Composition of Functions

Intuition

You already know how to compose two functions: to build $g(f(x))$ you drop $f(x)$ into $g$ wherever its input goes, and you can find a formula for the result. This lesson does not repeat that procedure; it assumes it. Instead it treats composition the way arithmetic treats addition, as an operation on its own objects, functions, with laws you can prove. Addition of numbers is associative, has the identity $0$ , and is commutative. Composition of functions is associative and has an identity too, but it is not always commutative, and its domain hides a subtlety the substitution procedure quietly skips. By the end you will see that the transformation rules of the previous lesson were compositions the whole time. You will also see that asking someone to decompose a function has many correct answers, not one.

Learning goals

Read $g \circ f$ right to left, with $f$ acting first
Use associativity to drop parentheses from a chain
Check the domain against the range of the inner function
Show that composition need not commute, but sometimes does
Decompose a function, knowing the split is a choice
Write $f^2$ for $f \circ f$ , never for the square

Reading a composition right to left

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

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

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, and $g \circ f$ names a new function built from the two you started with.

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; they agree only where $2x = 0$ , that is at $x = 0$ . 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

The associative law

Addition of three numbers needs no parentheses: $(a + b) + c$ and $a + (b + c)$ are equal, so we simply write $a + b + c$ . Composition has the same property, and it is a genuine theorem worth proving, because it is what makes a chain of three or more functions unambiguous.

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 and agree at every input, the equality test from Function Notation, so we check the values and the domains.

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 have the same domain as well as the same values. 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 same holds for any number of functions, which is why an iterated composition like $f \circ f \circ f$ is unambiguous and needs no bracketing to say what it means.

The identity function

Addition has the number $0$ : adding it changes nothing, since $a + 0 = a$ . Composition has a function that plays the same role, the identity function

\operatorname{id}(x) = x,

which hands back its input untouched. Composing with it changes nothing, but stating that carefully means being honest about the domain, because the equality test demands more than a matching formula.

The identity function is neutral for composition#

Let $f$ be a function, and let $\operatorname{id}(x) = x$ . We show $f \circ \operatorname{id} = f$ and $\operatorname{id} \circ f = f$ , taking $\operatorname{id}$ on the correct set each time.

Take $\operatorname{id}$ with the same domain as $f$ . Then for every $x \in \operatorname{dom} f$ ,

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

and the domain of $f \circ \operatorname{id}$ is $\{x \in \operatorname{dom} f : \operatorname{id}(x) \in \operatorname{dom} f\}$ , which is all of $\operatorname{dom} f$ because $\operatorname{id}(x) = x$ already lies there. Same domain, same values, so $f \circ \operatorname{id} = f$ .

Now take $\operatorname{id}$ on a set that contains the outputs of $f$ , such as its codomain. Then for every $x \in \operatorname{dom} f$ ,

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

and the domain is $\{x \in \operatorname{dom} f : f(x) \in \operatorname{dom} \operatorname{id}\}$ , which is again all of $\operatorname{dom} f$ because every output of $f$ is in that set. So $\operatorname{id} \circ f = f$ .

The care is the whole point. Sharing the formula $x$ is not enough for two functions to be equal; the equality test also demands the same domain. That is why $\operatorname{id}$ must be taken on the right set, the domain of $f$ on one side and a set holding its outputs on the other. Take it on too small a set and the composition would lose inputs and become a different function.

Order matters, but not never

The one law that addition has and composition 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 opposite orders can produce different outputs. With $f(x) = x + 1$ and $g(x) = x^2$ ,

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

and these differ by $2x$ , which is zero only at the single point $x = 0$ . So $g \circ f \ne f \circ g$ here.

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. For a general pair $g \circ f \ne f \circ g$ , while for special pairs the two are equal. Writing “composition is never commutative” is simply false, as the translations show; commuting is the exception, not the rule.

A little vocabulary ties the laws together. An operation that is associative and has an identity, but need not supply an inverse for every element or commute, gives its objects the structure algebraists call a monoid. Under composition the functions 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.

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. Writing that down gives the domain of the composition.

The domain of $g \circ f$ #

By definition $(g \circ f)(x) = g(f(x))$ , so this value exists exactly when every step inside it is legal. Reading from the inside out, first $f(x)$ must be defined, which requires $x \in \operatorname{dom} f$ . Given that, $g$ is then applied to the number $f(x)$ , which is legal exactly when $f(x) \in \operatorname{dom} g$ . Both conditions must hold, and if they do the value $g(f(x))$ exists, 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 requires knowing which numbers $f$ actually produces, 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 $\operatorname{dom} g$ . The reverse also happens: a value of $x$ outside $\operatorname{dom} g$ can still belong, as long as $x \in \operatorname{dom} f$ and $f(x) \in \operatorname{dom} g$ .

So computing the domain of a composition is where the range of the inner function finally earns its keep. Two examples make the point, and the second is the payoff of the whole chapter’s insistence that the domain is part of a function’s data.

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 throw out $x = 2$ (the value $g$ forbids) and keep $x = 4$ . 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.

Worked example 3 When the formula lies about the domain

Let $f(x) = \sqrt{x}$ and $g(x) = x^2$ . Find $g \circ f$ , and decide whether it is the 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 is just $x$ , so it is tempting to say $g \circ f$ is the identity. It is not. Its domain is

\operatorname{dom}(g \circ f) = \{\, x \in \operatorname{dom} f : \sqrt{x} \in \operatorname{dom} g \,\} = \{\, x \ge 0 \,\},

since $f$ requires $x \ge 0$ and $g$ accepts every real output. So $g \circ f$ is the rule $x \mapsto x$ on the domain $x \ge 0$ , and the identity function has domain all real numbers. By Function Notation’s equality test, which demands the same domain and not merely the same formula, these are different functions. The simplified $x$ threw away the record of where $\sqrt{\phantom{x}}$ refused to run, and 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 $x$ because $x^2 \ge 0$ always, so $\operatorname{dom}(f \circ g)$ is all real numbers while $\operatorname{dom}(g \circ f)$ is only $x \ge 0$ . Neither composition is the identity on all of $\mathbb{R}$ , and the two even have different domains, a reminder that $\operatorname{dom}(f \circ g)$ and $\operatorname{dom}(g \circ f)$ are in general different sets. Whether these two rules “undo” each other, and what that requires, is the business of the next lesson; here we only read off the domains.

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

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 is the move that 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). \checkmark

But this is not the only way, and that is the important part. Here are two more decompositions of the very same $h$ :

f(x) = x,\ g(x) = \sqrt{3x + 1} \quad\Rightarrow\quad g(f(x)) = \sqrt{3x + 1},

f(x) = 3x,\ g(x) = \sqrt{x + 1} \quad\Rightarrow\quad g(f(x)) = \sqrt{3x + 1}.

All three are correct, since each composes to $h$ . So “decompose $h$ ” has many right answers, not one, exactly as the previous lesson’s “which transformation produced this graph” had many. When a problem asks you to decompose a function it is asking for a valid split. Among the valid splits there is usually a most natural one, peel off the outermost operation, but no split is the unique correct answer. A question that treats one decomposition as “the” answer has no single answer, so we never phrase it that way.

Worked example 4 Two decompositions of one function

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

The natural split peels off 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). \checkmark

A different split absorbs the doubling into the inner step and leaves the subtraction with the cube. Let $f(x) = 2x$ and $g(x) = (x - 5)^3$ :

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

Both are valid decompositions of the same $h$ . Neither is more correct than the other; they simply draw the line between “inner step” and “outer step” in different places. A later course leans on exactly this idea, composing the rates of change of the pieces along a decomposition to differentiate $h$ , the chain rule. The point here is only that the decomposition that course walks along is a choice, not a fixed feature of $h$ .

Transformations were compositions all along

The previous lesson wrote every transformed graph in the shape

y = a\,f\big(b(x - h)\big) + k,

and split it into an inside that acts on the input and an outside that acts on the output. Name those two pieces as functions. Let

B(x) = b(x - h), \qquad A(y) = a y + k.

Then $B$ is the input map, $A$ is the output map, and the transformed rule is a composition of three functions:

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

The transformation was a composition the whole time. Reading right to left, $B$ acts first, then $f$ , then $A$ , which is exactly the story the previous lesson told: the inside happens before $f$ , and the outside after. Two of that lesson’s hardest points are now single facts about composition.

Its discovery that “stretch then shift” differs from “shift then stretch” is the non-commutativity of composition. Take the output maps “stretch by $2$ ,” written $S(y) = 2y$ , and “shift up $3$ ,” written $T(y) = y + 3$ . Stretching then shifting is $T \circ S$ , and shifting then stretching is $S \circ T$ :

(T \circ S)(y) = 2y + 3, \qquad (S \circ T)(y) = 2(y + 3) = 2y + 6.

These are different because $S$ and $T$ do not commute, the same reason $g \circ f \ne f \circ g$ in general. The order of two transformations is nothing but the order of a composition, and composition does not let you swap that order for free.

Its other rule, that input maps “run backward because you solve for $x$ ,” is the statement that $B$ is applied first. 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 $x$ . Solving is undoing the first-applied map, and on the input side the first-applied map is $B$ . So the whole previous lesson was a special case of this one. Transforming a graph is composing $f$ with an input map on one side and an output map on the other, and every ordering subtlety there was composition’s non-commutativity here.

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.

Now check that composing them rebuilds the rule. Applying $B$ first, then $f$ , then $A$ :

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

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. The input map $B$ sits on the right because it acts first, exactly the “solve for $x$ ” order of the previous lesson.

Iterating a function, and a warning about $f$ squared

Because composition is associative, composing a function with itself repeatedly is unambiguous, and it has its own notation. Write

f^2 = f \circ f, \qquad f^3 = f \circ f \circ f,

and so on, 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. By Function Notation’s equality test, coinciding at a stray input is not the same as being equal. Here the two functions meet only at $x = 1$ , where both equal $1$ , and differ everywhere else. 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, the function that undoes it, and it will never mean $\dfrac{1}{f}$ . A superscript on a function name is about composition, not multiplication.

Worked example 6 Iterating a linear function

Let $f(x) = 2x - 1$ . Compute $f^2$ and $f^3$ .

Apply $f$ to $f(x)$ , wrapping the whole input in the rule:

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

Apply $f$ once more, now to $f^2(x)$ :

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

A pattern is already visible: each step doubles the coefficient and the constant follows, giving $f^n(x) = 2^n x - (2^n - 1)$ , which reads $2x - 1$ , $4x - 3$ , $8x - 7$ for $n = 1, 2, 3$ . None of these equals 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, and it should not be confused with the arithmetic ways. You can add, subtract, multiply, or divide two functions by combining their outputs at each input, $(f + g)(x) = f(x) + g(x)$ , $(fg)(x) = f(x)\,g(x)$ , and $\left(\tfrac{f}{g}\right)(x) = \tfrac{f(x)}{g(x)}$ . These evaluate both functions at the same $x$ and then 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 $\tfrac{f}{g}$ additionally removes the inputs where $g(x) = 0$ . These arithmetic combinations are a lesson of their own, studied in an earlier course. The point here is only that combining outputs and chaining inputs are not the same operation, and composition is the one this lesson is about.

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: $g \circ f$ and $f \circ g$ differ for a general pair but are equal for special pairs. Any two translations $f(x) = x + a$ and $g(x) = x + b$ are such a pair, since both orders give $x + a + b$ .
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 $f$ . For $f(x) = \sqrt{x}$ and $g(x) = \tfrac{1}{x - 2}$ it excludes $x = 4$ , not $x = 2$ .
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 not the identity function. By Function Notation’s equality test the domain is part of the function.
Assuming a function decomposes in only one way. $h(x) = \sqrt{3x + 1}$ is $g \circ f$ for $f(x) = 3x + 1$ , $g(x) = \sqrt{x}$ , and also for $f(x) = 3x$ , $g(x) = \sqrt{x + 1}$ , so “decompose $h$ ” has many correct answers.
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 $(2x - 1)^2$ . And $f^{-1}$ in the next lesson will mean the inverse under composition, never $\tfrac{1}{f}$ .

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 associative, $(h \circ g) \circ f = h \circ (g \circ f)$ , so $h \circ g \circ f$ needs no parentheses. The identity function $\operatorname{id}(x) = x$ is neutral, $f \circ \operatorname{id} = f$ and $\operatorname{id} \circ f = f$ when $\operatorname{id}$ is taken on the right set, since the equality test demands equal domains.
Composition is not always commutative: $g \circ f$ and $f \circ g$ differ in general ( $f(x) = x + 1$ , $g(x) = x^2$ give $x^2 + 2x + 1$ versus $x^2 + 1$ ) but agree for special pairs like translations. Never say “never.”
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 $f$ . So $(g \circ f)(x) = x$ for $f(x) = \sqrt{x}$ , $g(x) = x^2$ has domain $x \ge 0$ and is not the identity.
Decomposition is not unique: $h(x) = \sqrt{3x + 1}$ splits as $g \circ f$ in several ways, so a decomposition is a choice, not a fixed feature of $h$ .
Every graph transformation $y = a\,f(b(x - h)) + k$ is the composition $A \circ f \circ B$ with $B(x) = b(x - h)$ and $A(y) = a y + k$ , so “order matters” there is exactly non-commutativity here.
Composing a function with itself is written $f^2 = f \circ f$ (not $(f(x))^2$ ). The functions under composition form a monoid: associative, with an identity, and with inverses only sometimes, which the next lesson takes up.

Composition of Functions

Learning goals

Reading a composition right to left

The associative law

The identity function

Order matters, but not never

The domain of a composition

Decomposing a function, and why the answer is not unique

Transformations were compositions all along

Iterating a function, and a warning about $f$ squared

Combining functions without chaining them

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)

Composition of Functions

Learning goals

Reading a composition right to left

The associative law

The identity function

Order matters, but not never

The domain of a composition

Decomposing a function, and why the answer is not unique

Transformations were compositions all along

Iterating a function, and a warning about fff squared

Combining functions without chaining them

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Free Response Questions (FRQ)

Iterating a function, and a warning about $f$ squared