A function assigns exactly one element of a set to each element of the other set. Functions are the rules that assigns one input to one output. The function can be represented as f:A→B. A is called the domain, and B is called the co-domain.
Figure 1: Relation map between domain A, and co-domain B:
Range
The elements of the co-domain that are participating in the function is called range. For example, in the following figure:
Figure 2: Relation map between domain A, and co-domain B:
The range of the function are only those elements participating in the function. The range R is:
R={a,b}
It is important to note that, any range is a subset of the co-domain, i.e.,
R⊆B
where, R is the range, and B is the co-domain.
Example 1: State if the following are functions or not.
f(x)=x1, where x∈R.
This is not a function, as for x=0, the function f(x)=f(0) is not defined.
f(x)=x, where x∈R.
This is not a function, as for any negative value of x, the function f(x) is not defined.
f(x)=±x2−1, where x∈R.
This is not a function, as there are two possible values for f(x), A function should only have, or point to one value.
Example 2: Given two sets, A={p,q,r,s}A=p,q,r,sA={p,q,r,s}A=p,q,r,s, and B={a,b,c}B=a,b,cB={a,b,c}B=a,b,c. Draw the function diagram and determine whether, the relation is a function or not. If it is a function, determine the range, domain and co-domain, for the following:
(1) {(p,a),(q,a),(r,a),(s,a)},
(2) {(p,a),(q,b),(p,b),(r,a),(s,b)}.
Solution
The relation map for the function is:
It is a function.
Range, R={a}
Domain, D={p,q,r,s}
Co-domain, D′={a,b,c}
The relation map for the function is:
It is not a function, since p returns two different values.
Types of Functions
One to one function: Injective function
A function f:A→B is said to be one to one function if different elements of A have different images in B.
Figure 3: Relation map between an one to one function of A and B:
Onto function: Subjective function
A function f:A→B is said to be onto function if every element of B is an image of some element of A, i.e., the range of f is equal to its co-domain.
Figure 4: Relation map between an onto function of A and B:
One to one correspondent function: Bijective function
A function f:A→B is called one to one correspondent function if the function is both one to one and onto function.
Figure 5: Relation map between an one to one correspondent function of A and B:
Composition of a Function
Let f:A→B and g:B→C be two function such that range of f is a subset of domain of g, then a new function gof:A→C, called the composition of f and g may be defined as:
gof(a)=g(f(a))∀a∈A
Figure 6: Relation map for composition of a function between A, B, and C:
For fog:
f(g(a))=f(c)=a
f(g(b))=f(b)=b
f(g(c))=f(c)=c
For gof:
g(f(a))=g(c)=a
g(f(b))=g(b)=b
g(f(c))=g(a)=c
Example 3: Let f:R→R be a function defined by f(x)=x3, and g:R→R be a function defined by g(x)=x−1. Compute fog and gof and find if they are equal or not.
Solution:
fog=f(g(x))=f(x−1)=(x−1)3=x3−3x2+3x−1
and,
gof=g(f(x))=g(x3)=x3−1
It is found that:
fog=gof
Example 4: Let f(x)=3x+1, and g(x)=x2. Find out fog, and gof.
Solution:
fog=f(g(x))=f(x2)=3x2+1
and,
gof=g(f(x))=g(3x+1)=(3x+1)2=9x2+6x+1
Example 5: Let f(x)=3x2. Find fof.
Solution:
fof=f(f(x))=f(3x2)=3(3x2)2=27x4
Example 6: Let f(x)=x, g(x)=2x, and h(x)=3x. Find fogoh, when x=−1.