Critical point of two variable function

gertrudethegr
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Homework Statement




Each of the functions f have a critical point at (0,0), however at this point the second derivatives are all zero. Determine te type of critical point at (0,0) in this case

1) f(x,y)=x2y+xy2
2)f(x,y) = x4+2*x3y+x2y2+y4





The Attempt at a Solution




For the first part by plotting it on Wofram I saw that it was a saddle, and i substituted the values x=y and x=-y to show that there is a saddle point in one direction and a straight line in the other, hence a saddle point, is this sufficient

On 2) I know I must use taylors expansion, but up to how many terms, and what do I do once i get there?
 
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gertrudethegr said:

Homework Statement




Each of the functions f have a critical point at (0,0), however at this point the second derivatives are all zero. Determine te type of critical point at (0,0) in this case

1) f(x,y)=x2y+xy2
2)f(x,y) = x4+2*x3y+x2y2+y4





The Attempt at a Solution




For the first part by plotting it on Wofram I saw that it was a saddle, and i substituted the values x=y and x=-y to show that there is a saddle point in one direction and a straight line in the other, hence a saddle point, is this sufficient
I doubt it very much. I don't see anything in your work that indicates you took partial derivatives. Your textbook should have some examples of categorizing critical points by the use of partials.
gertrudethegr said:
On 2) I know I must use taylors expansion, but up to how many terms, and what do I do once i get there?
I don't think a Taylor series has anything to do with this problem. The comment I made before applies here as well.
 
The only examples I found using partials considered the second derivative at the given point, however as the questions states the second derivative at the points are all 0, hence all the examples I have found give an inconclusive solution... that is why I have become unstuck, but if I take third derivatives at the point (0,0) there are some non zero solutions, but I am unsure how to identify them as max, min, saddle...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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