已知f(x)为奇函数,f(x+2)=f(x),当x属于[0,1]时f(x)=2^x-1则f(log2 1/24)=?

∵f(x)是奇函数
∴f[log(2)(1/24)]=f[-log(2)(24)]
=-f[log(2)(24)]
=-f[log(2)(16)+log(2)(3/2)]
=-f[4+log(2)(3/2)].f(x+4)=f(x+2)=f(x)
=-f[log(2)(3/2)]
∵0=log(2)(1)＜log(2)(3/2)＜log(2)(2)=1
∴f[log(2)(1/24)]=-2^[log(2)(3/2)]+1=-3/2+1=-1/2
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