高数微分证明题



就这道,感觉不难,就是那里卡住了,高数大神们求解啊

F(x) = (x-1)^2f(x)
F(1)=0
F(2) = f(2)=0

F'(x) = (x-1)^2 f'(x) + 2(x-1)f(x)
F'(1) = 0

expands F(x) about 1
F(x) = F(1)+ F'(1)(x-1) +F''(a)(x-1)^2/2!
= F''(a)(x-1)^2/2!
put x=2
F(2) = F''(a)/2 =0
F''(a)=0 展开追问 追问 F(x) = F(1)+ F'(1)(x-1) +F''(a)(x-1)^2/2!
= F''(a)(x-1)^2/2!
没看懂,哪儿出来的,f(x)去哪儿了 回答 这是Taylor expansion(泰勒展开) of F(x) about x=1
F(x) 在【0,1】上有二阶导数
=>存在ξ∈(1,2)
F(x) = F(1)+ F'(1)(x-1)/1! +F''(ξ)(x-1)^2/2!
代入 x=2
F(2) = F(1)+ F'(1) +F''(ξ)/2 =0
=>F''(ξ)=0
用两次Rolle定理即可,注意到F(1)=F(2)=0→F'(x)=0,x∈(1,2),
F'(1)=F'(x)=0→F''(ξ)=0,ξ∈(1,2)
对F(x)在[1,2]上用罗尔定理,得,存在c∈(1,2)使得F’(c)=0
对F’(x)在[1,c]上用罗尔定理,得,存在ξ∈(1,c)使得F’(ξ)=0


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