A-level · Numerical methods

The Newton–Raphson methodExam-style questions

A fast iteration that uses the tangent — x_(n+1) = x_n − f(x_n)/f'(x_n) — its geometric meaning, why it converges so quickly, and the cases where it fails (a horizontal tangent or a poor starting value).

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23 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAll Level 2 Level 3 Level 4 Level 5

1
In this question you must show detailed reasoning.

The diagram shows part of the curve $y = \dfrac{x + 6}{(x + 2)^2}$ , for $x > -2$ . At the point $P$ on the curve, the tangent is parallel to the line $x + 2y = 5$ .(a) Show that the $x$ -coordinate of $P$ is a root of the equation $x^3 + 6x^2 + 10x - 12 = 0$ .

(5 marks)
(b) Show by calculation that this equation has a root between $0.75$ and $0.8$ .

(2 marks)
(c) The iterative formula

$x_{n+1} = \frac{12 - x_n^3 - 6x_n^2}{10}$

is suggested as a way of finding the $x$ -coordinate of $P$ . Show that, whatever starting value is chosen, this formula will not converge to the $x$ -coordinate of $P$ .

(2 marks)
(d) Starting from $x_0 = 1$ , apply the Newton-Raphson method to the equation in part (a). Hence determine both coordinates of $P$ , each to $5$ decimal places.

(5 marks)
Level 5 · 14 marksTry it
2
The curve $C$ has equation $y = k\cos x + 2x^2 - 9x$ , for $0 \le x \le \pi$ , where $x$ is in radians and $k$ is a constant. The point on $C$ where $x = \tfrac13\pi$ is a point of inflection.
(a) Show that $k = 8$ .

(5 marks)

$C$ meets the $x$ -axis at only one point, $Q$ .
(b) Show that $y$ changes sign between $x = 0.7$ and $x = 0.8$ , and explain what this tells you about $Q$ .

(2 marks)
(c) Starting with $x_0 = 0.8$ , use the Newton-Raphson method to find the $x$ -coordinate of $Q$ correct to $5$ decimal places. Show the result of each iteration to $6$ decimal places.

(3 marks)
(d) Use a change of sign to confirm the accuracy of your answer to part (c).

(1 mark)
Level 5 · 11 marksTry it