Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 34

Given the function g(x)=sec(x) in the interval [-pi/6, pi/3]
We have to find the absolute extrema of the function on the closed interval.
So first let us find the derivative of the function and equate it to zero.
g'(x)=sec(x)tan(x)=0
sec(x) cannot be equal to 0 because its value ranges between 1 to +infinity and -1 to infinity only.
Therefore, tan(x)=0 which implies,
x=
So here the critical points are x=-pi/6, 0, pi/3 which falls in the interval
[-pi/6, pi/3].
So now substituting the critical points in the given function we get the absolute extremas.
g(0)=sec(0)=1
g(-pi/6)=sec(-pi/6)=2/sqrt(3)
g(pi/3)=sec(pi/3)=2

So the absolute maximum value is g(x)=2 at x=pi/3(critical as well as end point) and the absolute minimum value is g(x)=1 at x=0 (critical point)