15 Composite Functions and Domain

Domain Review

This format set-builder notation: {x|10 ≤ x < 30} describes the behavior of x in set-builder notation. The braces { } are read as “the set of,” and the vertical bar | is read as “such that,” so we would read {x | 10 ≤ x < 30} as “the set of x-values such that 10 is less than or equal to x, and x is less than 30.”

Previously we looked at finding domain of basic functions. But what about those cases where you have a function that is in the form of a fraction or has a square root?

Combining Functions Using Algebraic Operations

Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function.

Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If w(y) is the wife’s income and h(y) is the husband’s income in year y, and we want T to represent the total income, then we can define a new function.

If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write

Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.

For two functions f(x) and g(x) with real number outputs, we define new functions f + g, f − g, f g, and by the relations

Types	Algebraic
Sum	(f + g)(x) = f(x) + g(x)
Difference	(f – g) (x) = f(x) – g(x)
Product	(f g)(x) = f(x)g(x)
Quotient	= where g(x) ≠ 0

Find and Evaluate Composite Functions

Before we introduce the functions, we need to look at another operation on functions called composition. In composition, the output of one function is the input of a second function. For functions fand g, the composition is written f ∘ g and is defined by:

We read the left-hand side as “f composed with g at x,” and the right-hand side as “f of g of x.” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol ∘ is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases f(g(x)) ≠ f(x)g(x).

To do a composition, the output of the first function, g(x), becomes the input of the second function, f, and so we must be sure that it is part of the domain of f.

It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function g takes the input x first and yields an output g(x). Then the function f takes g(x) as an input and yields an output f(g(x)).

In general, f ∘ g and g∘ f are different functions. In other words, in many cases f(g(x)) ≠ g(f(x)) for all x. We will also see that sometimes two functions can be composed only in one specific order.

For example, if f(x) = x² and g(x) = x + 2, then

but

These expressions are not equal for all values of x, so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value x = −.

Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.

composite functionThe next example will demonstrate that (f∘ g)(x), (g∘ f)(x) and (f · g)(x) usually result in different outputs.

In the next example we will evaluate a composition for a specific value.

Finding the Domain of a Composite Function

As we discussed previously, the domain of a composite function such as f∘ g is dependent on the domain of g and the domain of f. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as f∘ g. Let us assume we know the domains of the functions f and g separately. If we write the composite function for an input x as f(g(x)), we can see right away that x must be a member of the domain of g in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that g(x) must be a member of the domain of f, otherwise the second function evaluation in f(g(x)) cannot be completed, and the expression is still undefined. Thus the domain of f∘ g consists of only those inputs in the domain of g that produce outputs from g belonging to the domain of f. Note that the domain of f composed with g is the set of all x such that x is in the domain of g and g(x) is in the domain of f.

Try it! – Finding the Domain of a Composite Function

Find the domain of

(f∘ g)(x) where and

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Solution:

The domain of g(x) consists of all real numbers except x = , since that input value would cause us to divide by 0. Likewise, the domain of f consists of all real numbers except 1. So we need to exclude from the domain of g(x) that value of x for which g(x) = 1.

= 1

4 = 3x – 2

6 = 3x

x = 2

So the domain of f∘ g is the set of all real numbers except and 2. This means that

x ≠ or x ≠ 2

We can write this in interval notation as

(−∞, ) ∪ (, 2) ∪ (2, ∞)

Try it! – Finding the Domain of a Composite Function Involving Radicals

Find the domain of

(f∘ g)(x) where f(x) = and g(x) =

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Solution:

Because we cannot take the square root of a negative number, the domain of g is (−∞, 3].Now we check the domain of the composite function

(f∘ g)(x) =

For (f∘ g)(x) = , ≥ 0, since the radicand of a square root must be positive. Since square roots are positive, ≥ 0, or, ≥ 0, which gives a domain of (-∞, 3]

Analysis:

This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of f∘ g can contain values that are not in the domain of f, though they must be in the domain of g.

3. Find the domain of

(f∘ g)(x) where f(x)= and g(x) =

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Solution:

[−4, 0) ∪ (0, ∞)

Decomposing a Composite Function into its Component Functions

In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient.

Decomposing a Function

Write f(x)= as the composition of two functions.

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Solution:

We are looking for two functions, g and h,so f(x) = g(h(x)). To do this, we look for a function inside a function in the formula for f(x).As one possibility, we might notice that the expression 5-x² is the inside of the square root. We could then decompose the function as

h(x) = 5 − x² and g(x) =

We can check our answer by recomposing the functions.

g(h(x)) = g(5−x²) =

2. Write f(x) =  as the composition of two functions.

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Solution:

Possible answer:

g(x) =

h(x) =  

f = h∘ g