作业帮设f(x)在[0,3]上连续,在(0,3)内可导,且f(0)+f(1)+f(2)=3,f(3)=1,试证明.必存在ξ∈(0,3),f'(ξ)=0
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![设f(x)在[0,3]上连续,在(0,3)内可导,且f(0)+f(1)+f(2)=3,f(3)=1,试证明.必存在ξ∈(0,3),f'(ξ)=0](/uploads/image/z/5303777-41-7.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%5B0%2C3%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%280%2C3%29%E5%86%85%E5%8F%AF%E5%AF%BC%2C%E4%B8%94f%280%29%2Bf%281%29%2Bf%282%29%3D3%2Cf%283%29%3D1%2C%E8%AF%95%E8%AF%81%E6%98%8E.%E5%BF%85%E5%AD%98%E5%9C%A8%CE%BE%E2%88%88%280%2C3%29%2Cf%27%28%CE%BE%29%3D0)
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设f(x)在[0,3]上连续,在(0,3)内可导,且f(0)+f(1)+f(2)=3,f(3)=1,试证明.必存在ξ∈(0,3),f'(ξ)=0
设f(x)在[0,3]上连续,在(0,3)内可导,且f(0)+f(1)+f(2)=3,f(3)=1,试证明.必存在ξ∈(0,3),f'(ξ)=0
设f(x)在[0,3]上连续,在(0,3)内可导,且f(0)+f(1)+f(2)=3,f(3)=1,试证明.必存在ξ∈(0,3),f'(ξ)=0
首先证明存在a∈(0,3),使得f(a)=1.
由此,f(x)在[0,3]上连续,（0,3）上可导,且f（a）=f(3)=1
利用罗尔定理,知道必存在ξ∈(a,3)包含于(0,3),使得f'(ξ)=0.
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