Find Relative Max and Relative Min

Question

F(x) = 18cos(2x) + 18x  on the interval (0,2pi)

Answer 1

F'(x) = -36(sin 2x) + 18. Setting this equal to 0, we have
-1/2 = sin 2x
2x = 11pi/6 = 2pi - pi/6 = 330 degrees, or 2x = 7pi/6 = 2pi - 5pi/6 =  210 degrees
x = 11pi/12 = 165 degrees, or x = 2pi - pi/12 = 23pi/12 = 345 degrees

or x = 7pi/12 = 105 degrees, or x = 2pi - 5pi/12 = 19pi/12 = 285 degrees

These are four critical points.

 

F''(x) = -72 cos 2x

F''(11pi/12) = F''(23pi/12) < 0   ==> relative max's

F''(7pi/12) = F''(19pi/12) > 0 ==> relative min's

 

Note we use the periodicity of sin and cos with period 2pi when obtaining the 2 solutions on (0,2pi) to 2x = theta for each of the two angles theta. I.e. there are two solutions to 2x = 11pi/6 = -pi/6 on the interval: 11pi/12 and 23pi/12