Topic 9 is about equations you cannot solve exactly. It covers locating roots by a change of sign, solving x = g(x) by iteration (with staircase and cobweb diagrams), the Newton–Raphson method and the ways all of these can fail, plus numerical integration with the trapezium rule.
Work through each screen, answer the questions as you go (some are wordy, most are calculations) and collect ⭐ stars. Watch for the Exam trap notes — that is where the marks get thrown away. Press Start when you are ready.
Numerical methods · sign change
Locating a root by change of sign
If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then f(x) = 0 has at least one root in (a, b).
Worked example — f(x) = x³ − 3x − 5
f(2) = 8 − 6 − 5 = −3 (negative)
f(3) = 27 − 9 − 5 = 13 (positive)
f is a polynomial, so it is continuous. The sign changes, therefore there is a root between x = 2 and x = 3. ∎
To show a root is 1.879 correct to 3 d.p., test the interval endpoints 1.8785 and 1.8795 and show the sign changes across them.
Exam trap: the word continuous is a mark. f(x) = 1/(x − 2) changes sign between 1 and 3 but has no root — it has an asymptote. Always say “f is continuous on the interval”.
Calculate
Your turn — evaluate for a sign change
1f(x) = x³ − 3x − 5. Calculate f(2).
Set-up
f(2) = 8 − 6 − 5 = −3.
Since f(3) = 27 − 9 − 5 = +13 and f is continuous, there is a root between 2 and 3.
Hint: 2³ − 3(2) − 5.
Quick check
Why continuity matters
?f(1) = −2 and f(3) = 4. A student concludes there must be a root between 1 and 3. When is this conclusion wrong?
Numerical methods · iteration
Iteration: x = g(x)
Rearrange f(x) = 0 into the form x = g(x), then iterate: x_(n+1) = g(x_n) starting from x₀.
Worked example — x³ − 3x − 5 = 0
Rearrange: x³ = 3x + 5 ⇒ x = (3x + 5)^(1/3). So g(x) = (3x + 5)^(1/3).
x₃ = (3 × 2.2684 + 5)^(1/3) = (11.8051)^(1/3) = 2.2770 … converging on the root 2.2790.
Exam trap: keep the full accuracy in your calculator between iterations (use the ANS key). Rounding each step to 4 d.p. and feeding that back in will drift your final answer.
Calculate
Your turn — first iterate
2Using x_(n+1) = (3x_n + 5)^(1/3) with x₀ = 2, find x₁ to 4 decimal places.
Set-up
3(2) + 5 = 11.
x₁ = 11^(1/3) = 2.22398… = 2.2240 (4 d.p.)
x₁ =
Hint: x₁ = (3 × 2 + 5)^(1/3) = ∛11.
Calculate
Your turn — second iterate
3Continue the iteration x_(n+1) = (3x_n + 5)^(1/3). Using x₁ = 2.223980, find x₂ to 4 decimal places.
Set-up
3(2.223980) + 5 = 11.671940.
x₂ = 11.671940^(1/3) = 2.26837… = 2.2684 (4 d.p.)
The sequence is converging towards the root 2.2790 (4 d.p.).
x₂ =
Hint: 3 × 2.223980 + 5 = 11.671940; now take the cube root.
Numerical methods · diagrams
Staircase and cobweb diagrams
Plot y = g(x) and y = x. The iteration bounces between them. The gradient of g near the root decides everything.
Positive gradient ⇒ the iterates approach from one side (staircase). Negative gradient ⇒ they alternate around the root (cobweb).
|g′(x)| < 1 near the root ⇒ the iteration converges.
|g′(x)| > 1 near the root ⇒ it diverges, however good the starting value.
A different rearrangement of the same equation can converge where another diverges — that is why the question tells you which one to use.
Quick check
Staircase or cobweb?
?Near the root, g′(x) = −0.6. What will the iteration x_(n+1) = g(x_n) do?
Numerical methods · Newton–Raphson
The Newton–Raphson method
Take the tangent at (x₀, f(x₀)) and slide down it to the x-axis. Where it lands is your next estimate.
x_(n+1) = x_n − f(x_n) / f′(x_n)usually much faster than a simple iteration — it roughly doubles the number of correct digits each step
Hint: f(2.333333) = 0.703704 and f′(2.333333) = 13.333333.
Sort it
Which method is being described?
Tap a description, then the numerical method it belongs to.
🔀 Change of sign
🔁 Iteration x = g(x)
📐 Newton–Raphson
Numerical methods · failure
When these methods fail
Every 9MA0 series asks about failure. Know the three headline reasons.
Newton–Raphson fails if f′(x₀) = 0 — the tangent is horizontal and never meets the x-axis (you divide by zero). Even a near-zero f′ throws the next estimate wildly far away.
Newton–Raphson can converge to a different root, or diverge, if x₀ is chosen badly or is near a turning point.
Iteration diverges if |g′(x)| > 1 near the root — try a different rearrangement.
Change of sign fails when the function is discontinuous on the interval, or when there is a repeated root (the curve touches the axis without crossing, so no sign change occurs), or when two roots sit inside the same interval and the signs cancel out.
How to answer: “Explain why the method fails for x₀ = 1” deserves a calculation. For f(x) = x³ − 3x − 5: f′(1) = 3 − 3 = 0, so the Newton–Raphson formula divides by zero. Show the number.
Quick check
Newton–Raphson breakdown
?For f(x) = x³ − 3x − 5, why does Newton–Raphson fail with the starting value x₀ = 1?
Calculate
Your turn — Newton–Raphson for a square root
6Use Newton–Raphson on f(x) = x² − 7 (so f′(x) = 2x) with x₀ = 3 to estimate √7. Find x₂ to 4 decimal places.
To improve accuracy, increase the number of strips (h gets smaller).
Integration is formally defined as the limit of a sum of rectangle areas as their width tends to zero — the trapezium rule is that idea, stopped early.
Exam trap: quote the number of strips, not ordinates. Six ordinates means five strips, and h is the interval width divided by five.
Quick check
How many ordinates?
?The trapezium rule is used with 5 strips to estimate an integral from x = 1 to x = 6. What is h, and how many ordinates are needed?
Recap
The big ideas to take away
Change of sign: f continuous and f(a), f(b) opposite signs ⇒ a root lies between them
Iteration: x_(n+1) = g(x_n); converges if |g′| < 1 near the root