# Composition of Functions Examples

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 \[ (f_o g)(x) = f(g(x)) \] 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 ValuesFunctions \( f \) and \( g \) are defined by their tables as follows
1) Find the values of: a) \( (f_o g)( 2 )\) , b) \( (f_o g)( 6 )\) , c) \( (f_o g)( 7 )\) , d) \( (f_o g)( 8 )\) e) \( (g_o f)( 3 ) \) , f) \( (g_o f)( 5 ) \) , g) \( (g_o f)( 6 ) \) , h) \( (g_o f)( 8 ) \) i) \( (g_o g)( 2 ) \) 2) Find the domain of \( f_o g \) and \( g_o f \) ## Solution to Example 11)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 GraphsFunctions \( f \) and \( g \) are given by their graphs shown belowFind 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 2a) 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 FormulasFunctions \( f \) 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 3a) 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 DomainsFunctions \( f \) 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 DomainsFunctions \( f \) 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 51)\( (f_o g)( x ) = f(g(x)) = (g(x))^2 + 1 = (\sqrt{4-x^2})^2 + 1 = 5 - x^2 \) 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 1In 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 commutativeLet \( f(x) = x^2 - 1 \) 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 2Let \( f, g \) 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 associativeShow that \( (f_o (g_o h))(x) = ((f_o g)_o h)(x) \)## Solution to Example 71) Left sideUse 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 3If \( f \) and \( g \) are invertible, then \( (f_o g)^{-1} = g^{-1}_o f^{-1} \) ## Example 8: The inverse of the composition of FunctionsLet \( f(x) = \dfrac{1}{x-1} \) and \( g(x) = - x + 5 \)Show that \( (f_o g)^{-1} = g^{-1}_o f^{-1} \) ## Solution to Example 8We 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 \( (g^{-1}_o f^{-1})(x) = (f_o g)^{-1}(x) \) Function composition |