2.3: Composition of Functions
Suppose we wanted to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The temperature depends on the day, and the cost depends on the temperature. Using descriptive variables, we can notate these two functions.
The first function, \(C(T)\), gives the cost \(C\) of heating a house when the average daily temperature is \(T\) degrees Celsius, and the second, \(T(d)\), gives the average daily temperature on day \(d\) of the year in some city. If we wanted to determine the cost of heating the house on the 5 \({}^{th}\) day of the year, we could do this by linking our two functions together, an idea called composition of functions. Using the function \(T(d)\), we could evaluate \(T(5)\) to determine the average daily temperature on the 5 \({}^{th}\) day of the year. We could then use that temperature as the input to the \(C(T)\) function to find the cost to heat the house on the 5 \({}^{th}\) day of the year: \(C(T(5))\).
Definition: Composition of Functions
When the output of one function is used as the input of another, we call the entire operation a composition of functions . We write \(f(g(x))\), and read this as “\(f\) of \(g\) of \(x\)” or “\(f\) composed with \(g\) at \(x\)”.
An alternate notation for composition uses the composition operator: \(\circ\)
\((f \circ g)(x)\) is read “\(f\) of \(g\) of \(x\)” or “\(f\) composed with \(g\) at \(x\)”, just like \(f(g(x))\).
Example \(\PageIndex{1}\)
Suppose \(c(s)\) gives the number of calories burned doing s sit-ups, and \(s(t)\) gives the number of sit-ups a person can do in t minutes. Interpret \(c(s(3))\).
Solution
When we are asked to interpret, we are being asked to explain the meaning of the expression in words. The inside expression in the composition is \(s(3)\). Since the input to the s function is time, the 3 is representing 3 minutes, and \(s(3)\) is the number of sit-ups that can be done in 3 minutes. Taking this output and using it as the input to the \(c(s)\) function will gives us the calories that can be burned by the number of sit-ups that can be done in 3 minutes.
Note that it is not important that the same variable be used for the output of the inside function and the input to the outside function. However, it is essential that the units on the output of the inside function match the units on the input to the outside function, if the units are specified.
Example \(\PageIndex{2}\)
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
The expression \(g(y)\) takes miles as the input and outputs a number of gallons. The function \(f(x)\) is expecting a number of hours as the input; trying to give it a number of gallons as input does not make sense. Remember the units must match, and number of gallons does not match number of hours, so the expression \(f(g(y))\) is meaningless.
The expression \(f(x)\) takes hours as input and outputs a number of miles driven. The function \(g(y)\) is expecting a number of miles as the input, so giving the output of the \(f(x)\) function (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 give the number of gallons of gas used, \(g\), driving a certain number of miles, \(f(x)\), in \(x\) hours.
Exercise \(\PageIndex{1}\)
In a department store you see a sign that says 50% off clearance merchandise, so final cost \(C\) depends on the clearance price, \(p\), according to the function \(C(p)\). Clearance price, \(p\), depends on the original discount, \(d\), given to the clearance item, \(p(d)\). Interpret \(C(p(d))\).
- Answer
-
The final cost, \(C\), depends on the clearance price, \(p\), which is based on the original discount, \(d\). (Or the original discount \(d\), determines the clearance price and the final cost is half of the clearance price.)
Composition of Functions using Tables and Graphs
When working with functions given as tables and graphs, we can look up values for the functions using a provided table or graph, as discussed in section 1.1. We start evaluation from the provided input, and first evaluate the inside function. We can then use the output of the inside function as the input to the outside function. To remember this, always work from the inside out.
Example \(\PageIndex{3}\)
Using the tables below, evaluate \(f(g(3))\)and \(g(f(4))\)
Solution
To evaluate \(f(g(3))\), we start from the inside with the 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 \(f\) function, so \(g(3)\) is replaced by the equivalent value 2 and we can evaluate \(f(2)\). Then using the table that defines the function \(f\), we find that \(f(2) = 8\).
\[f(g(3))=f(2)=8. \nonumber\]
To evaluate \(g(f(4))\), we first evaluate the inside expression \(f(4)\)using the first table: \(f(4) = 1\). Then using the table for \(g\) we can evaluate:
\[g(f(4))=g(1)=3. \nonumber\]
Exercise \(\PageIndex{2}\)
Using the tables from the example above, evaluate \(f(g(1))\) and \(g(f(3))\).
- Answer
-
\(f(g(1)) = f(3) = 3\) and \(g(f(3)) = g(3) = 2\)
Example \(\PageIndex{4}\)
Using the graphs below, evaluate \(f(g(1))\).
To evaluate \(f(g(1))\), we again start with the inside evaluation. We evaluate \(g(1)\) using the graph of the g(x) function, finding the input of 1 on the horizontal axis and finding the output value of the graph at that input. Here, \(g(1) = 3\).
Using this value as the input to the f function, \(f(g(1))=f(3)\). We can then evaluate this by looking to the graph of the \(f(x)\) function, finding the input of 3 on the horizontal axis, and reading the output value of the graph at this input.
\(f(3) = 6\), so \(f(g(1))=6\).
Exercise \(\PageIndex{3}\)
Using the graphs from the previous example, evaluate \(g(f(2))\).
- Answer
-
\(g(f(2)) = g(5) = 3\)
Composition using Formulas
When evaluating a composition of functions where we have either created or been given formulas, the concept of working from the inside out remains the same. First, we evaluate the inside function using the input value provided, then use the resulting output as the input to the outside function.
Example \(\PageIndex{5}\)
Given \(f(t)=t^{2} -t\) and \(h(x)=3x + 2\), evaluate \(f(h(1))\).
Solution
Since the inside evaluation is \(h(1)\) we start by evaluating the \(h(x)\) function at 1:
\[h(1) = 3(1) + 2 = 5\nonumber \]
Then \(f(h(1))=f(5)\), so we evaluate the f(t) function at an input of 5:
\[f(h(1)) = f(5) = 5^{2} - 5 = 20\nonumber \]
Exercise \(\PageIndex{4}\)
Using the functions from the example above, evaluate \(h(f(-2))\).
- Answer
-
\(h(f(-2)) = h(6) = 20\) ( did you remember to insert your input values using parentheses? )
While we can compose the functions as above for each individual input value, sometimes it would be really helpful to find a single formula which 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 put whatever value is inside the parentheses after the function name into the formula wherever we see the input variable.
Example \(\PageIndex{6}\)
Given \(f(t) = t^{2} - t\), evaluate \(f(3)\) and \(f(-2)\).
Solution
\[ \begin{align*} f(3) &= 3^2 - 3 \\[4pt] f(-2) &= (-2)^2 - (-2)\end{align*} \]
We could simplify the results above if we wanted to
\[ \begin{align*} f(3) &= 3^2 - 3 = 9 - 3 = 6 \\[4pt] f(-2) &= (-2)^2 - (-2) = 4 + 2 = 6 \end{align*}\]
We are not limited, however, to using a numerical value as the input to the function. We can put anything into the function: a value, a different variable, or even an algebraic expression, provided we use the input expression everywhere we see the input variable.
Example \(\PageIndex{7}\)
Using the function from the previous example, evaluate f(a).
Solution
This means that the input value for \(t\) is some unknown quantity \(a\). As before, we evaluate by replacing the input variable \(t\) with the input quantity, in this case \(a\).
\[f(a)=a^{2} - a\nonumber \]
The same idea can then be applied to expressions more complicated than a single letter.
Example \(\PageIndex{8}\)
Using the same \(f(t)\) function from above, evaluate \(f(x + 2)\).
Solution
Everywhere in the formula for f where there was a t , we would replace it with the input \((x+2)\). Since in the original formula the input t was squared in the first term, the entire input \(x+2\) needs to be squared when we substitute, so we need to use grouping parentheses . To avoid problems, it is advisable to always use parentheses around inputs.
\[f(x+2)=(x+2)^{2} -(x+2)\nonumber \]
We could simplify this expression further to \(f(x+2)=x^{2} +3x+2\) if we wanted to:
\[f(x+2)=(x+2)(x+2)-(x+2)\nonumber \]
Use the “FOIL” technique (first, outside, inside, last)
\[f(x+2)=x^{2} +2x+2x+4-(x+2)\nonumber \]
distribute the negative sign
\[f(x+2)=x^{2} +2x+2x+4-x-2\nonumber \]
combine like terms
\[f(x+2)=x^{2} +3x+2\nonumber \]
Example \(\PageIndex{9}\)
Using the same function, evaluate \(f(t^{3} )\).
Solution
Note that in this example, the same variable is used in the input expression and as the input variable of the function. This doesn’t matter – we still replace the original input t in the formula with the new input expression, \(t^{3}\).
\[f(t^{3} )=(t^{3} )^{2} -(t^{3} )=t^{6} - t^{3}\nonumber \]
Exercise \(\PageIndex{5}\)
Given \(g(x) = 3x - \sqrt{x}\), evaluate \(g(t - 2)\).
- Answer
-
\[g(t - 2) = 3(t - 2) - \sqrt{(t - 2)}\nonumber \]
This now allows us to find an expression for a composition of functions. If we want to find a formula for \(f(g(x))\), we can start by writing out the formula for \(g(x)\). We can then evaluate the function \(f(x)\) at that expression, as in the examples above.
Example \(\PageIndex{10}\)
Let \(f(x)=x^{2}\) and \(g(x)=\dfrac{1}{x} -2x\), find \(f(g(x))\) and \(g(f(x))\).
Solution
To find \(f(g(x))\), we start by evaluating the inside, writing out the formula for \(g(x)\).
\[g(x)=\dfrac{1}{x} -2x\nonumber \]
We then use the expression \((\dfrac{1}{x} -2x)\) as input for the function \(f\).
\[f(g(x)) = f(\dfrac{1}{x} - 2x)\nonumber \]
We then evaluate the function \(f(x)\) using the formula for \(g(x)\) as the input.
Since \(f(x)=x^{2}\),
\[f(\dfrac{1}{x} -2x) = (\dfrac{1}{x} - 2x)^{2}\nonumber \]
This gives us the formula for the composition:
\[f(g(x)) = (\dfrac{1}{x} - 2x)^{2}\nonumber \]
Likewise, to find g(f(x)), we evaluate the inside, writing out the formula for f(x)
\[g(f(x))=g(x^{2})\nonumber \]
Now we evaluate the function g(x) using \(x^{2}\) as the input.
\[g(f(x))=\dfrac{1}{x^{2} } -2x^{2}\nonumber \]
Exercise \(\PageIndex{6}\)
Let \(f(x) = x^{3} +3x\) and \(g(x)=\sqrt{x}\), find \(f(g(x))\) and \(g(f(x))\).
- Answer
-
\[f(g(x) = f(\sqrt{x}) = (\sqrt{x})^3 + 3(\sqrt{x})\nonumber \]
\[g(f(x) = g(x^3 + 3x) = \sqrt{(x^3 + 3x)}\nonumber \]
Example \(\PageIndex{11}\)
A city manager determines that the tax revenue, \(R\), in millions of dollars collected on a population of \(p\) thousand people is given by the formula \(R(p) = 0.03p + \sqrt{p}\), and that the city’s population, in thousands, is predicted to follow the formula \(p(t) = 60 + 2t + 0.3t^{2}\), where t is measured in years after 2010. Find a formula for the tax revenue as a function of the year.
Solution
Since we want tax revenue as a function of the year, we want year to be our initial input, and revenue to be our final output. To find revenue, we will first have to predict the city population, and then use that result as the input to the tax function. So we need to find \(R(p(t))\). Evaluating this,
\[R(p(t)) = R(60 + 2t + 0.3t^{2} ) = 0.03(60+2t+0.3t^{2}) + \sqrt{60 + 2t + 0.3t^{2} }\nonumber \]
This composition gives us a single formula which can be used to predict the tax revenue during a given year, without needing to find the intermediary population value.
For example, to predict the tax revenue in 2017, when t = 7 (because t is measured in years after 2010),
\[R(p(7))=0.03(60 + 2(7) + 0.3(7)^{2})+ \sqrt{60 + 2(7) + 0.3(7)^{2} } \approx 12.079\text{ million dollars}\nonumber \]
Domain of Compositions
When we think about the domain of a composition \(h(x)=f(g(x))\), we must consider both the domain of the inner function and the domain of the composition itself. While it is tempting to only look at the resulting composite function, if the inner function were undefined at a value of \(x\), the composition would not be possible.
Example \(\PageIndex{12}\)
Let \(f(x)=\dfrac{1}{x^{2} -1}\) and \(g(x)=\sqrt{x-2}\). Find the domain of \(f(g(x))\).
Since we want to avoid the square root of negative numbers, the domain of \(g(x)\) is the set of values where \(x - 2 \ge 0\). The domain is \(x \ge 2\).
The composition is \[f(g(x))=\dfrac{1}{(\sqrt{x-2})^{2} - 1} = \dfrac{1}{(x - 2)-1} = \dfrac{1}{x - 3}\nonumber \].
The composition is undefined when \(x = 3\), so that value must also be excluded from the domain. Notice that the composition doesn’t involve a square root, but we still have to consider the domain limitation from the inside function.
Combining the two restrictions, the domain is all values of x greater than or equal to 2, except \(x = 3\).
In inequalities, the domain is: \(2 \le x<3\) or \(x > 3\)
In interval notation, the domain is: \([2, 3) \cup (3, \infty)\).
Exercise \(\PageIndex{7}\)
Let \(f(x) = \dfrac{1}{x-2}\) and \(g(x)=\dfrac{1}{x}\). Find the domain of \(f(g(x))\).
- Answer
-
\(g(x) = \dfrac{1}{x}\) is undefined at \(x = 0\).
The composition, \[f(g(x)) = f(\dfrac{1}{x}) = \dfrac{1}{\dfrac{1}{x} - 2} = \dfrac{1}{\dfrac{1}{x} - \dfrac{2x}{x}} = \dfrac{1}{\dfrac{1 - 2x}{x}} = \dfrac{x}{1 - 2x}\nonumber \] is undefined when \(1 - 2x = 0\), when \(x = \dfrac{1}{2}\).
Restricting these two values, the domain is \((-\infty, 0) \cup (0, \dfrac{1}{2}) \cup (\dfrac{1}{2}, \infty)\)
Decomposing Functions
In some cases, it is desirable to decompose a function – to write it as a composition of two simpler functions.
Example \(\PageIndex{13}\)
Write \(f(x)=3+\sqrt{5 - x^{2} }\) as the composition of two functions.
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 \(5 - x^{2}\) is the inside of the square root. We could then decompose the function as:
\[h(x)=5-x^{2}\nonumber \]
\[g(x)=3+\sqrt{x}\nonumber \]
We can check our answer by recomposing the functions:
\[g(h(x)) = g(5 - x^{2})=3 + \sqrt{5-x^2 }\nonumber \]
Note that this is not the only solution to the problem. Another non-trivial decomposition would be \(h(x) = x^{2}\) and \(g(x) = 3 + \sqrt{5-x}\)
Composition of Functions
Suppose we wanted to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and 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 first function, \(C(T)\), gives the cost \(C\) of heating a house when the average daily temperature is \(T\) degrees Celsius, and the second, \(T(d)\), gives the average daily temperature on day \(d\) of the year in some city. If we wanted to determine the cost of heating the house on the \(5^{th}\) day of the year, we could do this by linking our two functions together, an idea called composition of functions. Using the function \(T(d)\), we could evaluate \(T(5)\) to determine the average daily temperature on the 5 \({}^{th}\) day of the year. We could then use that temperature as the input to the \(C(T)\) function to find the cost to heat the house on the 5 \({}^{th}\) day of the year: \(C(T(5))\).
Definition: Composition of Functions
The composition of the function \(f\) with \(g\) is denoted by \((f \circ g)(x)\), read “\(f\) of \(g\) of \(x\)”, and is defined by the equation
\((f \circ g)(x) = f(g(x))\)
In general, \(f{\circ}g\) and \(g{\circ}f\) are different functions.
It is important to realize that the product of functions \((fg)(x)\) is not the same as the function composition \((f \circ g)(x) \).
The domain of the composite function \(f \circ g\) is the set of all \(x\) such that
- \(x\) is in the domain of \(g\) and \( \qquad \qquad \leftarrow\) so that the first (inner) operation is defined
- \(g(x)\) is in the domain of \(f\). \( \qquad \qquad \leftarrow\) so that the final resulting operation is defined
Example \(\PageIndex{16}\): 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).
Note that it is not important that the same variable be used for the output of the inside function and the input to the outside function. However, it is essential that the units on the output of the inside function match the units on the input to the outside function, if the units are specified.
Example \(\PageIndex{17}\): 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 these expressions is meaningful: \(f(g(y))\) or \(g(f(x))\)?
Solution
The function \(y=f(x)\) is a function that produces the number of miles driven given the number of hours driven:
\[\text{number of miles } =f (\text{number of hours}) \nonumber\]
The function \(g(y)\) is a function that produces the number of gallons used given the number of miles driven:
\[\text{number of gallons } =g(\text{number of miles}) \nonumber\]
Consider the composition \(f(g(y))\). 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 into \(f\) does not make sense. The expression \(f(g(y))\) is meaningless.
Consider the composition \(g(f(x))\). 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.
Try It \(\PageIndex{18}\)
In a department store you see a sign that says 50% off clearance merchandise, so final cost \(C\) depends on the clearance price, \(p\), according to the function \(C(p)\). Clearance price, \(p\), depends on the original discount, \(d\), given to the clearance item, \(p(d)\). Interpret \(C(p(d))\).
- Answer
-
The final cost, \(C\), depends on the clearance price, \(p\), which is based on the original discount, \(d\). (Or the original discount \(d\), determines the clearance price and the final cost is half of the clearance price.)
Try It \(\PageIndex{19}\)
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.
- Answer
-
A gravitational force is still a force, so \(a(G(r))\) makes sense as the acceleration of a planet at a distance \(r\) from the Sun (due to gravity), but \(G(a(F))\) does not make sense.
Evaluating Composite Functions
When working with functions given as tables, graphs, and formulas we can look up values for the functions using a provided table, graph, or formula. Start evaluation with the inside function. Then use the output of the inside function as the input to the outside function. To remember this, always work from the inside out - just like with order of operations.
Evaluate 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.
Example \(\PageIndex{20}\): Using a Table to Evaluate a Composite Function
|
Using the tables below, evaluate \(f(g(3))\) and \(g(f(3))\).
|
Solution To evaluate \(f(g(3)),\) start from the inside with the input value 3. Then evaluate the inside expression \(g(3)\) using the table that defines the function \(g: g(3)=2\). Use that result as the input to the function \(f\), so \(g(3)\) is replaced by 2 to get \(f(2)\). Then, using the table that defines the function \(f\), we find that \(f(2)=8\). Step \(1: \; g(3)=2, \quad\) Step \(2: \; f(g(3))=f(2)=8\) To evaluate \(g(f(3)),\) first evaluate the inside expression \(f(3)\) using the first table: \(f(3)=3\). Then, using the table for \(g\), we can evaluate Step \(1: \; f(3)=3, \quad\) Step \(2: \; g(f(3))=g(3)=2\) |
Try It \(\PageIndex{21}\)
Use the table in Example \(\PageIndex{20}\) above to evaluate \(f(g(1))\), \(g(f(4))\) and \(f(g(4))\).
- Answer
-
\(f(g(1))=f(3)=3\), \(g(f(4))=g(1)=3\), and \(f(g(4)) = f(7) = \text{undefined}\)
Notice the composition is not a commutative operation!
Evaluate 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.
How to: Use graphs of individual functions to evaluate a composite function.
-
Locate the given input to the inner function on the \(x\)-axis of its graph.
Read off the output of the inner function from the \(y\)-axis of its graph. -
Locate the inner function output on the \(x\)-axis of the graph of the outer function.
Read the output of the outer function from the \(y\)-axis of its graph. This is the output of the composite function.
Example \(\PageIndex{22}\): Using a Graph to Evaluate a Composite Function
Using Figure \(\PageIndex{22}\), evaluate \(f(g(1))\).
Solution
To evaluate \(f(g(1))\), we start with the inside evaluation. See Figure \(\PageIndex{22}\).
Step 1. \(f(g(1))=f(3)\): 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 \(f\).
Step 2. \(f(3)=6\): 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 \(f(g(1))=6\).
Try It \(\PageIndex{23}\)
Using Figure \(\PageIndex{22}\), evaluate \(g(f(2))\).
- Answer
-
\(g(f(2))=g(5)=3\)
Evaluate 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.
How to: Evaluate a Composition \( (f \circ g)(c) \).
- Write the definition for the composition function: \((f \circ g)(c) = f(g(c)) \).
- Substitute \(c\) in the function definition for \(g\) to evaluate the inside function \(g(c)\). Let \(k\) be this result: \((f \circ g)(c) = f(g(c))=f(k) \).
- Substitute \(k\) in the function definition for \(f\) to evaluate the outside function
Example \(\PageIndex{24}\): Evaluate a Composition of Functions Expressed as Formulas with a Numerical Input
Given \(f(t)=t^2−t\) and \(h(x)=3x+2\), evaluate \( (f \circ h)(1))\).
Solution
Rewrite \( (f \circ h)(1))\) using the definition: \( (f \circ h)(1)) = f(h(1))\)
The inside expression is \(h(1)\). The argument of \(h\) is \(1\), so substitute \(1\) for \(x\) in the formula for \(h\) and evaluate it.
\[ \begin{align*} h(1)&=3(1)+2 \\[4pt] h(1)&=5 \end{align*} \]
Now \(f(h(1))=f(5)\). The argument of \(f\) is \(5\) so substitute \(5\) for \(t\) in the formula for \(f\) and evaluate it.
\[ \begin{align*} f(h(1)) &=f(5) \\[5pt] f(h(1))&=5^2−5 \\[5pt] f(h(1))&=20 \end{align*} \]
Analysis
It makes no difference what the input variables \(t\) and \(x\) were called in this problem because we evaluated for specific numerical values.
Example \(\PageIndex{25}\): Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input
Given \(f(x) = x^2-4x\), \(g(x) = 2-\sqrt{x+3}\), and \(h(x) = \dfrac{2x}{x+1}\) evaluate the following.
- \((g \circ f)(1) \qquad \) b. \((f \circ g)(1) \qquad \) c. \((g \circ g)(6) \qquad \) d. \((g \circ f)(2) \qquad \) e. \((g \circ h)(-1) \qquad \)
Solution
a. Use the definition: \((g \circ f)(1) = g(f(1))\).
\(\quad\) Evaluate the inner function: \(f(1) = -3\)
\(\quad\) Evaluate the outer function using this result: \( g(f(1)) = g(-3) = 2 \)
b. Use the definition to write \((f \circ g)(1) = f(g(1))\).
\(\quad\) Evaluate the inner function: \(g(1) = 0\),
\(\quad\) Evaluate the outer function using this result: \(f(g(1)) = f(0) = 0 \)
c. Use the definition \((g \circ g)(6) = g(g(6))\).
\(\quad\) Evaluate the inner function: \(g(6) = -1\)
\(\quad\) Evaluate the outer function using this result: \( g(g(6)) = g(-1) = 2-\sqrt{2} \)
d. Use the definition: \((g \circ f)(2) = g(f(2))\)
\(\quad\) Evaluate the inner function: \(f(2) = (2)^2-4(2)=4-8=-4\)
\(\quad\) Evaluate the outer function using this result: \(g(f(2)) = g(-4) = 2-\sqrt{-4+3} =2-\sqrt{-1} = 2-i\). This result is not a real number, so we say \((g \circ f)(2) \) is undefined.
e. Use the definition: \((g \circ h)(-1) = g(h(-1))\)
\(\quad\) Evaluate the inner function: \(h(-1) = \dfrac{2(-1)}{(-1)+1}= \dfrac{-2}{0} = \text{undefined}\)
\(\quad\) Evaluate the outer function: \( g(h(-1)) =g(\text{undefined}) \) cannot be evaluated! So \((g \circ h)(-1) = \text{undefined}\).
Try It \(\PageIndex{26}\)
Given \(f(t)=t^2−t\) and \(h(x)=3x+2\), evaluate
a. \(h(f(2))\)
b. \(h(f(−2))\)
- Answers:
-
a. 8 \( \qquad \) b. 20
Finding the Formula and Domain of a Composite Function
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 \((f \circ g)(x)\). To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function with a formula like \(f(t)=t^2−t\), we substitute the value inside the parentheses (called the argument) into the formula wherever we see the input variable.
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 \(f{\circ}g\). 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. Secondly, the outer function evaluation, \(f(g(x))\), must also be defined. Thus the restrictions to the domain of \(f{\circ}g\) consist of not only the values of \(x\) that make \(g(x)\) undefined, but also the values of \(x\) that make the expression \(f(g(x))\) undefined.
How to: Determine a Function Composition \( (f \circ g)(x) \) and its Domain.
- Write the definition for the composition function: \((f \circ g)(x) = f(g(x)) \).
-
Substitute the function definition for \(g\) in for the argument of \(f\).
Determine the domain of \(g\) because the argument to \(f\) must be defined during the composition process. - Substitute the argument expression \(g(x)\) into the function definition for \(f\) to determine the formula for \((f \circ g)(x)\).
- Simplify the formula for \((f \circ g)(x)\).
- The domain of \((f \circ g)(x)\) combines both the domain restrictions to the inner function and the domain restrictions apparent in the resulting function. It is important to include the domain restrictions to \(g\) because sometimes they get lost in the simplification process (step 4).
Example \(\PageIndex{27}\): Finding a composition and its domain
Given \(f(x)=5x−1 \) and \( g(x)=\dfrac{4}{3x−2}\) find \((f∘g)(x)\) and its domain.
Solution: 1. \((f∘g)(x) = f(g(x))\)
2. \(f(g(x)) = f \Big(\dfrac{4}{3x−2} \Big) \). 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.
3. \( f \Big(\dfrac{4}{3x−2} \Big) = 5\Big(\dfrac{4}{3x−2} \Big) -1 \). Notice in this form the domain restriction \(x \ne \frac{2}{3}\)
4. \(5\Big(\dfrac{4}{3x−2} \Big) -1 = 5\Big(\dfrac{4}{3x−2} \Big) - \dfrac{3x−2}{3x−2} = \dfrac{20 -3x+2}{3x−2} =\dfrac{22-3x}{3x−2} \)
5. The domain restriction of the simplified expression is still \(x \ne \frac{2}{3}\).
Answer: \((f∘g)(x) = \dfrac{22-3x}{3x−2}\) and its domain is \( (−\infty,\dfrac{2}{3}) \cup (\dfrac{2}{3},\infty) \)
Example \(\PageIndex{28}\)
Given \(f(x)=\sqrt{x+2}\) and \(g(x)=\sqrt{3−x}\) find \((f∘g)(x)\) and its domain.
Solution: 1. \((f∘g)(x) = f(g(x))\)
2. \(f(g(x)) = f (\sqrt{3−x})\). The domain restriction for \(g\) is \(3-x \ge 0 \longrightarrow 3 \ge x \longrightarrow x \le 3\)
3. \(f (\sqrt{3−x})= \sqrt{\sqrt{3−x}+2}\)
4. This expression cannot be simplified.
5. The domain restriction for this expression is \(\sqrt{3−x}+2 \ge 0\). As long as \(\sqrt{3-x}\) is a real number, \(\sqrt{3−x} \ge 0\) so \(\sqrt{3−x}+2 \ge 2\). The expression \(\sqrt{3-x}\) is a real number whenever \(3-x \ge 0\) or \(x \le 3\).
Answer: \((f∘g)(x)=\sqrt{ \sqrt{3−x}+2}\) and its domain is \((-∞,3]\).
Example \(\PageIndex{29}\)
Let \(f(x) = x^2-4x\), \(g(x) = 2-\sqrt{x+3}\), and \(h(x) = \dfrac{2x}{x+1}\). Find and simplify the indicated composite functions. State the domain of each.
- \((f \circ g)(x)\)
- \((h \circ g)(x)\)
- \((h \circ h)(x)\)
Solution
a. \(\quad \;\)1. \((f \circ g)(x)= f(g(x))\)
2. \(f(g(x)) = f(2-\sqrt{x+3})\). The domain of \(g\) is \( x+3 \ge 0 \longrightarrow x \ge -3\)
\( \begin{array}{rrll}
3.& f(2-\sqrt{x+3}) &= (2-\sqrt{x+3})^2 - 4(2-\sqrt{x+3}) \\
4.& &= 4 - 4 \sqrt{x+3} + (\sqrt{x+3})^2 - 8 + 4\sqrt{x+3} \\
&&= 4 + x + 3 - 8\\
&&= x-1
\end{array}\)
5. The resulting expression has no domain restrictions, but the domain restriction found for \(g\) still applies to the composition process.
Answer: \((f \circ g)(x)= x-1\) with a domain of \([-3, \infty)\)
b. \(\quad \;\)1. \((h \circ g)(x) = h(g(x))\)
2. \(h(g(x)) = h (2-\sqrt{x+3})\). The domain of \(g\) here is \(x+3 \ge 0 \longrightarrow x \ge -3\).
\( \begin{array}{rrll}
3.& h (2-\sqrt{x+3})&=\dfrac{2(2-\sqrt{x+3})}{(2-\sqrt{x+3})+1} \\
4.& &= \dfrac{4-2\sqrt{x+3}}{3-\sqrt{x+3}} &\text{Simplify} \\
& &= \dfrac{4-2\sqrt{x+3}}{3-\sqrt{x+3}}\cdot \dfrac{3+\sqrt{x+3}}{3+\sqrt{x+3}} &\text{Rationalize denominator} \\
& &= \dfrac{12+4\sqrt{x+3}-6\sqrt{x+3}-2(x+3)}{9-(x+3)} &\text{Multiply out and simplify} \\
& &= \dfrac{6-2x-2\sqrt{x+3}}{6-x} \\
\end{array}\)
5. The resulting expression has two domain restrictions: \(x+3 \ge 0 \) and \( 6-x \ne 0 \). Therefore, \(x \ge -3\) and \(x \ne 6\). Both of these restrictions apply to the composition process. The domain restriction to \(g\) found earlier is repeated here.
Answer: \((h \circ g)(x) = = \dfrac{4-2\sqrt{x+3}}{3-\sqrt{x+3}}\) with a domain of \([-3,6) \cup (6, \infty)\).
c. \(\quad \;\)1. \((h \circ h)(x) = h(h(x))\).
2. \( h(h(x)) = h \Big( \dfrac{2x}{x+1}\Big) \). The domain of \(h\) here is \(x + 1 \ne 0 \longrightarrow x \ne -1\).
\(\begin{array}{rrll}
3. & h \left( \dfrac{2x}{x+1} \right) &= \dfrac{2\Big( \dfrac{2x}{x+1} \Big)}{\Big( \dfrac{2x}{x+1} \Big) + 1 }\\
\end{array}\)
\(\begin{array}{rrll}
4.& \qquad \qquad \qquad \qquad &= \dfrac{ \Big( \dfrac{4x}{x+1} \Big)}{\Big( \dfrac{2x}{x+1} \Big) + 1 }\cdot \dfrac{x+1}{x+1} &\text{Simplify complex fraction}\\
&&= \dfrac{ \Big( \dfrac{4x}{x+1} \Big) \cdot (x+1)}{\Big( \dfrac{2x}{x+1} \Big)\cdot (x+1) + 1\cdot (x+1) }\\
&&= \dfrac{4x}{2x+x+1} = \dfrac{4x}{3x+1}
\end{array}\)
5. The resulting expression has one domain restriction, \( 3x+1 \ne 0\). Therefore, \(x \ne \frac{1}{3}\). The domain of the composition process includes not only this domain but also the domain restriction to \(h\) found earlier: \(x \ne -1\)
Answer: \((h \circ h)(x) =\dfrac{4x}{3x+1}\) and its domain is \((-\infty, -1) \cup \left(-1, -\frac{1}{3}\right) \cup \left(-\frac{1}{3}, \infty\right)\).
Try It \(\PageIndex{30}\)
Find and simplify \((f∘g)(x)\) and find its domain given
- \(f(x)=\dfrac{1}{x−2} \text{ and } g(x)=\sqrt{x+4}\)
- \( f(x)=\dfrac{4}{3x−2}\) and \(g(x)=\dfrac{1}{x-1}\)
- Answers
-
a. \((f∘g)(x) = \dfrac{2+\sqrt{x+4}}{x}\) with domain \([−4,0)∪(0,∞)\) \(\qquad\) b. \((f∘g)(x) = \dfrac{4(x-1)}{5-2x}\) with domain \((-∞, 1)∪(1, 2.5)∪(2.5,∞)\)
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 \(\PageIndex{31}\): Decomposing a Function
Write \(f(x)=\sqrt{5−x^2}\) as the composition of two functions.
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 \(5−x^2\) 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 our answer by recomposing the functions.
\(g(h(x))=g(5−x^2)=\sqrt{5−x^2}\)
Try It \(\PageIndex{32}\)
Write \(f(x)=\dfrac{4}{3−\sqrt{4+x^2}}\) as the composition of two functions.
- Answer
-
Possible answers:
\(g(x)=\sqrt{4+x^2}\)
\(h(x)=\dfrac{4}{3−x}\)
\(f=h{\circ}g\)
Important Topics of this Section
- Definition of Composition of Functions
- Compositions using:
- Words
- Tables
- Graphs
- Equations
- Domain of Compositions
- Decomposition of Functions