Homework Help: Calculus

Posted by Calculus on Monday, October 11, 2010 at 3:32pm.

f(x) = 6(cos(x))^2 - 12sin(x)
0 Ём x Ём 2(PI)
I have to find where the interval is increasing, decreasing, local mins and maxs, inflection points, and concave up and down. I understand how to do these type of problems. I am just getting messed up with the sin, cos, and pi stuff. So far I have
f'(x)= -12cos(x)sin(x)-12cos(x) and i have the critical numbers being -pi/2 and pi/2. and i am lost from there...

Calculus - Calculus, Monday, October 11, 2010 at 3:46pm
that should be less than or equal to for both. 0 (<=) x (<=) 2pi
Calculus - Anonymous, Wednesday, November 17, 2010 at 7:20pm
sdf

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