Roots in Intervals (Edexcel IAL Further Maths)

Spec point — 3.1

Roots in Intervals

What is a root of an equation?

• A root of the equation $\mathrm{f} ( x ) = 0$ is a solution

  • If $x=α$ is a root of $\mathrm{f} ( x ) = 0$ then $\mathrm{f} ( \\alpha ) = 0$

    • The Greek letter alpha is often used

• To find $α$ 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 $x=a$ and $x=b$, where $a<b$, then

  • $a<x<b$ is an interval containing the root

• Intervals can be written using bracket notation

  • $( a , b )$ is $a<x<b$
  • $[ a , b ]$ is $a\leq x\leq b$

How do I show that an interval contains a root?

• Use the sign-change and continuity test to show that the interval$a<x<b$ contains a root to the equation $\mathrm{f} ( x ) = 0$

• Show that $\mathrm{f} ( a )$ and $\mathrm{f} ( b )$ have different signs and check that $\mathrm{f} ( x )$ is continuous across $a<x<b$

  • Then the interval $a<x<b$ 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:

    • "$\mathrm{f} ( x )$ has a sign change in the interval $a<x<b$ and $\mathrm{f} ( x )$ is continuous in the interval, so a root must lie in the interval $a<x<b$"

Exam Hint When writing your conclusion in the exam, don't forget to mention $\mathrm{f} ( x )$ being continuous in the interval!

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

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

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

    • $1.385<x<1.395$

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

• A suitable conclusion would be

  • $\mathrm{f} ( x )$ has a sign change in the interval $1.385<x<1.395$
  • and $\mathrm{f} ( x )$ is continuous in the interval
  • so a root must lie in the interval $1.385<x<1.395$
  • 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 $y = \mathrm{f} ( x )$ only touches the $x$-axis at the root, you will not see a sign change (even though there is a root)

  • If the curve $y = \mathrm{f} ( x )$ has an asymptote in the interval, then you may see a sign change (but there is no root)

    • The asymptote means $\mathrm{f} ( x )$ 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)

Worked Example The equation $\mathrm{f} ( x ) = 0$ where $\mathrm{f} ( x ) = x - x^{4} + \frac{1}{2 x - 1}$ has exactly one positive root.

(a) Show that the intervals $[ 0 , 1 ]$ and $[ 1 , 2 ]$ both have a change of sign in $\mathrm{f} ( x )$.

Substitute $x=0$, $x=1$ and $x=2$ into $\mathrm{f} ( x )$

$\begin{gathered} \mathrm{f} ( 0 ) & = 0 - 0^{4} + \frac{1}{2 \times 0 - 1} = - 1 \\ \mathrm{f} ( 1 ) & = 1 - 1^{4} + \frac{1}{2 \times 1 - 1} = 1 \\ \mathrm{f} ( 2 ) & = 2 - 2^{4} + \frac{1}{2 \times 2 - 1} = - \frac{41}{3} \end{gathered}$

Comment that the signs are different for each interval

Final answer: $\mathrm{f} ( 0 ) = - 1 < 0$and$\mathrm{f} ( 1 ) = 1 > 0$so there is a sign change in$[ 0 , 1 ]$ $\mathrm{f} ( 1 ) = 1 > 0$and$\mathrm{f} ( 2 ) = - \frac{41}{3} < 0$so there is a sign change in$[ 1 , 2 ]$

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

An interval contains a root if $\mathrm{f} ( x )$ has a sign change and is continuous in that interval Check to see if $\mathrm{f} ( x )$ is continuous in $[ 0 , 1 ]$ You need to look for any asymptotes

$\frac{1}{2x-1}$ is undefined when $x=\frac{1}{2}$

The asymptote $x=\frac{1}{2}$ lies in the interval $[ 0 , 1 ]$* * There are no other asymptotes and only one root

Final answer: The interval$[ 1 , 2 ]$contains the root as there is a sign change and$\mathrm{f} ( x )$is continuous The interval$[ 0 , 1 ]$contains the asymptote$x=\frac{1}{2}$so$\mathrm{f} ( x )$is not continuous