The minimum value of the function 2cos2x-cos4x in 0

We have
f(x) = 2cos2x - cos4x
f ''(x) = -4sin2x + 4sin4x
= 4(sin4x - sin2x)
= 8cos3x.sinx   
Now, 
f ''(x) = 0
8cos3xsinx= 0
cos3xsinx = 0
cos3x = 0 or sinx = 0
or x = 0, .
or x = 0, .   

(As 0  x  )
Now,
f(0) = 2cos0 - cos0 = 2 - 1 = 1  

f = 2cos- cos2= -2 - 1 = -3 
    
and f() = 2cos2- cos4= 2 - 1 = 1
Thus, the maximum value of f(x) is  and the minimum value of f(x) is -3.

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Regards

Ramesh

Hi pallavi

2cos2x-cos4x

=2 cos2x -2cos²2x +1

=-2(cos2x-1/2)² +3/2

minimum value occur when cos2x-1/2 will be maximum

  =-2(-1-1/2)² +3/2

  =-3

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