Calculus

Set Theory

Set Theory Logo

Composition of Functions

Definition and Properties

Similarly to relations, we can compose two or more functions to create a new function. This operation is called the composition of functions.

Let \(g: A \to B\) and \(f: B \to C\) be two functions such that the range of \(g\) equals the domain of \(f.\) The composition of the functions \(f\) and \(g,\) denoted by \(f \circ g,\) is another function defined as

\[y = \left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right).\]

Composition of functions
Figure 1.

Since functions are a special case of relations, they inherit all properties of composition of relations and have some additional properties. We list here some of them:

  1. The composition of functions is associative. If \(h: A \to B,\) \(g: B \to C\) and \(f: C \to D,\) then \(\left( {f \circ g} \right) \circ h = f \circ \left( {g \circ h} \right).\)
  2. The composition of functions is not commutative. If \(g: A \to B\) and \(f: B \to C,\) then, as a rule, \(f \circ g \ne g \circ f.\)
  3. Let \(g: A \to B\) and \(f: B \to C\) be injective functions. Then the composition of the functions \(f \circ g\) is also injective.
  4. Let \(g: A \to B\) and \(f: B \to C\) be surjective functions. Then the composition of the functions \(f \circ g\) is also surjective.
  5. It follows from the last two properties that if two functions \(g\) and \(f\) are bijective, then their composition \(f \circ g\) is also bijective.

Examples

Example 1. Composition of Functions Defined on Finite Sets

Consider the sets \(A = \left\{ {1,2,3,4} \right\},\) \(B = \left\{ {a,b,c,d} \right\}\) and \(C = \left\{ \alpha, \beta, \gamma, \delta \right\}.\) The functions \(g: A \to B\) and \(f:B \to C\) are defined as

\[{g = \left\{ {\left( {1,b} \right),\left( {2,b} \right),\left( {3,a} \right),\left( {4,d} \right)} \right\},\;\;}\kern0pt{f = \left\{ {\left( {a,\gamma } \right),\left( {b,\alpha } \right),\left( {c,\delta } \right),\left( {d,\alpha } \right)} \right\}.}\]

It is convenient to illustrate the mapping between the sets in an arrow diagram:

A composite function defined on finite sets.
Figure 2.

Given the mapping, we see that

\[{\left( {f \circ g} \right)\left( 1 \right) = f\left( {g\left( 1 \right)} \right) }={ f\left( b \right) = \alpha ;}\]

\[{\left( {f \circ g} \right)\left( 2 \right) = f\left( {g\left( 2 \right)} \right) }={ f\left( b \right) = \alpha ;}\]

\[{\left( {f \circ g} \right)\left( 3 \right) = f\left( {g\left( 3 \right)} \right) }={ f\left( a \right) = \gamma ;}\]

\[{\left( {f \circ g} \right)\left( 4 \right) = f\left( {g\left( 4 \right)} \right) }={ f\left( d \right) = \alpha.}\]

Hence, the composition of functions \(f \circ g\) is given by

\[{f \circ g \text{ = }}\kern0pt{\left\{ {\left( {1,\alpha } \right),\left( {2,\alpha } \right),\left( {3,\gamma } \right),\left( {4,\alpha } \right)} \right\}.}\]

This is represented in the following diagram:

The arrow diagram for composition of functions f and g.
Figure 3.

Example 2. Composition of Functions Defined on Infinite Sets

Let \(g: \mathbb{R} \to \mathbb{R}\) and \(f: \mathbb{R} \to \mathbb{R}\) be two functions defined as

\[{g\left( x \right) = {x^2} + 3x + 1,\;\;}\kern0pt{f\left( x \right) = \cos x.}\]

Determine the composite functions \(f \circ g,\) \(g \circ f,\) \(f \circ f,\) \(g \circ g.\)

The first composite function \(\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right)\) is formed when the inner function \({g\left( x \right)}\) is substituted for \(x\) in the outer function \({f\left( x \right)}.\) This yields:

\[{\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) }={ \cos \left( {g\left( x \right)} \right) }={ \cos \left( {{x^2} + 3x + 1} \right).}\]

Similarly we find the other composite functions:

\[{\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) }={ {f^2}\left( x \right) + 3f\left( x \right) + 1 }={ {\cos ^2}x + 3\cos x + 1.}\]

\[{\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) }={ \cos \left( {f\left( x \right)} \right) }={ \cos \left( {\cos x} \right).}\]

\[{\left(g \circ g\right)\left(x\right) = g\left( {g\left( x \right)} \right) }={ {g^2}\left( x \right) + 3g\left( x \right) + 1 }={ {\left( {{x^2} + 3x + 1} \right)^2} }+{ 3\left( {{x^2} + 3x + 1} \right) + 1 }={ \left( {\color{magenta}{x^4} + \color{red}{9{x^2}} + 1 + \color{green}{6{x^3}} + \color{red}{2{x^2}} + \color{blue}{6x}} \right) }+{ \left( {\color{red}{3{x^2}} + \color{blue}{9x} + 3} \right) + 1 }={ \color{magenta}{x^4} + \color{red}{9{x^2}} + 1 + \color{green}{6{x^3}} + \color{red}{2{x^2}} }+{ \color{blue}{6x} + \color{red}{3{x^2}} }+{ \color{blue}{9x} + 3 + 1 }={ \color{magenta}{x^4} + \color{green}{6{x^3}} + \color{red}{14{x^2}} + \color{blue}{15x} + 5.}\]

Compositions Involving Inverse Functions

Let \(f: A \to B\) be a bijective function from domain \(A\) to codomain \(B.\) Then it has an inverse function \({f^{-1}}\) that maps \(B\) back to \(A.\) Then

\[{{f^{ – 1}} \circ f = {I_A}\;\text{ or }\;}\kern0pt{{f^{ – 1}}\left( {f\left( x \right)} \right) = x,}\]

where \({I_A}\) is the identity function in the domain \(A\) and \(x\) is any element of \(A.\)

Similarly,

\[{f \circ {f^{ – 1}} = {I_B}\;\text{ or }\;}\kern0pt{f\left( {{f^{ – 1}}\left( y \right)} \right) = y,}\]

where \({I_B}\) is the identity function in the codomain \(B\) and \(y\) is any element of \(B.\)


Solved Problems

Click or tap a problem to see the solution.

Example 1

The functions \(f,g\) are defined as sets of ordered pairs \[f = \left\{ {\left( {3,7} \right),\left( {4,1} \right),\left( {5,8} \right),\left( {6,1} \right)} \right\},\] \[g = \left\{ {\left( {5,4} \right),\left( {9,6} \right),\left( {7,3} \right),\left( {2,6} \right)} \right\}.\] Find the composition of functions \(f \circ g.\) State its domain and range.

Example 2

The functions \(f,g\) are defined as \(f\left( x \right) = \large{\frac{2}{{3x – 4}}}\normalsize\) and \(g\left( x \right) = \large{\frac{3}{{4x – 5}}}\normalsize.\) Find the domain of the composition of functions \(f \circ g.\)

Example 3

Consider the functions \(g: \mathbb{R} \to \mathbb{R}\) and \(f: \mathbb{R} \to \mathbb{R}\) defined as \[{g\left( x \right) = {x^2} – x,\;\;}\kern0pt{f\left( x \right) = {x^2} + 2x.}\] Find the compositions of functions:
  1. \(f \circ g\)
  2. \(g \circ f\)
  3. \(f \circ f\)
  4. \(g \circ g\)

Example 4

Let the functions \(g: \mathbb{R} \to \mathbb{R}\) and \(f: \mathbb{R} \to \mathbb{R}\) be defined as \[{g\left( x \right) = \sqrt{x},\;\;}\kern0pt{f\left( x \right) = {x^2} + 1.}\] Find the compositions of functions:
  1. \(f \circ g \circ f\)
  2. \(g \circ f \circ g\)
  3. \(g \circ f \circ f\)

Example 5

The function \(f:\left[ { – 1,\infty } \right) \to \left[ {0,\infty } \right)\) is defined as \(f\left( x \right) = \sqrt {{x^3} + 1} .\) Show that the composition \(f^{-1} \circ f\) is the identity function.

Example 6

The functions \(f, g: \mathbb{R} \to \mathbb{R}\) are defined as \(f\left( x \right) = 2{x^2} + 1,\) \(g\left( x \right) = 1 – x.\) Solve the equation \[f \circ g + g \circ f + 5 = 0.\]

Example 1.

The functions \(f,g\) are defined as sets of ordered pairs \[f = \left\{ {\left( {3,7} \right),\left( {4,1} \right),\left( {5,8} \right),\left( {6,1} \right)} \right\},\] \[g = \left\{ {\left( {5,4} \right),\left( {9,6} \right),\left( {7,3} \right),\left( {2,6} \right)} \right\}.\] Find the composition of functions \(f \circ g.\) State its domain and range.

Solution.

Draw the arrow diagram for the composition of functions \(f \circ g:\)

A composition of discrete functions
Figure 4.

It follows from the diagram:

\[{g\left( 5 \right) = 4,\;\;f\left( 4 \right) = 1,}\;\; \Rightarrow {\left( {f \circ g} \right)\left( 5 \right) = 1;}\]

\[{g\left( 7 \right) = 3,\;\;f\left( 3 \right) = 7,}\;\; \Rightarrow {\left( {f \circ g} \right)\left( 7 \right) = 7;}\]

\[{g\left( 9 \right) = 6,\;\;f\left( 6 \right) = 1,}\;\; \Rightarrow {\left( {f \circ g} \right)\left( 9 \right) = 1;}\]

\[{g\left( 2 \right) = 6,\;\;f\left( 6 \right) = 1,}\;\; \Rightarrow {\left( {f \circ g} \right)\left( 2 \right) = 1.}\]

Hence, the composite function \(f \circ g\) is given by

\[f \circ g = \left\{ {\left( {5,1} \right),\left( {7,7} \right),\left( {9,1} \right),\left( {2,1} \right)} \right\}.\]

The domain of \(f \circ g\) is the set \(\left\{ {5,7,9,2} \right\},\) and the range is the set \(\left\{ {7,1} \right\}.\)

Example 2.

The functions \(f,g\) are defined as \(f\left( x \right) = \large{\frac{2}{{3x – 4}}}\normalsize\) and \(g\left( x \right) = \large{\frac{3}{{4x – 5}}}\normalsize.\) Find the domain of the composition of functions \(f \circ g.\)

Solution.

By definition, \(\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right).\)

The inner function \({g\left( x \right)}\) is defined for all \(x\) except \(x = \large{\frac{5}{4}}\normalsize.\) Hence, the point \(x = \large{\frac{5}{4}}\normalsize\) must be excluded from the domain of the composition \(f \circ g.\)

The outer function \({f\left( x \right)}\) is defined for all \(x\) except \(x = \large{\frac{4}{3}}\normalsize.\) So, we also need to exclude those values of \(x\) for which the inner function \({g\left( x \right)}\) is equal to \(\large{\frac{4}{3}}\normalsize.\) Determine these values.

\[{g\left( x \right) = \frac{3}{{4x – 5}} = \frac{4}{3},}\;\; \Rightarrow {4\left( {4x – 5} \right) = 9, }\;\;\Rightarrow {16x – 20 = 9,}\;\; \Rightarrow {16x = 29,}\;\; \Rightarrow {x = \frac{{29}}{{16}}.}\]

Thus, the domain of \(f \circ g\) consists of all real values of \(x\) except the points \(x = \large{\frac{5}{4}}\normalsize, \large{\frac{29}{16}}\normalsize.\) This can be written in the form

\[{\text{dom}\left( {f \circ g} \right) \text{ = }}\kern0pt{\left( { – \infty ,\frac{5}{4}} \right) \cup \left( {\frac{5}{4},\frac{{29}}{{16}}} \right) \cup \left( {\frac{{29}}{{16}},\infty } \right).}\]

Example 3.

Consider the functions \(g: \mathbb{R} \to \mathbb{R}\) and \(f: \mathbb{R} \to \mathbb{R}\) defined as \[{g\left( x \right) = {x^2} – x,\;\;}\kern0pt{f\left( x \right) = {x^2} + 2x.}\] Find the compositions of functions:
  1. \(f \circ g\)
  2. \(g \circ f\)
  3. \(f \circ f\)
  4. \(g \circ g\)

Solution.

  1. The composite function \(\left(f \circ g\right)\left(x\right)\) is written as \(f\left( {g\left( x \right)} \right).\) We take the outer function \(f\left( x \right)\) and substitute the inner function \({g\left( x \right)}\) for \(x:\) \[{\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) }={ {g^2}\left( x \right) + 2g\left( x \right) }={ {\left( {{x^2} – x} \right)^2} + 2\left( {{x^2} – x} \right) }={ \color{magenta}{x^4} – \color{green}{2{x^3}} + \color{red}{x^2} + \color{red}{2{x^2}} -\color{blue}{2x} }={ \color{magenta}{x^4} -\color{green}{2{x^3}} + \color{red}{3{x^2}} – \color{blue}{2x}.}\]
  2. Calculate the composition \(g \circ f:\) \[{\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) }={ {f^2}\left( x \right) – f\left( x \right) = {\left( {{x^2} + 2x} \right)^2} – \left( {{x^2} + 2x} \right) }={ \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{4{x^2}} – \color{red}{x^2} – \color{blue}{2x} }={ \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{3{x^2}} – \color{blue}{2x}.}\] Notice that \(f \circ g \ne g \circ f,\) that is the composition of functions is not commutative.
  3. The composition of functions \(f \circ f\) is given by \[{\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) }={ {f^2}\left( x \right) + 2f\left( x \right) }={ {\left( {{x^2} + 2x} \right)^2} + 2\left( {{x^2} + 2x} \right) }={ \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{4{x^2}} + \color{red}{2{x^2}} + \color{blue}{4x} }={ \color{magenta}{x^4} + \color{green}{4{x^3}} + \color{red}{6{x^2}} + \color{blue}{4x}.}\]
  4. Similarly we determine the function \(g \circ g:\) \[\require{cancel}{\left(g \circ g\right)\left(x\right) = g\left( {g\left( x \right)} \right) }={ {g^2}\left( x \right) – g\left( x \right) }={ {\left( {{x^2} – x} \right)^2} – \left( {{x^2} – x} \right) }={ \color{magenta}{x^4} – \color{green}{2{x^3}} + \cancel{\color{red}{x^2}} – \cancel{\color{red}{x^2}} + \color{blue}{x} }={ \color{magenta}{x^4} – \color{green}{2{x^3}} + \color{blue}{x}.}\]

Example 4.

Let the functions \(g: \mathbb{R} \to \mathbb{R}\) and \(f: \mathbb{R} \to \mathbb{R}\) be defined as \[{g\left( x \right) = \sqrt{x},\;\;}\kern0pt{f\left( x \right) = {x^2} + 1.}\] Find the compositions of functions:
  1. \(f \circ g \circ f\)
  2. \(g \circ f \circ g\)
  3. \(g \circ f \circ f\)

Solution.

  1. The find the composite function \(f \circ g \circ f,\) we first determine the composition of inner functions \(g \circ f:\) \[{\left(g \circ f\right)\left(x\right) = g\left( {f\left( x \right)} \right) }={ g\left( {{x^2} + 1} \right) }={ \sqrt {{x^2} + 1} .}\] Then, using the associative law, the triple composition is given by \[{\left(f \circ g \circ f\right)\left(x\right) = \left(f \circ \left({g \circ f}\right)\right)\left(x\right) = f\left[ {g\left( {f\left( x \right)} \right)} \right] }={ f\left( {\sqrt {{x^2} + 1} } \right) }={ {\left( {\sqrt {{x^2} + 1} } \right)^2} + 1 }={ {x^2} + 1 + 1 }={ {x^2} + 2.}\]
  2. Here we first find the composition \(f \circ g:\) \[{\left(f \circ g\right)\left(x\right) = f\left( {g\left( x \right)} \right) }={ f\left( {\sqrt x } \right) }={ {\left( {\sqrt x } \right)^2} + 1 }={ x + 1.}\] Hence, \[{\left(g \circ f \circ g\right)\left(x\right) = \left(g \circ \left( {f \circ g} \right)\right)\left(x\right) }={ g\left[ {f\left( {g\left( x \right)} \right)} \right] }={ g\left( {x + 1} \right) }={ \sqrt {x + 1} .}\]
  3. Calculate the composition of functions \(g \circ f \circ f\) using the same approach. Since \[{\left(f \circ f\right)\left(x\right) = f\left( {f\left( x \right)} \right) }={ f\left( {{x^2} + 1} \right) }={ {\left( {{x^2} + 1} \right)^2} + 1 }={ {x^4} + 2{x^2} + 1 + 1 }={ {x^4} + 2{x^2} + 2,}\] the triple composition \(g \circ f \circ f\) is given by \[{\left(g \circ f \circ f\right)\left(x\right) = \left(g \circ \left( {f \circ f} \right)\right)\left(x\right) }={ g\left[ {f\left( {f\left( x \right)} \right)} \right] }={ g\left( {{x^4} + 2{x^2} + 2} \right) }={ \sqrt {{x^4} + 2{x^2} + 2} .}\]

Example 5.

The function \(f:\left[ { – 1,\infty } \right) \to \left[ {0,\infty } \right)\) is defined as \(f\left( x \right) = \sqrt {{x^3} + 1} .\) Show that the composition \(f^{-1} \circ f\) is the identity function.

Solution.

Find the inverse function \(f^{-1}\) by solving the equation \(y = \sqrt {{x^3} + 1}\) for \(x:\)

\[{y = \sqrt {{x^3} + 1} ,}\;\; \Rightarrow {{x^3} + 1 = {y^2},}\;\; \Rightarrow {{x^3} = {y^2} – 1,}\;\; \Rightarrow {x = \sqrt[3]{{{y^2} – 1}}.}\]

To calculate the composition of functions \(f^{-1} \circ f,\) we substitute \(y = \sqrt {{x^3} + 1}\) in the equation \(x = \sqrt[3]{{{y^2} – 1}}.\) For any \(x\) in the domain of \(f,\) we have

\[{\left({f^{ – 1}} \circ f\right)\left(x\right) = {f^{ – 1}}\left( {f\left( x \right)} \right) }={ {f^{ – 1}}\left( y \right) = \sqrt[3]{{{y^2} – 1}} }={ \sqrt[3]{{{{\left( {\sqrt {{x^3} + 1} } \right)}^2} – 1}} }={ \sqrt[3]{{{x^3} + \cancel{1} – \cancel{1}}} }={ \sqrt[3]{{{x^3}}} }={ x.}\]

Thus, \({f^{ – 1}} \circ f = I,\) where \(I\) is the identity function in the domain of \(f.\)

Example 6.

The functions \(f, g: \mathbb{R} \to \mathbb{R}\) are defined as \(f\left( x \right) = 2{x^2} + 1,\) \(g\left( x \right) = 1 – x.\) Solve the equation \[f \circ g + g \circ f + 5 = 0.\]

Solution.

Calculate the compositions of functions \(f \circ g\) and \(g \circ f:\)

\[{\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right) }={ f\left( {1 – x} \right) }={ 2{\left( {1 – x} \right)^2} + 1 }={ 2\left( {1 – 2x + {x^2}} \right) + 1 }={ 2 – 4x + 2{x^2} + 1 }={ 2{x^2} – 4x + 3;}\]

\[{\left( {g \circ f} \right)\left( x \right) }={ g\left( {f\left( x \right)} \right) }={ g\left( {2{x^2} + 1} \right) }={ 1 – \left( {2{x^2} + 1} \right) }={\cancel{1} – 2{x^2} – \cancel{1} }={ – 2{x^2}.}\]

Plug both expressions into the original equation and solve it for \(x:\)

\[{f \circ g + g \circ f + 5 = 0,}\;\; \Rightarrow {\cancel{2{x^2}} – 4x + 3 – \cancel{2{x^2}} + 5 = 0,}\;\; \Rightarrow {8 – 4x = 0,}\;\; \Rightarrow {x = 2.}\]