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Verify Rolle's theorem for a log function

Short answer

f(x) = log(x²+ab) − log(a+b) − log x. f(a)=0, f(b)=0. f'(x) = (x²−ab)/(x(x²+ab)). Setting f'(c)=0 gives c = √(ab) which lies in (a,b). Thus Rolle's theorem is verified. Rolle's theorem requires f(a)=f(b) and yields f'(c)=0; Lagrange's MVT does not require equal endpoints and gives f'(c)=(f(b)−f(a))/(b−a).

Verify Rolle's Theorem for f(x) = log((x²+ab)/((a+b)x)); x ∈ [a,b]. How does Rolle's Theorem differ from Lagrange's Mean value theorem?

Verification of Rolle's theorem

f(x) = log[(x² + ab)/((a + b)x)] = log(x² + ab) − log(a + b) − log x on [a, b], with 0 < a < b.

  1. Continuity/differentiability: x² + ab > 0 and x > 0 on [a, b], so f is continuous on [a, b] and differentiable on (a, b).

  2. End values:

    • f(a) = log[(a² + ab)/((a+b)a)] = log[a(a+b)/((a+b)a)] = log 1 = 0
    • f(b) = log[(b² + ab)/((a+b)b)] = log[b(a+b)/((a+b)b)] = log 1 = 0

    So f(a) = f(b).

  3. Derivative:

f′(x) = 2x/(x² + ab) − 1/x = (2x² − x² − ab)/(x(x² + ab))
      = (x² − ab)/(x(x² + ab))

f′(c) = 0 gives c² = ab, so c = √(ab). Since a < √(ab) < b (geometric mean), c lies in (a, b).

Rolle's theorem is verified with c = √(ab).

Rolle's vs Lagrange's theorem

Rolle's theorem needs f(a) = f(b) and gives f′(c) = 0 (tangent parallel to the x-axis). Lagrange's mean value theorem drops the condition f(a) = f(b) and gives f′(c) = (f(b) − f(a))/(b − a) (tangent parallel to the chord). Rolle's theorem is the special case of Lagrange's theorem when f(a) = f(b).

✓ Answer checked by Claude Opus 5.5 (high effort)

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