- #1
Johnnycab
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Homework Statement
Find the absolute max and min of h(x) = cos(2x) - 2sinx in the closed interval [(pie)/2, 2pie]
The Attempt at a Solution
I got
h(x) = cos(2x) - 2sinx
h'(x) = -sin(2x)*2 - 2cosx <---chain rule i did :rofl:
0=-2sin(2x) - 2cosx
0=[-2sin(2x)/-2] - [2cosx/-2] i divided both sides by -2
0=2sinxcosx+cosx
0=cosx(2sinx+1) <---- factored out cosx
cosx = 0
and 2sinx+1 = (-1/2)
The critical numbers i found out where 0, and -0.5
I don't know if I am right or wrong
The book said the critical numbers are at [3(pie)/2], [7(pie)/2], and [11(pie)/2]
^--- i don't know how the book got those answers, can u help me out please
i understand that all the critical numbers are zero
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