Composition of Functions: Free Response

Build the two inner composites first. For $g \circ f$, substitute the whole rule for $f$ into $g$:

$(g \circ f)(x) = g(4x + 3) = \frac{1}{4x + 3}.$

For $h \circ g$, substitute the whole rule for $g$ into $h$:

$(h \circ g)(x) = h\left(\frac{1}{x}\right) = \frac{1}{x} - 5.$

Now assemble the two groupings. Feeding $f$ into the composite $h \circ g$ gives

$\big((h \circ g) \circ f\big)(x) = (h \circ g)(4x + 3) = \frac{1}{4x + 3} - 5,$

and feeding the composite $g \circ f$ into $h$ gives

$\big(h \circ (g \circ f)\big)(x) = h\left(\frac{1}{4x + 3}\right) = \frac{1}{4x + 3} - 5.$

The two rules are identical. Both also refuse exactly the input that makes $4x + 3 = 0$, namely $x = -\frac{3}{4}$, and accept every other real number, so the two groupings agree in domain as well as in rule and are the same function. As a spot check at $x = 1$, both routes give $\frac{1}{7} - 5 = -\frac{34}{7}$.

Notice that the intermediate composites are genuinely different functions, $\frac{1}{4x+3}$ against $\frac{1}{x} - 5$. It is only the completed chains that coincide.

Two functions are equal only when they share a domain and agree at every input of it, so each composite needs its domain worked out as well as its rule.

For $p \circ j$ the inner function is $j$, so an input must first be legal for $j$, which means $x \ge 0$, and then $j(x) = x$ must be legal for $p$, which accepts everything. The rule is $p(j(x)) = 2x + 9$, so

$p \circ j: \quad x \mapsto 2x + 9 \quad \text{on } x \ge 0.$

For $j \circ p$ the inner function is $p$, which accepts every real number, and then the output $p(x)$ must be legal for $j$, which needs it to be at least $0$:

$2x + 9 \ge 0 \iff x \ge -\frac{9}{2}.$

The rule is $j(p(x)) = 2x + 9$, so

$j \circ p: \quad x \mapsto 2x + 9 \quad \text{on } x \ge -\frac{9}{2}.$

All three functions share the formula $2x + 9$, and all three have different domains, so no two of them are equal. The lesson's identity law is not damaged by this: it says the identity is neutral when it is taken on the right set. On the right of $p$ that set must accept every input $p$ accepts, and on the left it must accept every output $p$ produces. The rule $j$ here is too small for both jobs, and the composition loses inputs to prove it. Taken on all of $\mathbb{R}$ instead, the identity does its job: both $\mathrm{id} \circ p$ and $p \circ \mathrm{id}$ have domain $\mathbb{R}$ and rule $2x+9$, so both equal $p$.

Part A tested associativity: the two ways of bracketing a chain of three functions give the same function. That is exactly what licenses the unbracketed name. Writing

$h \circ g \circ f$

is meaningful only because the two readings it could carry, $(h \circ g) \circ f$ and $h \circ (g \circ f)$, are never different functions. Without the law the string would be an abbreviation for two possibly different things, and a chain of four functions would carry five readings instead of one.

Part B tested the second law. There is an identity for composition, and composing with it changes nothing, provided it is taken on a set large enough to lose no inputs.

Now name the structure. An operation on a collection of objects that is associative and has an identity, with no requirement that it commute and no requirement that every object have an inverse, gives that collection the structure algebraists call a monoid. Which collection is not a free choice: two functions can be composed only when the outputs of the first are inputs the second accepts, so take the rules that run from one fixed set back to that same set, whose identity is $\operatorname{id}$ on that set. Composition gives that collection exactly the structure just described. Part A supplied associativity, and Part B showed what the identity law needs and that it is met once the identity is taken on a large enough set. Commutativity is not available for a general pair, and an inverse is not available for every function. Two laws demanded, two left optional.

Substitute the whole rule for $f$ into $g$ and tidy the denominator:

$(g \circ f)(x) = g(x^2 + 2) = \frac{1}{(x^2 + 2) - 6} = \frac{1}{x^2 - 4}.$

Now the domain, in the two stages the rule asks for. First, $f$ must accept $x$, and $f(x) = x^2 + 2$ is defined for every real number, so this stage throws nothing away. Second, the output $f(x)$ must be a number $g$ accepts, and $g$ refuses exactly one input, $6$. So the inputs to remove are the ones $f$ sends to $6$:

$x^2 + 2 = 6 \iff x^2 = 4 \iff x = 2 \ \text{ or } \ x = -2.$

The domain is every real number except $2$ and $-2$. The simplified formula agrees, since $x^2 - 4$ is zero at exactly those two inputs. Note where the work happened: finding the exclusions meant asking which inputs $f$ sends to a particular value, not which inputs $f$ will accept.

This time $g$ is the inner function, so substitute its rule into $f$:

$(f \circ g)(x) = f\left(\frac{1}{x - 6}\right) = \left(\frac{1}{x - 6}\right)^2 + 2 = \frac{1}{(x - 6)^2} + 2.$

For the domain, run the two stages again. First $g$ must accept $x$, which rules out $x = 6$. Then the output $g(x)$ must be a number $f$ accepts, and $f$ accepts every real number, so the second stage removes nothing further:

$\operatorname{dom}(f \circ g) = \{\, x : x \ne 6 \,\}.$

Set that beside Part A, whose domain was every real except $2$ and $-2$. The same two functions, composed the two possible ways, produce different domains: this order keeps $2$ and $-2$ and drops $6$, while the other order does exactly the reverse. So $\operatorname{dom}(g \circ f)$ and $\operatorname{dom}(f \circ g)$ are in general different sets, and neither can be read off the other.

Apply the proposed rule first. Here $\operatorname{dom} f$ is every real number and $\operatorname{dom} g$ is every real number except $6$, so

$\operatorname{dom} f \cap \operatorname{dom} g = \{\, x : x \ne 6 \,\}.$

That is exactly the Part B domain, and it is not the Part A domain. So the proposal gets one order of this very pair right and the other wrong. That is enough to sink it: a rule you have to check against the truth before trusting it is doing no work, and here it even reverses the correct answer, keeping the two inputs that fail and discarding one that is perfectly good.

The reason for the failure is in what each stage tests. Intersecting the two domains only asks which numbers each function will accept as an input. The second stage of a composition asks something different: whether the number $f$ produces is acceptable to $g$. That is a question about which inputs $f$ maps to $6$, the preimage of $6$, and it cannot be answered by looking at domains at all.

The shape of the two answers makes the gap visible. Since $f$ accepts every real number and $g$ refuses a single input, the intersection can delete at most that one point, no matter what. Part A had to delete two:

$x^2 + 2 = 6 \iff x = 2 \ \text{ or } \ x = -2,$

and a quadratic reaching one value from two places is exactly the kind of thing an intersection of domains can never record.

Substitute one whole rule into the other, in each order. With $v$ inside, the cube acts on all of $7x$, coefficient included:

$(u \circ v)(x) = u(7x) = (7x)^3 = 343x^3.$

With $u$ inside, the multiplication by $7$ acts on the cube:

$(v \circ u)(x) = v(x^3) = 7x^3.$

These are different functions. To find where their values coincide, set them equal:

$343x^3 = 7x^3 \iff 336x^3 = 0 \iff x = 0.$

So the two agree at $x = 0$ and nowhere else. Agreeing at one input is not the same as being equal, so this pair does not commute.

Keep the exponents as letters, so that one calculation settles every pair at once. With $p_n$ inside,

$(p_m \circ p_n)(x) = p_m(x^n) = (x^n)^m = x^{nm},$

by the power of a power rule. With $p_m$ inside,

$(p_n \circ p_m)(x) = p_n(x^m) = (x^m)^n = x^{mn}.$

Multiplication of whole numbers commutes, so $nm = mn$ and the two rules are the same rule. Both composites are also defined for every real number, since a whole number power of a real number is always a real number, so the domains agree too and the equality test is satisfied:

$p_m \circ p_n = p_n \circ p_m = p_{mn}.$

Nothing in that calculation depended on which whole numbers $m$ and $n$ were, so it covers all of them at once: every pair of power maps commutes, and there are infinitely many such pairs. Taking $m = 3$ and $n = 4$, for instance, both orders give $x^{12}$.

It is worth seeing why Part A was different, since a cube appeared there too. The partner was $7x$, a scaling, which is not a power map. In one order the $7$ was cubed and in the other it was not, and that asymmetry is precisely what stacking two exponents does not produce.

The conclusion is too strong. Part A does show one pair whose two orders differ, and a single such pair is enough to establish that composition is not always commutative. It is nowhere near enough to establish that composition is never commutative, which is a claim about every pair at once. Part B settles that direction outright by exhibiting a whole family of pairs that do commute:

$p_3 \circ p_4 = p_4 \circ p_3 = p_{12}.$

So the claim the evidence supports is this: composition is not always commutative. Read carefully, that says two things. For a general pair you may not assume $u \circ v = v \circ u$, so an argument must never quietly swap the two. And for a particular pair you may not assume they differ either, so a claim that a specific pair fails to commute has to be established, usually by naming one input where the two values disagree, as Part A's calculation delivers: the two agree only at $x = 0$, so every other input is a witness that they differ.

Treating never and not always as interchangeable throws the second half away. It turns a correct warning about the general case into a false claim about every case, and it would deliver a wrong verdict on every commuting pair there is, including all of Part B's. The difference between the two claims is the difference between a rule you may rely on and a rule that is simply false.

Follow what happens to an input, in order: it is multiplied by $3$ and has $8$ subtracted, then the result is squared, then the reciprocal of that is taken. Give each stage a letter:

$a(x) = 3x - 8, \qquad b(u) = u^2, \qquad c(v) = \frac{1}{v}.$

Check the chain by substituting one rule into the next, working from the right, since the rightmost function acts first:

$c\big(b(a(x))\big) = c\big((3x - 8)^2\big) = \frac{1}{(3x - 8)^2} = h(x).$

Now the input the chain refuses. The stages $a$ and $b$ accept every real number, but $c$ refuses $0$, so the chain refuses exactly the inputs that $b \circ a$ sends to $0$:

$(3x - 8)^2 = 0 \iff 3x - 8 = 0 \iff x = \frac{8}{3}.$

That is precisely where $h$ itself is undefined, so the chain matches $h$ on the set it is defined on as well as in its formula, which is what rebuilding a function asks for.

Cut the chain somewhere else. Squaring and taking a reciprocal can be done in either order on a nonzero number, since

$\left(\frac{1}{u}\right)^2 = \frac{1}{u^2},$

so the last two stages can trade places:

$a_2(x) = 3x - 8, \qquad b_2(u) = \frac{1}{u}, \qquad c_2(v) = v^2,$

and indeed $c_2\big(b_2(a_2(x))\big) = \left(\dfrac{1}{3x - 8}\right)^2 = h(x)$, on the same set $x \ne \frac{8}{3}$. A third chain moves a cut on the input side instead, absorbing the multiplication into the first stage:

$a_3(x) = 3x, \qquad b_3(u) = (u - 8)^2, \qquad c_3(v) = \frac{1}{v},$

which rebuilds $h$ as well.

None of these has a better claim than the others. Each composes back to $h$ on the same domain, which is the only test there is, and they differ only in where the author chose to draw the lines between stages. A decomposition records how a rule was read, not a property the rule itself carries, which is why a question can ask for a valid chain but never for the chain. The first chain is the most natural in the sense that each stage performs one familiar operation in the order you would say them aloud, but natural is not the same as correct, and here every chain listed is correct.

The quoted line squared the value of $f$. It computed

$(f(x))^2 = (3x + 4)^2 = 9x^2 + 24x + 16,$

which multiplies the output of $f$ by itself. This lesson reads a superscript on a function name as counting applications of the function, so $f^2$ means $f \circ f$, and the entire rule goes back into $f$:

$f^2(x) = f(f(x)) = f(3x + 4) = 3(3x + 4) + 4 = 9x + 16.$

One is linear and one is quadratic, so they are certainly different functions. They do agree somewhere, though, and it is worth knowing where. Setting them equal:

$9x^2 + 24x + 16 = 9x + 16 \iff 9x^2 + 15x = 0 \iff 3x(3x + 5) = 0,$

so they agree at $x = 0$, where both give $16$, and at $x = -\frac{5}{3}$, where both give $1$. Two shared values out of infinitely many inputs, which is exactly why coinciding somewhere is no evidence at all of being the same function.

Each command takes a width and returns a width, so running one after the other is a composition. The command that runs second is the outer function, because it acts on what the first one produced.

Running Scale first puts $S$ inside and $T$ outside, which is the composite $T \circ S$:

$(T \circ S)(w) = T(4w) = 4w + 5.$

Running Frame first puts $T$ inside, and Scale then enlarges everything on the canvas, the new frame included:

$(S \circ T)(w) = S(w + 5) = 4(w + 5) = 4w + 20.$

The two rules have the same coefficient and different constants, so they are different functions for every $w$. The order of the commands is not a matter of workflow taste: it changes the finished width.

Put $w = 12$ through each rule.

$(T \circ S)(12) = 4(12) + 5 = 53, \qquad (S \circ T)(12) = 4(12 + 5) = 68.$

The finished widths are $53$ centimetres and $68$ centimetres, and

$68 - 53 = 15.$

The $15$ centimetres is the frame's own growth. Framing first draws a $5$ centimetre frame and then lets Scale multiply it by $4$, turning it into $20$ centimetres of frame; scaling first leaves the frame at the $5$ centimetres it was drawn at. The artwork itself finishes at $4(12) = 48$ centimetres under either order, so the entire difference sits in the border.

The classmate is wrong, and this question is already the counterexample. The outer map $A(y) = ay + k$ scales the output and then shifts it, which is the composite of $y \mapsto ay$ followed by $y \mapsto y + k$, exactly the shape of Scale followed by Frame. The proposed alternative shifts and then scales. Expanding it,

$a(y + k) = ay + ak,$

which differs from $ay + k$ by

$ak - k = k(a - 1).$

That gap is zero only when $k = 0$ or $a = 1$, which is to say only when one of the two operations does nothing at all. With this question's numbers, $a = 4$ and $k = 5$, the gap is $5(4 - 1) = 15$, the same $15$ centimetres separating the two orders in Part B. Two different outer maps produce two different final rules, so for the same $a$ and $k$ the two formulas describe different graphs.

The order inside the master formula is therefore forced rather than chosen. The formula commits to scaling the output first and shifting it second, and shifting first is a genuinely different function, so the two cannot be swapped for free. Non-commutativity is the entire reason the formula has to fix an order at all, and the same reason lies behind every warning that stretching then shifting differs from shifting then stretching.

The input map is written to the right of $f$ for the reading rule that governs the whole lesson. In

$y = (A \circ f \circ B)(x)$

the rightmost function acts first, so $B$ changes the input before $f$ ever sees it, which is precisely the job $b(x - h)$ does inside the parentheses.