Find the absolute extrema of function on closed interval

Question

g(x)=7 sec x [=pi/6,pi/3]

Answer 1

Take the derivative of g(x) and set it equal to zero.

 

g'(x) = 0

 

7secxtanx = 0

 

7(1/cosx)(sinx/cosx) = 0

 

7(sinx/cos²x) = 0

 

 

Since there is no angle between pi/6 and pi/3 that make sinx equal to zero, we just evaluate g(x) at the intervals.

 

g(π/6) = 7(1/cos(π/6)) = 7 / (√(3)/2)) = 14 / √3

 

g(π/3) = 7(1/cos(π/3)) = 7 / (1/2) = 14

 

 

Now the question is, do you want absolute max or absolute min extrema?