Posted by Calculus on .
f(x) = 6(cos(x))^2 - 12sin(x)
0 ¡Ü x ¡Ü 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...
that should be less than or equal to for both. 0 (<=) x (<=) 2pi