet another way of producing a new function from two given functions.

Definition 2.10: (Composition of functions)

Given two functions f (x ) and g( x) , the composition of the two functions is denoted by
f ∘ g and is defined by:
(f ∘ g)( x )=f [ g( x )] .
(f ∘ g)( x ) is read as f} {¿ composed with g of f ∘g
x} {¿ . The domain of
consists of those x s in the domain of g whose range values are in the domain of f ,
'

i.e. those x s for which g( x) is in the domain of f .

'

Example 2.26:
1. Suppose f ={(2 , z),(3 , q )} and g={(a ,2 ),(b ,3),(c, 5)} . The function
(f ∘ g)( x )=f ( g( x )) is found by taking elements in the domain of g and evaluating
as follows: (f ∘ g)(a )=f ( g(a ))=f (2 )=z , (f ∘g )(b )=f ( g(b ))=f (3 )=q

If we attempt to find f (g (c)) we get f (5 ) , but 5 is not in the domain of f (x ) and so we

cannot find (f ∘ g)(c ) . Hence, f ∘ g={( a , z),(b , q )} . The figure below illustrates this
situation.

66
 g
f
2
a 3 z
Domain
b of f q
c
5
Domain of g Range of g Range of f
2
2. Given f (x )=5 x −3 x +2 and g( x)=4 x+3 , find
a) (f ∘ g)(−2) b) (g ∘ f )(2 ) c) (f ∘ g)( x ) d)
(g ∘ f )( x )
Solution:
a) (f ∘ g)(−2)=f ( g (−2 )) …… First evaluate g(−2)=4(−2 )+3=−5
=f (−5)
=5(−5 )2−3 (−5 )+2=142
2
b) (g ∘ f )(2 )=g( f (2)) …….First evaluate f (2 )=5(2 ) −3 (2)+2=16
=g(16)
=4(16 )+3=67
c) (f ∘ g)( x )=f ( g( x )) ……. But g( x)=4 x+3
=f (4 x+3 )
=5( 4 x +3)2 −3( 4 x+3)+2
2
=80 x +108 x +38
2
d) (g ∘ f )( x )=g (f ( x )) ……. But f (x )=5 x −3 x +2
2
=g(5 x −3 x +2 )
2
=4(5 x −3 x +2 )+3
2
=20 x −12 x +11
x 2
f (x )= g( x )=
3. Given x +1 and x−1 , find
a) (f ∘ g)( x ) and its domain b) (g ∘ f )( x ) and its domain

2
2 x −1 2
( f ∘ g)( x )=f
x−1
=
2( )+1
=
x +1
Solution: a) x−1 . Thus, Dom(f ∘ g )={x : x≠±1} .

67
 2
(g ∘ f )( x )=g (f ( x ))= =−2 x−2
x
−1
b) x +1 . Since x must first be an input into
f (x ) and so must be in the domain of f , we see that
Dom( g∘ f )={x : x≠−1} .
6x
f (x )=
4. Let x 2 −9 and g( x)= √3 x . Find (f ∘ g)(12 ) and (g ∘ f )( x ) and its
domain.
36 4
Solution: We have ( f ∘ g)( 12 )=f ( g (12))=f ( √ 36 )=f ( 6)= 27 = 3 .
6 √3 x 6 3 x 2 3x
(f ∘ g)( x )=f ( g( x ))=f ( √ 3 x )= 2
= √ = √
( √ 3 x ) −9 3 x−9 x−3 .
The domain of f ∘ g is [0,3 )∪(3 ,∞) .
We now explore the meaning of equality of two functions. Let f : A→ B and g : A →B
be two functions. Then, f and g are subsets of A×B . Suppose f =g . Let x be
any element of A . Then, ( x , f (x ))∈ f =g and thus ( x , f (x ))∈ g . Since g is a
function and ( x,f (x )), ( x, g( x))∈ g , we must have f (x )=g( x ). Conversely, assume
that g( x )=f ( x ) for all x ∈ A . Let ( x , y)∈ f . Then, y=f ( x )=g( x ) . Thus,
( x, y)∈ g , which implies that f ⊆ g . Similarly, we can show that g ⊆f . It now
follows that f =g . Thus two functions f : A→ B and g : A →B are equal if and only
if f (x )=g( x ) for all x ∈ A . In general we have the following definition.

Definition 2.11: (Equality of functions)

Two functions are said to be equal if and only if the following two conditions hold:
i) The functions have the same domain;
ii) Their functional values are equal at each element of the domain.

Example 2.27:
+ + 2
1. Let f : Z →Z ∪{0} and g :Z →Z ∪{0} be defined by f ={(n , n ): n∈ Z }
2 2 2
and g={(n,|n| ):n∈ Z} . Now, for all n∈Z , f (n)=n =|n| =g( n) . Thus,
f =g .

68
 x 2 −25
f (x )= , x ∈ ℜ{5 ¿¿
2. Let x−5 , and g( x )=x +5 , x ∈ ℜ . The function f and
g are not equal because Dom(f )≠Dom( g ).

Exercise 2.4
2
2 g( x )=
1. For f (x )=x + x and x+3 , find each value:
2
a) (f −g)(2 ) c) g (3 ) e) (g ∘ f )(1 )
f
b)
()
g
(1)
d) (f ∘ g)(1 ) f) (g ∘ g)(3)
2
3 g( x )=
2. If f (x )=x +2 and x−1 , find a formula for each of the following and state
its domain.
g
a) (f +g )( x ) c)
()
f
(x)

b) (f ∘ g)( x ) d) (g ∘ f )( x )
2
3. Let f (x )=x and g( x)= √ x .
a) Find (f ∘ g)( x ) and its domain.
b) Find (g ∘ f )( x ) and its domain
c) Are (f ∘ g)( x ) and (g ∘ f )( x ) the same functions? Explain.
4. Let f (x )=5 x−3 . Find g( x) so that (f ∘ g)( x )=2 x +7 .
5. Let f (x )=2 x+1 . Find g( x) so that (f ∘ g)( x)=3 x −1 .
x −1 3 f ( x )+1
f (x )= f (2 x )=
6. If f is a real function defined by x+1 . Show that f (x )+3 .
7. Find two functions f and g so that the given function h( x )=(f ∘ g )(x ) , where
1
3 h( x )= +6
a) h( x )=( x+3) c) x
1
h( x )=
b) h( x)=√ 5 x−3 d) x +6
1
f (x )=4 x−3 , g ( x )= 2
8. Let x and h( x )=x −x . Find
a) f (5 x +7 ) c) f (g (h(3))) e) f (x +a )
b) 5 f ( x )+7 d) f (1 )⋅g(2)⋅h(3) f) f (x )+ a

69
 2.5 Types of functions

After completing this section, the student should be able to:

 define one to oneness and ontoness of a function

 check invertibility of a function
 find the inverse of an invertible function
In this section we shall study some important types of functions.

 One to One functions

Definition 2.12: A function f : A→ B is called one to one, often written 1 – 1, if and only
if for all x 1 , x 2 ∈ A , f (x 1 )=f ( x 2 ) implies x 1=x 2 . In words, no two elements of
A are mapped to one element of B .

Example 2.28:
1. If we consider the sets A={1,2,3 ,⋯,6} and B={7,a,b,c,d ,8 ,e} and if f =¿ ¿
(2,a), (3,b) , (4 ,b),(5,c),(6,8)¿¿ and g={(1,7 ),(2 ,a ),(3, b),( 4, c),(5,8),(6 , d)} ,
then both f and g are functions from A into B . Observe that f is not a 1
– 1 function because f (3 )=f (4 ) but 3≠4 . However, g is a 1 – 1 function.

2. Let A={1,2,3,4} and B={1,4,7,8} . Consider the functions

i) f : A→ B defined as f (1 )=1 , f (2 )=4 , f (3 )=4 , f (4 )=8
ii) g : A →B defined as f (1 )=4 , f (2 )=7 , f (3)=1 , f (4 )=8

Then, f is not 1 – 1, but g is a 1 – 1 function.

 Onto functions

Definition 2.13: Let f be a function from a set A into a set B . Then f is called
an onto function(or f maps onto B) if every element of B is image of some
element in A , i.e, Range(f )=B .

Example 2.29:

70
 1. Let A={1,2,3} and B={1,4,5} . The function f : A→ B defined as f (1 )=1 ,
f (2 )=5 , f (3 )=1 is not onto because there is no element in A , whose image
under f is 4. The function g : A →B given by g={(1,4 ),(2,5),(3,1 )} is onto
because each element of B is the image of at least one element of A .

Note that if A is a non-empty set, the function iA: A → A defined by i A ( x )=x

for all x ∈ A is a 1 – 1 function from A onto A . i A is called the identity map

on A .

2
2. Consider the relation f from Z into Z defined by f (n)=n for all n∈Z .

Now, domain of f is Z . Also, if n2 =(n ' )2 , i.e. f ( n)=f (n ' ) .

n=n' , then

Hence, f is well defined and a function. However, f (1 )=1=f (−1) and 1≠−1 ,
which implies that f is not 1 – 1. For all n ∈ Z , f (n) is a non-negative integer.
This shows that a negative integer has no preimage. Hence, f is not onto. Note that
f is onto {0,1,4,9,⋯} .
3. Consider the relation f from Z into Z defined by f (n)=2n for all n ∈ Z .
As in the previous example, we can show that f is a function. Let n , n' ∈ Z and
suppose that
'
f ( n)=f (n ) . Then 2 n=2 n' and thus n=n' . Hence, f is 1 – 1.
Since for all n∈Z , f (n) is an even integer; we see that an odd integer has no
preimage. Therefore, f is not onto.

 1 – 1 Correspondence

Definition 2.14: A function f : A→ B is said to be a 1 – 1 correspondence if f is both

1 – 1 and onto.

Example 2.30:
1. Let A={0, 1, 2, 3, 4, 5} and B={0, 5, 10, 15, 20 , 25} . Suppose f : A→ B given by
f (x )=5 x for all x ∈ A . One can easily see that every element of B has a
preimage in A and hence f is onto. Moreover, if f (x )=f ( y ) , then 5 x=5 y ,

71
 i.e. x= y . Hence, f is 1 – 1. Therefore, f is a 1 – 1 correspondence between
A and B .
2. Let A be a finite set. If f : A→ A is onto, then it is one to one.

Solution: Let
A={a 1 , a2 ,⋯, a n } . Then Range(f )={f (a1 ), f (a2 ),⋯, f ( an )} . Since f

is onto we have Range(f )= A .Thus, A={f (a1 ), f (a2 ),⋯, f (an )} , which implies that
f (a1 ) , f (a2 ) , ⋯ , f (an ) are all distinct. Hence,
ai ≠a j implies f (ai )≠f (a j )
for all 1≤i , j≤n . Therefore, f is 1 – 1.

 Inverse of a function
−1
Since a function is a relation , the inverse of a function f is denoted by f and is defined
by:
f −1={( y , x ):( x , y )∈ f }
−1
For instance, if f ={(2,4 ),(3,6 ),(1,7 )} , then f ={( 4,2),(6,3),(7,1)} . Note that the inverse
of a function is not always a function. To see this consider the function f =¿ ¿
−1
(5,4)¿¿ . Then, f ={( 4,2),(6,3),( 4,5)} , which is not a function.
As we have seen above not all functions have an inverse, so it is important to determine whether
or not a function has an inverse before we try to find the inverse. If the function does not have an
inverse, then we need to realize that it does not have an inverse so that we do not waste our time
trying to find something that does not exist.
A one to one function is special because only one to one functions have inverse. If a function is
one to one, to find the inverse we will follow the steps below:
1. Interchange x and y in the equation y=f ( x )
2. Solving the resulting equation for y , we will obtaining the inverse function.

Note that the domain of the inverse function is the range of the original function and the range of
the inverse function is the domain of the original function.

Example 2.31:
3 −1
1. Given y=f ( x )=x . Find f and its domain.

Solution: We begin by interchanging x and y , and we solve for y .

3
y=x Interchange x and y
3
x= y Take the cube root of both sides
3 This is the inverse of the function
√ x= y
72
 −1 3 −1
Thus, f ( x )= √ x . The domain of f is the set of all real numbers.
x
y=f ( x )= −1
2. Let x+ 2 . Find f ( x) .

Solution: Again we begin by interchanging x and y , and then we solve for y .

x
y=
x+ 2 Interchange x and y
y
x=
y+2 Solving for y
2x
x ( y+ 2)= y ⇔ xy +2 x= y ⇔ 2 x = y (1−x ) ⇔ y=
1−x
2x
f −1 ( x )=
Thus, 1−x .

Remark: Even though, in general, we use an exponent of −1 to indicate a reciprocal, inverse

−1
function notation is an exception to this rule. Please be aware that f ( x) is not the reciprocal
of f . That is,
1
f −1 ( x )≠
f (x )
If we want to write the reciprocal of the function f (x ) by using a negative exponent,
we must write
1 −1
=[ f ( x ) ]
f ( x) .
Exercise 2.5
2
1. Consider the function f ={( x , x ): x ∈ S } from S={−3,−2,−1,0,1,2,3} into Z . Is
f one to one? Is it onto?

2. Let A={1,2,3} . List all one to one functions from A onto A .

¿
3. Let f : A→ B . Let f be the inverse relation, i.e.
f ¿={( y , x )∈ B× A :f (x )= y } .
¿
a) Show by an example that f need not be a function.
¿
b) Show that f is a function from Range(f ) into A if and only if f is 1 –
1.

73
 ¿
c) Show that f is a function from B into A if and only if f is 1 – 1 and
onto.
¿
d) Show that if f is a function from B into A , then f
−1
=f
¿
.

4. Let A={x ∈ ℜ:0≤x≤1} and B={x ∈ ℜ:5≤x≤8} . Show that f : A→ B

defined by f (x )=5+( 8−5 )x is a 1 – 1 function from A onto B .

5. Which of the following functions are one to one?

a) f : ℜ→ℜ defined by f (x )=4 , x∈ ℜ
b) f : ℜ→ℜ defined by f (x )=6 x−1 , x∈ ℜ
c) f : ℜ→ℜ defined by f (x )=x 2 +7 , x ∈ ℜ
d) f : ℜ→ℜ defined by f (x )=x 3 , x ∈ ℜ
2 x +1
f (x )= , x ∈ ℜ{7 ¿ ¿
e) f : ℜ{7 ¿→ℜ ¿ defined by x−7

6. Which of the following functions are onto?

a) f : ℜ→ℜ defined by f (x )=115 x +49 , x ∈ ℜ
b) f : ℜ→ℜ defined by f (x )=|x|, x∈ℜ
2
c) f : ℜ→ℜ defined by f (x )=√ x , x ∈ ℜ
d) f : ℜ→ℜ defined by f (x )=x 2 +4 , x ∈ ℜ
−1
7. Find f ( x) if
4−x
f (x )= 2
a) f (x )=7 x−6 d) 3x g) f (x )==−( x+2) −1
2 x−9 5 x +3 2x
f (x )= f (x )= f (x )=
b) 4 e) 1−2 x h) 1+x
3
f (x )=1− 3
c) x f) f (x )=√ x+1

2.6 Polynomials, zeros of polynomials, rational functions and their graphs

After completing this section, the student should be able to:

 define polynomial and rational functions

74
  apply the theorems on polynomials to find the zeros of polynomial functions
 use the division algorithm to find quotient and remainder
 apply theorems on polynomials to solve related problems
 sketch and analyze the graphs of rational functions

The functions described in this section frequently occur as mathematical models of real-life
situations. For instance, in business the demand function gives the price per item, p , in terms
of the number of items sold, x . Suppose a company finds that the price p (in Birr) for its
model GC-5 calculator is related to the number of calculators sold, x (in millions), and is
2
given by the demand function p=80−x .
The manufacturer’s revenue is determined by multiplying the number of items sold ( x ) by the
price per item ( p ). Thus, the revenue function is
R=xp=x (80−x 2 )=80 x−x 3
These demand and revenue functions are examples of polynomial functions. The major aim of
this section is to better understand the significance of applied functions (such as this demand
function). In order to do this, we need to analyze the domain, range, and behavior of such
functions.

 Polynomial functions

Definition 2.15: A polynomial function is a function of the form

y=an x n + an−1 x n−1 +⋯+a1 x +a0 , an ≠0 .
a
Each i is assumed to be a real number, and n is a non-negative integer,
an 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.
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.
3 2
i.e p( x )=a3 x +a 2 x + a1 x +a0 , a3 ≠0 .

Example 2.32: p( x )=2 x +1 ,

2
q( x )=√ 3 x 4 +2 x−π and f (x )=2 x3 are examples of
polynomial functions.

 Properties of polynomial functions

75
 1. The graph of a polynomial is a smooth unbroken curve. The word smooth means that the
graph does not have any sharp corners as turning points.
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

The zeros of a polynomial function provide valuable information that can be helpful in sketching
its graph. One can find the zeros by factorizing the polynomial. However, we have no general
method for factorizing polynomials of degree greater than 2. In this subsection, we turn our
attention to methods that will allow us to find zeros of higher degree polynomials. To do this, we
first need to discuss about the division algorithm.

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)
dividend divisor quotient remainder , where either R( x )=0 or the degree of R( x) is less than
degree of d( x) .

4
x −1
Example 2.33: Divide x 4 +2 x .

Solution: Using long division we have

76
 2 x 2 −2 x+4
x +2 x 4
|x +0 x 3 +0 x 2 +0 x+1
−( x 4 +2 x 3 )
−2 x 3 +0 x 2
3 2
−(−2 x −4 x )
2
4 x +0 x
−( 4 x 2 +8 x )
−8 x−1
4
x⏟ x 2 +2 x ). (⏟
−1=(⏟ x2 −2 x+4 ) +(−8
⏟ x−1 )
This long division means dividend divisor quotient remainder .

With the aid of the division algorithm, we can derive two important theorems that will allow us
to recognize the zeros of polynomials.
If we apply the division algorithm where the divisor, d( x) , is linear (that is of the form
x−r ), we get

p( x )=( x−r)q (x )+R

Note that since the divisor is of the first degree, the remainder R , must be a constant. If we
now substitute x=r , into this equation, we get
P(r)=(r−r)q(r )+R=0⋅q (r)+R
Therefore, p(r)=R .

The result we just proved is called the remainder theorem.

The Remainder Theorem

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

p(r) .

3 2
Example 2.34: The remainder when P( x )=x −x +3 x−1 is divided by x−2 is
p(2)=9 .
As a consequence of the remainder theorem, if x−r is a factor of p( x) , then the remainder
must be 0. Conversely, if the remainder is 0, then x−r , is a factor of p( x) . This is known
as the Factor Theorem.

77
 The Factor Theorem
x−r is a factor of p( x) if and only if p(r)=0 .

The next theorem, called location theorem, allows us to verify that a zero exists somewhere
within an interval of numbers, and can also be used to zoom in closer on a value.

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

The Factor and Remainder theorems establish the intimate relationship between the factors of a
polynomial p( x) and its zeros. Recall that a polynomial of degree n can have at most n zeros.
Does every polynomial have a zero? Our answer depends on the number system in which we are
working. If we restrict ourselves to the set of real number system, then we are already familiar
with the fact that the polynomial p( x)=x 2 +1
has no real zeros. However, this polynomial
does have two zeros in the complex number system. (The zeros are i and −i ). Carl
Friedrich Gauss (1777-1855), in his doctoral dissertation, proved that within the complex number
system, every polynomial of degree ¿ 1 has at least one zero. This fact is usually referred to as
the Fundamental theorem of Algebra.

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.

Note that since all real numbers are complex numbers, a polynomial with real coefficients also
satisfies the Fundamental theorem of Algebra. As an immediate consequence of the Fundamental
theorem of Algebra, we have

The linear Factorization Theorem

n n−1
If p( x )=an x +an−1 x +⋯+a1 x +a0 , where n≥1 and
an ≠0 , then

p( x )=an ( x−r 1 ) ( x−r 2 )⋯( x−r n ) , where the r i are complex numbers (possible real and
not necessarily distinct).

78
From the linear factorization theorem, it follows that 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 2.35: Express each of the polynomials in the form described by the Linear
Factorization Theorem. List each zero and its multiplicity.
3 2
a) p( x )=x −6 x −16 x
b) q( x )=3 x 2−10 x+8
c) f (x )=2 x 4+8 x 3 +10 x 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 )=3 x 2−10 x +8 =(3 x−4 ) ( x−2 )
4
=3( x− )( x−2)
3
4
Thus, the zeros of q( x ) 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 x2 ( x 2 +4 x+5 )
=2 x 2 ( x−(−2+i) )( x−(−2−i ))
Thus, the zeros of f(x) are 0 with multiplicity two and −2+i and −2−i each with
multiplicity one.

Example 2.36:
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 (a) such that f(1) = 32.

79
Solution:
1. Based on the Factor Theorem, we may write the polynomial as:
3 4 2 3 4 2
p( x ) =( x−(−1)) ( x−2 ) ( x−5) =(x +1 ) ( x−2 ) (x −5)
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.
3 4 2
2. Based on part (1), we know that f (x ) =k (x +1 ) ( x−2 ) ( x−5) . Since we want
f (x )=32 , we have
f (1 )=k (1+1 )3 (1−2)4 (1−5 )2
32=k (8 )(1)(16 ) ⇒ k=14
1 3 4 2
Thus, f (x ) = 4 ( x +1) ( x−2) ( x−5 ) .

Our experience in using the quadratic formula on quadratic equations with real coefficients has
shown us that complex roots always appear in conjugate pairs. For example, the roots of
2
x −2 x +5=0 are 1+2i and 1−2 i . In fact, this property extends to all polynomial
equations with real coefficients.

Conjugate Roots Theorem

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

and b are real numbers) is a zero of p( x) , then so is its conjugate a−bi .
4 3 2
Example 2.37: Let r( x )=x +2 x −9 x +26 x−20 . Given that 1−√ 3 i is a zero, find the
other zero of r( x) .

Solution: According to the Conjugate Roots Theorem, if 1−√ 3 i is a zero, then its conjugate,
1+√ 3 i x−(1−√3 i ) and x−(1+ √ 3 i ) are both factors
must also be a zero. Therefore,

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 2−2 x +4 is a 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.

80
The theorems we have discussed so far are called existence theorems because they ensure the
existence of zeros and linear factors of polynomials. These theorems do not tell us how to find
the zeros or the linear factors. The Linear Factorization Theorem guarantees that we can factor a
polynomial of degree at least one into linear factors, but it does not tell us how.
We know from experience that if p( x) happens to be a quadratic function, then we may find

the zeros of p( x )= Ax 2 +Bx +C by using the quadratic formula to obtain the zeros
−B±√ B 2−4 AC
x= .
2A

The rest of this subsection is devoted to developing some special methods for finding the zeros
of a polynomial function.

As we have seen, even though we have no general techniques for factorizing polynomials of
degree greater than 2, if we happen to know a root, say r , we can use long division to divide
p( x) by x−r and obtain a quotient polynomial of lower degree. If we can get the quotient
polynomial down to a quadratic, then we are able to determine all the roots. But how do we find
a root to start the process? The following theorem can be most helpful.

The Rational Root Theorem

Suppose that f (x )=a n xn +a n−1 x n−1 +⋯+ a1 x+ a0 , where n≥1 , an ≠0 is an n

th

p
degree polynomial with integer coefficients. If q f (x )=0 , where
is a rational root of
p and q have no common factor other than ±1 , then p is a factor of a 0 and q

is a factor of
an .

3
To get a feeling as to why this theorem is true, suppose 2 is a root of
a3 x 3 + a2 x 2 +a 1 x+ a0 =0 .
3 3 3 2 3
Then,
a3
2 () () ()
+ a2
2
+a1
2
+a 0=0
which implies that
27 a3 9 a2 3 a1
+ + +a0 =0
8 4 2 multiplying both sides by 8

81
 27a3 +18a2 +12a1=−8 a0 ...................................................(1)

27a3 =−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,
a0 must be divisible by 3. From
equation (2),
a3 must be divisible by 2.

3 2
Example 2.38: Find all the zeros of the function p( x )=2 x +3 x −23 x−12 .
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 ±1, ±2, ±3, ±4 , ±6, ±12
p :
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 intermediate value theorem,
1 1
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 obtain
3 2 1 2
p( x )=2 x +3 x −23 x−12=( x + 2 )( 2 x +2 x−24 )
=2(x + 12 )( x +4 )( x−3 )
1
Therefore, the zeros of p(x) are − 2 , −4 and 3.

 Rational Functions and their Graphs

n( x)
f ( x) 
A rational function is a function of the form d ( x) where both n(x) and d(x) are
polynomials and d ( x )≠0 .
3 x−1
f ( x )= 2
x 5 +2 x 3 −x +1
f (x )= f (x )=
Example 2.39: The functions x +5 , x −4 and x +5 x
are examples of rational function.
n( x)
f ( x) 
d ( x) is {x :d ( x )≠0 }
Note that the domain of the rational function

82
 3 x−5
f ( x )= 2
Example 2.40: Find the domain and zeros of the function x −x−12 .
2
Solution: The values of x for which x −x−12=0 are excluded from the domain of f .
Since x 2−x−12=( x−4 )( x+3 ) , we have Dom(f )={x : x≠−3,4} . To find the zeros of
f (x ) , we solve the equation
n ( x)
=0 ⇔n (x )=0 ∧ q ( x )≠0
d( x)
5 5
x=
Therefore, to find the zeros of f (x ) , we solve 3 x−5=0 , giving 3 . Since 3 does
not make the denominator zero, it is the only zero of f (x ) .
The following terms and notations are useful in our next discussion.
Given a number a,
 x approaches a from the right means x takes any value near and near to a but x  a. This
is denoted by: xa+ (read: ‘x approaches a from the right’ ).
For instance, x 1+ means x can be 1.001, 1.0001, 1.00001, 1.000001, etc.
 x approaches a from the left means x takes any value near and near to a but x  a.
This is denoted by: xa– (read: ‘x approaches a from the left’ ).
For instance, x1– means x can be 0.99, 0.999, 0.9999, 0.9999, etc.
 x (read: ‘x approaches or tends to infinity’) means the value of x gets indefinitely larger
and larger in magnitude (keep increasing without bound). For instance, x can be 106, 1010,
1012, etc.
 x – (read: ‘x approaches or tends to negative infinity’) means the value of x is negative
and gets indefinitely larger and larger negative in magnitude (keep decreasing without bound).
For instance, x can be –106, –1010, –1012, etc.
The same meanings apply also for the values of a function f if we wrote f(x) or f(x).
The following figure illustrates these notion and notations.

y y f(x),
f(x),
asxa
asx

x– xa– xa+ x y =f(x)

x a
a

f(x) –,
y asx– f(x) –, asxa+

83
 Fig. 2.1. Graphical illustration of the idea of xa+, f(x), etc.
We may also write f(x)b (read: ‘f(x) approaches b’) to mean the function values, f(x),
becomes arbitrarily closer and closer to b (i.e., approximately b) but not exactly equal to b. For
1 1
f ( x) 
instance, if x , then f(x)0 as x; i.e., x is approximately 0 when x is arbitrarily large.
The following steps are usually used to sketch (or draw) the graph of a rational function f(x).
1. Identify the domain and simplify it.
2. Find the intercepts of the graph whenever possible. Recall the following:
 y–intercept is the point on y-axis where the graph of y = f(x) intersects with the y-axis. At
this point x=0. Thus, y = f(0), or (0, f(0) ) is the y-intercept if 0Dom(f).
 x–intercept is the point on x-axis where the graph of y = f(x) intersects with the x-axis. At
this point y=0. Thus, x=a or (a, 0) is x-intercept if f(a)=0.
3. Determine the asymptotes of the graph. Here, remember the following.

 Vertical Asymptote: The vertical line x=a is called a vertical asymptote(VA) of f(x) if
i) adom(f), i.e., f is not defined at x=a; and
ii) f(x) or f(x) – when xa+ or xa– . In this case, the graph of f is almost
vertically rising upward (if f(x)) or sinking downward (if f(x)) along with the
vertical line x=a when x approaches a either from the right or from the left.

1
f ( x)  n ,
Example 2.41: Consider ( x  a) where a  0 and n is a positive integer.
Obviously aDom(f). Next, we investigate the trend of the values of f(x) near a. To do this, we
consider two cases, when n is even or odd:
Suppose n is even: In this case (x – a)n  0 for all x\{a}; and since (x – a)n 0 as xa+ or
1
f ( x)  n 
xa– . Hence, ( x  a) as xa+ or xa– . Therefore, x=a is a VA of f(x). Moreover,
y= 1/an or (0, 1/an ) is its y-intercept since f(0)=1/an. However, it has no x-intercept since f(x)
0 for all x in its domain (See, Fig. 2.2 (A)).
Suppose n is odd: In this case (x – a)n 0 for all xa and 1/ (x – a)n  when xa+ as in the
above case. Thus, x=a is its VA. However, 1/(x–a)n – when xa– since (x – a)n< 0 for xa.
Moreover, y= –1/an or (0, –1/an ) is its y-intercept since f(0) = –1/an. However, it has no x-
intercept also in this case. (See, Fig. 2.2 (B)).
1
f ( x)  n 0
Note that in both cases, ( x  a) as xor x –.

84
 y 1 y 1
y n
y n
( x  a) ( x  a)
1/an

a x a x
1/an
x=a x=a
VA VA

Fig. 2.2 (A) Fig. 2.2 (B)

n( x )
f ( x) 
Remark: Let d ( x ) be a rational function. Then,

1. if d ( a)  0 and n(a)  0 , then x=a is a VA of f .

2. if d (a )  0  n( a) , then x=a may or may not be a VA of f . In this case, simplify f(x) and look
for VA of the simplest form of f.
 Horizontal Asymptote: A horizontal line y=b is called horizontal asymptote (HA) of f(x) if the
value of the function becomes closer and closer to b (i.e., f(x)b)as x or as x –.
In this case, the graph of f becomes almost a horizontal line along with (or near) the line y=b
1
f ( x)  n
as x and as x–. For instance, from the above example, the HA of ( x  a) is
y=0 (the x-axis) , for any positive integer n (See, Fig. 2.2).

n( x)
f ( x) 
Remark: A rational function d ( x ) has a HA only when degree(n(x)) degree(d(x)).

In this case, (i) If degree(n(x)) degree(d(x)), then y = 0 (the x-axis) is the HA of f.

n 1
n
an x  an 1 x    a1 x  a0
f ( x)  n 1
(ii) If degree(n(x)) =degree(d(x))=n, i.e.,
n
bn x  bn 1 x    b1 x  b0 ,
an
y
bn
then is the HA of f.
 Oblique Asymptote: The oblique line y=ax+b, a0, is called an oblique asymptote (OA) of f
if the value of the function, f(x), becomes closer and closer to ax+b(i.e., f(x) becomes
approximately ax+b) as either x or x –. In this case, the graph of f becomes almost a
straight line along with (or near) the oblique line y=ax+b as x and as x –.

85
 n( x)
f ( x) 
Note: A rational function d ( x ) has an OA only when degree(n(x)) = degree(d(x)) + 1. In

this case, using long division, if the quotient of n(x) ÷d(x) is ax +b, then y=ax+b is the OA of
f.
x2 x2  3x  2
(a) f ( x)  (b) g ( x) 
Example 2.42: Sketch the graphs of x 1 x2  1

Solution: (a) Since x1=0 at x=1, dom(f) = \{1}.

 Intercepts: y-intercept: x=0 y=f (0) = –2. Hence, (0, – 2) is y-intercept.
x-intercept: y=0 x+2=0 x= –2. Hence, (–2, 0) is x-intercept.
 Asymptotes:
 VA: Since x1=0 atx=1 and x+20 at x=1, x=1 is VA of f. In fact, if x1+ , then x+2
3 but the denominator x–1 is almost 0 (but positive).
Consequently, f(x) as x1+.
Moreover, f(x) – as x1– (since , if x1– then x–1 is almost 0 but negative ) .
(So, the graph of f rises up to + at the right side of x=1, and sink down to  at the left
side of x=1)
 HA: Note that if you divide x+2 by x–1, the quotient is 1 and remainder is 3. Thus,
x2 3 3
f ( x)   1
x 1 x  1 . Thus, if x (or x –), then x  1 0 so that f(x)1.
Hence, y=1 is the HA of f.
Using these information, you can sketch the graph of f as displayed below in Fig. 2.3 (A).

(b) Both the denominator and numerator are 0 at x=1. So, first factorize and simplify them:
x2+3x+2=(x+2)(x+1) and x2–1 = (x –1)( x+1) . Therefore,
x 2  3x  2 ( x  2)( x  1 )
g ( x)  
x 1
2
( x  1)( x  1 ) , x –1
x2

x 1 . (So, dom(g) = \{1, –1} )
This implies that only x=1 is VA.
x2 x2
g ( x)  , x  1, f ( x) 
Hence, the graph of x 1 is exactly the same as that of x  1 except
that g(x) is not defined at x= –1. Therefore, the graph of g and its VA are the same as that of f
except that there should be a ‘hole’ at the point corresponding to x= –1 on the graph of g as
shown on Fig. 2.3(B) below.

86
 x2 x2
y y , x  1
x 1 x 1

y=1 (HA) y=1

1
2 2
2 2

x=1 ‘hole’ x=1

VA atx=1

f ( x) 
x  2 x 2  3x  2 x  2
x 1
g ( x)   , x  1
x2  1 x 1

Exercise 2.6
1. Perform the requested divisions. Find the quotient and remainder and verify the
Remainder Theorem by computing p(a) .
2
a) Divide p( x )=x −5 x+8 by x +4
3 2
b) Divide p( x )=2 x −7 x + x+4 by x−4
c) Divide p( x )=1−x 4 by x−1
d) Divide p( x )=x 5 −2 x 2 −3 by x +1
2. Given that p( 4)  0 , factor p( x )=2 x 3−11 x 2 +10 x +8 as completely as possible.
r ( x )=4 x 3 −x 2−36 x +9 and r ( 4 )=0 , find the remaining zeros of
1
3. Given that
r( x) .
4 3 2
4. Given that 3 is a double zero of p( x )=x −3 x −19 x +87 x−90 , find all the zeros
of p( x) .
5. a) Write the general polynomial p( x) whose only zeros are 1, 2 and 3, with
multiplicity 3, 2 and 1 respectively. What is its degree?
b) Findp( x) described in part (a) if p(0 )=6 .
3 2
6. If 2−3i is a root of p( x )=2 x −5 x +14 x+39 , find the remaining zeros of p(x).
7. Determine the rational zeros of the polynomials
a) p( x )=x 3 −4 x 2 −7 x +10
3 2
b) p( x )=2 x −5 x −28 x+15
87
 3 2
c) p( x )=6 x +x −4 x+1
8. Find the domain and the real zeros of the given function.
2
3 x−3 ( x−3 )
f ( x )= g( x )= f (x )=
a)
2
x −25 b)
2
x 4 x−12 c) x −3 x 2 +2 x
3
d)
2
x −16
f (x )=
x 2 +4
9. Sketch the graph of
2 2
1−x x +1 1 x
f (x )= f (x )= f (x )= +2 f (x )= 2
a) x −3 b) x c) x d) x −4
x 3 −8 x−3
f (x )=
10. Determine the behavior of x−3 when x is near 3.
11. The graph of any rational function in which the degree of the numerator is exactly one
more than the degree of the denominator will have an oblique (or slant) asymptote.
a) Use long division to show that
2
x −x +6 8
y=f ( x )= =x+1+
x−2 x−2
b) Show that this means that the line y=x +1 is a slant asymptote for the graph and
sketch the graph of y=f ( x ) .

2.7 Definition and basic properties of logarithmic, exponential, and

trigonometric functions and their graphs

After completing this section, the student should be able to:

 define exponential, logarithmic and trigonometric functions

 understand the relationship between exponential and logarithmic functions
 sketch the graph of exponential, logarithmic, and trigonometric functions
 use basic properties of logarithmic, exponential and trigonometric functions to solve
problems

 Exponents and radicals

n
Definition 2.16: For a natural number n and a real number x , the power x , read “
th
the n power of x ” or “ x raised to n ”, is defined as follows:
x n= x⋅x⋅⋯⋅x
⏟
n factors each equal to x

88
 n
In the symbol x , x is called the base and n is called the exponent.

5
For example, 2 =2×2×2×2×2=32 .
n
Based of the definition of x , n must be a natural number. It does not make sense for n
to be negative or zero. However, we can extend the definition of exponents to include 0 and
negative exponents.

Definition 2.17: (Zero and Negative Exponents)

Definition of zero Exponent Definition of Negative Exponent
1
0 x−n = n ( x≠0 )
x =1 ( x≠0 ) x
0
Note: 0 is undefined.

1
−n
=x n
As a result of the above definition, we have x . We have the following rules of
exponents for integer exponents:

Rules for Integer Exponents

n m n+m n n n
1. x ⋅x =x 4. ( xy) =x y
n
x
=x n−m
2. ( x n ) m=x nm 5. x m

n n
x x
3.
() y
= n ( y≠0 )
y

Next we extend the definition of exponents even further to include rational number exponents.
To do this, we assume that we want the rules for integer exponents also to apply to rational
exponents and then use the rules to show us to define a rational exponent. For example, how do
1 1

we define a 2 ? Consider 92 .

1 2 1

If we apply rule 2 and square

1

9 2 , we get ( 9 ) =9
2 2
1
=9 . Thus, 9 2 is a number that, when
squared, yields 9. There are two possible answers: 3 and – 3, since squaring either number will
1

yield 9. To avoid ambiguity, we define a 2 (called the principal square root of a ) as the non-
1

negative quantity that, when squared, yield a . Thus, 9 =3 .

2

89
 1 1

We will arrive at the definition of a in the same way as we did for a . For example, if we
3 2

1 3 3

cube 8
1
3
, we get (8 ) =8 =8 3 3
. Thus,
1

83 is the number that, when cubed, yields 8. Since

1
1 1
3
2 =8 we have 8  2 . Similarly, (−27 ) =−3 . Thus, we define
3
3
a 3 (called the cube
root of a ) as the quantity that, when cubed yields a .

Definition 2.18: (Rational Exponent a )

n

1
n
If n is an odd positive integer, then a =b if and only if b =a
n

1
n
is an even positive integer and a≥0 , then a =|b| if and only if b =a
n
If n

1 1
th
We call an the principal n root of a . Hence, an is the real number (nonnegative
th
when n is even) that, when raised to the n power, yields a . Therefore,
1
2
( 16 ) =4 since 4 =16
2

1
3
(−125 ) 3 =−5 since (−5 ) =−125
1
1 4 1 14 1
( )
81
=
3 since () =
3 81
1
3
27 3 =3 since 3 =27
1

(−16 ) 4 is not a real number

1

Thus far, we have defined a n , where n is a natural number. With the help of the second
m

rule for exponent, we can define the expression a n , where m and n are natural numbers
m
and n is reduced to lowest terms.

Definition 2.19: (Rational Exponent an )

m 1 m

If an
1

is a real number, then a = a

n n ( ) (i.e. the n
th
root of a raised to the m
th
power)

We can also define negative rational exponents:

m
−n 1
a = m ( a≠0 )
a n

Example 2.43: Evaluate the following

90
 2 1 3
−2 −5
a) 27 3 b) 36 c) (−32 )

Solution: We have
2 1 2

a) ( )
27 3 = 27 3 =3 =9
2

1
− 1 1
36 2 = =
1
6
b) 36 2

3
− 1 1 1 1
(−32 ) 5 = = 3
= 3
=−
3 1
(−2 ) 8
c) (−32 ) 5
((−32) ) 5

Radical notation is an alternative way of writing an expression with rational exponents. We

th
define for real number a , the n root of a as follows:

1
th n
Definition 2.20 ( n root of a ): √a = a n , where n is a positive integer.

n th th
The number √ a is also called the principal n root of a . If the n root of a exists,
we have:
For a a real number and n a positive
integer,
n n 3
√ 53=5
4
√(−3)4=3
√a =¿ {|a|, if n is even¿¿¿¿ For example, and .

 Exponential Functions

In the previous sections we examined functions of the form f (x )=x n , where n is a

x
constant. How is this function different from f (x )=n .

x
Definition 2.21: A function of the form y=f ( x )=b , where b>0 and b≠1 , is called
an exponential function.
x
1
Example 2.44: The functions f (x )=2
x
, g( x)=3
x
and
h( x)=
2 () are examples of
exponential functions.

As usual the first question raised when we encounter a new function is its domain. Since rational
exponents are well defined, we know that any rational number will be in the domain of an

91
exponential function. For example, let f (x )=3 x . Then as x takes on the rational values
1 4
x=4 , – 2 , 2 and 5 , we have
1 1
f (4 )=3 4 =3⋅3⋅3⋅3=81 f (−2 )=3−2 =
32
=
9
1 4
1 4 5 4 5
f ( 2 )=3 2 = √ 3 f ( 5 )=3 5 = √ 3 =√ 81

5
Note that even though we do not know the exact values of √3 and √ 81 , we do know
exactly what they mean. However, what about f (x ) for irrational values of x ? For
√2
instance, f ( √ 2 )=3 =?

We have not defined the meaning of irrational exponents. In fact, a precise formal definition of
x
b where x is irrational requires the ideas of calculus. However, we can get an idea of what
3√ 2 should be by using successive rational approximations to √2 . For example, we have
1. 414 < √ 2<1. 415
Thus, it would seem reasonable to expect that 31. 414 <3√ 2 <31 . 415 . Since 1.414 and 1.415 are
1.414 1.415
rational numbers, 3 and 3 are well defined, even though we cannot compute their
values by hand. Using a calculator, we get 4 . 7276950<3 √2 <4 .7328918 . If we use better
approximations to √2 , we get 31. 4142 <3√2 <3 1. 4143 . Using a calculator again, we get
4 . 7287339<3√ 2 <4 .7292535 . Computing 3√ 2 directly on a calculator gives
3√ 2≈4 . 7288044 . This numerical evidence suggests that as x approaches √ 2 , the values
x √2
of 3 approach a unique real number that we designate by 3 , and so we will accept
without proof, the fact that the domain of the exponential function is the set of real numbers.

x
The exponential function y=b , where b>0 and b≠1 , is defined for all real values
of x . In addition all the rules for rational exponents hold for real number exponents as
well.

Before we state some general facts about exponential functions , let’s see if we can determine
what the graph of an exponential function will look like.

Example 2.45:
x
1. Sketch the graph of the function y=2 and identify its domain and range.

92
 Solution: To aid in our analysis, we set up a short table of values to give us a frame of
reference.
x y y
−3 −3 1
2 =8 y = 2x
−2 −2 1
2 = 4
−1 −1 1
2 =2
2 (1,2)
0
0 2 =1 1
1 21=2 O 1 x
2 22 =4
3
3 2 =8
With these points in hand, we draw a smooth curve through the points obtaining the graph
x
appearing above. Observe that the domain of y=2 is IR , the graph has no x−
intercepts, as
x →+∞ , the y values are increasing very rapidly, whereas as x →−∞ , the y values
are getting closer and closer to 0. Thus, x−axis is a horizontal asymptote, the y− intercept
x
is 1 and the range of y=2 is the set of positive real numbers.
x
1
2. Sketch the graph of
y=f ( x )=
2 () .

Solution: It would be instructive to compute a table of values as we did in example 1 above (you
are urged to do so). However, we will take a different approach. We note that
1 x 1 −x
y=f ( x )=
2() = x =2
2
x
. If f (x )=2 , then f (−x )=2
−x
. Thus by the graphing
−x x
principle for f (−x) , we can obtain the graph of y=2 by reflecting the graph of y=2
about the y−axis .

y  12  x
(1,2) 2
1

1 O 1 x

93
Here again the x−axis is a horizontal asymptote, there is no x− intercept, 1 is y−
intercept and the range is the set of positive real numbers. However, the graph is now decreasing
rather than increasing.

The following box summarizes the important facts about exponential functions and their graphs.

x
The Exponential function y=f ( x )=b
1. The domain of the exponential function is the set of real numbers
2. The range of the exponential function is the set of positive real numbers
x
3. The graph of y=b exhibits exponential growth if b>1 or exponential decay if
0<b <1 .
4. The y− intercept is 1.
5. The x− intercept is a horizontal asymptote
x y
6. The exponential function is 1 – 1. Algebraically if b =b , then x= y

Example 2.46: Sketch the graph of each of the following. Find the domain, range, intercepts, and
asymptotes.
x x+1 −x
a) y=3 +1 b) y=3 c) y=−9 +3
Solution:
x x
a) To get the graph of y=3 +1 . We start with the graph of y=3 , which is the basic
exponential growth graph, and shift it up 1 unit.

From the graph we see that

y=3x+1
10 - Dom(f )=ℜ
- Range(f )=(1, ∞ )
- The y− intercept is 2

The line y=1

2
y=1 - is a horizontal
1
asymptote
1 2
x+1 x
b) To get the graph of y=3 , we start with the graph of y=3 , and shift 1 unit to the
left.

94
 From the graph we see that
y=3x+1
- Dom(f )=ℜ
9 - Range(f )=(0 ,∞ )
- The y− intercept is 3

- The line y=0 is a horizontal

asymptote
1
−x −x
c) To get the graph of y=−9 +3 , we start with the basic exponential decay y=9 .
−x
We then reflect it with respect to the x−axis , which gives the graph of y=−9 .
−x
Finally, we shift this graph up 3 units to get the required graph of y=−9 +3 .

y y
y
(1,9) 9 1 3
1 y=3
1 x
1 2 y = 9 x +3
y=9x 1
1 y=9 x 1
x
1 O 1 x (1,9) 9

−x
From the graph of y=−9 +3 , we can see that Dom(h )=ℜ , Range(h)=(−∞ ,3) , the
1
line y=3 is a horizontal asymptote, 2 is the y− intercept and x=− 2 is the x−
intercept.
x
Remark: When the base b of the exponential function f (x )=b equals to the number e
, where e=2 .7182⋯ , we call the exponential function the natural exponential function.

 Logarithmic Functions
x
In the previous subsection we noted that the exponential function f (x )=b (where b>0
and b≠1 ) is one to one. Thus, the exponential function has an inverse function. What is the
x
inverse of f (x )=b ?
To find the inverse of f (x )=b x , let’s review the process for finding an inverse function by
3
comparing the process for the polynomial function y=x and the exponential function
95
 x
y=3 . Keep in mind that x is our independent variable and y is the dependent variable
and so whenever possible we want a function solved explicitly for y .

3 x
To find the inverse of y=x To find the inverse of y=3
3 x
y=x Interchange x and y y=3 Interchange x and y
3 y
x= y solve for y x=3 solve for y
y=√ x
3 y=??

y
There is no algebraic procedure we can use to solve x=3 for y . By introducing radical
3 3
notations we could express the inverse of y=x explicitly in the form y=√ x . In words,
3 3
y =x and y=√ x both mean exactly the same thing: y is the number whose cube is
y
x . Similarly, if we want to express x=3 explicitly as a function of x , we need to
y
invent a special notation for this. The key idea is to take the equation x=3 and express it
verbally.

y
x=3 means y is the exponent to which 3 must be raised to yield
x

We introduce the following notation, which expresses this idea in a much more compact form.

Definition 2.22: For b>0 and b≠1 , we write y=log b x to mean y is the
exponent to which b must be raised to yield x . In other words,
x=b y ⇔ y=log b x

We read
y=log b x as “ y equals the logarithm of x to the base b ”.

REMEMBER:
y=log b x is an alternative way of writing
y
x=b
y
When an expression is written in the form x=b , it is said to be in exponential form. When
an expression is written in the form b y=log x
, it is said to be in logarithmic form. The table
below illustrates the equivalence of the exponential and logarithmic forms.

96
 Exponential form Logarithmic form
2 log 4 16=2
4 =16
4
2 =16 log 2 16=4
1
5−3 = 125 1
log 5 125 =−3
1
1
6 2 =√ 6 log 6 √6= 2
0
7 =1 log7 1=0

Example 2.47:
1. Write each of the following in exponential form.
log 1 =−2 1
a) 39 b) log 16 2= 4
log 1 =−2 −2 1
Solution: We have a) 39 means 3 = 9 .
1 1

b) log 16 2= 4
means 16 4 =2
2. Write each of the following in logarithmic form.
2
−3
a) 10 =0 . 001 b) 27 3 =9

Solution: We have
−3
a) 10 =0 . 001 means
log10 0 .001=−3
2 2
b) 27 3 =9 means log 27 9= 3
3. Evaluate each of the following.
1
a)
log 3 81 b) log 8 64

Solution:
a) To evaluate
log 3 81 , we let t=log 3 81 , and then rewrite the equation in
t
exponential form, 3 =81 . Now, if we can express both sides in terms of the same
base, we can solve the resulting exponential equation, as follows:
Let
t=log 3 81 Rewrite in exponential form
t
3 =81 Express both sides in terms of the same base
t 4
3 =3 Since the exponential function is 1 – 1
t=4
Therefore,
log 81=4
3 .
b) We apply the same procedure as in part (a).
1
Let t =log 8 64 Rewrite in exponential form

97
 1
8t = 64 Express both sides in terms of the same base
t −2
8 =8 Since the exponential function is 1 – 1
t=−2
1
Therefore, log8 64 =−2 .

As was pointed out at the beginning of this subsection, logarithm notation was invented to
express the inverse of the exponential function. Thus,
log b x is a function of x . We usually

write f (x )=log b x rather than writing f (x )=log b ( x ) and use parenthesis only when
needed to clarify the input to the log function. For example,

If f (x )=log 5 (4−x ) , then f (−1 )=log 5 (4−(−1 ))=log 5 5=1 , whereas if

f (x )=4−log5 x , then f (−1 )=4−log 5 (−1) , which is undefined.

Example 2.48: Given f (x )=log 5 x , find

1
a) f (25 ) b) f ( 25 ) c) f (0) d) f (125)

Solution:
a) f (25 )=log 5 25=2 2
(since 5 =25 )
1 1 −2 1
b) f ( 25 )=log 5 25 =−2 (since 5 = 25 )
c) f (0)=log 5 0 is not defined (what power of 5 will yield 0?). We say that 0 is not in the

domain of f .
d) f (−125 )=log5 (−125) is not defined (what power of 5 will yield -125?). We say that

-125 is not in the domain of f .

Acknowledging that the logarithmic and exponential functions are inverses, we can derive a
great deal of information about the logarithmic function and its graph from the exponential
function and its graph.

Example 2.49: Sketch the graph of the following functions. Find the domain and range of each.
y=log 3 x y=log 1 x
a) b) 2

y=log 3 x x
Solution: a) Since is the inverse of y=3 , we can obtain the graph of
y=log 3 x by reflecting the graph of y=3 x about the line y=x , as shown below.

98
 y
y = 3x
y=x

y = log3x
1

1 x

y=log 1 x 1 x
y=( 2 )
b) To get the graph of 2 , we reflect the graph of about the line
y=x as shown below. y

y  12  x y=x

1 x

y  log 1 x
2

Taking note of the features of the two graphs we have the following important informations
about the graph of the log function:
The Logarithmic Function
y=log x b
1. Its domain is the set of positive real numbers
2. Its range is the set of real numbers.
3. Its graph exhibits logarithmic growth if b>1 and logarithmic decay if 0<b <1 .
4. The x− intercept is 1. There is no y− intercept.
5. The y−axis is a vertical asymptote.
Example 2.50:
1. Sketch the graph of f (x )=1+log 3 ( x−2) . Find the domain, range, asymptote and
intercepts.
Solution: We can obtain the graph of y=1+log 3 ( x−2 ) by applying the graphing
principle to shift the basic logarithmic growth graph 2 units to the right and 1 unit up.

99
 y
x= 2

y = 1+ log3(x2)
1

1 2 3 x

We have Dom(f )={x : x >2} , Range(f )=ℜ and the graph has the line x=2
as a vertical asymptote. To find the intercept, we set y=0 and solve for x . Setting
7 7
y=0 and solving for x , we will obtain x= 3 . Thus, the x− intercept is 3 .

2. Find the inverse function for

a) y=f ( x )=3 x +4 b) y=g( x )=log 3 ( x−2)

Solution: Following the procedure for finding an inverse function, we have

x y=log3 ( x−2 )
(a) y=3 +4 Interchange x and (b) Interchange x and
y y
y
x=3 +4 solve explicitly for y x=log 3 ( y−2 ) Write in logarithmic
y
x−4=3 Write in logarithmic form
x
form y−2=3 solve explicitly for y
y=log 3 ( x−4 ) x
y=3 +2
−1
Thus, f ( x )=log 3 ( x−4 ) −1 x
Thus, g ( x )=3 +2

The following table contains the basic properties of logarithm:

Properties of logarithm
Assume that b, u and v are positive and b≠1 . Then
1. log b (uv )=log b u+log b v
In words, logarithm of a product is equal to the sum of the logs of the factors.
u
2. log b ( v )=log b u−log b v

In words, the log of a quotient is the log of the numerator minus the log of the
denominator.

100
 3. log b ( ur )=r log b u
In words, the log of a power is the exponent times the log.

4. log b (b x )=x log b b=x

log b x
5. b =x
Example 2.51:
1. Express in terms of simpler logarithms.
log b ( x 3 y ) 3 log b √ xy3
( )
a) b) log b ( x + y ) c) z

Solution:
3 3
a) log b ( x y )=log b x + log b y=3 log b x+ log b y
b) Examining the properties of logarithms, we can see that they deal with log of a

product, quotient and power. Thus, log 3 ( x3 + y ) which is the log of a sum cannot
be simplified using log properties.
c) We have
1
1
√ xy 3 log ( xy ) 2
−3 log z= ( log b x +log b y )−3 log b z
log b ( z3 ) b
=log √ xy −log b ( z )
=
b b
2 .
1
2. Show that log b 2 =−log b 2 .
1
Solution: We have log b 2 =log b 1−log b 2=0−log b 2=−log b 2 .

The logarithmic function was introduced without stressing the particular base chosen. However,
there are two bases of special importance in science and mathematics, namely, b=10 and
b=e .

Definition 2.23: (Common Logarithm)

f (x )=log 10 x is called the common logarithm function. We write log 10 x=log x .

The inverse of the natural exponential function is called the natural logarithmic function and has
its own special notation.

Definition 2.24: (Natural Logarithm)

f (x )=log e x is called the natural logarithmic function. We write log e x=ln x .

Example 2.52:
1. Evaluate log 1000

101
 3
Solution: Let a=log 1000 . Then, a=log 10 1000=log 10 (10 )=3 .
x
2. Find the inverse function of f (x )=e +1 .
x
Solution: Let y=e +1 Interchange x and y
y
x=e +1 Solve for y
y
x−1=e Rewrite in logarithmic form
y=ln( x−1)
−1
Thus, f ( x )=ln( x−1) .
 Trigonometric functions and their graphs

For the functions we have encountered so far, namely polynomial, rational and exponential
functions, as the independent variable goes to infinity the graph of each of these three functions
either goes to infinity(very quickly) for exponential functions or approaches a finite horizontal
asymptote. None of these functions can 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. The
basic functions for studying regular periodic behaviour are the trigonometric functions. The
domain of the trigonometric functions is more naturally the set of all geometric angles.

Angle Measurement

An angle is the figure formed by two half-lines or rays with a common end point. The common
end point is called the vertex of the angle.

In forming the angle, one side remains fixed and 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, we call the angle positive angle, and if the terminal side rotates in a
clockwise direction, we call the angle negative angle.

B
B
What attribute of an angle are we trying
to measure when we measure the size of an angle? A
moment of thought will lead us to the conclusion that when we measure an angle we are trying to
answer the question: Through what part of a complete rotation has the terminal side rotated?
We will use degree () as the unit of measurement for angles. Recall that the measure of a full
102
round angle (full circle) is 360, straight angle is 180, and right angle is 90.

An alternative unit of measure for angles which will indicate their size is the radian measure. To
see the connection between the degree measure and radian measure of an angle, let us consider
an angle θ and draw a circle of radius r with the vertex of θ at its center O . Let
s represent the length of the arc of the circle intercepted by ∠θ (as shown below).

O 
r

Basic geometry tells us that the central angle θ will be the same fractional part of one
1
complete rotation as s will be of the circumference of the circle. For example, if θ is 10
1
of a complete rotation, then s will be 10 of the circumference of the circle. In other words,
we can set up the following proportion:
θ s s
= =
1 complete rotation circumference of circle 2πr

Thus, we have the following conversion formula:

θ in deg rees θ in radians
=
180∘ π

Example 2.53:
1. Convert each of the following radian measures to degrees.
π 3π
a) 6 b) 5

π
θ 6
=
Solution: a) By the conversion formula, we have 180∘ π , which implies that θ=30∘ .
3π
θ 5
=
b) Again using the conversion formula, we get 180∘ π , which implies that
∘
θ=108 .

103
 2. Convert to radian measures
a) 90∘ b) 270∘
Solution: a) Let θ represent the radian measure of 90∘ . Using the conversion formula, we
θ 90∘ π
= ∘ θ=
obtain: π 180 , which implies that 2 .

∘ ∘
b) Rather than using the conversion formula, we notice that 270 =3(90 ) . In part (a) we
π 3π
90∘= 270∘=
found that 2 , and so we have 2 .

To define the trigonometric functions, we will view all angles in the context of a Cartesian
coordinate system: that is, given an angle θ , we begin by putting θ in standard position,
meaning that the vertex of θ is placed at the origin and initial side of θ is placed along the
positive x−axis . Thus the location of the terminal side of θ will, of course, depend on the
size of θ .
Y Y
P(x,y)

ϴ r

X X

We then locate a point (other than the origin) on the terminal side of θ and identify its
coordinates ( x, y) and its distance to the origin, dented by r . Then, r is positive.

With  in standard position, we define the six trigonometric functions of  as follows:

Definition 2.25
Name of function Abbreviation Definition
y
sin  
Sine  sin  r
x
cos  
Cosine  cos r
y
tan  
Tangent  tan  x

104
 r
csc  
Cosecant  csc  y
r
sec  
Secant  sec  x
x
cot  
Cotangent  cot  y

s

Recall that the radian measure of an angle is defined as r , where  is angle in radians
s is the length of the arc intercepted by  and r is the length of the radius. Since s and r are
s
both lengths, the quotient r is a pure number without any units attached. Thus, any angle can be
interpreted as a real number. Conversely, any real number can be interpreted as an angle. Thus,
we can describe the domains of the trigonometric functions in the frame work of the real number
systems. If we let f ( )  sin  , the domain consists of all real numbers  for which sin  is
y
sin  
defined. Since r and r is never equal to zero, the domain for sin  is the set of all real
x
f ( )  cos  
numbers. Similarly, the domain of r is also the set of all real numbers.
 The graph of y  sin 

To analyze f ( )  sin  , we keep in mind that once we choose a real number  , we draw the
angle, in standard position, that corresponds to  . To simplify our analysis, we choose the point
( x, y ) on the terminal side so that r  1 . That is, ( x, y ) is a point on the unit circle x 2  y 2  1 .
y
sin   y
Note that 1 .

(0,1)
(x,y)

θ
(-1,0) (1,0)

(0,-1)

105
As the terminal side of  moves through the first quadrant, y 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
3 3
 increases from  to 2 , sin  decreases from 0 to – 1; and as  increases from 2 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 

Applying the same type of analysis to f ( )  cos , we will able to get a good idea of what its
graph looks like. The figure below shows the angle corresponding to  as it increases through
quadrant I, II, III and IV.
x
cos   x
Keeping in mind that 1 , we have the following:

1. As  increases from 0 to 2 , x  cos  decreases from 1 to 0.

2. As  increases from 2 to  , x  cos  decreases from 0 to – 1.

106
 3
3. As  increases from  to 2 , x  cos  increases from – 1 to 0.
3
4. As  increases from 2 to 2 , x  cos  increases from 0 to 1.

Based on this analysis, we have the graph of f ( )  cos as shown below:

 The graph of y  tan 

y
tan  
Since x is undefined whenever x  0 , tan  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 . Thus,

domain of tan  is
{ :   2  n } , where n is an integer.

1. As  increases from 0 to 2 , x decreases from 1 to 0 and y increases from 0 to 1;

therefore, tan  
y
x increases from 0 to  .

2. As  increases from 2 to  , x decreases from 0 to – 1 and y decreases from 1 to 0;

therefore, tan  
y
x increases from   to 0.
3
3. As  increases from  to 2 , x increases from – 1 to 0 and y decreases from 0 to – 1;

therefore, tan  
y
x increases from 0 to  .
3
4. As  increases from 2 to 2 , x increases from 0 to 1 and y increases from – 1 to 0;

therefore, tan  
y
x increases from   to 0.

You may want to add some more specific values to this analysis. In any case, we get the
following as the graph of the tangent function.

107
 Definition 2.26: (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 2.27: (Amplitude of a periodic function)

The amplitude of a periodic function f (x ) is
1
A [
2 maximum value of f (x )  minimum value of f (x )]

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

The portion of the graph of a sine or cosine function over one period is called a complete cycle
of the graph. In other words, the minimal portion of a sine or cosine graph that keeps repeating
itself is called a complete cycle of the graph.

Definition 2.28: (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.

The frequency of the basic sine curve y  sin x and the basic cosine curve y  cos x is 1,
because each graph makes 1 complete cycle in the interval [0,2 ] .

108
 
If a sine function has period of 2 (see the figure below), then the number of complete cycles its

graph will make in an

Y
interval of length 2
2
4

is 2 .
X

7π
4


Thus if a sine function has a period of 2 , its frequency is 4 and its graph will make 4 complete
cycles in an interval of length 2 .

Example 2.54: Sketch the graph of y  sin 2 x and find its amplitude, period and frequency.

Solution: We can obtain this graph by applying our knowledge of the basic sine graph. For the
basic curve, we have
sin 0  0 sin 2  1 sin   0 sin 32  1 sin 2  0
These quadrantal values serve as guide points, which help us draw the graph. To obtain similar
guide points for y  sin 2 x , we ask for what values of x is
2x  0 2 x  2 2x   2x  3
2 2 x  2
and we get
x0 x  4 x  2 x 3
2 x 
x  0, , , , and  , respectively. The
  3
Thus, y  sin 2 x will have the values 0, 1, 0,  1 , 0 at 4 2 4

graph of y  sin 2 x will thus complete one cycle in the interval [0, ] , and will repeat the same
values in the interval [ ,2 ] .

109
X
From this graph we see that y  sin 2 x has an amplitude of 1, a period  , and a frequency of 2.

For convenience we summarize our discussion on the domains of the trigonometric functions in
the table.
1. f ( x )  sin x Domain = All real numbers
Domain = All real numbers
2. f ( x )  cos x
{x : x  2  n }
3. f ( x )  tan x Domain =
f ( x )  csc x Domain = { x : x  n }
4.
{x : x  2  n }
Domain =
5. f ( x )  sec x
Domain = {x : x  n }
6. f ( x )  cot x
where n is an integer

We have the following trigonometric identities

1. sin x  cos x  1
2 2

2. tan x  1  sce x
2 2

3. 1  cot x  csc x
2 2

Exercise 2.7

1. Find the domain of the given function.

1 1
f ( x)  x f ( x) 
b) g ( x )  3  1 c) h( x )  2  8
x x
a) 6 d) 2 2
3x

2. Sketch the graph of the given function. Identify the domain, range, intercepts, and
asymptotes.
x x x 2
a) y  5 b) y  9  3 c) y  1  e d) y  e
x

3. Solve the given exponential equation.

3a  2
x 1
a) 2  8 b) 3  243
2x
c) 8  2
x
d) 16  1
4

4. Let f ( x )  2 . Show that f ( x  3)  8 f ( x ) .

x

110
 1
g ( x  2)  g ( x)
5. Let g ( x )  5 . Show that
x
25 .
f ( x  2)  f ( 2 )
 4( 3 x )
6. Let f ( x )  3 x
. Show that 2 .
7. Evaluate the given logarithmic expression (where it is defined).
a) log 2 32 c) log 3 ( 9) e) log 5 (log 3 243)
log 1 9 log 6 16 log 5
b) 3
d) f) 2 2
8. If f ( x )  log 2 ( x  4) , find f (6) and the domain of f .
2

9. If g ( x )  log 3 ( x  4 x  3) , find f ( 4) and the domain of g .

2

log 1 x   log 6 x
10. Show that 6

11. Sketch the graph of the given function and identify the domain, range, intercepts and
asymptotes.
a) f ( x )  log 2 ( x  3) b) f ( x )  3  log 2 x c) f ( x )   log 3 (  x ) d) f ( x )  3 log 5 x
( 3 x 1)
12. Find the inverse of f ( x )  e .

13. Let f ( x )  e . Find a function so that ( f  g )( x )  ( g  f )( x )  x .

x

14. Convert the given angle from radians to degrees

b)  2 c)  3
 5 4
a) 3
15. Convert the given angle from degrees to radians
a) 315 b)  40 c) 330
  

16. Sketch the graph of

a) f ( )  sec  c) f ( )  csc  e) f ( )  cot 
d) f ( x )  sin( x  2 )

b) f ( x )  1  cos x f) f ( x )  tan 2 x
17. Verify the following identities:
a) (sin x  cos x )(csc x  sec x )  tan x  cot x
b) sec x  csc x  tan x  cot x
2 2 2 2

18. Given tan   2 and sin   0 , find cos  .

1

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