Absolute Extrema on Closed Interval for f(x) = x-2cosx

[loadsy]

The question I'm currently working on now is to find the absolute maximum and absolute minimum values of f on the given interval.

f(x) = x-2cosx, [-pi, pi]
f'(x) = 1+2sinx = 0

From here I've found that -30 degrees(-pi/6) and -150(-5pi/6) degrees are the two intercepts in the equation. However I'm just having some trouble figuring out how to solve for this type of question. Any sort of hints would be great, thanks :P

 

[quasar987]

Inside the open interval (-pi,pi), you can calculate the derivative of f. The zeroes of the derivative are the critical values of f (either a min, a max or a saddle point). Calculate the value of f at each of these points. Also calculate f directly at -pi and pi. Now compare all the values of f you've calculated. The bigger is the absolute max and the smallest is the absolute min.

 

[loadsy]

Ahhh alrighty that makes sense. So pretty much just find the derivatives of f(-pi/6) and f(-5pi/6) along with f(-pi) and f(pi). Because I'm finding that the derivative of -pi/6 and -5pi/6 are 0 thus they are both critical numbers of f on (-pi,pi).

 

[loadsy]

Okay boo yeah, I think I figured it out, f(-5pi/6) = -5pi/6 - root3 is the absolute minimum and the absolute maximum is f(pi) = (pi) + 2