NM1 - Locating roots by change of sign

NM1 - Locating roots by change of sign

Some equations have awkward roots even when the equation itself is short. NM1 gives you a dependable way to trap a root between two numbers, narrow the gap and justify a rounded answer. The signs of the answers matter more than their size.

Why a change of sign traps a root

Where does the curve y=x3−x−4y=x^3-x-4 cross the xx-axis between 11 and 22? Neither integer is a solution, and an accurate-looking graph still leaves us guessing about the decimal places. We can turn that guess into a justified answer.

You already know

A root of f(x)=0\mathrm{f}(x)=0 is the xx-coordinate of a point where y=f(x)y=\mathrm{f}(x) meets the xx-axis. Rearrange an equation so one side is zero before testing signs.

For the opening curve, f(1)=1−1−4=−4\mathrm{f}(1)=1-1-4=-4, below the axis, whereas f(2)=8−2−4=2\mathrm{f}(2)=8-2-4=2, above it. A curve that stays connected between these points must pass through height zero along the way. It cannot get from below the floor to above the floor without crossing the floor.

[DIAGRAM: asset_name: A root trapped by opposite signs; asset_slug: nm1-sign-bracket; caption: Opposite signs at the ends of a continuous piece of curve trap a root in between; description: The graph y equals x cubed minus x minus 4 from x equals 1 to 2, with the points (1, minus 4) and (2, 2) marked and a root near x equals 1.796. A shaded vertical band highlights the interval from 1 to 2]
Opposite signs at the ends of a continuous piece of curve trap a root in between

The condition about staying connected has a mathematical name.

Continuous on an interval

A function is continuous on an interval if its graph has no break, hole or jump there. Polynomial functions are continuous for every real xx. An exponential function axa^x, with a>0a>0, is also continuous.

A root need not be a whole number or a terminating decimal. The sign argument finds an interval containing it without first knowing what sort of number it is.

Change-of-sign test

f(a)f(b)<0⟹at least one root lies between a and b\mathrm{f}(a)\mathrm{f}(b)<0\quad\Longrightarrow\quad \text{at least one root lies between }a\text{ and }b

This requires a<ba<b and f\mathrm{f} continuous throughout [a,b][a, b]. If an endpoint value is zero, that endpoint is already an exact root.
Not on the formulae sheet: learn it.

Here is a useful surprise: repeatedly halving a bracket only 1010 times makes it 10241024 times narrower, because 210=10242^{10}=1024. Quite modest arithmetic can replace a very large graph. You do not have to halve the interval; testing convenient tenths and hundredths often suits an exam question better.

Key idea

Opposite signs prove that at least one root is inside a continuous interval. They do not prove there is exactly one root, and the endpoint values are not the roots.

Read the conclusion as carefully as the calculation.

Narrowing the interval

To improve an estimate, test new values inside the bracket and keep endpoints with opposite signs.

A table makes the evidence easy to follow. Write the function you are using, keep enough digits to show each sign, then state what the signs establish. A tiny negative answer is still negative: rounding it to zero can throw away the evidence.

Worked example 1: narrowing to hundredths

Question. By change of sign, locate the root of x3−x−4=0x^3-x-4=0 between consecutive hundredths.

Plan. Start with [1,2][1, 2], test tenths and then test hundredths inside the smaller bracket.

f(1.7)=1.73−1.7−4=−0.787negativef(1.8)=1.83−1.8−4=0.032positivef(1.79)=1.793−1.79−4=−0.054661test inside the bracket\begin{aligned} \mathrm{f}(1.7)&=1.7^3-1.7-4=-0.787 &&\text{negative}\\ \mathrm{f}(1.8)&=1.8^3-1.8-4=0.032 &&\text{positive}\\ \mathrm{f}(1.79)&=1.79^3-1.79-4=-0.054661 &&\text{test inside the bracket} \end{aligned}

Why test 1.791.79? We already know 1.81.8 gives a positive answer. A negative answer at 1.791.79 would trap the root between consecutive hundredths without a long table.

Answer. There is a change of sign between 1.791.79 and 1.801.80. Since f\mathrm{f} is continuous, a root lies in (1.79,1.80)(1.79, 1.80).

Check. The new interval has width 0.010.01, one hundredth of the original width 11.

The second example changes only the constant in the equation. The bookkeeping stays the same, but the useful interval moves.

Worked example 2: the same search at a different root

Question. By change of sign, locate the root of x3−x−5=0x^3-x-5=0 between consecutive hundredths.

Plan. Evaluate g(x)=x3−x−5\mathrm{g}(x)=x^3-x-5 near the positive crossing and retain opposite signs.

g(1)=−5,g(2)=1initial bracketg(1.9)=1.93−1.9−5=−0.041negativeg(1.91)=1.913−1.91−5=0.057871positive\begin{aligned} \mathrm{g}(1)&=-5,\qquad \mathrm{g}(2)=1 &&\text{initial bracket}\\ \mathrm{g}(1.9)&=1.9^3-1.9-5=-0.041 &&\text{negative}\\ \mathrm{g}(1.91)&=1.91^3-1.91-5=0.057871 &&\text{positive} \end{aligned}

Why stop there? The endpoints are consecutive hundredths and have opposite signs, exactly what the question requests.

Answer. Continuity and the change of sign show that a root lies in (1.90,1.91)(1.90, 1.91).

Check. The function has risen from negative to positive across the stated interval; neither endpoint is claimed as a root.

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