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Locating roots and interval bisection

Edexcel IAL Further Mathematics · WFM01 Further Pure Mathematics 1

Many equations f(x) = 0 cannot be solved exactly. Numerical methods home in on a root. The first step is to show that a root lies in an interval; interval bisection then halves that interval again and again.

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Sign change

If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then f(x) = 0 has at least one root between a and b. Always write a conclusion: "sign change and f continuous, so there is a root in [a, b]".

Care: f(x) = 1/x changes sign between -1 and 1 but has no root, because it is not continuous at x = 0. An even number of roots in an interval can give no sign change.

Interval bisection

1. Find the midpoint m = (a + b)/2 and evaluate f(m).

2. Keep the half where the sign changes: [a, m] if f(a) and f(m) differ in sign, otherwise [m, b].

3. Repeat. Each step halves the width, so after n steps the width is (b - a)/2n.

abmf(m)
232.55.6250
22.52.251.8906
22.252.1250.3457

f(x) = x3 - 2x - 5, starting with [2, 3]

Good practice

Set out a table with a, b, the midpoint and the sign (or value) of f at the midpoint. Use radians for any trigonometric function. Bisection always works once you have a sign change, but it is slow.

Worked example

Show that x3 - 2x - 5 = 0 has a root in [2, 3] and use interval bisection three times to find a smaller interval containing it.

  1. f(2) = -1 < 0 and f(3) = 16 > 0. f is continuous, so there is a root in [2, 3].
  2. f(2.5) = 5.6250 > 0, so the root is in [2, 2.5].
  3. f(2.25) = 1.8906 > 0, so the root is in [2, 2.25].
  4. f(2.125) = 0.3457 > 0, so the root is in [2, 2.125].

Answer: The root lies in [2, 2.125]

Key idea: Sign change of a continuous function locates a root. Bisection keeps the half-interval with the sign change; the width halves each time.

Check yourself

  1. f(x) = x3 - 2x - 5. Find f(2).
    Show answer

    -1. 8 - 4 - 5 = -1.

  2. f(x) = x3 - 2x - 5 has a root in [2, 3]. Find f(2.5), the first value used in interval bisection.
    Show answer

    5.625. f(2.5) = 15.625 - 5 - 5 = 5.625.

  3. Continue the bisection of x3 - 2x - 5 = 0 from [2, 2.125]. Evaluate f(2.0625), to 4 decimal places.
    Show answer

    -0.3513. f(2.0625) = -0.3513, which is negative, so the root is in [2.0625, 2.125].

  4. Interval bisection on x2 - 7 = 0 starts with [2, 3]. Find the upper end of the interval after three bisections.
    Show answer

    2.75. f(2.5) = -0.75 < 0: [2.5, 3]. f(2.75) = 0.5625 > 0: [2.5, 2.75]. f(2.625) = -0.109375 < 0: [2.625, 2.75].

  5. Bisection starts on an interval of width 1. What is the least number of bisections needed to make the interval width less than 0.01?
    Show answer

    7. After n steps the width is 1/2n. 1/26 = 0.015625 and 1/27 = 0.0078125, so 7 bisections.

  6. f(x) = 1/x satisfies f(-1) < 0 and f(1) > 0. Why does this not show a root in [-1, 1]?
    • The interval is too wide
    • f(0) is positive
    • f is not continuous on [-1, 1]
    • f is an even function
    Show answer

    f is not continuous on [-1, 1]. The sign change rule only works for a continuous function. 1/x has a vertical asymptote at x = 0 and never equals 0.

  7. f(a) and f(b) are both positive. Which statement is correct?
    • There could be an even number of roots in [a, b]
    • There is definitely no root in [a, b]
    • There is exactly one root in [a, b]
    • f must be discontinuous on [a, b]
    Show answer

    There could be an even number of roots in [a, b]. The curve could dip below the axis and come back up, crossing twice, or touch the axis.

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Last updated 8 November 2026. Spotted a mistake? Tell us.