This mini-lesson covers AQA section I — Numerical methods: locating roots by change of sign (and when that fails), fixed-point iteration xn+1 = g(xn) with staircase and cobweb diagrams, the Newton–Raphson method and its failure cases, and numerical integration.
Work through each screen, answer the questions as you go (a few are conceptual, most are calculations) and collect ⭐ stars. Everything here is A-level standard, drawn from the AQA 7357 subject content. Press Start when you are ready.
Numerical methods · change of sign
Locating roots by change of sign
Most equations cannot be solved exactly. But if a continuous function changes sign across an interval, it must have crossed zero somewhere inside it.
If f is continuous on [a, b] and f(a) × f(b) < 0, then f(x) = 0 has at least one root in (a, b)this is the intermediate value theorem doing the work
Worked example — our running function
Let f(x) = x³ − 2x − 5.
f(2) = 8 − 4 − 5 = −1 (negative)
f(3) = 27 − 6 − 5 = 16 (positive)
f is a polynomial, hence continuous, and the sign changes. So there is a root between 2 and 3. (It is 2.0946 to 4 d.p.)
Say the magic words. Full marks require you to state that f is continuous on the interval and that there is a change of sign. Just quoting two numbers is not enough.
Numerical methods · failure
When change of sign fails
The method is a one-way test: a sign change guarantees a root, but no sign change does not guarantee no root.
Repeated root: the curve touches the axis without crossing (e.g. f(x) = (x − 2)²). There is a root, but no sign change.
Two roots in the interval: the function dips below and comes back, so the endpoint signs agree — the two crossings cancel out of the test.
Discontinuity: f(x) = 1/(x − 2) changes sign between 1 and 3, but has no root at all — it has an asymptote. This is why continuity is essential.
sign change ⇒ root (for continuous f)but: no sign change ⇏ no root
Fix: use a smaller interval, or sketch the curve first. If a question asks "explain why the method fails here", the answer is nearly always a repeated root or an asymptote.
Quick check
Why a sign change works
?f is continuous on [a, b] and f(a) × f(b) < 0. What can you conclude?
Numerical methods · iteration
Fixed-point iteration x = g(x)
Rearrange f(x) = 0 into the form x = g(x), then iterate from a starting value:
xn+1 = g(xn)a root of f is a fixed point of g — a point where the curve y = g(x) meets the line y = x
Then x₂ = (2 × 2.0801 + 5)1/3 = 2.0924, x₃ = 2.0942, x₄ = 2.0945 — converging to the root 2.0946.
The rearrangement matters enormously. The same equation also rearranges to x = (x³ − 5)/2 — and that iteration diverges. Convergence requires |g′(x)| < 1 near the root.
Sort it
Which numerical method?
Tap a description, then tap the method it belongs to.
± Change of sign
🔁 Iteration x = g(x)
📏 Newton–Raphson
Numerical methods · diagrams
Staircase and cobweb diagrams
Iterations are visualised on a graph of y = g(x) together with the line y = x. The root is where they cross. From x₀ you go vertically to the curve, then horizontally to the line, and repeat.
Staircase — the path climbs (or descends) towards the root in steps, all from one side. This happens when g′(x) is positive near the root.
Cobweb — the path spirals around the root, alternating from one side to the other. This happens when g′(x) is negative.
If |g′(x)| > 1 near the root, the staircase or cobweb moves away — the iteration diverges.
Vertical to the curve y = g(x), horizontal to the line y = x, repeat. A staircase approaches from one side; a cobweb spirals in.Quick check
The formula
?What is the Newton–Raphson iteration formula?
Numerical methods · Newton–Raphson
The Newton–Raphson method
Newton–Raphson uses the tangent at xn and takes the point where it cuts the x-axis as the next estimate. It converges very fast — typically doubling the number of correct digits each step.
xn+1 = xn − f(xn) / f′(xn)you must differentiate f — that is what the tangent needs
Two steps have already reached the true root 2.09455… to 4 decimal places.
Calculate
Your turn — evaluate f(2)
1Let f(x) = x³ − 2x − 5. Calculate f(2).
Hint: f(2) = 2³ − 2(2) − 5 = 8 − 4 − 5.
Calculate
Your turn — evaluate f(3)
2For the same f(x) = x³ − 2x − 5, calculate f(3). (Together with f(2), the sign change proves there is a root between 2 and 3.)
Hint: f(3) = 3³ − 2(3) − 5 = 27 − 6 − 5. Since f(2) is negative and f(3) is positive and f is continuous, a root lies in (2, 3).
Numerical methods · failure
When Newton–Raphson fails
Speed comes at a price: the method is fragile if the starting value is poorly chosen.
f′(xn) = 0 — the tangent is horizontal and never meets the x-axis. The formula divides by zero and the method breaks down immediately.
x₀ near a turning point — f′(x₀) is tiny, so f(x₀)/f′(x₀) is enormous and the tangent flings the next estimate far away, possibly converging to a completely different root.
x₀ at a discontinuity or where f is not differentiable — the tangent does not exist.
the danger is dividing by a very small f′(xn)always sketch, and choose x₀ close to the root and away from turning points
In the exam: "Explain why the Newton–Raphson method fails with this starting value" is nearly always answered by "f′(x₀) = 0, so the tangent is parallel to the x-axis" — and a sketch showing the tangent earns the mark.
Quick check
When it fails
?Newton–Raphson is applied with a starting value x₀ very close to a turning point of f. What is likely to happen?
Calculate
Your turn — first Newton–Raphson step
3For f(x) = x³ − 2x − 5 we have f′(x) = 3x² − 2. Starting from x₀ = 2, find x₁.
Hint: x₁ = x₀ − f(x₀)/f′(x₀). Here f(2) = −1 and f′(2) = 3(4) − 2 = 10, so x₁ = 2 − (−1)/10 = 2 + 0.1.
Numerical methods · integration
Numerical integration
The same spirit applies to areas: when ∫f(x) dx cannot be found exactly, approximate it. The trapezium rule is the method AQA requires.
∫ab y dx ≈ h/2 [y₀ + 2(y₁ + … + yn−1) + yn], h = (b − a)/nn strips, n + 1 ordinates; only the interior ones are doubled
Convex curve (f″ > 0, bending upwards) — chords lie above the curve, so the rule overestimates.
Concave curve (f″ < 0) — the rule underestimates.
More strips → smaller h → a better estimate. Doubling n roughly quarters the error.
Improving the estimate: the standard answer to "how could the estimate be improved?" is "increase the number of strips (decrease h)". Saying "use a better calculator" earns nothing.
Match it
Match the calculation to its value (f(x) = x³ − 2x − 5)
Tap an item on the left, then its partner on the right.
Calculation
Value
Numerical methods · comparing
Choosing and comparing methods
Three tools, three purposes. Know which is which.
Change of sign — locates a root in an interval, but does not refine it. Slow, and it needs continuity.
Fixed-point iteration — refines a root, requires no derivative, but only converges if |g′(x)| < 1, and the rearrangement is a genuine choice you have to make.
Newton–Raphson — refines a root very fast, but needs f′(x) and a good starting value.
accuracy claim: if f(2.0945) < 0 and f(2.0947) > 0, the root is 2.0946 to 4 d.p.this "show that" step is how you justify a stated accuracy
To prove an answer is correct to n decimal places, evaluate f at the two endpoints of the rounding interval and show a change of sign. Simply saying "my iteration gave 2.0946" does not demonstrate anything.
Quick check
Staircase or cobweb?
?A convergent iteration x_{n+1} = g(x_n) produces a cobweb diagram (the estimates alternate either side of the root). What does this tell you about g′(x) near the root?
Calculate
Your turn — second Newton–Raphson step
4Continue from x₁ = 2.1 to find x₂, to 4 decimal places.