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Function is a relation on two sets by a rule. It is a special mapping between two sets. It emerges that it is possible to combine two functions, provided co-domain of one function is domain of another function. The composite function is a relation by a new rule between sets, which are not common to the functions.
We can understand composition in terms of two functions. Let there be two functions defined as :
$$f:A\to B\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}\text{f(x) for all}\phantom{\rule{1em}{0ex}}x\in A$$
$$g:B\to C\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}\text{g(x) for all}\phantom{\rule{1em}{0ex}}x\in B$$
Observe that set “B” is common to two functions. The rules of the functions are given by “f(x)” and “g(x)” respectively. Our objective here is to define a new function $h:A\to C$ and its rule.
Thinking in terms of relation, “A” and “B” are the domain and co-domain of the function “f”. It means that every element “x” of “A” has an image “f(x)” in “B”.
Similarly, thinking in terms of relation, “B” and “C” are the domain and co-domain of the function “g”. In this function, "f(x)" – which was the image of pre-image “x” in “A” – is now pre-image for the function “g”. There is a corresponding unique image in set "C". Following the symbolic notation, "f(x)" has image denoted by "g(f(x))" in "C". The figure here depicts the relationship among three sets via two functions (relations) and the combination function.
For every element, “x” in “A”, there exists an element f(x) in set “B”. This is the requirement of function “f” by definition. For every element “f(x)” in “B”, there exists an element g(f(x)) in set “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 :
$$h:A\to C\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}\text{g(f(x)) for all}\phantom{\rule{1em}{0ex}}x\in A$$
By convention, we call this new function as “gof” and is read as "g circle f" or "g composed with f".
$$gof\left(x\right)=g\left(f\left(x\right)\right)\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in A$$
The two symbolical representations are equivalent.
Problem 1: Let two sets be defined as :
$$h:R\to R\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}{x}^{2}\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in R$$
$$k:R\to R\phantom{\rule{1em}{0ex}}\text{by}\phantom{\rule{1em}{0ex}}x+1\phantom{\rule{1em}{0ex}}\text{for all}\phantom{\rule{1em}{0ex}}x\in R$$
Determine “hok” and “koh”.
Solution : According to definition,
$$hok\left(x\right)=h\left(k\left(x\right)\right)$$
$$\Rightarrow hok\left(x\right)=h\left(x+1\right)$$
$$\Rightarrow hok\left(x\right)={\left(x+1\right)}^{2}$$
Again, according to definition,
$$koh\left(x\right)=k\left(h\left(x\right)\right)$$
$$\Rightarrow koh\left(x\right)=k\left({x}^{2}\right)$$
$$\Rightarrow koh\left(x\right)=\left({x}^{2}+1\right)$$
Importantly note that $hok\left(x\right)\ne koh\left(x\right)$ . It indicates that composition of functions is not commutative.
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” :
$$\text{Range of \u201cf\u201d}\subset \text{Domain of \u201cg\u201d}$$
Clearly, if this condition is met, then composition “gof” exists. Following this conclusion, “fog” will exist, if
$$\text{Range of \u201cg\u201d}\subset \text{Domain of \u201cf\u201d}$$
And, if both conditions are met simultaneously, then we can conclude that both “gof” and “fog” exist. Such possibility is generally met when all sets involved are set of real numbers, “R”.
Problem 2: Let two functions be defined as :
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