Statement and proof of the Intermediate Value Theorem

Statement of the theorem

Let f be a continuous function on a closed interval [a, b], where a ≤ b. For any real number k between f(a) and f(b) (i.e. f(a) ≤ k ≤ f(b) or f(b) ≤ k ≤ f(a)), there exists at least one c in [a, b] such that f(c) = k.

In other words: a continuous function cannot ‘jump’ from one value to another without passing through all the intermediate values.

Proof Strategy (Dichotomy)

Suppose f(a) ≤ k ≤ f(b). We construct a sequence of nested intervals [a_n, b_n] by the dichotomy method: we divide [a, b] in two, keep the half where the change in sign of f(x) – k persists, and repeat the process. The sequences (a_n) and (b_n) converge to the same limit c, and by the continuity of f, f(c) = k. This constructive method also provides a numerical algorithm for approximating c.

A concrete example

Let f(x) = x³ - x - 1 on [1, 2]. We have f(1) = -1 and f(2) = 5. Since f is continuous (a polynomial) and 0 lies between -1 and 5, the TVI guarantees the existence of a c in [1, 2) such that f(c) = 0, that is, a real root of the equation x³ - x - 1 = 0.

Classic pitfall

The TVI requires continuity over an interval: it does not apply if f is discontinuous between a and b, even if f(a) and f(b) lie on either side of k. Counterexample: f(x) = 1/x on [-1, 1], where f(-1) = -1 and f(1) = 1, never takes the value 0, because f is not continuous at 0 (it is not even defined there).