Composition of Functions in Precalculus
These are my notes on composition of functions in precalculus.
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 all 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.
\(T(y)=h(y)+w(y)\)
If this holds true for every year, then we can focus on the relation between
the functions without reference to a year and write:
\(T=h+w\)
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 kind 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, fg, and f/g by the relations:
\((f+g)(x)=f(x)+g(x)\)
\((f-g)(x)=f(x)-g(x)\)
\((fg)(x)=f(x)g(x)\)
\((f/g)(x)=f(x)/g(x)\)
Example 1
Performing Algebraic Operations on Functions
Find and simplify the functions \((g-f)(x)\) and \((g/f)(x)\), given
\(f(x)=x-1\) and
\(g(x)=x^2-1\). Are they the same function?
Solution:
begin by writing the general form, and then substitute the given functions.
\((g-f)(x) = g(x)-f(x)\)
\((g-f)(x) = x^2-1-(x-1)\)
\((g-f)(x) = x^2-x\)
\((g-f)(x) = x(x-1)\)
\((g/f)(x) = \frac{g(x)}{f(x)}\)
\((g/f)(x) = \frac{x^2-1}{x-1}\)
\((g/f)(x) = \frac{(x+1)(x-1)}{x-1}\)
\((g/f)(x) = x+1\)
No, the functions are not the same.
Note, for (g/f)(x), the condition \(x \not = 1\) is necessary because when
x=1, the denominator is equal to 0, which makes the function undefined.
Create a Function by Composition of Functions
Performing algebraic operations on functions combines them into a new function,
but we can also create functions by composing functions. When we wanted
to compute a heating cost from a day of the year, we created a new function
that takes a day as input and yields a cost as output. The process of combining
functions so that the output of one function becomes the input of another is
known as the composition of functions. The resulting function is known as a
composition function. We represent this combination by the following notation:
\(f \circ g)(x)=f(g(x))\)
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 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
\(f(g(x))\not=f(x)g(x)\).
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)).
g(x), the output of g is the input of f
In general, \(f \circ g\) and \(g \circ f\) are different functions. in other
words, in many cases, \(f(g(x)) \not= g(f(x))\) for all x. We will also see that
sometimes two functions can be composed only in one specific order.
For example, if \(f(x)=x^2\) and \(g(x)=x+2\), then:
\(f(g(x))=f(x+2) = (x+2)^2 = x^2+4x+4\)
but:
\(g(f(x))=g(x^2) = x^2+2\)
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 \(x=1/2\).
Note that the range of the inside function (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.
Composition of Functions
When the output of one function is used as the input of another function, we
call the entire operation a composition of functions. For any input x and
functions f and g, this action defines a composite function, which we write as
\(f \circ g\) such that \((f \circ g)(x)=f(g(x))\).
The domain of the composite function \(f \circ g\) is all x such that x is in
the domain of g and g(x) is in the domain of f. It is important to realize that
the product of functions fg is not the same as the function composition f(g(x)),
because , in general, \(f(x)g(x)\not=f(g(x))\).
Example 2
Determining Whether Compositions of Functions is Commutative
Using the functions provided, find f(g(x)) and g(f(x)). Determine whether the
composition of the functions is commutative.
\(f(x)=2x+1\) and \(g(x)=3-x\)
Solution:
Let's begin by substituting g(x) into f(x).
\(f(g(x))=2(3-x)+1 = 6-2x+1 = 7-2x\)
Now, we can substitute f(x) into g(x).
\(g(f(x))=3-(2x+1) = 3-2x-1 = -2x+2\)
We find that \(g(f(x)) \not= f(g(x))\), so the operation of function composition
is not commutative.
Example 3
Interpreting Composite Functions
The function c(s) gives the number of calories burned completing s sit-ups, and
s(t) gives the number of sit-ups a person can complete in t minutes. Interpret
c(s(3)).
Solution:
The inside expression in the composition is s(3). Because the input to the
s-function is time, t=3 represents 3 minutes, and s(3) is the number of sit-ups
completed in 3 minutes.
Using s(3) as the input to the function c(s) gives us the number of calories
burned during the number of sit-ups that can be completed in 3 minutes, or
simply the number of calories burned in 3 minutes by doing sit-ups.
Example 4
Investigating the Order of Function Composition
Suppose f(x) gives miles that can be driven in x hours and g(y) gives the
gallons of gas used in driving y miles. Which of the expressions is meaningful,
f(g(y)) or g(f(x))?
Solution:
The function \(y=f(x)\) is a function whose output is the number of miles driven
corresponding to the number of hours driven.
number of miles=f(number of hours)
The function g(y) is a function whose output is the number of gallons used
corresponding to the number of miles driven. This means:
number of gallons = g(number of miles)
The expression g(y) takes miles as the input and a number of gallons as the
output. The function f(x) requires a number of hours as the input. Trying to
input a number of gallons does not make sense. The expression f(g(y)) is
meaningless.
The expression f(x) takes hours as the input and a number of miles driven as the
output. The function g(y) requires a number of miles as the input. Using
f(x)(miles driven) as an input value for g(y), where gallons of gas depends on
miles driven, does make sense. The expression g(f(x)) makes sense, and will
yield the number of gallons of gas used, g, driving a certain number of miles,
f(x), in x hours.
Are there any situations where f(g(y)) and g(f(x)) would both be meaningful or
useful expressions?
Yes. For many pure mathematical functions, both compositions make sense, even
though they usually produce different new functions. In real-world problems,
functions whose inputs and outputs have the same units also may give
compositions that are meaningful in either order.
Evaluating Composite Functions
Once we compose a new function from two existing functions, we need to be able
to evaluate it for any input in its domain. We will do this with specific
numerical inputs for functions expressed as tables, graphs, and formulas and
with variables as inputs to functions expressed as formulas. In each case, we
evaluate the inner function using the starting input and then use the inner
function's output as the input for the outer function.
When working with functions given as tables, we read input and output values
from the table entries and always work from the inside to the outside. We
evaluate the inside function first and then use the output as of the inside
function as the input to the outside function.
Example 5
Using a Table to Evaluate a Composite Function
Evaluate f(g(3)) and g(f(3)).
x f(x) g(x)
1 6 3
2 8 5
3 3 2
4 1 7
Solution:
To evaluate f(g(3)), we start from the inside with the input value 3. We then
evaluate the inside expression g(3) using the table that defines the function
g:g(3)=2. We can then use that result as the input to the function f, sp g(3) is
replaced by 2 and we get f(2). Then using the table that defines the function f,
we find that f(2)=8.
\(g(3)=2\)
\(f(g(3))=f(2)=8\)
To evaluate g(f(3)), we first evaluate the inside expression f(3) using the
first table: f(3)=3. Then, using the table for g, we can evaluate:
\(g(f(3))=g(3)-2\)
Evaluating Composite Functions Using Graphs
When we are given individual functions as graphs, the procedure for evaluating
composite functions is similar to the process we use for evaluating tables. We
read the input and output values, but this time, from the x and y axes of the
graphs.
Given a composite function and graphs of its individual functions, evaluate it
using the information provided by the graphs.
1. Locate the given input to the inner function on the x-axis of its graph
1. Read off the input of the inner function from the y-axis of its graph
3. Locate the inner function output on the x-axis of the graph of the outer
function.
4. Read the output of the outer function from the y-axis of its graph. This is
the output of the composite function.
Example 6
Using a Graph to Evaluate a Composite Function
Evaluate f(g(1))
Solution:
We evaluate g(1) using the graph of g(x), finding the input of 1 on the x-axis
and finding the output value of the graph at that input. Here, g(1)=3. We
use this value as the input to the function f.
\(f(g(1))=f(3)\)
We can then evaluate the composite function by looking to the graph of f(x),
finding the input of 3 on the x-axis and reading the output value of the graph
at this input. Here, \(f(3)=6, \text{ so } f(g(1))=6\).
Evaluating Composite Functions Using Formulas
When evaluating a composite function where we have either created or been
given formulas, the rule of working from the inside out remains the same. The
input value to the outer function will be the output of the inner function,
which may be a numerical value, a variable name, or a more complicated
expression.
While we can compose the functions for each individual input value, it is
sometimes helpful to find a single formula that will calculate the result of a
composition \(f9g(x))\). To do this, we will extend our idea of the function
evaluation. Recall that, when we evaluate a function like \(f(t)=t^2-t\), we
substitute the value inside the parentheses into the formula wherever we see
the input variable.
Given a formula for a composite function, evaluate the function.
1. Evaluate the inside out function using the input value or variable
provided.
2. use the resulting output as the input to the outside function.
Example 7
Evaluating a Composition of Functions Expressed as Formulas with a numerical
Input
Given \(f(t)=t^2-t\) and \(h(x)=3x+2\), evaluate f(h(1)).
Solution:
Because the inside expression is h(1), we start by evaluating h(x) at 1.
\(h(1)=3(1)+2 = 5\)
Then \(f(h(1))=f(5)\), so we evaluate f9t) at an input of 5.
\(f(h(1))= f(5) = 5^2-5 = 20\)
It makes no difference what the input variables t and x were called because we
evaluated for specific numerical values.
Finding the Domain of a Composite Function
As we discussed previously, the domain of a composite function such as \(f
\circ g\) 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 \(f \circ g\). 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 can see right away that 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 \(f \circ
g\) 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 f composed with g is the
set of all x such that x is in the domain of g and g9x) is in the domain of
f.
Domain of a Composite Function
The domain of a composite function f(g(x)) is the set of those inputs x in the
domain of g for which g(x) is in the domain of f.
Given a function composition f(g(x)), determine its domain
1. Find the domain of g
2. Find the domain of f
3. Find those inputs x in the domain of g for which g(x) is in the domain of
f. That is, exclude those inputs x from the domain of g for which g(x) is not
in the domain of f. The resulting set is the domain of \(f \circ g\).
Example 8
Finding the Domain of a Composite Function
Find the domain of \((f \circ g)(x)\) where \(f(x)=5/(x-1) \text{ and } g(x) =
4/(3x-2)\).
Solution:
The domain of g(x) consists of all real numbers except \(x=2/3\), 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 g9x)=1.
\(\frac{4}{3x-2} = 1\)
\(4=3x-2\)
\(6=3x = 2\)
So the domain of \(f \circ g\) is the set of all real numbers except 2/3 and 2.
This means that :
\(x \not= \frac{2}{3} \text{ or } x\not=2\)
We can write this in interval notation as:
\((-\infty,\frac{2}{3}) \cup (\frac{2}{3},2) \cup (2,\infty)\)
Example 9
Finding the Domain of a Composite Function Involving Radicals
Find the domain of :
\((f \circ g)(x) \text{ where } f(x) = \sqrt{x+2} \text{ and } g(x) =
\sqrt{3-x}\)
Solution:
Because we cannot take the square root of a negative number, the domain of g
is \(-\infty,3]. Now we can check the domain of the composite function:
\((f \circ g)(x) = \sqrt{\sqrt{3-x}+2}\)
This expression must \(\geq 0\), since the radicand of a square root must be
positive. Since square roots are positive, \(\sqrt{3-x} \geq 0\) which gives
the domain of \((-\infty,3].
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 \(f \circ g\) can contain values
that are not in the domain of f, though they must be in the domain of g.
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.
Example 10
Decomposing a Function
Write \(f(x)=\sqrt{5-x^2}\) as the composition of two functions.
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 \(5-x^2\) is the inside of
the square root. We could then decompose the function as \(h(x)=5-x^2 \text{
and } g(x) = \sqrt{x}\)
We can check out answer by recomposing the functions.
\(g(h(x))=g(5-x^2)=\sqrt{5-x^2}\)