Page 193 - DMTH201_Basic Mathematics-1

Basic Mathematics – I

Notes “B”. This is the requirement of function “g” by definition. It follows, then, that for every element
“x” in “A”, there exists an element g(f(x)) in set “C”. This concluding statement is definition of
a new function:
h : A C by g(f(x)) for all x A
By convention, we call this new function as “g o f” and is read as “g circle f” or “g composed with
f”. g o f(x) = g(f(x)) for all x A
The two symbolical representations are equivalent.

Example: Let two sets be defined as :
h : R Rby x for all x R
2
k : R Rby x + 1 for all x R
Determine “h o k” and “k o h”.

Solution:
According to definition,
h o k(x) = h(k(x))
h o k(x) = h(x + 1)

h o k(x) = (x + 1) 2
Again, according to definition,
k o h(x) = k(h(x))
2
k o h(x) = k(x )
k o h(x) = (x + 1)
2
Importantly note that h o k(x) k o h(x). It indicates that composition of functions is not
commutative.

6.5.2 Existence of Composition Set

In accordance with the definition of function, “f”, the range of “f” is a subset of its co-domain “B”.
But, set “B” is the domain of function “g” such that there exists image g(f(x)) in “C” for every “x”
in “A”. This means that range of “f” is subset of domain of “g”:
Range of “f” Domain of “g”.
Clearly, if this condition is met, then composition “g o f” exists. Following this conclusion,
“f o g” will exist, if
Range of “g” Domain of “f”
And, if both conditions are met simultaneously, then we can conclude that both “g o f” and “f o
g” exist. Such possibility is generally met when all sets involved are set of real numbers, “R”.

Example: Let two functions be defined as:

f = {(1,2),(2,3),(3,4),(4,5)}
g = {(2,4),(3,2),(4,3),(5,1)}
Check whether “g o f” and “f o g” exit for the given functions.

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