stationary points revisited

**billym** · May 5th 2008, 07:30 AM

I have this function and I am looking for the stationary points:

$ f(x,y) = x^4 + 2x^2y^2 + y^4 + 2x^2 - 2y^2 + 6 $

I have the two partial derivatives:

$ 4x(x^2 + y^2 + 1) = 0, (E.1) $
$ 4y(y^2 + x^2 - 1) = 0, (E.2) $

I have already found the stationary points: $(0,0), (0,1)$ and $(0,-1)$

For $(E.2)$ can we set:

$ y^2 + x^2 = 1 $

and thus have:

$ ( \sqrt{1/2}, \sqrt{1/2} ) $

or is this useless because it doesn't satisfy $(E.1)$ ?

**TKHunny** · May 5th 2008, 07:49 AM

Two things:

1) I'd have to agree that it is of no value because it does not satisfy (E.1)

2) Why didn't you wonder about the other three really obvious suspects (not to mention the infinitely many others) from that last errant exercise?

**billym** · May 5th 2008, 07:56 AM

1. Yeah i figured it was a dumb question, just making sure.

2. Huh?

**TKHunny** · May 5th 2008, 07:58 AM

Not a dumb question. It shows you were on the right track.

$x^{2} + y^{2} = 1$ defines a circle. How many points are there on a circle?

**billym** · May 5th 2008, 08:04 AM

oooohh... does that mean (0,0) isn't a solution?

**billym** · May 5th 2008, 08:05 AM

or wait no. sorry. whoops.

**billym** · May 5th 2008, 08:14 AM

now i am confused

**TKHunny** · May 5th 2008, 08:40 AM

Please be unconfused. You got it right in the first place. It was only talking about the circle that became confusing. Rule out the circle and you're done.

Of course, (0,1) and (0,-1) were identified before you started playing with the circle, so keep those two.

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