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I have a question which asks me to calculate the global maximum and global minimum of
y=e^(-x)cos(x)
Well i read a maths book explaining how to calculate them (by working out the local maxima and minima, and then putting these values and the ends of the domain back into f(x) to work out the greatest value(global maxima) and least value(global minima). The domain in my case is
.So the domain is greater than
and less then
so i was wondering can i include
in the calculation of the global minima?Thanks (sorry for the mistakes in my post!)
Last edited by technique09 (2008-11-06 11:30:58)
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The nature of your domain is important. If it's -π < x < 3π/2, then the endpoints aren't included in it. If the value of the function at an endpoint is a 'better' extreme than any local one, then the function doesn't have that type of extreme at all.
If the inequalities are weak then the endpoints are included and you don't get that problem.
eg. Say you want to find the extremes for x²+1, in the range -1<x<2.
Differentiate to find any local extremes: f'(x) = 2x = 0 ∴ x=0.
This is the only critical point, so now we investigate the function at -1, 0 and 2.
f(-1) = 2
f(0) = 1
f(2) = 5
Clearly, x=0 is the minimum point and x=2 is the maximum.
However, since the domain has a strict inequality, 2 isn't included and so the function has no maximum.
Edit - And if you're interested, you can click this equation to see the code for what you were trying to do:
Why did the vector cross the road?
It wanted to be normal.
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Excellent Mathsyperson, extremely helpful as usual, you helped clear that up for me! Thank you very much! Just to clarify, in your example, x=o is a global minimum and there is no global maximum?
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You're very welcome!
That's correct, the global minimum is 0 and there isn't a maximum.
One important point I forgot - the minimum is 0 because x=0, not because f(x) = 0.
I'll edit my example a bit to make that less ambiguous.
Why did the vector cross the road?
It wanted to be normal.
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Yeah x=0 makes sense, its the x value for which dy/dx=0. Well thanks again Mathyperson always at hand to give a well explained answer.
Last edited by technique09 (2008-11-06 11:48:40)
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