已知f(x)為奇函數(shù),f(x+2)=f(x),當x屬于[0,1]時f(x)=2^x-1則f(log2 1/24)=?
已知f(x)為奇函數(shù),f(x+2)=f(x),當x屬于[0,1]時f(x)=2^x-1則f(log2 1/24)=?
數(shù)學人氣：405 ℃時間：2020-03-12 03:43:49
優(yōu)質(zhì)解答
∵f(x)是奇函數(shù)∴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...-f[log(2)(24)] =-f[log(2)(16)+log(2)(3/2)]這步怎么轉(zhuǎn)換 為什么后面是3/2就是log(2)(24)=log(2)(16×3/2)=log(2)(16)+log(2)(3/2)你看我后面的變換就知道為什么我要這樣做-2^log(2)(2/3)怎么算先不看負號，是2^log(2)(3/2)吧？設(shè)x=log(2)(3/2)∴2^x=3/2。。。這是定義∴2^log(2)(3/2)=2^x=3/2
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