已知函数y=1/2sin(3x+π/6)+1 ,若x∈(-π/3,π/6)求函数的值域
问题描述:
已知函数y=1/2sin(3x+π/6)+1 ,若x∈(-π/3,π/6)求函数的值域
答
∵-π/3<x<π/6
∴-π<3x<π/2
-5π/6<3x+π/6<2π/3
∴当3x+π/6=π/2时,即x=π/9时,[f(x)]max=1/2x1+1=3/2
当3x+π/6=-5π/6时,即x=-π/3时,[f(x)]min=1/2x(-1)+1=1/2
∴函数的值域为(1/2,3/2)
答
∵x∈(-π/3,π/6)
∴3x∈(-π,π/2)
∴3x+π/6∈(-5π/6,2π/3)
∴-1≤sin(3x+π/6)≤1
∴-1/2≤1/2sin(3x+π/6)≤1/2
∴1/2≤1/2sin(3x+π/6)+1≤3/2
∴函数的值域为[1/2,3/2]