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Domain and image of a function


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:

a) the domain (D) of f;
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 12=1, (-3)2=9, 32= 9 and 22= 4, we have y = x2.

2) Given the function f: IRIR (ie domain and contradiction are real numbers) defined by f (x) = x2-5x + 6, calculate:

a) f (2), f (3) and f (0);
b) the value of x whose image is worth 2.

Resolution:

a) f (2) = 22-5(2)+6 = 4-10+6 = 0
f (3) = 32-5(3)+6 = 9-15+6 = 0
f (0) = 02-5(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 x2-5x + 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.

Next: Obtaining the Domain