设f(x)在闭区间[a,b]上连续,x1,x2,...,xn是区间[a,b]上的点,求证在区间[a,b]上至少存在一点t,使得f(t)=(1/n)f(x1)+(1/n)f(x2)+...+(1/n)f(xn).
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![设f(x)在闭区间[a,b]上连续,x1,x2,...,xn是区间[a,b]上的点,求证在区间[a,b]上至少存在一点t,使得f(t)=(1/n)f(x1)+(1/n)f(x2)+...+(1/n)f(xn).](/uploads/image/z/1826024-32-4.jpg?t=%E8%AE%BEf%28x%29%E5%9C%A8%E9%97%AD%E5%8C%BA%E9%97%B4%5Ba%2Cb%5D%E4%B8%8A%E8%BF%9E%E7%BB%AD%2Cx1%2Cx2%2C...%2Cxn%E6%98%AF%E5%8C%BA%E9%97%B4%5Ba%2Cb%5D%E4%B8%8A%E7%9A%84%E7%82%B9%2C%E6%B1%82%E8%AF%81%E5%9C%A8%E5%8C%BA%E9%97%B4%5Ba%2Cb%5D%E4%B8%8A%E8%87%B3%E5%B0%91%E5%AD%98%E5%9C%A8%E4%B8%80%E7%82%B9t%2C%E4%BD%BF%E5%BE%97f%28t%29%3D%281%2Fn%29f%28x1%29%2B%281%2Fn%29f%28x2%29%2B...%2B%281%2Fn%29f%28xn%29.)
设f(x)在闭区间[a,b]上连续,x1,x2,...,xn是区间[a,b]上的点,求证在区间[a,b]上至少存在一点t,使得f(t)=(1/n)f(x1)+(1/n)f(x2)+...+(1/n)f(xn).
设f(x)在闭区间[a,b]上连续,x1,x2,...,xn是区间[a,b]上的点,求证在区间[a,b]上至少存在一点t,使得f(t)=(1/n)f(x1)+(1/n)f(x2)+...+(1/n)f(xn).
设f(x)在闭区间[a,b]上连续,x1,x2,...,xn是区间[a,b]上的点,求证在区间[a,b]上至少存在一点t,使得f(t)=(1/n)f(x1)+(1/n)f(x2)+...+(1/n)f(xn).
记m=min{f(x1)...,f(xn)},M=max{f(x1),f(x2),...,f(xn)},则
m
因为f(x)在[a,b]上连续,必可在这区间上取得最大值M有最小值m,即对一切x∈[a,b],有m≤f(x)≤M
所以m≤f(xi)≤M(i=1,2,…,n)
因为m=nm/n≤[f(x1)+f(x2)+…+f(xn)]/n≤nM/n=M
由介值定理,存在一点t,使得f(t)=(1/n)f(x1)+(1/n)f(x2)+...+(1/n)f(xn)
证:设所需证明的等式右端的值为L,并记f(xl)=max(1≦j≦n){f(xj)},
f(xm)=min(1≦j≦n){f(xj)},
则易证f(xm)≦L≦f(xl).,
于是由连续函数的介值定理知:
在区间[a,b]上必有点t,使f(t)=1/nf(x1)+1/nf(x2)+1/3f(x3)+……+1/nf(n).
根据闭区间上连续函数的中间值定理,闭区间上连续函数一定能取到最大值和最小值之间的任何一个值,由于
min(x∈[a,b]){f(x)}<=1/n (f(x1)+f(x2)+···+f(xn))<=max(x∈[a,b]){f(x)}
所以在[a,b]上有f(t)=1/n *(f(x1)+f(x2)+···+f(xn))成立
望采纳!
fuck fuck fuck 我不会.