I am having trouble determining the extrema of the above function. I think the derivative is 2 cos x - 2 sin 2x At 2 cos x= 0 ; cos is pi/2 and/or 3 pi/2. At - 2 sin 2x = 0; - 2 sin 2x = pi and/or 2 pi. 2 cos x = pi, ---> cos x = pi/2 and/or 3 pi/4. - 2 sin 2x = -pi/2 ---> - pi/4 or - pi/2. If what I did so far is correct, I do not understand how to use the information to determine the extrema of x a 0 less than or equal than x less than or equal than 2 pi.
I came up with critical numbers of pi/4, 7 pi/6 and 11 pi/6. How do these numbers relate to 0 less than or equal to x less than or equal to 2 pi? The smallest is pi/4 or 45 degrees, the largest is 11 pi/6 or 330 degrees. I appreciate your help. I truly need to understand how to solve this problem
I came up with critical numbers of pi/4, 7 pi/6 and 11 pi/6. How do these numbers relate to 0 less than or equal to x less than or equal to 2 pi? The smallest is pi/4 or 45 degrees, the largest is 11 pi/6 or 330 degrees. I appreciate your help. I truly need to understand how to solve this problem
cos x = 0
OR
1 - 2 sin x = 0
In the interval (0, 2Pi), the first condition gives x = Pi/2 or x = 3Pi/2
The second condition says sin x = 1/2 which happens at 30 degrees and 150 degrees. These two angles are: Pi/6 and 5Pi/6, respectively.
So, the critical numbers in your open interval (0, 2Pi) are, in increasing order:
Pi/6, Pi/2, 5Pi/6, and 3Pi/2
at Pi/6: the f' changes sign from + to - (graph the first derivative to see this!) so it is a max
at Pi/2: the f' changes sign from - to + so it is a min
at 5Pi/6: the f' changes sign from + to - so it is a max
and finally:
at 3Pi/2: the f' changes sign from - to + so it is a min.
I hope this helps!!