求函数y=2sinx(2x+π/3)(-π/6≤x≤π/6)的值域.怎么求,
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![求函数y=2sinx(2x+π/3)(-π/6≤x≤π/6)的值域.怎么求,](/uploads/image/z/13260030-6-0.jpg?t=%E6%B1%82%E5%87%BD%E6%95%B0y%3D2sinx%282x%2B%CF%80%2F3%EF%BC%89%EF%BC%88-%CF%80%2F6%E2%89%A4x%E2%89%A4%CF%80%2F6%EF%BC%89%E7%9A%84%E5%80%BC%E5%9F%9F.%E6%80%8E%E4%B9%88%E6%B1%82%2C)
求函数y=2sinx(2x+π/3)(-π/6≤x≤π/6)的值域.怎么求,
求函数y=2sinx(2x+π/3)(-π/6≤x≤π/6)的值域.怎么求,
求函数y=2sinx(2x+π/3)(-π/6≤x≤π/6)的值域.怎么求,
∵-π/6≤x≤π/6
∴-π/3≤2x≤π/3
0≤2x+(π/3)≤2π/3
又2π/3=(π/2)+(π/6)
∴sin0=0,sin(π/2)=1
∴函数y=2sin(2x+π/3)的值域为[0,2]
你的题目可能有问题,应该是y=2sin(2x+π/3)
因为-π/6≤x≤π/6,所以-π/3≤2x≤π/3,所以0≤2x+π/3≤2π/3,所以2sinx(2x+π/3)值域[0,2]
∵ -π/6≤x≤π/6 ∴ - π/3≤2x≤π/3 0 ≤2x+π/3≤2π/3
∴ 0≤ sinx(2x+π/3 )≤√3/2 ∴ 0≤ 2sinx(2x+π/3 )≤√3/2 *2=√3
sin0=0 sinπ/2=1
∴ 0≤ y≤2