Composition of Functions

By the end of this lesson, you will be able to compute $(f \circ g)(x)$ and $(g \circ f)(x)$, determine the domain of a composition, decompose a function into simpler components, and recognize composition as a key tool in calculus (chain rule).

The Big Idea

Function composition applies one function to the output of another: $(f \circ g)(x) = f(g(x))$. This "function of a function" concept is everywhere in mathematics — every complex formula in science is a composition. In calculus, the chain rule differentiates compositions. Understanding composition now makes chain rule intuitive later.

Notation and Definition

$$(f \circ g)(x) = f(g(x))$$

Apply $g$ first, then $f$. Think of it as feeding $x$ into $g$, then feeding $g$'s output into $f$.

Order matters: $(f \circ g)(x) \neq (g \circ f)(x)$ in general.

Domain of a Composition

$\text{Dom}(f \circ g)$ requires:

1. $x$ must be in the domain of $g$
2. $g(x)$ must be in the domain of $f$

Process: restrict the domain of $g$ to only those $x$ where $g(x)$ is in $\text{Dom}(f)$.

Example 1 — Computing Compositions

$f(x) = x^2 + 1$ and $g(x) = 3x - 2$:

$(f \circ g)(x) = f(3x-2) = (3x-2)^2 + 1 = 9x^2 - 12x + 5$

$(g \circ f)(x) = g(x^2+1) = 3(x^2+1) - 2 = 3x^2 + 1$

These are different — confirming composition is not commutative.

Example 2 — Domain of a Composition

$f(x) = \sqrt{x}$ and $g(x) = 2x - 8$:

$(f \circ g)(x) = \sqrt{2x - 8}$

Domain: $2x - 8 \geq 0 \implies x \geq 4$ → domain is $[4, \infty)$.

$(g \circ f)(x) = 2\sqrt{x} - 8$

Domain: $x \geq 0$ → domain is $[0, \infty)$.

Example 3 — Evaluating at a Point

$f(x) = x^3$ and $g(x) = \sin x$:

$(f \circ g)(\pi/2) = f(\sin(\pi/2)) = f(1) = 1$

$(g \circ f)(1) = g(1^3) = \sin(1) \approx 0.841$

Decomposing a Function

Given $h(x)$, find $f$ and $g$ such that $h(x) = f(g(x))$.

$h(x) = \sqrt{x^2 + 3}$: let $g(x) = x^2 + 3$ (inner) and $f(x) = \sqrt{x}$ (outer).

$h(x) = \sin^3(x)$: let $g(x) = \sin x$ and $f(x) = x^3$.

Decomposition is not unique — there are multiple valid answers.

Example 4 — Triple Composition

$f(x) = x+1$, $g(x) = x^2$, $h(x) = 3x$:

$(f \circ g \circ h)(x) = f(g(h(x))) = f(g(3x)) = f(9x^2) = 9x^2 + 1$

Apply innermost first: $h$, then $g$, then $f$.