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函数f(x)的定义域为D,若对于X1,X2∈D,当X1<X2时,都有f(X1)≤f(X2),则称f(x)在D上为非减函数.设函数f(x)在[0,1]上为非减函数,且满足
①f(0)=0 ②f(x/3)=(1/2)*f(x) ③f(1-x)=1-f(x).
则f(1/3)+f(1/8)=
人气:444 ℃ 时间:2020-03-25 05:11:50
解答
解,根据(3),取x = 0,得f(1)=f(1-0) = 1-f(0) = 1-0 = 1再根据(2)f(1/3) = 1/2 * f(1) = 1/2根据(3)f(1-x) = 1-f(x)所以f(2/3) = 1-f(1/3) = 1- 1/2 = 1/2即f(1/3) = f(2/3)因为f(x)是非减函数,所以当1/3...
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