1.3: The Arithmetic and Composition of Functions
- Refresher Topics
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Simplifying Expressions
- Multiplying and Dividing Rational Expressions
- Adding and Subtracting Rational Expressions
- Simplifying Compound Rational Expressions
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Factoring Techniques
- Factoring Difference of Squares
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Solving Equations
- Solving Rational Equations (involving a single rational expression)
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Simplifying Expressions
In this section, you will:
- Perform arithmetic on functions
- Combine functions using algebraic operations.
- Create a new function by composition of functions.
- Evaluate composite functions.
- Find the domain of a composite function.
- Decompose a composite function into its component functions.
- Compute the Difference Quotient for a given function.
Suppose we want to calculate how much it costs to heat a house on a particular day of the year. This cost will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.
Using descriptive variables, we can notate these two functions. The function \( C ( T ) \) gives the cost \( C \) of heating a house for a given average daily temperature in \( T \) degrees Celsius. The function \( T ( d ) \) gives the average daily temperature on day \( d \) of the year. For any given day, \( \text{Cost\} = C ( T ( d ) ) \) means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature \( T ( d ) \). For example, we could evaluate \( T ( 5 ) \) to determine the average daily temperature on the 5 th day of the year. Then, we could evaluate the cost function at that temperature. We would write \( C ( T ( 5 ) ) \).
By combining these two relationships into one function, we have performed function composition, which is the focus of this section.
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.\[ T ( y ) = h ( y ) + w ( y ). \nonumber \]If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write\[ T = h + w. \nonumber \]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 \( \frac{f}{g} \) by the relations\[ \begin{array}{rcl}
( f + g ) ( x ) & = & f ( x ) + g ( x ) \\[6pt]
( f - g ) ( x ) & = & f ( x ) - g ( x ) \\[6pt]
( f g ) ( x ) & = & f ( x ) g ( x ) \\[6pt]
\left( \dfrac{f}{g} \right) ( x ) & = & \dfrac{f ( x )}{g ( x )} \\[6pt]
\end{array} \nonumber \]
Find and simplify the functions \( ( g − f ) ( x ) \) and \( \left( \frac{g}{f}\right) ( x ) \), given \( f ( x ) = x − 1 \) and \( g ( x ) = x^2 − 1 \). Are they the same function?
- Solution
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Begin by writing the general form, and then substitute the given functions. For \( (g - f)(x) \), we have\[ \begin{array}{rclcl}
(g - f)(x) & = & g(x) - f(x) & & \\[6pt]
& = & x^2 - 1 - (x - 1) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & x^2 - 1 - x + 1 & \quad & \left( \text{distributing} \right) \\[6pt]
& = & x^2 - x & \quad & \left( \text{combining like terms} \right) \\[6pt]
& = & x(x - 1) & \quad & \left( \text{factoring out the GCF} \right) \\[6pt]
\end{array} \nonumber \]For \( \left( \frac{g}{f} \right)(x) \), we have\[ \begin{array}{rclcl}
\left( \dfrac{g}{f} \right)(x) & = & \dfrac{g(x)}{f(x)} & & \\[6pt]
& = & \dfrac{x^2 - 1}{x - 1} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{(x - 1)(x + 1)}{x - 1} & \quad & \left( \text{factoring} \right) \\[6pt]
& = & \dfrac{\cancelto{1}{(x - 1)}(x + 1)}{\cancelto{1}{(x - 1)}} & \quad & \left( \text{canceling like terms and }\textbf{assuming }x \neq 1 \right) \\[6pt]
& = & x + 1 & \quad & \left( \text{where }x \neq 1 \right) \\[6pt]
\end{array} \nonumber \]Therefore, the functions are not the same.
In Example \( \PageIndex{ 1 } \), for \( \left( \frac{g}{f} \right) ( x ) \), the condition \( x \neq 1 \) is necessary because when \( x = 1\), the denominator is equal to 0, which makes the function undefined.
Find and simplify the functions \( ( f g ) ( x ) \) and \( ( f − g ) ( x ) \) given\[ f ( x ) = x − 1 \text{ and } g ( x ) = x^2 − 1.\nonumber \]Are they the same function?
Focusing on Calculus - Difference Quotients
Now that we know how to perform arithmetic with functions, it's time to turn our attention to the primary building block of a huge concept in Calculus. This primary building block is called the difference quotient .
Given a function \(f\), the difference quotient of \(f\) is the expression\[\dfrac{f(x+h)-f(x)}{h}. \label{dq} \]
You might see this written online and in other texts as\[ \dfrac{f(a + h) - f(a)}{h}, \nonumber \]but fear not - this is the same expression (just with \( a \) instead of \( x \)).
We will revisit this concept in more detail in a future section, but for now we use it as a way to practice function notation and function arithmetic. For reasons that will become clear in Calculus, "simplifying" a difference quotient means rewriting it in a form where the \(h\) in Equation \( \ref{dq} \) cancels from the denominator. Once that happens, we consider our work to be done.
Find and simplify the difference quotient for the following function.\[f(x)=x^{2}-x-2\nonumber \]
- Solution
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To find \(f(x+h)\), we replace every occurrence of \(x\) in the formula \(f(x)=x^{2}-x-2\) with the quantity \((x + h)\) to get\[ \begin{array}{rclcl}
f(x+h) & = & (x+h)^{2}-(x+h)-2 & \quad & \left( \text{substituting} \right)\\[6pt]
& = & x^{2}+2 x h+h^{2}-x-h-2 & \quad & \left( \text{distributing} \right) \\[6pt]
\end{array} \nonumber \]So the difference quotient is\[ \begin{array}{rclcl}
\dfrac{f(x+h)-f(x)}{h} & = & \dfrac{\left(x^{2}+2 x h+h^{2}-x-h-2\right)-\left(x^{2}-x-2\right)}{h} & & \\[6pt]
& = & \dfrac{x^{2}+2 x h+h^{2}-x-h-2-x^{2}+x+2}{h} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{2 x h+h^{2}-h}{h} & \quad & \left( \text{combining like terms} \right) \\[6pt]
& = & \dfrac{h(2 x+h-1)}{h} & \quad & \left( \text{factoring out the GCF} \right)\\[6pt]
& = & \dfrac{\cancel{h}(2 x+h-1)}{\cancel{h}} & \quad & \left( \text{canceling like factors} \right)\\[6pt]
& = & 2x + h - 1. \\[6pt]
\end{array} \nonumber \]We know we are done once the original \( h \) in the denominator is canceled out.
A word of caution before we continue: watch your negatives! Notice that at some point in the solution of Example \( \PageIndex{ 2 } \) we had the expression \( \left(x^{2}+2 x h+h^{2}-x-h-2\right)-\left(x^{2}-x-2\right) \) in the numerator. Those parentheses on the second group are critically important and necessary. We are subtracting
the entire function
\( f(x) \). As such, we needed \( f(x + h) - f(x) = f(x + h) - \left( x^2 - x - 2 \right) \). If we didn't have those parentheses, we would (incorrectly) have written \( f(x + h) - f(x) = f(x + h) - x^2 - x - 2 \), which I hope you would agree is not the same thing.
Let's state that one more time for clarity -
the biggest mistake made on a difference quotient problem is not using grouping symbols properly!
You have been warned.
Find and simplify the difference quotient for \( g(x) = 1 - 3x \).
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 a composition of functions . The resulting function is known as a composite function . We represent this combination by the following notation:\[ ( f \circ g ) ( x ) = f ( g ( x ) ).\nonumber \]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 \( \circ \) 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 take two numbers and give a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases \( f ( g ( x ) ) \neq f ( x ) g ( x )\).
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 \( f ( g ( x ) ) \).
In general, \( f \circ g \) and \( g \circ f \) are different functions. In other words, in many cases \( f ( g ( x ) ) \neq g ( f ( x ) ) \) 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\[ \begin{array}{rclcl}
f(g(x)) & = & f(x + 2) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & (x + 2)^2 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & x^2 + 4x + 4 & \quad & \left( \text{distributing} \right) \\[6pt]
\end{array} \nonumber \]but\[ \begin{array}{rclcl}
g(f(x)) & = & g(x^2) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & x^2 + 2 & \quad & \left( \text{substituting} \right) \\[6pt]
\end{array} \nonumber \]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 \( x = −\frac{1}{2} \).
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.
When the output of one function is used as the input of another, 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\[ ( f \circ g ) ( x ) = f ( g ( x ) ). \nonumber \]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 \( f g \) is not the same as the function composition \( f ( g ( x ) ) \), because, in general, \( f ( x ) g ( x ) \neq f ( g ( x ) ) \).
Using the functions provided, find \( f ( g ( x ) ) \) and \( g ( f ( x ) ) \). Determine whether the composition of the functions is commutative .\[ f ( x ) = 2 x + 1, \quad g ( x ) = 3 − x. \nonumber \]
- Solution
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Let's begin by substituting \( g(x) \) into \( f ( x ) \).\[ \begin{array}{rclcl}
f(g(x)) & = & f(3 - x) + 1 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 2(3 - x) + 1 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 6 - 2x + 1 & \quad & \left( \text{distributing} \right) \\[6pt]
& = & 7 - 2x & \quad & \left( \text{combining like terms} \right) \\[6pt]
\end{array} \nonumber \]Now we can substitute \( f(x) \) into \( g ( x ) \).\[ \begin{array}{rclcl}
g(f(x)) & = & g(2x + 1) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 3 - (2x + 1) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 3 - 2x - 1 & \quad & \left( \text{distributing} \right) \\[6pt]
& = & 2 - 2x - 1 & \quad & \left( \text{combining like terms} \right) \\[6pt]
\end{array} \nonumber \]We find that \( g ( f ( x ) ) \neq f ( g ( x ) ) \), so the operation of function composition is not commutative.
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
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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).
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 these expressions is meaningful: \( f ( g ( y ) ) \) or \( g ( f ( x ) ) \)?
- Solution
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The function \( y = f ( x ) \) is a function whose output is the number of miles driven corresponding to the number of hours driven.\[ \text{number of miles} = f ( \text{number of hours} ). \nonumber \]The function \( g(y) \) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means:\[ \text{number of gallons} = g ( \text{number of miles} ). \nonumber \]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 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.
The gravitational force on a planet a distance \( r \) from the sun is given by the function \( G ( r ) \). The acceleration of a planet subjected to any force \( F \) is given by the function \( a ( F ) \). Form a meaningful composition of these two functions, and explain what it means.
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.
Evaluating Composite Functions Using Tables
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 of the inside function as the input to the outside function.
Using the following table, 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
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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 \( f \), so \( g ( 3 ) \) is replaced by 2 and we get \( f ( 2 ) \). Then, using the table that defines the function \( f \), we find that \( f ( 2 ) = 8 \).\[ \begin{array}{rrcl}
& g(3) & = & 2 \\[6pt]
\implies & f(g(3)) & = & f(2) = 8 \\[6pt]
\end{array} \nonumber \]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\[ g(f(3)) = g(3) = 2. \nonumber \]The following table shows the composite functions \( f \circ g \) and \( g \circ f \) as tables.\( x \) \( g(x) \) \( f ( g ( x ) ) \) \( f(x) \) \( g ( f ( x ) ) \) 3 2 8 3 2
Use the table from Example \( \PageIndex{ 6 } \) to evaluate \( f ( g ( 1 ) ) \) and \( g ( f ( 4 ) ) \).
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.
Use the following figure to evaluate \( f ( g ( 1 ) ) \).
- Solution
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To evaluate \( f ( g ( 1 ) ) \), we start with the inside evaluation.
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 \( f \).\[ f ( g ( 1 ) ) = f ( 3 )\nonumber \]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, \( f ( 3 ) = 6 \), so \( f ( g ( 1 ) ) = 6 \). - Analysis
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The following figure shows how we can mark the graphs with arrows to trace the path from the input value to the output value.
Use the graph from Example \( \PageIndex{ 7 } \) to evaluate \( g ( f ( 2 ) ) \).
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 \( f ( g ( x ) ) \). To do this, we will extend our idea of 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 \( f ( t ) = t^2 − t \) and \( h ( x ) = 3 x + 2\), evaluate \( f ( h ( 1 ) ) \).
- Solution
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Because the inside expression is \( h ( 1 ) \), we start by evaluating \( h ( x ) \) at 1.\[ \begin{array}{rclcl}
h(1) & = & 3(1) + 2 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 5 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]Then \( f ( h ( 1 ) ) = f ( 5 ) \), so we evaluate \( f ( t ) \) at an input of 5.\[ \begin{array}{rclcl}
f(h(1)) & = & f(5) & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 5^2 - 5 & \quad & \left( \text{substituting} \right) \\[6pt]
& = & 25 - 5 & \quad & \left( \text{squaring} \right) \\[6pt]
& = & 20 & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \] - Analysis
- It makes no difference what the input variables \( T \) and \( x \) were called in this problem because we evaluated for specific numerical values.
Given \( f ( t ) = t^2 − t \) and \( h ( x ) = 3 x + 2 \), evaluate
- \( h ( f ( 2 ) ) \)
- \( h ( f ( − 2 ) ) \)
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, it is important 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. 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 \( f \circ g \) 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 \( g \) and \( g(x) \) is in the domain of \( f \).
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 \( f \).
To find the domain of the composite function \( f(g(x)) \), we start by finding the domain of \( g \). Those values of \( x \) for which \( g \) is defined will be the largest possible domain of the composition; however, this set might get restricted once we take the function \( f \) into consideration. That said, note the domain of \( f \). Our goal is to find the elements, \( x \), 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 \( f \circ g \).
Find the domain of \( ( f \circ g ) ( x ) \) where \( f ( x ) = \frac{5}{x − 1} \) and \( g ( x ) = \frac{4}{3 x − 2} \).
- Solution
-
The domain of \( g(x) \) consists of all real numbers except \( x = \frac{2}{3} \), 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 \( g ( x ) = 1 \).\[ \begin{array}{rrclcl}
& g(x) & = & 1 & & \\[6pt]
\implies & \dfrac{4}{3x - 2} & = & 1 & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & 4 & = & 1(3x - 2) & \quad & \left( \text{multiplying both sides by the LCD} \right) \\[6pt]
\implies & 4 & = & 3x - 2 & \quad & \left( \text{distributing} \right) \\[6pt]
\implies & 6 & = & 3x & \quad & \left( \text{adding }2\text{ to both sides} \right) \\[6pt]
\implies & 2 & = & x & \quad & \left( \text{dividing both sides by }3 \right) \\[6pt]
\end{array} \nonumber \]So the domain of \( f \circ g \) is the set of all real numbers except \( \frac{2}{3} \) and \( 2 \). This means that \( x \neq \frac{2}{3} \) o \( x \neq 2 \).We can write this in interval notation as\[ \left( − \infty , \dfrac{2}{3} \right) \cup \left( \dfrac{2}{3} , 2 \right) \cup ( 2 , \infty ). \nonumber \]
Find the domain of \( ( f \circ g ) ( x ) \) where \( f ( x ) = \sqrt{ x + 2 } \) 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 check the domain of the composite function\[ ( f \circ g ) ( x ) = \sqrt{\sqrt{ 3 − x } + 2 }. \nonumber \]Since the radicand of a square root must be positive, \( \sqrt{ 3 − x } + 2 \geq 0 \). Hence,\[ \begin{array}{rrclcl}
& \sqrt{ 3 − x } + 2 & \geq & 0 & & \\[6pt]
\implies & \sqrt{ 3 − x } & \geq & -2 & \quad & \left( \text{subtracting }2\text{ from both sides} \right) \\[6pt]
\end{array} \nonumber \]Since square roots never return negative numbers, the inequality \( \sqrt{3 - x} \geq -2 \) is always true. Thus, our only restriction on this composition is the original restriction from the domain of \( g \) - namely that the values of \( x \) must be in the interval \( ( − \infty , 3 ] \). This gives a domain of \( ( - \infty , 3 ] \). - 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 \( f \circ g \) can contain values that are not in the domain of \( f \), though they must be in the domain of \( g \).
If we change the function \( f \) very slightly in Example \( \PageIndex{ 10 } \) to \( f(x) = \sqrt{x - 2} \), then the composition, \( (f \circ g)(x) \) becomes\[ (f \circ g)(x) = \sqrt{\sqrt{3 - x} - 2}. \nonumber \]Again, since the radicand is never negative, we know\[ \begin{array}{rrclcl}
& \sqrt{ 3 − x } - 2 & \geq & 0 & & \\[6pt]
\implies & \sqrt{ 3 − x } & \geq & 2 & \quad & \left( \textbf{adding }2\text{ to both sides} \right) \\[6pt]
\end{array} \nonumber \]This makes the problem of finding the domain a more complex. If \( \sqrt{3 - x} \geq 2 \), then\[ \left( \sqrt{3 - x} \right)^2 \geq 2^2 \implies 3 - x \geq 4 \implies 3 \geq 4 + x \implies -1 \geq x. \nonumber \]Hence, we would require \( x \leq 3 \) (from the domain of \( g \)) and, simultaneously, \( -1 \geq x \). The common intersection of these two domain restrictions is \( x \leq -1 \). Thus, the domain (in this adjusted case) would be \( ( -\infty,-1] \).
Find the domain of \( ( f \circ g ) ( x ) \) where \( f ( x ) = \frac{1}{x − 2} \) and \(g ( x ) = \sqrt{ x + 4 } \).
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.
Write \( f ( x ) = \sqrt{ 5 − x^2 } \) as the composition of two functions.
- Solution
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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\[ h ( x ) = 5 − x^2 \text{ and } g ( x ) = \sqrt{ x }. \nonumber \]We can check our answer by recomposing the functions.\[ g ( h ( x ) ) = g ( 5 − x^2 ) = \sqrt{ 5 − x^2 }. \nonumber \]
Write \( f ( x ) = \frac{4}{3 − \sqrt{ 4 + x^2 }} \) as the composition of two functions.
Access these online resources for additional instruction and practice with composite functions.