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Bisection — Definition, Formula & Examples

Bisection is a root-finding method that repeatedly cuts an interval in half to zero in on where a continuous function equals zero. You start with two x-values where the function has opposite signs, then check the midpoint to decide which half still contains the root.

Given a continuous function ff on an interval [a,b][a, b] with f(a)⋅f(b)<0f(a) \cdot f(b) < 0, the bisection method computes the midpoint c=a+b2c = \frac{a + b}{2} and replaces [a,b][a, b] with either [a,c][a, c] or [c,b][c, b] depending on the sign of f(c)f(c), iterating until ∣b−a∣|b - a| is smaller than a desired tolerance. By the Intermediate Value Theorem, each sub-interval is guaranteed to contain at least one root.

Key Formula

c=a+b2c = \frac{a + b}{2}
Where:
  • aa = Left endpoint of the current interval
  • bb = Right endpoint of the current interval
  • cc = Midpoint where the function is evaluated

How It Works

Pick an interval [a,b][a, b] where f(a)f(a) and f(b)f(b) have opposite signs — one positive, one negative. Compute the midpoint c=a+b2c = \frac{a+b}{2} and evaluate f(c)f(c). If f(c)=0f(c) = 0, you found the root. Otherwise, replace whichever endpoint shares the same sign as f(c)f(c), so the new interval still brackets the root. Repeat until the interval is narrow enough for your desired accuracy. Each iteration cuts the possible error in half, so after nn steps the error is at most b−a2n\frac{b - a}{2^n}.

Worked Example

Problem: Use bisection to approximate a root of f(x)=x2−2f(x) = x^2 - 2 on [1,2][1, 2] (this root is 2\sqrt{2}). Perform three iterations.
Iteration 1: Compute the midpoint. Since f(1)=−1<0f(1) = -1 < 0 and f(2)=2>0f(2) = 2 > 0, the root lies in [1,2][1, 2].
c=1+22=1.5,f(1.5)=0.25>0c = \frac{1+2}{2} = 1.5, \quad f(1.5) = 0.25 > 0
Iteration 2: Now f(1)<0f(1) < 0 and f(1.5)>0f(1.5) > 0, so the root is in [1,1.5][1, 1.5].
c=1+1.52=1.25,f(1.25)=−0.4375<0c = \frac{1+1.5}{2} = 1.25, \quad f(1.25) = -0.4375 < 0
Iteration 3: Now f(1.25)<0f(1.25) < 0 and f(1.5)>0f(1.5) > 0, so the root is in [1.25,1.5][1.25, 1.5].
c=1.25+1.52=1.375,f(1.375)=−0.109375<0c = \frac{1.25+1.5}{2} = 1.375, \quad f(1.375) = -0.109375 < 0
Answer: After three iterations the root is bracketed in [1.375,1.5][1.375, 1.5], giving an estimate of about 1.43751.4375 with a maximum error of 0.06250.0625. The actual value 2≈1.4142\sqrt{2} \approx 1.4142.

Why It Matters

The bisection method is often the first algorithm taught in precalculus or introductory numerical analysis because it always converges for continuous functions. Engineers and scientists use it as a reliable fallback when faster methods like Newton's method fail to converge. Understanding bisection also reinforces the Intermediate Value Theorem in a practical, computational way.

Common Mistakes

Mistake: Choosing an interval where f(a)f(a) and f(b)f(b) have the same sign.
Correction: The method requires f(a)⋅f(b)<0f(a) \cdot f(b) < 0. Without a sign change, you have no guarantee a root exists in the interval, and the algorithm will not work correctly.