O **domain** of a function from A to B is always the starting set itself, ie D = A. If an x element A is associated with a y element B, we say y is a **Image** x (indicate y = f (x) and read “y equals f of x”).

Notice the domain and image in the function below.

*Another example*: if f is a function of IN in IN (this means that domain and contradiction are the natural numbers) defined by y = x + 2, then we have to:

- The image of 1 through f is 3, ie f (1) = 1 + 2 = 3;
- The image of 2 through f is 4, ie f (2) = 2 + 2 = 4;

In general, the image of x through f is x + 2, ie: f (x) = x + 2.

In a function f from A to B, the elements of B which are images of the elements of A by applying f form the image set of f. According to the concept of function, there are two conditions for a relation f to be a function:

**1ª)**The domain must always match the starting set, ie

**every element of A**It is arrow starting point. If we have an element of

**THE**from which no arrow departs, the relationship is not a function.

**2ª)**From each element of

**THE**must leave

**Single**arrow. If of an element of

**THE**From more than one arrow, the relationship is not a function.

Comments:

- Since x and y have their values varying in sets A and B, they are called variables.
- The variable x is called the independent variable and the variable y, the dependent variable, because to get the value of y we depend on a value of x.
- A function f is defined when given its domain (set A), its contradiction (set B) and the law of association y = f (x).

## Exercises solved

**1) Consider the function f: A B represented by the following diagram:**

Determine:

b) f (1), f (-3), f (3) and f (2);

c) the image set (Im) of f;

d) the law of association

Resolution:

a) The domain is equal to the starting set, ie D = A.b) f (1) = 1, f (-3) = 9, f (3) = 9 and f (2) = 4.

c) The image set is formed by all images of the domain elements, therefore:

Im = {1,4,9}.

d) As 1

^{2}=1, (-3)

^{2}=9, 3

^{2}= 9 and 2

^{2}= 4, we have y = x

^{2}.

**2) Given the function f: IRIR (ie domain and contradiction are real numbers) defined by f (x) = x ^{2}-5x + 6, calculate:**

a) f (2), f (3) and f (0);

b) the value of x whose image is worth 2.

Resolution:

a) f (2) = 2^{2}-5(2)+6 = 4-10+6 = 0

f (3) = 3

^{2}-5(3)+6 = 9-15+6 = 0

f (0) = 0

^{2}-5(0)+6 = 0-0+6 = 6

b) Calculating the value of x whose image is worth 2 equals solving the equation f (x) = 2, ie x^{2}-5x + 6 = 2. Using Bhaskara's formula we find roots 1 and 4. Therefore the values of x that have image 2 are 1 and 4.