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# The minimum value of the function 2cos2x-cos4x in 0 12 years ago

We have
f(x) = 2cos2x - cos4x f ''(x) = -4sin2x + 4sin4x
= 4(sin4x - sin2x)
= 8cos3x.sin
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.

--

Regards

Ramesh Badiuddin askIITians.ismu Expert
147 Points
12 years ago

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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