定义域[0,1],f(0)=0,f(x)+f(1-x)=1,f(1/5x)=1/2f(x),且当0≤x1≤x2≤1时,f(x1)≤f(x2),求f(1/2008)
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![定义域[0,1],f(0)=0,f(x)+f(1-x)=1,f(1/5x)=1/2f(x),且当0≤x1≤x2≤1时,f(x1)≤f(x2),求f(1/2008)](/uploads/image/z/14996907-27-7.jpg?t=%E5%AE%9A%E4%B9%89%E5%9F%9F%EF%BC%BB0%2C1%EF%BC%BD%2Cf%280%29%3D0%2Cf%28x%29%2Bf%281-x%29%3D1%2Cf%281%2F5x%29%3D1%2F2f%28x%29%2C%E4%B8%94%E5%BD%930%E2%89%A4x1%E2%89%A4x2%E2%89%A41%E6%97%B6%2Cf%28x1%29%E2%89%A4f%28x2%29%2C%E6%B1%82f%281%2F2008%29)
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定义域[0,1],f(0)=0,f(x)+f(1-x)=1,f(1/5x)=1/2f(x),且当0≤x1≤x2≤1时,f(x1)≤f(x2),求f(1/2008)
定义域[0,1],f(0)=0,f(x)+f(1-x)=1,f(1/5x)=1/2f(x),且当0≤x1≤x2≤1时,f(x1)≤f(x2),求f(1/2008)
定义域[0,1],f(0)=0,f(x)+f(1-x)=1,f(1/5x)=1/2f(x),且当0≤x1≤x2≤1时,f(x1)≤f(x2),求f(1/2008)
f(0)=0,f(x)+f(1-x)=1 ==> f(1/2)=1/2,f(1)=1
f(1)=1,f(x/5)=1/2f(x),f(x)+f(1-x)=1 ==> f(1/5)=1/2,f(4/5)=1/2
f(1/5)=1/2,f(1/2)=1/2,f(4/5)=1/2,f(x1)≤f(x2){0≤x1 [1/5,4/5]区间f(x)=1/2
f(x/5)=1/2f(x) ==> [1/5^n,4/5^n]区间上f(x)=1/2^n
1/2020 属于 [1/5^5,4/5^5] ==> f(1/2008) = 1/2^5 =1/32