Learn/Algebra/Function Composition

Algebra • Topic 6

Function Composition

Function composition involves applying one function to the result of another. In competitions, this concept appears in iterative processes and complex functional equations.

Definition

Definition (Composition)

Given two functions

f : B \to C

and

g : A \to B

, the composite function

f \circ g

(read "

f

g

") is defined by:

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

Warning

Composition is not commutative. In general,

f (g (x)) \neq = g (f (x))

Domain of Composite Functions

The domain of

f \circ g

consists of all

x

such that:

$x$ is in the domain of $g$ .
$g (x)$ is in the domain of $f$ .

Example

Let

f (x) = x

and

g (x) = x - 2

. Find the domain of

f \circ g

Proof. $(f \circ g) (x) = x - 2$ .

Domain of $g$ : All real numbers.
Domain of $f$ : Inputs $\geq 0$ . So we need $g (x) \geq 0 ⟹ x - 2 \geq 0 ⟹ x \geq 2$ .

Domain:

[2, \infty)

. ∎

Iterated Functions

We denote applying

f

repeatedly as

f^{n} (x)

$f^{1} (x) = f (x)$
$f^{2} (x) = f (f (x))$
$f^{n} (x) = f (f^{n - 1} (x))$

Fixed Points

A value

c

is a fixed point if

f (c) = c

Geometrically, this is where the graph $y = f (x)$ intersects the line $y = x$ .
If $f (c) = c$ , then $f (f (c)) = f (c) = c$ . Fixed points are invariant under iteration.

Example (Solving Iterated Equations)

Solve

f (f (x)) = x

where

f (x) = \frac{1}{x}

Proof. $f (f (x)) = f (\frac{1}{x}) = \frac{1}{1/ x} = x$ . Since the identity holds for all valid $x$ , the solution is the entire domain of $f$ : $x \neq = 0$ . ∎

Practice Problems

Exercise (Problem 1)

Let

f (x) = x^{2} + 1

and

g (x) = x - 1

. Find

f (g (x))

and

g (f (x))

. Are they equal?

Exercise (Problem 2)

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

, calculate

f (f (f (x)))

Exercise (Problem 3)

Find all linear functions

f (x) = a x + b

such that

f (f (x)) = 4 x + 3