设函数f(x)在【0,1】上连续,在(0,1)内可导,且f(0)=0,f(1)=1,试证在(0,1)内至少存在一点ξ， 设函数f(x)在【0,1】上连续,在(0,1)内可导,且3∫...

\u8bbe\u51fd\u6570f(x)\u5728\u30140,1\u3015\u4e0a\u8fde\u7eed\uff0c\u5728\uff080,1\uff09\u5185\u53ef\u5bfc\uff0c\u4e14f(0)=f(1)=0,\u8bc1\u660e

F=f(x)e^(x/2),F\u5728\u533a\u95f4[0,1]\u6e80\u8db3\u7f57\u5c14\u5b9a\u7406\u7684\u6761\u4ef6.\u7531\u7f57\u5c14\u5b9a\u7406,\u5728\uff080,1\uff09\u5185\u81f3\u5c11\u6709\u4e00\u70b9\u03be\uff0c\u4f7fF'(\u03be)=0,\u4f46F'(x)=f'(x)e^(x/2)+(1/2)f(x)e^(x/2),\u4ee3\u5165\u5373\u5f97\u7ed3\u8bba

\u56e0\u4e3a3\u222bf(x)dx=3f\uff08k)(1-2/3)=f(k)\u5176\u4e2dk\u2208(2/3,1)(\u8fd9\u91cc\u7528\u7684\u662f\u5b9a\u79ef\u5206\u7684\u4e2d\u503c\u5b9a\u7406\uff09
\u6240\u4ee5f(0)=f(k)
\u6545\u6839\u636e\u7f57\u5c14\u5b9a\u7406\uff0c\u53ef\u77e5\u9053\uff0c\u5728\uff080\uff0ck\uff09\u4e0a\u5b58\u5728\u4e00\u70b9c\u4f7f\u5f97\uff0cf\u2018\uff08c\uff09=0
\u56e0\u6b64\u5728\uff080\uff0c1\uff09\u5185\u81f3\u5c11\u5b58\u5728\u4e00\u70b9C\u4f7ff\u2019\uff08C\uff09\uff1d0
\u4e0d\u77e5\u662f\u5426\u660e\u767d\u4e86O(\u2229_\u2229)O\u54c8\uff01

即证： 1/2f（1）^2＞＝1/4f（1）^4
即 ：2f（1）＞＝f（1）^4
因为f（x）的导数大于0小于等于1 所以f（1）大于0小于等于1
所以得证~~

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