Root location and fixed-point iteration

What you’ll learn

• Locate a root using a sign change and continuity.
• Carry out a stated iterative procedure.
• Justify a rounded numerical root.

Understand the idea

If a continuous function has opposite signs at two endpoints, it has at least one zero between them. Continuity matters: a sign change across a vertical asymptote does not establish a root. Smaller sign-changing intervals locate a root more precisely but do not automatically show it is unique.

Rearranging an equation as $x=g(x)$ gives an iteration $x_{n+1}=g(x_n)$. Start from the prescribed value, retain calculator precision and record the requested approximations. Some rearrangements converge and others do not; approaching a stable value is evidence to inspect, not a reason to assume every rearrangement works.

To justify an answer to a given number of decimal places, evaluate the original continuous function at the two rounding boundaries. Opposite signs establish a root inside the interval that rounds to the stated answer. Substituting the rounded answer alone only checks that its residual is small.

Key formulae

$$f(a)f(b)<0\text{ and }f\text{ continuous on }[a,b]\Rightarrow\exists c\in(a,b):f(c)=0$$

$$x_{n+1}=g(x_n)$$

$$\alpha\in\left(r-\frac12\times10^{-d},\ r+\frac12\times10^{-d}\right)\Rightarrow\alpha\text{ rounds to }r\text{ at }d\text{ d.p.}$$

Worked example

For $f(x)=x^3-x-3$, show that there is a root between $1$ and $2$, then calculate $x_1,x_2$ using $x_{n+1}=\sqrt[3]{x_n+3}$ and $x_0=1.5$.

1. $f(1)=-3$ and $f(2)=3$. As a polynomial, $f$ is continuous, so a root lies between them.
2. $x_1=\sqrt[3]{4.5}=1.650963\ldots$.
3. Using the unrounded value, $x_2=\sqrt[3]{x_1+3}=1.66922\ldots$.

Answer: A root lies in $(1,2)$; $x_1\approx1.65096$, $x_2\approx1.66922$.