State and prove Lagrange's mean value theorem

Statement

If a function f is (i) continuous on the closed interval [a, b] and (ii) differentiable on the open interval (a, b), then there is at least one point c in (a, b) such that

f′(c) = (f(b) − f(a)) / (b − a).

Geometrically: somewhere on the curve between A(a, f(a)) and B(b, f(b)) the tangent is parallel to the chord AB.

Proof

Rolle's theorem is used. Define

φ(x) = f(x) − f(a) − k(x − a), where k = (f(b) − f(a)) / (b − a).

1. φ is continuous on [a, b] and differentiable on (a, b), because f is, and k(x − a) is a polynomial.
2. φ(a) = f(a) − f(a) − 0 = 0.
3. φ(b) = f(b) − f(a) − k(b − a) = f(b) − f(a) − (f(b) − f(a)) = 0.

So φ(a) = φ(b) = 0. By Rolle's theorem there is c in (a, b) with φ′(c) = 0.

Now φ′(x) = f′(x) − k, so f′(c) − k = 0, that is

f′(c) = (f(b) − f(a)) / (b − a).

This proves the theorem. Writing b = a + h and c = a + θh (0 < θ < 1), it takes the form f(a + h) = f(a) + h f′(a + θh).