What is 9.1: Locating roots by change of sign?

If a continuous function has values of opposite sign at the endpoints of an interval, the intermediate value theorem guarantees at least one root inside that interval.

How many marks are 9.1 questions worth?

9.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 9.1 appeared in?

We have not yet indexed a verified past-paper appearance for 9.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

9.1

Locate roots of f(x) = 0 by considering changes of sign of f(x) in an interval on which f(x) is sufficiently well behaved; understand how change of sign methods can fail.

Draft — not yet indexed

Locating roots by change of sign

Worked answers and methods for 9.1 on Edexcel A-level Maths 9MA0.

Explanation

If a continuous function has values of opposite sign at the endpoints of an interval, the intermediate value theorem guarantees at least one root inside that interval.
Evaluate $f(a)$ and $f(b)$ accurately, record their signs, and conclude that a root lies in $(a,b)$ only after checking that $f$ is continuous there.
For example, $f(1)<0$ and $f(2)>0$ for a continuous $f$ , so the graph must cross the $x$ -axis at least once between $1$ and $2$ .
A sign change across a discontinuity need not contain a root, while a repeated root can touch the axis without changing sign; a common error is to treat the sign test as an equivalence.

Worked example

For $f(x)=x^3+x-5$ , show that there is exactly one root in $(1,2)$ .

1.Calculate $f(1)=1+1-5=-3$ and $f(2)=8+2-5=5$ .
2.Continuity gives at least one root in $(1,2)$ .
3.Also $f'(x)=3x^2+1$ , which is positive for every real $x$ , so $f$ is strictly increasing and cannot cross the axis more than once.

Answer: $f(1)=-3$ and $f(2)=5$ , so a root lies in $(1,2)$ .; Since $f'(x)=3x^2+1>0$ , $f$ is strictly increasing and the root is unique.

Common mistakes

Don't conclude that an even-multiplicity root cannot exist because endpoint values have the same sign.
Don't claim a root from a sign change without checking continuity, or claim uniqueness without a monotonicity argument.

Exam tip

To establish exactly one root, combine a sign change on a continuous interval with a separate uniqueness argument.

Worked practice

Tier 1 · Easy

Let

f(x)=x^2-x-1

. Use endpoint values to establish that

(1,2)

brackets a zero of

f

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$f(1)=-1$ and $f(2)=1$ , so continuity and the change of sign give a root in $(1,2)$ .	2
Notes
Substitute the endpoints: $f(1)=1-1-1=-1$ and $f(2)=4-2-1=1$ . A polynomial is continuous, and the endpoint values have opposite signs, so the intermediate value theorem guarantees at least one root between $1$ and $2$ .

(2 marks)

Tier 2 · Standard

The function

f(x)=(x-1)^2

has a root in the interval

[0,2]

. Explain why a search based only on a change of sign between the endpoints does not detect this root.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$f(0)=f(2)=1$ , so there is no endpoint sign change, although $f(1)=0$ . The graph touches the axis at the repeated (even-multiplicity) root instead of crossing it; any equivalent explanation that the sign stays positive on both sides is valid.	3
Notes
At the endpoints, $f(0)=(-1)^2=1$ and $f(2)=1^2=1$ , so their product is positive and the usual sign-change test gives no evidence of a root. Nevertheless, $f(1)=0$ . Because the root is repeated, the non-negative graph touches the $x$ -axis at $x=1$ and turns back without changing sign.

(3 marks)

Tier 3 · Hard

Two sign tests are proposed. For

h(x)=\frac{1}{x-2}

, the values

h(1)

and

h(3)

have opposite signs. For

k(x)=(x-2)^2

, the values

k(1.9)

and

k(2.1)

have the same sign. Explain why neither test gives the correct conclusion about roots in the stated intervals.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$h$ changes sign across $x=2$ but has no root because it is discontinuous there. $k$ has the repeated root $x=2$ but does not change sign across it.	4
Notes
For $h$ , $h(1)=-1$ and $h(3)=1$ , but $h$ is undefined at $x=2$ , so continuity fails and the sign change occurs across a vertical asymptote rather than a root. For $k$ , $k(2)=0$ , but squaring makes $k(x)\geq0$ on both sides of $2$ , so the graph touches the axis at a repeated root without crossing it.

(4 marks)

Tier 1 · Easy

Let

p(x)=x^3-4x+1

. Show that

p(x)=0

has a root in the interval

(0,1)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$p(0)=1$ and $p(1)=-2$ , so continuity and the change of sign give a root in $(0,1)$ .	2
Notes
The polynomial $p$ is continuous. Also $p(0)=1$ and $p(1)=1-4+1=-2$ . Since these values have opposite signs, the intermediate value theorem guarantees at least one root in $(0,1)$ .

(2 marks)

Tier 2 · Standard

For

-1\leq x\leq0

, let

f(x)=x+\cos x

. Show that

f(x)=0

has exactly one root in

(-1,0)

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$f(-1)=\cos1-1<0$ and $f(0)=1>0$ , so a root lies in $(-1,0)$ ; since $f'(x)=1-\sin x>0$ on this interval, the root is unique.	4
Notes
$f$ is continuous on $[-1,0]$ . Its endpoint values are $f(-1)=\cos1-1<0$ and $f(0)=1>0$ , so there is at least one root in $(-1,0)$ . On this interval $\sin x\leq0$ , so $f'(x)=1-\sin x\geq1>0$ . Thus $f$ is strictly increasing and can have at most one root, proving uniqueness.

(4 marks)

Tier 3 · Hard

Let

g(x)=x^3-3x+1

. By considering changes of sign on three disjoint intervals, prove that

g(x)=0

has exactly three real roots.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	There is one root in each of $(-2,-1)$ , $(0,1)$ and $(1,2)$ ; these are exactly the three real roots.	5
Notes
$g$ is continuous. The values $g(-2)=-1$ and $g(-1)=3$ have opposite signs, as do $g(0)=1$ and $g(1)=-1$ , and $g(1)=-1$ and $g(2)=3$ . Hence there is a root in each of the three disjoint intervals $(-2,-1)$ , $(0,1)$ and $(1,2)$ . A non-zero cubic polynomial has at most three real roots, so these three roots are all the real roots of $g$ .

(5 marks)

Tier 2 · Standard

For

f(x)=x^3+x-a

, where

a

is a constant, find the range of values of

a

for which the sign-change test on the interval

(1,2)

guarantees a root of

f(x)=0

in that interval.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$2<a<10$	4
Notes
The polynomial is continuous. At the endpoints, $f(1)=2-a$ and $f(2)=10-a$ . The sign-change test guarantees a root when $(2-a)(10-a)<0$ , which holds for $2<a<10$ . The endpoints are excluded because $a=2$ or $a=10$ places the root at an endpoint rather than inside $(1,2)$ .

(4 marks)

Tier 3 · Hard

Let

h(x)=e^x-3x

. By considering endpoint values and the sign of

h'(x)

, show that

h(x)=0

has exactly two roots in the interval

(0,2)

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	There is one root in $(0,1)$ and one root in $(1,2)$ , and these are the only roots in $(0,2)$ .	5
Notes
$h$ is continuous, with $h(0)=1>0$ , $h(1)=e-3<0$ and $h(2)=e^2-6>0$ . Hence sign changes give a root in each of $(0,1)$ and $(1,2)$ . Also $h'(x)=e^x-3$ , which is negative for $x<\ln3$ and positive for $x>\ln3$ . Thus $h$ decreases and then increases, so it has at most one root on each side of its single turning point; the turning point itself is not a root, since $h(\ln3)=3-3\ln3<0$ . The two bracketed roots are therefore the only roots in $(0,2)$ .

(5 marks)

Tier 3 · Hard

Let

f(x)=x^3+x-1

. Establish the existence and uniqueness of a root in

(0.6,0.7)

. By evaluating

f(0.68)

and

f(0.685)

, give this root to

2

decimal places.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$f(0.6)=-0.184$ and $f(0.7)=0.043$ , so a root lies in $(0.6,0.7)$ . $f'(x)=3x^2+1>0$ , so the root is unique. $f(0.68)=-0.005568$ and $f(0.685)=0.006419125$ , so the root is $0.68$ to $2$ decimal places.	6
Notes
The polynomial $f$ is continuous. Since $f(0.6)=0.6^3+0.6-1=-0.184$ and $f(0.7)=0.7^3+0.7-1=0.043$ , a change of sign gives a root in $(0.6,0.7)$ . Also $f'(x)=3x^2+1>0$ for every real $x$ , so $f$ is strictly increasing and the root is unique. Now $f(0.68)=0.68^3+0.68-1=-0.005568$ , while $f(0.685)=0.685^3+0.685-1=0.006419125$ . Hence the root lies in $(0.68,0.685)$ and is $0.68$ to $2$ decimal places.

(6 marks)

Q10

Tier 3 · Hard

10.

A function

g

is continuous on

[-3,2]

. Some of its values are shown:

g(-3)=2

g(-2)=-1

g(-1)=4

g(0)=0

g(1)=3

and

g(2)=-2

. Determine the least number of distinct roots that

g(x)=0

must have in

[-3,2]

. Identify where they occur and explain why the information does not establish the exact number of roots.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	At least four distinct roots: one in each of $(-3,-2)$ , $(-2,-1)$ and $(1,2)$ , together with the root $x=0$ . The data give no uniqueness or monotonicity information, so additional roots may occur between the sampled points.	5
Notes
Continuity and the sign changes between $-3$ and $-2$ , between $-2$ and $-1$ , and between $1$ and $2$ guarantee a root in each of those three disjoint open intervals. The stated value $g(0)=0$ supplies a fourth root, distinct from the three bracketed roots. Thus four is the least number forced by the data. A continuous graph could cross the axis additional times inside any sampled interval, so these values alone cannot prove that there are exactly four roots.

(5 marks)

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