I: Numerical methods: Locate roots of f x = 0 by considering changes of sign of f x in an interval of x on which f x is sufficiently well-behaved. Understand how change of sign methods can fail.

The idea

To locate a root of f(x) = 0, evaluate f at two points a and b. If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then by the intermediate value theorem there is at least one root between a and b. You then repeatedly bisect the interval, testing the midpoint, to narrow down the root. However, change of sign methods can fail: if f is not continuous, a sign change may not indicate a root; if there is an even number of roots in the interval, the signs at the endpoints may be the same; or if the interval is too large, roots may be missed. Also, a root may exist without a sign change if the graph touches the axis.

What earns the marks

• Evaluates f(a) and f(b) correctly and states that a change of sign indicates a root in (a, b) provided f is continuous.
• Performs at least one iteration of the bisection method correctly, computing the midpoint and evaluating f there.
• States the new interval containing the root after an iteration, ensuring the sign change is maintained.
• Explains a situation where the change of sign method fails, such as a discontinuity, an even number of roots, or a repeated root.
• Uses appropriate accuracy and rounding when evaluating f, and continues until the interval is sufficiently small.
• Justifies the existence of a root using the intermediate value theorem, referring to continuity.

Common mistakes and how to put them right

• Assuming a sign change always means a root without checking continuity.

  Put it right: For example, f(x) = 1/x has a sign change at x = 0 but no root.

• Stopping after one iteration without narrowing the interval sufficiently.

  Put it right: Continue until the required accuracy is reached.

• Failing to maintain the sign change when choosing the new interval.

  Put it right: Check which half contains the sign change.

• Ignoring the possibility of multiple roots in the interval.

  Put it right: If there are two roots, the signs at the endpoints may be the same, so no sign change is detected.

Exam tips

• Show the values of f(a) and f(b) clearly, and state that there is a change of sign, so a root lies between a and b.
• When using bisection, present your working in a table with columns for a, b, midpoint, f(midpoint), and the new interval.
• If asked to explain why the method might fail, give a specific example or describe a scenario such as a discontinuity or an even number of roots.
• Use the language of the intermediate value theorem correctly: mention continuity on the interval and the sign change.