Composition of Functions

Learning goals

Compute $g \circ f$ , reading it right to left with $f$ acting first
Decide whether $g \circ f$ equals $f \circ g$ for a given pair
Find the domain of $g \circ f$ from the range of $f$
Decompose a function as $g \circ f$ , knowing the split is a choice
Write $f^2$ for $f \circ f$ , never for the squared output

Reading a composition right to left

Give the chaining its symbol. The composition of $f$ and $g$ , written $g \circ f$ , is the function

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

Check it against the numbers above: with $f(x) = x + 1$ and $g(x) = x^2$ ,

(g \circ f)(2) = g(f(2)) = g(3) = 9,

exactly the $2 \to 3 \to 9$ chain from the opening. You feed $x$ to $f$ , and then feed the result $f(x)$ to $g$ ; the circle is the operation sign, exactly as $+$ is the sign for addition.

The one thing to fix in your reading, because it causes more confusion than anything else in this lesson, is the order. In $g \circ f$ the function on the right, $f$ , acts first, and the function on the left, $g$ , acts second. The notation runs right to left, opposite to the way you read a sentence. The reason is built into the definition: $g(f(x))$ has $f$ in the inner parentheses, and the inner operation always happens before the outer one. So $g \circ f$ means “do $f$ , then $g$ ,” never “do $g$ , then $f$ .”

The composite g of f as a two-machine pipeline. The input x enters f and comes out as f(x); that value enters g and comes out as g(f(x)). The machines run left to right, but we write g of f because g acts on the output of f, so the name reads right to left and f goes first.

Worked example 1 Both orders of the same two functions

Let $f(x) = x + 1$ and $g(x) = x^2$ . Compute $g \circ f$ and $f \circ g$ , keeping track of which function acts first.

For $g \circ f$ , the inner function is $f$ , so substitute $f(x) = x + 1$ into $g$ :

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

For $f \circ g$ , the inner function is now $g$ , so substitute $g(x) = x^2$ into $f$ :

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

The two results, $x^2 + 2x + 1$ and $x^2 + 1$ , are different functions. Swapping the order changed the answer, which is the first sign that composition, unlike addition, does not always let you reverse the two inputs.

Check your understanding

For $f(x) = 2x$ and $g(x) = x - 3$ , which expression is $(g \circ f)(x)$ ?

2x - 6

2x - 3

2x + 3

x - 6

Order matters, but not always

The one law that addition has and composition often lacks is commutativity: for numbers, $a + b = b + a$ always. For functions, $g \circ f$ and $f \circ g$ are usually different, and Worked Example 1 already showed why: running the same two machines in the opposite order gave $x^2 + 2x + 1$ instead of $x^2 + 1$ . These differ by $2x$ , which is zero only at $x = 0$ .

But “usually different” is not “never equal,” and the difference is worth stating exactly, because overshooting it is the classic error. Some pairs genuinely commute. Take the two translations $f(x) = x + 1$ and $g(x) = x + 2$ . Then

(f \circ g)(x) = (x + 2) + 1 = x + 3 = (g \circ f)(x),

and the same works for any two translations $f(x) = x + a$ and $g(x) = x + b$ , since both orders give $x + a + b$ and addition of the shifts does not care about order. So the correct statement names both cases: composition is not always commutative, but it is not never commutative either; commuting is the exception, not the rule.

Composition has two more things in common with addition, each worth a name. It is associative: for any three functions, grouping does not matter, $(h \circ g) \circ f = h \circ (g \circ f)$ , so you can write $h \circ g \circ f$ with no parentheses.

And it has an identity: a function $\operatorname{id}(x) = x$ that changes nothing when composed with $f$ , the way $0$ changes nothing under addition. Which set the identity is defined on matters, so when we need to be specific we write $\operatorname{id}_S$ for the identity on the set $S$ . Throughout this lesson, when a function’s codomain is not stated separately, take it to be that function’s range, the values it actually produces. Every domain and identity claim below relies on that convention.

Check your understanding

For which pair do the two compositions agree, $(f \circ g)(x) = (g \circ f)(x)$ for every $x$ ?

f(x) = x^2,\ g(x) = x + 1

f(x) = 2x,\ g(x) = x + 1

f(x) = x + 5,\ g(x) = x - 2

f(x) = x^2,\ g(x) = 2x

The domain of a composition

Here is the idea the substitution procedure hides, and it is the most important thing in the lesson. When you build $(g \circ f)(x) = g(f(x))$ , two things must be true for the answer to exist: $f$ must accept $x$ , and $g$ must accept the number $f(x)$ that comes out.

The domain of $g \circ f$ #

By definition $(g \circ f)(x) = g(f(x))$ , so this value exists exactly when both steps inside it are legal. First $f(x)$ must be defined, which requires $x \in \operatorname{dom} f$ . Then $g$ is applied to the number $f(x)$ , which is legal exactly when $f(x) \in \operatorname{dom} g$ . So

\operatorname{dom}(g \circ f) = \{\, x \in \operatorname{dom} f : f(x) \in \operatorname{dom} g \,\}.

This is not $\operatorname{dom} f \cap \operatorname{dom} g$ . The second condition constrains $f(x)$ , an output of $f$ , so testing it needs the range of $f$ , not its domain. A value of $x$ can lie in both $\operatorname{dom} f$ and $\operatorname{dom} g$ yet be excluded, because $f(x)$ escapes $\operatorname{dom} g$ . Or it can lie outside $\operatorname{dom} g$ and still be fine, as long as $x \in \operatorname{dom} f$ and $f(x) \in \operatorname{dom} g$ .

Worked example 2 A square root inside a reciprocal

Let $f(x) = \sqrt{x}$ and $g(x) = \dfrac{1}{x - 2}$ . Find a formula for $g \circ f$ and its domain.

The formula is a direct substitution:

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

Now the domain. First $f$ must accept $x$ , and $\sqrt{x}$ needs $x \ge 0$ . Second $g$ must accept the output $\sqrt{x}$ , and $g$ rejects only the input $2$ , so we need $\sqrt{x} \ne 2$ . Since $\sqrt{x} = 2$ exactly when $x = 4$ ,

\operatorname{dom}(g \circ f) = \{\, x \ge 0 \text{ and } x \ne 4 \,\}.

Compare that with the wrong shortcut. The intersection $\operatorname{dom} f \cap \operatorname{dom} g$ would keep $x = 4$ and throw out $x = 2$ , the value $g$ forbids. The truth is the reverse: $x = 2$ is fine, because $\sqrt{2} \ne 2$ , while $x = 4$ is excluded, because $\sqrt{4} = 2$ is the one input $g$ cannot take. The excluded input is found through the range of $f$ , not by intersecting the two domains.

The domain of g of f is the part of dom f whose image lands where g is defined. Here f(x) = sqrt(x) sends the inputs x greater than or equal to 0 (top, and 4 is one of them) into the values g accepts (bottom, everything except 2). The inputs 0 and 9 map to 0 and 3 and survive; the input 4 is in dom f but maps to the barred value 2 (dashed), so 4 is removed from dom(g of f), not from dom f. Hence dom(g of f) is x greater than or equal to 0 with x not equal to 4.

Check your understanding

Let $f(x) = \sqrt{x}$ and $g(x) = \dfrac{1}{x - 3}$ . What is the domain of $g \circ f$ ?

A all real

x

except

x = 3

x \ge 0

x \ge 0

and

x \ne 9

x \ge 0

and

x \ne 3

Worked example 3 Same formula, smaller domain

Let $f(x) = \sqrt{x}$ and $g(x) = x^2$ . Find $g \circ f$ , and check its domain before deciding whether it is an identity function.

Substituting gives a formula that simplifies all the way down:

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

The formula alone looks like the identity. Check the domain before trusting that. First $f$ needs $x \ge 0$ ; then $g$ accepts every real number that comes out, so nothing more is excluded. So

\operatorname{dom}(g \circ f) = \{\, x \ge 0 \,\}.

So $g \circ f$ is the rule $x \mapsto x$ with domain $x \ge 0$ : that is $\operatorname{id}_{[0,\infty)}$ , the identity on $[0, \infty)$ , but not $\operatorname{id}_{\mathbb{R}}$ , the identity on all of $\mathbb{R}$ . The two share a formula but not a domain, so by the equality test they are different functions. The simplified $x$ threw away the record of where $\sqrt{\phantom{x}}$ refused to run; only the domain remembers it.

The other order behaves differently still. Here $(f \circ g)(x) = \sqrt{x^2} = \lvert x \rvert$ , defined for every real $x$ , so $\operatorname{dom}(f \circ g)$ is all of $\mathbb{R}$ while $\operatorname{dom}(g \circ f)$ is only $x \ge 0$ , a reminder that $\operatorname{dom}(f \circ g)$ and $\operatorname{dom}(g \circ f)$ are in general different sets.

Decomposing a function, and why the answer is not unique

Composition also runs backward as a question. Given a function $h$ , can you write it as $g \circ f$ for simpler pieces $f$ and $g$ ? This is decomposition, and it lets you see a complicated rule as a short pipeline of easy steps.

Take $h(x) = \sqrt{3x + 1}$ . The natural split peels off the outermost operation, the square root, and calls the inside a separate step: let $f(x) = 3x + 1$ and $g(x) = \sqrt{x}$ . Then

(g \circ f)(x) = g(3x + 1) = \sqrt{3x + 1} = h(x).

But this is not the only way. A different split absorbs the $3$ elsewhere: $f(x) = 3x$ and $g(x) = \sqrt{x + 1}$ also gives $g(f(x)) = \sqrt{3x + 1} = h(x)$ . Both are correct, since each composes to $h$ . So “decompose $h$ ” has more than one right answer. Among the valid splits there is usually a most natural one, peel off the outermost operation, but no split is the unique correct answer.

Worked example 4 Decomposing a cube

Write $h(x) = (2x - 5)^3$ as a composition $g \circ f$ .

Peel off the outermost operation, the cube. Let $f(x) = 2x - 5$ be the inside and $g(x) = x^3$ be the cube:

(g \circ f)(x) = g(2x - 5) = (2x - 5)^3 = h(x).

Check by recomposing: substitute $f$ ‘s formula into $g$ and confirm it rebuilds $h$ exactly, which it does. As with the square-root example, this is one valid decomposition among several; the checkpoint below asks you to find another.

Check your understanding

The lesson decomposes $h(x) = (2x - 5)^3$ as $f(x) = 2x - 5$ , $g(x) = x^3$ . Which pair gives a genuinely different valid decomposition, $h = g \circ f$ ?

f(x) = 2x,\ g(x) = (x - 5)^3

f(x) = 2x - 5,\ g(x) = x^3

f(x) = 3x - 5,\ g(x) = x^3

f(x) = 2x - 10,\ g(x) = x^3 + 5

Composing a function with itself

Because composition is associative, composing a function with itself repeatedly is unambiguous, and it has its own notation. Write $f^2 = f \circ f$ , the small superscript counting how many times $f$ is applied. Read $f^2(x) = f(f(x))$ : apply $f$ , then apply $f$ again to the result.

Here is the notation warning. That superscript clashes with the other thing a superscript can mean, squaring the output, $(f(x))^2 = f(x) \cdot f(x)$ . The two are different functions. For $f(x) = 2x - 1$ ,

f^2(x) = f(f(x)) = 2(2x - 1) - 1 = 4x - 3,

while

(f(x))^2 = (2x - 1)^2 = 4x^2 - 4x + 1,

and these are different functions: one is linear, the other is not. Which meaning $f^2$ carries is a matter of convention, and this lesson uses $f^2$ for the composition $f \circ f$ ; when we mean the square of the output we write $(f(x))^2$ . One more warning for the next lesson: there $f^{-1}$ will mean the inverse of $f$ under composition, never $\dfrac{1}{f}$ . In this lesson’s convention, a superscript on a function’s name is about composition, not multiplication.

Check your understanding

For $f(x) = 3x + 2$ , which expression equals $f^2(x)$ , meaning $f \circ f$ ?

9x + 8

9x^2 + 12x + 4

6x + 4

3x^2 + 2

Transformations were compositions all along

The previous lesson wrote every transformed graph as $y = a\,f\big(b(x - h)\big) + k$ , split into an inside that acts on the input before $f$ and an outside that acts on $f$ ‘s output afterward. Name those two pieces as functions, an input map and an output map. The transformed rule is then exactly a composition of three functions: input map, then $f$ , then output map. That is also why “stretch then shift” gave a different graph from “shift then stretch”: the two output maps did not commute, the same reason $g \circ f \ne f \circ g$ in general.

One more distinction is worth a sentence. Composition chains two functions, feeding one’s output into the other’s input; adding, subtracting, multiplying, or dividing two functions instead combines their outputs at the same input, $(f + g)(x) = f(x) + g(x)$ . They are different operations that happen to both start from two functions.

Common mistakes

Reading $g \circ f$ left to right. In $g \circ f$ the right function $f$ acts first, then $g$ , so $(g \circ f)(x) = g(f(x))$ means “do $f$ , then $g$ .” It is usually not the same as $f \circ g$ .
Saying composition is never commutative. It is not always commutative, but it is not never commutative either: any two translations $f(x) = x + a$ and $g(x) = x + b$ give the same result in either order.
Using $\operatorname{dom} f \cap \operatorname{dom} g$ for the domain of $g \circ f$ . The right set is $\{x \in \operatorname{dom} f : f(x) \in \operatorname{dom} g\}$ , which depends on the range of $f$ , not the domain of $g$ .
Trusting a simplified formula over the domain. For $f(x) = \sqrt{x}$ and $g(x) = x^2$ , $(g \circ f)(x)$ simplifies to $x$ , but its domain is $x \ge 0$ , so it is $\operatorname{id}_{[0,\infty)}$ , not $\operatorname{id}_{\mathbb{R}}$ .
Assuming a function decomposes in only one way. $h(x) = \sqrt{3x + 1}$ is $g \circ f$ for more than one valid choice of $f$ and $g$ .
Reading $f^2$ as the square of the output. In this lesson $f^2$ means $f \circ f$ , so for $f(x) = 2x - 1$ , $f^2(x) = 4x - 3$ , not $(2x - 1)^2$ .

Key takeaways

Composition $g \circ f$ is defined by $(g \circ f)(x) = g(f(x))$ and reads right to left: $f$ acts first, then $g$ .
Composition is not always commutative ( $g \circ f \ne f \circ g$ in general), though it is associative and has an identity $\operatorname{id}_S$ on any set $S$ .
The domain is $\operatorname{dom}(g \circ f) = \{x \in \operatorname{dom} f : f(x) \in \operatorname{dom} g\}$ , not $\operatorname{dom} f \cap \operatorname{dom} g$ : it needs the range of $f$ .
A simplified formula can hide a restricted domain: $(g \circ f)(x) = x$ for $f(x) = \sqrt{x}$ , $g(x) = x^2$ has domain $x \ge 0$ , so it is $\operatorname{id}_{[0,\infty)}$ , not $\operatorname{id}_{\mathbb{R}}$ .
Decomposition is not unique: $h(x) = \sqrt{3x + 1}$ splits as $g \circ f$ in more than one valid way.
Composing a function with itself is written $f^2 = f \circ f$ , never $(f(x))^2$ .
Every graph transformation $y = a\,f(b(x - h)) + k$ is the composition $A \circ f \circ B$ , so “order matters” there is exactly non-commutativity here.

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

3.4 Composition of Functions (OpenStax)
10.1 Finding Composite and Inverse Functions (OpenStax)
Composing when Functions are Sets of Points (Purplemath)
Operations on Functions (Purplemath)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why composition is associative

The main lesson states this fact. Here is the proof, which is also why an iterated composition like $f \circ f \circ f$ needs no parentheses to say what it means.

Composition is associative#

Let $f$ , $g$ , and $h$ be functions, and consider the two ways to group their composition, $(h \circ g) \circ f$ and $h \circ (g \circ f)$ . Two functions are equal exactly when they have the same domain, codomain, and values, the equality test from Function Notation, so we check the values, the domains, and the codomains.

For the values, apply each grouping to an input $x$ and unfold the definition one step at a time. The left grouping gives

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

and the right grouping gives

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

Both collapse to the single expression $h(g(f(x)))$ , so the two functions agree wherever they are defined.

For the domains, read off what each grouping requires. Building $h(g(f(x)))$ is legal exactly when $x \in \operatorname{dom} f$ so that $f(x)$ exists, then $f(x) \in \operatorname{dom} g$ so that $g(f(x))$ exists, then $g(f(x)) \in \operatorname{dom} h$ so that the outer $h$ applies. These three conditions are the same for both groupings, so the two functions share the same domain. Both also share the same codomain, whatever codomain we declared for $h$ , since both groupings end by applying $h$ . By the equality test they are the same function.

Because the grouping never matters, we drop the parentheses and write $h \circ g \circ f$ .

The identity function, defined carefully

The main lesson uses the identity function informally. Here is why composing with it changes nothing, done carefully: the equality test demands the same domain, codomain, and values, not just a matching formula.

The identity function is neutral for composition#

Let $f: A \to B$ be a function. For any set $S$ , let $\operatorname{id}_S(x) = x$ be the identity function on $S$ , the function with domain $S$ , codomain $S$ , and rule “return the input unchanged.” We show $f \circ \operatorname{id}_A = f$ and $\operatorname{id}_B \circ f = f$ , taking the identity on the correct set each time.

For every $x \in A$ ,

(f \circ \operatorname{id}_A)(x) = f(\operatorname{id}_A(x)) = f(x),

and $f \circ \operatorname{id}_A$ has domain $A$ and codomain $B$ , matching $f$ exactly. So $f \circ \operatorname{id}_A = f$ .

Likewise, for every $x \in A$ ,

(\operatorname{id}_B \circ f)(x) = \operatorname{id}_B(f(x)) = f(x),

since $f(x) \in B$ , the domain of $\operatorname{id}_B$ , so the composition is legal for every $x \in A$ . And $\operatorname{id}_B \circ f$ has domain $A$ and codomain $B$ , again matching $f$ . So $\operatorname{id}_B \circ f = f$ .

The two sides are not symmetric. On the left, the identity’s codomain is forced: $\operatorname{id}_C \circ f = f$ only when $C = B$ exactly, since a composite’s codomain always matches its outer function’s, and that has to equal $B$ . On the right, $f \circ \operatorname{id}_A = f$ , and so does $f \circ \operatorname{id}_T$ for any set $T$ that contains $A$ . The domain of that composite is $\{x \in T : x \in A\} = A$ either way, since $f$ ‘s own domain restriction already confines things to $A$ . So $\operatorname{id}_A$ is simply the smallest, most natural choice on that side, not the only one that works; the care that matters is on the left.

Composition also has a name for what these two facts give it, together with the fact that it is not generally commutative. When $f$ , $g$ , and $h$ all map one fixed set $X$ back into itself, composing any two of them gives another function $X \to X$ , $\operatorname{id}_X$ serves as the identity, and composition of such functions is associative. Associative, with an identity, but not generally commutative and not guaranteed an inverse for every element: that structure is what algebraists call a monoid. The self-maps of $X$ under composition form one such monoid. The missing piece, an inverse that undoes a function, is what the next lesson supplies for the functions that have one.

Writing a transformation as a composition

The main lesson states the connection in one paragraph. Here it is in full, with the input and output maps named and a worked example.

Write $B(x) = b(x - h)$ for the input map and $A(y) = ay + k$ for the output map, so $B$ acts on the input before $f$ and $A$ acts on $f$ ‘s output afterward. Then

y = A\big(f(B(x))\big) = (A \circ f \circ B)(x).

Reading right to left, $B$ acts first, then $f$ , then $A$ , exactly the story the previous lesson told: the inside happens before $f$ , and the outside after. Its rule that input maps “run backward because you solve for $x$ ” is also explained: in $A \circ f \circ B$ the rightmost map $B$ acts before $f$ . So to find which $x$ produces a given input to $f$ , you must undo $B$ , that is, solve $B(x) = u$ for $x$ .

Worked example 5 A transformation read as a composition

Write $y = 2f(3x - 6) + 1$ in the form $A \circ f \circ B$ , and identify $A$ and $B$ .

Factor the inside so it matches $b(x - h)$ , and read the outside directly:

3x - 6 = 3(x - 2), \qquad \text{so} \qquad B(x) = 3(x - 2), \quad A(y) = 2y + 1.

Check that composing them rebuilds the rule:

(A \circ f \circ B)(x) = A\big(f(3(x - 2))\big) = 2\,f(3x - 6) + 1.

So the transformation is the composition $A \circ f \circ B$ with $B(x) = 3(x - 2)$ acting on the input and $A(y) = 2y + 1$ acting on the output.

Iterating further:

f^3

and a general pattern

The main lesson computes $f^2$ for $f(x) = 2x - 1$ . Here is $f^3$ , and the general pattern it starts.

Apply $f$ once more, now to $f^2(x) = 4x - 3$ :

f^3(x) = f(f^2(x)) = 2(4x - 3) - 1 = 8x - 7.

A pattern is visible: each application doubles the coefficient of $x$ and adjusts the constant to match, giving $f^n(x) = 2^n x - (2^n - 1)$ , which reads $2x - 1$ , $4x - 3$ , $8x - 7$ for $n = 1, 2, 3$ . The pattern continues because each new application does the same two things to whatever came before it: double the result and subtract $1$ . So the coefficient of $x$ doubles again, and the constant term absorbs one more copy of that same rule. None of these equal the output square $(f(x))^2 = 4x^2 - 4x + 1$ , which is not even linear: iterating a function and squaring its value are unrelated operations that happen to share a symbol.

Combining functions without chaining them

Composition is not the only way to build a new function from two others. You can add, subtract, multiply, or divide two functions by combining their outputs at each input:

\begin{aligned} (f + g)(x) &= f(x) + g(x), \qquad (fg)(x) = f(x)\,g(x), \\ \left(\frac{f}{g}\right)(x) &= \frac{f(x)}{g(x)}. \end{aligned}

These evaluate both functions at the same $x$ and combine the two numbers, so both functions see the original input. Composition is different in kind: it chains them, feeding the output of one as the input of the other, so only $f$ sees $x$ and $g$ sees $f(x)$ . The domain rules differ to match. A sum or product is defined where both functions are, $\operatorname{dom} f \cap \operatorname{dom} g$ , the very intersection that is the wrong answer for composition. A quotient $\dfrac{f}{g}$ additionally removes the inputs where $g(x) = 0$ .

A bit of history (optional)

For most of the nineteenth century, mathematicians who studied groups only ever meant one thing: groups of permutations, the different ways to rearrange a list. Combining two rearrangements meant doing one after the other, which is composition and nothing else. Nobody thought of it as an example of something more general, because nothing else looked like it.

Arthur Cayley, an English mathematician who earned his living as a lawyer, changed that in 1854. His short papers on groups kept the laws an operation obeys and let go of the objects obeying them. In place of a recipe for shuffling a list, he printed a table of results. The same laws turned out to describe things with nothing else in common, such as square blocks of numbers under their own rule for multiplying.

This lesson takes the identical step with composition, a procedure you already knew. It names the laws that procedure obeys: associative, carrying an identity, and, among functions from one set to itself, not generally commutative.