Consider the function f(x)=x+2cos(x), 0

Question

Consider the function f(x)=x+2cos(x), 0<x<2π. 
f(x) is increasing on the interval(s):
f(x) is decreasing on the interval(s):
f(x) has a local minimum at:
f(x) has a local maximum at:
f(x) is concave up on the interval(s):
f(x) is concave down on the interval(s):
f(x) has 2 points of inflection at:

anonymous · Answer

i would start by getting the first derivative and setting it to zero

anonymous · Answer

inorder to get the critical points

anonymous · Answer

I have both derivatives f'(x): 1-2sinx and f''(x): -2cosx Then the rest I don't know how to find.

anonymous · Answer

set them to zero and solve for x

anonymous · Answer

keeping in mind, the interval your are working within

anonymous · Answer

For the first one I got sinx=1/2 and the second 0. Is that right?

anonymous · Answer

oaky, so at what angle(s) within 0<x<2pi is sin(x), 1/2?

at what angle(s) withing that interval is cos(x), 0?

anonymous · Answer

60 and 30. I don't think I'm doing this right.

anonymous · Answer

Trig throws me off.

anonymous · Answer

actually, the critical points come from the first derivative, so if you said the first derivative is 1-2sin(x)=0
sin(x)=1/2
so the occurs at  30 deg or pi/6