P9.2bThe Newton–Raphson method

Edexcel A level Maths (9MA0) · Pure mathematics › Numerical methods

Practise The Newton–Raphson method. 1 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.

Build a paper on this topic

▶ Watch videos on The Newton–Raphson method (TLMaths on YouTube) · Practise all of Numerical methods

Sample questions

Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.

Question 1Easy4 marks

\(g(x) = x^{3} - x^{2} - 6\).
The equation \(g(x) = 0\) has a root \(\alpha\) close to \(x = 2\).
You are given that \(g'(x) = 3x^{2} - 2x\).

(a) Taking \(x_0 = 2\) as a first approximation to \(\alpha\), apply the Newton–Raphson method once to \(g(x)\) to find a second approximation to \(\alpha\). Give your answer to 3 decimal places.[2]

(b) Apply the Newton–Raphson method a second time to find \(x_2\), giving your answer to 4 decimal places.[2]

Show the answer and mark scheme

(a) Answer: \(x_1 = 2.250\)

M1 for \(x_1 = 2 - \frac{g(2)}{g'(2)}\) with their values
A1 for \(x_1 = 2.250\) (awrt)

Worked solution: \(g(2) = -2\) and \(g'(2) = 8\), so
\(x_1 = 2 - \frac{-2}{8} = 2.25 = 2.250\) (3 d.p.).

(b) Answer: \(x_2 = 2.2193\)

M1 for \(x_2 = x_1 - \frac{g(x_1)}{g'(x_1)}\) using their unrounded \(x_1\)
A1 for \(x_2 = 2.2193\) (awrt)

Worked solution: \(g(x_1) = 0.328125\), \(g'(x_1) = 10.6875\), so \(x_2 = 2.25 - \frac{0.328125}{10.6875} = 2.219298\ldots = 2.2193\) (4 d.p.).

Question 2Medium6 marks

(a) Show that applying the Newton–Raphson method to \(\mathrm{f}(x) = x^2 - 7\) gives the iteration
\[x_{n + 1} = \frac{1}{2}\left(x_n + \frac{7}{x_n}\right)\][2]

(b) Starting with \(x_1 = 3\), find \(x_2\) and \(x_3\), giving each as an exact fraction.[2]

(d) A student starts with \(x_1 = -3\) and says that the iteration will again converge to \(\sqrt{7}\). Explain whether the student is correct.[1]

Show the answer and mark scheme

(a) Answer: \(x_{n + 1} = x_n - \frac{x_n^2 - 7}{2x_n} = \frac{2x_n^2 - x_n^2 + 7}{2x_n} = \frac{1}{2}\left(x_n + \frac{7}{x_n}\right)\)

M1 for \(x_{n + 1} = x_n - \frac{x_n^2 - 7}{2x_n}\)
A1* for simplifying to the given form

Worked solution: \(\mathrm{f}'(x) = 2x\), so \(x_{n + 1} = x_n - \dfrac{x_n^2 - 7}{2x_n} = \dfrac{x_n^2 + 7}{2x_n} = \dfrac{1}{2}\left(x_n + \dfrac{7}{x_n}\right)\).

(b) Answer: \(x_2 = \frac{8}{3}\), \(x_3 = \frac{127}{48}\)

M1 for \(x_2 = \frac{1}{2}\left(3 + \frac{7}{3}\right) = \frac{8}{3}\)
A1 for \(x_3 = \frac{127}{48}\)

Worked solution: \(x_2 = \frac{1}{2}\left(3 + \frac{7}{3}\right) = \frac{8}{3}\). \(x_3 = \frac{1}{2}\left(\frac{8}{3} + \frac{21}{8}\right) = \frac{1}{2} \times \frac{127}{24} = \frac{127}{48}\).

(c) Answer: \(\mathrm{f}'(0) = 0\): the tangent at \(x = 0\) is horizontal and never meets the \(x\)-axis (division by zero).

B1 for \(\mathrm{f}'(0) = 0\) (the tangent is horizontal), so the formula divides by zero

Worked solution: \(\mathrm{f}'(0) = 0\), so the tangent at \(x = 0\) is horizontal and does not cross the \(x\)-axis; the formula would require dividing by zero.

(d) Answer: No: if \(x_n \lt 0\) then \(x_{n + 1} \lt 0\) (e.g. \(x_2 = -\frac{8}{3}\)), so the iteration converges to \(-\sqrt{7}\).

B1 for not correct: the iterates stay negative (\(x_2 = -\frac{8}{3}\), …) and converge to \(-\sqrt{7}\)

Worked solution: If \(x_n\) is negative, then \(x_n + \frac{7}{x_n}\) is negative, so every iterate is negative: \(x_2 = -\frac{8}{3}\), \(x_3 = -\frac{127}{48}\), … The iteration converges to the other root, \(-\sqrt{7}\), so the student is wrong.

Question 3Hard8 marks

\(f(x) = x^{3} - 20x\). The equation \(f(x) = 0\) has a positive root \(\beta\).

(a) Show that, when the Newton–Raphson method is applied to \(f(x)\) with \(x_0 = 2\), the next approximation is \(x_1 = -2\).[3]

(b) Explain what happens if the Newton–Raphson method is continued from this starting value.[2]

(c) Using \(x_0 = 5\), apply the Newton–Raphson method twice to find an approximation to \(\beta\), giving your answer to 4 decimal places.[3]

Show the answer and mark scheme

(a) Answer: Shown

B1 for \(f'(x) = 3x^{2} - 20\)
M1 for \(x_1 = 2 - \frac{f(2)}{f'(2)}\) with \(f(2) = -32\) and \(f'(2) = -8\)
A1* for \(x_1 = -2\)

Worked solution: \(f(2) = -32\) and \(f'(2) = -8\), so \(x_1 = 2 - \frac{-32}{-8} = 2 - 4 = -2\).

(b) Answer: The approximations alternate between \(2\) and \(-2\) for ever

M1 for using the symmetry of \(f\) (an odd function, so \(f(-x) = -f(x)\) and \(f'(-x) = f'(x)\)) or calculating \(x_2 = 2\)
A1 for concluding that the approximations cycle between \(2\) and \(-2\) and never converge to a root

Worked solution: Since \(f(-2) = 32\) and \(f'(-2) = -8\), \(x_2 = -2 - \frac{32}{-8} = 2\). The values alternate \(2, -2, 2, -2, \ldots\) for ever, so the method never converges.

M1 for a correct first application: \(x_1 = 5 - \frac{25}{55}\)
A1 for \(x_1 = 4.5455\) (awrt)
A1 for \(x_2 = 4.4739\) (awrt)

Worked solution: \(x_1 = 5 - \frac{25}{55} = 4.545455\ldots\); \(x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 4.473873\ldots = 4.4739\). (The root is \(\beta = \sqrt{20} = 4.4721\ldots\).)

Related subtopics

Stuck? Get 1-to-1 help. Chhetri Academy tutors GCSE and A level Maths and Science online, with a free 30-minute trial lesson.

Book a free trial