Locating Roots by Sign Change, and Iterative Methods

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.

Why a sign change forces a crossing — the pen that never lifts, before any letters

In plain terms

Picture drawing one line on a piece of paper without ever lifting your pen off it — you drag the pen smoothly from a starting point to a finishing point, no picking it up in between. Draw a horizontal line across the middle of the page first. Now start your pen 5 centimetres BELOW that middle line, and drag it across the page until you finish 9 centimetres ABOVE it. Because the pen never left the paper — no jumps, no skips, no gaps — there was no way for it to get from 5 cm below the line to 9 cm above it except by physically dragging across the middle line at some point. It doesn't matter how wobbly the path was, or whether it crossed the middle line three times instead of once — as long as it started below, finished above, and never once left the page, it crossed that line at least once. Now do the opposite on purpose: start drawing 5 cm below the middle line, lift the pen clean off the page partway across, and put it back down 9 cm above the line somewhere else. Same start height, same finish height, same 'sign change' from below to above — but this time the pen genuinely never touched the middle line, because it wasn't drawing anything during the gap. That single difference — pen down the whole way, versus pen lifted even once — is the entire reason the crossing is guaranteed in the first case and not in the second.

What you were tracing is the curve $y = f(x)$ , and the middle line you kept crossing is the x-axis, the set of points where the height is zero. 'Never lifting the pen' is exactly what continuity means: no jumps, breaks, or missing pieces in the curve anywhere across the interval you're drawing it on. Starting below the middle line and finishing above it is a sign change — $f(a)$ negative, $f(b)$ positive (or the reverse) at the two ends $a$ and $b$ of the interval. And crossing the middle line is a root: a point where $f(x) = 0$ . 'Lifting the pen' is what a discontinuity looks like on a real curve — a vertical asymptote or a genuine break — and it is exactly the situation where a sign change can happen with no crossing at all, because the curve simply isn't there at the gap.

Formally

If $f$ is continuous on the interval $[a, b]$ and $f(a)$ , $f(b)$ have opposite signs, then there exists at least one value $c$ strictly between $a$ and $b$ such that $f(c) = 0$ . Continuity is what forces every height between $f(a)$ and $f(b)$ — zero included — to actually be attained somewhere on the interval; remove it and the guarantee fails, as it does for $f(x) = \frac{1}{x}$ across $x = 0$ , where $f(-1) = -1$ and $f(1) = 1$ is a genuine sign change with no root anywhere, because the curve is discontinuous at $x = 0$ .

What 'locating a root' means, and the condition that makes the test valid

A root of $f(x) = 0$ , a zero of $f$ , and an x-intercept of $y = f(x)$ are the same set of numbers, under three different names — the x-values where the height of the curve is zero. Spec 6.1 gives one method for proving one exists somewhere, without solving for it exactly: considering changes of sign of $f(x)$ in an interval of $x$ in which $f(x)$ is continuous (the spec's own wording). Evaluate $f$ at two values $a < b$ . If $f(a)$ and $f(b)$ have opposite signs — one negative, one positive — and $f$ is across the whole of $[a, b]$ , then somewhere strictly between $a$ and $b$ , $f(x) = 0$ .

Both conditions are load-bearing, and the second is the one the paper checks for explicitly and repeatedly. 'Continuous' does not mean 'smooth-looking' or 'a nice function' — it means the curve has no breaks, jumps, or vertical asymptotes anywhere across the interval you are testing on. For a polynomial this is always true, everywhere, with no exceptions to check — which is exactly why it is safe to state in one clause rather than argue at length. For a function built from division, a root, a logarithm, or a trig function with a restricted domain, it is a genuine claim that needs to actually be true of the specific interval being used, not assumed because the function looks well-behaved on a calculator screen.

Sign change and continuity together prove that AT LEAST one root exists in the interval — not exactly one, and not where in the interval it sits. A curve can cross the x-axis three times between $a$ and $b$ and still leave $f(a)$ and $f(b)$ with opposite signs, because an odd number of crossings always flips the overall sign. (An even number of crossings would restore the original sign and hide the crossings from this test entirely — which is why the test can miss roots but can never invent one that isn't there.) Getting a precise decimal value for whichever root you've located is a separate job, covered by 6.2 below.

The exact write-up the mark scheme wants — and where iteration takes over

A real mark-scheme requirement for this exact question type, quoted directly, names three components: correct values of $f(a)$ and $f(b)$ ; a reason that mentions continuity and states or indicates the sign change; and a conclusion — 'hence root'. All three, not two. Examiner reports across essentially every series with this question type record the same specific gap: the values and the conclusion both present, and the middle sentence — continuity — missing. The fix costs one clause: 'f is a polynomial and so is continuous' is a complete, sufficient reason for almost every function this paper will give you.

Sign change narrows a root down to an interval. It cannot, on its own, produce a decimal value — and that is precisely the gap 6.2 fills: approximate solution of equations using simple iterative methods, including recurrence relations of the form $x_{n+1} = f(x_n)$ (the spec's own wording, using $f$ generically for 'whatever function is on the right' — this course writes it as $x_{n+1} = g(x_n)$ below, to keep it visually distinct from the original equation's own $f$ ). Rearrange $f(x) = 0$ into the form $x = g(x)$ , pick a starting value $x_0$ , then generate $x_1 = g(x_0)$ , $x_2 = g(x_1)$ , $x_3 = g(x_2)$ , and so on — feeding each output back in as the next input.

The spec's own guidance for 6.2 is explicit: iterative procedures are examined 'for which leads will be given' — meaning you are not asked to invent a working rearrangement from a blank page. What you ARE asked to do, routinely, is verify algebraically that a stated rearrangement is a valid one — a 'show that' question turning $f(x) = 0$ into the given $x = g(x)$ form — and then carry out the iteration itself from a given starting value.

Two habits protect every mark in this half of the topic, and both are named directly by examiner reports below: show each substitution as its own line — $x_{n+1} = g(x_n)$ with the actual number written in, not a value that appears from nowhere via a calculator's ANS key — and give every iterate to exactly the number of decimal places the question asks for, computed from the previous FULL-precision value rather than a rounded-off one. Rounding early doesn't just look untidy; it changes the answer by the time it compounds through two or three more steps.

Mechanism

Why a sign change guarantees a root — continuity, not luck

This is not a rule to memorise; it follows from what continuity actually means. Picture tracing the curve $y = f(x)$ from the point $(a, f(a))$ to the point $(b, f(b))$ without lifting the pen — that is what 'continuous on $[a, b]$ ' means, concretely: the curve has no jumps, so every height between the start and the finish is reached at some point along the way. If $f(a)$ is negative (the curve starts below the x-axis) and $f(b)$ is positive (it finishes above it), then zero is a height strictly between the start and the finish — and since the pen never left the paper, it must have passed through height zero at least once on the way from one side to the other. That single observation — a continuous path between two heights of opposite sign must cross every height in between, zero included — is the whole theorem; sign change is just the easiest version of 'opposite sign' to check by substitution. Nothing here depends on the function being a polynomial, or being differentiable, or having any particular shape: it depends only on there being no break in the curve across the interval, which is exactly why continuity is not an optional extra clause in the write-up. Remove it and the argument has a hole exactly the size of $\frac{1}{x}$ across $x = 0$ : a genuine sign change, with no crossing anywhere, because the pen was lifted.

Diagram — A genuine crossing, next to a genuine counterexample

x-axis: x · y-axis: y = f(x)

C1 · a continuous curve, f(a) < 0, f(b) > 0: One unbroken line from a point below the x-axis at x = a to a point above it at x = b, drawn with no gaps, jumps, or vertical segments anywhere in between. Because the curve is never lifted from the paper, it is forced to pass through height zero somewhere strictly inside the interval — at least once, possibly more (an odd number of times in total).
C2 · y = 1/x across x = 0 — the counterexample: Two entirely separate branches: one below the axis for all x < 0, one above it for all x > 0, with a vertical asymptote at x = 0 splitting them. f(−1) = −1 and f(1) = 1 is a genuine sign change by the numbers, but there is no single unbroken curve joining those two points — the pen was lifted at x = 0, and 1/x is never actually zero anywhere.
C2 · y = 1/x across x = 0 — the counterexample (cont.): Two entirely separate branches: one below the axis for all x < 0, one above it for all x > 0, with a vertical asymptote at x = 0 splitting them. f(−1) = −1 and f(1) = 1 is a genuine sign change by the numbers, but there is no single unbroken curve joining those two points — the pen was lifted at x = 0, and 1/x is never actually zero anywhere.

The interval (a, b) itself: Sign change plus continuity proves a root exists somewhere inside — it says nothing about where. Locating the decimal value is a separate job, done by the iteration in the second half of this lesson.
Possible extra roots: A sign change proves AT LEAST one root, never exactly one. A curve can cross the axis three times between a and b and still show opposite signs at the two ends, because three crossings — an odd number — still flips the overall sign.
The break in C2: This is what a genuine discontinuity looks like on a curve, as opposed to a curve that merely looks complicated: not a sharp corner or a steep gradient, but a point the curve simply never passes through at all.

Common error: Treating any sign change between f(a) and f(b) as proof of a root, without checking or stating that f is continuous across the whole interval.

Correct: State continuity explicitly before drawing the conclusion. For a polynomial this is a one-clause justification ('f is a polynomial, so continuous'), but it still has to be written — and for a function built from division, roots, logs, or restricted trig, it is a genuine claim to check against that specific interval, not a formality to skip.

examiner-report · Jan 2022 · Q5(a)

Mechanism

Why the same test proves a root to a stated number of decimal places

Once an approximate value has been found by , the exact same sign-change idea proves it correct to whatever precision is being claimed — the trick is choosing the right width of interval. Suppose an iteration is converging towards a value that looks like it rounds to $2.589$ (to 3 decimal places). 'Correct to 3 dp' means: the true root is closer to $2.589$ than to any other number with 3 decimal places — that is, the true root lies strictly between $2.5885$ and $2.5895$ , the two numbers exactly halfway to $2.589$ 's neighbours ( $2.588$ and $2.590$ ). Those two numbers, $2.5885$ and $2.5895$ , are not an arbitrary choice of test points; they are the precise boundary of what 'rounds to $2.589$ ' means. So: evaluate $f$ at both. If $f(2.5885)$ and $f(2.5895)$ have opposite signs, and $f$ is continuous on that tiny interval (true here, as it was on the wide one, because $f$ is still the same polynomial), then the root lies strictly inside $(2.5885, 2.5895)$ — which is exactly the set of numbers that round to $2.589$ . The claimed accuracy is proved, by the identical argument used to locate the root in the first place, just applied on an interval of width $10^{-3}$ instead of width $1$ . Get the width wrong — test $2.585$ and $2.595$ instead, half a unit either side of the wrong decimal place — and the same correct arithmetic proves a real but weaker claim: accuracy to 2 dp, not 3.

Mechanism

Why iteration converges to a root — the fixed point IS the root

Rearranging $f(x) = 0$ into $x = g(x)$ and iterating $x_{n+1} = g(x_n)$ is not a different technique from ; it is a way of turning the SAME equation into a machine that homes in on a decimal answer. Suppose the sequence $x_0, x_1, x_2, \ldots$ genuinely converges — the terms settle down and stop changing, to whatever precision is being tracked — and call that settled value $L$ . Since $x_{n+1} = g(x_n)$ holds at every single step of the sequence, it must also hold in the limit, where $x_{n+1}$ and $x_n$ have both become $L$ : $L = g(L)$ . But $g$ was built by rearranging $f(x) = 0$ — so $L = g(L)$ is, algebraically, exactly the same statement as $f(L) = 0$ . The value the sequence settles on is not just SOME number the process happens to produce; it is forced, by the algebra of the rearrangement itself, to satisfy the original equation. That is the entire justification for why repeatedly substituting a number into a formula and getting a fresh number out ever finds a root at all, rather than just producing an arbitrary sequence of numbers that happens to stop moving.

Worked, in full

Full process: locate, iterate, and confirm — for $x^3 - 5x - 3 = 0$

01
Evaluate the function at the two ends of a sensible interval. $f(x) = x^3 - 5x - 3$ : $f(2) = 8 - 10 - 3 = -5$ , and $f(3) = 27 - 15 - 3 = 9$ .
Earns: M1 — the method mark for attempting both values. Real mark schemes are lenient about what this needs: substitution shown, or just ONE value correct to 1 significant figure, is enough — verified verbatim, Jan 2022 Q5(a): 'Attempts to find y at −1.25 and −1.2 with one correct to 1sf'; Oct 2023 Q1(a): 'Attempts f(1) and f(2) with substitution seen or at least one correct to 1 d.p.'. This mark does not by itself bank credit for correct values independent of what follows — see stage 2.
02
State the reason and reach the conclusion. $f$ is a polynomial, so it is continuous everywhere, including on $[2, 3]$ . $f(2) = -5$ is negative and $f(3) = 9$ is positive, so $f$ changes sign across the interval. Hence $f(x) = 0$ has a root $\alpha$ between $x = 2$ and $x = 3$ .
Earns: A1 — DEPENDENT on the M1 above, and it bundles THREE things into this one mark, not three separate marks: both values correct, the reason (naming continuity and stating the sign change), and the conclusion. Verified verbatim, Oct 2023 Q1(a): 'Must have • both f(1) and f(2) correct... • reason, which must mention continuity and state or indicate sign change in some way • conclusion, "hence root"'. Get both values exactly right but skip the reason, and this WHOLE mark is lost — there is no separate mark that correct arithmetic alone can bank.
03
The rearrangement $x = \sqrt[3]{5x + 3}$ is given (verified: cube both sides and it returns $x^3 = 5x + 3$ , the same equation as $f(x) = 0$ rearranged). Substitute the given $x_0 = 2.5$ , writing the substitution explicitly rather than jumping to the answer: $x_1 = \sqrt[3]{5(2.5) + 3} = \sqrt[3]{15.5} = 2.4933$ (4 dp).
Earns: M1 — correct method: the previous value embedded directly in the formula and shown, not read silently off a calculator's stored-answer key. This is the exact distinction examiner reports draw between an answer that demonstrates the method and one that doesn't.
04
Repeat twice more, each time substituting the previous iterate: $x_2 = \sqrt[3]{5(2.4933) + 3} = 2.4915$ (4 dp); $x_3 = \sqrt[3]{5(2.4915) + 3} = 2.4910$ (4 dp).
Earns: A1 A1 — one accuracy mark per additional iterate ( $x_2$ , then $x_3$ ), each correct to the requested 4 dp, not one mark spanning both. Real mark schemes credit each freshly-produced iterate its own dependent accuracy mark rather than bundling several into one: verified, Jan 2022 Q5(b) gives 'M1 A1' for the first produced value and then a further separate 'B1' for the next; Oct 2023 Q1(b) gives 'M1A1' for the first value and then a further separate 'A1: awrt 1.7340 provided M1 has been scored' for the next; Jan 2025 Q1(c) shows the identical pattern ('A1: ... following the award of the M mark'). Getting exactly the number of decimal places asked for matters as much as getting the digits right — examiner reports record marks lost for giving fewer.
05
Choose an interval of the right WIDTH for the accuracy being claimed. $x_3 = 2.4910$ suggests the root rounds to $2.491$ (3 dp) — so test the boundary of that claim: half a unit either side of the third decimal place, i.e. $2.4905$ and $2.4915$ .
Earns: dM1 — a further method mark, dependent on the iteration above having produced a sensible candidate value to test around. Choosing an interval width matched to the stated degree of accuracy (not an arbitrary width) is itself part of the method being assessed.
06
Evaluate at both ends: $f(2.4905) = (2.4905)^3 - 5(2.4905) - 3 = -0.0049$ (negative, 4 dp); $f(2.4915) = (2.4915)^3 - 5(2.4915) - 3 = 0.0087$ (positive, 4 dp).
Earns: A1 — both values correct, with the correct signs.
07
Conclude: $f$ is continuous (still a polynomial), and $f(2.4905) < 0$ while $f(2.4915) > 0$ , a sign change — hence there is a root inside $(2.4905, 2.4915)$ , and every number in that interval rounds to $2.491$ . So $\alpha = 2.491$ , correct to 3 decimal places.
Earns: A1 — the full conclusion, correct answer only: continuity is restated, the sign change is named, and the claimed degree of accuracy is exactly the one the tested interval actually proves.

Source — Examiner report, Oct 2022

"the omission of reference to 'continuity' or 'continuous' being the most common error"

Complete it yourself

Complete the chain — locating and confirming the root of $x^3 + 2x - 6 = 0$ near $x = 1.5$

01
$f(x) = x^3 + 2x - 6$ . $f(1) = 1 + 2 - 6 = -3$ , $f(2) = 8 + 4 - 6 = 6$ . $f$ is a polynomial, so continuous on $[1, 2]$ ; the sign changes from negative to positive, hence $f(x) = 0$ has a root $\alpha$ between $x = 1$ and $x = 2$ .
02
Given: the equation rearranges to $x = \sqrt[3]{6 - 2x}$ , and the iterative formula $x_{n+1} = \sqrt[3]{6 - 2x_n}$ is to be used with $x_0 = 1.5$ .

Marked, line by line

$f(x) = x^3 - 4x - 7$ . (a) Show that $f(x) = 0$ has a root $\alpha$ between $x = 2$ and $x = 3$ . (2) (b) Show that the equation $f(x) = 0$ can be rearranged as $x = \sqrt[3]{4x + 7}$ . (2) (c) Using the iterative formula $x_{n+1} = \sqrt[3]{4x_n + 7}$ with $x_0 = 2.5$ , find, to 4 decimal places, the values of $x_1$ , $x_2$ and $x_3$ . (3) (d) By choosing a suitable interval, show that $\alpha = 2.589$ , correct to 3 decimal places. (3) — VERIDIAN-original question, inspired by the real four-part shape of WMA13's numerical-methods items (sign-change 'show that', a rearrangement 'show that', an iteration table, then a decimal-place confirmation) documented across the facts bank. Not a reproduction of any past-paper question, and the per-line mark allocations below are modelled on verified mark-scheme conventions rather than copied from a real scheme.

10 marks available

(a) — 2 marks

01
$f(2) = 2^3 - 4(2) - 7 = 8 - 8 - 7 = -7$ ; $f(3) = 3^3 - 4(3) - 7 = 27 - 12 - 7 = 8$ .
Method mark for attempting both values. Real mark schemes are lenient about what this needs: substitution shown, or just one value correct to 1 significant figure, is enough — the full accuracy of both values is checked at the next, dependent mark, not this one.
M1
02
$f$ is a polynomial, so it is continuous on $[2, 3]$ . $f(2) = -7$ is negative and $f(3) = 8$ is positive, so $f$ changes sign — hence there is a root $\alpha$ between $x = 2$ and $x = 3$ .
Accuracy mark, dependent on the M1 above — and it bundles three separate requirements into this one mark, verified verbatim from a real mark scheme (Oct 2023 Q1(a)): both f(2) and f(3) correct; a reason that mentions continuity and states or indicates the sign change; and a conclusion, 'hence root'. All three together earn this mark; any one missing — including a candidate's perfectly correct arithmetic with the continuity clause left out — loses the WHOLE mark, not a fraction of it. There is no separate mark for 'values right, write-up missing'.
A1

(b) — 2 marks

101
$x^3 - 4x - 7 = 0 \Rightarrow x^3 = 4x + 7$
Method mark for a correct rearrangement step: isolating the cubic term on one side, with the goal of cube-rooting visibly in sight.
M1
102
Cube-root both sides: $x = \sqrt[3]{4x + 7}$ , as required.
Accuracy mark, correct solution only. This is a 'show that' question with a printed target — the working must land on exactly the stated form, not merely something equivalent to it.
A1

(c) — 3 marks

201
$x_1 = \sqrt[3]{4(2.5) + 7} = \sqrt[3]{17} = 2.5713$ (4 dp)
Method mark: correct substitution of x₀ into the given formula, shown explicitly with the embedded value visible — not a value produced silently via a calculator's stored-answer key.
M1
202
$x_2 = \sqrt[3]{4(2.5713) + 7} = 2.5856$ (4 dp)
Accuracy mark for x₂, correct to the requested 4 dp.
A1
203
$x_3 = \sqrt[3]{4(2.5856) + 7} = 2.5884$ (4 dp)
Accuracy mark for x₃, correct to the requested 4 dp. Note: computed from the full-precision x₂, not the rounded display value — the two happen to agree here to 4 dp, but that agreement is not guaranteed in general, which is exactly why premature rounding is a real risk rather than a cosmetic one.
A1

(d) — 3 marks

301
$x_3 = 2.5884$ suggests $\alpha \approx 2.589$ (3 dp). Test the interval $[2.5885, 2.5895]$ — half a unit either side of the third decimal place.
Method mark: an interval chosen at the width that actually matches the accuracy being claimed (3 dp), not an arbitrary or mismatched width.
M1
302
$f(2.5885) = (2.5885)^3 - 4(2.5885) - 7 = -0.0102$ (4 dp); $f(2.5895) = (2.5895)^3 - 4(2.5895) - 7 = 0.0059$ (4 dp)
Accuracy mark for both values, correct with the correct signs.
A1
303
$f$ is continuous (a polynomial), $f(2.5885) < 0$ and $f(2.5895) > 0$ — a sign change — so $\alpha \in (2.5885, 2.5895)$ , and every number in that interval rounds to $2.589$ . Hence $\alpha = 2.589$ , correct to 3 decimal places.
Accuracy mark, correct solution only. The printed target (2.589 to 3 dp) is reached exactly, with continuity, the sign change, and the specific claimed precision all stated.
A1

In your own words

In one sentence: why does a sequence that stops changing at $x_n = L$ have to be sitting at a root of the original equation, rather than at some arbitrary steady value the process happens to produce?

Named traps

continuity-statement-omitted: The single most repeated specific error across the entire numerical-methods topic, confirmed independently in general summary AND question-specific commentary across at least five series. "the omission of reference to 'continuity' or 'continuous' being the most common error" (Oct 2022, general summary); "there was a failure to also comment on the fact that the function was continuous" (Jan 2022, Q5(a)); "students often lost a mark in question 2(a) for not making a reference to the continuity of the function" (Jan 2024, general summary, referring to Q2(a)). Encouragingly, the most recent series checked shows this improving: "The mention of continuity in some form... now exceeds the lack of mention as candidates have adapted to this requirement being needed" (Jan 2025, Q1(a)) — but the fix costs one clause ("f is a polynomial, so continuous") and there is no reason to be part of the group still losing it.
hence-root-conclusion-missing: A real mark scheme's stated requirements for the sign-change accuracy mark name three separate components, verbatim: both function values correct, "reason, which must mention continuity and state or indicate sign change in some way," and "conclusion, 'hence root'" (Oct 2023, Q1(a)). Values and reasoning with no final sentence tying them together is, by this wording, still an incomplete answer — the write-up needs to explicitly land on "hence there is a root in this interval," not leave the reader to infer it.
iterate-given-to-fewer-decimal-places-than-asked: Confirmed across at least two series: "there were candidates who did not give the value of their iterate to the required number of decimal places" (Oct 2020, Q6(c)); "A common error in part (c)(ii) was an error in rounding or premature rounding of intermediate steps" (Jan 2024, Q2(c)(ii)). This is two failure modes wearing one name: stating fewer decimal places than requested (loses the mark even if the digits shown are a fair rounding), and rounding an INTERMEDIATE value before using it in the next substitution (which changes the final answer, not just its presentation).
ans-key-substitution-not-shown: Confirmed verbatim: "it is even more important that candidates show the method of embedding the values in the iterative formula to demonstrate they understand how to generate the values" (Jun 2022, Q8(c)) — the report explicitly frames a correct final iterate with no substitution shown as "an answer not implying a correct method to solving had been shown." A calculator's stored-answer key can chain a whole sequence of iterates silently; the mark scheme is checking for the written line that proves the chain, not just its last link.
calculator-solver-used-despite-the-no-calculator-instruction: Verified verbatim from an actual WMA13 question paper, printed directly above a question of this type: "In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable." (Jan 2023, Q5). An examiner report independently corroborates the consequence: "there were a number of parts which stated that either relying on or entirely relying on the use of calculator technology was not allowed" (Jun 2022) — and this warning recurs across multiple series specifically on numerical-methods-style questions, where a calculator's built-in equation solver can produce the final answer directly, bypassing exactly the method the question is testing.

Beyond the spec

Spec 6.2 is examined with the rearrangement always supplied — 'for which leads will be given' — so this course does not need you to determine, from scratch, whether a given $x = g(x)$ form will converge. But understanding roughly why some rearrangements of the exact same equation converge and others don't makes the 'leads will be given' framing make sense, rather than feel arbitrary, and it explains what's actually happening if you ever experiment with a rearrangement and watch the numbers run away instead of settling.

Take $x^3 - 5x - 3 = 0$ and rearrange it two different, both algebraically valid, ways: $x = \sqrt[3]{5x+3}$ (used in the worked chain above, and it converges from $x_0 = 2.5$ ), or $x = \frac{x^3 - 3}{5}$ (rearranging the $5x$ term to the other side and dividing by 5 instead — equally valid algebra, same equation). Try the second one from the same $x_0 = 2.5$ : $x_1 = \frac{2.5^3 - 3}{5} = \frac{15.625 - 3}{5} = 2.525$ , $x_2 = \frac{2.525^3 - 3}{5} = 2.6197...$ , and the values climb away from the root rather than towards it. Two algebraically equivalent rearrangements of the identical equation, from the identical starting point — one settles down, the other runs off. The informal reason is about how sensitively $g$ responds to small changes in its input near the root: picture plotting $y = g(x)$ and the line $y = x$ on the same axes and stepping between them (across to the curve, then across to the line, then up or down to the curve again) — a 'staircase' or 'cobweb' path. If the curve is shallower than the line $y = x$ at the root, each step lands closer to the intersection than the last, and the staircase spirals or steps in towards a fixed point; if the curve is steeper, each step overshoots by more than the last, and it spirals away. This is exactly why the exam always hands you the rearrangement that behaves the first way — someone has already checked it — rather than asking you to discover, under time pressure, which of several valid-looking algebraic rearrangements is the one that actually works.

Retrieval — with feedback on every choice

Question 1

2 marks

$f(x) = \tan x$ ( $x$ in radians). $f(1) \approx 1.5574$ and $f(2) \approx -2.1850$ — a sign change. Does the sign-change test prove that $\tan x = 0$ has a root between $x = 1$ and $x = 2$ ?

Question 2

2 marks

$g(x) = x^3 + x - 9$ . A candidate writes: ' $g(1) = -7$ , $g(2) = 1$ , so there's a root between 1 and 2.' Which single addition earns full marks?

Question 3

1 mark

A question asks for $x_n$ values 'to 4 decimal places'. A candidate's working shows $x_1 = 2.571$ . What happens to the accuracy mark for $x_1$ ?

Question 4

2 marks

Two candidates both write $x_3 = 2.5884$ as their final answer. Candidate A's working shows ' $x_3 = \sqrt[3]{4(2.5856) + 7} = 2.5884$ '. Candidate B's working shows only ' $x_3 = 2.5884$ ', with no intermediate line. Are both certain to score the method mark?

Question 5

2 marks

Using $x^3 - 4x - 7 = 0$ , iteration gives $x_3 = 2.5884$ (4 dp). Which interval correctly tests whether $\alpha = 2.589$ , correct to 3 decimal places?

Question 6

1 mark

A question is printed with: 'In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.' A candidate solves it entirely with their calculator's equation solver and writes down only the correct final answer. What does the general marking guidance imply happens?

Reference — not a study method, a lookup

Sign change: f(a), f(b) opposite signs + f continuous on [a,b] ⇒ root exists between a and b. At least one — could be more.
Full marks needs three parts: correct f(a), f(b); reason naming continuity AND sign change; conclusion 'hence root'.
Confirm to n dp: test [value − 0.5×10⁻ⁿ, value + 0.5×10⁻ⁿ]. Sign change there ⇒ correct to n dp.
Iteration xₙ₊₁ = g(xₙ): show every substitution, never via ANS silently. Round only at the end, to the dp asked.
A converged limit L satisfies L = g(L) — algebraically the same equation as the original f(x) = 0.

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, what a sign-change accuracy mark requires) rather than transcribed from a real mark scheme, which for an original question does not exist.

Question 12 marks

$f(x) = \tan x$ ( $x$ in radians). $f(1) \approx 1.5574$ and $f(2) \approx -2.1850$ — a sign change. Does the sign-change test prove that $\tan x = 0$ has a root between $x = 1$ and $x = 2$ ?

No — $f$ has a discontinuity (a vertical asymptote) at $x = \frac{\pi}{2} \approx 1.5708$ , which lies inside $(1, 2)$ , so the test cannot be applied
Correct. The interval (1, 2) contains a point where tan x is undefined, so continuity fails and nothing can be concluded from the sign change. In fact tan x = 0 only at integer multiples of π, and π ≈ 3.14159 does not lie between 1 and 2 — there is no root here at all, despite the genuine sign change.
BYes — a sign change is always sufficient on its own
This ignores the continuity condition entirely, and this is exactly the case built to show why that condition can't be skipped: the sign change here is caused by an asymptote, not a crossing.
CYes, and the root is at $x = \frac{\pi}{2}$ , since that's where the sign flips
tan x is undefined at π/2, not zero there. The curve doesn't cross the axis at the asymptote; it diverges to +∞ approaching from one side and comes from −∞ on the other.
DNo — because $f(1)$ and $f(2)$ don't have the same sign
Opposite signs are exactly what the test looks for, not the same sign. Continuity, not the signs, is what fails here.

Traps tested: Sign change assumed sufficient without continuity · Asymptote mistaken for a root · Same sign required confused with sign change

Question 22 marks

$g(x) = x^3 + x - 9$ . A candidate writes: ' $g(1) = -7$ , $g(2) = 1$ , so there's a root between 1 and 2.' Which single addition earns full marks?

' $g$ is a polynomial, so continuous on $[1, 2]$ ; $g$ changes sign from negative to positive, hence there is a root in the interval.'
Correct — this is the missing middle component, matching the real mark-scheme's three-part requirement: values, a reason naming continuity and sign change, and a conclusion.
BNothing — the values and the conclusion are already both present
A value and a conclusion with no reasoning connecting them skips exactly the component examiner reports most often find missing.
CA sketch of the curve, since a picture proves continuity
A sketch can illustrate a fact already established by argument; it cannot itself establish that fact. The mark scheme wants a stated reason, not a drawing.
DA narrower interval, such as checking that $g(1.5)$ is also negative
This relocates the same missing reasoning to a smaller interval rather than supplying it.

Traps tested: Continuity treated as implied by the values · Sketch substituted for a continuity statement · Narrower search substituted for required reasoning

Question 31 mark

A question asks for $x_n$ values 'to 4 decimal places'. A candidate's working shows $x_1 = 2.571$ . What happens to the accuracy mark for $x_1$ ?

It's lost — even though $2.571$ is a correct rounding of the true value, the mark is written for the value stated to the requested precision, and 3 dp doesn't meet a 4 dp requirement
Correct. Precision is part of what the mark checks, independently of whether the digits shown are consistent with the true value.
BIt's awarded, since $2.571$ rounds correctly from the true value $2.5713$
Correct digits at the wrong precision still don't meet a stated precision requirement — the question specified 4 dp, and this gives 3.
CIt's awarded, with a one-mark follow-through penalty applied elsewhere in the question
There is no such penalty mechanism in the general marking guidance for a precision shortfall — the mark for this specific value is either earned or not, on its own terms.
DIt depends on whether $x_2$ is still correct afterwards
The mark for x₁ is assessed on x₁ itself, not made conditional on a later value in the sequence.

Traps tested: Correct rounding treated as sufficient regardless of requested precision · Ft invented for a precision shortfall · Current mark made conditional on a later one

Question 42 marks

Two candidates both write $x_3 = 2.5884$ as their final answer. Candidate A's working shows ' $x_3 = \sqrt[3]{4(2.5856) + 7} = 2.5884$ '. Candidate B's working shows only ' $x_3 = 2.5884$ ', with no intermediate line. Are both certain to score the method mark?

No — only Candidate A is. The general marking guidance is explicit that the substitution needs to be embedded and shown; a value produced silently is exactly the pattern examiner reports mark as not demonstrating a correct method
Correct. This is directly grounded in verified examiner-report commentary on this exact scenario — the same correct final number, credited differently depending on whether the substitution itself is visible.
BYes — both give the same correct final number, and that's all that's checked
The final answer being correct is necessary but not sufficient — the method mark specifically checks for evidence of the method, which an unexplained final number doesn't provide.
CNo — neither scores, because $x_1$ and $x_2$ are missing from both
The method mark for this particular step doesn't require every prior iterate to be re-shown, only that THIS substitution — the one producing x₃ — is itself visible.
DYes, because the iterative formula is already printed on the question paper
A printed formula doesn't substitute for showing it applied to a specific value — the method mark is for the application, not for the formula's existence.

Traps tested: Final answer treated as sufficient evidence of method · Overpenalises a genuinely sufficient single line · Printed formula assumed to excuse showing substitution

Question 52 marks

Using $x^3 - 4x - 7 = 0$ , iteration gives $x_3 = 2.5884$ (4 dp). Which interval correctly tests whether $\alpha = 2.589$ , correct to 3 decimal places?

$[2.5885, 2.5895]$
Correct — half a unit either side of the third decimal place, which is exactly the boundary of what 'rounds to 2.589' means.
B $[2.588, 2.590]$
This is a whole unit either side of the third decimal place, not half — too wide. It would still pass even for a true root that rounds to, say, 2.5875 or 2.5905, neither of which is 2.589.
C $[2.5883, 2.5885]$
Centred near x₃ itself rather than the value being confirmed, and far too narrow to reliably bracket the true root even if it happened to work here.
D $[2.58, 2.60]$
This interval width tests accuracy to 1 dp, not 3 — three decimal places too coarse for what the question asks.

Traps tested: Interval width not halved · Interval centred on the wrong value · Interval width matches wrong decimal place

Question 61 mark

A question is printed with: 'In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.' A candidate solves it entirely with their calculator's equation solver and writes down only the correct final answer. What does the general marking guidance imply happens?

The method mark(s) are at risk regardless of the correct answer — examiner reports record exactly this pattern being penalised on questions carrying this instruction
Correct, and directly grounded: this exact instruction, and this exact consequence, are both independently verified against real question papers and examiner reports.
BNothing — a correct final answer is always credited in full
The printed instruction exists specifically to override the usual benefit of the doubt a correct answer might otherwise get — it removes 'the answer is right, so the method must have been fine' as a valid inference.
COnly the final accuracy mark is lost; method marks are unaffected since the calculator performed the correct method internally
A calculator's internal process is not visible working — it cannot earn a method mark that specifically requires the method to be shown on the page.
DThe whole question must be left blank to avoid any penalty
Overcorrecting by abandoning the question throws away marks that are still genuinely available — showing the working, even after realising the calculator-only approach was risky, still earns what it earns.

Traps tested: Correct answer assumed to override a no calculator instruction · Calculator internal process treated as shown method · Overreacts by abandoning the question

Practice this for real

This site teaches the mechanism; the exam is sat on Pearson's own real questions. Go find and attempt these yourself — nothing here substitutes for actually sitting a timed paper.

Examiner report: Jan 2022 · Q5(a) — cited directly in this lesson
Examiner report: Oct 2022 — cited directly in this lesson

Pearson's official past-papers portal

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

Locating Roots by Sign Change, and Iterative Methods

Why a sign change forces a crossing — the pen that never lifts, before any letters

What 'locating a root' means, and the condition that makes the test valid

The exact write-up the mark scheme wants — and where iteration takes over

Why a sign change guarantees a root — continuity, not luck

Why the same test proves a root to a stated number of decimal places

Why iteration converges to a root — the fixed point IS the root

Full process: locate, iterate, and confirm — for x3−5x−3=0x^3 - 5x - 3 = 0x3−5x−3=0

Complete the chain — locating and confirming the root of x3+2x−6=0x^3 + 2x - 6 = 0x3+2x−6=0 near x=1.5x = 1.5x=1.5

Deep dive

Practice this for real

Full process: locate, iterate, and confirm — for $x^3 - 5x - 3 = 0$

Complete the chain — locating and confirming the root of $x^3 + 2x - 6 = 0$ near $x = 1.5$