Using two given functions, we can create a third function, called the composition.
If \(f\) and \(g\) are the names of the two component functions, then \(f \circ g\) is the symbol for the composition of \(f\) and \(g\).
We compose \(f\) and \(g\) to form \((f \circ g)\).
The order makes a difference.
We compose \(g\) and \(f\) to form \((g \circ f)\).
Most of time these are not equal functions.
Let \(F\) and \(G\) be two functions.
The composition of \(F\) and \(G\) is another function called the composition of \(F\) and \(G\). A circle between the function names denotes a composition.
The composition applies \(G\) to an element of the domain of \(G\) and then uses the resulting function value from \(G\) as a domain member of \(F\) obtaining a function value of \(F\).
This ultimate value of \(F\) resulting from this process is the value of \(F \circ G\).
Composition is our way of representing an ordered application of processes.
Our first version of algebraic representations for functions is formulas (there will be other versions).
Each component function has its own formula, using its own variable.
The composition is evaluated by first evaluating the inner function at the domain number, and then evaluating the outer function at the value of the inner function.
Two Component Functions:
Let \(H\) and \(K\) be two functions defined by the following formulas.
From the formula for \(H\), we can see that \(H(0) = -5\).
From the formula for \(K\), we can see that \(K(-5) = 6\).
These two values tell us that \((K \circ H)(0) = 6\)
From the formula for \(K\), we can see that \(K(2) = -1\).
From the formula for \(H\), we can see that \(H(-1) = -3\).
These two values tell us that \((H \circ K)(2) = -3\)
From the formula for \(H\), we can see that \(H(-3) = 13\).
From the formula for \(K\), we can see that \(K(13) = -12\).
These two values tell us that \((K \circ H)(-3) = -12\)
From the formula for \(K\), we can see that \(K(-9) = 10\).
From the formula for \(H\), we can see that \(H(10)\) is undefined.
We cannot include \(10\) in the domain of \((H \circ K)\).
Let \(a\) and \(b\) be two functions defined by the following formulas.
From the formula for \(b\), we can see that \(b(1) = \answer {-1}\).
From the formula for \(a\), we can see that \(a(-1) = \answer {1}\).
These two values tell us that \((a \circ b)(1) = \answer {1}\)
Let \(a\) and \(b\) be two functions defined by the following formulas.
From the formula for \(a\), we can see that \(a(3) = \answer {-3}\).
From the formula for \(b\), we can see that \(b(-3) = \answer {-9}\).
These two values tell us that \((b \circ a)(3) = \answer {-9}\)
Let \(a\) and \(b\) be two functions defined by the following formulas.
Is \(4\) in the domain of \((b \circ a)\)?
Is \(4\) in the domain of \((a \circ b)\)?
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more examples can be found by following this link
More Examples of Formulas