Question

已知函数fx=3tan(2x-π/3)

已知函数f(x)=3tan(2x-π/3),求f(x)的定义域；比较f(π/2)与f(-π/8)的大小。

Answer 1

解:(1) 由2x-π/3≠kπ+π/2 得 x≠kπ/2+5π/12,k∈Z
所以f(x)=3tan(2x-π/3)的定义域是{x∈R|x≠kπ/2+5π/12,k∈Z}

(2) f(π/2)=3tan(2(π/2)-π/3)=-3tan(π/3)<0
f(-π/8)=3tan(2(-π/8)-π/3)=3tan(-7π/12)=3tan(5π/12)>0

所以f(π/2)<f(-π/8)

希望对你有点帮助！