Locating a root | CAIE 9709 Pure Mathematics 2 Notes

Pure Mathematics 2

Mathematics9709

Connect an equation to its graph

Key idea

A root

\alpha

f(x)=0

is a value for which

f(\alpha)=0

. The Greek letter

\alpha

, read “alpha”, is a name for the root before its value is known accurately.

On the graph of $y=f(x)$ , a root is an $x$ -coordinate where the curve meets the $x$ -axis. The curve may cross the axis or only touch it. If an equation is written as $g(x)=h(x)$ , its solutions are the $x$ -coordinates where the graphs of $y=g(x)$ and $y=h(x)$ meet. At an intersection, both graphs have the same $y$ -value for the same $x$ -value.

A root is an intersection

y = 1 + e²ˣ

y = |x − 4|

same x-value: x ≈ 0.465

The full V-shape and the rising exponential show one intersection.

In 9709/22/O/N/24 Q4, the graphs $y=1+e^{2x}$ and $y=|x-4|$ meet once. Their shared $x$ -coordinate is the root used later in the iteration. (9709/22/O/N/24 Q4(a))

Choose the pair of graphs named in the question, or split the equation into two sides that are easy to sketch.
Show the key features that identify each curve, including any intercept, turning point or asymptote needed in the stated interval. An asymptote is a line that a curve approaches without meeting in the part being considered.
Mark and count the intersections only inside the stated interval.
A sketch locates a root; it does not give its final accuracy.

Use a change of sign to locate a root

A function is continuous on an interval if its graph has no gap, jump or vertical asymptote there. If a continuous curve is below the $x$ -axis at one end and above it at the other, it must cross the axis between them.

Move every term to one side and define one function $f(x)$ , so the equation becomes $f(x)=0$ .
Calculate $f(p)$ and $f(q)$ . Opposite signs mean one value is positive and the other is negative; this is also written $f(p)f(q)<0$ .
Finish the argument: “There is a change of sign, so at least one root lies between $p$ and $q$ .”

Worked example

Locate a stationary point between 2.5 and 3

A stationary point is a point where $dy/dx=0$ . In 9709/22/M/J/24 Q6, its equation can be written as:

\begin{aligned}f(x)&=x+0.5\\&\quad-\frac{x+3}{\ln(2x+1)}\end{aligned}

f(2.5)=-0.0696\ldots

f(3)=0.4166\ldots

The values have opposite signs and $f$ is continuous on this interval. Therefore at least one root lies between $2.5$ and $3$ . (9709/22/M/J/24 Q6(c))

Examiner note

The two calculated values are evidence. The change-of-sign conclusion completes the argument.

State only what the sign test proves

What the sign test can prove

opposite signs

crossing root found

same sign

a touching root still exists

No sign change does not prove that there is no root.

For a continuous curve, a sign change proves a crossing root; a touching root may have no sign change.

Opposite signs for a continuous function prove at least one root between the endpoints. They do not prove that the root is unique.
A graph or another argument is needed to show that there is exactly one root.
A curve can touch the axis and turn back. The signs then stay the same even though a root exists.
Across a gap or vertical asymptote, signs can change without a root. This is why continuity must be checked.

Your turnIndependent check [3 marks]

Show by calculation that $x^3+x-4=0$ has a root between $1$ and $2$ . State exactly what the calculation proves.

Show worked answer

Let $f(x)=x^3+x-4$ . This polynomial is continuous.

f(1)=-2,\qquad f(2)=6

There is a change of sign, so at least one root lies between 1 and 2. The calculation alone does not prove that the root is unique.