find the max and min of the first and second derivative from the function sec(2x)

f(x)  = sec(2x)

f ' (x)  = 2sec(2x)tan(2x)   set to 0

2sec(2x)tan(2x) = 0      the critical points on [0, 2pi)   are   0, pi/2, pi , 3pi/2 

f'  " (x)  = 4sec(2x)tan^2(2x) + 4sec^3(2x)

At 0 and pi , f '' (x) > 0, so the curve is concave up at these values, indicating relative minimums

At pi/2 and 3pi/2, f  " (x) < 0, so the curve is concave down at these values, indicating relative maximims

So....relative minimums occur at  (0,1), (pi, 1)  on the interval

And relative max's occur at (pi/2, -1), (3pi/2, -1) on the interval

In a general form.....minimums occur at 0 ± n pi   .... where n is an integer

And max's occur at  pi/2 + n pi   ....... where  n is an integer

Here's the graph of  sec(2x)........https://www.desmos.com/calculator/kws9rfyi0r

Nice one Chris :)