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
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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)=
①f(0)=0 ②f(x/3)=(1/2)*f(x) ③f(1-x)=1-f(x).
则f(1/3)+f(1/8)=
![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](/uploads/image/z/5421473-17-3.jpg?t=%E5%87%BD%E6%95%B0f%EF%BC%88x%EF%BC%89%E7%9A%84%E5%AE%9A%E4%B9%89%E5%9F%9F%E4%B8%BAD%2C%E8%8B%A5%E5%AF%B9%E4%BA%8EX1%2CX2%E2%88%88D%2C%E5%BD%93X1%EF%BC%9CX2%E6%97%B6%2C%E9%83%BD%E6%9C%89f%EF%BC%88X1%EF%BC%89%E2%89%A4f%EF%BC%88X2%EF%BC%89%2C%E5%88%99%E7%A7%B0f%EF%BC%88x%EF%BC%89%E5%9C%A8D%E4%B8%8A%E4%B8%BA%E9%9D%9E%E5%87%8F%E5%87%BD%E6%95%B0)
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
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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