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suju101
how do i solve f(x)=secx at [-pi/3,pi/6] to get global extreme values??
There are two methods:- 1)Graph Do you know nature of sec x graph? 2) theory: f'(x)=sec x tan x you need to see in (-pi/6,pi/3) if it is 0,+ve or _ve.
Here f'(x)<0 when x<0 as sec x>0 and tan x<0 and f'(x)>0 for x>0 so function is decreasing for -ve x and increasing for +ve x hence max will be attained in the given interval at max possible x=pi/6
What you're stating is contradictory: You're asking to find the global values of a function... within an interval?
it is a closed interval so yes..
@across they're not contradictory You could have mutiple extemums in a given interval the question wants the maximun of those local maximums
Perhaps I'm misunderstanding the meaning of "global" here.
err yeah global in this case is the given domain....
Anyways suju this question is easier by graph method... Do you know graph of sec x?
Sorry, I'm a nitpick when it comes to word usage. :P Wouldn't those be local extrema instead?
Not really in a given interval what if there are two local maximums, The maximum of the two maximums is assumed to be global in this case...
Oh and in the theory part you also need to check the least value of x as f(x) is decreasing when x<0
I see. Thanks for clarifying. :)
@shankee i don't know abt the graph and i am supposed to solve this theoritically.
Err okay so f'(x)<0 at x<0 means function is initially decreasing. Then at x=F'(x)=0 as tanx=0 so tangent is horizontal and f'(x)>0at x>0 implies function increases afterwards. Now theoratically the maximum could occur at two points which are the two end points THis is bcoz All values in between them in their nieghbourhood will be less than them. So just plug in the end points whichever gives greater value is the answer
|dw:1325938871185:dw| Just to make it clear initially it is decreasing then increasing.