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
Show that the equation $e^x = 3 - x$ has a root between $x = 0$ and $x = 1$ .
Level 3 · 2 marksTry it
2
$f(x) = x^3 + x - 1$ .
(a) Show that $f(x) = 0$ has a root in the interval $[0, 1]$ .
(b) Show that this root is $0.7$ correct to 1 decimal place.
Level 3 · 4 marksTry it
3
The function $f(x) = \dfrac{1}{x - 3}$ satisfies $f(2) = -1$ and $f(4) = 1$ . Explain why this does not show that $f(x) = 0$ has a root between $x = 2$ and $x = 4$ .
Level 3 · 2 marksTry it
4
Show that the equation $\ln x + x = 2$ has a root between $x = 1$ and $x = 2$ .
Level 3 · 2 marksTry it
5
$f(x) = x^3 - 3x + 1$ . Show that $f(x) = 0$ has three roots, one in each of the intervals $[-2, -1]$ , $[0, 1]$ and $[1, 2]$ .
Level 3 · 3 marksTry it

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