设f(x)在(0,1)内有定义,且exf(x)与e-f(x)在(0,1)内都是单调增函数,证明:f(x)在(0,1)内连续.

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问题 设f(x)在(0,1)内有定义,且exf(x)与e-f(x)在(0,1)内都是单调增函数,证明:f(x)在(0,1)内连续.

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答案对任意的c∈(0,1), 当x<c时,由exf(x)≤ecf(c)及e-f(x)≤e-f(c)得f(c)≤f(x)≤ec-xf(c), 令x→c得f(c-0)=f(c); 当x>c时,由exf(x)≥ecf(c)及e-f(x)≥e-f(x)得f(c)≥f(x)≥ec-xf(c), 令x→c+得f(c+0)=f(c), 因为f(c-0)=f(c+0)=f(c),所以f(x)在x=c处连续,由c的任意性得f(x)在(0,1)内连续.

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