大一高数题:设f(x)在闭区间[0,1]上连续,f(0)=0,f(1)=1,证明:存在ξ∈(0,1),使得f(ξ-1/3)=f(ξ)-1/3.
问题描述:
大一高数题:设f(x)在闭区间[0,1]上连续,f(0)=0,f(1)=1,证明:存在ξ∈(0,1),使得f(ξ-1/3)=f(ξ)-1/3.
答
设F(x)=f(x-1/3)-f(x)+1/3F(1/3)=f(0)-f(1/3)+1/3=-f(1/3)+1/3F(2/3)=f(1/3)-f(2/3)+1/3F(1)=f(2/3)-f(1)+1/3=f(2/3)-2/3F(1/3)+F(2/3)=-f(2/3)+2/3 ,由介值性定理,至少存在a,(1/3《a《2/3),使:F(a)=(F(1/3)+F(2/3)...