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

Sep 12, 2015

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

#### Explanation:

$f \left(x\right) = \sec x$ on the closed interval $\left[- \frac{\pi}{6} , \frac{\pi}{3}\right]$

$f ' \left(x\right) = \sec x \tan x = 0$ nlyat $0 + \pi k$ 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.

$f \left(- \frac{\pi}{6}\right) = \sec \left(- \frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}$

$f \left(0\right) = \sec \left(0\right) = 1$

$f \left(\frac{\pi}{3}\right) = \sec \left(\frac{\pi}{3}\right) = 2$

The minimum is $1$ (at $0$)
The maximum is $2$ (at $\frac{\pi}{3}$)