Question

已知函数f(x)=2cos2x+sin平方x-4cosx

1。求f=(π/3)的值 2 求f(x) 最大最小值

Answer 1

f(x)=2cos2x+sin^x-4cosx
=2(2cos^x-1)+(1-cos^x)-4cosx
=3cos^x-4cosx-1
f(π/3)=3cos^(π/3)-4cosπ/3-1=3x(1/2)^-4x1/2-1=(-9)/4;
令t=cosx,t∈[-1,1]. F(t)=3(t^2)-4t-1
F'(t)=6t-4 令F'(t)=0,t=2/3;
t∈[-1,2/3], F(t)递减，t∈[2/3,1],F(t)递增。
f(x)(min)=F(2/3)=(-7)/3;
F(-1)=6,F(1)=-2, f(x)(max)=max{F(-1),F(1)}=6;

Answer 2

f(x)=2(2cos2x-1)+sin2x-4cosx=4cos2x-2(sin2x+cos2x)+sin2x-4cosx
=2cos2x-sin2x+4cosx=2cos2x-(1-cos2x)+4cosx
=3(cosx+2/3)2-7/3
1)f(π/3)=11/12
2)最大值为6