Composition of Functions
We can build new functions from old ones by composition.
Given functions f and g with domains and target sets consisting of real numbers, the composite of g and f is the new function g o f defined as follows.
(g o f)(x) = g(f(x))
We must be a little careful in determining the domain of the composite g o f: a real number x lies in the domain of g o f if and only if it is in the domain of f AND the value f(x) is in the domain of g. We sometimes refer to f as the "inner"function of the composite g o f; the function g is the "outer" function in this case. The composite function accepts a domain value, applies the inner function to obtain a new value, and then applies the outer function to this new value.
Example. If f(x) = x+1 and g(x) = 3x, then
g(f(x)) = g(x+1) = 3(x+1) = 3x+3.
In this case, the inner function g is the one that adds one; the outer function f simply multiplies by 3. We can also form the composition in the other order, exchanging the roles of the inner and outer function. In this case we compute
f(g(x)) = f(3x) = 3x+1.
Choosing a special domain value of x=2, notice that f(g(2)) = 3(2)+1 = 7 and that g(f(2)) = 3(2)+3 = 9. Thus the composites f o g and g o f are sometimes different. Order makes a difference in composition; in fact it is very rare to find two functions f and g for which f o g and g o f agree.
In the previous example, we illustrated the process of forming composites of given functions. An essential skill for us in calculus it to be able to go the other way:
|Given a function H, find functions f and g such that
H(x) = g(f(x)).
Sample Problems. Decompose each of the following functions as a composite.
- H(x) = (2x+1)3
- J(x) = 2x3 + 1
- K(x) = sqrt(6x-5)
- L(x) = sin(7x2-4)
You should write some solutions to these exercises on paper and then check to see if your answers agree with ours.
More sample problems of this sort are to be found in the Practice area. There are many possible answers to some of these questions. The "best" answers will be the ones that help you differentiate the given functions H, J, K, and L. For more about differentiation of composite functions, read on!!
A composite of differentiable functions is differentiable. Our next general differentiation rule is the chain rule; it shows us how to differentiate a composite of differentiable funcitons.
Why is THAT true?
Notice that the roles of f and g are NOT symmetric in the chain rule formula---it is essential for us to identify the "inside" and the "outside" function before we apply the chain rule.
Looking more closely at the formula, notice that the chain rule involves multiplication of a derivative value for the outside function times a derivative value for the inside function. However, when we differentiate the composite function y=g(f(x)) at x, we multiply the derivative value of the inside function at x times the derivative value of the outside function at f(x)....NOT at x!!!
On the next page, we apply the chain rule to differentiate the functions from our sample decomposition exercises above.
All rights reserved---1996
William A. Bogley