设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c
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![设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c](/uploads/image/z/8712147-3-7.jpg?t=%E8%AE%BEf%EF%BC%88x%EF%BC%89%E5%9C%A8%E3%80%900%2C1%E3%80%91%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%EF%BC%880%2C1%EF%BC%89%E5%8F%AF%E5%AF%BC.f%EF%BC%880%EF%BC%89%3D0+%2Cf%EF%BC%881%EF%BC%89%3D1.%E8%AF%81%E6%98%8E%EF%BC%9A%E5%AD%98%E5%9C%A8C%E5%B1%9E%E4%BA%8E%EF%BC%880%2C1%EF%BC%89%E4%BD%BFf%EF%BC%88c%EF%BC%89%3D1-c%E8%AE%BEf%EF%BC%88x%EF%BC%89%E5%9C%A8%E3%80%900%2C1%E3%80%91%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%EF%BC%880%2C1%EF%BC%89%E5%8F%AF%E5%AF%BC.f%EF%BC%880%EF%BC%89%3D0++%2Cf%EF%BC%881%EF%BC%89%3D1.%E8%AF%81%E6%98%8E%EF%BC%9A%E5%AD%98%E5%9C%A8C%E5%B1%9E%E4%BA%8E%EF%BC%880%2C1%EF%BC%89%E4%BD%BFf%EF%BC%88c%EF%BC%89%3D1-c)
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设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c
设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c
设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c
设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c设f(x)在【0,1】上连续,(0,1)可导.f(0)=0 ,f(1)=1.证明:存在C属于(0,1)使f(c)=1-c
证明:
令F(x)=f(x)+x-1
因为f(x)在[0,1]上连续,在(0,1)可导,
所以F(x)在[0,1]上连续,在(0,1)可导,
又
F(0)=f(0)+0-1=-1<0
F(1)=f(1)+1-1=1>0
F(x)在[0,1]必有零点 所以存在f(c)+c-1=0即f(c)=1-c
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