Domain, Co-Domain And Range Of Function/ Graphical Representation Function
In this article, we will discuss about domain, co-domain, range and graphical representation of function. Let : A → B (f be function from A to B), then
Set A is referred to as the domain of the function ‘f’
Set B is referred to as the co-domain of the function ‘f’
Set of all f-images of all the elements of A is referred to as the range of f. Therefore, range of f is denoted by f(A).
Note:
Range ε co-domain
Example on Domain, co-domain and range of function:
1. Which of the arrow diagrams below illustrates a mapping? Provide reasons to back up your answer.
Solution:
(a) a has exclusive image p.
(b) has exclusive image q.
(c) has exclusive image q.
(d) has exclusive image r.
Therefore, every element of A possess a unique image in B.
Thus, the given arrow diagram showcases a mapping.
(b) In the specified arrow diagram, the element ‘a’ of set A is connected with two elements, i.e., q and r of set B. Therefore, every element of set A does not possess a unique image in B.
Consequently, the given arrow diagram does not illustrate a mapping.
(c) The element ‘b’ of set A is not connected with any element of set B. Therefore, b ε A does not have any image. For a mapping from A to B, every element of set A ought to have a unique image in set B which is not characterized by this arrow diagram. Therefore, the given arrow diagram does not showcase a mapping.
(d) a has an exclusive image p. b has an exceptional image q. c has an exclusive image r. Therefore, every element in set A has an exclusive image in set B.
Thus, the given arrow diagram illustrates a mapping.
2. Find out if R is a mapping from A to B.
(i) Let A = {3, 4, 5} and B = {6, 7, 8, 9} and R = {(3, 6) (4, 7) (5, 8)}
Solution:
Since, R = {(3, 6); (4, 7); (5, 8)} then Domain (R) = {3, 4, 5} = A
We observe that no two ordered pairs in R possess the same first constituent.
Thus, R is a mapping from A to B.
(ii) Let A = {1, 2, 3} and B= {7, 11} and R = {(1, 7); (1, 11); (2, 11); (3, 11)}
Solution:
Given that, R = {(1, 7); (1, 11); (2, 11); (3, 11)} then Domain (R) = {1, 2, 3} = A
Nevertheless the ordered pairs (1, 7) (1, 11) possess the same first component.
Thus, R is not a mapping from A to B.
3. Let A = {1, 2, 3, 4} and B = {0, 3, 6, 8, 12, 15}
Consider a rule f (x) = x2 – 1, x ε A, then
(a) show that f is a mapping from A to B.
(b) draw the arrow diagram to illustrate the mapping.
(c) Illustrate the mapping in the roster form.
(d) write the domain and range of the mapping.
Solution:
Using f (x) = x2 – 1, x ε A we have
f(1) = 0,
f(2) = 3,
f(3) = 8,
f(4) = 15
We notice that each element in set A has exclusive image in set B.
Thus, f is a mapping from A to B.
(c) Mapping can be illustrated in the roster form as
f = {(1, 0); (2, 3); (3, 8); (4, 15)}
(d) Domain (f) = {1, 2, 3, 4} Range (f) = {0, 3, 8, 15}
Representation of a function by an arrow diagram:
In this form of representation, we illustrate the sets by closed figures and the elements are represented by points in the closed figure.
The mapping f : A → B is illustrated by arrow which emanates from elements of A and ends at the elements of B.
Every element of A possess a distinctive image in B
Function as a special type of relation:
If A and B are two non-empty sets, A relation f from A to B is known as a function from A to B if all the element of A (say x) possess one and only one image ( y, for instance) in B. The f-image of x is represented by f (x) and therefore, we write y = f (x). The element x is known as the pre-image of y under ‘f’.
Real valued function of a real variable:
If the domain and range of a function ‘f’ are subsets of R (set of real numbers), in that case, f is said to be the real valued function of real variable or just a real function. It may be defined as
A function f A → B is known as a real valued function if B is a subset of R. If A and B are subsets of R then f is known as a real function.
A few more examples on domain, co-domain and range of function:
1. Let N be the set of natural number if f: N → N by f (x) = 3x +2, then find f (1), f (2), f (-3), f (-4).
Solution:
Given that for f(x) = 3x + 2
thus f(1) = 3 × 1 + 2 = 3 + 2 = 5
f(2) = 3 × 2 + 2 = 6 + 2 = 8
thus f(-3) = 3 × (-3) + 2 = -9 + 2 = -7
f(-4) = 3 × -4 + 2 = -12 + 2 = -10
2. Let A = {a, b, c, d} and B= {c, d, e, f, g}
Let R1 = {(a, c) (b, d) (c, e)}
R2 = {(a, c) (a, g) (b, d) (c, e) (d, f)}
R3 = {(a, c) (b, d) (c, e) (d, f)}
Justify which of the given relation is a function from A to B.
Solution:
We will obtain,
(i) Domain R1 {a, b, c} ≠ A
Thus, R1 is not a function from A to B.
(ii) Two dissimilar ordered pairs (a, c) (a, g) possess the same first component.
Thus, R2 is not a function from A → B.
(iii) Domain R3 = {a, b, c, d} = A and not two different ordered pair possess same first component.
Thus, R3 is a function from A to B.
Inverse Trigonometric Functions
This topic will be of fewer words, but more formulas. It is essential that you are able:
1. Commit to memory the useful graphs, identities and formulas.
2. Take time to learn how to spend your time trying to derive all the identities. When you have done with learning these two points, you would be able to score what is required of you in this topic which is usually about sketching graphs, or differentiating and integrating them
If sin y = x. An inverse trigonometric function inverses the trigonometric function, and is written as y = sin-1 x.
Observe that there is a difference between sin-1x and (sin x)-1. This is only one of the 6 inverse trigonometric functions, the remaining 5 of them are cos-1 x, tan-1 x, sec-1 x, csc-1 x, and cot-1 x.
Forward-Inverse Identities
It is not difficult to prove this one. Make x = cos y, and make use of the identity cos2 x + sin2 x = 1. The remaining follows too. The one of tan(cos-1 x) will likely be a bit harder to prove. See below:
Inverse Sum Identities
Prove the first one by making x = cos (π/2 – y) = sin y. Try to discover how the rest was worked out.
sin-1 (-x) = –sin-1 x
csc-1 (-x) = –csc-1 x
cos-1 (-x) = π – cos-1 x
sec-1 (-x) = π – sec-1 x
tan-1 (-x) = –tan-1 x
cot-1 (-x) = –cot-1 x
Graphical Representation of a Function
A method of expressing a function in graphical system is known as Graphical representation of a function.
Suppose, there are two sets and each set is made up of a few elements. The two sets are named P and Q. The elements in set P are a, b, c and d. The elements in set Q are p,q,r and s. Assume, the elements in Set P associate elements in set Q. It signifies that the elements in Set P have relation with elements in set Q. In other words, the elements a,b,c and d correlate elements p,q,r and s correspondingly.
In graphical representation method, every set is written as a closed figure; in which elements are displayed. With regards to our supposition, the elements in set P correlate elements in Set Q. The set whose elements correlate elements in other set ought to be displayed in left side and the set whose elements are connected by the elements set ought to be displayed in right hand side. Thus, Set P and Set Q are displayed in left and right side correspondingly. In other words, the set shown in left side is referred to as domain and the set shown in right side is referred to as Codomain (or Range).
The arrow bearing from an element to another element stands for the direction of correlation. The arrow direction from an element towards another element tells, the elements a,b,c and d correlate elements p,q,r and s correspondingly.
The elements in Set P have a relation with elements in set Q. In both sets are related each other. The relation between both sets is referred to as association, the direction of association is shown by an arrow direction symbol by linking both sets with it.
The relation between both sets depends on working principle and the working principle in mathematical language is referred to as a function. In graphical representation, function is represented by a symbol f as displayed in the picture.
As explained here, every element, set, direction of association and functionality are articulated in graphical representation system.
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