Learning Objectives:

At the end of this lesson you will be able to:

• Define function, domain, co-domain and range of function.
• Identify the given relation is function or not.
• Determine the domain and range of function.
• Find the sum, difference, product and quotient of functions.
• Find the domain of the sum, difference, product and quotient of functions.
• The set of all possible values which qualify as inputs to a function.
• The entire set of values possible for independent variables.

Key Terms

• Domain of a function
• Co-domain of a function
• Range of a function

Brainstorming Questions

Consider the following relations

R₁ = {(1, 2), (3, 4), (2, 5), (5, 6), (4, 7)}

R₂ = {(1, 2), (3, 2), (2, 5), (6, 5), (4, 7)}

R₃ = {(1, 2), (1, 4), (2, 5), (2, 6), (4, 7)}

a. What differences do you see between these relations?

b. How are the first elements of the coordinates paired with the second elements of the coordinates?

c. In each relation, are there ordered pairs with the same first coordinate? 1 Consider the following relations

2.1. The notion of function

Definition:

ü  A function is relation in which no two ordered pairs have the same first entry.

·        I.e. A function 𝑓 is a set of ordered pairs with the property that whenever (x, y) and (x, z) belong to 𝑓, then x = 𝑧.

·        It is a relation in which no two distinct ordered pairs have the same first element.

Example 1

Consider the relation R = {(5, 2), (6, 8), (4, 2), (6, 3)}

Since 6 is paired with both 8 and 3 the relation R is not a function.

Example 2

Let R = {(3, 2), (7, 8), (4, 3)}. This relation is a function because no first-coordinate is paired (mapped) with more than one element of the second coordinate.

Example 3

The relation R = {(x, y): x is the brother of y} is not a function because a person may have more than one brother.

Example 4

Consider the following arrow diagrams.

Which of these relations are functions?

Solution:

R₁ is a function. (Why?)

R₂ is not a function because 1 and 3 are both mapped onto two numbers.

R₃ is a function. (Why?)

2.1.1 Domain co-domain and range of a function

Domain of function

• The set of all possible values which qualify as inputs to a function.
• The entire set of values possible for independent variables.

Co-Domain:

• The set of all the outputs of a function is known as the range of the function.
• After substituting the domain, the entire set of all values possible as outcomes of the dependent variable.
• The range of the function 𝑓 is the set of all values 𝑓(x ) which corresponds to the values on the graph in the xy -plane.

Notation:

• If x is an element in the domain of a function 𝑓, then the element in the
  range that is associated with is denoted by 𝑓(x).
• This is called an image of x under the function 𝑓.
• The notation 𝑓(x) is referred to function value and we read as ‘ 𝑓 𝑜𝑓 x ’ and x is the pre-image of 𝑓(x).

Example 4

For each of the following functions, find the domain, co-domain and the range.

a. 𝐹 is a set of ordered pairs such that (3, -2), (5, 4), (1, 2) and (-3, 7) are members of 𝐹.
b. 𝑅 is a set of ordered pairs (x, y) of real numbers such that y =2x.

Solution:

a. Domain of 𝐹 is a collection of -3, 1, 3 and 5 , Co-domain of F is a

collection of -2, 2, 4 and 7, and the range of 𝐹 is a collection of -2, 2, 4 and 7.

b. Domain of 𝑅 is a set containing all real numbers and the co-domain and range is the set containing all real numbers.

Note:

• A function from 𝐴 𝑡𝑜 𝐵 can sometimes be denoted as 𝑓: 𝐴 → 𝐵, where the domain of the function 𝑓 is 𝐴 and the range of the function 𝑓 is contained in 𝐵.
• 𝐵 contains the image of the elements of 𝐴 under the function of 𝑓.

Example 5

Is the relation R = {x, y): x = y²} a function?

Solution:

This is not a function because numbers for x are paired with more than one number in y.

For example, (9, –3) and (9, 3) satisfy the relation with 9 being mapped to both -3 and 3.

Example 6

Let A = {1, 2, 3, 4} and B = {3, 4, 5, 7, 9}

If f : A → B is the function given by f (x) = 2x + 1, then find the domain and the range of f.

Solution:

Since f(1) = 3 ∈ B, f(2) = 5 ∈ B, f(3) = 7 \in B ), and ( f(4) = 9 ∈ B ), the domain of ( f ) is [ D = { 1, 2, 3, 4 }]
and the range of ( f ) is[ R = { 3, 5, 7, 9 }.

Example 7:

Let f(x) = $\sqrt{x+1}$, then find:

1.  f (3)
2.  f (12)

Solution:   

Take f (x) = $\squrt{x+1}$ and evaluate:

1.  f (3) = 3−3 = 0 = 0
2.  f (x) = 12 − 3 = 9 = 3.

Example 8:

For the function f (x) = 1– x²

1. Find the domain and the range.
2. Evaluate f (2) and f (–1).

Solution:

1. The domain of the function is D = {x: x ∈ R}, since it is defined for all real numbers. The range is R ={y: y ∈ R}, y ≤ 1}
2. f (2) = 1– (2)^{2} = 1– 4 = –3 and

f (–1) = 1 – (–1)^{2} = 1–1 = 0.

Example 9

Take f (x) = $\sqrt{x – 3}$ : Find the domain and the range of f(x).

Solution:

Since the expression in the radical must be non-zero, x – 3 ≥ 0. This implies x ≥ 3. Hence, the domain of the function is the set of real numbers greater than or equal to 3.
The range of the function is greater than or equal to zero.

2.2. Combination of functions

A. Sum and Difference of functions

Let 𝑓 and 𝑔 be two functions with common domains. Then, for all x common to both domains:

1. the sum of 𝑓 and 𝑔 defined as follows:

(𝑓 + 𝑔)(x) = 𝑓(x) + 𝑔(x)

ii. the difference of 𝑓and 𝑔 is defined as follows:

(𝑓 – 𝑔)(x) = 𝑓(x) – 𝑔(x)

Product and Quotient of functions

Let 𝑓 and 𝑔 be two functions with overlapping domains. Then, for all common to
both domains:

iii. the product of 𝑓and 𝑔 is defined as follows:

(𝑓 ∙ 𝑔)(x) = 𝑓(x) ∙ 𝑔(x)

iv. the quotient of 𝑓and 𝑔 is defined as follows:

$\frac{f}{g}(x)$ = $\frac{f(x)}{g(x)}, \quad g(x) \neq 0$

If f (x) = 2 – x and g(x) = 3x + 2 then the sum of these functions is given

Example 1

by (f + g)(x) = (2 – x) + (3x + 2) = 2x + 4, which is also a function.

Example 2

If f (x) = 3x + 2 and g (x) = x – 4, then the difference of these functions is

(f – g)(x) = f (x) – g (x) = (3x + 2) – (x – 4) = 2x + 6.

Example 3

If f (x) = 2x and g (x) = 3 – x then the product of these functions

(fg)(x) = f (x) g (x) = (2x) (3 – x) = 6x – 2x²

Example 4

If f (x) = 3 and g (x) = 2 + x, then the quotient of these functions

$\frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{3}{2+x}, \quad x \neq -2$

The domain of combination of functions

Note:

• The domain of the sum, difference, product and quotient of functions f and 𝑔 $(f+g, f-g, fg\; and \;\frac{f}{g})$ consists of real numbers common to the domain of f and 𝑔.
• For the quotient of function , there is further restriction that 𝑔(x) is not equal to zero.

Example 1

Let $f(x) = x \quad \text{and} \quad g(x) = \sqrt{x}$. Determine

1. the sum f + g
2. the domain of (f + g)

Solution:

a. $(f + g)(x) = f(x) + g(x) = x + \sqrt{x}$

b. Domain of $f + g = {x : x \geq 0}$.

Example 2

Let $f(x) = \frac{x}{x-2} \quad \text{and} \quad g(x) = \frac{x-3}{2x}$

1.  Find

a. $f+g$(x)

b. $f-g$(x)

c. $f \times g$(x)

d. $\frac{f}{g}$(x)

ii. Determine the domain of each function.

Solution

i. a. $(f + g)(x) = f(x) + g(x) = \frac{x}{x-2} + \frac{x-3}{2x} = \frac{3x^2 – 5x + 6}{2x(x-2)}$

b. $(f – g)(x) = f(x) – g(x) = \frac{x}{x-2} – \frac{x-3}{2x} = \frac{3x^2 + 5x – 6}{2x(x-2)}$

c. $(f \cdot g)(x) = f(x) \cdot g(x) = \frac{x}{x-2} \cdot \frac{x-3}{2x} = \frac{x(x-3)}{2x(x-2)} = \frac{x-3}{2(x-2)}$

d. $\frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{\frac{x}{x -2}}{\frac{x-3}{2x}} = \frac{x}{x-3} \cdot \frac{2x}{x-2} = \frac{2x^2}{x^2 – 5x + 6}$

ii. Domain of f + g = Domain of f – g = Domain of f g: $R\setminus \{0, 2\} \;or\; (–\infty, 0) U (0, 2) U (2, \infty)$.

Domain of $\frac{f}{g}$ is   $ R\setminus \{0, 2, 3\} \;or\; (-\infty, 0) U (0, 2) U (2, 3)U (3, \infty)$.

Note:

A numerical relation 𝑅 is a function if and only if no vertical line in the plane
intersects the graph of 𝑅 in more than one point.

Vertical Line Test:

• If any curve cuts a vertical line at more than one point then the curve is not a function.
• Any vertical lines cross the graph of function at most once.

Example:

The figure below describes the graph is not a function.