Function & Relation

Functions and Relations

Understanding functions and relations is essential for solving problems in arithmetic aptitude, especially for BBA students. Here’s a detailed explanation of these concepts:

1. Relations

Definition: A relation from a set $A$ to a set $B$ is a subset of the Cartesian product $A \times B$. It consists of ordered pairs where the first element is from $A$ and the second element is from $B$.

Notation: If $A = \{1, 2\}$ and $B = \{a, b\}$, a relation $R$ from $A$ to $B$ can be written as $R \subseteq A \times B$. For example, $R = \{(1, a), (2, b)\}$.

Example: If $A = \{1, 2\}$ and $B = \{a, b\}$, possible relations include:

• $R = \{(1, a), (2, b)\}$
  
• $R = \{(1, a), (1, b), (2, a)\}$
  

2. Functions

Definition: A function $f$ from a set $A$ to a set $B$ is a special type of relation where each element in $A$ is associated with exactly one element in $B$. It is denoted as $f:A\to B$

Domain, Codomain, and Range:

• Domain: The set of all possible inputs for the function.
• Codomain: The set of all possible outputs (not necessarily the actual outputs).
• Range: The set of all actual outputs of the function.

Example: Let $A = \{1, 2, 3\}$ and $B = \{a, b\}$. A function $f$ from $A$ to $B$ could be $f = \{(1, a), (2, b), (3, a)\}$.

• Domain: $\{1, 2, 3\}$
  
• Codomain: $\{a, b\}$
  
• Range: $\{a, b\}$
  

Function Notation: If $f(x) = y$, then $x$ is an element from the domain and $y$ is the corresponding element in the codomain.

3. Types of Functions

One-to-One (Injective) Function: A function $f: A \to B$ is injective if different elements in $A$ map to different elements in $B$. No two elements in the domain map to the same element in the codomain.

• Example: For $f(x) = 2x$ where $A = \{1, 2, 3\}$ and $B = \{2, 4, 6\}$, $f$ is injective because:
  • $f(1) = 2$
    
  • $f(2) = 4$
    
  • $f(3)=6$

Onto (Surjective) Function: A function $f: A \to B$ is surjective if every element in $B$ is mapped by at least one element in $A$. Every element in the codomain has a pre-image in the domain.

• Example: For $f(x) = x^2$ where $A = \{1, 2, 3\}$ and $B = \{1, 4, 9\}$, $f$ is surjective because:
  • $f(1) = 1$
    
  • $f(2)=4$
  • $f(3) = 9$
    

One-to-One Correspondence (Bi-jective) Function: A function $f: A \to B$ is bijective if it is both injective and surjective. Every element in $A$ maps to a unique element in $B$, and every element in $B$ is mapped by exactly one element in $A$.

• Example: For $f(x) = x + 1$ where $A = \{1, 2, 3\}$ and $B = \{2, 3, 4\}$, $f$ is bijective because:

• $f(1)=2$
  
• $f(2)=3$
  
• $f(3)=4$

4. Domain, Codomain, and Range

Domain: The set of all possible input values for the function.

• Example: For $f(x) = x^2$where $x \in \mathbb{R}$, the domain is all real numbers, $\mathbb{R}$.

Codomain: The set of all possible output values that a function can produce.

• Example: For $f(x) = x^2$, if we define $f: \mathbb{R} \to \mathbb{R}$, the codomain is $\mathbb{R}$.

Range: The set of all actual output values of the function. For $f(x) = x^2$, the range is $[0, \infty)$ if $x \in \mathbb{R}$.

5. Composite Functions

Definition: The composition of two functions $f$ and $g$, denoted $g \circ f$, is a function where the output of $f$ is the input to $g$. Formally, $(g \circ f)(x) = g(f(x))$.

Example: Let $f(x) = x + 1$ and $g(x) = 2x$. The composite function $g \circ f$ is: $(g\circ f)(x)=g(f(x))=g(x+1)=2(x+1)$

=2𝑥+2

6. Inverse Functions

Definition: A function $f$ has an inverse $f^{-1}$ if $f(f^{-1}(x)) = x$ and $f^{-1}(f(x)) = x$. The inverse function $f^{-1}$ reverses the mapping of $f$.

Example: If $f(x) = 2x + 3$, solving for $x$ in terms of $y$ yields: $y = 2x + 3 \implies x = \frac{y - 3}{2}$

So, the inverse function $f^{-1}(x) = \frac{x - 3}{2}$.

Example Problems

1. Identifying a Function: Given $A = \{1, 2, 3\}$ and $B = \{a, b\}$, and $R = \{(1, a), (2, b), (3, a)\}$, determine if $R$ is a function.

   • Solution: Yes, $R$ is a function because each element in $A$ maps to exactly one element in $B$.

2. Finding Function Values: If $f(x) = 3x + 2$, find $f(4)$.

   • Solution: $f(4) = 3(4) + 2 = 12 + 2 = 14$.

3. Composite Functions: For $f(x) = x^2$ and $g(x) = x + 1$, find $(f \circ g)(2)$.

• Solution: $(f\circ g)(2)=f(g(2))=f(2+1)<br>$

$=𝑓(3)=3^2=9$.

1. Inverse Functions: Find the inverse of $f(x) = \frac{x + 5}{3}$.

   • Solution: Let $y = \frac{x + 5}{3}$. Solving for $x$: $y = \frac{x + 5}{3} \implies x = 3y - 5$
     

     So, $f^{-1}(x) = 3x - 5$.

By mastering these concepts of functions and relations, BBA students will be better equipped to tackle problems in arithmetic aptitude and apply these ideas in various business and analytical contexts.