Pre-Calculus Functions (advanced)

Function decomposition (writing as a composite)

Q: What does it mean to decompose a function?

To rewrite a single function h of x as a composition f of g of x by choosing an inner function g and an outer function f. It is the reverse of composing.

Q: How do you find the inner and outer functions?

Find the expression you would evaluate first, the part inside a root, power, or denominator. That is the inner function g; the operation applied to it is the outer function f.

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Function Decomposition

Common Core Pre-Calculus • Standard F-BF.1c • Functions

Function Decomposition is a topic in Functions in the Common Core State Standards. It is aligned to Standard F-BF.1c, which requires students to build a function from simpler functions, including expressing it as a composition.

Function decomposition is the reverse of composition: it rewrites a single function \(h(x)\) as \(f(g(x))\) by identifying an inner function \(g\) and an outer function \(f\).

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Theory

Decomposing a function is the reverse of composing: you start with a single function \(h\) and rewrite it as \(h=f\circ g\), i.e.

\[h(x)=f\big(g(x)\big),\]

by choosing an inner function \(g\) and an outer function \(f\).

The trick is to spot the “inside” block — the expression tucked under a root, raised to a power, or sitting in a denominator. That block becomes \(g(x)\); whatever is done to it becomes \(f(u)\).

Why it matters: recognizing a function as a composite is the first step of the chain rule in calculus, so decomposition is a skill you reuse constantly later.

\(h(x)=(3x-1)^4\) splits into \(g(x)=3x-1\) then \(f(u)=u^4\).

Read the inner block (base) and the outer action (the power) straight off the expression.

A decomposition satisfies

\[h(x)=f\big(g(x)\big),\qquad g=\text{inner},\quad f=\text{outer}\]

h of x equals f of g of x, with g inner and f outer

There is no single correct answer — any pair \((f,g)\) that rebuilds \(h\) when composed is valid. The most useful choice is usually the one that makes \(g\) the expression you would evaluate first.

Check every decomposition by composing. If \(f(g(x))\ne h(x)\), the split is wrong.

How to decompose a function

Find the inside. Look for the expression under a root, inside a power, or in a denominator — the part you would compute first.
Set \(g(x)=\) that inside expression.
Set \(f(u)=\) whatever is done to it, written in terms of a placeholder \(u\).
Verify by forming \(f(g(x))\) — it must reproduce \(h(x)\).

Tip: if \(h\) has several layers (a power of a root of a linear expression), you can decompose into three or more functions the same way.

Example 1 — Decompose a power

Write \(h(x)=(3x-1)^4\) as a composition \(f\circ g\).

Solution

The expression inside the power is \(3x-1\); the outer action is “raise to the 4th power.” So take

\(g(x)\)	\(=\)	\(3x-1\)
\(f(u)\)	\(=\)	\(u^4\)

Check by composing:

\(f(g(x))\)

\(=\)

\((3x-1)^4=h(x)\)

inner g of x is 3x minus 1; outer f of u is u to the fourth

Example 2 — Decompose a radical

Decompose \(h(x)=\sqrt{x^2+9}\) into an inner and an outer function.

Solution

The radicand \(x^2+9\) sits inside the square root, so let it be the inner function and let the square root be the outer function.

\(g(x)\)	\(=\)	\(x^2+9\)
\(f(u)\)	\(=\)	\(\sqrt{u}\)
\(f(g(x))\)	\(=\)	\(\sqrt{x^2+9}=h(x)\)

inner is x squared plus 9; outer is the square root

Example 3 — A denominator inside

Express \(h(x)=\dfrac{1}{(x+2)^2}\) as \(f\circ g\).

Solution

Take the inner function to be the base \(x+2\); the outer function squares and reciprocates.

\(g(x)\)	\(=\)	\(x+2\)
\(f(u)\)	\(=\)	\(\dfrac{1}{u^2}\)
\(f(g(x))\)	\(=\)	\(\dfrac{1}{(x+2)^2}=h(x)\)

inner is x plus 2; outer is 1 over u squared

Example 4 — Decomposition is not unique

Give two different decompositions of \(h(x)=|2x-7|\).

Solution

One natural split puts everything under the bars inside:

\(g(x)=2x-7,\ \ f(u)\)

\(=\)

\(|u|\)

Another keeps only \(2x\) inside and lets the outer function handle the shift:

\(g(x)=2x,\ \ f(u)\)

\(=\)

\(|u-7|\)

Both give \(f(g(x))=|2x-7|\), so a composite can be decomposed in more than one valid way.

two valid decompositions of the absolute value of 2x minus 7

Common pitfalls

Decomposition is not unique. Several \((f,g)\) pairs can be correct; don't expect one “official” answer.

Keep the placeholder honest. Write the outer function in terms of \(u\) (or \(x\)) — \(f(u)=u^4\), not \(f(x)=(3x-1)^4\), which just copies \(h\).

Don't split into a product. \(h=f\circ g\) is a composition, not \(f(x)\cdot g(x)\).

Frequently asked questions

What does it mean to decompose a function?

To rewrite a single function \(h(x)\) as a composition \(f(g(x))\) by choosing an inner function \(g\) and an outer function \(f\). It is the reverse of composing.

How do you find the inner and outer functions?

Find the expression you would evaluate first — the part inside a root, power, or denominator. That is the inner function \(g\); the operation applied to it is the outer function \(f\).

Is a decomposition unique?

No. A function can be written as a composite in more than one valid way; any pair \((f,g)\) with \(f(g(x))=h(x)\) works.

Why is decomposition useful?

It is the first step of the chain rule in calculus: to differentiate a composite you must first see it as an inner and an outer function.

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Inverse functions (verifying with composition, domain restrictions)