f(x)=2sin(2x+π/6)+1,求函数f(x)在【-π/6,π/3】上的值域

题目

f(x)=2sin(2x+π/6)+1,求函数f(x)在【-π/6,π/3】上的值域

答案

-π/6<=x<=π/3
-π/3<=2x<=2π/3
-π/6<=2x+π/6<=5π/6
sinx在(-π/2,π/2)递增
(π/2,3π/2)递减
所以2x+π/6=π/2是最大
2x+π/6=-π/6是最小
所以最大=2*1+1=3
最小=2*（-1/2）+1=0
值域[0,3]