Question

设f(x)为奇函数,且当x>0时,f(x)=log(1/2) (x)

求当x＜0时，f（x）的解析式

Answer 1

x<0则-x>0
所以f(-x)适用f(x)=log(1/2) (x)
所以f(-x)=log(1/2) (-x)
奇函数
所以x<0
f(x)=-f(-x)=-log(1/2) (-x)

Answer 2

y=1/2cosπxcosπ/3-1/2sinπxsinπ/3-sinπxcos5π/6-cosπxsin5π/6
=1/4cosπx-√3/4sinπx+√3/2sinπx-1/2cosπx
=-1/4cosπx+√3/4sinπx
=1/2(sinπxcosπ/6-cosπxsinπ/6)
=1/2sin(πx-π/6)
sin增区间是(2kπ-π/2,2kπ+π/2)
所以即2kπ-π/2<πx-π/6<2kπ+π/2
2k-1/3<x<2k+2/3