求函y=2sin(2x-π⼀4),x∈{-π⼀6,π⼀3}的值域

2025-02-26 22:26:46
推荐回答(3个)
回答1:

∵x∈(-π/6,π/3)
∴2x-π/4∈(-7π/12,5π/12)
设2x-π/4为A则是求sinA的值域
该函数图象是先减后增,则当A=-π/2时有最小值,Y=-2
当X=5π/12时有最大值,Y=(√6+√2)/2

回答2:

{-π/6,π/3}这样表示什么定义域,开域还是闭域

回答3:

∵x∈(-π/6,π/3)
∴2x-π/4∈(-7π/12,5π/12)
∴sin(2x-π/4)∈{-(√6+√2)/4,(√6+√2)/4}
y∈{-(√6+√2)/2,(√6+√2)/2}