Composing Functions, foundations for calculus lesson

A square garden is fenced with a known length of fence. The fence length tells you the side, and the side tells you the area. Two small rules run one after the other, and chaining them gives one rule that goes straight from fence length to area. That chaining is composition, and calculus uses it constantly, most of all in the chain rule.

Key terms

Output used as input

Definition Let $g$ and $f$ be functions and let $x$ be in the domain of $g$ . The output $g(x)$ is a number, and if that number lies in the domain of $f$ , then $f$ can be evaluated at it, giving the number $f(g(x))$ . If $g(x)$ is not in the domain of $f$ , the second step is undefined for that $x$ .

In plain words The first function produces a number, and the second function treats that number as its own input.

Example Let $g(x) = x + 3$ and $f(u) = u^2$ . Follow a few inputs through both steps.

$x$	$g(x)$	$f(g(x))$
$-3$	$0$	$0$
$1$	$4$	$16$
$2$	$5$	$25$

Every middle value is visible, so you can see exactly where each final output came from.

Function composition

Definition The composition of $f$ with $g$ is the function $f \circ g$ , read ” $f$ composed with $g$ ” or ” $f$ of $g$ ”, given by

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

Its domain is the set of all $x$ in the domain of $g$ for which $g(x)$ is in the domain of $f$ . The small circle is not multiplication. In general $(f \circ g)(x)$ and $f(x) \cdot g(x)$ are different numbers.

In plain words Apply $g$ first, then apply $f$ to the result. The function written on the right acts first.

Example With $f(x) = \frac{1}{x}$ and $g(x) = x^2 + 1$ ,

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

Here $g(x) \ge 1$ for every real $x$ , so $g(x)$ is never $0$ and the composite is defined for all real numbers. Compare the product $f(x) \cdot g(x) = x + \frac{1}{x}$ , which is a different function with domain $x \ne 0$ .

Inner function

Definition In the composite $f(g(x))$ , the function $g$ is the inner function. It is applied first, directly to $x$ , and its output $g(x)$ is called the intermediate value. Writing a given formula as $f(g(x))$ is called decomposing it. A formula can usually be decomposed in more than one way, and the useful choice is the one whose pieces are simple.

In plain words The inner function is the part that touches $x$ first, the expression sitting inside the parentheses, the radical or the exponent.

Example Take $h(x) = (3x - 1)^5$ . The first thing done to $x$ is to compute $3x - 1$ , so a natural choice is the inner function $g(x) = 3x - 1$ with outer function $f(u) = u^5$ . At $x = 1$ the intermediate value is $g(1) = 2$ , and then $h(1) = 2^5 = 32$ .

Outer function

Definition In the composite $f(g(x))$ , the function $f$ is the outer function. It is applied last, to the intermediate value $g(x)$ . To form the composite formula, replace every occurrence of the input variable in the rule for $f$ by the whole expression $g(x)$ , written inside parentheses so its grouping is kept.

In plain words The outer rule does not care what its input is called. Whatever you feed it goes into every slot where its variable appeared, wrapped in parentheses.

Example Let $f(x) = x^2 - 2x$ and $g(x) = x + 1$ . Each $x$ in the rule for $f$ becomes $(x + 1)$ .

\begin{aligned} f(g(x)) &= (x + 1)^2 - 2(x + 1) \\[0.4em] &= x^2 + 2x + 1 - 2x - 2 \\[0.4em] &= x^2 - 1. \end{aligned}

Dropping the parentheses would give $x + 1^2 - 2x + 1 = -x + 2$ , which is wrong. At $x = 2$ the correct composite gives $f(3) = 9 - 6 = 3$ , and indeed $2^2 - 1 = 3$ .

Composite function

Definition The composite function $f \circ g$ is a single function in its own right, with one input, one output and its own domain. When a quantity $q$ depends on a quantity $s$ through $q = f(s)$ , and $s$ depends on $x$ through $s = g(x)$ , the composite $q = f(g(x))$ expresses $q$ directly in terms of $x$ . The units must match along the chain, so the output unit of $g$ has to be the input unit that $f$ expects.

In plain words Once the two steps are merged you get one formula, yet you can still say what each piece measures.

Example A square has perimeter $P$ centimeters, with $P > 0$ . All four sides are equal, so the side length in centimeters is

s(P) = \frac{P}{4}.

The area of a square with side $s > 0$ centimeters is $A(s) = s^2$ square centimeters. A squared length is an area, so the unit becomes square centimeters. The composite gives area from perimeter.

\begin{aligned} A(s(P)) &= \left(\frac{P}{4}\right)^2 \\[0.4em] &= \frac{P^2}{16} \text{ square centimeters}, \\[0.4em] &P > 0. \end{aligned}

For $P = 20$ the side is $5$ centimeters and the area is $25$ square centimeters, and the single formula agrees since $\frac{400}{16} = 25$ . If the perimeter were measured in meters and the area rule expected centimeters, you would first have to convert.

Composite domain

Definition The domain of $f \circ g$ is

\begin{aligned} \{\, x : \ & x \in \operatorname{dom}(g) \ \text{and} \\[0.4em] & g(x) \in \operatorname{dom}(f) \,\}. \end{aligned}

Both conditions are required. The domain is decided by the two original rules, not by how the final formula looks after simplifying.

In plain words An input is allowed only if the inner function accepts it and the outer function accepts what comes out.

Example Let $f(x) = \frac{1}{x - 2}$ and $g(x) = \sqrt{x}$ . The inner function needs $x \ge 0$ . The outer function needs its input to differ from $2$ , so we need $\sqrt{x} \ne 2$ , which rules out $x = 4$ . The composite

f(g(x)) = \frac{1}{\sqrt{x} - 2}

has domain $[0, 4) \cup (4, \infty)$ . A simplified formula can hide a condition. With $f(x) = x^2$ and $g(x) = \sqrt{x}$ the composite simplifies to $x$ , yet its domain is still $[0, \infty)$ , because $g(-1)$ is not a real number.

Order of composition

Definition Composition is generally not commutative. For most pairs of functions $f \circ g$ and $g \circ f$ have different formulas, and they can also have different domains and different outputs. The order of operations is read from the inside out, so in $f(g(x))$ the function $g$ acts first.

In plain words Putting on socks and then shoes is not the same as shoes and then socks. Which function acts first matters.

Example Let $f(x) = x^2$ and $g(x) = x + 3$ . Then

\begin{gathered} f(g(x)) = (x + 3)^2, \\[0.4em] g(f(x)) = x^2 + 3. \end{gathered}

At $x = 1$ the first gives $16$ and the second gives $4$ . Domains can differ too. With $f(x) = \sqrt{x}$ and $g(x) = x - 4$ , the composite $f(g(x)) = \sqrt{x - 4}$ has domain $[4, \infty)$ , while $g(f(x)) = \sqrt{x} - 4$ has domain $[0, \infty)$ .

Check yourself

Let $f(x) = 2x - 1$ and $g(x) = x^2 + 1$ . Find $f(g(x))$ and $g(f(x))$ , and evaluate both at $x = 2$ .

Show the answer

Substituting the whole inner expression each time gives

\begin{aligned} f(g(x)) &= 2(x^2 + 1) - 1 \\[0.4em] &= 2x^2 + 1, \\[0.4em] g(f(x)) &= (2x - 1)^2 + 1 \\[0.4em] &= 4x^2 - 4x + 2. \end{aligned}

At $x = 2$ the first is $9$ and the second is $10$ , so the order changes the output.

Let $f(x) = \sqrt{x - 1}$ and $g(x) = \frac{1}{x}$ . Find the domain of $f(g(x))$ .

Show the answer

The inner function needs $x \ne 0$ . The outer function needs $\frac{1}{x} - 1 \ge 0$ , that is $\frac{1}{x} \ge 1$ , which holds exactly when $0 < x \le 1$ . The domain is $(0, 1]$ .

A circle has circumference $C$ centimeters, with $C > 0$ . Write its area as a function of $C$ by composing the radius rule with the area rule, and find the area when $C = 10$ .

Show the answer

The radius is $r = \frac{C}{2\pi}$ centimeters and the area is $\pi r^2$ square centimeters, so

\begin{aligned} A &= \pi \left(\frac{C}{2\pi}\right)^2 \\[0.4em] &= \frac{C^2}{4\pi} \text{ square centimeters}, \\[0.4em] &C > 0. \end{aligned}

For $C = 10$ the area is $\frac{100}{4\pi} = \frac{25}{\pi} \approx 7.96$ square centimeters.

Composing Functions

Key terms

Output used as input

Function composition

Inner function

Outer function

Composite function

Composite domain

Order of composition

Check yourself

How this lesson connects

Builds on

Used later in

Where this leads in Calculus