Pure · Numerical methods

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Locating roots by change of sign

Trapping a root of f(x) = 0 in an interval using the change-of-sign rule — why continuity is essential, what a sign change does and doesn't guarantee, and refining a root to a given number of decimal places.

Pure · Numerical methods

Locating roots by change of sign

Why it works

Most equations cannot be solved exactly. There is no neat formula for the root of

x^3 - 2x - 5 = 0

, and an equation like

e^x = 3 - x

mixes a curve and a line in a way no rearrangement untangles. So instead of solving, you locate the root: first trap it in an interval, then squeeze that interval down.

A root of

f(x) = 0

is exactly where the graph

y = f(x)

crosses the $x$ -axis — where

y

changes from negative to positive (or back). That is the whole idea behind the change-of-sign rule:If

f

is continuous on

[a, b]

and

f(a)

and

f(b)

have opposite signs, then

f(x) = 0

has a root somewhere between

a

and

b

.The reasoning is simple: to travel from a value below the axis to a value above it without lifting your pen, the graph has no choice but to cross zero on the way. So to show a root lies in an interval you just evaluate

f

at each end and check the signs differ — a quick, certain test.Continuity is not optional — always state it. The "no lifting your pen" picture breaks if the graph jumps. Take

f(x) = \dfrac{1}{x}

[-1, 1]

f(-1) = -1

and

f(1) = 1

have opposite signs, yet the equation

\frac1x = 0

has no root at all. The rule fails because

f

is discontinuous at

x = 0

(an asymptote, not a crossing). For polynomials,

e^x

\ln x

and the like there is no such trouble — but you should still say "

f

is continuous on this interval" as part of the argument.

What a sign change does and doesn't tell you. A change of sign guarantees an odd number of roots in the interval — at least one. But no sign change does not guarantee no root: a curve can touch the axis without crossing (a repeated root such as

(x-2)^2

), or dip across and back, giving an even number of roots with the same sign at both ends. So sign change ⇒ root; same sign ⇒ inconclusive.

Refining to decimal places. Once a root is trapped, shrink the interval. To show a root is

2.1

to 1 decimal place, test the half-interval

[2.05, 2.15]

: a sign change there pins the root between

2.05

and

2.15

, so it rounds to

2.1

. Repeating this halving is interval bisection; the iteration and Newton–Raphson methods that follow refine far faster.