设函数f(x)=3sin(ωx+π/6),ω＞0,且以π/2为最小正周期

题目

设函数f(x)=3sin(ωx+π/6),ω＞0,且以π/2为最小正周期
当x∈[-π/12,π/6]时,求f(x)的最值

答案

以π/2为最小正周期,那显然ω=4.
x∈[-π/12,π/6],则ωx+π/6∈[-π/6,5π/6].
最大值是3,当ωx+π/6=π/2时取到
最小值是-1.5,当ωx+π/6=-π/6时取到