A-level · Numerical methods

Locating roots by change of signExam-style questions

Trapping a root of f(x) = 0 in an interval using the change-of-sign rule — why continuity is essential, what a sign change does and doesn't guarantee, and refining a root to a given number of decimal places.

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22 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
$\mathrm{f}(x) = x^3 + x - 5$
(a) Show that the equation $\mathrm{f}(x) = 0$ has a root $\alpha$ between $1.5$ and $1.6$ . (2 marks)
(b) Show that the equation $\mathrm{f}(x) = 0$ can be rearranged into the form $x = \sqrt[3]{5 - x}$ . (1 mark)
(c) Use the iterative formula

$x_{n+1} = \sqrt[3]{5 - x_n}$

with $x_0 = 1.5$ to find $x_1$ , $x_2$ and $x_3$ , giving each answer to $4$ decimal places. (3 marks)
(d) Show that $\alpha = 1.516$ correct to $3$ decimal places. (2 marks)
Level 4 · 8 marksTry it
2
$\mathrm{f}(x) = x^3 + 4x - 7$
(a) Show that the equation $\mathrm{f}(x) = 0$ has a root $\alpha$ between $1.2$ and $1.3$ . (2 marks)
(b) Taking $x_0 = 1.3$ , apply the Newton–Raphson procedure twice to $\mathrm{f}(x)$ , giving $x_2$ to $5$ decimal places. (4 marks)
(c) Write down the value of $\alpha$ correct to $4$ significant figures. (1 mark)
Level 4 · 7 marksTry it