Bisection Method
Computational Math
Bracketing
No Derivatives
Guaranteed Convergence
Root Finding
NoteConvergence plots
NoteIteration table
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.