Help finding critical points of sec(x)?

Question

The function is f(x) = sec(x), [-2pi , 2pi]

i got the derivative, which is sec(x)tan(x) so how do you find the critical points?

Answer 1

The critical points are where the derivative is 0.

sec(x) tan(x) = 0.

sec(x) = 1 / cos(x), and since cos(x) always has a value between - 1 and 1, sec(x) always has a value greater than 1 or less than - 1.

sec(x) is therefore never 0, and the derivative is 0 when tan(x) = 0, which is at - 2pi, - pi, 0, pi and 2pi in [- 2pi, 2pi].

When x is just below 0, tan(x) < 0.

When x is just above 0, tan(x) > 0.

x = 0 is a minimum.

Considering all the above points in this way, you find:

- 2pi, 0, 2pi give a minimum, with sec(x) = 1;

- pi, pi give a maximum, with sec(x) = - 1.

There are no points of inflecton.

cos(x) = 0 when x = - 3pi /2, - pi / 2, pi / 2, 3pi /2. That means sec(x) has vertical asymptotes at those points.

You can plot the graph here:

http://www.walterzorn.com/grapher/graphe...