已知函數(shù)f(x)=3tan(2x-π/3),求f(x)的定義域;比較f(π/2)與f(-π/8)的大小.
已知函數(shù)f(x)=3tan(2x-π/3),求f(x)的定義域;比較f(π/2)與f(-π/8)的大小.
其他人氣:720 ℃時間:2020-05-14 02:18:32
優(yōu)質(zhì)解答
解:
f(x)=3tan(2x-π/3)
所以
①定義域為2x-π/3≠π/2+kπ
x≠5π/12+kπ/2
②
f(π/2)=3tan(2π/3)=-3根號3<0
f(-π/8)=-3tan(π/4+π/3)=-3(1+根號3)/(1-根號3)>0
所以f(π/2)