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).

Start practising Read the lesson

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
The equation $\ln x + x = 2$ has a root near $x = 1.5$ . Use the Newton–Raphson method with $x_0 = 1.5$ to find $x_1$ to 4 decimal places.
Level 4 · 2 marksTry it
2
$f(x) = x^3 - 4x + 1$
(a) Show that the equation $f(x) = 0$ has a root $\alpha$ such that $1.8 < \alpha < 1.9$ . (2 marks)
(b) Show that the Newton–Raphson iteration for this equation can be written as

$x_{n+1} = \dfrac{2x_n^3 - 1}{3x_n^2 - 4}$ (2 marks)
(c) Using $x_0 = 2$ , find $\alpha$ correct to 4 decimal places. Show the result of each iteration. (3 marks)
(d) Explain why the Newton–Raphson method cannot be used to find $\alpha$ starting from $x_0 = \dfrac{2}{\sqrt{3}}$ . (1 mark)
Level 4 · 8 marksTry it
3
$\text{f}(x) = x - 3\ln x \qquad x > 0$
(a) Show that the equation $\text{f}(x) = 0$ has a root $\alpha$ in the interval $[4.5,\ 5]$

(2 marks)
(b) Find $\text{f}'(x)$

(2 marks)
(c) Using $x_0 = 5$ as a first approximation for $\alpha$ , apply the Newton–Raphson method once to obtain a second approximation for $\alpha$ .

Give your answer to 3 decimal places.

(2 marks)

A student attempts to use the Newton–Raphson method with $x_0 = 3$ as a first approximation for $\alpha$ .
(d) Explain why the method fails.

(1 mark)
Level 4 · 7 marksTry it
4
$\text{f}(x) = 5\sin 2x - 3x + 2 \qquad 0 \le x \le \pi$

where $x$ is measured in radians.

The curve with equation $y = \text{f}(x)$ has a single maximum turning point in the interval $0 \le x \le \pi$
(a) Using calculus, find the $x$ coordinate of this maximum turning point, giving your answer to 3 decimal places.

(4 marks)

The equation $\text{f}(x) = 0$ has a single root $\alpha$ in the interval $0 \le x \le \pi$

It is given that $\text{f}(1.3) = 0.678$ and $\text{f}(1.4) = -0.525$ , each correct to 3 decimal places.
(b) Explain how these two values show that $1.3 < \alpha < 1.4$

(1 mark)
(c) Using $x_0 = 1.4$ as a first approximation for $\alpha$ , apply the Newton–Raphson method once to obtain a second approximation for $\alpha$ .

Show your method and give your answer to 3 significant figures.

(2 marks)
Level 4 · 7 marksTry it
5
For $x > 0$ , the curve $y = (x + 3)\text{e}^{-x} + x^2$ has a single stationary point, $S$ , whose $x$ -coordinate is approximately $0.7$ .
(a) Show that the $x$ -coordinate of $S$ is a solution of the equation $2x\text{e}^{x} - x - 2 = 0$ .

(2 marks)
(b) Show that applying the Newton-Raphson method to the equation in part (a) leads to the iterative formula

$x_{n+1} = \frac{2x_n^2\text{e}^{x_n} + 2}{2(x_n + 1)\text{e}^{x_n} - 1}.$

(4 marks)
(c) Use this formula, with $x_1 = 0.7$ , to find $x_2$ and $x_3$ .

(1 mark)
(d) Hence find the coordinates of $S$ , each correct to $3$ decimal places.

(2 marks)
Level 4 · 9 marksTry it