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#1 2006-09-17 00:21:56

gossen
Member
Registered: 2006-09-17
Posts: 3

Def.amount of a function with 2 variables (hard :( )

f(x,y)=ln(1-x^2-y^2)-x-y

a) What is the functions def.amount meaning what allowed x-values are there?

b) Find the functions stationary dots (meaning highest, lowest) and classify them.

I have tried following the examples in the book but no advance since maths really isn't my strongest side. Since I´m in university I have noone to turn to atm and this assignment is due to get handed in tomorrow. I have tried all weekend without succes, any help or pointers would be appreciated. Thanks in advance /Brian

Last edited by gossen (2006-09-17 01:34:20)

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#2 2006-09-17 01:34:53

gossen
Member
Registered: 2006-09-17
Posts: 3

Re: Def.amount of a function with 2 variables (hard :( )

*Bump* , anyone? sad

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#3 2006-09-17 04:28:27

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Def.amount of a function with 2 variables (hard :( )

To figure out what x values are allowed, first you must figure out what x values aren't allowed.  Can we take the ln of anything less than 0?  What value of x would make the stuff inside the ln always less than 0, no matter what y is?

As for finding max and mins, do you know the chain rule?  If you do, find the derivative of f(x, y) with repect to both x and y, set that to 0 to find the critial points, then find which of these points is the lowest and which is the highest.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#4 2006-09-17 11:24:20

gossen
Member
Registered: 2006-09-17
Posts: 3

Re: Def.amount of a function with 2 variables (hard :( )

Wow thanks Ricky, I think it worked :DD

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#5 2006-09-17 12:54:25

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Def.amount of a function with 2 variables (hard :( )

No problem.  If you would share your answer with us, we could check it over if you like.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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