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

1 Answer
Sep 12, 2015

Answer:

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#)