Introduction to Functions and Their Types

Definition of a Function

A relation 'f' from a set A to a set B is said to be a function if every element of set A has one and only one image in set B. This is one of the most important concepts in mathematics.

For a relation to be a function, it must satisfy two conditions:

1. The domain of the function must be the entire set A.
2. No two distinct ordered pairs in the function can have the same first element.

Function Notation

• We denote a function 'f' from A to B as $f: A \to B$.
• If $(a, b) \in f$, we write $f(a) = b$. Here, 'b' is the image of 'a' under 'f', and 'a' is the preimage of 'b'.

Classification of Functions (Mappings)

• One-one function (Injection): A function $f: A \to B$ is one-one if every element in the domain A has a distinct image in the codomain B. Mathematically, $f(x_1) = f(x_2) \implies x_1 = x_2$. Graphically, this can be checked with the Horizontal Line Test (a horizontal line should never intersect the graph more than once).

• Onto function (Surjection): A function $f: A \to B$ is onto if every element in the codomain B is the image of at least one element in the domain A. In other words, the Range must be equal to the Codomain.

• Bijective function (One-one and Onto): A function is a bijection if it is both injective (one-one) and surjective (onto).

Standard Function Types

• Identity Function: $f(x) = x$.
• Constant Function: $f(x) = c$, where c is a constant.
• Modulus Function: $f(x) = |x|$.
• Signum Function: $f(x) = \text{sgn}(x)$.
• Greatest Integer Function (GIF): $f(x) = [x]$.

Example 1: Identifying Functions from Relations

Question: Examine if the relation $R = \{(2,2), (2,4), (3,3)\}$ is a function.

Solution:

This is not a function. The first element, 2, corresponds to two different images (2 and 4). This violates the 'one and only one image' rule.

Example 2: Finding Domain of a Radical Function

Question: Find the domain of the real function $f(x) = \sqrt{9-x^2}$.

Solution:

For the square root to be defined, the expression inside must be non-negative.

$$9 - x^2 \ge 0 \implies x^2 \le 9 \implies -3 \le x \le 3$$

So, the Domain is $[-3, 3]$.

Example 3: Finding Domain of a Rational Function

Question: Find the domain of the function $f(x) = \frac{x^2+3x+5}{x^2-5x+4}$.

Solution: The function is defined for all real numbers except where the denominator is zero. We set the denominator to zero: $x^2-5x+4 = 0 \implies (x-4)(x-1) = 0$. The roots are $x=1$ and $x=4$. The domain is all real numbers except these two values. Domain = $\mathbb{R} - \{1, 4\}$.

Example 4: Finding Range of a Modulus Function

Question: Find the range of the function $f(x) = -|x|$.

Solution: The standard modulus function, $|x|$, has a range of $[0, \infty)$. Since our function is multiplied by -1, all the output values will be negative or zero. Therefore, the range of $f(x) = -|x|$ is $(-\infty, 0]$.

Example 5: Evaluating the Greatest Integer Function

Question: Find the values of (i) $[7.3]$ (ii) $[-4.8]$.

Solution:

(i) The greatest integer less than or equal to 7.3 is 7. So, $[7.3] = 7$.

(ii) The integers less than or equal to -4.8 are -5, -6, -7, … The greatest among these is -5. So, $[-4.8] = -5$.

Example 6: Classifying Functions (Injective/Surjective)

Question: Classify the following functions as injective, surjective, bijective, or neither.

(i) $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = 2x+1$.

(ii) $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = x^2$.

Solution:

(i) For $f(x) = 2x+1$:

• Injective Check: Let $f(x_1) = f(x_2)$. Then $2x_1+1 = 2x_2+1 \implies 2x_1 = 2x_2 \implies x_1 = x_2$. It is injective.

• Surjective Check: The range of a linear function is $\mathbb{R}$, which is equal to the codomain $\mathbb{R}$. It is surjective.

• Conclusion: Since it is both injective and surjective, it is a bijection.

(ii) For $f(x) = x^2$:

• Injective Check: Let $f(x_1) = f(x_2)$. Then $x_1^2 = x_2^2 \implies x_1 = \pm x_2$. For example, $f(2)=4$ and $f(-2)=4$. Since different inputs can give the same output, it is not injective.
• Surjective Check: The range of $f(x)=x^2$ is $[0, \infty)$. The codomain is $\mathbb{R}$. Since the range is not equal to the codomain, it is not surjective.
• Conclusion: It is neither injective nor surjective.