Roots in Intervals (Edexcel International AS Further Maths: Further Pure 1): Revision Note

Roots in Intervals

What is a root of an equation?

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

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

    • 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 $x=a$ and $x=b$, where $a<b$, then

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

• Intervals can be written using bracket notation

  • $\left(a,b\right)$ is $a<x<b$

  • $\left[a,b\right]$ is $a\le x\le b$

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 $\mathrm{f}\left(x\right)=0$

• Show that $\mathrm{f}\left(a\right)$ and $\mathrm{f}\left(b\right)$ have different signs and check that $\mathrm{f}\left(x\right)$ 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:

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

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

  • $\mathrm{f}\left(x\right)$ has a sign change in the interval $1.385<x<1.395$

  • and $\mathrm{f}\left(x\right)$ 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 $y=\mathrm{f}\left(x\right)$ only touches the $x$-axis at the root, you will not see a sign change (even though there is a root)

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

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