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College Algebra/Composite and Inverse Functions/Decomposing Composite Functions

Composite and Inverse Functions

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Decomposing Composite Functions

Composite functions can be decomposed into two or more functions using methods such as the properties of operations, the properties of exponents, and the word form of the function.

Decomposing functions can be useful when working with complicated functions. When the complicated function is written as a composition of simpler functions, it can be easier to understand things about the function. There is often more than one way to decompose a function.

The function h(x)=(x+1)12h(x)=(x+1)^{12} can be decomposed as f(g(x))f(g(x)) in many ways. For example:

  • f(x)=x12f(x)=x^{12} and g(x)=x+1g(x)=x+1
  • f(x)=x2f(x)=x^2 and g(x)=(x+1)6g(x)=(x+1)^6
  • f(x)=x4f(x)=x^4 and g(x)=(x+1)3g(x)=(x+1)^3
  • f(x)=(x+2)12f(x)=(x+2)^{12} and g(x)=x−1g(x)=x-1

One method for decomposing a function is to focus on how its equation can be written in words or spoken aloud. The word form can help to identify the parts of the function rule and the order of these parts.


Step-By-Step Example
Decomposing a Radical Function
Decompose the given function into two functions:
h(x)=x2−53h(x)=\sqrt[3]{x^{2}-5}
Step 1

State the function in words, and identify two separate parts.

Word phrase: the cube root of {\color{#0047af} \text{the cube root of }} x squared minus 5 {\color{#c42126} x \text{ squared minus 5 }}

First part: the cube root of {\color{#0047af}\text{the cube root of }}
Second part: x squared minus 5 {\color{#c42126} x \text{ squared minus 5 }}

Step 2
Translate the second part back into symbols, and set it equal to g(x)g(x). This will be the inside function of the composition.
x squared minus 5=x2−5g(x)=x2−5\begin{aligned}{\color{#c42126} x \text{ squared minus 5}}&=x^{2}-5\\g(x)&=x^2-5\end{aligned}
Step 3
In the full word phrase, replace the second part with xx. Then translate the remaining phrase back into symbols, and set it equal to f(x)f(x).
the cube root of x=x3f(x)=x3\begin{aligned}{\color{#0047af}\text{the cube root of }}x&=\sqrt[3]{x}\\ f(x)&=\sqrt[3]{x}\end{aligned}
Solution
Check that the functions can be composed into h(x)h(x).
h(x)=x2−53h(x)=\sqrt[3]{x^{2}-5}
Remember that:
f(x)=x3g(x)=x2−5\begin{aligned}f(x) &= \sqrt[3]{x}\\g(x) &= x^{2}-5\end{aligned}
Use the values to check the function:
f(g(x))=x2−53=h(x)f(g(x)) = \sqrt[3]{x^{2}-5} = h(x)
So, the function can be decomposed as f(g(x))f(g(x)).
Step-By-Step Example
Decomposing a Rational Function
Decompose the given function into two functions.
h(x)=1x+3h(x)=\frac{1}{x+3}
Step 1

State the function in words, and identify two separate parts.

Word phrase: one divided by{\color{#0047af}\text{one divided by}} the quantity{\color{#c42126} \text{the quantity}} x{\color{#c42126}x} plus 3{\color{#c42126} \text{plus 3}}

First part: one divided by{\color{#0047af}\text{one divided by}}
Second part: the quantity{\color{#c42126} \text{the quantity}} x{\color{#c42126}x} plus 3{\color{#c42126} \text{plus 3}}

Another way to state the function in words is to replace divided by into reciprocal.

Word phrase: the reciprocal of{\color{#0047af}\text{the reciprocal of}} the quantity{\color{#c42126} \text{the quantity}} x{\color{#c42126} x} plus 3{\color{#c42126} \text{plus 3}}

First part: the reciprocal of{\color{#0047af}\text{the reciprocal of}}
Second part: the quantity{\color{#c42126} \text{the quantity}} x{\color{#c42126} x} plus 3{\color{#c42126} \text{plus 3}}

Step 2
Translate the second part back into symbols, and set it equal to g(x)g(x). This will be the inside function of the composition.
the quantity x plus 3=x+3g(x)=x+3\begin{aligned}{\color{#c42126}\text{the quantity } x \text{ plus 3}}&=x+3\\g(x)&=x+3\end{aligned}
Step 3
In the full word phrase, replace the second part with xx. Then translate the remaining phrase back into symbols and set it equal to f(x)f(x).
one divided by x or the reciprocal of x=1xf(x)=1x\begin{aligned}{\color{#0047af}\text{one divided by }x} \text{ or } {\color{#0047af} \text{the reciprocal of }x}&=\frac{1}{x}\\ f(x) &=\frac{1}{x}\end{aligned}
Solution
Check that the functions can be composed into h(x)h(x).
h(x)=1x+3h(x)=\frac{1}{x+3}
Remember that:
f(x)=1xg(x)=x+3\begin{aligned}f(x)&= \frac{1}{x} \\ g(x)&=x+3\end{aligned}
Use the values to check the functions:
f(g(x))=1x+3=h(x)f(g(x)) = \frac{1}{x+3} = h(x)
So, the given function can be decomposed as f(g(x))f(g(x)).

Associative Property

Since composition of functions is associative, the way in which three or more functions are grouped in a composite function does not change the result.
Composition of functions is associative, which means that the way in which functions are grouped when they are composed does not matter.
(f∘(g∘h))(x)=((f∘g)∘h)(x)(f \circ (g \circ h))(x)=((f \circ g) \circ h)(x)
Step-By-Step Example
Confirming That Composition Is Associative
Consider the given values for f(x)f(x), g(x)g(x), and h(x)h(x):
f(x)=x+6g(x)=8xh(x)=x2\begin{aligned}f(x)&=x+6\\g(x)&=8x\\h(x)&=\frac{x}{2}\end{aligned}
Identify (f∘(g∘h))(x)(f \circ (g \circ h))(x) and ((f∘g)∘h)(x)((f \circ g) \circ h)(x), and then determine if they are equivalent.
Step 1
Identify the composition of the two inside functions of (f∘(g∘h))(x)(f \circ (g \circ h))(x).
g(h(x))=g(x2)=8(x2)=4x\begin{aligned}g(h(x))&=g\left(\frac{x}{2}\right)\\&=8\left(\frac{x}{2}\right)\\&=4x\end{aligned}
Step 2
Use the result to find (f∘(g∘h))(x)(f \circ (g \circ h))(x).
(f∘(g(h(x)))=f(4x)=(4x)+6=4x+6\begin{aligned}(f \circ(g(h(x)))&=f(4x)\\&=(4x)+6\\&=4x+6\end{aligned}
Step 3
Identify the composition of the two inside functions of ((f∘g)∘h)(x)((f \circ g) \circ h)(x).
f(g(x))=f(8x)=(8x)+6=8x+6\begin{aligned}f(g(x))&=f(8x)\\&=(8x)+6\\&=8x+6\end{aligned}
Step 4
Use the result to find ((f∘g)∘h)(x)((f \circ g) \circ h)(x).
f(g(x))∘h(x)=f(g(x2))=8(x2)+6=4x+6\begin{aligned}f(g(x))\circ h(x)&=f\left(g\left(\frac{x}{2}\right)\right)\\&=8\left(\frac{x}{2}\right)+6\\&=4x+6\end{aligned}
Solution
Compare the two composite functions.
4x+6=4x+6(f∘(g∘h))(x)=((f∘g)∘h)(x)\begin{aligned}4x+6&=4x+6\\(f \circ (g \circ h))(x)&=((f \circ g) \circ h)(x)\end{aligned}
Since the composite functions are equal, the composition of the functions can be written without grouping, such as:
(f∘g∘h)(x)(f \circ g \circ h)(x)
The meaning will not change.

Composition Is Not Commutative

Composition is not commutative, so composing functions in a different order does not necessarily yield the same result.
In general, the composition of functions is not commutative, which means that if two functions are composed in a different order, the composite functions are not necessarily equal.
(f∘g)(x)≠(g∘f)(x)(f \circ g)(x) \neq (g \circ f)(x)
Step-By-Step Example
Determining Whether a Composition is Non-Commutative

Let f(x)=2xf(x)=2x and g(x)=x2g(x)=x^2. Identify (f∘g)(x)(f \circ g)(x) and (g∘f)(x)(g \circ f)(x), and determine whether they are equivalent.


Step 1
Identify (f∘g)(x)(f \circ g)(x). Substitute the inside function into the outside function, and simplify.
(f∘g)(x)=f(g(x))=2(x2)=2x2\begin{aligned}(f \circ g)(x)&=f(g(x))\\&=2\left(x^2\right)\\&=2x^2\end{aligned}
Step 2
Identify (g∘f)(x)(g \circ f)(x). Substitute the inside function into the outside function, and simplify.
(g∘f)(x)=g(f(x))=(2x)2=4x2\begin{aligned}(g \circ f)(x)&=g(f(x))\\&=(2x)^2\\&=4x^2\end{aligned}
Solution
Compare the two composite functions.
2x2≠4x2(f∘g)(x)≠(g∘f)(x)\begin{aligned}2x^2&\neq 4x^2\\(f \circ g)(x) &\neq (g \circ f)(x) \end{aligned}
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