MATH 705
Set Theory and Logic
Module 7
F
UNCTIONS
Objectives
After studying this module, you should be able to:
1. define function;
2. determine the domain, co-domain, image, and range of a function;
3. differentiate equal, one-one, onto, identity, and constant functions;
4. solve composition of functions;
5. find inverse of a function and inverse function; and
6. graph functions.
Introduction
One of the most important concepts in mathematics is that of a function. The
term “map”, “mapping”, “transportation”, and many others mean the same thing. The
choice of which word to use in a given situation is usually determined by tradition and
the mathematical background of the person using the term.
Functions
Suppose that to each element in a set A there is assigned, by some manner or
other, a unique element of a set B. We call such assignments a function. If we let f
denote these assignments, we write
f:A→B
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which reads “f is a function of A into B”. The set A is called the domain of the function f,
and B is called the co-domain of f. Further, if a ε A then the element in B which is
assigned to a is called the image of a and is denoted by f (a), which reads “f of a”.
We list a number of instructive examples of functions.
(1) Let f assign to each real number its square, that is, for every real numbers x
let f(x)= x2. The domain and co- domain of f are both the real numbers, so we can write
f : R# → R#
the image of -3 is 9; hence we can also write f (-3) = 9, or f : -3 → 9.
(2) Let f assign to each country in the world its capital city. Here the domain of f
is the set of countries in the world; the co-domain of f is the list of capital cities in the
world. The image of Philippines is Manila, that is, f (Philippines) = Manila.
(3) Let A = {a,b,c,d} and B = {a,b,c}. Define a function f of A into B by the
correspondence f (a) = b, f (b) = c and f (d) = b. By this definition, the image, for
example, of b is c.
(4) Let A = {-1,2}. Let f assign to each rational number in R# the number 1, and to
each rational number in R# the number -1. Then f :R# → A, and f can be defined
concisely by
1 if x is rational
f (x) =
-1 if x is irrational
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(5) Let A= {a,b,c,d} and B = {x,y,z}. Let f: A → B de defined by the diagram:
A
x
B
y
C
z
D
Notice that the functions in Examples (1) and (4) are defined by specific
formulas. But this need not always be the case, as is indicated by the other examples.
The rules of correspondence which define functions can be diagrams as in Example (5)
can be geographical as in Example (2) or, when the domain is finite, the correspondence
can be listed for each elements in the domain as in Example (3).
Equal Functions
If f and g are functions defined on the same domain D and if f (a) = g (a) for every
a  D, then the functions f and g are equal and we write
f=g
Examples
(1) Let f (x) =x² where x is a real number. Let g (x) = x² where x is a complex
number. Then the function f is not equal to g since they have different domains.
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(2) Let the function f be defined by the diagram
1
1
2
2
3
4
Let a function g be defined by the formula g (x) = x2 where the domain of g is set
{1,2}. Then f = g since they both have the same domain and since f and g assign the
same image to each element in the domain.
Example 2.3: Let f: R# → R# and g: R# → R#. Suppose f is defined by f (x) = x²and g by
g(y)= y² Then f and g are equal function, that is, f = g. Notice that x and y are merely
dummy variables in the formulas defining the functions.
Range of a Function
Let f be a mapping of A into B, that is, let f: A B. Each element in B need not
appear as the image of an element in A. We define the range of f to consist precisely of
those elements in B which appear as the image of at least one element in A. We denote
the range of f: A → B by
f (A)
Notice that f (A) is a subset of B.
Examples
(1) Let the function f: R# → R# be defined by the formula f(x)= x² Then the range
of f consists of the positive real numbers and zero.
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(2) Let f: A → B be the function in example (3) of function. Then f (A) = {b,c}.
Wait! … pause for a while, answer first the following question.
SAQ1
Let V = {-2, -1, 0, 1, 2}. Let the function g : V  R# be defined by the formula
g(x) = x2 + 1.
Find the range of g.
ASAQ1
We compute the image of each element of V.
g(-2) = (-2)2 + 1 = 4 + 1 = 5
g(-1) = (-1)2 + 1 = 1 + 1 = 2
g(0) = (0)2 + 1 = 0 + 1 = 1
g(1) = (1)2 + 1 = 1 + 1 = 2
g(2) = (2)2 + 1 = 4 + 1 = 5
Thus the range of g is the set of image points {5, 2, 1, 2, 5}, i.e. the set {5, 2, 1}.
One-one Function
Let f map A into B. Then f is called a one-one function if different elements in B
are assigned to different elements in A, that is if no two different elements in A have the
same image. More briefly, f: A → B is one-one if f(a) = f(a’) implies a = a’ or, equivalently,
f(a)  f(a’).
Examples
(1) Let the function f: R# → R# be defined by the formula f(x) = x² Then f is not a
one-one function since f(2) = 4, that is, since the image of two different real numbers, 2
and -2, is the same number, 4.
(2) Let the function f: R# → R# be defined by the formula f(x) =x³ Then f is a oneone mapping since the cubes of two different real numbers are themselves different.
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(3) The function f which assigns to each country in the world its capital city is
one-one since different countries have different capitals, that is, no city is the capital of
two different countries.
Onto Functions
Let f be a function of A into B. Then the range f(A) of the function f is a subset of
B, that is, f(A)  B, if f(A) = B, that is, if every member of B appears as the image of at
least one element of A, then we say “f is a function of A onto B” or “f maps A onto B” or “f
is an onto function”
Examples
(1) Let the function f : R# → R# be defined by the formula f(x) =x² Then f is not an
onto function since the negative numbers do not appear in the range of f, that is, no
negative number is the square of a real number.
(2) Let f : A → B be the function in example 1.3. Notice that f(A) = {b,c}. Since B =
{a,b,c}, the range of f does not equal the co-domain, i.e. f is not onto.
(3) Let f: A → B be the function in example 1.5. Notice that
f(A) = {x,y,z} = B
that is, the range of f is equal to the co- domain B. Thus f maps A onto B, i.e, f is an
onto mapping.
Identity Function
Let A be any set. Let the function f: A → A be defined by the formula f(x) = x, that
is, let f assign to each element in A the element itself. Then f is called the identity
function.
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Constant Functions
A function f of A into B is called constant function if the same element b  B is
assigned to every element in A. In other worlds, f : A→ B is a constant function if the
range of f consists of only one element.
Examples
(1) Let the function f be defined by the diagram
a
1
b
2
c
3
Then f is not a constant function since the range of f consists of both 1 and 2.
(2) Let the function f be defined by the diagram:
a
1
b
2
c
3
Then f is a constant function since 3 is assigned to every element in A.
(3) Let f : R# → R# be defined by the formula f (x) = 5. Then f is a constant
function since 5 is assigned to every element.
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Product Function
Let f be a function of A into B and let g be a function of B, the co-domain of f, into C.
We illustrate the functions below.
A
f
B
g
C
Let a  A; then its image f (a) is in B which is the domain of g. Accordingly, we can find
the image of f (a) under the mapping g, that is, we can find g (f (a)). Thus we have a rule
which assigns to each element a  A a corresponding element g(f(a))  C. In other
worlds, we have a function of A into C. This new function is called the product function or
composition function of f and g and it is denoted by
(g ο f ) or (g f )
More briefly, if f: A → B and g: B → C then we define a function (g ο f ) : A → C by
(g ο f )(a)  g(f (a))
Here  is used to mean equal by definition. We can now complete our diagram:
A
f
B
g
C
(gοf)
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Examples
(1) Let f: A → B and g: B → C be defined by the diagrams
f
a
b
c
g
x
r
y
s
z
t
We compute (g ο f ) : A → C by its definition:
(g ο f )(a)  g(f(a)) = g(y) = t
(g ο f )(b)  g(f(b)) = g(z) = r
(g ο f )(c)  g(f(c)) = g(y) = t
Notice that the function (g ο f ) is equivalent to “following the arrows” from A to C
into the diagram of the function f and g.
(2) To each real number let f assign it square, i.e. Let f(x) = x². To each real
number let g assign the number plus 3, i.e. let g(x) = x+3. Then
(f ο g)(2) = f (g(2)) = f (5) = 25,
(g ο f)(2) = g(f (2)) = g (4) = 7
Notice that the product functions (f ο g) and (g ο f) are not the same function.
We compute a general formula for these product functions:
(f ο g)(a) ≡ f (g(2)) = f(x+3) = (x+3)² = x²+ 6x + 9
(g ο f)(x) ≡ g (f(x)) = g(x²) = x²+3
Note: Let f: A → B. then
1B ο f =f and f ο 1A = f
that is, the product of any function and the identity function is the function itself.
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Associativity of Products of Functions
Let f: A → B, g : B→ C and h : C→ D. Then, as illustrated in Figure 4-1, we can
form the product function g ο f : A → C, and then the function h ο ( g ο f ) : A → D.
A
f
B
g
C
h
D
(gοf)
h ο (gοf)
Similarly, as illustrated in Figure 4-2, we can form the product function h ο g : B → D and
then the function (h ο g) ο f : A → D.
f
A
g
h
B
C
D
hοg
(h ο g) ο f
g)
Both h ο (g ο f) and (h ο g) ο f are functions A into D. A basic theorem on functions
states that these functions are equal. Specifically,
Theorem: Let f: A → B, g : B → C and h: C → D. Then (h ο g) ο f = h ο (g ο f). In view of
this theorem, thus, we can write h ο g ο f : A → D without any parenthesis.
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Wait! … pause for a while, answer first the following question.
SAQ2
Let the functions f : R#  R# and g : R#  R# be defined by
f(x) = 2x + 1 and g(x) = x2 – 2.
Find formulas which define the product functions g ο f and f ο g.
ASAQ2
We first compute g ο f : R#  R#. Essentially we want to substitute the formula for f
inside the formula for g. We use the definition of the product function as follows:
(g ο f ) (x)  g (f(x)) = g(2x + 1) = (2x + 1)2 – 2 = 4x2 + 4x – 1
Perhaps you are now familiar with the process if the functions are defined as:
y = f(x) = 2x + 1, z = g(y) = y2 – 2
Then y is eliminated from the two formulas:
z = y2 – 2 = (2x – 1)2 – 2 = 4x2 + 4x – 1
You should become familiar with the first method. It is necessary to realize that x is
merely a dummy variable. We now compute f ο g : R# R#:
(f ο g) (x)  f (g(x)) = f(x2 – 2) = 2(x2 – 2) + 1 = 2x2 – 3
Inverse of a Function
Let f be a function of A into B, and let b ε B. Then the inverse of b, denoted by
f -1 (b)
consists of those elements in A which are mapped onto b, that is, those elements in A
which have b as their image. More briefly, if f : A → B then
f
-1
(b) = {x │x  A, f (x) = b}
Notice that f -1 (b) is always a subset of A. We read f -1 as “f inverse”.
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Examples
(1) Let the function f : A → B be defined by the diagram
f
Then f
-1
a
x
b
y
c
z
A
B
(x) = {b,c}, since both b and c have x as their image point. Also, f -1 (y)
= {a}, as only a is mapped into y. The inverse of z, f -1(z), is the null set , since no
element of A is mapped into z.
(2) Let f : R# → R#, the real numbers, be defined by the formula f(x) = x². Then f -1
(4) = {2,-2}, since 4 is the image of both 2 and -2 and there is no other real number
whose square is four. Notice that f -1 (-3) = , since there is no elements in R# whose
square is -3.
(3) Let f be a function of the complex numbers into the complex numbers, where f
is defined by the formula f(x) = x². Then f -1(-3) = { 3 i, -
3 i}, as the square of each of
these numbers is -3.
Notice that the functions in Example 8.2 and 8.3 are different although they are
defined by the same formula.
We now extend the definition of the inverse of a function. Let f : A → B and let D be a
subset of B, that is, D  B. Then the inverse of D under the mapping f, denoted by f -1
(D), consists of those elements in A which are mapped onto some element in D. More
briefly,
f -1 (D) = {x │ x  A, f(x)  D)
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Examples
(1) Let the function f : A → B be defined by the diagram
x
r
y
s
z
t
Then f -1 ({r, s}) = {y}, since only y is mapped into r or s Also f -1 ({r, t}) = {x, y, z}
= A, since each element in A has its image r or t.
(2) Let f : R# → R# be defined by f(x) = x ², and let
D = [4, 9] = {x │ 4 ≤ x ≤ 9}
Then
f
-1
(D) = {x │ -3 ≤ x ≤ -2 or 2 ≤ x ≤ 3}
(3) Let f : A → B be any function. Then f -1 (B) = A, since every element in A has
its image in B. If f(A) denotes the range of the function f, then
f -1 (f(A)) = A
Further, if b  B, then
f -1 (b) = f -1 ({b})
Here f -1 has two meanings, as the inverse of an element B and as the inverse of a
subset of B.
Inverse Function
Let f be a function of A into B . In general, f -1 (b) could consist of more than one
element or might even be the empty set . Now if f: A → B is a one-one function and an
onto function, then for each b  B the inverse f -1 (b) will consist of a single element in
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A . We therefore have a rule that assigns to each b  B a unique element f -1 (b) in A.
Accordingly, f -1 is a function of B into A and we can write
f -1 : B → A
In this situation, when f : A → B is one-one and onto, we call f -1 the inverse function of f.
Examples
(1) Let the function f : A → B be defined by the diagram
f
a
x
b
y
c
z
A
B
Notice that f is one-one and onto. Therefore f -1, the inverse function, exists.
We describe f : B → A by the diagram
f -1
x
a
y
b
z
c
B
A
Notice, further, that if we send the arrows in the opposite direction in the first diagram of f
we essentially have the diagram of f -1.
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(2) Let the function f : A → B be defined by the diagram
A
B
f
a
x
b
y
c
z
Since f(a) = y and f(c) = y, the function f is not one-one. Therefore the inverse function f 1
does not exist. As f -1 (y) = {a, c}, we cannot assign both a and c to the element y  B.
(3) Let R# → R#, the real numbers be defined by f(x) = x³. Notice that f is one-
one and onto. Hence f -1 : R# → R# exists. In fact we have a formula which defines the
inverse function, f -1(x) = 3 x .
Wait! … pause for a while, answer first the following question.
SAQ3
Let A = R# - {3} and B = R# - {1}. Let the function f : A  B be defined by f(x) =
x-2
.
x-3
Then f is one-one and onto. Find a formula that defines f -1.
ASAQ3
2 - 3y
x-2
for x in terms of y, we obtain x =
. Hence the inverse function is
1- y
x-3
2 - 3x
f -1 (x) =
.
1- x
Solving y =
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Graph of a Function
Let f be a function of A into B, that is, let f : A → B. The graph f* of the function
f consists of all ordered pairs in which a  A appears as a first element and its image
appears as its second element. In other words,
f* = {(a, b)} │ a  A, b = f (a)}
Notice that f* , the graph of f: A → B, is a subset of A × B .
Examples
(1) Let the function f : A → B be defined by the diagram
A
B
a
1
b
2
c
3
d
Then f(a) = 2, f(b) = 3, f(c) =
2 and
f(d) = 1. Hence the graph of
f
is
f * = {(a, 2), (b, 3), (c, 2), (d, 1)}.
(2) Let W = {1, 2, 3, 4}. Let the function f : W → R# be defined by
f(x) = x + 3.
Then the graph of f is
f * ={(1, 4), (2, 5), (3, 6), (4, 7)}.
(3) Let N be the natural numbers 1, 2, 3, . . . . Le the function g : N → N be
defined by
g(x) = x³.
Then the graph of g is
g * = {(1, 1), (2, 8), (3,27), (4,64), . . .}
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Properties of the Graph of a Function
Let f : A → B. We recall two properties of the function f . First, for each element
a  A there is assigned an element in B. Secondly, there is only one element in B which
is assigned to each a  A. In view of these properties of f, the graph f * of f has the
following two properties:
Property 1: For each a  A, there is an ordered pair (a, b)  f *.
Property 2: Each a  A appears as the first element in only one ordered pair in f * , that is,
(a, b) a  f * , (a, c)  f * implies b = c
In the following examples, let A = {1, 2, 3, 4} and B = {3, 4, 5, 6}.
Examples
(1) The set of ordered pairs {(1, 5), (2, 3), (4, 6)} cannot be the graph of a
function of A into B since it violates Property 1. Specifically, 3  A and there is no
ordered pair in which 3 is a first element.
(2) The set of ordered pairs {(1, 5), (2, 3), (3, 6), (4, 6), (2, 4)} cannot be the
graph of a function of A into B since it violates Property 2, that is, the element 2  A
appears as the first element in two different ordered pairs (2, 3) and (2, 4).
Wait! … pause for a while, answer first the following question.
SAQ4
1. Let the function h : R#  R# be defined by h(x) = x + 3. State whether or not each of
the following ordered pairs belongs to the graph h* of the function h:
a. (2, 6)
b. (8, 11)
c. (10, 12)
d. (4, 7)
e. (-6, -9)
f. (-1, 2).
2. Let S = { a, e, I, o, u}. To each letter in S let the function g assign the letter which
follows it in the alphabet. Find the graph g* of the function g.
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ASAQ4
1. a. h(2) = 2 + 3 = 5; so (2, 6) does not belong to h*.
b. h(8) = 8 + 3 = 11; so (8, 11) does not belong to h*.
c. h(10) = 10 + 3 = 13; so (10, 12) does not belong to h*.
d. h(4) = 4 + 3 = 7; so (4, 7) belongs to h*.
e. h(-6) = -6 + 3 = -3; so (-6, -9) does not belong to h*.
f.
h(-1) = -1 + 3 = 2; so (-1, 2) belongs to h*.
2. First g(a) = b, g(e) = f, g(i) = j, g(o) = p and g(u) = v. Thus g* = {(a, b), (e, f), (i, j),
(o, p), (u, v)}.
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Activity No. 7
1. Let the function f: R#  R# be defined by
3x – 1 if x > 3
f(x) =
x2 – 2 if -2 < x < 3
2x + 3 if x < -2
Find a) f(2), b) f(4), c) f(-1), d) f(-3) .
2. Each of the following formulas defines a function from R# into R#. Find the range
of each function.
a) f(x) = x3
c) h(x) = x2 + 1
b) g(x) = sin x
3. Let the functions f, g, and h be defined by
f(x) = x2 where 0 < x < 1
g(x) = y2 where 2 < y < 8
h(z) = z2 where z  R#
Which of these functions are equal?
4. Let A = {1, 2, 3, 4, 5} and let the functions f : A  A and g : A  A be defined by:
f(1) = 3
f(2) = 5
f(3) = 3
f(4) = 1
f(5) = 2
g(1) = 4
g(2) = 1
g(3) = 1
g(4) = 2
g(5) = 3
Find the composition functions f ο g and g ο f.
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5 Let f: R#  R# be defined by f(x) = 2x – 3. Note that f is one-one and onto, so f has
an inverse function f-1: R# R#. Find a formula that defines the inverse function f-
5. Let the function g assign to each name in the set { Jenny, Ashlene May, Keith
Lorenz} the number of different letters needed to spell the name. Find the graph
of g.
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