设函数f(x)在[0,1]上连续,且f(0)=f(1),证明:一定存在x属于【0,1/2】,使得f(x)=f(x+1/2)
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![设函数f(x)在[0,1]上连续,且f(0)=f(1),证明:一定存在x属于【0,1/2】,使得f(x)=f(x+1/2)](/uploads/image/z/5217523-43-3.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%E5%9C%A8%5B0%2C1%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E4%B8%94f%280%29%3Df%281%29%2C%E8%AF%81%E6%98%8E%EF%BC%9A%E4%B8%80%E5%AE%9A%E5%AD%98%E5%9C%A8x%E5%B1%9E%E4%BA%8E%E3%80%900%2C1%2F2%E3%80%91%2C%E4%BD%BF%E5%BE%97f%28x%29%3Df%28x%2B1%2F2%29)
设函数f(x)在[0,1]上连续,且f(0)=f(1),证明:一定存在x属于【0,1/2】,使得f(x)=f(x+1/2)
设函数f(x)在[0,1]上连续,且f(0)=f(1),证明:一定存在x属于【0,1/2】,使得f(x)=f(x+1/2)
设函数f(x)在[0,1]上连续,且f(0)=f(1),证明:一定存在x属于【0,1/2】,使得f(x)=f(x+1/2)
令F(x)=f(x)-f(x+1/2)
有 F(0)=f(1)-f(1/2)
F(1/2)=f(1/2)-f(0)=f(1/2)-f(1)=-F(0)
所以F(0)与F(1/2)异号
所以一定存在t∈[0,1/2]使得F(t)=f(t)-f(t+1/2)=0
所以原命题得证
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