设f(x)在[0,1]上有定义,且exf(x)与e-f(x)在[0,1]上单调增加.证明:f(x)在[0,1]上连续.

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问题 设f(x)在[0,1]上有定义,且exf(x)与e-f(x)在[0,1]上单调增加.证明:f(x)在[0,1]上连续.

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答案对任意的x0∈[0,1],因为exf(x)与e-f(x)在[0,1]上单调增加, 所以当x<x0时,有[*],故f(x0)≤f(x)≤ex0-xf(x0), 令x→x0,由夹逼定理得f(x0一0)=f(x0); 当x>x0时,有[*]故ex0-xf(x0)≤f(x)≤f(x0), 令x→x0,由夹逼定理得f(x0+0)=f(x0),故f(x0一0)=f(x0+0)=f(x0), 即f(x)在x=x0处连续。由x0的任意性得f(x)在[0,1]上连续.

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