What must you state to make a change-of-sign argument complete?

The values of f at both ends, that f is continuous, and that there is a change of sign so a root lies in the interval.

What does interval bisection do?

Halves an interval containing a root by testing the midpoint and keeping the half where the sign changes.

f(0.6)=0.0221, f(0.7)=-0.0862. What is the conclusion?

f is continuous and changes sign, so there is a root in 0.6<x<0.7.

Why does f(x)=\frac{1}{x-1} change sign between x=0 and x=2 without a root?

It is not continuous at x=1 (an asymptote), and \frac{1}{x-1} is never zero.

How can a root be missed when there is no sign change?

A repeated root (the graph touches the axis), or an even number of roots in the interval such as two.

For f(x)=(x-2)^2(x+2), why is f(1)=3 and f(3)=5 misleading?

There is a repeated root at x=2 in [1,3]; f does not change sign there.

A sign change over a continuous interval shows what about the number of roots?

An odd number of roots, so at least one, but not necessarily exactly one.

Which type of function often causes a false sign change?

One with an asymptote, such as \tan x or \frac{1}{x-a}.

How can missing two close roots be avoided?

Use a smaller interval, or evaluate f at more points in the interval.

Locating roots by change of signAQA A-Level Maths: Flashcards

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State the change-of-sign rule.

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State the change-of-sign rule.: If $f$ is continuous on $[a,b]$ and $f(a)$ , $f(b)$ have opposite signs, then $f(x)=0$ has a root in $a<x<b$ .
What must you state to make a change-of-sign argument complete?: The values of $f$ at both ends, that $f$ is continuous, and that there is a change of sign so a root lies in the interval.
Write $e^x=3x$ in the form $f(x)=0$ .: $f(x)=e^x-3x=0$
What does interval bisection do?: Halves an interval containing a root by testing the midpoint and keeping the half where the sign changes.
$f(0.6)=0.0221$ , $f(0.7)=-0.0862$ . What is the conclusion?: $f$ is continuous and changes sign, so there is a root in $0.6<x<0.7$ .
Why does $f(x)=\frac{1}{x-1}$ change sign between $x=0$ and $x=2$ without a root?: It is not continuous at $x=1$ (an asymptote), and $\frac{1}{x-1}$ is never zero.
How can a root be missed when there is no sign change?: A repeated root (the graph touches the axis), or an even number of roots in the interval such as two.
For $f(x)=(x-2)^2(x+2)$ , why is $f(1)=3$ and $f(3)=5$ misleading?: There is a repeated root at $x=2$ in $[1,3]$ ; $f$ does not change sign there.
A sign change over a continuous interval shows what about the number of roots?: An odd number of roots, so at least one, but not necessarily exactly one.
Which type of function often causes a false sign change?: One with an asymptote, such as $\tan x$ or $\frac{1}{x-a}$ .
How can missing two close roots be avoided?: Use a smaller interval, or evaluate $f$ at more points in the interval.
What happens to the interval width in each bisection step?: It halves.

Exam questions on Locating roots by change of sign

The function $f(x)=x^3-5x+3$ is continuous for all real $x$ .
Show that $f(x)=0$ has a root between $x=0.6$ and $x=0.7$ .2 marks
Two functions are defined for $x\ne1$ by $f(x)=\frac{1}{x-1}$ and for all real $x$ by $g(x)=(x-1)^2$ .
Explain why the change of sign of $f$ between $x=0$ and $x=2$ does not show that $f(x)=0$ has a root in that interval.2 marks
The equation $e^x=3x$ is written as $f(x)=0$ , where $f(x)=e^x-3x$ .
Show that $f(x)=0$ has a root $\alpha$ in the interval $0.6<x<0.7$ .3 marks

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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).