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:
$\mathbf\small{\rm{\left(\frac{f}{g} \right)(x) =\frac{f(x)}{g(x)}=\frac{1}{\sqrt x},\;x \ne 0, \;x\in X}}$
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)
In
the next
section, we will see a few more solved examples.
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