Question

f(x)=2根号3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π),将f（x）的图像向右平移π/6个单位求在[0,π]上

的最大值和最少值

Answer 1

f(x)=2√3sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)
=√3*23sin(x/2+π/4)cos(x/2+π/4)-sin(x+π)
=√3sin(x+π/2)-sin(x+π)
=√3cosx+sinx
=2cos(x-π/6)
将f（x）的图像向右平移π/6个单位
则g(x)=f(x-π/6)=2cos(x-π/3)
∴在[0，π]上，当x=π/3时有最大值2，当x=π时有最小值-1

追问

怎样知道x=π/3是最大值 x=π是最少值

追答

画出函数f(x)=2cos(x-π/6)的大致曲线

追问

怎样画的 我也不会画- -！求教

Answer 2

f（X）=√3 sin(x+π/2)+sin x
=√3 cos x +sin x
=2[√3/2* cos x +1/2*sin x]
=2[ sin π/3* cosx + cos π/3* sinx]
=2sin(x+π/3)
将f（x）的图像向右平移π/6个单位，画出图形得知，在[0,π]上，π/6处取最大值为2，π处最小值为-1。