Composition of Functions Examples (2024)

The definition and the properties of the composition of functions are discussed through examples with detailed solutions and explanations.

Definition of Composition of Functions

Let \( f \) and \( g \) be two functions. If we let function \( f \) take as input function \( g \) , see diagram below, the resulting function is called the composite function or composition of \( f \) and \( g \) denoted by \( f_o g \) and is defined by

\[ (f_o g)(x) = f(g(x)) \]

This composite function is defined if \(x \) is in the domain of \( g \) and \( g(x) \) is in the domain of \( f \). (see digram below).

Examples with Solutions

Example 1: Composition of Functions Given Tables of Values

Functions \( f \) and \( g \) are defined by their tables as follows

Solution to Example 1

1)
a)Use definition of composition of functions to write
\( (f_o g)( 2 ) = f(g(2)) \)
Use table to find the value of \( g(2) = 6 \)
Substitute back in \( f(g(2)) \) to write
\( (f_o g)( 2 ) = f(g(2)) = f(6)\)
Use table to evaluate \(f(6)\)
\( (f_o g)( 2 ) = f(g(2)) = f(6) = 8\)
Use similar steps as above
b) \( (f_o g)( 6 ) = f(g(6)) = f(3) = 4 \)
c) \( (f_o g)( 7 ) = f(g(7)) = f(9) = \) undefined
d) \( (f_o g)( 8 ) = f(g(8)) = f(5) = 7\)
e) \( (g_o f)( 3 ) = g(f(3)) = g(4) = \) undefined
f) \( (g_o f)( 5 ) = g(f(5)) = g(7) = 9\)
g) \( (g_o f)( 6 ) = g(f(6)) = g(8) = 5\)
h) \( (g_o f)( 8 ) = g(f(8)) = g(11) = \) undefined
i) \( (g_o g)( 2 ) = g(g(2)) = g(6) = 3\)
2)
Use the results in part 1) to write the ordered pairs defining \( g_o f \) and \( f_o g \) and then deduce the domain.
\( g_o f : \{(2,6),(6,4),(8,7)\} \) , hence the domain of \( g_o f \) is given by: {2,6,8}
\( f_o g : \{(5,9),(6,5)\} \) , hence the domain of \( f_o g \) is given by: {5,6}

Example 2: Composition of Functions Given by their Graphs

Functions \( f \) and \( g \) are given by their graphs shown below
Find the values of:
a) \( (f_o g)( -2 )\) b) \( (f_o g)( 0 )\) c) \( (f_o g)( 4 )\) d) \( (f_o g)( 8 )\) e) \( (g_o f)( 2 )\) f) \( (g_o f)( 0 )\)

Solution to Example 2

a) Use definition of composition of functions to write
\( (f_o g)( -2 ) = f(g(-2)) \)
Use the graph of g to find the value of g(-2) = - 3
Substitute back in \( f(g(-2)) \) to write
\( (f_o g)( -2 ) = f(g(-2)) = f(-3) \)
Use the graph of f to evaluate \( f(-3) \)
\( (f_o g)( -2 ) = f(g(-2)) = f(-3) = 9 \)
Use similar steps as above to evaluate the following
b) \( (f_o g)( 0 ) = f(g(0)) = f(-2) = 4\)
c) \( (f_o g)( 4 ) = f(g(4)) = f(0) = 0\)
d) \( (f_o g)( 8 ) = f(g(8)) = f(2) = 4\)
e) \( (g_o f)( 2 ) = g(f(2)) = g(4) = 0\)
f) \( (g_o f)( 0 ) = g(f(0)) = g(0) = -2 \)

Example 3: Composition of Functions Given by their Formulas

Functions \( f \) and \( g \) are defined by the formulas: \( f(x) = 2x + 1 \) and \( g(x) = - x + 1 \)
a) Find the composite function \( (g_o f)( x ) \).
b) Find the composite function \( (f_o g)( x ) \).

Solution to Example 3

a)
Use the definition of composition of functions to write
\( (g_o f)( x ) = g(f(x)) \)
Express \( g(f(x)) \) in terms of \( f(x) \)
\( (g_o f)( x ) = g(f(x)) = - ( f(x) ) + 1 \)
Substitute \( f(x) \) by its formula
\( (g_o f)( x ) = g(f(x)) = - ( 2x + 1 ) + 1 \)
Simplify
\( (g_o f)( x ) = g(f(x)) = - ( 2x + 1 ) + 1 = - 2 x \)
b)
Use definition of composition of functions to write
\( (f_o g)( x ) = f(g(x)) \)
Express \( f(g(x)) \) in terms of \( g(x) \)
\( (f_o g)( x ) = f(g(x)) = 2 g(x) + 1\)
Substitute \( g(x) \) by its formula and simplify
\( (f_o g)( x ) = f(g(x)) = 2 g(x) + 1 = 2(- x + 1) + 1 = -2x + 3\)
Note that \( (g_o f)( x ) \ne (f_o g)( x ) \) , which means that the composition of functions is not commutative.

Example 4: Composition of Functions and their Domains

Functions \( f \) and \( g \) are defined by the formulas: \( f(x) = 2x + 1 \) and \( g(x) = \sqrt{x - 1} \)
Find the composite function \( (g_o f)( x ) \) and and its domain.

Solution to Example 4

\( (g_o f)( x ) = g(f(x)) = \sqrt{f(x) - 1} = \sqrt{2x+1 - 1} =\sqrt{2x}\)
Two conditions for the domain of \( (g_o f) \)
1) \( x \) must be in the domain of \( f \) which is given by the interval : \( (-\infty , +\infty) \)
2) \( f(x) \) must be in the domain of \( g \) which is the domain of \( g(f(x)) \) .
The domain of \( g(f(x)) = \sqrt{2x} \) is found by solving the inequality: \( 2x \ge 0 \) wihich has the solution set given by the interval: \( [0 , + \infty) \)
The domain of \( (g_o f) \) is given by the intersection (red) of the sets in 1) and 2) ( in blue) : \( [0 , + \infty) \)

Example 5: Composition of Functions and their Domains

Functions \( f \) and \( g \) are defined by the formulas: \( f(x) = x^2 + 1\) and \( g(x) = \sqrt{4-x^2} \)
1) Find the composite function \( (f_o g)( x ) \) and and its domain.
2) Graph function \( f \), \( g \) and \( f_o g \) in the same system of coordinates.

Solution to Example 5

1)
\( (f_o g)( x ) = f(g(x)) = (g(x))^2 + 1 = (\sqrt{4-x^2})^2 + 1 = 5 - x^2 \)
Two conditions for the domain of \( (f_o g) \)
1) \( x \) must be in the domain of g which is found by solving the inequality \( 4-x^2 \ge 0 \). The domain of g is given by the interval : \( [-2, + 2] \)
2) \( g(x) \) must be in the domain of f which is the domain of \( f(g(x)) \) which is the interval \( (-\infty , +\infty) \).
The domain of \( (f_o g) \) is given by the intersection (red) of the sets in 1) and 2) (blue) : \( [-2 , + 2] \)

2)
Below are shown the graphs of \( f \), \( g \) and \( f_o g \).

Properties of Composite Functions

Property 1

In general \( (f_o g)(x) \ne (g_o f)(x) \) and therefore the composition of functions is

not commutative

.
Example 3 above already shows that the composition of functions is not commutative.

Example 6: The composition is not commutative

Let \( f(x) = x^2 - 1 \) and \( g(x) = 2x \)
Show that \( (f_o g)(x) \ne (g_o f)(x) \)

Solution to Example 6

\( (f_o g)(x) = f(g(x)) = (g(x))^2 - 1 = (2x)^2 - 1 = 4x^2 - 1 \)
\( (g_o f)(x) = g(f(x)) = 2 f(x) = 2 (x^2 - 1) = 2x^2 - 2 \)
Therefore \( (f_o g)(x) \ne (g_o f)(x) \) and the composition of functions is not commutative.

Property 2

Let \( f, g \) and \( h \) be three functions, \( f_o (g_o h) = (f_o g)_o h \) and therefore the composition of funtions is

associative

Example 7: The composition of Functions is associative

Show that \( (f_o (g_o h))(x) = ((f_o g)_o h)(x) \)

Solution to Example 7

1) Left side
Use definition of composition to write
\( (f_o (g_o h))(x) = f((g_o h)(x)) \)
Use definition of composition again to write
\( = f(g(h(x)) \)
2) Right side
Use definition of composition to write
\( ((f_o g)_o h)(x) = (f_o g)(h(x)) \)
Use definition of composition to write
\( = f(g(h(x))) \)
Therefore
\( (f_o (g_o h))(x) = ((f_o g)_o h)(x) \)

Property 3

If \( f \) and \( g \) are invertible , then \( (f_o g)^{-1} = g^{-1}_o f^{-1} \)

Example 8: The inverse of the composition of Functions

Let \( f(x) = \dfrac{1}{x-1} \) and \( g(x) = - x + 5 \)
Show that \( (f_o g)^{-1} = g^{-1}_o f^{-1} \)

Solution to Example 8

We first calculate \( (f_o g)(x) \) and then its inverse \( (f_o g)^{-1}(x) \)
\( (f_o g)(x) = f(g(x)) = \dfrac{1}{g(x)-1} = \dfrac{1}{-x + 5 -1} = \dfrac{1}{-x + 4} \)
\( (f_o g)^{-1}(x) = - \dfrac{1}{x} + 4 \)
We now calculate the inverses \( f^{-1}(x) \) , \( g^{-1}(x) \) and then calculate the composition \( (g^{-1}_o f^{-1})(x) \).
\( f^{-1}(x) = \dfrac{1}{x} + 1 \)
\( g^{-1}(x) = - x + 5 \)
\( (g^{-1}_o f^{-1})(x) = g^{-1}(f^{-1}(x)) = - (f^{-1}(x)) + 5 = - ( \dfrac{1}{x} + 1) + 5 = - \dfrac{1}{x} + 4 \)
We conclude that
\( (g^{-1}_o f^{-1})(x) = (f_o g)^{-1}(x) \)