3 Functions 52
SECTION F Graphs of Functions
By the end of this section you will be able to



 understand what is meant by an ordered pair define a graph in terms of ordered pairs check for a function by examining the graph test a given function is injective and/or surjective by graphical and algebraic methods
F1 Ordered Pair
What does the term’ ordered pair’ mean?
Pair means two – pair of shoes, pair of socks, pair of gloves etc.
What do you think ordered pair means?
Ordered pair is two elements where order matters . One of the elements call it x is the first element and the other element say y is the second element. An ordered pair where x is the first element and y the second is denoted by

,

For example, points on the Cartesian coordinate system are ordered pairs. y
This is the ordered pair (1,2)
5
2
1
4
3
This is the ordered pair (2,1)
1 2 3 4 5 x
Fig 35
Note that (1, 2) is different from (2,1), we say (1, 2) is an example of an ordered pair which is different from the ordered pair (2,1).
The set
  not an ordered pair. Why not?
Because
 
Remember the curly brackets represent a set and order does not matter in sets whilst for ordered pair it does matter.
In 1914 Norbert Wiener defined the ordered pair

,

as

(3.17)
 
,
  where the element x is the first element and y the second. Therefore

1, 2
    
When are two ordered pairs such as
   
(3.18)

,
 
,

if and only if x
 a and y
 b
N Wiener was born in Columbia, USA. Both his parents were Jewish immigrants from Europe.
Wiener was well ahead of his years in education receiving his PHD by the age of 18 from Harvard.
After his PHD he studied in Europe under the most eminent mathematicians of that era, around 1914, such as Landau, Hilbert and Hardy.
After the first World War he started as an instructor at MIT where he remained for the rest of his working life and becoming a professor.
3 Functions 53
Fig 36
Norbert Wiener 1894 to 1964
We can visualize a given function by plotting it on the Cartesian Coordinate
System . This system allows you to plot a point in the two dimensional plane
2
by assigning two real values generally labelled as x and y . The x value is called the x coordinate and y value is the y coordinate of the uniquely determined point in the plane. The coordinates are defined by two axes at right angles to each other with the common point called the origin and nominated O .
A point in the Cartesian coordinate system is defined by an ordered pair, real numbers. The ordered pairs (1, 3),
 
1, 2

,

0,

1

and


,

, of
1, 1

are illustrated below: y
3
This point is (-1,2)
-3 -2 -1
This point is ( -1 , -1 )
-2
-3
2
1
This point is (1,3)
This is the origin ( 0 , 0 )
1 2 3 x
-1
This point is ( 0 , -1 )
Fig 37

3 Functions 54
Note that the horizontal axis is labelled x and is called the x -axis. Similarly the vertical axis is called the y -axis.
Also note that the ordered pair

,

on the Cartesian coordinate system needs to be real numbers, that is x

and y

. The Cartesian coordinate system is an example of the plane
2  
.
Fig 38
Rene Descartes
1596 to 1650
The name Cartesian is derived from the French mathematician and philosopher
Rene Descartes born in 1596. Descartes attended a Jesuit college and because of his poor health he was allowed to remain in bed until 11 o’clock in the morning, a habit he continued until his death in 1650.
In 1616 Descartes obtained a law degree from University of Poitiers in France.
After this he travelled through Europe but settle in Holland for 21 years between 1628 to 1649. During this 21 year period he lived a secluded life and contributed to philosophy and mathematics. In 1649 he moved to Sweden but did not survive the cold winter there, dying of pneumonia in 1650.
F2 Graph of a Function
Let :

B be a function. We define
Graph of f

  x ,
   x

A

What does this mean?
In words we say the ‘Graph of function f is the set of ordered pairs
 x ,
  
such that x is in the domain A
’.
Generally we let y
  
and so
Graph of f

 
,
 x

A and y
   
For example let f :

be a function given by the formula
   x
2 
1
We can describe the graph as any one of the following:
1.
Graph of f

 
,
 x

2.
Graph of
3.
Graph of f f


 
  x ,
   x
, 2 
1
 x

and y
 x 2 
1

or
and
   x 2 
1

or


In each case the graph is the elements in the set of ordered pairs

,

where y
    x 2 
1 and x is in the domain

3 Functions 55
The domain (start) of the function is plotted on the horizontal
 
axis and codomain
(destination) on the vertical
 
axis.
For a graph on the Cartesian coordinate system we need both the domain (start) and codomain (arrival) to be subsets of real numbers. Next we check whether a given graph on the Cartesian system is a function or not.
Remember :

B is a function if every element in A is assigned a unique element in B . How can we check whether a given graph is a function or not?
Assuming A B because for the Cartesian coordinate system the domain and codomain need to be subsets of . Let y
  
then for each x value we have only one y value. How can we check this on a graph?
Each vertical line will intersect the graph at only one point . W hich of the following are graphs of functions? y
10
Graph (a)
8
6
4
2 y = x Vertical Line for Testing
Function
20 40
1 y
60 80 100 x
Graph (b)
-1 -0.5
0.5
x
2
+ y
2
= 1
0.5
-0.5
-1
1 x
Vertical Line for Testing
Function
2 y
1
Graph (c) 1 y
2
= x
2 3 4 5 x
Vertical Line for Testing
Function
-1
-2
Fig 39

3 Functions 56
Only graph (a) is a function because for graphs (b) and (c) we have a vertical line which cuts the graph at two points. That means for an x value we have two different y values, hence not a function. (Each x has two different destinations).
F3 Injection and Surjection
We can check a given function is injective or surjective by examining its graph. How do we check whether a function is injective?
Injective means that the function is one to one so for each y there is only one x such that y
  
. In this case we use a horizontal line test.
For the function to be injective, each horizontal line will intersect at only one point of the graph of the function.
Which of the following graphs are injective? y
10
Graph (a)
4
2
8 y = x
2
6
Horizontal Line for
Testing Injection
-3 -2 -1
10 y
1 2 3 x
Horizontal Line for
Testing Injection
5
Graph (b)
-2 -1
-5
1 y = x
3
2 x
y = |x|+ 1
-10
5 y
4
Graph (c) 3
2
Horizontal Line for
Testing Injection
1 Fig 40
-4 -2 2 4 x
Clearly ONLY graph (b) is injective because a horizontal line intersects only at one point of the graph. In graphs (a) and (c) the horizontal line crosses the graph at two different points therefore it is not injective .

3 Functions 57
In graph (a) for y

0 (greater than 0) we have two domain values x ,
 x with the same y value: y
    f
 
Similarly for graph (c) for y

1 we have two domain values x ,
 x with the same y value: y
    f
 
In both these cases the graphs cannot be injective (cannot be one to one).
How do we check the graph of a given function is surjective?
We examine the codomain and range of the function and if they are equal then we conclude that the function is surjective. For example if :

B and the range of f does not equal B then the function is not surjective because the codomain is B .
Remember the codomain is the vertical
 
axis.
Example 27
Let f :

be given by
   x
2 
1 . Show that the function is not surjective nor injective by graphical and algebraic methods.
Solution
Graphically
We can show this by plotting the graph of y
    x
2 
1 : y y = x
2 – 1
5
4
3
2
1 Horizontal Line for
Testing Injection
-2 -1 1 2 x
-1
-2
Fig 41
Since f :

therefore the codomain is all the real numbers, . What is the range of this function
   x
2 
1 ?
By examining the vertical axis in Fig 41 we can see that the range is all the real numbers greater than or equal to

1 . Therefore the codomain, , does not equal the range so we conclude the function is not surjective (not onto).
Is it injective?
No because the horizontal line shows that the graph cuts the line in two different points. For all y
 
1 we have two domain values x and
 x which correspond to this y value.
Algebraically
We have tested functions for surjection and injection in previous sections by algebraic methods and only repeat the procedure below.
How do we show the given function is not surjective algebraically?
Let y
  
and solve this equation for x . Let y

then

3 Functions 58 y
 x
2 
1 x
2    x
  y
 
Only real values are possible for value corresponds to y 1 0 which gives y
 
1 . Hence no real x y
 
1 therefore the function is not surjective (not onto).
How do we show that the given function is not injective?
Assume , are in the domain and show that
    
    
 x
 y : x
2
1 y
2 
1 x
2  y
2 x
  y means that x
 y or x
  y x
  y
. Hence the given function is not injective (not one to one).
Which method, algebraic or graphic, is easier?
If you know what the graph of the given function looks like then the graphical method is straightforward.
To show a given function is not injective or surjective it is generally easier to give a counter example. In the above example (27) to show the given function is not injective it is enough to say f
   f
 
4
Therefore f is not one to one or not injective.
To show that f is not surjective, consider
   x
2 y
 
2
1
then there is no real x such that
2
Hence there is no x in the domain with
   
2 . The function f is not surjective.
Example 28
Let f :
 
be given by

1 x
. Check the function for surjection and injection by graphical and algebraic methods.
Solution
Graphically
The graph of y
   
1 x
for x

0 (because domain is y

) is:
5
4
3
2
1 y = 1 x
Horizontal Line for
Testing Injection
Fig 42
1 2 3 4 5 x

3 Functions 59
What is the codomain of f ?
Since f :
 
therefore the codomain is all the real numbers, . W hat is the range of f ?
By examining the graph in Fig 42 we can see that the range of the function
 y

0

is all the positive real numbers,

. Since the codomain, , and range,

, are not equal, therefore the given function is not surjective. Is the function injective?
Yes because the horizontal line crosses the graph at just one point.
Algebraically
How do we show the given function is not surjective algebraically?
Let y be in the codomain, . Solve the equation y
  
for x . y

1 x
Because y
   x

1 y

If y is negative then x

1
is negative and so there is no x in the domain, y which corresponds to this negative y . Why not?

,
Because x
 
(is a positive real number) therefore y

1 cannot be negative. x
Hence the given function is not surjective.
How do we check that the function is injective?
Assume , are in the domain and show that
      x
 y :
 
Hence the given function
   
1

1 x y
which gives x
 y

1 x
is injective.
Sometimes the graph of a function is a set of isolated points. For example, graphs where the domain is the set of integers, , will be a set of isolated points.
The graph of the function f :

given by the formula
   x is the set of ordered pairs

,

where x is an integer and y
 x y
    x . What ordered pairs are members of this set?

2,
     
is the same integer because
    
etc
The graph of these ordered pairs is: y
2
1
2 x
-2 -1 1
-1
-2
Fig 43

3 Functions 60
Example 29
Let f :

be given by
   x
3
. Test the function for injection and surjection.
Solution
By plotting the graph of y
    x
3
for selected integer values of x we have: y
30
20
10
-3 -2 -1 1 2 3 x
-10
-20
Horizontal Line for
Testing Injection
Fig 44
-30
Notice that we only have isolated points because the domain is the set of integers, .
The graph of this function will be the set of ordered pairs given by

,
 x
3

where x is an integer
A few selected ordered pairs of this graph are

2, 8

, (0, 0) and (3, 27).
The function is not surjective (onto) because by examining the graph we can see that there is no integer x such that
   x
3 
20 . There is no integer in the domain, , which corresponds to
  
20 therefore it is not surjective. Of course we could have selected most integers such as 5, 14, 22 etc which are not transformed to by the given function f . That is, there is no x in the domain such that
  
5,

14,

22 etc
Is the function injective?
Yes because if we draw a horizontal line the graph only cuts the line at one point.
Hence the function is injective but not surjective.
SUMMARY
Let :

B where

be a function. We define
Graph of f

  x ,
   x

A
 x ,
  
is an ordered pair. Ordered pair are two elements x and y where
 one element say x is the first element and y the second. It is normally denoted by
,

.
We can check a given graph is a function by the vertical line test:
Each vertical line will intersect at only one point of the graph for it to be a function.
For a function to be injective each horizontal line will intersect at only one point of the graph of the function.
If the codomain of the function and the range of the graph are equal then we conclude that the function is surjective.