Finding and Verifying Saddle Points for f(x,y) = e^(1/x^2*y)

[pbialos]

I have to find the critical points of a function f, and verify they are saddle points
$$f(x,y) =\left\{\begin{array}{cc}e^{\frac {1} {x^2*y}},&\mbox{ if }x*y\neq 0\\0, & \mbox{x*y=0}\end{array}\right$$.
I found that the partial derivatives are 0 only on the axis, where f(x,y) is equal to 0, and i know that for every point outside the axis f(x,y)>0. My question is if the axis are saddle points and why?.

Any help would be much appreciated.
Regards, Paul.

 

[whozum]

Test the partials at points around the axis to see its behavior there. If it behaves something like f(x) = x^3, its a saddle, if it goes something like f(x) = x^2 or -x^2, its a minimum/maximum.

 

[pbialos]

I don't know what you mean, by test the partials. What information should i get from that?
The problem is that i think the axis should be minimums, but they are supposed to be saddle points acording to the exercise statement. I think they should be minimums just because de function is always positive except on the axis where it is 0.

Many Thanks, Paul.

 

[AKG]

To find if a point is a point of inflection for a single-variable function, you use the second derivative test, and there's an analogous test for functions of several variables. It involves the eigenvalues of the Hessian matrix, where the Hessian is the 2x2 matrix whose entries are the second order partials. It is the matrix:

(d²f/dx² d²f/dxdy)
(d²f/dxdy d²f/dy²)

Find the eigenvalues of this matrix and evaluate for each (x,y) where xy = 0 (note you'll have to check that these partial derivatives are "nice" here since the function is defined piecewise and might not be smooth). If both eigenvalues are positive, you have a minimum, if both are negative, you have a maximum, and if you have 1 positive and 1 negative, then you have a saddle of type (1-1).

 

[whozum]

They are saddle points. I've attached x and y slices of the function for you to see.
Testing points around the function means find values of the function around x = 0 and y = 0 to see what shape the graph takes.

http://www.public.asu.edu/~hyousif/xslice.JPG
http://www.public.asu.edu/~hyousif/yslice.JPG

 

[AKG]

Wait, are you sure you have the function written correctly? Say you take y = 1, and let x approach 0. Then 1/yx² will increase greatly, and exp(1/yx²) will be huge. At x = 0, it will suddenly drop to 0. You're right, you'll get minima on the axes since the function will be positive everywhere else. Imagine the graph y = |1/x| with y(0) = 0. Then 0 is obviously a minimum, but it's not going to be a "nice" minimum. As far as I can tell, if you took a slice of the graph of your function, then it would look something similar to the graph of the 1-variable function y = |1/x| with y(0) = 0, can you picture it?

 

[pbialos]

misunderstanding

Whozum misunderstood my function for what i can see on maple graphics. You are right AKG, my function exp(1/yx²) will tend to +infinite when y is positive and x tends to 0 or when y tends to 0+(0 by left).But it will tend to 0 in any other case(any other way to approach to the axis).
I still think the axis are minimums and not saddle points, am i right?

Many Thanks, Paul.

 

[AKG]

Yes, I think you're right. Approaching the x-axis from the positive-y side will give infinity, and approaching from the negative-y side will give 0. As you approach the negative half of the y-axis you get 0, and as you approach the positive half of the y-axis you get infinity. The points on the axes are minima. I'm attaching what I think the graph would look like.