Functions

 

Let \( A \) and \( B \) be two non-empty sets, and let \( f \) be a relation defined from \( A \) to \( B \). If every element in \( A \) is mapped to exactly one element in \( B \) by the relation \( f \), then the relation \( f \) is called a function from \( A \) to \( B \).

It is denoted as:
\[ f: A \longrightarrow B \quad \text{or} \quad A \rightarrow{f} B \]

Here, \( A \) is called the domain and \( B \) is called the codomain of the function. Furthermore, the set consisting of the elements in \( B \) that are actually mapped to by the elements of \( A \) is called the image set (or range) of \( A \), which is denoted by \( f(A) \).

 

Example:

 

\( A = \{ a, b, c, d \} \) is the domain,

\( B = \{ 1, 2, 3, 4, 5, 6, 7 \} \) is the codomain,

\( f(A) = \{ 4, 5, 6 \} \) is the range, where \( f(A) \subseteq B \).

If \( f: A \to B \) and \( (x,y) \in f \), it is expressed as:
\[ f: x \to y \quad \text{or} \quad y = f(x) \]
where \( y \) is said to be the image of \( x \) under \( f \).

 

Example:

 

Consider the sets \( A = \{ -1, 0, 1, 2 \} \) and \( B = \{ -1, 0, 1, 2, 3, 4 \} \). Let a function \( f \) from \( A \) to \( B \) be defined as:

\[
f = \{ (x, y) \mid y = x^2 \}
\]

In this case, the mapping rule is given by:
\[
f(x) = x^2
\]

Evaluating each element:

\[ \text{For } x = -1 \quad \implies \quad f(-1) = (-1)^2 = 1 \]

\[ \text{For } x = 0 \quad \implies \quad f(0) = 0^2 = 0 \]

\[ \text{For } x = 1 \quad \implies \quad f(1) = 1^2 = 1 \]

\[ \text{For } x = 2 \quad \implies \quad f(2) = 2^2 = 4 \]

Thus, the function in roster form is \( f = \{ (-1, 1), (0, 0), (1, 1), (2, 4) \} \), and it can be visualized using an arrow diagram. The range is found to be:

\[ f(A) = \{ 0, 1, 4 \} \]

 

Note:

 

For a relation \( f \) defined from \( A \) to \( B \) to qualify as a function, it must satisfy two structural criteria:

\( 1) \) No element in the domain \( A \) can be left unmapped (unpaired); however, there may be elements left over in the codomain \( B \).
\( 2) \) Every element in the domain \( A \) must map to exactly one image in the codomain \( B \).

 

Examples:

 

Consider the sets \( A = \{ a, b, c \} \) and \( B = \{ 1, 2, 3 \} \). Let us determine whether the following relations defined from \( A \) to \( B \) represent functions.

\[
\bullet \quad f_1 = \{ (a, 1), (b, 3) \} \quad \text{Let us model this relation with an arrow diagram.}
\]

 

Since the element \( c \in A \) is left unmapped, the relation \( f_1 \) is not a function.

\[
\bullet \quad f_2 = \{ (a, 1), (b,2), (b, 3), (c, 3) \} \quad \text{Let us model this relation with an arrow diagram.}
\]

\[
\bullet \quad f_3 \quad \text{Let us model this relation with an arrow diagram.}
\]

Since the relation \( f_3 \) satisfies all structural criteria, it is a valid function.

\[
\bullet \quad f_4 \quad \text{Let us model this relation with an arrow diagram.}
\]

 

Examples:

 

Let us investigate whether the following algebraic relations define functions on their specified sets.

1) Defined from \( \mathbb{R} \) to \( \mathbb{R} \):
\[
f_1 = \{(x,y) \mid y = \frac{x}{x^2 – 1} \}
\]

For the rational expression \[ y = \frac{x}{x^2 – 1} \] the denominator becomes zero when \( x = \pm1 \), rendering the expression undefined. Consequently, the elements \( -1 \) and \( 1 \) in the domain are left without a valid mapping, which implies that the relation \( f_1 \) is not a function.

2) Defined from \( \mathbb{R} \) to \( \mathbb{R} \):
\[
f_2 = \{(x,y) \mid |y| = x^2 + 1 \}
\]

Given the absolute value equation \[ |y| = x^2 + 1 \] every input value in the domain yields two distinct outputs. For example, setting \( x = 1 \) results in \( y = \pm 2 \). Since domain elements map to multiple values, the relation \( f_2 \) is not a function.

3) Defined from \( \mathbb{R} \) to \( \mathbb{R} \):

\[
f_3 = \{(x,y) \mid y = \frac{x}{x^2 + 1} \}
\]

Since the quadratic expression in the denominator is positive for all real numbers (\( x^2 + 1 \neq 0 \)), every real input \( x \) maps uniquely to exactly one real output \( y \). Therefore, the relation \( f_3 \) is a valid function.

4) Defined from \( \mathbb{Z} \) to \( \mathbb{Z} \):
\[
f_4 = \{(x,y) \mid y = \frac{x – 1}{3} \}
\]

 

Strategy (The Vertical Line Test):

 

 

2)

  \(\Rightarrow \)       

 

 

3)

\(\Rightarrow \)

Every vertical line intersects the curve exactly once across the entire domain, with no unmapped areas. Thus, the relation \( f_3 \) is a valid function.

4)

\(\Rightarrow \)

 

 

5)

 

 

 

Question 9