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 is the first element and the other element say is the second element. An ordered pair where is the first element and the second is denoted by
For example, points on the Cartesian coordinate system are ordered pairs.
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 is 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 is the first element and the second. Therefore
When are two ordered pairs such as and equal?
(3.18) if and only if and
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 by assigning two real values generally labelled as and . The value is called the coordinate and value is the 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 .
A point in the Cartesian coordinate system is defined by an ordered pair, , of real numbers. The ordered pairs (1, 3), , and are illustrated below:
Note that the horizontal axis is labelled and is called the -axis. Similarly the vertical axis is called the -axis.
Also note that the ordered pair on the Cartesian coordinate system needs to be real numbers, that is and . The Cartesian coordinate system is an example of the plane .
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 be a function. We define
Graph of
What does this mean?
In words we say the ‘Graph of function is the set of ordered pairs such that is in the domain ’.
Generally we let and so
Graph of
For example let be a function given by the formula
We can describe the graph as any one of the following:
- Graph of or
- Graph of or
- Graph of
In each case the graph is the elements in the set of ordered pairs where
and is in the domain
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 is a function if every element in is assigned a unique element in . How can we check whether a given graph is a function or not?
Assuming because for the Cartesian coordinate system the domain and codomain need to be subsets of . Let then for each value we have only one value. How can we check this on a graph?
Each vertical line will intersect the graph at only one point. Which of the following are graphs of functions?
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 value we have two different values, hence not a function. (Each 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 there is only one such that . 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?
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.
In graph (a) for (greater than 0) we have two domain values with the
same value:
Similarly for graph (c) for we have two domain values with the
same value:
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 and the range of does not equal then the function is not surjective because the codomain is . Remember the codomain is the vertical axis.
Example 27
Let be given by . 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 :
Fig 41
Since therefore the codomain is all the real numbers, . What is the range of this function ?
By examining the vertical axis in Fig 41 we can see that the range is all the real numbers greater than or equal to . 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 we have two domain values and which correspond to this 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 and solve this equation for . Let then
Only real values are possible for . Hence no real value corresponds to 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 :
means that . 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
Therefore is not one to one or not injective.
To show that is not surjective, consider then there is no real such that
Hence there is no in the domain with . The function is not surjective.
Example 28
Let be given by . Check the function for surjection and injection by graphical and algebraic methods.
Solution
Graphically
The graph of for (because domain is ) is:
What is the codomain of ?
Since therefore the codomain is all the real numbers, . What is the range of ?
By examining the graph in Fig 42 we can see that the range of the function 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 be in the codomain, . Solve the equation for .
If is negative then is negative and so there is no in the domain, , which corresponds to this negative . Why not?
Because (is a positive real number) therefore cannot be negative.
Hence the given function is not surjective.
How do we check that the function is injective?
Assume are in the domain and show that :
Hence the given function 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 given by the formula is the set of ordered pairs where is an integer and is the same integer because . What ordered pairs are members of this set?
etc
The graph of these ordered pairs is:
Example 29
Let be given by . Test the function for injection and surjection.
Solution
By plotting the graph of for selected integer values of we have:
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
where is an integer
A few selected ordered pairs of this graph are , (0, 0) and (3, 27).
The function is not surjective (onto) because by examining the graph we can see that there is no integer such that . There is no integer in the domain, , which corresponds to 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 . That is, there is no in the domain such that
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 be a function. We define
Graph of
where is an ordered pair. Ordered pair are two elements and where one element say is the first element and 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.