Locating roots

The idea

If $f(x)$ is continuous and changes sign between $a$ and $b$ , there is a root between them. A sign change can mislead if the graph has an asymptote in the interval, and a repeated root can touch the axis with no sign change at all.

Rearrange to $f(x) = 0$ .
Work out $f(a)$ and $f(b)$ and show their signs.
Conclude: sign change and $f$ continuous, so a root lies between $a$ and $b$ .

Watch out Say that $f$ is continuous: without it, a sign change proves nothing.

See it worked

Worked example

Show that $x^3 - 2x - 5 = 0$ has a root between $2$ and $3$ .

$f(2) = 8 - 4 - 5 = -1 < 0$ .
$f(3) = 27 - 6 - 5 = 16 > 0$ .
$f$ is continuous and changes sign, so there is a root in $(2, 3)$ .

Answer $f(2) = -1$ , $f(3) = 16$ : sign change, so a root lies between

Your turn

Question 1Calculator allowed3 marks

$f(x) = x^3 + x - 3$ . Find $f(1.2)$ .

Question 2Non-calculator3 marks

$f(x) = \frac{1}{x - 2}$ has $f(1) = -1$ and $f(3) = 1$ . Explain why there is no root between $1$ and $3$ .

Write your answer, then show the working

$f(x)$ is undefined at $x = 2$ .
The graph jumps from $-\infty$ to $+\infty$ at the asymptote.
$f(x)$ is never $0$ , so the sign change does not show a root.

Answer $f$ is not continuous at $x = 2$ : the sign change comes from the asymptote

Question 3Calculator allowed3 marks

$f(x) = \cos x - x$ , with $x$ in radians. Find $f(0.8)$ to 3 decimal places.

In past papers

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