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...

1 answer

that should be less than or equal to for both. 0 (<=) x (<=) 2pi
Similar Questions
  1. Find f'(x) if f(x)=sin^3(4x)A. 4cos^3(4x) B. 3sin^2(4x)cos(4x) C. cos^3(4x) D. 12sin^2(4x)cos(4x) E. None of these I got D using
    1. answers icon 1 answer
  2. ``(please show the steps to how the problems were solved)Using exact values, find the value of: A.12sin 30 degree - 6 tan 45
    1. answers icon 1 answer
  3. 8-12SIN^2X=4COS^2XSOLVE FOR X
    1. answers icon 1 answer
    1. answers icon 1 answer
more similar questions