The definition and the properties of the composition of functions are discussed through examples with detailed step-by-step solutions and explanations hidden in collapsible dropdowns.
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 \( f \circ g \) and defined by:
\[ (f \circ g)(x) = f(g(x)) \]This composite function is defined if \( x \) is in the domain of \( g \) and \( g(x) \) is in the domain of \( f \).
Functions \( f \) and \( g \) are defined by their tables as follows:
| \( x \) | \( f(x) \) | \( x \) | \( g(x) \) | |
| 3 | 4 | 2 | 6 | |
| 5 | 7 | 6 | 3 | |
| 6 | 8 | 7 | 9 | |
| 8 | 11 | 8 | 5 |
1) Find the values of:
a) \( (f \circ g)(2) \), b) \( (f \circ g)(6) \), c) \( (f \circ g)(7) \), d) \( (f \circ g)(8) \)
e) \( (g \circ f)(3) \), f) \( (g \circ f)(5) \), g) \( (g \circ f)(6) \), h) \( (g \circ f)(8) \)
i) \( (g \circ g)(2) \)
2) Find the domain of \( f \circ g \) and \( g \circ f \).
1)
a) \( (f \circ g)(2) = f(g(2)) = f(6) = 8 \)
b) \( (f \circ g)(6) = f(g(6)) = f(3) = 4 \)
c) \( (f \circ g)(7) = f(g(7)) = f(9) = \text{undefined} \)
d) \( (f \circ g)(8) = f(g(8)) = f(5) = 7 \)
e) \( (g \circ f)(3) = g(f(3)) = g(4) = \text{undefined} \)
f) \( (g \circ f)(5) = g(f(5)) = g(7) = 9 \)
g) \( (g \circ f)(6) = g(f(6)) = g(8) = 5 \)
h) \( (g \circ f)(8) = g(f(8)) = g(11) = \text{undefined} \)
i) \( (g \circ g)(2) = g(g(2)) = g(6) = 3 \)
2)
Using the results in part 1 to write the ordered pairs defining each composition:
\( f \circ g : \{(2,8), (6,4), (8,7)\} \), hence the domain of \( f \circ g \) is: \(\{2, 6, 8\}\)
\( g \circ f : \{(5,9), (6,5)\} \), hence the domain of \( g \circ f \) is: \(\{5, 6\}\)
Functions \( f \) and \( g \) are given by their graphs shown below:
Find the values of:
a) \( (f \circ g)(-2) \) b) \( (f \circ g)(0) \) c) \( (f \circ g)(4) \) d) \( (f \circ g)(8) \) e) \( (g \circ f)(2) \) f) \( (g \circ f)(0) \)
a) \( (f \circ g)(-2) = f(g(-2)) = f(-3) = 9 \)
b) \( (f \circ g)(0) = f(g(0)) = f(-2) = 4 \)
c) \( (f \circ g)(4) = f(g(4)) = f(0) = 0 \)
d) \( (f \circ g)(8) = f(g(8)) = f(2) = 4 \)
e) \( (g \circ f)(2) = g(f(2)) = g(4) = 0 \)
f) \( (g \circ f)(0) = g(f(0)) = g(0) = -2 \)
Functions \( f \) and \( g \) are defined by the formulas: \( f(x) = 2x + 1 \) and \( g(x) = -x + 1 \).
a) Find the composite function \( (g \circ f)(x) \).
b) Find the composite function \( (f \circ g)(x) \).
a)
\( (g \circ f)(x) = g(f(x)) = -(2x + 1) + 1 = -2x \)
b)
\( (f \circ g)(x) = f(g(x)) = 2(-x + 1) + 1 = -2x + 3 \)
Note that \( (g \circ f)(x) \neq (f \circ g)(x) \), demonstrating that function composition is not commutative.
Functions \( f \) and \( g \) are defined by the formulas: \( f(x) = 2x + 1 \) and \( g(x) = \sqrt{x - 1} \).
Find the composite function \( (g \circ f)(x) \) and its domain.
Two conditions for the domain of \( (g \circ f) \):
The domain of \( (g \circ f) \) is the intersection of these sets: \( [0, +\infty) \).
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 \( (f \circ g)(x) \) and its domain.
2) Graph functions \( f \), \( g \), and \( f \circ g \) in the same system of coordinates.
1)
\[ (f \circ g)(x) = f(g(x)) = (g(x))^2 + 1 = (\sqrt{4 - x^2})^2 + 1 = 5 - x^2 \]Conditions for the domain:
The domain of \( (f \circ g) \) is: \( [-2, 2] \).
2)
In general, \( (f \circ g)(x) \neq (g \circ f)(x) \), meaning function composition is not commutative.
Let \( f(x) = x^2 - 1 \) and \( g(x) = 2x \). Show that \( (f \circ g)(x) \neq (g \circ f)(x) \).
Therefore, \( (f \circ g)(x) \neq (g \circ f)(x) \).
Let \( f, g, \) and \( h \) be three functions. Then \( f \circ (g \circ h) = (f \circ g) \circ h \), meaning function composition is associative.
Show that \( (f \circ (g \circ h))(x) = ((f \circ g) \circ h)(x) \).
Left side:
\( (f \circ (g \circ h))(x) = f((g \circ h)(x)) = f(g(h(x))) \)
Right side:
\( ((f \circ g) \circ h)(x) = (f \circ g)(h(x)) = f(g(h(x))) \)
Both sides yield identical expressions, proving associativity.
If \( f \) and \( g \) are invertible, then \( (f \circ g)^{-1} = g^{-1} \circ f^{-1} \).
Let \( f(x) = \dfrac{1}{x - 1} \) and \( g(x) = -x + 5 \). Show that \( (f \circ g)^{-1} = g^{-1} \circ f^{-1} \).
First, calculate \( (f \circ g)(x) \) and its inverse:
\[ (f \circ g)(x) = f(g(x)) = \dfrac{1}{(-x + 5) - 1} = \dfrac{1}{-x + 4} \]Finding the inverse \( (f \circ g)^{-1}(x) = -\dfrac{1}{x} + 4 \).
Next, calculate the individual inverses and their composition:
\( f^{-1}(x) = \dfrac{1}{x} + 1 \)
\( g^{-1}(x) = -x + 5 \)
\[ (g^{-1} \circ f^{-1})(x) = g^{-1}(f^{-1}(x)) = -\left(\dfrac{1}{x} + 1\right) + 5 = -\dfrac{1}{x} + 4 \]We conclude that \( (g^{-1} \circ f^{-1})(x) = (f \circ g)^{-1}(x) \).