Composition of Functions — AP® Precalculus Notes

Many processes happen in stages. A store takes a discount and then adds tax. A thermostat reads a temperature in Celsius and a display converts it to Fahrenheit. Each stage is a function, and running one stage after another is called composition. This page shows how to evaluate a composite from a table, a graph or a formula, why order matters, which inputs a composite can take, and how to split a function into stages.

What a composite function does

The composite function $f(g(x))$ takes an input $x$ , applies $g$ , and then uses that output as the input of $f$ . It is also written $(f \circ g)(x)$ , read "f of g of x". The function written closest to $x$ acts first. Think of an assembly line: $g$ is the first machine and $f$ is the second.

Because the output of $g$ becomes an input of $f$ , the work always goes from the inside out. On the exam, show the inner value: graders look for it.

Worked example 1 (from a table). A table gives $g(2) = 0$ and $f(0) = 4$ . Find $f(g(2))$ .

Inner value first: $g(2) = 0$ .

Then the outer function at that value: $f(g(2)) = f(0) = 4$ .

Order matters

Back at the store, take $d(p) = p - 15$ (the discount) and $t(p) = 1.08p$ (the tax). Discount first gives $t(d(p))$ $= 1.08(p - 15)$ $= 1.08p - 16.2$ . Tax first gives $d(t(p)) = 1.08p - 15$ . They differ by $1.2$ dollars, whatever the price. Composition is not commutative: $f(g(x))$ and $g(f(x))$ are usually different functions. There are exceptions, but never assume one.

Composites from formulas

To write $f(g(x))$ as a formula, replace every $x$ in the formula of $f$ by the whole expression $g(x)$ , in parentheses.

Worked example 2. Let $f(x) = x^2 + 3$ and $g(x) = 2x - 1$ . Find $f(g(x))$ and $g(f(x))$ .

$f(g(x)) = (2x - 1)^2 + 3$ . The parentheses matter: the whole of $2x - 1$ is squared.

$g(f(x))$ $= 2(x^2 + 3) - 1$ $= 2x^2 + 5$ .

At $x = 1$ : $f(g(1)) = 1 + 3 = 4$ , but $g(f(1)) = 2 \cdot 4 - 1 = 7$ . Different functions.

Composition also explains transformations. With $g(x) = x - 2$ , $f(g(x)) = f(x - 2)$ is the graph of $f$ moved $2$ units right. With $f(x) = 5 \cdot 2^x$ and $g(x) = x + 3$ , $f(g(x))$ $= 5 \cdot 2^{x + 3}$ $= 40 \cdot 2^x$ : for an exponential function, a horizontal shift is the same as a vertical dilation.

Composites from graphs

A graph works like a table: read the inner function at $x$ , then read the outer function at that output. To solve an equation such as $g(f(x)) = 2$ , work from the outside in: first find every input of $g$ that gives $2$ , then every $x$ with $f(x)$ equal to one of those inputs. Several inputs can share an output, so check for more than one.

Worked example 3. The parabola $g$ reaches $2$ at the inputs $-1$ and $3$ . The line $f$ has $f(0) = -1$ and $f(2) = 3$ . Solve $g(f(x)) = 2$ .

Outer step: $g(u) = 2$ for $u = -1$ or $u = 3$ .

Inner step: $f(x) = -1$ gives $x = 0$ ; $f(x) = 3$ gives $x = 2$ .

So $x = 0$ and $x = 2$ . Stopping at $-1$ and $3$ gives the values of $f(x)$ , not of $x$ .

The domain of a composite

An input $x$ belongs to the domain of $f(g(x))$ when two things hold: $x$ is in the domain of $g$ , and the output $g(x)$ is in the domain of $f$ .

Worked example 4. Let $f(x) = \sqrt{x}$ and $g(x) = 6 - 2x$ . Find the domain of $f(g(x)) = \sqrt{6 - 2x}$ .

$f$ needs a non-negative input, so $6 - 2x \ge 0$ .

Then $-2x \ge -6$ , and dividing by $-2$ flips the inequality: $x \le 3$ .

The identity function

The identity function $I(x) = x$ returns every input unchanged. Composing with it does nothing: $f(I(x)) = f(x)$ and $I(f(x)) = f(x)$ . It plays the role for composition that $0$ plays for addition and $1$ plays for multiplication. You will need it in the next topic.

Taking a function apart

Reading a formula as stages helps you evaluate it, graph it and later undo it. Ask what is done to $x$ first (the inner function) and what is done to that result (the outer function).

Function	Inner $g(x)$	Outer $f(x)$
$h(x) = (3x + 1)^4$	$3x + 1$	$x^4$
$h(x) = \sqrt{x^2 + 9}$	$x^2 + 9$	$\sqrt{x}$
$h(x) = 2^{x - 5}$	$x - 5$	$2^x$

A decomposition is not unique, but the order of the stages is fixed by the formula. Check a pair by composing it: $f(g(x))$ must give back $h(x)$ .