Question

设函数f(x)=2cos^2x+√3 sin2x,当x属于[0,π/3]时求f(x)的最大值

十万火急啊！！

Answer 1

f(x)=2cos²x+根号下3sin2x
=2cos²x-1+1+根号下3sin2x
=cos2x+根号下3sin2x+1
=2sin(2x+π/6)+1
当 x∈[0,π/3] 2x+π/6∈[π/6,5π/6]
1<= 2sin(2x+π/6)<=2
2<=2sin(2x+π/6)+1<=3
所以 f(x)最大值=3

Answer 2

f(x)=2cos^2x+√3sin2x
=2cos^2x-1+1+√3sin2x
=cos2x+√3sin2x+1
=2（1/2cos2x+√3/2sin2x）+1
=2sin（2x+π/3）+1
x∈[0,π/3]
2x∈[0,2π/3]
2x+π/3∈[π/3,π]
当2x+π/3=π/2时，函数f(x)的值最大，最大值为：3

Answer 3

f(x)=2cos^2x+√3 sin2x
=1+cos2x+√3 sin2x
=2sin(2x+π/6)+1
x∈【0，π/3】
所以
当x=π/6时
函数取最大值=2+1=3