Functions: Domain, Range, Composition and Inverses

Before you read on

Two or three questions on exactly what this lesson teaches. Being wrong here is fine — it's the fastest way to find out what to pay attention to next.

What a function actually is, and the one distinction the whole topic runs on

A function $f$ is a rule that assigns to every input exactly one output. This course's own transcription of the spec's own guidance puts it as "the concept of a function as a one-one or many-one mapping from $\mathbb{R}$ (or a subset of $\mathbb{R}$ ) to $\mathbb{R}$ " — every input to a genuine function has exactly one output, which is why $y^2 = x$ is not a function of $x$ (the input $x=4$ gives two outputs, $y=2$ and $y=-2$ ) even though $y = x^2$ is one (every $x$ gives exactly one $y$ ). Written as $f: x \mapsto x^2$ or $f(x) = x^2$ , the two notations mean the same thing; the spec expects both to be recognised.

The domain is the set of inputs the function is actually applied to — what you are allowed to put in. The range is the set of outputs that actually come out — not a set you choose, but a consequence of the domain and the rule together. This is the distinction the third prequestion question above was built to expose: an unrestricted domain (" $x$ can be anything") says nothing on its own about whether the range is unrestricted too. A quadratic with domain all of $\mathbb{R}$ still has a range bounded on one side, because its graph has a turning point; a fraction with domain all of $\mathbb{R}$ except one value has a range that is missing a value too, for a related reason met later in this lesson.

The distinction the spec singles out by name — "one-one or many-one" — is about how many inputs share an output. A function is one-one if every output comes from exactly one input: no two different $x$ -values ever give the same $f(x)$ . A function is many-one if at least one output is shared by more than one input. $f(x) = x^2$ on all of $\mathbb{R}$ is , because $f(-3) = f(3) = 9$ : two different inputs, one output. $f(x) = 2x - 1$ on all of $\mathbb{R}$ is : solving $2a - 1 = 2b - 1$ forces $a = b$ , so no two different inputs can ever share an output. This is not a side classification — it is the single fact the rest of this lesson is built to use, because whether a function has an inverse (as a function, not just as a reversed rule) depends on nothing else.

A graphical shortcut for one-one-ness, standard in this topic though not named in the spec's own wording: sketch the curve $y = f(x)$ and draw a horizontal line at any height. If every horizontal line meets the curve at most once, the function is one-one; if some horizontal line meets it twice, it is many-one. For $y = x^2$ , the line $y = 9$ meets the curve at both $x=-3$ and $x=3$ — visibly many-one. This is the same fact as the algebraic test above, seen from the picture instead of the equation, and it is worth having both: the picture is faster to check, the algebra is what a mark scheme actually credits.

Composition — building one function out of two

Given two functions $f$ and $g$ , the composite function $fg$ is formed by applying $g$ first and then applying $f$ to the result. This course's own research transcribes the spec's own guidance verbatim: "students should know that $fg$ means 'do g first, then f'." So $fg(x)$ means $f(g(x))$ — read the notation right to left, applied left to right: the letter closest to the bracket, $g$ , acts on $x$ first.

Concretely, with $f(x) = 2x - 1$ and $g(x) = x^2$ : $fg(x) = f(g(x)) = f(x^2) = 2x^2 - 1$ — take $g$ 's output, $x^2$ , and feed it into $f$ . Going the other way, $gf(x) = g(f(x)) = g(2x-1) = (2x-1)^2 = 4x^2 - 4x + 1$ . These are genuinely different functions, not two ways of writing the same thing — $fg(2) = 2(4)-1 = 7$ , while $gf(2) = (3)^2 = 9$ . Composition order is not a notational nicety to get right for its own sake; $fg$ and $gf$ are different rules with, in general, different domains, different ranges, and different values everywhere except where they happen to coincide by accident.

The domain of a is not automatically the domain of its outer function. $fg(x) = f(g(x))$ is only defined where $g(x)$ is itself a valid input to $f$ — so if $f$ 's own domain is restricted, that restriction becomes a condition on $g(x)$ , not on $x$ directly, and has to be translated back into a condition on $x$ . Take $f(x) = x^2 - 4x + 7$ restricted to $x \geq 2$ (met properly, with its inverse, later in this lesson), composed with $g(x) = 2x - 1$ on all of $\mathbb{R}$ : $fg(x)$ only makes sense where $g(x) \geq 2$ , i.e. where $2x - 1 \geq 2$ , i.e. $x \geq 1.5$ — a restriction on $x$ that came entirely from $f$ 's own restricted domain, filtered through what $g$ does to $x$ first. Forgetting this is a real and easy trap: a composite built from a domain-restricted function inherits a domain restriction of its own, translated through the inner function, not copied unchanged.

Why some machines can't be run backwards — before the algebra

In plain terms

Picture a machine that squares whatever number you post into it. Post in 3, it hands you back 9. Post in −3, it also hands you back 9 — two different starting numbers, one and the same output. Now walk round to the OUTPUT slot, holding a 9, and try to run the machine in reverse: what number was originally posted in? There is no single right answer — it could have been 3, it could have been −3, and both are equally true. That is the whole problem with reversing this machine: reversing means handing back one number, and here there are genuinely two correct ones. Now change the rule at the input slot: only non-negative numbers allowed in — 0, 1, 2, 3, 4, and so on, nothing negative ever accepted. Post in 3, get 9. Post in 0, get 0. Post in 4, get 16. Every number you're now allowed to post gives a different output from every other one you're allowed to post, because the negative twin that used to share its output (−3, −4, ...) has been locked out at the door. Walk round to the output slot again, holding a 9: this time exactly one number could have produced it and still have been allowed in — 3. Reversing now works, every time, with no ambiguity. There's a second thing worth noticing while the numbers are still in front of you. Reversing the machine means: take an output (say 9), hand back its non-negative square root (3). The numbers the reversed machine is allowed to accept as input are exactly the numbers the original machine could actually produce as output — 0, 1, 4, 9, 16, and so on, never anything negative, because the squaring machine never outputs a negative number either. And the numbers the reversed machine hands back are exactly the numbers the original machine was allowed to accept — 0, 1, 2, 3, 4, .... The 'allowed in' list and the 'comes out' list have swapped jobs completely between the two machines.

Name what just happened. A machine (a function) is many-one when two different inputs — like 3 and −3 — share a single output, like 9; a many-one machine cannot be run in reverse, because reversing it would have to hand back two different numbers for the one input it was just given, and a machine that gives more than one answer for the same input has stopped being a function at all, reversed or not. Banning the negative inputs turned the squaring machine one-one: every input left allowed now owns its output alone, nobody sharing it with anyone else. The list of allowed inputs is the machine's ; the list of outputs it can actually produce is its . Building the reverse machine swaps the two lists over completely — whatever could go IN to the original becomes what comes OUT of the reverse, and whatever used to come OUT becomes what's now allowed IN.

Formally

An inverse function $f^{-1}$ exists only where $f$ is one-one on its domain: if $f(a) = f(b)$ for some $a \neq b$ , then $f^{-1}$ would have to map that shared output to both $a$ and $b$ , which violates the one-input-one-output requirement that defines a function in the first place. Restricting a many-one function's domain to a single branch — $x \geq 0$ for $f(x) = x^2$ , exactly as above — makes it one-one and gives it an inverse, $f^{-1}(x) = \sqrt{x}$ . Because finding $f^{-1}$ means swapping $x$ and $y$ in $y = f(x)$ , this domain restriction has a direct consequence: the domain of $f^{-1}$ equals the range of $f$ , and the range of $f^{-1}$ equals the domain of $f$ . The mechanism below derives this rigorously and applies it to the restricted quadratics this course actually examines.

Mechanism

Why an inverse needs one-one — and why domain and range swap places

An $f^{-1}$ is meant to undo $f$ : feed it an output of $f$ , and it should hand back the input that produced it. Follow that requirement to its logical end and one-one-ness stops being a rule to memorise and becomes unavoidable. Suppose $f$ is many-one: some output $y_0$ is shared by two different inputs, $a$ and $b$ , with $f(a) = f(b) = y_0$ and $a \neq b$ . Feed $y_0$ into $f^{-1}$ and ask what comes out. It has to be $a$ , because $f^{-1}$ is meant to undo $f(a)$ . It also has to be $b$ , because $f^{-1}$ is meant to undo $f(b)$ . One input, $y_0$ , with two required outputs — but that is precisely the property that disqualifies something from being a function at all, the same one-input-one-output requirement the very first teach block opened with. So $f^{-1}$ exists as a function if and only if $f$ is one-one: not a separate condition bolted onto the definition, but the definition of "function" applied to the reversed rule. This is exactly why domain restriction matters so much in this topic — $f(x) = x^2$ on all of $\mathbb{R}$ is many-one and has no inverse function, but $f(x) = x^2$ restricted to $x \geq 0$ is one-one (every non-negative output now comes from exactly one non-negative input), and that restricted version does have an inverse: $f^{-1}(x) = \sqrt{x}$ . Nothing about the formula changed; what changed is which inputs are on offer. Now follow the mechanics of actually finding $f^{-1}$ , and a second fact falls out for free. The standard method is: write $y = f(x)$ , swap $x$ and $y$ (this is the "walk it backwards" step — the new equation asks "what $x$ produces this $y$ ?" using the old equation's own rule), then rearrange to make $y$ the subject again. Because the swap literally exchanges the two letters' roles, whatever set of values $x$ was allowed to range over becomes the set $y$ ranges over in the new equation, and vice versa. Translated back into function language: the domain of $f^{-1}$ is the range of $f$ , and the range of $f^{-1}$ is the domain of $f$ . This is not a separate fact to memorise alongside "swap $x$ and $y$ " — it is what swapping $x$ and $y$ *means*. It is also why the spec's own guidance, transcribed verbatim during this course's own research pass, states $f^{-1}f(x) = ff^{-1}(x) = x$ : applying $f$ and then $f^{-1}$ (or the other way round) walks forward along the rule and then immediately back along it, landing exactly where it started, for every $x$ in the appropriate domain.

Diagram — f and f⁻¹ as reflections in y = x — and why the swap makes domain and range change places

x-axis: x · y-axis: y

C1 · y = f(x), a one-one restricted quadratic: An upward parabola shown only on its increasing branch, from its vertex rightward — the domain restriction (x ≥ the vertex's x-coordinate) is what makes this branch one-one, since no horizontal line crosses it twice. The curve starts at the vertex and rises without bound to the right.
C2 · y = f⁻¹(x), the reflection of C1 in the line y = x: The mirror image of C1 in the diagonal line y = x. Every point (p, q) on C1 becomes the point (q, p) on C2 — literally swapping the coordinates, which is the geometric picture of algebraically swapping x and y. It starts at the reflected vertex and rises, but now reading left to right along the OTHER axis from where C1's growth was unbounded.
C3 · y = x, the mirror line itself: Drawn as a thin diagonal reference line. C1 and C2 are symmetric about it — fold the page along this line and the two curves land exactly on each other. Neither curve needs to touch this line anywhere for the reflection relationship to hold.

Vertex of f, (v, m), and its mirror image (m, v) on f⁻¹: Coordinates literally swap. If the vertex of the restricted quadratic is at (v, m) — v is where the domain restriction starts, m is the minimum output — the corresponding point on f⁻¹ is (m, v): m is now an input, v is now an output. This single swapped point is the fastest way to sanity-check a claimed inverse.
Domain of f = range of f⁻¹; range of f = domain of f⁻¹: The mechanism block derives this from the x/y swap directly. On the picture, it shows up as: whatever interval of the x-axis C1's domain covers, C2's range covers the same interval on the y-axis — because C2 is C1 with the axes' roles exchanged.
Only the ONE surviving branch reflects into a function: If the full parabola (both branches, unrestricted domain) were reflected in y = x, the mirror image would fail the vertical-line test — it would be many-valued, not a function. The domain restriction on f is what guarantees the reflection is a genuine function rather than just a mirrored curve.

Common error: Drawing f⁻¹ as a translation or a different-shaped curve altogether, rather than the exact mirror image of f in y = x — or omitting the y = x line entirely, so the reflection relationship is asserted rather than shown.

Correct: f⁻¹'s graph is not a new curve to invent; every point on it is a specific, mechanically determined reflection of a point on f. Sketching y = x first and reflecting f's key points across it (the endpoint, and one or two further points) is faster and safer than trying to picture the inverse shape from scratch.

Worked, in full

Find f⁻¹(x) and state its domain, for f(x) = x² − 4x + 7, x ≥ 2

01
Confirm f has an inverse before doing anything else. Complete the square: half of $-4$ is $-2$ , and $(x-2)^2 = x^2 - 4x + 4$ is $4$ too big, so $x^2 - 4x = (x-2)^2 - 4$ and $f(x) = (x-2)^2 - 4 + 7 = (x-2)^2 + 3$ . The vertex is at $(2, 3)$ , and the stated domain $x \geq 2$ starts exactly there — this is the increasing branch, so $f$ is one-one on this domain and does have an inverse. Without this domain restriction, $f$ would be the full, many-one parabola and no inverse function would exist at all.
Earns: B1 — a correct explanation that f is one-one on the given domain (increasing from the vertex, or equivalently: no repeated outputs on x ≥ 2). Independent of every mark below — an entirely separate B mark for the reasoning, per the general marking guidance's own definition of a B mark as needing no method attached to it.
02
Having confirmed the completed-square form is needed anyway, use it: state the range of $f$ directly from $(x-2)^2 + 3$ . The square is zero at $x=2$ (the left end of the domain) and grows without bound as $x$ increases, so $f(x) \geq 3$ .
Earns: B1 — states the range of f, f(x) ≥ 3. This mark matters beyond itself: the mechanism block above established that the domain of f⁻¹ IS the range of f, so this line is not a detour, it is the answer to the final stage of this chain, found early because the completed-square form makes it available for free.
03
Let $y = (x-2)^2 + 3$ and swap $x$ and $y$ : $x = (y-2)^2 + 3$ . This is the "walk it backwards" step from the mechanism block, applied mechanically.
Earns: M1 — attempts the swap-and-rearrange method for an inverse: writes y = f(x) (or works directly from f(x)) and swaps the roles of x and y, or equivalently rearranges directly for x in terms of y without the explicit swap. Either presentation earns this mark; what is being checked is the method, not the notation used to display it.
04
Rearrange for $y$ : $x - 3 = (y-2)^2$ , so $y - 2 = \pm\sqrt{x-3}$ . Two branches appear, and the original domain restriction is what decides between them: the stated domain was $x \geq 2$ , i.e. $y \geq 2$ once relabelled, so $y - 2 \geq 0$ — which selects the positive square root. $y = 2 + \sqrt{x-3}$ .
Earns: dM1 — correctly resolves the ± ambiguity using the original domain restriction, dependent on the method mark above. This is the step a rushed answer skips entirely: writing 2 ± √(x−3) and stopping is not yet an inverse function, since a function cannot return two values for one input — exactly the one-input-one-output requirement from the very first teach block, now applied to the answer itself.
05
State the result in function notation: $f^{-1}(x) = 2 + \sqrt{x-3}$ .
Earns: A1 — the correct expression for f⁻¹(x), cao. Check it against the mechanism block's own test: swapping x and y should reflect (2,3), the vertex of f, into (3,2) on f⁻¹ — and indeed f⁻¹(3) = 2 + √0 = 2. ✓
06
State the domain of $f^{-1}$ : it is the range of $f$ , already found in stage 2. Domain of $f^{-1}$ : $x \geq 3$ .
Earns: B1 — the domain of f⁻¹, independent of the algebra above (it is read from stage 2's range of f, not derived from the formula for f⁻¹ itself). This is the exact structure this course's own research verifies from a real mark scheme (Jan 2023 Q1(c), which finds g⁻¹(x) rather than f⁻¹(x) — the function tested there is a fraction, not a quadratic, but the mark structure is identical): M1 A1 for the algebra of the inverse, and a separate, independent B1 for its domain — and that same research pass's per-topic findings record this specific mark, across three separate series, as the one most often left unclaimed even by candidates who got everything above it right.

Complete it yourself

Complete the chain — f⁻¹(x) and its domain, for f(x) = (x + 1)² − 6, x ≤ −1

01
$f(x) = (x+1)^2 - 6$ is already in completed-square form: vertex $(-1, -6)$ . The domain $x \leq -1$ is the branch running from $-\infty$ up to the vertex — the LEFT arm, where the curve is decreasing as $x$ increases towards $-1$ . This branch is one-one (each output on it comes from exactly one input), so $f$ has an inverse here, even though the full unrestricted parabola would not.
02
Range of $f$ on this domain: the square is zero at $x=-1$ , the right-hand end of the domain, and grows without bound as $x$ decreases from there — so the minimum value $-6$ is still attained, and $f(x) \geq -6$ .

In your own words

In one sentence: why does swapping x and y to find an inverse automatically mean the domain of f⁻¹ has to equal the range of f, rather than this being two separate facts to remember?

Marked, line by line

$f(x) = 2x^2 - 12x + 19$ , $x \geq 3$ . (a) Express $f(x)$ in the form $a(x - p)^2 + q$ , and hence state the range of $f$ . (3) (b) Explain why $f$ has an inverse function, find $f^{-1}(x)$ , and state its domain. (4) (c) $g(x) = x - 1$ for $x \in \mathbb{R}$ . Find $fg(x)$ in the form $a(x-p)^2 + q$ , and hence write down the minimum value of $fg(x)$ . (3) — VERIDIAN-original question, inspired by the real structure this course's own research confirms for this sub-topic: a completing-the-square part feeding a 'hence' inverse-function part with its own separately-marked domain (the exact shape verified from Jan 2023 Q1(c)'s M1 A1 / B1 split), followed by a composition part. Not a reproduction of any past-paper question; the per-line mark allocations are modelled on verified mark-scheme conventions, not copied from a real scheme.

10 marks available

(a) — 3 marks

01
$a = 2$ : factor 2 out of the first two terms only. $f(x) = 2(x^2 - 6x) + 19$ .
Independent accuracy mark — a can be written down by inspection, needing no method, which is exactly the general marking guidance's own definition of a B mark.
B1
02
Half of $-6$ is $-3$ , and $(x-3)^2 = x^2 - 6x + 9$ is 9 too big, so $x^2 - 6x = (x-3)^2 - 9$ and $f(x) = 2\left[(x-3)^2 - 9\right] + 19$ .
Method mark for the completing-the-square structure: half the coefficient of x inside the bracket, its square subtracted outside. Earned for the structure, whatever the arithmetic does next.
M1
03
$f(x) = 2(x-3)^2 - 18 + 19 = 2(x-3)^2 + 1$ . Range: $(x-3)^2 \geq 0$ for all $x$ , with equality at the left end of the domain, $x=3$ , so $f(x) \geq 1$ .
Accuracy mark, cao, and unavailable without the M mark above. The −9 sat inside the bracket, so multiplying out by the 2 makes it −18, not −9 — this is the single most-lost mark in the part, and it is checkable in five seconds by expanding the answer back out.
A1

(b) — 4 marks

101
$f$ is defined on $x \geq 3$ , which is exactly where the completed-square form $2(x-3)^2+1$ is increasing (the branch to the right of the vertex), so no two values of $x$ in this domain give the same output — $f$ is one-one, and therefore has an inverse function.
Independent mark for a correct explanation of why f⁻¹ exists — needs no numerical method, so it survives untouched by any arithmetic error made anywhere else in the question, in either direction.
B1
102
Let $y = 2(x-3)^2 + 1$ , swap $x$ and $y$ : $x = 2(y-3)^2 + 1$ .
Method mark for the swap-and-rearrange approach to finding an inverse, applied to their part (a) answer.
M1
103
$\dfrac{x-1}{2} = (y-3)^2 \Rightarrow y - 3 = \pm\sqrt{\dfrac{x-1}{2}}$ . The domain $x \geq 3$ means $y \geq 3$ once relabelled, so $y - 3 \geq 0$ : take the $+$ root. $f^{-1}(x) = 3 + \sqrt{\dfrac{x-1}{2}}$ .
Accuracy mark, cao (oe — an equivalent surd form such as 3 + √(2x−2)/2 is accepted; the mark scheme abbreviation for this, verified against the primary mark scheme, is exactly 'oe' — or equivalent). Check: the vertex (3,1) on f reflects to (1,3) on f⁻¹, and f⁻¹(1) = 3 + 0 = 3. ✓
A1
104
Domain of $f^{-1}$ = range of $f$ (from part (a)) = $\{x : x \geq 1\}$ .
Independent mark for the domain — read directly from part (a)'s range, not derived from the formula for f⁻¹ itself. This is the exact real-verified structure (Jan 2023 Q1(c)): algebra earns M1 A1, domain earns its own separate B1.
B1

(c) — 3 marks

201
$fg(x)$ means "do $g$ first, then $f$ ": $fg(x) = f(g(x)) = f(x-1) = 2\left[(x-1) - 3\right]^2 + 1$ .
Method mark for substituting g(x) into f in the correct order — f(g(x)), not g(f(x)). Reversing this order is a real and different function; the mark is specifically for getting the order right, which is why the spec's own guidance is quoted verbatim in the teach block above.
M1
202
$fg(x) = 2(x-4)^2 + 1$ .
Accuracy mark for the simplified form, cao. Equally acceptable expanded: 2x² − 16x + 33 — the two are the same function, and the completed-square form is what makes the next line immediate rather than a fresh piece of work.
A1
203
$(x-4)^2 \geq 0$ always, so the minimum value of $fg(x)$ is $1$ , at $x=4$ .
Independent mark, follow through their completed-square form from this part — reading a minimum off a correctly-structured a(x−p)²+q form needs no further method, the same B-mark pattern used for reading a turning point in the previous lesson's marked solution.
B1 ft

Named traps

domain-of-inverse-omitted: The single most consistently reported error anywhere in the WMA13 archive for this sub-topic, confirmed across at least three separate series. Verified verbatim, Jun 2022 Q2: "Despite being a standard question, the majority of candidates still failed to state the domain for their inverse function and did not achieve the B mark." Verified verbatim, Jan 2022 Q6: "a majority of candidates were not aware that the domain was required, and therefore by far the most common score seen was 2/3" (out of 3) — independently confirmed again in Oct 2021 Q1(b). The mechanism is structural, not carelessness alone: the domain of f⁻¹ is an independent B mark (see the next trap), so getting every line of algebra correct still leaves it unclaimed unless it is written down as its own separate statement.
domain-of-inverse-derived-from-scratch-not-recognised-as-range-of-f: A distinct pattern from simply omitting the domain: some candidates DO attempt to state a domain for f⁻¹, but re-derive it from first principles on the new expression rather than recognising it is already sitting in the answer to an earlier part of the question, as the range of f. Verified, Oct 2022 Q2(b) — the report notes candidates deriving the inverse's domain "from scratch" instead of using the range of f they had, or could have had, already. This costs time even where it does not cost the mark outright, and it is the exact opposite of the efficient route the worked chain above demonstrates: find the range of f early, and the domain of f⁻¹ is already answered before the algebra for f⁻¹(x) itself has even begun.
inverse-algebra-and-its-domain-are-independently-marked: Not an error a candidate makes so much as a fact about the mark scheme that, misunderstood, produces one: the domain of an inverse function is not a bonus tacked onto the M1 A1 for finding its formula, and it is not lost automatically if the algebra goes wrong, or gained automatically if it goes right. This course's own research verifies this directly from a real mark scheme, Jan 2023 Q1(c) (there, finding g⁻¹(x) for a fraction g rather than f⁻¹(x) for a quadratic, but the structure is identical to every example in this lesson): the scheme awards M1 A1 for finding the inverse and a SEPARATE, independent B1 for the domain. Treating the domain as "the last part of finding the inverse" rather than its own distinct, separately-earned statement is the root cause behind both traps above.
inverse-rearrangement-shown-as-calculator-output-with-no-algebra: WMA13 carries an explicit, paper-wide, verbatim rubric on exactly this kind of algebra: certain questions state "In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable" (verified verbatim from the January 2023 question paper, and confirmed recurring across multiple series). Finding f⁻¹(x) is a rearrangement — precisely the kind of step a calculator or an equation solver can shortcut silently. Where a question carries this instruction, arriving at a correct f⁻¹(x) with no visible swap-and-rearrange working risks the method mark outright, whatever the final expression says.
inverse-answer-attached-to-the-wrong-function-letter: A distinct, purely notational failure mode, confirmed directly in a real mark scheme rather than an examiner report. When a question defines more than one function — exactly the situation in every worked example, chain drill and marked solution in this lesson, since composition and inverses are normally taught and tested together — a correctly-derived inverse can still lose its accuracy mark if it is labelled with the wrong function's letter. Verified verbatim, Jan 2023 Q1(c), which defines both f(x) and g(x) and asks specifically for g⁻¹(x): the scheme's note on the A1 mark reads "Condone y = (3−x)/(2x) o.e and even g⁻¹ = (3−x)/(2x) but NOT f⁻¹ = (3−x)/(2x) o.e." Right rearrangement, right final expression, zero marks — because f is the letter more often inverted in practice, and writing "f⁻¹(x) = ..." out of habit when the question asked for g⁻¹(x) is an easy, silent slip that the algebra itself never reveals.

Beyond the spec

The spec asks only that a student can find a composite function, find an inverse function of a one-one function, and state domains and ranges correctly (spec 1.2) — nothing in it requires knowing WHY some functions are their own inverse, or what property guarantees a composite is one-one. This is not needed to score full marks on any WMA13 question, but it reframes "swap x and y" from a memorised recipe into a visible symmetry, which is exactly the kind of thing that makes the domain-of-inverse trap above harder to fall into by accident.

A function that is its own inverse — $f^{-1}(x) = f(x)$ for every $x$ in its domain — is called self-inverse or an involution. Geometrically, this means its graph is symmetric about the line $y = x$ on its own, without needing a second curve reflected onto it: $f(x) = \frac{1}{x}$ (for $x \neq 0$ ) is the standard example, since swapping $x$ and $y$ in $y = \frac{1}{x}$ gives $x = \frac{1}{y}$ , i.e. $y = \frac{1}{x}$ again — the equation is unchanged by the swap. So is $f(x) = -x$ , and so is $f(x) = c - x$ for any constant $c$ (swap and rearrange: $x = c - y \Rightarrow y = c - x$ , identical to the start). What all three share is that reflecting them in $y=x$ produces the same curve, not a different one — which is a genuinely useful thing to notice fast in an exam: if a question's f(x) reflects into itself, f⁻¹(x) = f(x) is the whole of part (b) with no rearrangement needed at all, though the domain still has to be checked and stated separately exactly as usual. (This shortcut applies to the specific function given, not as a general labour-saving assumption — most functions on this paper are not self-inverse, and assuming one is without checking is its own way to lose the accuracy mark.)

Retrieval — with feedback on every choice

Question 1

3 marks

$f(x) = (x - 5)^2 + 2$ , $x \geq 5$ . Find $f^{-1}(x)$ and state its domain.

Question 2

2 marks

$f(x) = 3x + 2$ and $g(x) = x^2 - 1$ , both with domain $x \in \mathbb{R}$ . What is $gf(x)$ ?

Question 3

2 marks

Which of the following functions, each with domain $x \in \mathbb{R}$ , is one-one?

Question 4

2 marks

$f(x) = \dfrac{1}{x - 4}$ , $x \in \mathbb{R}$ , $x \neq 4$ . What is the domain of $f$ , and why?

Question 5

3 marks

$f(x) = x^2 + 6x + 4$ , $x \leq -3$ . Given that $f^{-1}$ exists on this domain, which statement about it is correct?

Reference — not a study method, a lookup

Function: one input → exactly one output. One-one: no two inputs share an output. Many-one: some do.
fg(x) = f(g(x)) — do g first, then f. fg ≠ gf in general.
f⁻¹ exists as a function only if f is one-one (restrict the domain if it isn't).
Swap x and y to find f⁻¹ ⇒ domain of f⁻¹ = range of f; range of f⁻¹ = domain of f. Always state both.
f⁻¹f(x) = ff⁻¹(x) = x. Graph of f⁻¹ is the reflection of f's graph in y = x.

Not affiliated with or endorsed by Pearson Edexcel. Every quotation and figure attributed to a mark scheme or examiner report in this lesson was independently verified against the primary Pearson document, not carried over from prior course material. Every question in this lesson — prequestion, worked chain, chain drill, marked solution and MCQ alike — is VERIDIAN-original wording, inspired by confirmed real question types, never a reproduction of a real Pearson question; and because the questions are original, the per-line mark allocations attached to them are modelled on verified mark-scheme conventions (what M, A and B marks mean, when follow-through applies, the independent-B-mark structure for a stated domain) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 13 marks

$f(x) = (x - 5)^2 + 2$ , $x \geq 5$ . Find $f^{-1}(x)$ and state its domain.

$f^{-1}(x) = 5 + \sqrt{x - 2}$ , domain $x \geq 2$
Correct. $y = (x-5)^2+2 \Rightarrow x = (y-5)^2+2 \Rightarrow y - 5 = \pm\sqrt{x-2}$ ; the domain $x \geq 5$ means $y-5 \geq 0$ , selecting the $+$ root. The domain of $f^{-1}$ is the range of $f$ : since the vertex $(5,2)$ is the left end of the domain, the range is $f(x) \geq 2$ , so the domain of $f^{-1}$ is $x \geq 2$ .
B $f^{-1}(x) = 5 + \sqrt{x - 2}$ , domain $x \in \mathbb{R}$
The algebra is entirely correct and the domain is missing — exactly the single most-attested error in this whole topic. The domain of $f^{-1}$ is a separate, independently-marked statement (the range of f, here $x \geq 2$ ), not something a correct formula for $f^{-1}(x)$ automatically supplies.
C $f^{-1}(x) = 5 - \sqrt{x - 2}$ , domain $x \geq 2$
The domain is right and the branch is wrong. The original domain $x \geq 5$ means $y - 5 \geq 0$ once the letters are swapped, which selects the POSITIVE root, not the negative one. Check: $f^{-1}(2)$ should equal the vertex's x-value, $5$ — this formula instead gives $5 - 0 = 5$ , which happens to still work at this one point, but $f^{-1}(11) = 5 - 3 = 2$ , and $f(2) = 9 + 2 = 11$ ... except $2$ is not in f's domain ( $x \geq 5$ ) at all, so $2$ cannot be a valid output of $f^{-1}$ here.
D $f^{-1}(x) = 5 + \sqrt{x + 2}$ , domain $x \geq -2$
The sign inside the square root is wrong. Rearranging $x = (y-5)^2 + 2$ gives $x - 2 = (y-5)^2$ — subtract 2 from both sides, since 2 is being added on the right, not subtracted. This is the same 'reverse the operation, not repeat it' error as undoing $y=3x+4$ by adding 4 instead of subtracting it.

Traps tested: Domain of inverse omitted · Wrong branch of square root selected · Order of undoing not reversed

Question 22 marks

$f(x) = 3x + 2$ and $g(x) = x^2 - 1$ , both with domain $x \in \mathbb{R}$ . What is $gf(x)$ ?

$9x^2 + 12x + 3$
Correct. $gf(x) = g(f(x)) = g(3x+2) = (3x+2)^2 - 1 = 9x^2 + 12x + 4 - 1 = 9x^2+12x+3$ — apply $f$ first (the letter next to $x$ ), then feed its output into $g$ .
B $3x^2 + 1$
This computes $fg(x)$ instead: $f(g(x)) = f(x^2-1) = 3(x^2-1)+2 = 3x^2 -1$ , not even matching this option exactly, which suggests the composition order AND the arithmetic have both gone wrong. The notation $gf$ means apply $f$ first, then $g$ — the letter closest to the bracket acts first.
C $3x^2 + 5$
This is closer to $fg(x)$ : $f(g(x)) = 3(x^2-1)+2 = 3x^2-3+2 = 3x^2-1$ — even this doesn't match, so this option applies the wrong function order (f applied to g's output) and still slips on the arithmetic. Whichever function sits closest to the bracket in $gf(x)$ is the one applied to $x$ first.
D $9x^2 + 4$
The order is right (f applied first, then g) but the middle term of $(3x+2)^2$ has vanished: $(3x+2)^2 = 9x^2 + 12x + 4$ , not $9x^2 + 4$ — the cross term $2 \times 3x \times 2 = 12x$ from expanding the bracket has been dropped.

Traps tested: Composition order reversed · Cross term dropped when expanding

Question 32 marks

Which of the following functions, each with domain $x \in \mathbb{R}$ , is one-one?

$f(x) = 2x - 7$
Correct. Suppose $f(a) = f(b)$ : $2a - 7 = 2b - 7 \Rightarrow 2a = 2b \Rightarrow a = b$ . Two inputs can only share an output if they were already the same input, so no two different inputs ever do — one-one. Graphically, every horizontal line meets this straight, non-horizontal line at most once.
B $f(x) = x^2 + 4$
Many-one: $f(-1) = 5$ and $f(1) = 5$ , two different inputs sharing one output, and this holds for every nonzero input in the same way. The horizontal-line test confirms it visually — any horizontal line above the vertex meets the curve twice.
C $f(x) = |x|$
Many-one: $f(-3) = 3$ and $f(3) = 3$ . The modulus function folds every negative input onto the positive value it started as the mirror image of, so every nonzero output (except the vertex) is shared by exactly two inputs.
D $f(x) = (x - 1)(x - 5)$
Many-one: this is a quadratic, so it has a turning point and the same two-inputs-one-output structure as option B once expanded. $f(1) = 0$ and $f(5) = 0$ already show two different inputs sharing an output directly, without needing to expand anything.

Traps tested: Many one mistaken for one one

Question 42 marks

$f(x) = \dfrac{1}{x - 4}$ , $x \in \mathbb{R}$ , $x \neq 4$ . What is the domain of $f$ , and why?

All real $x$ except $x = 4$ , because the denominator $x - 4$ would be zero there and division by zero is undefined
Correct — and it is worth noticing this domain was already given in the question ( $x \neq 4$ is stated explicitly). A function with a fraction excludes wherever its denominator is zero; here that is the single point $x=4$ , so the domain is every other real number.
BAll real $x$ except $x = 0$ , because a fraction is undefined at $x = 0$
This is the rule for a denominator that IS $x$ (like $\frac{1}{x}$ ), applied to the wrong function. Here the denominator is $x - 4$ , not $x$ itself, and $\frac{1}{0-4} = -\frac{1}{4}$ is perfectly well defined — the restriction has to be found from the actual denominator in front of you, not assumed by pattern.
C $x > 4$ only, because the function needs a positive denominator
There is no such requirement — a fraction is defined for a negative denominator just as well as a positive one; $\frac{1}{-1} = -1$ is a perfectly ordinary output. Only a denominator of exactly zero is disallowed, which happens once, at $x=4$ , not for an entire half-line either side of it.
DAll real $x$ , since $x = 4$ is only a single point and the function is defined almost everywhere
"Almost everywhere" is not the same as "everywhere," and a domain excludes every point where the rule genuinely fails, however few there are. $f(4)$ is not a small or negligible number here — it is not a number at all, because division by zero has no result to assign.

Traps tested: Denominator misidentified · Fraction assumed to need positive denominator · Single excluded point treated as negligible

Question 53 marks

$f(x) = x^2 + 6x + 4$ , $x \leq -3$ . Given that $f^{-1}$ exists on this domain, which statement about it is correct?

The domain of $f^{-1}$ is $\{x : x \geq -5\}$ , because that is the range of $f$
Correct. Completing the square: $x^2+6x+4 = (x+3)^2 - 5$ , vertex $(-3,-5)$ . The domain $x \leq -3$ is the left branch, running down from the vertex, so the minimum $-5$ is attained and the range is $f(x) \geq -5$ . The domain of $f^{-1}$ is always the range of $f$ — this is not a fact about this particular function, it is what swapping x and y always produces.
BThe domain of $f^{-1}$ is $\{x : x \leq -3\}$ , the same as the domain of $f$
This confuses the domain of $f^{-1}$ with the domain of $f$ itself — the domain of $f^{-1}$ is the RANGE of $f$ , not a repeat of $f$ 's own domain. $x \leq -3$ is in fact the correct range of $f^{-1}$ (since range of f⁻¹ = domain of f), just attached to the wrong one of the two functions.
C $f^{-1}$ does not exist, because $f$ is a quadratic and quadratics are always many-one
A quadratic on its full, unrestricted domain is many-one, but the domain here has been deliberately restricted to $x \leq -3$ — exactly one side of the vertex, which makes this branch one-one. The question states the inverse exists precisely because this restriction has already been applied; the general rule about unrestricted quadratics does not apply once a domain restriction is in force.
DThe domain of $f^{-1}$ cannot be determined without first finding the formula for $f^{-1}(x)$
It can, and finding it this way is faster: the domain of $f^{-1}$ is the range of $f$ , which is available directly from the completed-square form ( $f(x) \geq -5$ ) without swapping x and y or rearranging anything. The formula for $f^{-1}(x)$ and its domain are two separate pieces of information, earned by two separate marks, and the domain does not have to wait for the formula.

Traps tested: Range of f confused with domain of finverse · Restricted domain not accounted for · Domain of inverse derived from scratch not recognised as range of f

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Pearson's official past-papers portal

Select International Advanced Level → Mathematics → any series, then look for WMA13.

Functions: Domain, Range, Composition and Inverses

What a function actually is, and the one distinction the whole topic runs on

Composition — building one function out of two

Why some machines can't be run backwards — before the algebra

Why an inverse needs one-one — and why domain and range swap places

Find f⁻¹(x) and state its domain, for f(x) = x² − 4x + 7, x ≥ 2

Complete the chain — f⁻¹(x) and its domain, for f(x) = (x + 1)² − 6, x ≤ −1

Deep dive

Practice this for real