Question

已知函数f（x）=2cos2x+sin^2 x-4cosx

(1)求f=（π/3）的值
（2）求f(x)的最大值和最小值

Answer 1

解：f(x)=2(2cos²x-1)+(1-cos²x)-4cosx
=3cos²x-4cosx-1
=3[cos²x-(4/3)cosx]-1
=3[(cosx-2/3)²-4/9]-1
=3(cosx-2/3)²-7/3≧-7/3
(1).f(π/3)=3[(1/2)-(2/3)]²-7/3
=(1/12)-(7/3)
=-27/12
=-9/4
(2) 当cosx=2/3时f(x)获得最小值-7/3；
当cosx=-1 时f(x)获得最大值16/3.

Answer 2

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 3

f(x)=2cos2x+1+sin^2 x-4cosx=3cos^2 x-4cosx-1

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(1)把π/3带入f(x)=-9/4

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(2)把cosx看做是a
所以f(x)=3a^2-4a-1 a大于等于0小于等于1
配方得：f(x)=3（a-2/3)^2-13/9
所以f(x)min=-7/3
f(x)max=18/3