Mathematics 9709 · Pure Mathematics 2

Locating a Root: Sign Change and Sketching a Pair of Graphs

Some equations, like $x=e^x-3$ , have no algebra answer at all. You can't solve for the root, so you find roughly where it is instead.
The locate stage does two things. It shows a root exists, and it shows roughly where it is.
Two easy-mark tools do this. The words in the question tell you which one to use.

“Show there is only one root” → sketch a pair of graphs and count where they cross. “Show a root lies between p and q” → the change-of-sign test. Almost every iteration question starts with one of these two. Two easy marks.

Sketch a pair of graphs to count the roots

The curve $y=e^x-3-x$ is hard to draw. So split the equation into two curves you already know, and see where they meet.
Rewrite $e^x-3=x$ as $e^x=x+3$ , then sketch $y=e^x$ and the line $y=x+3$ on one diagram.
Each place they cross is one root.

y = eˣ and the line y = x + 3 cross just once, so there is only one root.

For the marks, draw both curves with the right shape and show the one meeting point clearly.
Split it so both pieces are graphs you already know how to draw (see the curve-sketching notes for the standard shapes).

Common mistake

Pick a sensible split.

e^x=x+3

is good because

y=e^x

and

y=x+3

are both shapes you know. A bad split leaves you with two curves you can't draw. Then the “only one root” argument falls apart.

The change-of-sign test: a root between p and q

Move everything to one side as $f(x)=0$ . Now check the two ends. If $f(p)$ and $f(q)$ have opposite signs (one plus, one minus), and $f$ has no break in between, a root lies between $p$ and $q$ .

Picture it: if the curve is below the axis at $p$ and above it at $q$ , it had to cross the axis somewhere between them. A crossing is a root.
First move everything to one side. Then you only test the sign of one $f(x)$ .

Worked example

Real question: show $x=e^x-3$ has a root between $1$ and $2$

9709/31 O/N 2023 Q8(b): the locate stage of the iteration question this whole chapter is built on.

Get a single function equal to zero: $f(x)=e^x-3-x$ .
Evaluate at each end: $f(1)=e^1-3-1=-1.28\ldots$ (negative), $f(2)=e^2-3-2=2.39\ldots$ (positive).
State the conclusion in words — this is the line that earns the mark.

Answer

f(1)<0,\ f(2)>0:\ \text{change of sign}\Rightarrow\text{root between }1\text{ and }2

Copy that conclusion sentence word for word. The two numbers alone score nothing. (9709/31/O/N/23 Q8)

Worked example

Real question: verify $\operatorname{cosec}(\tfrac12 x)=e^x-3$ has a root between $1$ and $2$

9709/31 M/J 2024 Q6(b). Same steps, but with a trig term. One trap: this only works in radians.

One side as a single function: $f(x)=\operatorname{cosec}(\tfrac12 x)-e^x+3$ .
In radians: $f(1)=2.37\ldots$ (positive), $f(2)=-3.20\ldots$ (negative). The sign flips the other way this time, but a flip is a flip.

Answer

f(1)>0,\ f(2)<0:\ \text{change of sign}\Rightarrow\text{root between }1\text{ and }2

Positive then negative is fine. The conclusion is exactly the same. (9709/31/M/J/24 Q6)

Common mistake

The most common dropped mark in this topic: writing

f(1)

and

f(2)

but forgetting the conclusion. Always finish with “change of sign, so there is a root between

p

and

q

”. If there is any trig, set your calculator to radians first. Degrees gives the wrong sign.

One catch: the curve must have no break across the gap

A sign change only means a root if the graph has no break across that gap. For example $\tan x$ flips sign as it jumps to $\pm\infty$ on either side of $x=\tfrac{\pi}{2}$ . That is a break, not a root. Every function in this chapter has no break. So this is just a warning to know about, not something they test. (PM2&3 coursebook Explore 6.1)

Your turn— tap to reveal the worked answer (9709/31/O/N/23 Q8(b))

For $f(x)=e^x-3-x$ you work out $f(1)=-1.28\ldots$ and $f(2)=+2.39\ldots$ . Write the conclusion the way the mark scheme wants it.

Answer

f(1)<0,\ f(2)>0:\ \text{change of sign}\Rightarrow\text{root between }1\text{ and }2

Choosing the tool from the command word

The question says…	Use	What earns the marks
“sketch a suitable pair of graphs” / “only one root”	Pair of graphs	both curves drawn, single intersection shown
“show a root lies between $p$ and $q$ ”	Change of sign	$f(p),\,f(q)$ + the conclusion sentence

Both are just the warm-up.
Once you have found a root, the iteration in §6.2 turns that rough range into an accurate answer.

PreviousThe trapezium rule NextRearranging f(x)=0 to x=F(x), and Proving It Converges to the Root

Locating a Root: Sign Change and Sketching a Pair of Graphs

Some equations, like $x=e^x-3$ , have no algebra answer at all. You can't solve for the root, so you find roughly where it is instead.
The locate stage does two things. It shows a root exists, and it shows roughly where it is.
Two easy-mark tools do this. The words in the question tell you which one to use.

Sketch a pair of graphs to count the roots

The curve $y=e^x-3-x$ is hard to draw. So split the equation into two curves you already know, and see where they meet.
Rewrite $e^x-3=x$ as $e^x=x+3$ , then sketch $y=e^x$ and the line $y=x+3$ on one diagram.
Each place they cross is one root.

y = eˣ and the line y = x + 3 cross just once, so there is only one root.

For the marks, draw both curves with the right shape and show the one meeting point clearly.
Split it so both pieces are graphs you already know how to draw (see the curve-sketching notes for the standard shapes).

Common mistake

Pick a sensible split.

e^x=x+3

is good because

y=e^x

and

y=x+3

are both shapes you know. A bad split leaves you with two curves you can't draw. Then the “only one root” argument falls apart.

The change-of-sign test: a root between p and q

Picture it: if the curve is below the axis at $p$ and above it at $q$ , it had to cross the axis somewhere between them. A crossing is a root.
First move everything to one side. Then you only test the sign of one $f(x)$ .

Worked example

Real question: show $x=e^x-3$ has a root between $1$ and $2$

9709/31 O/N 2023 Q8(b): the locate stage of the iteration question this whole chapter is built on.

Get a single function equal to zero: $f(x)=e^x-3-x$ .
Evaluate at each end: $f(1)=e^1-3-1=-1.28\ldots$ (negative), $f(2)=e^2-3-2=2.39\ldots$ (positive).
State the conclusion in words — this is the line that earns the mark.

Answer

f(1)<0,\ f(2)>0:\ \text{change of sign}\Rightarrow\text{root between }1\text{ and }2

Copy that conclusion sentence word for word. The two numbers alone score nothing. (9709/31/O/N/23 Q8)

Worked example

Real question: verify $\operatorname{cosec}(\tfrac12 x)=e^x-3$ has a root between $1$ and $2$

9709/31 M/J 2024 Q6(b). Same steps, but with a trig term. One trap: this only works in radians.

One side as a single function: $f(x)=\operatorname{cosec}(\tfrac12 x)-e^x+3$ .
In radians: $f(1)=2.37\ldots$ (positive), $f(2)=-3.20\ldots$ (negative). The sign flips the other way this time, but a flip is a flip.

Answer

f(1)>0,\ f(2)<0:\ \text{change of sign}\Rightarrow\text{root between }1\text{ and }2

Positive then negative is fine. The conclusion is exactly the same. (9709/31/M/J/24 Q6)

Common mistake

The most common dropped mark in this topic: writing

f(1)

and

f(2)

but forgetting the conclusion. Always finish with “change of sign, so there is a root between

p

and

q

”. If there is any trig, set your calculator to radians first. Degrees gives the wrong sign.

One catch: the curve must have no break across the gap

Your turn— tap to reveal the worked answer (9709/31/O/N/23 Q8(b))

For $f(x)=e^x-3-x$ you work out $f(1)=-1.28\ldots$ and $f(2)=+2.39\ldots$ . Write the conclusion the way the mark scheme wants it.

Answer

f(1)<0,\ f(2)>0:\ \text{change of sign}\Rightarrow\text{root between }1\text{ and }2

Choosing the tool from the command word

The question says…	Use	What earns the marks
“sketch a suitable pair of graphs” / “only one root”	Pair of graphs	both curves drawn, single intersection shown
“show a root lies between $p$ and $q$ ”	Change of sign	$f(p),\,f(q)$ + the conclusion sentence

Both are just the warm-up.
Once you have found a root, the iteration in §6.2 turns that rough range into an accurate answer.