Types of Functions in Mathematics

Types of Functions:

(1) One-to-one Functions- A function, f: A → B, is said to be one-to-one or injective if for distinct elements x₁, x₂ ∈ A, there exist f(x₁), f(x₂) respectively in B such that for x₁ ≠ x₂, f( x₁) ≠ f(x₂). Linear functions like f(x) = 2x – 1, f(x) = x + 2, f(x) = (2x + 1)/3 etc., are all one-to-one functions. The graphs of one-to-one functions denote straight lines. The mapping has been shown in the figure below.

(2) Many-to-one Functions- Functions that are not one-to-one are said to be many-to-one functions. Functions like y – x2, y = x3, y = | x | are many-to-one functions. The mapping has been shown in the figure below. The codomain of the function, in this case, is set B but the range is {y₁, y₂, y₃}.

(3) Onto-(Surjective) Functions- A function, f: A → B, is said to be an onto or surjective function if each element y of B has a pre-image x in A. The codomain in this case is called the Range of f. Set B in this case must also be exhausted as shown in the figure below.

Thus, Range of f = f(A) = {f(x): x ∈ A}
Obviously, f(A) ⊆ B

(4) Into Functions- A function, f: A → B, is said to be an into-function if there exists at least one element in B having no pre-image in A. In the figure below, there are two elements y₄ and y₅ in set B for which there is no pre-image in A. The function in the case is an into-function.

(5) Bijection- A function, f: A → B, is a bijection if-

It is one-one i.e. f(x₁) = f(x₂) ⇒ x₁ = x₂ for all x₁, x₂ ∈ A.
It is onto i.e. for all y ∈ B, there exists x ∈ A such f(x) = y.

(6) Identity Function- Let A be a non-void set. A function, f: A → A, is said to be the identity function on set A if f associates every element of set A to the element itself. Identity function is given by f(x) = x.

(7) Constant Function- A function, f: A → B, is a constant function if every member of A has the same image in B under the function f. Thus, in this case f(x) = c, for all x ∈ A, c ∈ B.

(8) Exponential Function- The function is given by f(x) = e^x. It lies completely above the X-axis and from left to right it rises exponentially.

(9) Logarithmic Function- It is given by f(x) = log_e x. The graph of the function lies completely on the right side of the Y-axis and it rises above from left to right as shown in figure (a).

When f(x) = e^xand f(x) = log_e x are drawn on the same scale, it would be clear that one is reflection of the others in the line y = x (the identity function) as shown in figure (b).

(10) The Greatest Integer Function- It is defined as f(x) = [x], where [x] refers to the greatest integer contained in x. It is sometimes referred to as the Integral value function or the Staircase function (as the graph of the function looks like a staircase). The graph of the function consists of many broken pieces, each being coincident with the graph of a constant function as shown in the figure below.

(11) Trigonometric Functions- The graphs of f(x) = sin x, f(x) = cos x and f(x) = tan x are shown below.

(12) Signum Function- The function is defined as f(x) = | x |/x, when x ≠ 0 and f(x) = 0, when x = 0. Clearly, (0, 0) is a point on the graph and | x |/x = 1, when x > 0, | x |/x = -1, when x < 0 are the two branches of the graph as shown in the figure below.

(13) Reciprocal Function- The function is given by f(x) = 1/x, x ≠ 0. It is symmetrical about the origin as shown in the figure below.

(14) Absolute Value Function-

It is also known as the Modulus function. The graph is composed of two rays originating from the same point and is symmetrical with respect to the Y-axis. It lies completely above the X-axis.

Example- Find whether the function, f: I → I defined by f(x) = x² + 5, for all x ∈ I is one-one or not.

Solution- Let x₁, x₂ ∈ I (domain) such that f(x₁) = f(x₂)
⇒ x₁² + 5 = x₂² + 5
⇒ x₁² = x₂²
⇒ x₁ = ± x₂
⇒ x₁ ≠ x₂

∴ f is not one-one.

Example- State whether the following functions are onto-functions:

(i) f: R → R given by f(x) = x² + 3 for all x ∈ R

Solution- Let y ∈ R (co-domain) such that f(x) = y
⇒ x² + 3 = y
⇒ x² = y – 3
⇒ x = √(y – 3)

For all values of y < 3, x ∉ R (domain).
∴ the function is into.

(ii) f: I → I defined by f(x) = 5x + 2 for all x ∈ I

Solution- Let y ∈ I (co-domain) such that f(x) = y
⇒ 5x + 2 = y
⇒ 5x = y -2
⇒ x = (y – 2)/5

For some values of y, x ∉ I.
∴ f is an into function.

(iii) f: Q → Q given by f(x) = 2x – 5 for all x ∈ Q

Solution- Let y ∈ Q (co-domain) such that f(x) = y
⇒ 2x – 5 = y
⇒ 2x = y + 5
⇒ x = (y + 5)/2

For each y ∈ Q (co-domain), x ∈ Q (domain).
∴ f is an onto function.

Example- Show that g: R → R defined by g(x) = 2x³ + 8 for all x ∈ R is a bijection.

Solution- g is one-one:
Let x₁, x₂ ∈ R (Domain) such that g(x₁) = g(x₂)
⇒ 2x₁³ + 8 = 2x₂³ + 8
⇒ 2x₁³ = 2x₂³
⇒ x₁³ = x₂³
⇒ x₁ = x₂

∴ g is one-one.

g is onto:
Let y ∈ R (co-domain) such that g(x) = y
⇒ 2x³ + 8 = y
⇒ 2x³ = y -8
⇒ x³ = (y -8)/2
⇒ x = [(y -8)/2]^1/3

For each y ∈ R (co-domain), x ∈ R (domain).
∴ g is onto.

Thus, g is a bijection.

Example- Show that the function, g: A → B defined by g(x) = (x – 3)/(x – 4) where A = {x: x ∈ R, x ≠ 3} and B = {y: y ∈ R, y ≠ 1}, is a bijection.

Solution- g is one-one:
Let x₁, x₂ ∈ A (Domain) such that g(x₁) = g(x₂)
⇒ (x₁ – 3)/(x₁ – 4) = (x₂ – 3)/(x₂ – 4)
⇒ x₁x₂ – 4x₁ – 3x₂ + 12 = x₁x₂ – 3x₁ – 4x₂ + 12
⇒ – 4x₁ – 3x₂ = – 3x₁ – 4x₂
⇒ -x₁ = -x₂
⇒ x₁ = x₂

∴ g is one-one.

g is onto:
Let y ∈ B (co-domain) such that g(x) = y
⇒ (x – 3)/(x – 4) = y
Apply Componendo and Dividendo:
⇒ 2x – 7 = (y + 1)/(y – 1)
⇒ 2x = (y + 1)/(y – 1) + 7
⇒ 2x = (y + 1 + 7y -7)/(y – 1)
⇒ 2x = (8y – 6)/(y – 1)
⇒ x = (4y – 3)/(y – 1)

For each y ∈ B = R – {1}, x ∈ A = R – {4}
∴ g is onto.

Thus, g is a bijection.

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Function (mathematics)– Wikipedia