In the previous section, we saw relations. In this section, we will see functions.
Some basics about functions can be written in 3 steps:
1. We have seen the method to define a relation from set A to set B. All we need to do is this:
♦ Find the ordered pairs which satisfy the relation.
♦ Write those ordered pairs as a set.
2. Now, to define a function, we need to check two conditions:
(i) We know that the first elements in R, will be from set A.
• We need to make sure that, every element of A is present as first elements in the R. No element of A should be left out.
(ii) Next we need to make sure that no element in A is present more than once in R.
3. If both the conditions in (2) are satisfied, that relation is a function.
Let us examine some of the relations that we saw in the previous section.
◼ In Example 1, we have:
R = {(1, Maths), (2, Maths), (2, Chemistry), (3, Chemistry), (3, Biology), and (4, Geography)}
• All the four students are present in R. So it may be a function.
• But student 2 appears more than once. So it is not a function.
(We agree that student 3 also appears more than once. But a single instance is sufficient to confirm that the R is not a function)
◼ In Example 2, we have:
R = {(5, 3), (6, 4), (7, 5)}
• All the three elements in A are present in R. So it may be a function.
• Each of those elements appear only once. So it is a function.
◼ In Example 3, we have:
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
• The element ‘6’ is present in set A. But it is not present in R. So it is not a function.
(If the first condition is not satisfied, we can confirm that it is not a function. There is no need to check the second condition)
◼ In solved example 2.15, we have:
R = {(1,3), (2,6), (3,9), (4,12)}
• The element ‘5’ is present in A. But it is not present in R. So it is not a function.
(We agree that, none of the elements coming after 4, are present in R. But a single instance is sufficient to confirm that the R is not a function.
◼ In solved example 2.16, (file available here) we have:
R = {(1,6), (2,7), (3,8)}
• This relation is on N. But all elements of N are not present in R. So it is not a function.
◼ In solved example 2.17, we have:
R = {(9,3), (9,-3), (4,2), (4,-2), (25,5), (25,-5)}
• All elements of P are present in R. So it may be a function.
• But ‘9’ appears more than once. So it is not a function.
(We agree that ‘4’ and ‘25’ also appear more than once. But a single instance is sufficient to confirm that the R is not a function)
◼ In solved example 2.21, 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)}
• All elements of A are present in R. So it may be a function.
• But ‘1’ appears more than once. So it is not a function.
(We agree that ‘2’ and ‘3’ also appear more than once. But a single instance is sufficient to confirm that the R is not a function)
◼ In solved example 2.22, we have:
R = {(0,5), (1,6), (2,7), (3,8), (4,9), (5,10)}
• All the elements in the given set are present in R. So it may be a function.
• Each of those elements appear only once. So it is a function.
◼ In solved example 2.25, we have:
R is a relation from Z to Z
• We found out that, domain of R is the set Z
♦ That means, all the elements in Z are present as first elements in R
• Since all elements of Z are present, it may be a function.
• But there will be many repetitions of the first element.
♦ For example, we can put 1 in the place of ‘a’
♦ and put infinite different integers for b.
♦ Every result will be an integer.
• Since the first elements in R appear more than once, it is not a function.
Let us see a solved example:
Solved example 2.26
Let N be the set of natural numbers and the relation R be defined on N such that R = {(x, y) : y = 2x, x, y ∈ N}.
What is the domain, codomain and range of R? Is this relation a function?
Solution:
1. In our present case, the relation R is a set which contains ordered pairs of
the form (x, y)
♦ 'x' should be from set N
♦ Since the relation is on N, 'y' should also be from set N
• The x and y in each ordered pair in R should satisfy the condition:
y = 2x
2. Recall that natural numbers are 1, 2, 3, 4, . . . (Details here)
• Let us take each possible value for ‘x’ from set N:
• Let x = 1
♦ Then y = 2x = (2 × 1) = 2
♦ So the first ordered pair in R is (1, 2)
• Let x = 2
♦ Then y = 2x = (2 × 2) = 4
♦ So the second ordered pair in R is (2, 4)
• Let x = 3
♦ Then y = 2x = (2 × 3) = 6
♦ So the third ordered pair in R is (3, 6)
3. In this way, we can obtain infinite number of ordered pairs in R.
• We will be using all the natural numbers as 'x'. That means, all elements of N will appear in R. So this may be a function
• What happens if we use a natural number (in the place of x) more than once?
Ans: we will be getting the same ordered pair more than once.
• There cannot be repetition
of ordered pairs in R. So we will be using every natural number (in the place of x) only once. So this is a function.
Solved example 2.27
Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) R = {(2,1),(3,1),(4,2)}
(ii) R = {(2,2),(2,4),(3,3),(4,4)}
(iii) R = {(1,2),(2,3),(3,4),(4,5),(5,6),(6,7)}
Solution:
• Usually, a relation is defined from a set A to set B. Or from a set A to itself.
• But here, we are not given A or B. So we will assume that, all the elements of A are present as first elements in R.
• If all elements are not present, we will be able to straight away say that, they are not functions.
• Thus in all the three questions, we need to check the second condition only.
Part (i):
All the first elements appear only once. So it is a function.
Part (ii):
The first element '2' appear more than once. So it is not a function.
Part (iii):
All the first elements appear only once. So it is a function.
◼ From the above discussion, it is clear that:
♦ All functions are relations.
♦ But all relations are not functions.
• We have seen how to confirm whether a relation is a function or not. Now we will see some technical terms related to functions. Use of technical terms will help us to describe the functions using minimum words. They can be written in 14 steps:
1. If a relation is a function, we use the letter ‘f’ instead of ‘R’.
2. We know that, a relation is defined from one set A to another set B.
• If that relation is a function, we write: f: A→B
3. Some times a relation is defined from one set A to itself.
• If that relation is a function, we write: f: A→A
4. We saw that all functions are relations.
• So the terms domain, codomain, range and image that we saw for relations are applicable to functions also.
• For example,
♦ We write domain of a relation.
♦ We can write domain of a function also.
5. We saw that, if it is to be a function, every element of A should be present as first elements.
♦ The set containing the first elements is the domain.
• So if it is a function, the domain will contain all the elements of A
• Thus it is clear that, if it is a function, the domain will be same as set A.
6. We know that, like R, the f is also a set of ordered pairs.
• If we denote those ordered pairs as (x,y), then:
♦ y is called the image of x under f.
♦ x is called the image of y under f.
7. Consider a relation that we saw in the previous section:
y = x + 1
• We input various values of x and calculated the corresponding 'y values'. Then we wrote them as ordered pairs in the form (x,y).
8. Note that, the 'y values' are obtained by inputting various 'x values'.
• The input x values are 'processed' according to the rule given by the relation y = x + 1
♦ Here, the rule says that, we must add '1' to the input value of x
9. If the relation is a function, we can write:
The 'x values' are 'processed' according to the rule given by the function.
10. Or we can simply write:
The 'x values' are 'processed' according to the function.
11. So we can write:
♦ 'y values' are obtained
♦ when the 'x values' are processed
♦ according to the function.
• This can be schematically represented as in fig.2.9 below:
 |
| Fig.2.9 |
12. We have done this type of 'processing' in our earlier classes. There we denoted the output as 'y'.
• For example, in y = x +1, if we add various values of x to '1', we will get various y values.
♦ Here, [addition of '1' to x] is the processing.
♦ So it is obvious that y is same as f(x).
♦ Then we can write: f(x) = x + 1
13. A function is a relation. So just like R, for f also, there will be a set of ordered pairs.
♦ The first values in that ordered pairs will form the domain of that function.
♦ The second values in that ordered pairs will form the range of that function.
◼ If all the elements in the range set are real numbers, then that function is called a real valued function.
◼ If in a real valued function, all the elements in the domain set are real numbers, then that function is called a real function.
14. Let us see an example:
Let N be the set of natural numbers. Define a real valued function
f : N→N by f (x) = 2x + 1.
Solution:
(i) Given that, it is a real valued function. That means, all values obtained after processing, must be real values.
• The function can be defined by writing the ordered pairs which satisfy that function.
• So our next task is to find those ordered pairs.
(ii) It is given that, f : N→N
• This indicates that,
♦ the first elements of the ordered pairs (input x values) should be taken from the set N.
♦ the second elements (resulting y values) must be present in the set N
✰ N is a subset of R. So indeed, it will be a real valued function
(iii) The set N is the set of natural numbers. That is., N = {1, 2, 3, 4, . . .}
• Let x = 1
♦ This x is processed as follows:
♦ f(1) = (2 × 1 + 1) = (2 + 1) = 3
♦ f(1) is the 'y value' when 'x value' is 1
♦ So the first ordered pair (x,y) is (1,3)
• Let x = 2
♦ This x is processed as follows:
♦ f(2) = (2 × 2 + 1) = (4 + 1) = 5
♦ f(2) is the 'y value' when 'x value' is 2
♦ So the second ordered pair (x,y) is (2,5)
• Let x = 3
♦ This x is processed as follows:
♦ f(3) = (2 × 3 + 1) = (6 + 1) = 7
♦ f(3) is the 'y value' when 'x value' is 3
♦ So the third ordered pair (x,y) is (3,7)
(iv) Proceeding like this, we will get infinite number of ordered pairs. All those ordered pairs should be included in the set f.
• So we can write: f = {(1,3), (2,5), (3,7), (4,9), (5,11), (6,13), (7,15), . . .}
(v) We can make a table using the x and y values in the set f. Such a table is convenient to draw the graph of the function.
 |
| Table 2.1 |
Let us see some common functions and their graphs
A. Identity function
This is a real valued function f: R→R defined by y = f (x) = x
Details can be written in 10 steps:
1. Given that, it is a real valued function. That means, all values obtained after processing, must be real values.
• The function can be defined by writing the ordered pairs which satisfy that function.
• So our next task is to find those ordered pairs.
2. It is given that, f : R→R
• This indicates that,
♦ the first elements of the ordered pairs (input x values) should be taken from the set R.
♦ the second elements (resulting y values) should be present in the set R
3. The set R is the set of real numbers. It will include integers, negative values, positive values, fractions, decimals, recurring decimals, numbers like √2, √5, π etc.,. In short, R will contain every value which can be plotted on a number line. Recall that we plotted √2, √5, π etc., in our previous classes.
• Since different types of numbers are present in R, we will choose some convenient numbers at random.
• Let x = -7
♦ This x is processed as follows:
♦ f(-7) = x = -7
♦ f(-7) is the 'y value' when 'x value' is -7
♦ So we get an ordered pair (x,y) as: (-7,-7)
• Let x = -3
♦ This x is processed as follows:
♦ f(-3) = x = -3
♦ f(-3) is the 'y value' when 'x value' is -3
♦ So we get another ordered pair (x,y) as: (-3,-3)
• Let x = 1.414
♦ This x is processed as follows:
♦ f(1.414) = x = 1.414
♦ f(1.414) is the 'y value' when 'x value' is 1.414
♦ So we get another ordered pair (x,y) as: (1.414,1.414)
• We see that, whatever be the value of x, the value of y will also be the same.
4. Proceeding like this, we will get infinite number of ordered pairs. All
those ordered pairs should be included in the set f.
• So we can write: f = {. . . , (-7,-7), (-3,-3), (1.414,1.414), (5,5), . . .}
5. The above set f is written in roster form. But we have to remember an important point. It can be written in 3 steps:
(i) Both elements of the ordered pairs are real numbers.
(ii) Since they are real numbers, there will be integers,
negative values, positive values, fractions, decimals, recurring
decimals, numbers like √2, √5, π etc.,. We cannot think of a definite sequence to write them.
(iii) So it is better to use set builder form to write f.
6. In the set builder form, we can write:
f = {(x,y) : x ∈ R, y = x}
• That means:
♦ The set f contains all ordered pairs such that,
♦ x is a real number,
♦ y is equal to x.
7. Once we write the set f, we can write the domain and range of f.
(i) First we will write the domain:
• Domain of f is the set containing all the first elements of the ordered pairs in f.
• In our present case, there are infinite number of ordered pairs. So there will be infinite number of first elements.
•
We saw that all the first elements are real numbers. Since they are
real numbers, there will be integers,
negative values, positive values, fractions, decimals, recurring
decimals, numbers like √2, √5, π etc.,. We cannot think of a definite
sequence to write them. So it is better to use set builder form rather
than the roster form.
• We can write:
♦ Domain of f = {x : x ∈ R}
• That means:
♦ The domain of f is the set of all x such that,
♦ x is a real number.
(ii) Next we will write the range:
• Range of f is the set containing all the second elements of the ordered pairs in f.
• In our present case, there are infinite number of ordered pairs. So there will be infinite number of second elements.
•
We saw that all the first elements are real numbers. Since the second
elements are equal to first elements, they are also real numbers.
• We can write:
♦ Range of f = {y : y ∈ R}
• That means:
♦ The range of f is the set of all y such that,
♦ y is a real number.
8. We can make a table using the x and y values in the set f. Such a table is convenient to draw the graph of the function.
• Note that, to input for x, we choose convenient numbers from the set R.
• It is better not to choose numbers with recurring decimals. They will be difficult to plot.
 |
| Table 2.2 |
9. The red line in fig.2.10(a) below, is the graph of this function.
 |
| Fig.2.10 |
• We can write some peculiarities of this red line. They can be written in 5 steps:
(i) The red line always passes through the origin (0,0)
(ii) If x axis and y axis are drawn to the same scale (Details here), the red line will make 45o degrees with the x axis.
• In other words, if the two axes are drawn to the same scale, the red line will bisect the angle between the two axes.
(iii) Mark any point on the red line. Note the coordinates of that point.
♦ The x coordinate will be same as the y coordinate.
♦ This is shown in fig.b
(iv) Mark any point on the x axis. For example, let us mark 3.5.
• Draw a vertical line through that point.
♦ Here, it is the green vertical dotted line in fig.b.
• That vertical line will meet the red line at a point.
• Through that meeting point, draw a horizontal line.
♦ Here, it is the green horizontal dotted line.
• This horizontal line will meet the y axis at a point which have the same x value (here it is 3.5) from where we started off.
• We will get this result even if the two axes are drawn in different scales.
(v) We see arrows at both ends of the red line.
• The arrow at the top end of the red line indicates that, the line can extend up to the point where x = +∞ and y = +∞.
• The arrow at the bottom end of the red line indicates that, the line can extend up to the point where x = -∞ and y = -∞.
10. The identity function has many applications in science and engineering.
In
the next
section, we will see a few more common functions.
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