Composition of Functions

Composing two functions means feeding the output of one as the input of the other. This operation, denoted $f \circ g$ , is fundamental to calculus (especially the Chain Rule) and to computer science, where it models the chaining of procedures.

Quick Reference

Concept	Formula / Rule
Composite function	$(f \circ g) (x) = f (g (x))$
Domain of $f \circ g$	${x \in Dom (g) ∣ g (x) \in Dom (f)}$
Order matters	$f \circ g \neq g \circ f$ in general
Associativity	$f \circ g \circ h = (f \circ g) \circ h = f \circ (g \circ h)$

What Is Composition?

If $f (x) = \sqrt{x + 1}$ and $g (x) = x^{2}$ , we can define a new function $h$ by substituting $g (x)$ into $f$ :

h (x) = f (g (x)) = f (x^{2}) = \sqrt{x^{2} + 1} .

In general, to form the composite $f \circ g$ , we start with $x$ in the domain of $g$ , apply $g$ to get $g (x)$ , and then apply $f$ to that result, obtaining $f (g (x))$ . This requires $g (x)$ to lie in the domain of $f$ .

Machine diagram showing input x through g then f to produce f(g(x)). — (a) Machine diagram: $x \to g \to g (x) \to f \to f (g (x))$ .

Arrow diagram showing the mapping from x to g(x) to f(g(x)). — (b) Arrow diagram for $f \circ g$ .

Let $f$ and $g$ be two functions. The composite function $f \circ g$ is defined by

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

for all $x$ in the domain of $g$ for which $g (x)$ lies in the domain of $f$ .

The domain of $f \circ g$ is:

Dom (f \circ g) = {x ∣ x \in Dom (g) and g (x) \in Dom (f)} .

Order matters: $(f \circ g) (x) = f (g (x))$ and $(g \circ f) (x) = g (f (x))$ are generally different functions.

Examples

Let $f (x) = \sqrt{x}$ and $g (x) = 1 + x^{2}$ . Find $f \circ g$ and $g \circ f$ with their domains.

Solution

Dom (f) = [0, \infty)

Dom (g) = ℝ

. $f \circ g$ :

(f \circ g) (x) = f (g (x)) = f (1 + x^{2}) = \sqrt{1 + x^{2}} .

Since

1 + x^{2} \geq 1 > 0

for all real

x

g (x)

always lies in

Dom (f)

. So

Dom (f \circ g) = ℝ

. $g \circ f$ :

(g \circ f) (x) = g (f (x)) = g (\sqrt{x}) = 1 + (\sqrt{x})^{2} = 1 + x (x \geq 0) .

f

is defined only for

x \geq 0

, and

f (x) = \sqrt{x} \geq 0

always lies in

Dom (g) = ℝ

. So

Dom (g \circ f) = [0, \infty)

If $f (x) = \frac{x + 1}{x - 1}$ , find $f \circ f$ and its domain.

Solution

(f \circ f) (x) = f (f (x)) = f (\frac{x + 1}{x - 1}) = \frac{\frac{x + 1}{x - 1} + 1}{\frac{x + 1}{x - 1} - 1} = \frac{\frac{2 x}{x - 1}}{\frac{2}{x - 1}} = \frac{2 x}{2} = x .

The formula simplifies to

x

, but the domain is not all of

ℝ

. We need:

$x \in Dom (f)$ : $x \neq 1$ .
$f (x) \in Dom (f)$ : $\frac{x + 1}{x - 1} \neq 1$ , i.e., $x + 1 \neq x - 1$ , i.e., $- 1 \neq 1$ (always true).

So the second condition imposes no additional restriction, and

Dom (f \circ f) = ℝ - {1}

Decomposing a Function

Any complex function can often be written as a composition of simpler ones. This skill is essential for the Chain Rule in calculus.

Find $f$ and $g$ such that $h = f \circ g$ where $h (x) = \cos (x^{2})$ .

Solution

To compute

h (x)

, we first square

x

, then take the cosine. So:

g (x) = x^{2} (square the input), f (x) = \cos x (take cosine) .

Check:

(f \circ g) (x) = f (x^{2}) = \cos (x^{2}) = h (x)

. ✓

Find $f$ , $g$ , $h$ such that $F = f \circ g \circ h$ where $F (x) = \sqrt{| x | + 3}$ .

Solution

To compute

F (x)

: first take the absolute value, then add 3, then take the square root.

h (x) = | x |, g (x) = x + 3, f (x) = \sqrt{x} .

Check:

(f \circ g \circ h) (x) = f (g (| x |)) = f (| x | + 3) = \sqrt{| x | + 3} = F (x)

. ✓

Composing Three or More Functions

Composition is associative: $f \circ g \circ h = (f \circ g) \circ h = f \circ (g \circ h)$ . You can group them in any order when composing more than two functions.

For example, $F (x) = \frac{1}{2 + x^{2}}$ can be written as $F = f \circ g \circ h$ where $h (x) = x^{2}$ , $g (x) = x + 2$ , $f (x) = \frac{1}{x}$ .

Frequently Asked Questions

Is composition commutative?

Generally no.

f \circ g

and

g \circ f

are usually different functions. For example, with

f (x) = x + 1

and

g (x) = x^{2}

(f \circ g) (x) = x^{2} + 1

but

(g \circ f) (x) = (x + 1)^{2}

. These are not equal.

How do I find the domain of a composite function?

First, find

Dom (g)

. Then, from those

x

-values, keep only those where

g (x)

falls inside

Dom (f)

. The intersection of these two restrictions is

Dom (f \circ g)

Why is composition important in calculus?

The Chain Rule in differential calculus says: if

h = f \circ g

, then

. To apply the Chain Rule, you need to recognize a function as a composition and identify its inner and outer parts.

Can a function be composed with itself?

Yes.

f \circ f

is called the iterate of

f

. For example, if

f (x) = x^{2}

, then

(f \circ f) (x) = (x^{2})^{2} = x^{4}

. Iterates appear in dynamical systems and the study of fractals.

Quick Reference

What Is Composition?

Examples

Decomposing a Function

Composing Three or More Functions

Frequently Asked Questions

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