Find the real roots of f(x) = 0 inside a search range you choose.
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From the article Solving f(x) = 0 Numerically — Bisection and Newton's Method.
Take f(x) = x³ − x − 2 on the interval [1, 2].
f(1) = 1 − 1 − 2 = −2 (negative)
f(2) = 8 − 2 − 2 = +4 (positive)
Signs are opposite, so a root lies between 1 and 2.
After five steps the root is trapped between 1.5 and 1.53125. The true root is 1.5214 (to 4 decimals). Notice the pattern: every step halves the uncertainty, and nothing can go wrong.
Worked in full in Solving f(x) = 0 Numerically — Bisection and Newton's Method.
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