设fx在〔a,b〕上连续,在(a,b)内可倒,则至少存在一点使e^fb-e^fa=

Question

设fx在〔a,b〕上连续,在(a,b)内可倒,则至少存在一点使e^fb-e^fa=

Answer 1

设g(x)=(x-b)f(x),则g(x)在[a,b]连续,在(a,b)可导,且g(a)=(a-b)f(a),g(b)=0.
由Lagrange中值定理,存在(a,b)内点t,使g'(t)=(g(a)-g(b))/(a-b)=f(a).
而g'(x)=((x-b)f(x))'=f(x)+(x-b)f'(x),于是有f(a)=f(t)+(t-b)f'(t).
整理即得f'(t)=(f(t)-f(a))/(b-t).