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Calculus min/max problem
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\[f(x) = \cos^2(x)-2\sin(x)\] \[x \epsilon [0,2\pi]\]
just need a confirmation on what you got for your local max / min. I got for my local max at x = pi/2 Local min at x = 3/2 pi
having contradictions with using the closed interval method and using the sign table. I am getting the exact opposit when I use either so not sure...
@jhonyy9
@julian25
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got for my local min at x = pi/2 Local max at x = 3/2 pi
why u dont test ur answers making the graph
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