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Linear interpolation

Edexcel IAL Further Mathematics · WFM01 Further Pure Mathematics 1

Linear interpolation improves on bisection by using the sizes of f(a) and f(b), not just their signs. It joins (a, f(a)) and (b, f(b)) with a straight line and takes the point where that line crosses the x-axis.

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The idea

If f(a) < 0 and f(b) > 0, the chord from (a, f(a)) to (b, f(b)) crosses the x-axis at an estimate x_1 of the root. If |f(a)| is much smaller than |f(b)|, the root is probably nearer a.

Similar triangles

The two triangles either side of x_1 are similar:

(x_1 - a)/|f(a)| = (b - x_1)/|f(b)|.

Solving gives x_1 = a + |f(a)|(b - a)/(|f(a)| + |f(b)|), or in signed form x_1 = a - f(a)(b - a)/(f(b) - f(a)). Show the similar triangles equation in an exam.

Repeating

Evaluate f(x_1). Its sign tells you which of [a, x_1] and [x_1, b] contains the root. Interpolate again on that interval. Give answers to the accuracy asked, and keep more figures in your working.

Worked example

Use linear interpolation once on [2, 3] to estimate the root of x3 - 2x - 5 = 0.

  1. f(2) = -1 and f(3) = 16.
  2. (x_1 - 2)/1 = (3 - x_1)/16, so 16x_1 - 32 = 3 - x_1.
  3. 17x_1 = 35.

Answer: x_1 = 35/17 = 2.0588

Key idea: Linear interpolation: (x - a)/|f(a)| = (b - x)/|f(b)| from similar triangles. Then check the sign of f at the estimate and repeat.

Check yourself

  1. f(x) = x3 - 2x - 5. Use linear interpolation once on [2, 3] to estimate the root, to 3 decimal places.
    Show answer

    2.059. (x - 2)/1 = (3 - x)/16 gives x = 35/17 = 2.059.

  2. f(2.0588) is negative. Apply linear interpolation again to x3 - 2x - 5 = 0, on [35/17, 3]. Give your estimate to 3 decimal places.
    Show answer

    2.081. f(35/17) = -0.3908. Interpolating between this and f(3) = 16 gives 2.081.

  3. Use linear interpolation once on [2, 3] to estimate sqrt(7), the positive root of x2 - 7 = 0. Give your answer to 2 decimal places.
    Show answer

    2.60. f(2) = -3 and f(3) = 2. (x - 2)/3 = (3 - x)/2 gives 2x - 4 = 9 - 3x, so 5x = 13 and x = 2.6. (sqrt(7) = 2.646.)

  4. f(x) = 2x - 3x has a root in [3, 4]. Use linear interpolation once to estimate it, to 3 decimal places.
    Show answer

    3.200. f(3) = 8 - 9 = -1 and f(4) = 16 - 12 = 4. (x - 3)/1 = (4 - x)/4 gives 5x = 16, x = 3.200.

  5. f(x) = sin x - 0.5x, with x in radians, has a root in [1.5, 2]. Use linear interpolation once to estimate it, to 3 decimal places.
    Show answer

    1.866. f(1.5) = 0.2475 and f(2) = -0.0907. Interpolating gives 1.866.

  6. f(1) = -0.2 and f(2) = 5, and f is continuous. Where does linear interpolation suggest the root is?
    • Much nearer 1 than 2
    • Much nearer 2 than 1
    • Exactly at 1.5
    • Outside [1, 2]
    Show answer

    Much nearer 1 than 2. |f(1)| is small compared with |f(2)|, so the chord crosses the axis close to x = 1: at 1 + 0.2/5.2.

  7. f(a) < 0 < f(b). Which equation gives the linear interpolation estimate x?
    • (x - a)/f(b) = (b - x)/(-f(a))
    • x = (a + b)/2
    • (x - a)/(-f(a)) = (b - x)/f(b)
    • (x - a)/(b - a) = f(a)/f(b)
    Show answer

    (x - a)/(-f(a)) = (b - x)/f(b). The triangle near a has height |f(a)| = -f(a) and base x - a; the one near b has height f(b) and base b - x. They are similar.

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