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The Newton-Raphson process

Edexcel IAL Further Mathematics · WFM01 Further Pure Mathematics 1

The Newton-Raphson process uses the tangent to the curve to improve an estimate of a root. When it works it converges very quickly, but it needs the derivative and a sensible starting value.

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The formula

Starting from x_0, the tangent to y = f(x) at (x_n, f(x_n)) crosses the x-axis at

x_(n+1) = x_n - f(x_n)/f'(x_n).

In F1, f'(x) only uses P1 and P2 differentiation: powers of x, including negative and fractional powers.

Setting out

Write down f'(x). Show the substitution for the first iteration, for example x_1 = 2 - f(2)/f'(2) = 2 - 5/20 = 1.75. Use your calculator's memory or ANS key to avoid rounding errors, and give each answer to the accuracy asked.

When it fails

If f'(x_n) = 0 the tangent is horizontal and never meets the x-axis: the formula divides by zero.
If x_0 is near a stationary point, f'(x_0) is small and x_1 can be thrown far away, possibly towards a different root.
Check a final answer with a sign change: for a root 1.879 to 3 d.p., show f(1.8785) and f(1.8795) have opposite signs.

Worked example

f(x) = x³ + 8x - 19. Taking x_0 = 2, apply Newton-Raphson twice, giving x_2 to 3 decimal places.

f'(x) = 3x² + 8.
f(2) = 5 and f'(2) = 20, so x_1 = 2 - 5/20 = 1.75.
f(1.75) = 0.3594 and f'(1.75) = 17.1875, so x_2 = 1.75 - (0.3594)/17.1875.

Answer: x_2 = 1.729

Key idea: x_(n+1) = x_n - f(x_n)/f'(x_n). It fails when f'(x_n) = 0 and can misbehave near a stationary point.

Check yourself

f(x) = x³ + 8x - 19. With x_0 = 2, find x_1 using Newton-Raphson.
Show answer
1.75. f(2) = 8 + 16 - 19 = 5 and f'(2) = 12 + 8 = 20, so x_1 = 2 - 5/20 = 1.75.
Continue: find x_2 for f(x) = x³ + 8x - 19, to 3 decimal places.
Show answer
1.729. f(1.75) = 0.3594 and f'(1.75) = 17.1875, so x_2 = 1.729.
Use Newton-Raphson on f(x) = x² - 5 with x_0 = 2 to find x_2, to 4 decimal places.
Show answer
2.2361. x_1 = 2 - (-1)/4 = 2.25. x_2 = 2.25 - 0.0625/4.5 = 2.2361. (sqrt(5) = 2.2361.)
f(x) = x² - 2 - 3/x. With x_0 = 2, find x_1 using Newton-Raphson, to 3 decimal places.
Show answer
1.895. f'(x) = 2x + 3/x². f(2) = 4 - 2 - 1.5 = 0.5 and f'(2) = 4.75, so x_1 = 2 - 0.5/4.75 = 1.895.
f(x) = 2sqrt(x) - x + 1. With x_0 = 5, find x_1 using Newton-Raphson, to 3 decimal places.
Show answer
5.854. f'(x) = x^-1/2 - 1. f(5) = 0.4721 and f'(5) = -0.5528, so x_1 = 5.854.
Newton-Raphson is applied to f(x) = x³ - 3x - 1 with x_0 = 1. What happens?
- It converges straight to the root x = 1
- It gives x_1 = 0
- It gives x_1 = 2
- It fails because f'(1) = 0
Show answer
It fails because f'(1) = 0. f'(x) = 3x² - 3, so f'(1) = 0. The tangent at x = 1 is horizontal, and the formula would divide by zero.
What is the geometric meaning of one Newton-Raphson step from x_n?
- x_(n+1) is the midpoint of the current interval
- x_(n+1) is where the tangent at x = x_n crosses the x-axis
- x_(n+1) is where a chord crosses the x-axis
- x_(n+1) is where the normal at x = x_n crosses the x-axis
Show answer
x_(n+1) is where the tangent at x = x_n crosses the x-axis. The tangent at (x_n, f(x_n)) has gradient f'(x_n). Setting y = 0 in its equation gives the formula.

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