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find all critical point of f(x)=cos^2+sqrt3sinx
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were you able to find f'(x)? :)
its -cos(x) (sqrt(3)+2 sin(x))
\[f'(x)=2cosx(-sinx)+\sqrt3 cosx=cosx(\sqrt3-2sinx)\] Hmm I don't think there was a negative on the sine term, no? :o
im not sure
(sqrqt3-2sinx)cos(x)
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Yah I think that's what you should get :D I think you had 1 too many negative signs in your original answer
So to find critical points, we set this derivative equal to zero and solve for the different x values, yes? :D
can we do it together
is it pi/3, and -pi/3
|dw:1351913764988:dw| Hmmm I think these are the values we get D:
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