1.9.2 Failure of sign change methods (A-level only)

What you'll learn

How a change of sign can be used to locate a root.
Why continuity is essential to the method.
How repeated roots, multiple roots and discontinuities can make sign tests fail.
How to recognise these failures and respond appropriately in an exam.

Prerequisites: roots and signs

A root of the equation $f (x) = 0$ is a value of $x$ at which the graph of $y = f (x)$ meets the $x$ -axis.

The value of a function is:

positive when its graph is above the $x$ -axis;
negative when its graph is below the $x$ -axis;
zero when its graph is on the $x$ -axis.

If $f (a)$ and $f (b)$ have opposite signs, then their product is negative:

f (a) f (b) < 0.

This is called a change of sign between $x = a$ and $x = b$ .

Why sign-change methods usually work

A function is continuous on an interval if its graph can be drawn across that interval without lifting your pen. Informally, the graph has no gaps, jumps or vertical asymptotes there.

If a continuous graph is below the $x$ -axis at one end of an interval and above it at the other, it must pass through the $x$ -axis somewhere in between.

Definition

Sign-change result

If $f$ is continuous on $[a, b]$ and

f (a) f (b) < 0,

then there is at least one root of $f (x) = 0$ in the interval $(a, b)$ .

This result is the basis of methods such as interval bisection and decimal search, where you repeatedly evaluate the function and reduce the size of an interval containing a sign change.

Example

Confirming that an interval contains a root

Show that $f(x)=x^3+x-1$ has a root between $x = 0.5$ and $x = 0.75$ .

Evaluate the function at the lower endpoint:
$f(0.5)=(0.5)^3+0.5-1=-0.375.$
Evaluate it at the upper endpoint:
$f(0.75)=(0.75)^3+0.75-1=0.171875.$
The values have opposite signs, so
$f (0.5) f (0.75) < 0.$
Since a polynomial is continuous, there must be at least one root in $(0.5, 0.75)$ .

Key Idea

A one-way test

For a continuous function, a sign change guarantees a root. However, the absence of a sign change does not guarantee that there is no root.

Three different graph behaviours

The diagrams compare a root where the graph crosses the axis, a repeated root where it only touches the axis, and a discontinuity that produces opposite signs without any root.

Three graphs comparing a crossing root, a repeated root that only touches the axis, and a discontinuity with no root

These cases show why you must consider the shape and continuity of the graph, rather than relying only on endpoint signs.

Failure 1: a root without a sign change

A graph can touch the $x$ -axis and turn around without crossing it. The function then has the same sign on both sides of the root.

This often happens at a repeated root.

Definition

Multiplicity of a root

If a polynomial contains a factor $x-r)^m$ , then $x = r$ is a root with multiplicity $m$ .

If $m$ is odd, the graph usually crosses the $x$ -axis and changes sign.
If $m$ is even, the graph usually touches the $x$ -axis and does not change sign.

For example, the function

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

has a root at $x = 2$ with multiplicity 2. Because a square is never negative, the function is positive on both sides of the root.

Example

A repeated root missed by a sign test

Test the signs of $f(x)=(x-2)^2$ on the interval $[1.9, 2.1]$ .

At the lower endpoint,
$f(1.9)=(1.9-2)^2=0.01.$
At the upper endpoint,
$f(2.1)=(2.1-2)^2=0.01.$
Both endpoint values are positive, so there is no change of sign. Nevertheless,
$f(2)=(2-2)^2=0,$
so the interval does contain a root. The sign test misses it because the graph touches the axis without crossing it.

Common Mistake

Treating the converse as true

Do not conclude that an interval contains no root just because $f (a)$ and $f (b)$ have the same sign. The sign-change result does not work in reverse.

Failure 2: an even number of crossing roots

An interval can contain two, four or another even number of roots where the graph crosses the axis. Each crossing reverses the sign, so an even number of reversals leaves the endpoint signs the same.

Example

Two roots hidden inside one interval

Consider $f (x) = (x - 1) (x - 2)$ on the interval $[0.5, 2.5]$ .

Evaluate the endpoints:
$f(0.5)=(0.5−1)(0.5−2)=0.75,f(2.5)=(2.5−1)(2.5−2)=0.75.\begin{aligned} f(0.5)&=(0.5-1)(0.5-2)=0.75,\\ f(2.5)&=(2.5-1)(2.5-2)=0.75. \end{aligned}$
Both values are positive, so the endpoints show no change of sign.
However, the factors show that
$\quad \text{and} \quad f(2)=0.$
The graph crosses the axis twice. The first crossing changes the sign from positive to negative, and the second changes it back to positive.

A coarse decimal search can therefore jump over two nearby roots. Using smaller intervals may reveal the negative values between them, although it still does not solve the repeated-root problem.

Tip

Use additional information

A graph, factorisation or knowledge of turning points can reveal roots that an endpoint sign test misses. Sign checks are most reliable when combined with information about the function's shape.

Failure 3: a sign change without a root

A sign change only guarantees a root when the function is continuous throughout the interval.

If the function has a vertical asymptote or another discontinuity, its values can switch from negative to positive without ever taking the value zero.

Example

A discontinuity mistaken for a root

Consider

f(x)=1x−3f(x)=\frac{1}{x-3}

on the interval $[2.9, 3.1]$ .

Evaluate the function on the left of 3:
$f(2.9)=12.9−3=−10.f(2.9)=\frac{1}{2.9-3}=-10.$
Evaluate it on the right of 3:
$f(3.1)=13.1−3=10.f(3.1)=\frac{1}{3.1-3}=10.$
There is a change of sign.
Solving $f (x) = 0$ would require
$1x−3=0,\frac{1}{x-3}=0,$
but a fraction with numerator 1 can never equal zero. Also, $f (3)$ is undefined, so the function is not continuous on the interval. The sign change is caused by a vertical asymptote, not a root.

Common Mistake

Check continuity first

A negative product $f (a) f (b) < 0$ is not enough by itself. Before claiming that a root exists, check that the function is continuous on the whole interval $[a, b]$ .

For a rational function, values excluded from the domain are especially important. If a denominator becomes zero inside the interval, the continuity condition fails.

What a sign test can and cannot prove

For a continuous function:

Opposite endpoint signs prove that at least one root lies between the endpoints.
Opposite signs do not prove that there is exactly one root.
Equal endpoint signs do not prove that there are no roots.

For a function that is not continuous:

Opposite endpoint signs may be caused by a discontinuity.
You cannot apply the sign-change result across the discontinuity.

Key Idea

Check the hypotheses

A numerical method is only as reliable as the conditions behind it. For a sign-change argument, the essential condition is continuity on the entire interval.

Dealing with possible failure

When a sign-change method appears to find no root, you can investigate by:

sketching or plotting the graph;
using smaller intervals;
factorising where possible;
checking whether the graph touches the axis at a turning point;
checking the domain for discontinuities;
considering whether more than one root lies in the interval.

A calculated value may also be displayed as zero because of rounding. Keep sufficient accuracy during numerical work and distinguish an exact zero from a small rounded value.

Exam technique

In the exam

Calculate $f (a)$ and $f (b)$ accurately and state their signs.
Before using a sign change to prove that a root exists, state that $f$ is continuous on the interval.
If there is no sign change, do not claim that there is no root; consider repeated roots or an even number of crossings.
Check for excluded values and vertical asymptotes before applying the method to a rational function.
Use a graph, factorisation or turning-point information when the endpoint signs alone are inconclusive.

Self review

Check yourself

Why does $f (a) f (b) > 0$ not prove that the interval $(a, b)$ contains no roots?
How can a vertical asymptote create an apparent change of sign?
What is the difference between the graph near a root of even multiplicity and one of odd multiplicity?

What you'll learn

How a change of sign can be used to locate a root.
Why continuity is essential to the method.
How repeated roots, multiple roots and discontinuities can make sign tests fail.
How to recognise these failures and respond appropriately in an exam.

Prerequisites: roots and signs

A root of the equation $f (x) = 0$ is a value of $x$ at which the graph of $y = f (x)$ meets the $x$ -axis.

The value of a function is:

positive when its graph is above the $x$ -axis;
negative when its graph is below the $x$ -axis;
zero when its graph is on the $x$ -axis.

If $f (a)$ and $f (b)$ have opposite signs, then their product is negative:

f (a) f (b) < 0.

This is called a change of sign between $x = a$ and $x = b$ .

Why sign-change methods usually work

A function is continuous on an interval if its graph can be drawn across that interval without lifting your pen. Informally, the graph has no gaps, jumps or vertical asymptotes there.

If a continuous graph is below the $x$ -axis at one end of an interval and above it at the other, it must pass through the $x$ -axis somewhere in between.

Definition

Sign-change result

If $f$ is continuous on $[a, b]$ and

f (a) f (b) < 0,

then there is at least one root of $f (x) = 0$ in the interval $(a, b)$ .

This result is the basis of methods such as interval bisection and decimal search, where you repeatedly evaluate the function and reduce the size of an interval containing a sign change.

Example

Confirming that an interval contains a root

Show that $f(x)=x^3+x-1$ has a root between $x = 0.5$ and $x = 0.75$ .

Evaluate the function at the lower endpoint:
$f(0.5)=(0.5)^3+0.5-1=-0.375.$
Evaluate it at the upper endpoint:
$f(0.75)=(0.75)^3+0.75-1=0.171875.$
The values have opposite signs, so
$f (0.5) f (0.75) < 0.$
Since a polynomial is continuous, there must be at least one root in $(0.5, 0.75)$ .

Key Idea

A one-way test

For a continuous function, a sign change guarantees a root. However, the absence of a sign change does not guarantee that there is no root.

Three different graph behaviours

The diagrams compare a root where the graph crosses the axis, a repeated root where it only touches the axis, and a discontinuity that produces opposite signs without any root.

Three graphs comparing a crossing root, a repeated root that only touches the axis, and a discontinuity with no root

These cases show why you must consider the shape and continuity of the graph, rather than relying only on endpoint signs.

Failure 1: a root without a sign change

A graph can touch the $x$ -axis and turn around without crossing it. The function then has the same sign on both sides of the root.

This often happens at a repeated root.

Definition

Multiplicity of a root

If a polynomial contains a factor $x-r)^m$ , then $x = r$ is a root with multiplicity $m$ .

If $m$ is odd, the graph usually crosses the $x$ -axis and changes sign.
If $m$ is even, the graph usually touches the $x$ -axis and does not change sign.

For example, the function

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

has a root at $x = 2$ with multiplicity 2. Because a square is never negative, the function is positive on both sides of the root.

Example

A repeated root missed by a sign test

Test the signs of $f(x)=(x-2)^2$ on the interval $[1.9, 2.1]$ .

At the lower endpoint,
$f(1.9)=(1.9-2)^2=0.01.$
At the upper endpoint,
$f(2.1)=(2.1-2)^2=0.01.$
Both endpoint values are positive, so there is no change of sign. Nevertheless,
$f(2)=(2-2)^2=0,$
so the interval does contain a root. The sign test misses it because the graph touches the axis without crossing it.

Common Mistake

Treating the converse as true

Do not conclude that an interval contains no root just because $f (a)$ and $f (b)$ have the same sign. The sign-change result does not work in reverse.

Failure 2: an even number of crossing roots

An interval can contain two, four or another even number of roots where the graph crosses the axis. Each crossing reverses the sign, so an even number of reversals leaves the endpoint signs the same.

Example

Two roots hidden inside one interval

Consider $f (x) = (x - 1) (x - 2)$ on the interval $[0.5, 2.5]$ .

Evaluate the endpoints:
$f(0.5)=(0.5−1)(0.5−2)=0.75,f(2.5)=(2.5−1)(2.5−2)=0.75.\begin{aligned} f(0.5)&=(0.5-1)(0.5-2)=0.75,\\ f(2.5)&=(2.5-1)(2.5-2)=0.75. \end{aligned}$
Both values are positive, so the endpoints show no change of sign.
However, the factors show that
$\quad \text{and} \quad f(2)=0.$
The graph crosses the axis twice. The first crossing changes the sign from positive to negative, and the second changes it back to positive.

A coarse decimal search can therefore jump over two nearby roots. Using smaller intervals may reveal the negative values between them, although it still does not solve the repeated-root problem.

Tip

Use additional information

A graph, factorisation or knowledge of turning points can reveal roots that an endpoint sign test misses. Sign checks are most reliable when combined with information about the function's shape.

Failure 3: a sign change without a root

A sign change only guarantees a root when the function is continuous throughout the interval.

If the function has a vertical asymptote or another discontinuity, its values can switch from negative to positive without ever taking the value zero.

Example

A discontinuity mistaken for a root

Consider

f(x)=1x−3f(x)=\frac{1}{x-3}

on the interval $[2.9, 3.1]$ .

Evaluate the function on the left of 3:
$f(2.9)=12.9−3=−10.f(2.9)=\frac{1}{2.9-3}=-10.$
Evaluate it on the right of 3:
$f(3.1)=13.1−3=10.f(3.1)=\frac{1}{3.1-3}=10.$
There is a change of sign.
Solving $f (x) = 0$ would require
$1x−3=0,\frac{1}{x-3}=0,$
but a fraction with numerator 1 can never equal zero. Also, $f (3)$ is undefined, so the function is not continuous on the interval. The sign change is caused by a vertical asymptote, not a root.

Common Mistake

Check continuity first

A negative product $f (a) f (b) < 0$ is not enough by itself. Before claiming that a root exists, check that the function is continuous on the whole interval $[a, b]$ .

For a rational function, values excluded from the domain are especially important. If a denominator becomes zero inside the interval, the continuity condition fails.

What a sign test can and cannot prove

For a continuous function:

Opposite endpoint signs prove that at least one root lies between the endpoints.
Opposite signs do not prove that there is exactly one root.
Equal endpoint signs do not prove that there are no roots.

For a function that is not continuous:

Opposite endpoint signs may be caused by a discontinuity.
You cannot apply the sign-change result across the discontinuity.

Key Idea

Check the hypotheses

A numerical method is only as reliable as the conditions behind it. For a sign-change argument, the essential condition is continuity on the entire interval.

Dealing with possible failure

When a sign-change method appears to find no root, you can investigate by:

sketching or plotting the graph;
using smaller intervals;
factorising where possible;
checking whether the graph touches the axis at a turning point;
checking the domain for discontinuities;
considering whether more than one root lies in the interval.

A calculated value may also be displayed as zero because of rounding. Keep sufficient accuracy during numerical work and distinguish an exact zero from a small rounded value.

Exam technique

In the exam

Calculate $f (a)$ and $f (b)$ accurately and state their signs.
Before using a sign change to prove that a root exists, state that $f$ is continuous on the interval.
If there is no sign change, do not claim that there is no root; consider repeated roots or an even number of crossings.
Check for excluded values and vertical asymptotes before applying the method to a rational function.
Use a graph, factorisation or turning-point information when the endpoint signs alone are inconclusive.

Self review

Check yourself

Why does $f (a) f (b) > 0$ not prove that the interval $(a, b)$ contains no roots?
How can a vertical asymptote create an apparent change of sign?
What is the difference between the graph near a root of even multiplicity and one of odd multiplicity?

What you'll learn

Prerequisites: roots and signs

Why sign-change methods usually work

Three different graph behaviours

Failure 1: a root without a sign change

Failure 2: an even number of crossing roots

Failure 3: a sign change without a root

What a sign test can and cannot prove

Dealing with possible failure

Review 1.9.2 Failure of sign change methods (A-level only) by teaching Genie

1.9.2 Failure of sign change methods (A-level only)

What you'll learn

Prerequisites: roots and signs

Why sign-change methods usually work

Three different graph behaviours

Failure 1: a root without a sign change

Failure 2: an even number of crossing roots

Failure 3: a sign change without a root

What a sign test can and cannot prove

Dealing with possible failure

Review 1.9.2 Failure of sign change methods (A-level only) by teaching Genie

1.9.2 Failure of sign change methods (A-level only)

What you'll learn

Prerequisites: roots and signs

Why sign-change methods usually work

Three different graph behaviours

Failure 1: a root without a sign change

Failure 2: an even number of crossing roots

Failure 3: a sign change without a root

What a sign test can and cannot prove

Dealing with possible failure

Review 1.9.2 Failure of sign change methods (A-level only) by teaching Genie

1.9.2 Failure of sign change methods (A-level only) Revision Guide

1.9.2 Failure of sign change methods (A-level only)

What you'll learn

Prerequisites: roots and signs

Why sign-change methods usually work

Three different graph behaviours

Failure 1: a root without a sign change

Failure 2: an even number of crossing roots

Failure 3: a sign change without a root

What a sign test can and cannot prove

Dealing with possible failure

Review 1.9.2 Failure of sign change methods (A-level only) by teaching Genie

1.9.2 Failure of sign change methods (A-level only) Revision Guide