Question

已知函数f(x)=2cos2x+sin^2x-4cosx,求f=(pi/3)并求最大值与最小值

Answer 1

解析：f(x)=2cos2x+(sinx)^2-4cosx

=2[2(cosx)^2-1]+1-(cosx)^2-4cosx

=3(cosx)^2-4cosx-1,

=3(cosx-2/3)^2-7/3,

∴f(π/3)=3[cos(π/3)-2/3]^2-7/3=3(1/2-2/3)^2-7/3=-9/4,

ymax=3(-1-2/3)^2-7/3=6,当cosx=-1时,

ymin=3(2/3-2/3)^2-7/3=-7/3,当cosx=2/3时,