当x属于【-π/2,π/2】时,函数f(x)=sinx+根号3cosx的最值?
问题描述：
当x属于【-π/2,π/2】时,函数f(x)=sinx+根号3cosx的最值?
答
f(x) = 2sin(x+pi/3)
x = -pi/3, f(x) = 0
x = pi/6, f(x) = 2
答
f(x)=sinx+√3cosx
=2﹙sinxcosπ/3+cosxsinπ/3﹚
=2sin﹙x＋π/3﹚
∵x属于【-π/2,π/2】
∴x＋π/3∈[-π/6,5π/6】
∴f(x)=sinx+√3cosx
在x＋π/3=-π/6即
x=-π/2时,取最小值2×﹙﹣½﹚=-1,在x＋π/3=π/2即
x=π/6时,取最大值2×1=2