Showing a root exists — change of sign

Why a sign change in a continuous function traps a root between two values, how to present the argument for full marks, and how the same idea justifies a root to a given accuracy.

Algebra · Iteration

Showing a root exists — change of sign

Why a sign change in a continuous function traps a root between two values, how to present the argument for full marks, and how the same idea justifies a root to a given accuracy.

Why it works

A root of an equation

f(x) = 0

is where the graph of

y = f(x)

crosses the

x

-axis. If

f

is negative at one value and positive at another, the graph starts below the axis and ends above it — and a smooth (continuous) curve cannot jump the axis without touching it. So:

f(a) < 0 \;\text{ and }\; f(b) > 0 \;\Rightarrow\; \text{a root lies between } a \text{ and } b.

The argument has three compulsory parts — and the marks track them:

Evaluate $f$ at both ends, showing the values:

for

f(x) = x^3 - 3x - 5

f(2) = 8 - 6 - 5 = -3

and

f(3) = 27 - 9 - 5 = 13

Name the sign change: one negative, one positive.
Conclude: "there is a change of sign, so a root lies between 2

and 3."

Numbers without the conclusion, or a conclusion without the numbers, both drop marks. Substituting into the WRONG thing — the iteration formula instead of

f

— proves nothing about

f

's roots.

The interval can be as narrow as you like. The same test with

f(2.2) = -0.952

and

f(2.3) = 0.267

pins the root between 2.2 and 2.3 — one decimal place of precision.

Justifying a rounded answer — the grade-9 twist. To show the root is

2.28

correct to 2 decimal places, test the two values that bound everything-that-rounds-to-2.28:

f(2.275) \approx -0.05 < 0

and

f(2.285) \approx +0.08 > 0

. The sign change traps the root inside

(2.275, 2.285)

— and every number in that interval rounds to

2.28