Location of rootsEdexcel International A Level Maths: Revision notes

Section 1

The change-of-sign test

A root of $f(x)=0$ is a value of $x$ where the graph crosses the $x$ -axis. If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f(x)=0$ has at least one root in $(a,b)$ . Example: $f(x)=x^3+x-5$ . $f(1)=-3<0$ and $f(2)=5>0$ , so there is a root between 1 and 2. A continuous graph cannot go from below the axis to above it without crossing it.

Key termsrootcontinuouschange of sign

Exam tip

Use your calculator in radians for trigonometric functions, unless told otherwise.

Section 2

Writing a full conclusion

A complete answer states four things:

the values of $f$ at both ends (to enough accuracy to show the sign);
that there is a change of sign;
that $f$ is continuous (for example, it is a polynomial or $e^x-3x$ );
the conclusion: a root lies in the interval. Example: $f(1.5)=-0.125$ and $f(1.6)=0.696$ . The sign changes and $f$ is continuous, so $1.5<\alpha<1.6$ .

Key termsconclusion

Common mistake

Quoting the two values but not stating continuity or the conclusion. Both are marked.

Section 3

Narrowing the interval

To locate a root more accurately, evaluate $f$ at values inside the interval and keep the sub-interval where the sign changes. Example: $f(x)=e^x-3x$ has $f(1.5)=-0.018$ and $f(1.6)=0.153$ . Then $f(1.51)=-0.003$ and $f(1.52)=0.012$ , so $1.51<\beta<1.52$ . To show a root is correct to $n$ decimal places, find a sign change over the interval $\pm0.5\times10^{-n}$ . For example, $g(3.45)<0$ and $g(3.55)>0$ show a root is 3.5 to 1 d.p.

Key termsinterval

Exam tip

Choose the test points at the rounding boundaries (such as $3.45$ and $3.55$ ) when asked for decimal places.

Section 4

When the test fails or is misused

Not continuous. $h(x)=\frac{1}{x-2}$ has $h(1)=-1$ and $h(3)=1$ but no root, because $h$ is undefined at $x=2$ . The same applies to $\tan x$ across $\frac\pi2$ . Same sign at the ends. This does not prove there is no root. An even number of roots gives no net sign change: $g(x)=x^3-6x^2+9x-1$ has $g(0)=-1$ and $g(3.5)=-0.125$ , both negative, yet there are two roots in $[0,3.5]$ . Repeated root. $(x-2)^2$ touches the axis at $x=2$ but does not change sign there.

Key termsasymptote

Common mistake

Saying 'there is a sign change so there must be a root' without checking that the function is continuous.

Section 5

Counting roots in an interval

A sign change shows at least one root, and an odd number of roots overall. To show several distinct roots, find a sign change in each of several non-overlapping intervals. Example: $g(0)=-1$ , $g(1)=3$ , $g(2)=1$ , $g(3)=-1$ , $g(4)=3$ . Sign changes in $[0,1]$ , $[2,3]$ and $[3,4]$ give three distinct roots. A cubic has at most three roots, so these are all of them.

Key termsdistinct roots

Exam tip

Tabulate $f(x)$ at integer values first to find where the sign changes.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Location of roots

The function $f$ is defined by $f(x)=x^3+x-5$ for $x\in\mathbb{R}$ . The equation $f(x)=0$ has exactly one real root $\alpha$ .
Given that $f(1.6)=0.696$ , explain how the values of $f(1.5)$ and $f(1.6)$ show that $1.5<\alpha<1.6$ .2 marks
The functions $h$ and $p$ are defined by $h(x)=\frac{1}{x-2}$ for $x\neq2$ , and $p(x)=x^2-2$ for $x\in\mathbb{R}$ .
A student notes that $h(1)<0$ and $h(3)>0$ , and concludes that $h(x)=0$ has a root in $[1,3]$ . Explain why this conclusion is wrong.2 marks
The function $f$ is defined by $f(x)=e^x-3x$ for $x\in\mathbb{R}$ . The equation $f(x)=0$ has two real roots, $\alpha$ and $\beta$ , with $\alpha<\beta$ .
Show that $\alpha$ lies in the interval $[0.5,0.7]$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).