Types of Functions JEE Notes | EduRev

FUNCTIONS
We can define a function as a special relation which maps each element of set A with one and only one element of set B. Both the sets A and B must be non -empty. A function defines a particular output for a particular input. Hence, f: A → B is a function such that for a ∈ A there is a unique element b ∈ B such that (a, b) ∈ f

• The set ‘A’ is called the domain of ‘f’.
• The set ‘B’ is called the co-domain of ‘f’.
• Set of images (outputs) of different elements of the set A is called the range of ‘f’. It is obvious that range could be a subset of the co-domain as we may have some elements in the co-domain which are not the images of any element of A (of course, these elements of the co-domain will not be included in the range). Range is also called domain of variation.
• If R is the set of real numbers and A and B are subsets of R, then the function f (A) is called a real−valued function or a real function.

Domain of a function ‘f’ is normally represented as Domain (f). Range is represented as Range (f). Note that some times domain of the function is not explicitly defined. In these cases domain would mean the set of values of ‘x’ for which f (x) assumes real values that is if y = f (x) then called Domain (f) = {x : f (x) is a real number}.

In other words, domain is defined as a set of all those values of x for which the given function is defined.

Q. Find the domain and range of the function 

Ans. The function

is defined for x ≥ 5
⇒ The domain is [5, ∞] .
Also, for any

⇒ The range of the function is [0, ∞] .

Q. Find the domain and the range of the function y = f (x), where f (x) is given by
(i) x² − 2x − 3


(iv) tan x
(v) tan⁻¹x
(vi) log₁₀ (x).
(vii) sin x
Ans. 
(i) Here y = (x − 3) (x + 1).
The function is defined for all real values of x
⇒ Its domain is R.
Also x² − 2x − 3 − y = 0 for real x
⇒ 4 + 4 (3 + y) ≥ 0 ⇒ − 4 ≤ y < ∞.
Hence the range of the given function is [- 4, ∞].
(ii) Here

⇒ (x − 3) (x + 1) ≥ 0 so that x ≥ 3 or x ≤ − 1.
Hence the domain is R − (− 1, 3) or (− ∞, - 1] ∪ [3, ∞].

Since f(x) is non−negative in the domain, the range of f (x) is the interval [0, ∞].
(iii) The given function is defined for all x ≥ 0
⇒ The domain is [0, ∞].
Moreover - 1 ≤ f (x) < ∞
⇒ The range is [- 1, ∞].
(iv) The function f (x) = tan x = sin x/cos x is not defined when
cos x = 0, or x = (2n + 1) π/2, n = 0, ± 1, ± 2, …….
Hence domain of tan x is R -{(2n + 1)π/2, n = 0, + 1, + 2,....}, and its range is R. 

(v) The function f (x) = tan⁻¹ x, is defined for all real values of x and -π/2 < tan^-1 < π/2.
Hence its domain is R and the range is (-π/2, π/2).
(vi) The function f (x) = log₁₀ x is defined for all x > 0. Hence its domain is (0, ∞) and range is R. 

(vii) The function f (x) = sin x, (x in radians) is defined for all real values of x
⇒ domain of f (x) is R. Also − 1 ≤ sin x ≤ 1, for all x,
so that the range of f (x) is [− 1, 1].

TYPES OF FUNCTIONS
We have already learned about some types of functions like Identity, Polynomial, Rational, Modulus, Signum, Greatest Integer functions. In this section, we will learn about other types of function.
One to One Function
A function f: A → B is One to One if for each element of A there is a distinct element of B. It is also known as Injective. Consider if a₁ ∈ A and a₂ ∈ B, f is defined as f: A → B such that f (a₁) = f (a₂) f is defined as f: A → B such that f(x) = y then

f: A → B is said to be one-to one function if 
f (a₁) = f (a₂) 
⇒ (a₁) = (a₂), (a₁)(a₂) ∈ A 
Let A = {(a₁)(a₂)(a₃)(a₄)} and B = {(b₁)(b₂)(b₃)(b₄)}

In other words if a₁, a₂ ∈ A such that a₁ ≠ a₂  then f(a₁) ≠ f(a₂)

Solved examples of one to one function
Q.1. Determine if the function given below is one to one. 
(i) To each state of India assign its Capital
Ans. This is not one to one function because each state of India has not different capital.

(ii) Function = {(2, 4), (3, 6), (-1, -7)} 
Ans. The above function is one to one because each value of range has different value of domain.

(iii) f(x) = |x|
Ans. Here to check whether the given function is one to one or not, we will consider some values of x (domain) and from the given function find the value of range(y).

From the above table we can see that an element in the range repeats, then this is not a 1 to 1 function.

Q.2. Without using graph prove that the function
F : R → R defiend by f (x) = 4 + 3x is one-to-one
Ans. For a function to be one-one function if
F (x₁) = F (x₁) ⇒ x₁ = x₂  ∀  x_1, x₂  ∈ domain
∴ Now f (x₁) = f (x₂) gives
4 + 3x₁ = 4 + 3x₂ or x_{1 =} x₂ 
∴ F is a one-one function.

Many to One Function
It is a function which maps two or more elements of A to the same element of set B. Two or more elements of A have the same image in B.

Q.3. Without using graph check the following function f: R → R defined by f (x) = x² is one-to-one or not ?
Ans. For the function to be one to one , different elements must have different images
But here if x = 1 then f(1) = (1)² =1
And if x = -1 then f(-1) =(-1)² = 1
Clearly, x = 1 and x = -1 both have same image
So given function is not one to one i.e many to one function
In other words, we can say that a function which is not one to one that can be known as many to one.

Onto Function
If there exists a function for which every element of set B there is (are) pre-image(s) in set A, it is Onto Function. Onto is also referred as Surjective Function.
i.e ∀ y ∈ B, ∃ at least one element x ∈ A such that f(x) = y

e.g., let A = {2, 3, 4}, B= {5, 6} . Then f : A → B defined by f = {(2, 5), (3, 6),
(4, 5)} is an onto function as each element of B is the image of at least one element of A.
Note: A function which is not onto is known as into function.

Q.4. If f: R → R is defined as f (x)= 3x + 7, x ∈ R, then show that f is an onto function.
Ans. We have f (x) = 3x + 7, x ∈ R . Let b ∈ R so that f (x) = b ⇒ 3x + 7 = b or
x = (b - 7)/3. Since b ∈ R, (b - 7)/3 ∈ R.

⇒ b is image of (b - 7)/3 where b is arbitrary.
Hence f (x) is an onto function.

Q.5. Let f: N → I be a function defined as f(x) = x – 1000. Determine whether f(x) is into or onto.
Ans.
Let y ∈ I such that f(x) = y ,where x∈ N
⇒ x – 1000 = y
⇒ x = y + 1000 = g(y)
Here g(y) is defined for each y ∈ I, but g(y) ∉ N for y ≤ – 1000.
Hence f(x) is into.

One – One and Onto Function
A function, f is One – One and Onto or Bijective if the function f is both One to One and Onto function. In other words, the function f associates each element of A with a distinct element of B and every element of B has a pre-image in A.

Q. Prove that
F : R → R defined by f (x) = 4x³ - 5 is a bijection
Ans. Now f (x₁) = f (₂) ∀ x_1, x₂ Domain
∴ 4x₁ ³ - 5 = 4x₂³ - 5
⇒ x₁ ³ = x₂³
⇒ x₁ ³ - x₂³ = 0 ⇒ (x₂ - x₁) (x₁² + x₁x₂ + x₂²) = 0
⇒ x₁ = x₂ or
x₁² + x₁x₂ + x₂² = 0 (rejected).
It has no real value of x₁ and x₂.
∴ F is a one-one function.
Again let y = (x) where y ∈ codomain, x ∈ domain.
We have y = 4x³ - 5
or  

∴ For each y ∈ codomain ∃ x ∈ domain such that f (x) = y.
Thus F is onto function.
∴ F is a bijection.

Other Types of Functions
A function is uniquely represented by its graph which is nothing but a set of all pairs of x and f(x) as coordinates. Let us get ready to know more about the types of functions and their graphs.

Q. Prove that F: R → R defined by F (x) = x² + 3 is neither one-one nor onto function.
Ans. We have F (x₁) = F (x₂) ∀ x_1, x₂ domain giving
x₁ ³ + 3 = x₂² + 3 ⇒ x₁² = x₂²
or x₁² - x₂² = 0 ⇒ x₁ = x₂  or x₁ = x₂
or F is not one-one function.
Again let y = F (x) where y ∈ codomain
x ∈ domain.
⇒ y = x² + 3

⇒ ∀ y < 3 no real value of x in the domain.
∴ F is not an onto function.

Identity Function
Let R be the set of real numbers. If the function f: R → R is defined as f(x) = y = x, for x ∈ R, then the function is known as Identity function. The domain and the range being R. The graph is always a straight line and passes through the origin.

Solved Example of Identity Functions
Example:
Let A = {1, 2, 3, 4, 5, 6}
Then Identity function on set A will be defined as
I_A : A → A , I_A = x , x ∈ A
for x = 1 , I_A(1) = x = 1
for x = 2 , I_A(1) = x = 2
for x = 3 , I_A(1) = x = 3
for x = 4 , I_A(1) = x = 4
for x = 5 , I_A(1) = x = 5
Domain, Range and co-domain will be Set A

Constant Function
If the function f: R→R is defined as f(x) = y = c, for x ∈ R and c is a constant in R, then such function is known as Constant function. The domain of the function f is R and its range is a constant, c. Plotting a graph, we find a straight line parallel to the x-axis.

Solved examples of constant functions
Q.1. which is below function is a constant function?
(a) y = x
(b) y = 11
(c) y = π
(d) x + y = 1
(e) y = x²
Ans.
For the function to be a constant function ,it should yield same value for each x
(a) This is constant function as output is different for each input
(b) This is constant function as output is same for each input. It is of the format y = c
(c) This is constant function as output is same for each input. It is of the format y = c
(d) This is constant function as output is different for each input
(e) This is constant function as output is different for each input

Q.2. which of the graph represent constant function?
Ans.
The graph should be parallel to x-axis for the function to be constant function
So C and D are constant function.

Polynomial Function
A polynomial function is defined by y = a₀ + a₁x + a₂x² + … + a_nxⁿ, where n is a non-negative integer and a₀, a₁, a₂,…, n ∈ R. The highest power in the expression is the degree of the polynomial function. Polynomial functions are further classified based on their degrees:
Constant Function: If the degree is zero, the polynomial function is a constant function (explained above).

Linear Function: The polynomial function with degree one. Such as y = x + 1 or y = x or y = 2x – 5 etc. Taking into consideration, y = x – 6. The domain and the range are R. The graph is always a straight line.

Quadratic Function: If the degree of the polynomial function is two, then it is a quadratic function. It is expressed as f(x) = ax² + bx + c, where a ≠ 0 and a, b, c are constant & x is a variable. The domain and the range are R. The graphical representation of a quadratic function say, f(x) = x² – 4 is

Cubic Function: A cubic polynomial function is a polynomial of degree three and can be denoted by f(x) = ax³ + bx² + cx +d, where a ≠ 0 and a, b, c, and d are constant & x is a variable. Graph for f(x) = y = x³ – 5. The domain and the range are R.

Rational Function: A rational function is any function which can be represented by a rational fraction say, f(x)/g(x) in which numerator, f(x) and denominator, g(x) are polynomial functions of x, where g(x) ≠ 0. Let a function f: R → R is defined say, f(x) = 1/(x + 2.5). The domain and the range are R. The Graphical representation shows asymptotes, the curves which seem to touch the axes-lines.

Modulus Function
The absolute value of any number, c is represented in the form of |c|. If any function f: R→ R is defined by f(x) = |x|, it is known as Modulus Function. For each non-negative value of x, f(x) = x and for each negative value of x, f(x) = -x, i.e.,
f(x) = {x, if x ≥ 0; – x, if x < 0.
Its graph is given as, where the domain and the range are R.

Signum Function
A function f: R→ R defined by
f(x) = {1, if x > 0; 0, if x = 0; -1, if x < 0}
Signum or the sign function extracts the sign of the real number and is also known as step function.

Greatest Integer Function
If a function f: R→ R is defined by f(x) = [x], x ∈ X. It round-off to the real number to the integer less than the number. Suppose, the given interval is in the form of (k, k + 1), the value of greatest integer function is k which is an integer. For example: [-21] = 21, [5.12] = 5. The graphical representation is

Solved Example
Q. Which of the following is a function?

Ans. Figure (iii) is an example of a function. Since the given function maps every element of A with that of B. In figure (ii), the given function maps one element of A with two elements of B (one to many). Figure (i) is a violation of the definition of the function. The given function does not map every element of A.

ALGEBRA OF REAL FUNCTIONS
Real-valued Mathematical Functions
In mathematics, a real-valued function is a function whose values are real numbers. It is a function that maps a real number to each member of its domain. Also, we can say that a real-valued function is a function whose outputs are real numbers i.e., f: R → R (R stands for Real).

Algebra of Real Functions
In this section, we will get to know about addition, subtraction, multiplication, and division of real mathematical functions with another.

Addition of Two Real Functions
Let f and g be two real valued functions such that f: X → R and g: X → R where X ⊂ R. The addition of these two functions (f + g) : X → R  is defined by:

(f + g) (x) = f(x) + g(x), for all x ∈ X

Subtraction of One Real Function from the Other
Let f : X → R and g : X → R be two real functions where X ⊂ R. The subtraction of these two functions (f – g): X → R  is defined by:

(f – g) (x) = f(x) – g(x), for all x ∈ X

Multiplication by a Scalar
Let f: X → R be a real-valued function and γ be any scalar (real number). Then the product of a real function by a scalar γf: X → R is given by:
(γf) (x) = γ f(x), for all x ∈ X.

Equal functions 
Let f and g be two functions defined from A to B. Then f , g : A → B are equal if f (x) = g(x), x ∈ A .
If the function f and g are equal, then the subsets, graph of f and graph of g, of A x B are equal.

Q.1. 

find the value of k
Ans. We need to consider only one equation
2k = 6
k = 3

Q.2. Find the values of x and y.

Ans:
2x – 6 = 5
2x = 11
x = 5.5
4 – y = 3
y = 1

Multiplication of Two Real Functions
The product of two real functions say, f and g such that f: X → R and g: X → R, is given by

(fg) (x) = f(x) g(x), for all x ∈ X

Division of Two Real Functions
Let f and g be two real-valued functions such that f: X → R and g: X → R where X ⊂ R. The quotient of these two functions (f  ⁄ g): X → R  is defined by:

(f / g) (x) = f(x) / g(x), for all x ∈ X, but g(x) ≠ 0 for all x ∈ X

Note: It is also called point wise multiplication.

Solved Example
Q. Let f(x) = x³ and g(x) = 3x + 1 and a scalar, γ= 6. Find
(a) (f + g) (x)
(b) (f – g) (x)
(c) (γf) (x)
(d) (γg) (x)
(e) (fg) (x)
(f) (f / g) (x)
Sol: We have,
(a) (f + g) (x) = f(x) + g(x) = x³ + 3x + 1.
(b) (f – g) (x) = f(x) – g(x) = x³ – (3x + 1) = x³ – 3x – 1.
(c) (γf) (x) = γ f(x) = 6x³ 
(d) (γg) (x) = γ g(x) = 6 (3x + 1) = 18x + 6.
(e) (fg) (x) = f(x) g(x) = x³ (3x +1) = 3x⁴ + x³.
(f) (f / g) (x) = f(x) / g(x) = x³ / (3x + 1), provided x ≠ – 1/3.