Answer: f'(c)/g'(c) = [f(b)-f(a)]/[g(b)-g(a)].
- A f'(c)/g'(c) = [f(b)-f(a)]/[g(b)-g(a)]
- B f'(c) = 0, the conclusion of Rolle's theorem instead
- C f(c) = g(c), assuming the two functions intersect at c
- D f'(c) = g'(c), assuming the two derivatives must be equal
Correct answer: A. f'(c)/g'(c) = [f(b)-f(a)]/[g(b)-g(a)]
Explanation: Cauchy MVT generalises Lagrange's MVT to two functions: for continuous f, g on [a,b], differentiable on (a,b) with g'(x) ≠ 0, there exists c ∈ (a,b) such that f'(c)/g'(c) = [f(b)−f(a)]/[g(b)−g(a)].
As point Q slides along the curve toward P, the secant line PQ rotates into the tangent line at P, whose slope is the derivative.
Concept context
Rate of change, differentiation rules, and applications