Chapter 2.7 - Algebra of Real Functions

In the previous section, we saw greatest integer function. In this section, we will see algebra of real functions. Algebra of real functions involves the following operations:
• Addition of two real functions.
• Subtraction of a real function from another real function.
• Multiply a real function by a scalar. (the scalar will be a real number)
• Multiply a real function by another real function.
• Divide a real function by another real function.

Addition of two real functions
This can be explained in 4 steps:
1. Consider a real function: f: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be a subset of set R
    ♦ For example, X can be the set of integers.
2. Consider another real function: g: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be the same X mentioned in 1
3. We can add the above two functions.
• The resulting function is denoted as: (f + g): X→R
• This resulting function is defined as:
(f + g) (x) = f(x) + g(x), for all x ∈ X
4. Let us see some examples:
Example 1:
Let f(x) = x² and g(x) = 2x + 1 be two real functions. Find (f + g) (x)
Solution:
• We have: (f + g) (x) = f(x) + g(x)
• So in our present case, we get:
(f + g) (x) = x² + 2x + 1
Example 2:
A real function f: X→R is defined as: f(x) = √x
Another real function g: X→R is defined as: g(x) = x
Where X is the set of non-negative real numbers. Find (f + g) (x)
Solution:
• We have: (f + g) (x) = f(x) + g(x)
• So in our present case, we get:
(f + g) (x) = √x + x
Note: In the question, it is specified that, X is the set of non-negative real numbers. So all  input x values will be non-negative real numbers only. If we input negative real numbers, √x will give imaginary numbers. So the function f(x) will not be a real function. The function (f + g) (x) will also be not a real function.

Subtraction of a real function from another real function
This can be explained in 4 steps:
1. Consider a real function: f: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be a subset of set R
    ♦ For example, X can be the set of integers.
2. Consider another real function: g: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be the same X mentioned in 1
3. We can subtract the second function from the first.
• The resulting function is denoted as: (f - g): X→R
• This resulting function is defined as:
(f - g) (x) = f(x) - g(x), for all x ∈ X
4. Let us see some examples:
Example 1:
Let f(x) = x² and g(x) = 2x + 1 be two real functions. Find (f - g) (x)
Solution:
• We have: (f - g) (x) = f(x) - g(x)
• So in our present case, we get:
(f - g) (x) = x² - 2x - 1
Example 2:
A real function f: X→R is defined as: f(x) = √x
Another real function g: X→R is defined as: g(x) = x
Where X is the set of non-negative real numbers. Find (f - g) (x)
Solution:
• We have: (f - g) (x) = f(x) - g(x)
• So in our present case, we get:
(f - g) (x) = √x - x
Note: In the question, it is specified that, X is the set of non-negative real numbers. So all  input x values will be non-negative real numbers only. If we input negative real numbers, √x will give imaginary numbers. So the function f(x) will not be a real function. The function (f - g) (x) will also be not a real function.

Multiplication of a real function by a scalar
This can be explained in 3 steps:
1. Consider a real function: f: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be a subset of set R
    ♦ For example, X can be the set of integers.
2. Consider a scalar 𝛼. Here 𝛼 is a real number
3. We can multiply the function in (1) by 𝛼.
• The resulting function is denoted as: (𝛼f): X→R
• This resulting function is defined as:
(𝛼f) (x) = 𝛼f(x), for all x ∈ X

Multiplication of two real functions
This can be explained in 5 steps:
1. Consider a real function: f: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be a subset of set R
    ♦ For example, X can be the set of integers.
2. Consider another real function: g: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be the same X mentioned in 1
3. We can multiply the above two functions.
• The resulting function is denoted as: (fg): X→R
• This resulting function is defined as:
(fg) (x) = f(x) g(x), for all x ∈ X
4. Let us see some examples:
Example 1:
Let f(x) = x² and g(x) = 2x + 1 be two real functions. Find (fg) (x)
Solution:
• We have: (fg) (x) = f(x) g(x)
• So in our present case, we get:
(fg) (x) = x² × (2x + 1) = 2x³ + x²
Example 2:
A real function f: X→R is defined as: f(x) = √x
Another real function g: X→R is defined as: g(x) = x
Where X is the set of non-negative real numbers. Find (fg) (x)
Solution:
• We have: (fg) (x) = f(x) g(x)
• So in our present case, we get:
(fg) (x) = √x × x = $\mathbf\small{\rm{x^{1/2} \times x = x^{1/2 + 1}=x^{3/2}}}$
Note: In the question, it is specified that, X is the set of non-negative real numbers. So all  input x values will be non-negative real numbers only. If we input negative real numbers, √x will give imaginary numbers. So the function f(x) will not be a real function. The function (fg) (x) will also be not a real function.
5. Multiplication of two real functions is also known as pointwise multiplication.

Quotient of two real functions
This can be explained in 4 steps:
1. Consider a real function: f: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be a subset of set R
    ♦ For example, X can be the set of integers.
2. Consider another real function: g: X→R
• From the notation, it is clear that, the function is defined from a set X to the set R
• Set X should be the same X mentioned in 1
3. We can divide the first function by the second function.
• The resulting function is denoted as: $\mathbf\small{\rm{\left(\frac{f}{g} \right):X\rightarrow R}}$
• This resulting function is defined as:
$\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) =\frac{f(x)}{g(x)}}}$, provided g(x) ≠ 0, x ∈ X
4. Let us see some examples:
Example 1:
Let f(x) = x² and g(x) = 2x + 1 be two real functions. Find $\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) }}$
Solution:
• We have: $\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) =\frac{f(x)}{g(x)}}}$
• So in our present case, we get:
$\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) =\frac{f(x)}{g(x)}=\frac{x^2}{2x+1}}}$
• Note that, in the denominator, we have (2x + 1). If the input x value is (-¹⁄₂), the whole denominator will become zero. So at x = (-¹⁄₂), the quotient function is not defined. We must mention this in the final answer.
• We can write:
$\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) =\frac{x^2}{2x+1},\;x\ne -\frac{1}{2},\;x\in X}}$
Example 2:
A real function f: X→R is defined as: f(x) = √x
Another real function g: X→R is defined as: g(x) = x
Where X is the set of non-negative real numbers. Find $\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) }}$
Solution:
• We have: $\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) =\frac{f(x)}{g(x)}}}$
• So in our present case, we get:
$\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) =\frac{f(x)}{g(x)}=\frac{\sqrt x}{x}=\frac{x^{1/2}}{x}=x^{(1/2 - 1)}=x^{-1/2}=\frac{1}{\sqrt x}}}$
• Note that, in the denominator, we have √x. If the input x value is zero, the whole denominator will become zero. So at x = 0, the quotient function is not defined. We must mention this in the final answer.
• We can write:
$\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) =\frac{f(x)}{g(x)}=\frac{1}{\sqrt x},\;x \ne 0, \;x\in X}}$

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Now we will see some solved examples related to the topic of functions as a whole.

Solved example 2.29
Which of the following relations are functions? Give reasons. If it is a function,
determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2,1), (4,2), (6,3), (8,4), (10,5), (12,6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}.
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.
    ♦ Domain of f = {2, 5, 8, 11, 14, 17}
    ♦ Range of f = {1}
Part (ii):
• All the first elements appear only once. So it is a function.
    ♦ Domain of f = {2, 4, 6, 8, 10, 12, 14}
    ♦ Range of f = {1, 2, 3, 4, 5, 6, 7}
Part (iii):
• The first element '1' appear more than once. So it is not a function.

Solved example 2.30
Find the domain and range of the following real functions:
(i) f(x) = - |x|
(ii) f(x) = √(9 - x²)
Solution:
Part (i): f(x) = - |x|
1. Given that, this is a real function. So all input x must be real numbers.
2. |x| on the right side indicates that, what ever be the input x, it's modulus will be taken. Modulus values are positive numbers.
3. But there is also a negative sign on the right side. So all the positive values derived from |x| will be converted to negative numbers. All those negative numbers will be real values.
4. Let us see some examples:
    ♦ Let input x be -5. Then f(x) = -|x| = -|-5| = -5
    ♦ Let input x be +4. Then f(x) = -|x| = -|4| = -4
5. It is clear that:
    ♦ Input x values can be negative or positive.
    ♦ Output f(x) values will be negative
6. Input x values can be any real number. So we can write:
Domain = {x : x ∈ R}
• That means,
    ♦ Domain is the set of all x such that,
    ♦ x is an element of R
• We can write in a simpler form also: Domain = R
7. Output f(x) values will be negative real numbers. So we can write:
Range = {y : y ∈ R^-}
• That means,
    ♦ Range is the set of all y such that,
    ♦ y is an element of R^-
• We can write in a simpler form also: Range = R^-
8. Another way to write range:
• We have seen intervals as sets. (See fig.1.3 of section 1.4)
• In our present case, the range does not contain all numbers on the real number line.
    ♦ The range contains only an interval in the real number line.
    ♦ That interval starts from -∞ and ends at 0
• This interval can be written in set form as: (-∞, 0]
    ♦ So we can write: Range = (-∞, 0]
• This indicates that:
    ♦ Every real number between -∞ and 0 will be an f(x) value.
    ♦ -∞ cannot be an f(x) value. (note the '(' on the left side of -∞)
    ♦ 0 can be an f(x) value. (note the ']' on the right side of 0)
9. Fig.2.18 below shows the graph of this function:

Fig.2.18

• We can mark any point on the two lines. It's y coordinate will be negative.

Part (ii): f(x) = √(9 - x²)
1. Given that, this is a real function. So all input x must be real numbers.
2. On the right side, we have a square root. The portion inside the square root symbol, should not become negative. Because, square root of negative numbers are imaginary numbers. They are not real numbers. So if the portion inside the square root symbol becomes negative, the function will not be real.
3. It is clear that, the portion inside the square root symbol must be zero or greater than zero. First we will see when it becomes zero. For that, we equate it to zero. We get:
9 - x² = 0
⇒ x² = 9
⇒ x = +3 or -3
So the portion becomes zero when x = +3 or -3
4. If x is greater than +3, the portion will become negative. So a value greater than +3 cannot be allowed.
• Also, if x is less than -3, the portion will become negative. So a value less than -3 cannot be allowed.
• Thus we get: Domain = [-3,3]
• That means:
    ♦ Every real number between -3 and +3 can be an input x value.
    ♦ -3 can be an input x value. (note the '[' on the left side of -3)
    ♦ +3 can be an input x value. (note the ']' on the right side of 3)
5. Next we will find the range.
• We see that, f(x) is obtained by subtracting a quantity from 9, and then taking the square root.
• If zero is subtracted from 9, then f(x) will be maximum. We get:
f(x) = √(9 - x²) = √(9 - 0²) = √(9 - 0) = √9 = +3
• If 9 is subtracted from 9, then f(x) will be zero. We get:
f(x) = √(9 - x²) = √(9 - 3²) = √(9 - 9) = √(9 - 9) = 0
• So the maximum value of f(x) possible is +3 and the minimum possible is 0.
• Thus we can write: Range = [0,3]
• That means:
    ♦ Every real number between 0 and +3 will be a f(x) value.
    ♦ 0 will be a f(x) value.
    ♦ +3 will be a f(x) value.
6. Fig.2.19 below shows the graph of this function:

Fig.2.19

 

Solved example 2.31
A function f is defined by f(x) = 2x –5. Write down the values of
(i) f(0), (ii) f(7), (iii) f(–3).
Solution:
(i) f(0) = (2 × 0) - 5 = (0 - 5) = -5
(ii) f(7) = (2 × 7) - 5 = (14 - 5) = 9
(iii) f(-3) = (2 × -3) - 5 = (-6 - 5) = -11

Solved example 2.32
The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by $\mathbf\small{\rm{t(C)=\frac{9C}{5}+32}}$
Find (i) t(0) (ii) t(28) (iii) t(–10) (iv) The value of C, when t(C) = 212.
Solution:
(i) $\mathbf\small{\rm{t(0)=\frac{9 \times 0}{5}+32}}$ = (0 + 32) = 32
(ii) $\mathbf\small{\rm{t(28)=\frac{9 \times 28}{5}+32}}$ = (50.4 + 32) = 82.4
(iii) $\mathbf\small{\rm{t(-10)=\frac{9 \times -10}{5}+32}}$ = (-18 + 32) = 14
(iv) $\mathbf\small{\rm{t(C)=200 =\frac{9 \times C}{5}+32}}$
⇒ 212 = 1.8C + 32
⇒ 1.8C = (212 - 32) = 180
⇒ C = ¹⁸⁰⁄_1.8  = 100

Solved example 2.33
Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x > 0.
(ii) f(x) = x² + 2, x is a real number.
(iii) f(x) = x, x is a real number.
Solution:
Part (i):
1. Given that: x ∈ R and x > 0.
    ♦ So the input x must be a real number,
    ♦ Also the input x must be positive and greater than zero.
2. Smallest possible input x will be that value which is very close to zero.
• Let us put x = 0. We will get: f(0) = (2 - (3 × 0)) = 2
    ♦ But this '2' cannot be accepted as a range value because we cannot use x = 0
• Let us put x = 1. We will get: f(1) = (2 - (3 × 1)) = (2 - 3) = -1
• Let us put x = 2. We will get: f(2) = (2 - (3 × 2)) = (2 - 6) = -4
• Let us put x = 3. We will get: f(3) = (2 - (3 × 3)) = (2 - 9) = -7
3. We see that, as input x increases, the value of f(x) decreases.
• Proceeding like this, we will reach f(x) = -∞ for a very large value of x.
4. So the range will be (-∞, 2)
• That means:
    ♦ Every real number between -∞ and +2 will be a f(x) value.
    ♦ -∞ will not be a f(x) value. (note the '(' on the left side of -∞)
    ♦ +2 will not be a f(x) value. (note the ')' on the right side of +2)

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In the next section, we will see a few more solved examples.

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Chapter 2.4 - Rational Function

In the previous section, we saw constant function and polynomial function. In this section, we will see rational function.

D. Rational function
• A rational function is a real valued function f: R→R defined by:
$\mathbf\small{\rm{f(x)=\frac{p(x)}{q(x)}}}$
• Details can be written in 11 steps:
1. Given that, it is a real valued function. That means, all values obtained after processing, must be real values.
2. Let us examine the numerator and denominator:
• The numerator is p(x). It must be a polynomial function.
• The denominator is q(x). It must also be a polynomial function.
   ♦ This q(x) should not be equal to zero.
   ♦ If it is zero, the function f(x) cannot be defined.
3. The function can be defined by writing the ordered pairs which satisfy that function.
• So our next task is to find those ordered pairs.
• We will take a simple rational function: f(x) = $\mathbf\small{\rm{\frac{1}{x}}}$
    ♦ Recall that all constants are polynomials. So '1' is a polynomial.
    ♦ For example, 4 can be written as: 4x⁰, which is (4 × 1) = 4
4. 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
5. 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) = $\mathbf\small{\rm{\frac{1}{(-7)}}}$ = -0.143
   ♦ f(-7) is the 'y value' when 'x value' is -7
   ♦ So we get an ordered pair (x,y) as: (-7, -0.143)
• Let x = -5
   ♦ This x is processed as follows:
   ♦ f(-5) = $\mathbf\small{\rm{\frac{1}{(-5)}}}$ = -0.2
   ♦ f(-5) is the 'y value' when 'x value' is -5
   ♦ So we get an ordered pair (x,y) as: (-5, -0.2)
• Let x = 1
   ♦ This x is processed as follows:
   ♦ f(1) = $\mathbf\small{\rm{\frac{1}{(1)}}}$ = 1
   ♦ f(1) is the 'y value' when 'x value' is 1
   ♦ So we get an ordered pair (x,y) as: (1, 1)
• Let x = 2
   ♦ This x is processed as follows:
   ♦ f(2) = $\mathbf\small{\rm{\frac{1}{(2)}}}$ = 0.5
   ♦ f(2) is the 'y value' when 'x value' is 2
   ♦ So we get an ordered pair (x,y) as: (2, 0.5)
• Let x = 5
   ♦ This x is processed as follows:
   ♦ f(5) = $\mathbf\small{\rm{\frac{1}{(5)}}}$ = 0.2
   ♦ f(5) is the 'y value' when 'x value' is 0.2
   ♦ So we get an ordered pair (x,y) as: (5, 0.2)
• Let x = 10
   ♦ This x is processed as follows:
   ♦ f(10) = $\mathbf\small{\rm{\frac{1}{(10)}}}$ = 0.1
   ♦ f(10) is the 'y value' when 'x value' is 10
   ♦ So we get an ordered pair (x,y) as: (10, 0.1)
• We see that, whatever be the value of x, the value of y will be the reciprocal of that x value.
6. 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,-0.143), (-5, -0.2), (1, 1), (2, 0.5), (5, 0.2), (10, 0.1), . . .}
7. 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.
8. In the set builder form, we can write:
f = {(x,y) : x ∈ R, y = $\mathbf\small{\rm{\frac{1}{x}}}$, $\mathbf\small{\rm{\frac{1}{x}}}$ ∈ R}
• That means:
    ♦ The set f contains all ordered pairs (x,y) such that,
    ♦ x is a real number,
    ♦ y is the reciprocal of x,
    ♦ reciprocal of x is a real number
• But there is a problem. We cannot put x = 0. The function is not defined at x = 0.
• This problem can be solved in 3 steps:
(i) Consider the set R - {0}.
    ♦ Here we are subtracting the 'set containing zero' from the set R
(ii) The resulting set will not contain zero.
    ♦ We must take x only from R - {0}
(iii) So we must modify the above set builder form. We get:
f = {(x,y) : x ∈ R - {0}, y = $\mathbf\small{\rm{\frac{1}{x}}}$, $\mathbf\small{\rm{\frac{1}{x}}}$ ∈ R}
• That means:
    ♦ The set f contains all ordered pairs (x,y) such that,
    ♦ x is an element of R - {0},
    ♦ y is the reciprocal of x,
    ♦ reciprocal of x is a real number
9. 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 (except zero). 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 - {0}}
• That means:
    ♦ The domain of f will contain all real numbers except zero.
(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 (except zero). Since the second elements are reciprocals, they will also be real numbers. So there will be integers, negative values, positive values, fractions, decimals, recurring decimals 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:
    ♦ Range of f = {$\mathbf\small{\rm{\frac{1}{x}}}$ : x ∈ R - {0}}
• That means:
    ♦ The range of f will contain reciprocals of all real numbers except zero.
10. 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 - {0}.
• It is better not to choose numbers with recurring decimals. They will be difficult to plot.

Table 2.6

11. The red curves in fig.2.14(a) below, is the graph of the function.

Fig.2.14

• The graph in fig.b is drawn to a different scale. This can be explained in 5 steps:
(i) The y value is the reciprocal of x value. So we will need many small values on the y axis.
(ii) To achieve this within limited space, we change the scale of the y axis.
(iii) In the fig.2.14(a), we have:
    ♦ Scale of x axis as: 1 cm = 1 unit
    ♦ Scale of y axis as: 1 cm = 1 unit
(iv) In the fig.2.13(b), we have:
    ♦ Scale of x axis as: 1 cm = 1 unit
    ♦ Scale of y axis as: 1 cm = 0.25 unit
(v) So the graph in fig.b is a bit distorted. But the coordinates do not change.
• The reader is advised to prepare a table and draw the graph in his/her own graph books.
• We can write some peculiarities of this red curve. They can be written in 3 steps:
(i) The red curve of f(x) = $\mathbf\small{\rm{\frac{1}{x}}}$ never passes through the origin (0,0)
(ii) Mark any point on the red curve. Note the coordinates of that point.
   ♦ The y coordinate will be the reciprocal of the x coordinate.
   ♦ This is shown in fig.b
   ♦ The green vertical dashed line shows 2
         ✰ The green horizontal dashed line shows the reciprocal 0.5
   ♦ The magenta vertical dashed line shows -5
         ✰ The magenta horizontal dashed line shows the reciprocal -0.2
(iii) We see arrows at both ends of the red curves. There are four arrows. We will call them I, II, III and IV.
I. The top arrow of the right side red curve:
• This arrow indicates that, the curve can go very high up.
• This can be explained as follows:
   ♦ As we move along this curve from right towards the left, x takes values closer and closer to zero.
         ✰ For example, 0.00001 is closer to zero than 0.0001
   ♦ Values closer to zero are very small.
         ✰ The reciprocals of such small values will be very large.
   ♦ The reciprocals are y values.
         ✰ That means y values will be very large.
         ✰ They are very high up in the graph.
         ✰ This arrow indeed indicates that the graph will go very high up (y = +∞).
   ♦ As the x takes values closer and closer to zero, we get the feeling that, this end of the curve is going to touch the y axis.
         ✰ It will touch the y axis when x = 0
         ✰ However, x cannot become equal to zero.
         ✰ If it is zero, then the function can not be defined.
         ✰ This is because, $\mathbf\small{\rm{\frac{1}{x}}}$ becomes $\mathbf\small{\rm{\frac{1}{0}}}$
   ♦ So this end will never touch the y axis
         ✰ When we draw this graph on a sheet of paper or the computer screen, it appears to touch.
         ✰ But if we zoom in, we will see that, there is a little gap.
II. The bottom arrow of the right side red curve:
• This arrow indicates that, the curve can reach to the point where (x = +∞).
• This can be explained as follows:
   ♦ As we move along this curve from left towards the right, x take values closer and closer to +∞.
   ♦ Values closer to +∞ are very large.
         ✰ The reciprocals of such small values will be very small.
   ♦ The reciprocals are y values.
         ✰ That means y values will be very small.
         ✰ They get closer and closer to zero.
   ♦ As the y becomes closer and closer to zero, we get the feeling that, this end of the curve is going to touch the x axis.
         ✰ It will touch the x axis when x = +∞
         ✰ However, x cannot become equal to +∞.
   ♦ So this end will never touch the x axis
         ✰ When we draw this graph on a sheet of paper or the computer screen, it appears to touch.
         ✰ But if we zoom in, we will see that, there is a little gap.
III. The top arrow of the left side red curve:
• This arrow indicates that, the curve can reach to the point where (x = -∞).
• This can be explained as follows:
   ♦ As we move along this curve from right towards the left, x take values closer and closer to -∞.
   ♦ Values closer to -∞ are very large (numerically).
         ✰ The reciprocals of such small values will be very small.
   ♦ The reciprocals are negative y values.
         ✰ That means negative y values will be very small.
         ✰ They get closer and closer to zero.
   ♦ As the y becomes closer and closer to zero, we get the feeling that, this end of the curve is going to touch the x axis.
         ✰ It will touch the x axis when x = -∞
         ✰ However, x cannot become equal to -∞.
   ♦ So this end will never touch the x axis
         ✰ When we draw this graph on a sheet of paper or the computer screen, it appears to touch.
         ✰ But if we zoom in, we will see that, there is a little gap.
IV. The bottom arrow of the left side red curve:
• This arrow indicates that, the curve can go very low bottom.
• This can be explained as follows:
   ♦ As we move along this curve from left towards the right, x takes values closer and closer to zero.
         ✰ For example, -0.0001 is closer to zero than -0.00001
   ♦ Values closer to zero are very small.
         ✰ The reciprocals of such small values will be very large.
   ♦ The reciprocals are negative y values.
         ✰ That means negative y values will be very large (numerically).
         ✰ They are very low down in the graph.
         ✰ This arrow indeed indicates that the graph will go very low bottom (y = -∞).
   ♦ As the x takes values closer and closer to zero, we get the feeling that, this end of the curve is going to touch the y axis.
         ✰ It will touch the y axis when x = 0
         ✰ However, x cannot become equal to zero.
         ✰ If it is zero, then the function can not be defined.
         ✰ This is because, $\mathbf\small{\rm{\frac{1}{x}}}$ becomes $\mathbf\small{\rm{\frac{1}{0}}}$
   ♦ So this end will never touch the y axis
         ✰ When we draw this graph on a sheet of paper or the computer screen, it appears to touch.
         ✰ But if we zoom in, we will see that, there is a little gap.
10. The rational function f(x) = $\mathbf\small{\rm{\frac{1}{x}}}$ has many applications in science and engineering.

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In the next section, we will see a few more common functions.

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Chapter 2.3 - Polynomial Function

In the previous section, we saw the basics about functions. We also saw the features of identity function. In this section, we will see constant function and polynomial function.

B. Constant function
This is a real valued function f: R→R defined by y = f (x) = c
Details can be written in 11 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) = c
   ♦ f(-7) is the 'y value' when 'x value' is -7
   ♦ So we get an ordered pair (x,y) as: (-7,c)
• Let x = -3
   ♦ This x is processed as follows:
   ♦ f(-3) = c
   ♦ f(-3) is the 'y value' when 'x value' is -3
   ♦ So we get another ordered pair (x,y) as: (-3,c)
• Let x = 1.414
   ♦ This x is processed as follows:
   ♦ f(1.414) = c
   ♦ f(1.414) is the 'y value' when 'x value' is 1.414
   ♦ So we get another ordered pair (x,y) as: (1.414,c)
• We see that, whatever be the value of x, the value of y will always be c.
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,c), (-3,c), (1.414,c), (5,c), . . .}
5. In step (2), we noted that, the second elements of the ordered pairs must also be from set R.
• Here we have only one second element, which is 'c'.
   ♦ But this 'c' must be from R.
   ♦ That means, 'c' must be a real number.
• Let us put c = 4. Then the function becomes f(x) = 4
• We can write: f = {. . . , (-7,4), (-3,4), (1.414,4), (5,4), . . .}
6. 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.
7. In the set builder form, we can write:
f = {(x,y) : x ∈ R, y = c, c ∈ R}
• That means:
    ♦ The set f contains all ordered pairs (x,y) such that,
    ♦ x is a real number,
    ♦ y is a constant c,
    ♦ c is a real number
8. 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 c. But in a set, we write repeating elements only once.
• We can write:
    ♦ Range of f = {c}
9. 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 2.3

 10. The red line in fig.2.11(a) below, is the graph of this function.

Fig.2.11

• The reader is advised to prepare a table and draw the graph in his/her own graph books.
• We can write some peculiarities of this red line. They can be written in 6 steps:
(i) The red line will be always horizontal and passes through y = c
(ii) Mark different points on the red line. Note the coordinates of those points.
   ♦ All y coordinates will be the same c.
   ♦ This is shown in fig.b
(iii) We see arrows at both ends of the red line.
• The arrow at the left end of the red line indicates that, the line can extend up to the point where x = -∞ and y = c.
• The arrow at the right end of the red line indicates that, the line can extend up to the point where x = +∞ and y = c.
11. The constant function has many applications in science and engineering.

C. Polynomial function
• We have seen some basics about polynomials in our earlier classes (Details here).
• A polynomial function is a real valued function f: R→R defined by:
y = f (x) = a₀ + a₁x + a₂x² + a₃x³ + . . . + a_nxⁿ
• Details can be written in 11 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.
• We will take a simple polynomial function: f(x) = x²
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) = (-7)² = 49
   ♦ f(-7) is the 'y value' when 'x value' is -7
   ♦ So we get an ordered pair (x,y) as: (-7,49)
• Let x = -3
   ♦ This x is processed as follows:
   ♦ f(-3) = (-3)² = 9
   ♦ f(-3) is the 'y value' when 'x value' is -3
   ♦ So we get an ordered pair (x,y) as: (-3,9)
• Let x = 1.414
   ♦ This x is processed as follows:
   ♦ f(1.414) = (1.414)² = 2
   ♦ f(1.414) is the 'y value' when 'x value' is 1.414
   ♦ So we get an ordered pair (x,y) as: (1.414,2)
• We see that, whatever be the value of x, the value of y will be the square of that x value.
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 = {. . . , (-7,49), (-3,9), (1.414,2), (5,25), . . .}
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 = x², x² ∈ R}
• That means:
    ♦ The set f contains all ordered pairs (x,y) such that,
    ♦ x is a real number,
    ♦ y is the square of x,
    ♦ square of x is a real number
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 will contain all real numbers.
(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 squares, there will not be any negative numbers. But there will be integers,  fractions, decimals, recurring decimals 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:
    ♦ Range of f = {x² : x ∈ R}
• That means:
    ♦ The domain of f will contain squares of all real numbers.
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 2.4

 9. The red curve in fig.2.12(a) below, is the graph of the function.

Fig.2.12

• The reader is advised to prepare a table and draw the graph in his/her own graph books.
• We can write some peculiarities of this red curve. They can be written in 3 steps:
(i) The red curve of f(x) = x² always passes through the origin (0,0)
(ii) Mark any point on the red curve. Note the coordinates of that point.
   ♦ The y coordinate will be the square of the x coordinate.
   ♦ This is shown in fig.b
   ♦ The green vertical dashed line shows 1.414
         ✰ The green horizontal dashed line shows the square 2
   ♦ The magenta vertical dashed line shows -4
         ✰ The magenta horizontal dashed line shows the square 16
(iii) We see arrows at both ends of the red curve.
• The arrow at the left end of the red curve indicates that, the curve can extend up to the point where x = -∞ and y = square of that x value.
• The arrow at the right end of the red curve indicates that, the curve can extend up to the point where x = +∞ and y = square of that x value.
10. The polynomial function f(x) = x² has many applications in science and engineering.
11. We wrote the general form of polynomial functions as:
y = f (x) = a₀ + a₁x + a₂x² + a₃x³ + . . . + a_nxⁿ
◼ It is important to ensure that:
a₀, a₁, a₂, a₃, . . . , a_n are real numbers
    ♦ In other words, a₀, a₁, a₂, a₃, . . . , a_n ∈ R
◼ It is important to ensure that:
No exponent of x is negative.
◼ It is important to ensure that:
All exponents of x are integers.
[Recall the conditions for a polynomial that we saw in our previous classes. (Details here)]
• Some examples of polynomial functions are:
    ♦ f(x) = x³ – x² + 2
    ♦ g(x) = x⁴ + √2 x
• The following function is not a polynomial function:
    ♦ $\mathbf\small{\rm{h(x)=x^{\frac{2}{3} }+2x}}$
    ♦ This is because, the exponent of x in the first term is not an integer

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Let us see a solved example:
Solved example 2.28
Draw the graph of the real valued function f :R → R defined by f(x) = x³, x ∈ R.
Solution:
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.
• The given function is: f(x) = x³
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) = (-7)³ = -343
   ♦ f(-7) is the 'y value' when 'x value' is -7
   ♦ So we get an ordered pair (x,y) as: (-7, -343)
• Let x = -3
   ♦ This x is processed as follows:
   ♦ f(-3) = (-3)³ = -27
   ♦ f(-3) is the 'y value' when 'x value' is -3
   ♦ So we get an ordered pair (x,y) as: (-3, -27)
• Let x = 1.414
   ♦ This x is processed as follows:
   ♦ f(1.414) = (1.414)³ = 2.8271
   ♦ f(1.414) is the 'y value' when 'x value' is 1.414
   ♦ So we get an ordered pair (x,y) as: (1.414, 2.8271)
(Note that 1.414 is √2. So 2.8271 will be 2√2)
• We see that, whatever be the value of x, the value of y will be the cube of that x value.
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, -343), (-3, -27), (1.414, 2.8271), (5, 125), . . .}
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 = x³, x³ ∈ R}
• That means:
    ♦ The set f contains all ordered pairs (x,y) such that,
    ♦ x is a real number,
    ♦ y is the cube of x,
    ♦ cube of x is a real number.
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 cubes, there will be negative numbers also. There will be integers,  fractions, decimals, recurring decimals 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:
    ♦ Range of f = {x³ : x ∈ R}
• That means:
    ♦ The domain of f is the set of all x³ such that,
    ♦ x 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 2.5

9. The red curve in fig.2.13(a) below, is the graph of the function.

Fig.2.13

• The graph in fig.b is drawn to a different scale. This can be explained in 5 steps:
(i) The y value is cube of x value. So we will need more values on the y axis.
(ii) To achieve this within limited space, we change the scale of the y axis.
(iii) In the fig.2.13(a), we have:
    ♦ Scale of x axis as: 1 cm = 1 unit
    ♦ Scale of y axis as: 1 cm = 1 unit
(iv) In the fig.2.13(b), we have:
    ♦ Scale of x axis as: 1 cm = 1 unit
    ♦ Scale of y axis as: 1 cm = 10 unit
(v) So the graph in fig.b is a bit distorted. But the coordinates do not change.
• The reader is advised to prepare a table and draw the graph in his/her own graph books.
• We can write some peculiarities of this red curve. They can be written in 3 steps:
(i) The red curve of f(x) = x³ always passes through the origin (0,0)
(ii) Mark any point on the red curve. Note the coordinates of that point.
   ♦ The y coordinate will be the cube of the x coordinate.
   ♦ This is shown in fig.b
   ♦ The green vertical dashed line shows 3
         ✰ The green horizontal dashed line shows the cube 27
   ♦ The magenta vertical dashed line shows -4
         ✰ The magenta horizontal dashed line shows the cube -64
(iii) We see arrows at top and bottom of the red curve.
• The arrow at the top end of the red curve indicates that, the curve can extend up to the point where x = +∞ and y = cube of that x value.
• The arrow at the bottom end of the red curve indicates that, the curve can extend up to the point where x = -∞ and y = cube of that x value.
10. The polynomial function f(x) = x³ has many applications in science and engineering.

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In the next section, we will see a Rational function.

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Chapter 2.1 - Relations

In the previous section, we saw Cartesian product. In this section, we will see relations.

We will see the basics about relations using three examples.
Example 1:
This can be written in 7 steps:
1. Consider two sets A and B
   ♦ A is a set of four students: Student 1, Student 2, Student 3, Student 4.
         ✰ A = {1, 2, 3, 4}
   ♦ B is a set of five available courses.
         ✰ B = {Maths, Physics, Chemistry, Biology, Geography}
2. The students have the liberty to learn one or more of the available courses.
• So there are twenty possible combinations. They are:
(1, Maths),           (2, Maths),         (3, Maths),         (4, Maths),
(1, Physics),         (2, Physics),       (3, Physics),        (4, Physics),
(1, Chemistry),    (2, Chemistry),   (3, Chemistry),   (4, Chemistry),
(1, Biology),         (2, Biology),       (3, Biology),        (4, Biology),
(1, Geography),   (2, Geography),  (3, Geography),   (4, Geography).
• They are the 20 ordered pairs.
3. Using the method that we saw in the previous section, the above twenty ordered pairs can be denoted by red dots. This is shown in fig.2.4(a) below:

Fig.2.4

4. The red dots only give us the various possibilities. For example, (3, Biology) indicates that, the student 3 can choose to do Biology.
• So the red dots are useful during the admission processes. They help students and authorities to select and allocate various courses.
• Once the admission processes are complete, the red dots have not much value. At that stage, we will be wanting to know which student chose which course.
5. So we need to pick the appropriate red dots from among the total 20 red dots. This is shown in fig.2.4(b) above. The selected dots are marked with green circles.
• Consider any one green circle. Say the one at the intersection of student 2 line and Chemistry line.
• This green circle indicates that, student 2 chose to do the chemistry course.
6. The ordered pairs with green circle are:
(1, Maths), (2, Maths), (2, Chemistry), (3, Chemistry), (3, Biology), and (4, Geography)
• These pairs give us the following information:
   ♦ Student 1 chose to do Maths.
   ♦ Student 2 chose to do Maths.
   ♦ Student 2 chose to do Chemistry.
   ♦ Student 3 chose to do Chemistry.
   ♦ Student 3 chose to do Biology.
   ♦ Student 4 chose to do Geography.
7. This information can be shown in an arrow diagram also. It is shown in fig.2.4(c) above.
• Note that:
   ♦ The number of green circles in fig.b is 6
   ♦ The number of green arrows in fig.c is also 6
• This is because, both the figs. b and c convey the same information.

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• Let us write this example in terms of sets and ordered pairs. The 8 steps given below will help us:
1. A is the set of students and B is the set of courses.
2. All the ‘possible combinations’ is given by the red dots in fig.2.4(a)
• As we saw in the previous section, all those red dots will be included in the set A × B
3. The green circles in fig.2.4(b) shows the relation between students and their chosen courses.
• We can make a set R which contains all the green circles in fig.2.4(b)
4. All the green circles are obtained from the red dots. So the set R will be a subset of A × B
• We can write: R ⊂ (A × B)
5. In the previous section, we saw that all elements of A × B are ordered pairs.
• Since R is a subset of A × B, all the elements of R will also be ordered pairs.
• In our present case,
R = {(1, Maths), (2, Maths), (2, Chemistry), (3, Chemistry), (3, Biology), and (4, Geography)}
6. The green circles in fig.b can be better visualized using the arrow diagram in fig.c
   ♦ The green circles in fig.b give us the ordered pairs in R.
   ♦ The green arrows in fig.c give us the same ordered pairs in R.
7. The usefulness of the arrow diagram will be clear from the following 3 steps:
(i) Take any ordered pair in R. Look at the corresponding green arrow in the arrow diagram.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
• We can work in the reverse also. It can be written in 4 steps:
(i) Take any green arrow in the arrow diagram. Corresponding to that arrow, there will be an ordered pair in R.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
(iv) All the green arrows must be included as ordered pairs in the set R.
8. In step (5), we wrote R in the roster form. We must be able to write it in the set builder form also. The following 3 steps will enable us to do so:
(i) We know that, set R contains ordered pairs. Let the general form of those ordered pairs be (x,y)
(ii) Then x will be the student and y will be the course chosen by that student.
(iii) So the set builder form will be:
R = {(x,y) : y is the course chosen by student x, x ∈ A, y ∈ B}

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Example 2:
This can be written in 6 steps:
1. Consider two sets A and B
   ♦ A = {5, 6, 7}
   ♦ B = {3, 4, 5}
2. n(A) = 3 and n(B) = 3
• So there are nine possible combinations. They are:
(5, 3),            (6, 3),           (7, 3),
(5, 4),            (6, 4),           (7, 4),
(5, 5),            (6, 5),           (7, 5).
• They are the 9 ordered pairs.
3. Using the method that we saw in the previous section, the above nine ordered pairs can be denoted by red dots. This is shown in fig.2.5(a) below:

Fig.2.5

4. The red dots give us the various possible combinations.
Suppose that, we want only those combinations in which:
   ♦ The element taken from B
   ♦ is 2 less than
   ♦ The element taken from A.
• Then we need to pick the appropriate red dots from among the total 9 red dots.
• The appropriate red dots can be determined using 2 steps:
(i) Let x be the element (which satisfies the relation) from A. Let y be the corresponding element (which satisfies the relation) from B
• Then the algebraic form of the relation is: x - 2 = y
(ii) Let us take each possible value for x from set A:
• When x = 5,
   ♦ y = (x-2) = (5-2) = 3
   ♦ '3' is available in B
   ♦ So the ordered pair (5, 3) satisfies the given relation.
         ✰ Note that (5, 3) is one among the red dots in fig.a
• When x = 6,
   ♦ y = (x-2) = (6-2) = 4
   ♦ '4' is available in B
   ♦ So the ordered pair (6, 4) satisfies the given relation.
         ✰ Note that (6, 4) is one among the red dots in fig.a
• When x = 7,
   ♦ y = (x-2) = (7-2) = 5
   ♦ '5' is available in B
   ♦ So the ordered pair (7, 5) satisfies the given relation.
         ✰ Note that (7, 5) is one among the red dots in fig.a
5. We need to select the above three ordered pairs from among the 9 ordered pairs. This is shown in fig.2.5(b) above. The selected dots are marked with green circles.
• The ordered pairs with green circle are:
(5, 3), (6, 4) and (7, 5)
6. This information can be shown in an arrow diagram also. It is shown in fig.2.5(c) above.
• Note that:
   ♦ The number of green circles in fig.b is 3
   ♦ The number of green arrows in fig.c is also 3
• This is because, both the figs. b and c convey the same information.

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• Let us write this example in terms of sets and ordered pairs. The 8 steps given below will help us:
1. A = {5, 6, 7} and B = {3, 4, 5}
2. All the ‘possible combinations’ is given by the red dots in fig.2.5(a)
• As we saw in the previous section, all those red dots will be included in the set A × B
3. The green circles in fig.2.5(b) shows those ordered pairs which satisfy a particular relation.
• The relation is this:
   ♦ The element taken from B
   ♦ is 2 less than
   ♦ The element taken from A.
• We can make a set R which contains all the green circles in fig.2.4(b)
4. All the green circles are obtained from the red dots. So the set R will be a subset of A × B
• We can write: R ⊂ (A × B)
5. In the previous section, we saw that all elements of A × B are ordered pairs.
• Since R is a subset of A × B, all the elements of R will also be ordered pairs.
• In our present case,
R = {(5, 3), (6, 4), (7, 5)}
6. The green circles in fig.b can be better visualized using the arrow diagram in fig.c
   ♦ The green circles in fig.b give us the ordered pairs in R.
   ♦ The green arrows in fig.c give us the same ordered pairs in R.
7. The usefulness of the arrow diagram will be clear from the following 3 steps:
(i) Take any ordered pair in R. Look at the corresponding green arrow in the arrow diagram.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
• We can work in the reverse also. It can be written in 4 steps:
(i) Take any green arrow in the arrow diagram. Corresponding to that arrow, there will be an ordered pair in R.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
(iv) All the green arrows must be included as ordered pairs in the set R.
8. In step (5), we wrote R in the roster form. We must be able to write it in the set builder form also. The following 3 steps will enable us to do so:
(i) We know that, set R contains ordered pairs. Let the general form of those ordered pairs be (x,y)
(ii) Then x will be the element from A and y will be the element from B.
(iii) So the set builder form will be:
R = {(x,y) : y = x - 2, x ∈ A, y ∈ B}

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Example 3:
This can be written in 6 steps:
1. Consider the set A
   ♦ A = {1, 2, 3, 4, 5, 6}
2. We want the possible combinations of A with itself.
n(A) = 6
• So there are 36 possible combinations. They are:
(1, 1),            (2, 1),           (3, 1),         (4, 1),            (5, 1),           (6, 1),
(1, 2),            (2, 2),           (3, 2),         (4, 2),            (5, 2),           (6, 2),
(1, 3),            (2, 3),           (3, 3),         (4, 3),            (5, 3),           (6, 3),
(1, 4),            (2, 4),           (3, 4),         (4, 4),            (5, 4),           (6, 4),
(1, 5),            (2, 5),           (3, 5),         (4, 5),            (5, 5),           (6, 5),
(1, 6),            (2, 6),           (3, 6),         (4, 6),            (5, 6),           (6, 6).
• They are the 36 ordered pairs.
3. Using the method that we saw in the previous section, the above 36 ordered pairs can be denoted by red dots. This is shown in fig.2.6(a) below:

Fig.2.6

4. The red dots give us the various possible combinations.
Suppose that, we want only those combinations in which:
   ♦ The element taken from B
   ♦ is 1 greater than
   ♦ The element taken from A.
• Then we need to pick the appropriate red dots from among the total 36 red dots.
• The appropriate red dots can be determined using 2 steps:
(i) Let x be the element (which satisfies the relation) from A. Let y be the corresponding element (which satisfies the relation) from B
• Then the algebraic form of the relation is: x + 1 = y
(ii) Let us take each possible value for x from set A:
• When x = 1,
   ♦ y = (x+1) = (1+1) = 2
   ♦ '2' is available in A
   ♦ So the ordered pair (1, 2) satisfies the given relation.
         ✰ Note that (1, 2) is one among the red dots in fig.a
• When x = 2,
   ♦ y = (x+1) = (2+1) = 3
   ♦ '3' is available in A
   ♦ So the ordered pair (2, 3) satisfies the given relation.
         ✰ Note that (2, 3) is one among the red dots in fig.a
• When x = 3,
   ♦ y = (x+1) = (3+1) = 4
   ♦ '4' is available in A
   ♦ So the ordered pair (3, 4) satisfies the given relation.
         ✰ Note that (3, 4) is one among the red dots in fig.a
• When x = 4,
   ♦ y = (x+1) = (4+1) = 5
   ♦ '5' is available in A
   ♦ So the ordered pair (4, 5) satisfies the given relation.
         ✰ Note that (4, 5) is one among the red dots in fig.a
• When x = 5,
   ♦ y = (x+1) = (5+1) = 6
   ♦ '6' is available in A
   ♦ So the ordered pair (5, 6) satisfies the given relation.
         ✰ Note that (5, 6) is one among the red dots in fig.a
• When x = 6,
   ♦ y = (x+1) = (6+1) = 7
   ♦ '7' is not available in A
   ♦ So the ordered pair (6, 7) does not satisfy the given relation.
         ✰ Note that (6, 7) is not among the red dots in fig.a
5. We need to select the above five ordered pairs from among the 36 ordered pairs. This is shown in fig.2.6(b) above. The selected dots are marked with green circles.
• The ordered pairs with green circle are:
(1, 2), (2, 3), (3, 4), (4, 5) and (5, 6)
6. This information can be shown in an arrow diagram also. It is shown in fig.2.6(c) above.
• Note that:
   ♦ The number of green circles in fig.c is 5
   ♦ The number of green arrows in fig.c is also 5
• This is because, both the figs. b and c convey the same information.

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• Let us write this example in terms of sets and ordered pairs. The 8 steps given below will help us:
1. A = {1, 2, 3, 4, 5, 6}
2. All the ‘possible combinations’ from A to A is given by the red dots in fig.2.6(a)
• As we saw in the previous section, all those red dots will be included in the set A × A
3. The green circles in fig.2.6(b) shows those ordered pairs which satisfy a particular relation.
• The relation is this:
   ♦ The element taken from A
   ♦ is 1 greater than
   ♦ The element taken from A.
• We can make a set R which contains all the green circles in fig.2.6(b)
4. All the green circles are obtained from the red dots. So the set R will be a subset of A × A
• We can write: R ⊂ (A × A)
5. In the previous section, we saw that all elements of A × A are ordered pairs.
• Since R is a subset of A × A, all the elements of R will also be ordered pairs.
• In our present case,
R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
6. The green circles in fig.b can be better visualized using the arrow diagram in fig.c
   ♦ The green circles in fig.b give us the ordered pairs in R.
   ♦ The green arrows in fig.c give us the same ordered pairs in R.
7. The usefulness of the arrow diagram will be clear from the following 3 steps:
(i) Take any ordered pair in R. Look at the corresponding green arrow in the arrow diagram.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
• We can work in the reverse also. It can be written in 4 steps:
(i) Take any green arrow in the arrow diagram. Corresponding to that arrow, there will be an ordered pair in R.
(ii) The tail end of the arrow will be the first element of that ordered pair.
(iii) The head end of the arrow will be the second element of that ordered pair.
(iv) All the green arrows must be included as ordered pairs in the set R.
8. In step (5), we wrote R in the roster form. We must be able to write it in the set builder form also. The following steps will enable us to do so:
(i) We know that, set R contains ordered pairs. Let the general form of those ordered pairs be (x,y)
(ii) Then x will be the element from A and y will be the element from A.
(iii) So the set builder form will be:
R = {(x,y) : y = x + 1, x ∈ A, y ∈ A}

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• The above three examples help us to understand the basics about relations.
• We will now see some important terms involved with relations. The important terms are:
(a) image
(b) domain
(c) range
(d) Codomain

(a) image:
This can be explained in 3 steps:
1. We know that, the set R will contain one or more ordered pairs.
2. Each of those ordered pairs will have two elements.
3. The second element is called the image of the first element.
◼ Consider the ordered pairs in R of our first example. We can write:
    ♦ Maths is the image of student 1
    ♦ Maths is the image of student 2
    ♦ Chemistry is the image of student 2
    ♦ Chemistry is the image of student 3
    ♦ Biology is the image of student 3
    ♦ Geography is the image of student 4
◼ Consider the ordered pairs in R of our second example. We can write:
    ♦ 3 is the image of 5
    ♦ 4 is the image of 6
    ♦ 5 is the image of 7
◼ Consider the ordered pairs in R of our third example. We can write:
    ♦ 2 is the image of 1
    ♦ 3 is the image of 2
    ♦ 4 is the image of 3
    ♦ 5 is the image of 4
    ♦ 6 is the image of 5

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(b) domain:
This can be explained in 5 steps:
1. We know that, the set R will contain one or more ordered pairs.
2. Each of those ordered pairs will have two elements.
3. Pick out all the first elements.
4. Make a set using those first elements.
5. This set is called the domain of the relation R.
◼ Consider the ordered pairs in R of our first example. We can write:
    ♦ The first elements are: 1, 2, 2, 3, 3, 4
    ♦ When we write them as a set, repeating elements should appear only once.
    ♦ So we get: domain = {1, 2, 3, 4}
◼ Consider the ordered pairs in R of our second example. We can write:
    ♦ The first elements are: 5, 6, 7
    ♦ Here there are no repeating elements.
    ♦ So we get: domain = {5, 6, 7}
◼ Consider the ordered pairs in R of our third example. We can write:
    ♦ The first elements are: 1, 2, 3, 4, 5
    ♦ Here there are no repeating elements.
    ♦ So we get: domain = {1, 2, 3, 4, 5}

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(c) range:
This can be explained in 5 steps:
1. We know that, the set R will contain one or more ordered pairs.
2. Each of those ordered pairs will have two elements.
3. Pick out all the second elements.
4. Make a set using those second elements.
5. This set is called the range of the relation R.
◼ Consider the ordered pairs in R of our first example. We can write:
    ♦ The second elements are: maths, maths, chemistry, chemistry, biology, geography.
    ♦ When we write them as a set, repeating elements should appear only once.
    ♦ So we get: range = {maths, chemistry, biology, geography}
◼ Consider the ordered pairs in R of our second example. We can write:
    ♦ The second elements are: 3, 4, 5
    ♦ Here there are no repeating elements.
    ♦ So we get: range = {3, 4, 5}
◼ Consider the ordered pairs in R of our third example. We can write:
    ♦ The second elements are: 2, 3, 4, 5, 6
    ♦ Here there are no repeating elements.
    ♦ So we get: domain = {2, 3, 4, 5, 6}

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(d) codomain:
We know that, the relation R is defined from set A to set B.
• The set B is also known as codomain of the relation R.
◼ Consider the relation R of our first example.
   ♦ The set B for this relation is: {Maths, Physics, Chemistry, Biology, Geography}
   ♦ So codomain of this R is : {Maths, Physics, Chemistry, Biology, Geography}
◼ Consider the relation R of our second example.
   ♦ The set B for this relation is: {3, 4, 5}
   ♦ So codomain of this R is : {3, 4, 5}
◼ Consider the relation R of our third example.
   ♦ The set B for this relation is: {1, 2, 3, 4, 5, 6}
   ♦ So codomain of this R is : {1, 2, 3, 4, 5, 6}

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From the above four definitions, following 4 points can be noted:
(i) domain will contain only those elements belonging to Set A.
    ♦ all elements of A may not be present in domain.
(ii) codomain will contain only those elements belonging to Set B.
    ♦ all elements of B will be present in codomain.
(iii) range will contain only those elements belonging to Set B.
    ♦ all elements of B may not be present in range.
(iv) From (ii) and (iii), it is clear that:
range ⊂ codomain.

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• Once we understand the basics, there will not be any need to write all the lengthy steps. We will be able to obtain the results using minimum steps.
• The solved examples given below will demonstrate the process

Solved example 2.15
Let A = {1, 2, 3,...,14}. Define a relation R from A to A by
R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and
range.
Solution:
1. The relation R is a set which contains ordered pairs of the form (x, y)
   ♦ 'x' should be from set A
   ♦ Since the relation is from A to A, 'y' should also be from set A
• The x and y in each ordered pair in R should satisfy the condition: 3x - y = 0
2. The given condition can be rearranged as: 3x = y
Let us take each possible value for x from set A:
• When x = 1,
   ♦ y = 3x = (3 × 1) = 3
   ♦ '3' is available in A
   ♦ So the first ordered pair in R is (1, 3)
• When x = 2,
   ♦ y = 3x = (3 × 2) = 6
   ♦ '6' is available in A
   ♦ So the second ordered pair in R is (2, 6)
• When x = 3,
   ♦ y = 3x = (3 × 3) = 9
   ♦ '9' is available in A
   ♦ So the third ordered pair in R is (3, 9)
• When x = 4,
   ♦ y = 3x = (3 × 4) = 12
   ♦ '12' is available in A
   ♦ So the fourth ordered pair in R is (4, 12)
• When x = 5,
   ♦ y = 3x = (3 × 5) = 15
   ♦ '15' is not available in A
   ♦ So the ordered pair (5, 15) does not satisfy the given relation.
         ✰ Note that (5, 15) will not be available in A × A also.
3. The set R will contain the four ordered pairs that we determined above. We can write:
R = {(1,3), (2,6), (3,9), (4,12)}
4. Domain is the set containing all the first elements in the ordered pairs of R. So we get:
Domain of R = {1, 2, 3, 4}
5. Codomain is the set from which we take the second elements of the ordered pairs in R. In effect, co domain is the set B.
• In our present case, since the relation is from A to A, we have A in place of B.
• So we get: codomain of R = {1, 2, 3,...,14}
6. Range is the set containing all the second elements in the ordered pairs of R.
• So we get: Range of R = {3, 6, 9, 12}

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• We know that, a relation is defined from one set A to another set B
• But some times, the relation is defined from one set A to the same set A
• In such situations, we can use any one of the two statements below:
   ♦ Relation R from A to A
   ♦ Relation R on A

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More solved examples are given at the link below:

Solved examples 2.16 to 2.25

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In the next section, we will see functions.

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Chapter 2.2 - Functions

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.

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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.

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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.

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◼ From the above discussion, it is clear that:
   ♦ All functions are relations.
   ♦ But all relations are not functions.

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• 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

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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 45^o 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.

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In the next section, we will see a few more common functions.

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