已知函数f(x)满足,x大於等於4,则f(x)=(1/2)^x,当x小於4时,f(x)=f(x+1),则f(2+log下面2上3)=?

题目

已知函数f(x)满足,x大於等於4,则f(x)=(1/2)^x,当x小於4时,f(x)=f(x+1),则f(2+log下面2上3)=?

答案

2<3<4
所以log2(2)13<2+log2(3)<4
所以f[2+log2(3)]=f{[2+log2(3)]+1}=f[3+log2(3)]
1所以4<3+log2(3)<5
则满足f(x)=(1/2)^x=2^(-x)
所以原式=2^-[3+log2(3)]
=2^(-3-log2(3))
=2^(-3)/2^log2(3)
=(1/8)/3
=1/24