ON INVERSE FUNCTIONS

Composition of Functions

Suppose the rule of function f(x) is and the rule of function g(x) is . Suppose now that you want to "leapfrog" the functions as follows: Take a 2 in the domain of f and link it to 9 with the f(x) rule, and then take the 9 and link it to 157 with the g(x) rule. This is a lot of work and you would rather just work with one function, a function that would link the 2 directly to the 157.

Since the g function operates on f(x), we can write the composition as g(f(x)). Letís call the new function h(x) = g(f(x)). You can simplify from the inside out or the outside in.

Inside Out:

Let's check to see if the above function will link 2 directly to 157.

It does.

Outside In:

You can see that it is the same as the function we derived from the inside out.

The following is an example of finding the composition of three functions.

Example 5: Given three functions , , and .

Find the composite function f(g(h(x))).

What is the domain of the composite function f(g(h(x)))?

Since all the functions are polynomials, the domain of each function and the composite function is the set of real numbers.

Letís check our answer with a number in the domain, say x = 1:

The function h(x) links the 1 to 0, the function g(x) links the 0 to - 5, and the function f(x) links the - 5 to 0. Letís see if the new function will take us directly from 1 to 0.

=

It does and our composite function is correct.

Go the beginning of this review on composite functions.

Review inverse functions where there is no restriction on the domain.

Review inverse functions where there is a restriction on the domain.

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Author:Nancy Marcus

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