+130591 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
