Question

已知函数f(x)=2cos2x+sin^2x

已知函数f(x)=2cos2x+sin^2x(1)求f(2派/3) (2)求函数f(x)的最大值和最小值求详细过程，快快快，谢谢了

Answer 1

y=2cos2x+(sinx)^2
=2cos2x+(sinx)^2-1/2+1/2
=2cos2x-(cos2x)/2+1/2
=3cos2x/2+1/2
(1)f(2π/3)=3cos(4π/3)/2+1/2
=-3cos(π/3)/2+1/2
=-3/4+1/2=-1/4
(2)最大值为3/2+1/2=2,当cos2x=1时取最大，即x=kπ，k为整数。
最小值为-3/2+1/2=-1,当cos2x=-1时取最小，即x=(k+1)π,k为整数。

Answer 2

f(x)=2cos2x+sin²x
=2cos2x+2sin²x/2
=2cos2x+ -（1-2sin²x-1）/2
=2cos2x-（cos2x-1）/2
=2cos2x-1/2cos2x+1/2
=3/2cos2x+1/2
f(2π/3）=3/2cos（2*2π/3）+1/2
=3/2cos（4π/3）+1/2
=3/2cos（π+π/3）+1/2
=3/2*（-1/2）+1/2
=-1/4

当cos2x取最大值1时，函数f(x)取最大值2
当cos2x取最小值-1时，函数f(x)取最小值-1

Answer 3

(1)f(x)=2(1-2sinx^2)+sinx^2=-3sinx^2+2
f(2π/3)=-3(3/4)+2=-1/4
(2)-1<=sinx<=1
0<=sinx^2<=1
-3<=-3sinx^2<=0
-1<=-3sinx^2+2<=2
所以f(x)max=2 f(x)min=-1