Bisection Method

Computational Math
Bracketing
No Derivatives
Guaranteed Convergence
Root Finding
Authors

Dhruv Azad

Apurva Nakade

Published

June 25, 2026

The bisection method finds a root of \(f(x)=0\) by repeatedly halving an interval \([a,b]\) where \(f\) changes sign, keeping whichever half still brackets the root.

Examples

The bisection method is slower than Newton’s method or the secant method, but it has one major advantage: if \(f\) is continuous and changes sign on the interval \([a,b]\), then a root is guaranteed to remain inside the bracket.

The examples below illustrate ordinary convergence, multiple possible roots, endpoint roots, invalid brackets, and roots near singular-looking behavior.

Click any example to load it into the app.

Standard guaranteed convergence
f(x) = x³ − x − 2,   [a,b] = [1,2]
A classic example with a sign change across the interval. Bisection steadily shrinks the bracket around the real root near 1.521.
Square root of 2
f(x) = x² − 2,   [a,b] = [1,2]
The method converges to √2. This is a simple example where the interval width halves at every step.
Transcendental equation
f(x) = cos(x) − x,   [a,b] = [0,1]
Finds the root of cos(x) = x. This is the same solution that appears in fixed point iteration, but here the bracket gives a guarantee.
Bracket around a negative root
f(x) = x³ − 2x + 2,   [a,b] = [−2,−1]
This function can cause surprising behavior for Newton's method, but bisection behaves predictably once a sign-changing bracket is provided.
Choosing one root among many
f(x) = (x−1)(x−2)(x−3),   [a,b] = [0.5,1.5]
The polynomial has three roots, but this bracket isolates the one near x = 1.
Same function, different bracket
f(x) = (x−1)(x−2)(x−3),   [a,b] = [2.5,3.5]
Changing only the bracket sends the method toward a different root of the same function, this time near x = 3.
Endpoint is already a root
f(x) = x − 1,   [a,b] = [1,3]
The left endpoint satisfies f(a) = 0, so the method has already found a root before any interval halving is needed.
Invalid bracket: no real root
f(x) = x² + 1,   [a,b] = [−1,1]
There is no sign change and no real root. Bisection correctly refuses to proceed because the guarantee does not apply.
Even-multiplicity root is missed
f(x) = x²,   [a,b] = [−1,1]
Although x = 0 is a root, f does not change sign across the interval. This shows a limitation of sign-change bracketing.
Sign change without a valid root
f(x) = 1/x,   [a,b] = [−1,1]
The function changes sign but is not continuous on the interval. The bisection theorem requires continuity, so the usual guarantee fails.