求函数y=2sin(2x-π/3),x∈[0,π/2]的值域

Question

求函数y=2sin(2x-π/3),x∈[0,π/2]的值域

Answer 1

设：t=2x-π/3, y=f(x)=2sin(2x-π/3)=2sint
因为x∈[0,π/2],所以 t∈[-π/3,2π/3]
将定义域分段讨论
1、t∈[-π/3,π/2]为增函数,所以当t=-π/3,y最小值=2sin(-π/3)=-3^(1/2),当yt=-π/2最大值为2sin(π/2)=2；
2、t∈(π/2,2π/3]为减函数,当t=2π/3,y最小值为=2sin(2π/3)=3^(1/2).
综上所述,原函数y=2sin(2x-π/3),x∈[0,π/2]的值域为[-3^(1/2）,2]
（完毕）