Question

How do you find the extrema for `f(x) = sec x` on the closed interval `[-pi/6, pi/3]`?

Answer 1

Evaluate `f` at the endpoints of the interval and at the critical point(s) in the interval.

Explanation:

`f(x) = sec x` on the closed interval `[-pi/6, pi/3]`

`f'(x) =secxtanx = 0` nlyat `0+pik` for integer `k`.

The only such value in the interval is `x=0` so there is only one critical number to consider in the interval.

`f(-pi/6) = sec(-pi/6) = 2/sqrt3`

`f(0) = sec(0) = 1`

`f(pi/3) = sec(pi/3) =2`

The minimum is `1` (at `0`)
The maximum is `2` (at `pi/3`)