Composite Functions

Section 7.4 Composite Functions

We combine two functions to get a new function by using function composition. Given two functions \(f\) and \(g\) we create a new function such that the image of \(a\) in the domain of \(f\) is \(g(f(a))\text{.}\) To compute \(g(f(a))\) we first apply \(f\) to determine \(f(a)\text{,}\) and then apply \(g\) to the result. This is only works if \(f(a)\) is in the domain of \(g\text{.}\)

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Definition 7.4.1.

Let \(f \colon A \to B\text{,}\) and let \(g \colon B \to C\text{.}\) The composite function \(g \circ f\text{,}\) is the function \(g \circ f \colon A \to C\) defined by

\begin{equation*} (g \circ f)(x) = g(f(x))\text{.} \end{equation*}

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We read \(g\circ f\) as “the composite of (the functions) \(g\) and \(f\text{.}\)” We read \((g\circ f)(x)\) as “the composite of \(g\) and \(f\) of \(x\)” or as “\(g\) of \(f\) of \(x\text{.}\)”

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We can use less strict conditions for the domain and codomain of \(f\) and \(g\text{.}\) We only require the codomain of \(f\) to be a subset of the domain of \(g\text{.}\)

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In the video in Figure 7.4.2 we motivate the composition of functions and give examples.

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Figure 7.4.2. Composition of Functions by Matt Farmer and Stephen Steward

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Evaluating a composite function is no harder than evaluating a function.

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Example 7.4.3. Evaluating the composite of two functions.

Let \(f:\{1,2,3,4,5\}\to\Z\) be given by \(f(x)=3\cdot x\) and \(g:\Z\to\Z\) be given by \(g(x)=x+2\text{.}\)

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Let \(h:=g\circ f\) be the composite of \(f\) and \(g\text{.}\) The domain of \(g\circ f\) is the domain of \(f\) namely \(\{1,2,3,4,5\}\) and the codomain of \(g\circ f\) is \(\Z\) which is the codomain of \(g\text{,}\) in short:

\begin{equation*} g\circ f:\{1,2,3,4,5\}\to \Z. \end{equation*}

We evaluate \(h\) at several integers.

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First we compute \(h(5)\text{.}\) By the definition of \(h\) as the composite of \(f\) and \(g\) we have

\begin{equation*} h(5)=(g\circ f)(5)=g(f(5)). \end{equation*}

We evaluate \(g(f(5))\) from the inside out. First we use the definition of \(f\) to find \(f(5)=3\cdot 5=15\text{.}\) Then we evaluate \(g\) at the result of this computation, namely 15. We get

\begin{equation*} h(5)=(g\circ f)(5)=g(f(5))=g(15)=15+2=17 \end{equation*}

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Similarly we get

\begin{align*} h(1)\amp=(g\circ f)(1)=g(f(1))=g(3\cdot 1)=g(3)=3+2=5\\ h(2)\amp=(g\circ f)(2)=g(f(2))=g(3\cdot 2)=g(6)=6+2=8\\ h(3)\amp=(g\circ f)(3)=g(f(3))=g(3\cdot 3)=g(9)=9+2=11\\ h(4)\amp=(g\circ f)(4)=g(f(4))=g(3\cdot 4)=g(12)=12+2=14 \end{align*}

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Example 7.4.4. Composite of \(\mathrm{grade}\) and \(\mathrm{studentid}\).

We use the functions

\begin{equation*} \mathrm{studentid} \colon N \to I \end{equation*}

and

\begin{equation*} \mathrm{grade} \colon I \to G \end{equation*}

from Example 7.1.4 and Example 7.1.6 given by the tables in Figure 7.1.3 and Figure 7.1.5 respectively.

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To find a student’s grade, first look up the student’s identification number in the table from Figure 7.1.3. Then use that identification number to look up the grade in the table from Figure 7.1.3.

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So to find Alice’s grade we first look up her identification number in Figure 7.1.3 and find that it is \(1001\text{.}\) From Figure 7.1.5 we get that the grade of the student with identification number \(1001\) is a \(\mathsf{B}\text{.}\) So Alice’s grade in MAT 112 is a \(\mathsf{B}\text{.}\)

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Now we formulate this process in terms of function composition The composite function

\begin{equation*} \mathrm{grade} \circ\mathrm{studentid} \end{equation*}

given a student’s name gives the student’s grade. The domain of \(\mathrm{grade} \circ\mathrm{studentid}\) is the set

\begin{equation*} N=\{\mathsf{Aaron},\mathsf{Alice},\mathsf{Bob},\mathsf{Eve},\mathsf{James},\mathsf{Nathan},\mathsf{Oscar},\mathsf{Sandi}\} \end{equation*}

of student names and the codomain of \(\mathrm{grade} \circ\mathrm{studentid}\) is the set

\begin{equation*} G=\{\mathsf{A},\mathsf{B},\mathsf{C},\mathsf{D},\mathsf{F}\} \end{equation*}

of grades. We get

\begin{align*} (\mathsf{grade} \circ\mathrm{studentid} )(\mathsf{ Alice } )\amp=\mathrm{grade} (\mathsf{studentid} (\mathsf{ Alice } ))\\ \amp =\mathrm{grade} (1001)=\mathsf{B}\text{.} \end{align*}

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In Figure 7.4.7, we give an example of the composite of two functions that are given by a diagram.

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Example 7.4.5. Composite of two functions algebraic.

Let \(s:\N\to\N\) be given by \(s(n):=n^2\) as in Example 7.1.9, and let \(m:\N\to \Z_5\) be given by \(m(a):=a\fmod 5\) as in Example 7.1.10.

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The composite function \(m\circ s\) is a function from \(\N\) to \(\Z_5\text{,}\) and we have that \((m\circ s)(n)=m\left(s(n)\right)=m(n^2)=n^2\fmod 5\) for each \(n \in \N\text{.}\) Notice that the algebraic rule for \(m\circ s\) is the same as the algebraic rule for the function \(g\) in Figure 7.1.8. But \(m \circ s \neq g\) since the domain of \(m \circ s\) is \(\N\) and the domain of \(g\) is \(\Z_5\text{.}\)

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The order in which the functions are composed matters, that is, there are functions \(f\) and \(g\) such that \(g \circ f \neq f \circ g\text{.}\)

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Example 7.4.6. Order of composition matters.

We show that the order of the composition of function matters. Let \(f:\N\to \N\) given by \(f(n):=2\cdot n\) and \(g:\N\to\N\) given by \(g(m):=m^2\text{.}\) The domains of \(f\) and \(g\) allow us to form the composites \(g\circ f\) and \(f\circ g\text{.}\) To show that \(f\circ g\) is not equal to \(g\circ f\) we only need to find an \(b\in\N\) with \((g\circ f)(b)\ne (f\circ g)(b)\text{.}\) For \(b=3\) we have

\begin{equation*} (g\circ f)(3)=g(f(3))=g(2\cdot 3)=g(6)=6^2=36 \end{equation*}

and

\begin{equation*} (f\circ g)(3)=f(g(3))=f(3^2)=f(9)=2\cdot 9=18\text{,} \end{equation*}

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So the functions \(g\circ f\) and \(f\circ g\) are not equal.

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Problem 7.4.8. Do \(h\circ g\) and \(g\circ h\) exits ?

Consider the two functions

\begin{align*} g:\amp \{-1,0,1\}\to \{0,1,2\} \text{ given by } g(x) := x^2, \text{ and }\\ h:\amp \{0,1,2\}\to \{2,3,4\} \text{ given by } h(y) := y+2\text{.} \end{align*}

Does the composite function \(h\circ g\) exist? If yes, give \((h\circ g)(x)\) for all x in \(\{-1,0,1\}\text{.}\)
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Does the composite function \(g\circ h\) exist? If yes, specify \((g\circ h)(x)\) for all x in \(\{0,1,2\}\text{.}\)
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Solution.

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The codomain of \(g\) is equal to the domain of \(h\text{,}\) so the composite \(h\circ g\) exists. We specify \(h\circ g:\{-1,0,1\}\to \{2,3,4\}\) by evaluating it at all elements of its domain. We have:

\begin{align*} (h\circ g)(-1)\amp =h(g(-1))=h(1)=3\\ (h\circ g)(0)\amp =h(g(0))=h(0)=2\\ (h\circ g)(1)\amp =h(g(1))=h(1)=3 \end{align*}

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The codomain of \(h\) is not equal to the domain of \(g\text{,}\) so \(g\circ h\) does not exist.
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