Question

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

Answer 1

设F(x)=f(x-1/3)-f(x)+1/3

F(1/3)=f(0)-f(1/3)+1/3=-f(1/3)+1/3

F(2/3)=f(1/3)-f(2/3)+1/3

F(1)=f(2/3)-f(1)+1/3=f(2/3)-2/3

F(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))/2=(-f(2/3)+2/3)/2

故：F(a)F(1)=(-f(2/3)+2/3)/2*(f(2/3)-2/3)《0,由根的存在性定理：

至少存在ξ,使得F(ξ)=0 ,即：f(ξ-1/3)=f(ξ)-1/3