Why must continuity be mentioned?

A discontinuous function such as \frac{1}{x-2} can change sign without crossing zero.

Can a root exist with no sign change?

Yes. A repeated root, such as x=2 in (x-2)^2, touches the axis without a sign change.

How does interval bisection work?

Evaluate f at the midpoint and keep the half where f changes sign.

What does linear interpolation assume?

That f is approximately a straight line between (a,f(a)) and (b,f(b)).

Linear interpolation formula with f(a)<0<f(b)?

x=a+\frac{|f(a)|}{|f(a)|+|f(b)|}(b-a)

How do you show a root is correct to 3 d.p.?

Show f changes sign between the two values half a unit in the last place either side, e.g. 2.2075 and 2.2085.

Which is usually faster, bisection or linear interpolation?

Linear interpolation, if the curve is nearly straight on the interval.

Interval bisection and linear interpolationEdexcel International A Level Further Maths: Flashcards

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State the change-of-sign test for a root.

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State the change-of-sign test for a root.: If $f$ is continuous on $[a,b]$ and $f(a)$ , $f(b)$ have opposite signs, then $f(x)=0$ has a root in $[a,b]$ .
Why must continuity be mentioned?: A discontinuous function such as $\frac{1}{x-2}$ can change sign without crossing zero.
Can a root exist with no sign change?: Yes. A repeated root, such as $x=2$ in $(x-2)^2$ , touches the axis without a sign change.
How does interval bisection work?: Evaluate $f$ at the midpoint and keep the half where $f$ changes sign.
Width of the interval after $n$ bisections of $[a,b]$ ?: $\frac{b-a}{2^n}$
Midpoint of $[a,b]$ ?: $\frac{a+b}{2}$
What does linear interpolation assume?: That $f$ is approximately a straight line between $(a,f(a))$ and $(b,f(b))$ .
Linear interpolation formula with $f(a)<0<f(b)$ ?: $x=a+\frac{|f(a)|}{|f(a)|+|f(b)|}(b-a)$
$f(1)=-3$ , $f(2)=5$ : first interpolation estimate?: $1+\frac{3}{8}=1.375$
How do you show a root is correct to 3 d.p.?: Show $f$ changes sign between the two values half a unit in the last place either side, e.g. $2.2075$ and $2.2085$ .
Which is usually faster, bisection or linear interpolation?: Linear interpolation, if the curve is nearly straight on the interval.
What mode must a calculator be in for $f(x)=\cos x-x$ ?: Radians.

Exam questions on Interval bisection and linear interpolation

Let $f(x)=x^3+2x-7$ for real $x$ .
Show that the equation $f(x)=0$ has a root in the interval $[1,2]$ .2 marks
Two functions are defined for real $x$ : $f(x)=\frac{1}{x-2}$ (for $x\neq2$ ) and $g(x)=(x-2)^2$ .
Explain why the change of sign of $f$ between $x=1$ and $x=3$ does not show that $f(x)=0$ has a root in $[1,3]$ .2 marks
The equation $x^3+x-5=0$ has a single real root $\alpha$ . Let $f(x)=x^3+x-5$ .
Use linear interpolation on the interval $[1,2]$ to find a first approximation to $\alpha$ .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).