Question

一道用中值定理证明的证明题.设f(x)在[a,b]上连续,在(a,b)内可导,f(a)=f(b)=1,证明：存在ξ,η∈

Answer 1

首先,由g(x) = e^x在[a,b]连续,在(a,b)可导,根据Lagrange中值定理,

存在ξ ∈ (a,b),使e^ξ = g'(ξ) = (g(b)-g(a))/(b-a) = (e^b-e^a)/(b-a).

其次,由h(x) = e^x·f(x)在[a,b]连续,在(a,b)可导,根据Lagrange中值定理,

存在η ∈ (a,b),使e^η(f(η)+f'(η)) = h'(η) = ((h(b)-h(a))/(b-a) = (e^b-e^a)/(b-a) (f(a) = f(b) = 1).

于是e^η(f(η)+f'(η)) = e^ξ,即有e^(η-ξ)(f(η)+f'(η)) = 1.