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Prove Lagrange's MVT using Rolle's theorem

Short answer

Rolle's theorem: if f is continuous on [a,b], differentiable on (a,b) and f(a)=f(b), then ∃c∈(a,b): f'(c)=0. To prove Lagrange's MVT, set k=(f(b)−f(a))/(b−a) and φ(x)=f(x)−f(a)−k(x−a). φ satisfies Rolle's conditions, giving φ'(c)=0 ⇒ f'(c)=k. Hence f'(c) = (f(b)−f(a))/(b−a).

State Rolle's theorem and use it to prove Lagrange's mean value theorem.

Rolle's theorem

If f is continuous on [a, b], differentiable on (a, b) and f(a) = f(b), then f′(c) = 0 for some c in (a, b).

Lagrange's mean value theorem (to prove)

If f is continuous on [a, b] and differentiable on (a, b), then for some c in (a, b),

f′(c) = (f(b) − f(a))/(b − a).

Proof using Rolle's theorem

Let k = (f(b) − f(a))/(b − a) and define the auxiliary function

φ(x) = f(x) − f(a) − k(x − a).

  1. φ is the difference of f and a linear function, so it is continuous on [a, b] and differentiable on (a, b).
  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), and Rolle's theorem applies to φ: there is c in (a, b) with φ′(c) = 0.

Since φ′(x) = f′(x) − k, we get f′(c) = k, that is

f′(c) = (f(b) − f(a))/(b − a). ∎

If in addition f(a) = f(b), then k = 0 and the result reduces to Rolle's theorem.

✓ Answer checked by Claude Opus 5.5 (high effort)

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