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Just wanted to chime is here, too. I also got (-1,0) and (1,0) for the maxima and I too noticed that this did not line up visually with the image on the exam. I go so hung up on this fact that I reworked the problem way too many times!

I finally noticed that the function for the image did not exactly match the function given in problem number 5. As I remember it, the problem in 5 was exactly what Justin has in his answer:

$$ f(x, y) = (x^2-y^2)e^{-(x^2+y^2)} $$

The image, 2a, I think it was, looked more like this, but I am not 100% sure about the numbers:

$$ f(x, y) = (x^2-y^2)e^{-(2x^2+3y^2)}. $$

This gave us an image that looked more like this:

I left my answer as (-1,0), (1,0) because I had already spent way to much time on this problem. I made a note of this and am hoping for the best...I am sure Dr. McClure will be fair...

Just wanted to chime is in here, too. I also got (-1,0) and (1,0) for the maxima and I too noticed that this did not line up visually with the image on the exam. I go so hung up on this fact that I reworked the problem way too many times!

I finally noticed that the function for the image did not exactly match the function given in problem number 5. As I remember it, the problem in 5 was exactly what Justin has in his answer:

$$ f(x, y) = (x^2-y^2)e^{-(x^2+y^2)} $$

The image, 2a, I think it was, looked more like this, but I am not 100% sure about the numbers:

$$ f(x, y) = (x^2-y^2)e^{-(2x^2+3y^2)}. $$

This gave us an image that looked more like this:

I left my answer as (-1,0), (1,0) because I had already spent way to much time on this problem. I made a note of this and am hoping for the best...I am sure Dr. McClure will be fair...

Just wanted to chime in here, too. I also got (-1,0) and (1,0) for the maxima and I too noticed that this did not line up visually with the image on the exam. I go got so hung up on this fact that I reworked the problem way too many times!

I finally noticed that the function for the image did not exactly match the function given in problem number 5. As I remember it, the problem in 5 was exactly what Justin has in his answer:

$$ f(x, y) = (x^2-y^2)e^{-(x^2+y^2)} $$

The image, 2a, I think it was, looked more like this, but I am not 100% sure about the numbers:

$$ f(x, y) = (x^2-y^2)e^{-(2x^2+3y^2)}. $$

This gave us an image that looked more like this:

I left my answer as (-1,0), (1,0) because I had already spent way to much time on this problem. I made a note of this and am hoping for the best...I am sure Dr. McClure will be fair...