How do you find the extrema for $f (x) = sec x$ $f(x) = sec x$ on the closed interval $[- \frac{π}{6}, \frac{π}{3}]$ $[-pi/6, pi/3]$ ?

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Answer 1 · 2015-09-12T00:55:25

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

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

$f'(x) =secxtanx = 0$ 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 $pi/3$ )

How do you find the extrema for f(x) = sec xf(x)=secxf(x) = sec x on the closed interval [-pi/6, pi/3][−π6,π3][-pi/6, pi/3]?