Today is our topic of discussion Function example.
Function example
Example 18.
Suppose A = {0, 1, 2, 3} is a set. The x < y relation existing between the elements of the set A can be described by subset of A × A which is S = {(0, 1), (0, 2), (0, 3), (1, 2), (1, 3), (2, 3)}, where for ordered pairs included under S (first entry)<(second entry). Here S is the <relation described under A.
Example 19.
Suppose, a is the father, b is the mother, c is the elder son, d is the younger son, e is the daughter, ƒ is the wife of the elder son in a family. Assuming the set of members of the family as F, we get F = {a, b, c, d, e, f}. The brother relation in set F, which means the brother relation between 2 and y can be described by B = {(c, d), (c, e), (d, c), (d, e)}, where the first entry of the ordered pairs under set B is the brother of the second entry of the same ordered pair. The set B denotes the brother relation of set F.
Example 20.
Relation between the radius and circumference of a circle is expressed as p = 2pi*r where variables r and p denote the radius and the circumference of the circle respectively. Here for every possible value of r, one and only one value of p is definite.
We say that, variable p is a function of variable r which is written as p = f(r) f(r) = 2pi*r In this functional relation the function has been assumed to be defined with values of r taken from set X and values of p from set Y. This function treated as relating X to Y \ (r, p) :r in X and p in Y p = 2pi*r \ . The idea of relations has been explained in the Mathematics book of class 9-10.
Example 21.
f x -> 2x + 1 ,x in Z; describes a function from the set of integers Z to Z. Under this function image of x is y = f(x) = 2x + 1
Example 22.
A set of ordered pairs
F = {(0,0), (1,1), (−1,1), (2,4), (−2,4), (3,9), (−3,9)} describes a function, whose domain is a set of first entries of the ordered pairs under F and range is set of second entries of the ordered pairs under F.
So, domain F = {0, 1, −1, 2, -2, 3, -3} and range F = {0, 1, 4,9}
If you look closely you will see in this case that under F, z dom of F, F(x) = r². To be noted that, set of an ordered pair will only define a function when first entries of different ordered pairs will be different.
Example 23.
Considering the domain and range of the function F described below as A and B respectively, the function can be described with a diagram, where one and only one arrow sign started from every point of set A and ends at one and only one point of set B(left sided diagram). To be noted that, a set Y can be regarded as the codomain of the function, whose subset can be drawn with B(right sided diagram).
Example 24.
Suppose, A = {2, 3, 5, 7} and B = {1, 2, 4, 7, 10} . Those elements of B which are divisible by the elements of A are shown below by drawing a diagram using arrow signs:
Here we now form the set D = {(2, 2), (2, 4), (2, 10), (5, 10), (7, 7)} of all ordered pairs which describe the relation of disivility. In the set D, the first entries of the included ordered pairs belongs to A while the second entries belong to B and are divisible by the first entries. So, D subset AB and D= (x, y) 😡 in A,y in B and y is divisible by x}, Here the set D is a relation from set A to set B.
Example 25.
Let’s consider the set of ordered pairs of real numbers L= (x, y) : x ∈ R y ∈ R and x < y . a < b for two real numbers a, b if and only if it is (a, b) in L. Therefore, the set L describes the relation of lesser or greater.
Example 26.
Which one of the following relations is not a function? Give reasons.
Solution:
The relation shown in upper left is not a function because 2 -> 4 2 -> 5 and 3 -> 4 , 3 -> 5 On the other hand, all other three relations are functions.
Example 27.
Find the range of the function f / x -> 2x ^ 2 + 1 where the domain is X = \{1, 2, 3\}
Solution: f(x) = 2x ^ 2 + 1 where x in X
f(1) = 2 * (1) ^ 2 + 1 = 3, f(2) = 2 * (2) ^ 2 + 1 = 9 and f(3) = 2 * (3) ^ 2 + 1 = 19 ..range set of \{1, 2, 3\} = \{3, 9, 19\} .
Example 28.
For the function f / x -> mx + c the images of 2 and 4 are respectively 7 and -1. So find out:
1) The values of m and c.
2) The image of 5 under f.
3) The preimage of 3 under ƒ.
Solution:
1) According to f(x) = mx + c
f : 2 -> 7So, f(2) = 7or ,2m+c=7………………(1)
f : 4 -> – S , f(4) = – 1 or, 4m + c =-1………………(2)
From (1) and (2) we get m = – 4 and c = 15
2) Under ƒ, image of 5 is f(5) = – 4 * 5 + 15 = – 5
3) If x is the preimage of 3, then f(x) = 3So , – 4x + 15 = 3 or x = 3
Example 29.
Find the domain of the function F(x) = √1-x. Find whichever is defined among F(−3), F(0), F (2), F(1), F(2).
Solution:
F(x) = √1- € R if and only if 1 x ≥ 0 or 1 x Therefore,
So, Dom. F = {x: x = R and x ≤ 1}
See more: