Cauchy mean value theorem

Imagine two travelers running along their own routes over the same time window. The ordinary mean value theorem compares one traveler's progress against the clock. Cauchy's version instead compares the two travelers against each other, and finds an instant where their speeds stand in the same ratio as their total displacements.

Statement: if f and g are continuous on [a, b] and differentiable on (a, b), then there exists c in (a, b) with (f(b) - f(a)) g'(c) = (g(b) - g(a)) f'(c). When g(b) is not g(a) and g'(c) is not 0, this rearranges to (f(b) - f(a)) / (g(b) - g(a)) = f'(c) / g'(c) — a ratio of derivatives equaling a ratio of total changes. Taking g(x) = x recovers the ordinary mean value theorem.

The standard proof applies Rolle's theorem to the auxiliary function h(x) = (f(b) - f(a)) g(x) - (g(b) - g(a)) f(x), which satisfies h(a) = h(b); then h'(c) = 0 is exactly the assertion. Note that the symmetric product form needs no nonvanishing hypothesis, whereas the fraction form requires g(b) not equal g(a); writing it as a single c for both functions (not two separate points) is what gives the theorem its power.