定义在R上的函数f(x)满足f(x)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/2008)=不好意思,打错了定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/20
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![定义在R上的函数f(x)满足f(x)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/2008)=不好意思,打错了定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/20](/uploads/image/z/2553985-1-5.jpg?t=%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%28x%29%3D0%2Cf%28x%29%2Bf%281-x%29%3D1%2Cf%28x%2F5%29%3Df%28x%29%2F2%2C%E4%B8%94%E5%BD%930%E2%89%A4a%EF%BC%9Cb%E2%89%A41%E6%97%B6%2Cf%28a%29%E2%89%A4f%28b%29%2C%E5%88%99f%281%2F2008%29%3D%E4%B8%8D%E5%A5%BD%E6%84%8F%E6%80%9D%2C%E6%89%93%E9%94%99%E4%BA%86%E5%AE%9A%E4%B9%89%E5%9C%A8R%E4%B8%8A%E7%9A%84%E5%87%BD%E6%95%B0f%28x%29%E6%BB%A1%E8%B6%B3f%280%29%3D0%2Cf%28x%29%2Bf%281-x%29%3D1%2Cf%28x%2F5%29%3Df%28x%29%2F2%EF%BC%8C%E4%B8%94%E5%BD%930%E2%89%A4a%EF%BC%9Cb%E2%89%A41%E6%97%B6%2Cf%28a%29%E2%89%A4f%28b%29%2C%E5%88%99f%281%2F20)
定义在R上的函数f(x)满足f(x)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/2008)=不好意思,打错了定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/20
定义在R上的函数f(x)满足f(x)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/2008)=
不好意思,打错了
定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/2008)=?
定义在R上的函数f(x)满足f(x)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/2008)=不好意思,打错了定义在R上的函数f(x)满足f(0)=0,f(x)+f(1-x)=1,f(x/5)=f(x)/2,且当0≤a<b≤1时,f(a)≤f(b),则f(1/20
令x=0
f(0)+f(1)=1
所以f(1)=1
f(1/5)=f(1)/2=1/2
f(1/25)=f(1/5)/2=1/4
以此类推可得f(1/3125)=1/32
令x=1/2
f(1/2)+f(1/2)=1
f(1/2)=1/2
f(1/10)=f(1/2)/2=1/4
与上面同样的方法可得
f(1/1250)=1/32
因为1/3125<1/2008<1/1250
又因为当0≤a<b≤1时,f(a)≤f(b)
即f(1/3125)≤f(1/2008)≤f(1/1250)
所以f(1/2008)=1/32
ryk6u,yi.