Chapter Four
Functions
4.1. Functions
Now , we define a function as a rule.
Definition 4.1. A function ( mapping) f from a set A into a set
B is a rule which assign to each element of A a unique element
in B .
The set A is called the domain , and we denote it by
D f  A , and the set B is called the co-domain of f , and we
denote it by CoDf  B .
If f is a function from a set A to a set B , we write
f : A  B . If a  A , then f (a ) is called the image of a in B .
The set Imf  f (A )  f (a ) a  A   R f is called the image of f
or the range of f . If f from a set A into A , then f is called an
operator or a transformation on the set A .
Now, we define a function as a rule.
Definition 4.2. Let A and B be sets. A function ( mapping)
from A to B is a relation f  A  B such that for every a  A and
b1 , b2  B if (a, b1 )  f and (a, b2 )  f , then b1  b2 .
If (a, b )  f , then we write f (a )  b .
To prove that the rule f from a set A into a set B is a well
defined function, we let x , y  A such that x  y , and we prove
that f (x )  f ( y ) , that is if the elements in A are equal, then the
images are equal in B , or equivalently , we let x , y  A such that
f (x )  f ( y ) , and we prove that x  y , that is if the images are not
equal in B , then the elements are not equal in A .
Example 4.1. The following rules define functions from
R :
R
into
33
1.
2.
3.
4.
5.
for all x  R . For if x , y  R such that x  y , then
x  1  y  1 , and hence f (x )  f ( y ) . Thus f is a well-defined
function.
f (x )  e x for all x  R . For if x , y  R , such that x  y , then
e x  e y . Thus f (x )  f ( y ) . Hence f
is a well-defined
function.
f (x )  sin x for all x  R . For if x , y  R , such that x  y , then
sin x  sin y . Thus f (x )  f ( y ) . Hence f is a well-defined
function.
f (x )  x 2  2 for all x  R . For if x , y  R , such that x  y ,
then x 2  y 2 , and so x 2  1  y 2  1 . Thus f (x )  f ( y ) . Hence f
is a well-defined function.
f (x )  3 x for all x  R . For if x , y  R , such that f (x )  f ( y ) ,
then 3 x  3 y , and thus x  y . Hence f is a well-defined
function.
f (x )  x  1
Example 4.2. Let A  1, 2,3, 4 and B  a,b ,c  . Define a rule f
from A into B by f (1)  b , f (2)  c , f (3)  c and f (4)  b . Then f is a
function, because it assigns to each element in A a unique
element in B . Thus Df  A , CoD f  B and Imf  Rf  b ,c .
Example 4.3. Let A  1, 2,3, 4 and B  a,b ,c  . Define a rule f
from A into B by f (1)  a, f (1)  b , f (2)  c , f (3)  c and f (4)  b . Then
f is not a function for a  f (1)  f (1)  b .
Example 4.4. Let A  1, 2,3, 4 . Then
1. f  (1, 2),(2,3),(3, 4),(4,1) is a relation on A and also is a
function of A into A . Thus f (1)  2, f (2)  3, f (3)  4 , f (4)  1 ,
and f (f (3))  f (4)  1 .
2. R 2  (1,1),(1, 2),(2,3),(3, 4),(4, 4) is a relation on A , but it is not a
function of A into A , for (1,1)  R 2 , (1, 2)  R 2 , but 1  2 , that
is 1 has two images 1 and 2 .
3. R3  (1, 2),(3, 4),(4,1) is a relation on A , but it is not a
function of A into A , for 2  A , but has no image in A ,
34
that is 2 does not appear as the first element in any order
pair in R 3 .
4. g  (1, 2),(2, 2),(3, 2),(4, 2) is a relation on A and also is a
function of A into A . Thus f (1)  2, f (2)  2, f (3)  2 , and
f (4)  2 .
Example 4.5.Let A  a,b ,c and B  b ,c , d ,e  . Then
1. f  (a,b ),(a,c ),(b , d ),(c ,e )  A  B is not a function from A into
B , for (a, b )  f and (a, c )  f , that is f (a )  b , and f (a )  c
which means a has two images b and c , where b  c .
2. g  (a,c ),(b , d ),(c ,e ) is a function from A into B . Thus
D g  A and Rang g  g (A )  Img  c , d , e  .
3. h  (a,b ),(c , d ) is not a function from A into B , for b  A
has no image in B under h .
Definition 4.3. Let f and g be functions defined on the same
domain , that is D f  D g . Then f and g are equal, we write f  g ,
if f (x )  g (x ) for all x  D f  D g .
Example 4.6. Let f (x )  x 2 for all x 1, 2,3 , and g (x )  x 2 for all
x 1, 2,5 . Then f  g , for D f  D g . But if D f  D g  R , then f  g .
Definition 4.4. Let A be any set . A function I A  I : A  A
defined by f (x )  x for all x  A is called the identity function.
Definition 4.5. A function f from a set A into a set B is called a
constant function if the element b  B is the image for all the
elements in A , that is f (a )  b for all a  A , and f (A )  b .
Example 4.7. Let A  a,b ,c and B  b ,c , d ,e  . Then the function
f : A  B defined by f (a )  d , f (b )  d , f (c )  d is a constant function
with f (A )  d  .
35
Example 4.8. Let A  a,b ,c and B  b , c , d , e . Then the function
f : A  B defined by f (a )  d , f (b )  d , f (c )  e is not a constant
function with f (A )  d ,e .
Example 4.9. The function f : R  R defined by
x  R is a constant function with f (R)  4 .
f (x )  4
for all
4.2. Properties of Functions
We have seen that a function f is often regarded as a means of
corresponding to each element of a set A an element in another
set B . We have several questions to ask by this point of view.
For example,
A
1. If
f : A  B , then do two distinct elements of
correspond to the same element of B under f ?
2. If f : A  B is a function, then does each element of B have
an element of A corresponding to it under f ?
Now, we discuss this questions.
Definition 4.6. A function f from a set A into a set B is
called a one to one or injective if for every x , y  A such that
f (x )  f ( y ) implies that x  y , that is if the images of two
elements are equal, then the elements are equal,
or
equivalently, if x  y , then f (x )  f ( y ) , that is that is images of
two different elements are different. If f is a one to one
function, then we write 1  1 .
Example 4.10. Define f : R  R by the rule f (x )  4x  3 for all
x  R . Let x , y  R such that x  y , then 4x  4 y and hence
4x  3  4 y  3 . Thus f (x )  f ( y ) . Therefore, f is well-defined
function . Next, let f (x )  f ( y ) , that is 4x  3  4 y  3 , and
then 4x  4 y . Thus x  y . Therefore, f is one to one function.
Example 4.11. Define f : R  R by the rule f (x )  x 2 for all
x  R . Let x , y  R such that x  y , then x 2  y 2 . Thus f (x )  f ( y ) .
36
is well-defined function . Next, 3,3  R , and
f (3)  f (3)  9 , that is the images of two different real numbers
3,3 is the same number 9 . Hence, f is not a one to one fuction.
Therefore,
f
Example 4.12. Define f : R  R by the rule f (x )  x 3  2 for all
x  R . Let x , y  R such that x  y , then x 3  y 3 , and x 3  2  y 3  2 .
Thus f (x )  f ( y ) . Therefore, f is well-defined function . If x  y ,
then x 3  y 3 , and x 3  2  y 3  2 . Hence, f (x )  f ( y ) . Therefore, f is
a 1 1 .
Definition 4.7. A function f from a set A into a set B is called
an onto or surjective if for every element y  B , there is an
element x  A , such that f (x )  y , that is every element of B
appears as the image of at least one element of A . Hence, f is
an onto function if R f  B . If f : A  B is an onto function, then
we say f maps A onto B .
Example 4.13. Let the function f : R  R defined by
f (x )  4x  3 for all x  R . Now, we show that f is an onto
function. Let y CoDf  R . Then there exist x  Df  R such that
f (x )  y
. Thus
f (x )  f (
4x  3  y
and
y 3
y 3
)  4(
)3 y
4
4
y 3
.
4
4x  y  3 x 
. Therefore,
f
Hence,
is an onto function.
Example 4.14. Let the function f : R  R defined by f (x )  x 2
for all x  R . Then f is not an onto function, for the negative
numbers do not appear as the image in range of f . Thus there is
no negative number is the square of a real number.
Example 4.15. Let A  1, 2,3 , and B  4,5 . Define f : A  B by
the rule, f (1)  f (3)  5 and f (2)  4 . Then f is a function and
R f  f (A )  4,5  B . Thus f is onto function. Also, f is not one
to one for 1  3 , and f (1)  f (3)  5 .
37
Example 4.16. Let A  1, 2,3, 4 and B  1, 2,3 . Then define a
f (1)  f (4)  2, f (2)  f (3)  3 .
function
Then
f : A  B by
R f  f (A )  2,3  B . Therefore, f is not an onto function.
Example 4.17. Let A  1, 2,3, 4 and B  5,6,7 . Then define a
function f : A  B by f (1)  f (4)  6 , f (2)  5 , and f (3)  7 . Then
R f  f (A )  B . Therefore, f is an onto function. Also, f is not
1  1 function.
Definition 4.7. A function f from a set A into a set B is called
bijective if f is one to one and onto, that is if f is injective and
surjective.
Example 4.18. Let A  1, 2,3 and B  5,6,7 . Define a function
f by f (1)  6, f (2)  5, and f (3)  7 . Then f is one to one and onto
function. Thus, f is a bijective function.
Example 4.19. (i) If f : R  R by the rule f (x )  4x  3 for all
x  R , then by Example 10, and Example 13, f is one to one
and onto function, that is, f is bijective.
(ii) If f : R  R by the rule f (x )  x 2 for all x  R , then f is not a
one to one and onto function. Thus f is not bijective.
Example 4.20. The identity function is a bijective function.
4.3.Compositions of Relations and Functions
Now, we introduce the composition of relations.
Definition 4.8. Let R be a relation from a set A to a set B , and
S a relation from B to a set C . Then the composition of S and
R is the relation from A to C defined by
S R  (a, c ) There exist b  B such that aRb and bSc   A C
Example 4.21. Let
A  a,b  , B  c , d 
and C  e , f , g  .
38
1. Let
R  (a, c ),(b , c ),(b , d )
S
2. Let
R  (a,c ),(b , c )
and S  (c ,e ),(d , g ) . Then
R  (a,e ),(b ,e ),(b , g )
and
S  (d ,e ),(d , f ),(d , g ) .
S R  ,
A to C ,
The empty relation from
exist a  A , and c C such that
Example 4.22. Let
Then
R
since no
Then
b B
(a, b )  R , and (b , c )  S
does there
.
A  a,b ,c  .
Consider the relation
R  (a, a),(a, b ),(a, c ),(b , a)
R  (a, a),(a, b ),(a, c ),(b , a),(b , b ),(b , c )  A  A .
Definition 4.9. Let f be a function from a set A to a set B , and
g a function from B to a set C . We define g f , the
composition of g and f by
( g f )(a )  g (f (a ))
for all a  A .
So if a  A , then b  f (a )  B , where, B  D g  CoD f . Then, we
can find g (b )  g (f (a )) C . To see that g f is a function, we let
x , y  A such that x  y . But f is a function, so f (x )  f ( y ) . Also,
we have g is a function. Then g (f (x ))  g (f ( y )) . Therefore,
( g f )(x )  ( g f )( y ) . Hence, g f is a function.
A  a,b ,c  , B  1, 2,3 ,
f (a )  2, f (b )  3, f (c )  2,
g (1)  x , g (2)  z , g (3)  x .
Example 4.23. Let
Then
and C  x , y , z  . Define
( g f )(a )  g (f (a ))  g (2)  z ,
( g f )(b )  g (f (b ))  g (3)  x ,
( g f )(c )  g (f (c ))  g (2)  z .
Example 4.24. Let
f : R  R, and g : R  R
be functions defined by
f (x )  x  2, and g (x )  x 2
for all
x A .
Then
( g f )(x )  g (f (x ))  g (x  2)  (x  2) 2
and
39
g )(x )  f ( g (x ))  g (x 2 )  x 2  2
(f
It is clear that, in general
(f
g g f
f
. Also
f )(x )  f (f (x ))  f (x  2)  (x  2)  2  x  4
and
( g g )(x )  g ( g (x ))  g (x 2 )  (x 2 ) 2  x 4
We have
( g f )(2)  (2  2)2  16, and (f
Example 4.25. Let
g )(2)  22  2  6 .
f : A  B be
(I B f
a functions between sets. Then
)(x )  I B (f (x ))  f (x ) ,
and
I A )(x )  f (I A (x ))  f (x ) ,
(f
for all
x A .
Thus
IB f  f
, and
f
IA  f
.
Example 4.26. Let e : R  R , defined by e : x  e x the exponential
function and
ln : R  R, defined by ln : x  ln(x ) the natural
logarithmic functions. Then ln(e x )  x for all x  R , and e ln( x )  x
x R . Thus e ln  I R , and ln e  I R .
Example 4.27. Let
f
and
g
be two functions such that:

f : R  R, defined by f (x )  x
and
g : R  R , defined by g (x )  x 2  1
then
( g f )(x )  g (f (x ))  g ( x )  x  1
and
(f
g )(x )  f ( g (x ))  f (x 2  1)  x 2  1
Theorem 4.1. Let f be a function from a set A to a set B , and
g a function from B to a set C .
1. If f and g are both one to one (injective) functions,
then g f is a one to one (injective) function.
2. If f and g are both onto (surjective) functions, then
g f is an onto ( surjective) function.
40
3. If f and g are both one to one and onto (bijective)
functions, then g f is a one to one , and onto (bijective)
function.
4. If g f is a one to one and an onto (bijective) function,
then f is a one to one (injective), and g is an onto
(surjective) function.
Proof.
1. Let f and g be both one to one (injective) functions,
and let x , y  A such that ( g f )(x )  ( g f )( y ) , and hence
g (f (x ))  g (f ( y )) . But g is a one to one function . Then
f (x )  f ( y ) . As f is a one to one function, we have x  y .
Therefore, g f is a one to one function.
2. Let f and g be both onto (surjective) functions, and let
c C . We want to show that there is a  A such that
( g f )(a )  c . Since c C , and g is onto, there is b  B
such that g (b )  c . Also, b  B , and f is onto, there is
a  A such that f (a )  b . Thus
( g f )(a )  g (f (a ))  g (b )  c .
Therefore, g f is an onto function.
3. It is clear from (1) and (2).
4. Let g f be a one to one and onto function.
(i) Let x , y  A such that f (x )  f ( y ) in B . Then
g (f (x ))  g (f ( y )) in C . Thus ( g f )(x )  ( g f )( y ) . But
g f is bijective and hence one to one. Thus
x y .
(ii) Exercise.
Theorem 4.2. Let A , B , C and D be sets, and let
g : B  C , and h :C  D be a functions . Then
f :A B ,
(h g ) f  h ( g f )
Proof. Two functions are equal if the action on an element in
the domain is the same. If x  A , then
((h g ) f )(x )  (h g )(f (x ))  h ( g (f (x )))  h (( g f )(x ))  (h ( g f ))(x )
Thus
(h g ) f  h ( g f )
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4.4. Inverse of a Function
Let f : A  B be a function between two sets
Then the inverse of an element b  B is the set
f
1
A
and
B
.
(b )  a  A f (a )  b   A
that is f 1 (b ) is the set consists of those elements in
mapped onto b . We read f 1 as f " inverse ".
A
which are
We notice that f 1 (b ) is a subset of A . In general, f 1 (b )
could consist of more than one element or may be empty set.
If f : A  B is a function between sets, and
define the inverse image of C under f is the set
f
1
C B
, the we
(C )  x  A f (x ) C   A
Which is called the preimage of
f 1 (b )  f 1 (a) .
C
under
f
. If
C  b  ,
we write,
If the function f : A  B is a one to one and onto function,
then for each b  B , then f 1 (b ) consist of a single element in A .
Hence we have a rule that assign to each element b  B a unigue
element f 1 (b ) in A . Therefore, there is a function f 1 : B  A ,
and f 1 is called the inverse of the function f . Hence ,
f (a )  b  a  f
1
(b )
Example 4.28. Let A  1, 2,3 , and B  a,b ,c  . Define a function
f : A  B such that f (1)  c , f (2)  a , and f (3)  b . Then f is a one to
one and onto function. Hence, f 1 : B  A exists and f 1 (a)  2 ,
f 1 (b )  3 , and f 1 (c )  1. We note that
f (1)  c  1  f
1
(c ),
f (2)  a  2  f
1
f (3)  b  3  f
1
(a ),
(b )
Example 4.29. Let A  1, 2,3 , and B  a,b ,c  . Define a function
f : A  B such that f (1)  a , f (2)  a , and f (3)  b . Then f is not a
one to one, and not onto. Thus f 1 : B  A does not exist.
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Example 4.30. Let f : R  R be defined by f (x )  2x  5 . We want
to find f 1 : R  R if it is exist. First, we prove that f is a function
and one to one. Let x , y  R . Then
x  y  2x  2 y  2x  5  2 y  5  f ( x )  f ( y )
Next, we show that f is onto. So, let b R . Then there is exist
an element x  R such that f (x )  b . Hence
2x  5  b  2x  b  5  x 
b 5
2
b 5
b 5
)  2(
)  5  b . Therefore, f 1 : R  R exist. To
2
2
1
find a formula for f , we set b  f (x )  f 1 (b )  x . Then b  2x  3 ,
and hence x  b  5 . Thus f 1 (b )  b  5 . Replace b by x to get
2
2
x 5
f 1 (x ) 
.
2
and
f (x )  f (
f : R  R be defined by f (x )  3x  4 ,
show that f 1 (x )  y  4 .
3
Example 4.31. Let f : R  R be defined by f (x )  x 3 .
Similarly, if
1
find f : R  R if it is exist. First, we prove that
and one to one. Let x , y  R . Then
f
the we can
We want to
is a function
x  y  x 3  y 3  f (x )  f ( y )
So, let b R . Then there is exist an element x  R such that
f (x )  b . Hence x  3 b , and f (x )  f ( 3 b )  ( 3 b )3  b . Thus f is onto.
Therefore, f 1 : R  R exist. To find a formula for f 1 , we set
b  f (x )  f 1 (b )  x . Then f 1 (x )  3 x .
Example 4.32. Let f : A  x  R x  1  R be defined by
f (x )  x  1 . First, we prove that f is a function and one to one.
Let x , y  A . Then
x  y  x  1  y  1  x  1  y  1  f (x )  f ( y )
Let b R . Then there is exist an element
Hence b  x  1 , and b 2  1  x . Thus
x A
such that
f (x )  b .
f (x )  f (b 2  1)  b 2  1  1  b 2  b
43
So
set
is onto. Therefore, f 1 exist. To find a formula for
b  f (x )  f 1 (b )  x . Then f 1 (x )  x 2  1 .
f
Example 4.33. Let
Let
x , y  R- 3
f : R  3  R- 1
be defined by
f (x ) 
f
1
, we
x 2
.
x 3
. Then
f (x )  f ( y )
x 2 y 2

x 3 y 3
 (x  2)( y  3)  (x  3)( y  2)

 xy  2 y  3x  6  xy  3 y  2x  6
x  y
Thus
f
is a function. Next, let
f (x )  f ( y )
x 2 y 2

x 3 y 3
 (x  2)( y  3)  (x  3)( y  2)

 xy  2 y  3x  6  xy  3 y  2x  6
x y
Thus f is one to one function.
Now, let Let b  R  1 . Then there is exist an element
such that
f (x )  b .
x  R  3
x 2
, and hence,
x 3
xb  3b  x  2
b
Thus
 xb  x  3b  2
 x (b  1)  3b  2
3b  2
b 1
and is onto. Therefore, f 1 (x )  3x  2 .
x 1
Example 4.34. Let A  1, 2,3, 4,5 . Definea
x 
f (1)  4, f (2)  1, f (3)  4, f
function
(4)  2 , and f (5)  4 . Then
f 1 (1)  2 ,
f 1 (2)  4 ,
f
f
f
f
f
1
(3)  ,
1
(4)  1, 2,5 ,
1
(5)  ,
1
(1, 2)  2, 4 ,
1
(2,3, 4)  4,1,3,5
f :A A
by
44
Theorem 4.3. Let A , and B be sets, and let f : A  B be a one to
one and onto (bijective) function. Then
1. f 1 : B  A is a one to one and onto (bijective) function.
2. (f 1 )1  f .
3. f 1 f  I A , and f f 1  I B .
Proof.
1. Now, we see that the function f 1 is a one to one
(injective) function. Let u ,v  B such that x  f 1 (u )  f 1 (v ) .
Then u  f (x )  v . Thus f 1 is a one to one (injective)
function. Next, to prove that f 1 is an onto (surjective)
function, we let a  A , then there exists an element b  B
such that f 1 (b )  a . But f is onto, hence for b  B , there is
a  A such that b  f (a ) , that is f 1 (b )  a . Hence f 1 is onto
(surjective).
2. We notice that (f 1 )1 : A  B is a function and its domain is
A . Let x  A . By the definition of the inverse of a function,
we have
(f
1 1
) (x )  y  x  f
1
( y )  f (x )  y
1 1
Thus, (f )  f .
3. Let x  A , and let
(f
1
x  f 1 ( y ) ,
f )(x )  f 1 (f (x ))  f 1 ( y )  x  I A (x )
f (x )  y
Thus, f 1 f  I A .
Similarly, we prove f
in A . Then f (x )  y , and
(f
f
1
f
in
1
)( y )  f (f
B
. Then
 IB .
1
Let
y B
and
, and let
f
1
(y )  x
( y ))  f (x )  y  I B ( y )
Example 4.35. Let Let A , and B be sets, and let f : A  B , and
g : B  A be functions such that g f  I A . Then
1. g  f 1 is true (see Theorem 4.3 (3)).
2. f is an onto function is false see Theorem 4.1 (4)).
3. f is a one to one function is true see Theorem 4.1 (4)).
4. g is an onto function is true see Theorem 4.1 (4)).
5. g is a one to one function is false see Theorem 4.1 (4)).
45
Theorem 4.4. Let A , and B be sets, and let f : A  B be a
function. If U  B , and V  B , then
1. f 1 (U V )  f 1 (U ) f 1 (V ) .
2. f 1 (U V )  f 1 (U ) f 1 (V ) .
Proof.
1. We show that x  f 1 (U V )  x  f 1 (U ) f 1 (V ) .
1
x f
(U V )  f (x ) U V
 f ( x ) U  f ( x ) V
1
x f
2. We show that
1
x f
 x f
1
 x f
1
(U )  x  f
(U ) f
1
(U V )  x  f
1
1
(V )
(V )
(U ) f
1
(V ) .
(U V )  f (x ) U V
 f ( x ) U  f ( x ) V
 x f
1
 x f
1
(U )  x  f
(U ) f
1
1
(V )
(V )
Theorem 4.5. Let A , and B be sets, and let f : A  B be a
function. If V i i  X , index set be a family of subset of B then
1. f 1 ( V i )  f 1 (V i ) .
i X
2.
f
1
3.
i X
Vi ) 
(
f
i X
i X
1
.
f
(b )  A
1
(V i ) .
b B
4. If b , b   B with b  b  , then f 1 (b )
Proof.
1. We show that x  f 1 ( V i )  x  f
i X
x f
1
1
(b )   .
(V i ).
Then
i X
V i )  f (x ) 
(
1
f
i X
Vi
i X
 f ( x ) V i , for each i  X
 x f
1
x 
(V i ), for each i  X
f
1
(V i ).
i X
Thus
f
1
Vi ) 
(
i X
f
1
(V i ) .
i X
2. We show that
x f
1
Vi )  x 
(
i X
f
1
(V i ).
Then
i X
46
x f
1
V i )  f (x ) 
(
i X
Vi
i X
 f ( x ) V i , for some i  X
 x f
1
x 
(V i ), for some i  X
f
1
(V i ).
i X
Thus,
f
1
Vi ) 
(
i X
f
1
(V i ) .
i X
3) and 4) is a special cases from 1) and 2).
47