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Question: MVT Sister Let f: [a,b] -> R be a continuous function, differentiable on (a,b) and nowhere zero on (a,b). Prove that there exists t s.t. a < t < b and f'(t)/f(t) = 1/(a - t) + 1/(b - t). Answer: Define g(x) := f(x)*(a-x)*(b-x). Differentiating yields g'(x) = f'(x)*(a-x)*(b-x) + f(x)*((x-a) + (x-b)). Note that g is cts on [a,b] and diff. on (a,b), so applying the MVT to g, we see that there exists a t in (a,b) such that: g`(t) = (g(b) - g(a)) / (b-a) = (0 - 0) / (b-a) = 0 <=> f'(t)*(a-t)*(b-t) + f(t)*((t-a) + (t-b)) = 0 <=> f'(t)*(a-t)*(b-t) = f(t)*((a-t) + (b-t)) <=> f'(t) / f(t) = 1/(a-t) + 1/(b-t), since f is nowhere 0 on (a,b). QED - Michael Viscardi |
