Roots in Intervals (Edexcel International AS Further Maths)

Revision Note

Mark Curtis

Written by: Mark Curtis

Reviewed by: Dan Finlay

Roots in Intervals

What is a root of an equation?

  • A root of the equation straight f open parentheses x close parentheses equals 0 is a solution

    • If x equals alpha is a root of straight f open parentheses x close parentheses equals 0 then straight f open parentheses alpha close parentheses equals 0

      • The Greek letter alpha is often used

  • To find alpha exactly, you need to solve the equation algebraically (analytically)

  • Some equations cannot be solved algebraically

    • In which case you can approximate a root to a given accuracy

      • For example, to 3 decimal places

    • This is called solving equations numerically

What is an interval?

  • If you don't know the root but know that it lies between x equals a and x equals b, where a less than b, then

    • a less than x less than b is an interval containing the root

  • Intervals can be written using bracket notation

    • open parentheses a comma space b close parentheses is a less than x less than b

    • open square brackets a comma space b close square brackets is a less or equal than x less or equal than b

How do I show that an interval contains a root?

  • Use the sign-change and continuity test to show that the interval a less than x less than b contains a root to the equation straight f open parentheses x close parentheses equals 0

  • Show that straight f open parentheses a close parentheses and straight f open parentheses b close parentheses have different signs and check that straight f open parentheses x close parentheses is continuous across a less than x less than b

    • Then the interval a less than x less than b must contains a root

    • Continuous means no jumps or asymptotes in the interval

      • Check the equation does not divide by zero within the interval

      • (Jumps or asymptotes outside of the interval are fine)

    • You must write a conclusion to the test in words, for example:

      • "straight f open parentheses x close parentheses has a sign change in the interval a less than x less than b and straight f open parentheses x close parentheses is continuous in the interval, so a root must lie in the interval a less than x less than b"

Examiner Tips and Tricks

When writing your conclusion in the exam, don't forget to mention straight f open parentheses x close parentheses being continuous in the interval!

How do I show that an equation has a root to a given accuracy?

  • If asked to show that x equals 1.39 is the root of an equation to 2 decimal places

    • write an interval using the lower and upper bound of the root

      • 1.385 less than x less than 1.395

    • use the sign-change and continuity test on this interval

How to use the sign change test to show that a root is true to a given accuracy
  • A suitable conclusion would be

    • straight f open parentheses x close parentheses has a sign change in the interval 1.385 less than x less than 1.395

    • and straight f open parentheses x close parentheses is continuous in the interval

    • so a root must lie in the interval 1.385 less than x less than 1.395

    • All values in the interval round to 1.39

    • so the root is 1.39 to 2 decimal places

How do I know when the sign-change test fails?

  • You need to know the three cases when the sign-change and continuity test fails to work properly:

    • If the curve y equals straight f open parentheses x close parentheses only touches the x-axis at the root, you will not see a sign change (even though there is a root)

    • If the curve y equals straight f open parentheses x close parentheses has an asymptote in the interval, then you may see a sign change (but there is no root)

      • The asymptote means straight f open parentheses x close parentheses is not continuous

    • If the interval is too large, there may be more than one root in it

      • You may see a sign change (an odd number of roots)

      • Or you may not see a sign change (an even number of roots)

A diagram showing the cases when the change of sign test fails

Worked Example

The equation straight f open parentheses x close parentheses equals 0 where straight f open parentheses x close parentheses equals x minus x to the power of 4 plus fraction numerator 1 over denominator 2 x minus 1 end fraction has exactly one positive root.

(a) Show that the intervals open square brackets 0 comma space 1 close square brackets and open square brackets 1 comma space 2 close square brackets both have a change of sign in straight f open parentheses x close parentheses.

Substitute x equals 0, x equals 1 and x equals 2 into straight f open parentheses x close parentheses

table row cell straight f open parentheses 0 close parentheses end cell equals cell 0 minus 0 to the power of 4 plus fraction numerator 1 over denominator 2 cross times 0 minus 1 end fraction equals negative 1 end cell row cell straight f open parentheses 1 close parentheses end cell equals cell 1 minus 1 to the power of 4 plus fraction numerator 1 over denominator 2 cross times 1 minus 1 end fraction equals 1 end cell row cell straight f open parentheses 2 close parentheses end cell equals cell 2 minus 2 to the power of 4 plus fraction numerator 1 over denominator 2 cross times 2 minus 1 end fraction equals negative 41 over 3 end cell end table

Comment that the signs are different for each interval

straight f open parentheses 0 close parentheses equals negative 1 less than 0 and straight f open parentheses 1 close parentheses equals 1 greater than 0 so there is a sign change in open square brackets 0 comma space 1 close square brackets
straight f open parentheses 1 close parentheses equals 1 greater than 0 and straight f open parentheses 2 close parentheses equals negative 41 over 3 less than 0 so there is a sign change in open square brackets 1 comma space 2 close square brackets

(b) Determine, with a reason, which of the intervals contains the root.

An interval contains a root if straight f open parentheses x close parentheses has a sign change and is continuous in that interval
Check to see if straight f open parentheses x close parentheses is continuous in open square brackets 0 comma space 1 close square brackets
You need to look for any asymptotes

fraction numerator 1 over denominator 2 x minus 1 end fraction is undefined when x equals 1 half

The asymptote x equals 1 half lies in the interval open square brackets 0 comma space 1 close square brackets
There are no other asymptotes and only one root

The interval open square brackets 1 comma space 2 close square brackets contains the root as there is a sign change and straight f open parentheses x close parentheses is continuous
The interval open square brackets 0 comma space 1 close square brackets contains the asymptote x equals 1 half so straight f open parentheses x close parentheses is not continuous

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Mark Curtis

Author: Mark Curtis

Expertise: Maths

Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

Dan Finlay

Author: Dan Finlay

Expertise: Maths Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.