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Prove Lagrange's MVT; verify for log x

Short answer

Lagrange's MVT: If f is continuous on [a,b] and differentiable on (a,b), ∃c∈(a,b) with f'(c) = (f(b)−f(a))/(b−a). Proof: define φ(x)=f(x)−f(a)−k(x−a), k=(f(b)−f(a))/(b−a); φ(a)=φ(b)=0; by Rolle's, φ'(c)=0 ⇒ f'(c)=k. Verify: f(x)=log x on [1,e], f(1)=0, f(e)=1, slope 1/(e−1); f'(x)=1/x, so c=e−1 ≈1.718 in (1,e).

State and prove Lagrange's mean value theorem and verify f(x) = log x, x ∈ [1, e].

Statement

If f is continuous on [a, b] and differentiable on (a, b), there exists c in (a, b) such that f′(c) = (f(b) − f(a))/(b − a).

Proof

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

  • φ is continuous on [a, b] and differentiable on (a, b).
  • φ(a) = 0 and φ(b) = f(b) − f(a) − k(b − a) = 0.

By Rolle's theorem there is c in (a, b) with φ′(c) = 0. But φ′(x) = f′(x) − k, so f′(c) = k = (f(b) − f(a))/(b − a). ∎

Verification for f(x) = log x on [1, e]

  1. log x is continuous on [1, e] and differentiable on (1, e), since x > 0.
  2. f(1) = 0, f(e) = 1, so the chord slope is (1 − 0)/(e − 1) = 1/(e − 1).
  3. f′(x) = 1/x, so f′(c) = 1/c = 1/(e − 1), giving c = e − 1 ≈ 1.718.

Since 1 < 1.718 < e ≈ 2.718, c lies in (1, e).

Lagrange's MVT is verified with c = e − 1.

✓ Answer checked by Claude Opus 5.5 (high effort)

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