设函数f(x)在[a,b]上连续,在(a,b)内可导,f(a)=0,证明:对于正整数n,存在ξ属于(a,b),使f(ξ)=[(b-ξ)f'(ξ)]/n
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![设函数f(x)在[a,b]上连续,在(a,b)内可导,f(a)=0,证明:对于正整数n,存在ξ属于(a,b),使f(ξ)=[(b-ξ)f'(ξ)]/n](/uploads/image/z/13418737-25-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2C%E5%9C%A8%EF%BC%88a%2Cb%EF%BC%89%E5%86%85%E5%8F%AF%E5%AF%BC%2Cf%28a%29%3D0%2C%E8%AF%81%E6%98%8E%3A%E5%AF%B9%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0n%2C%E5%AD%98%E5%9C%A8%CE%BE%E5%B1%9E%E4%BA%8E%28a%2Cb%29%2C%E4%BD%BFf%28%CE%BE%29%3D%5B%28b-%CE%BE%29f%27%28%CE%BE%29%5D%2Fn)
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设函数f(x)在[a,b]上连续,在(a,b)内可导,f(a)=0,证明:对于正整数n,存在ξ属于(a,b),使f(ξ)=[(b-ξ)f'(ξ)]/n
设函数f(x)在[a,b]上连续,在(a,b)内可导,f(a)=0,证明:对于正整数n,存在ξ属于(a,b),使f(ξ)=[(b-ξ)f'(ξ)]/n
设函数f(x)在[a,b]上连续,在(a,b)内可导,f(a)=0,证明:对于正整数n,存在ξ属于(a,b),使f(ξ)=[(b-ξ)f'(ξ)]/n
题目好像少了一个条件,即f(b)=0
若题目中结论正确,则应有f‘(ξ)=[-f'(ξ)]/n,所以f'(ξ)=0,则只要证明函数f(x)满足其在(a,b)上存在f'(x)=0即可.而由已知f(x)在(a,b)上可导,在[a,b]上连续,且f(a)=f(b)=0,显然存在f'(x)=0,得证.
如果没有f(a)=f(b)=0的已知,则题目中的结论不一定成立
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