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Find the local max/min or saddle points on the function $f(x, y) = xy$

My attempt I've got the first partial derivatives to be

$$f_x = y$$ $$f_y = x$$

Giving a critical point at (0,0)

For the second derivative test, I've got $f_{xx} = 0$, $f_{yy} = 0$, $f_{xy} = 1$

I know that if $$f_{xx} \times f_{yy} < (f_{xy})^2$$ implies a saddle point. But is this true if the second derivatives $f_{xx}$ or $f_{yy}$ is zero.

Thanks in advance.

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    $\begingroup$ Yes its true. Take a look at a plot of $f(x,y) = xy$ near 0 and you'll see the characteristic (horse) saddle shape $\endgroup$ – Gregory Oct 28 '17 at 15:47
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Yes, $(0,0)$ is a saddle point. The condition that you write in the question come from the general test of the Hessian determinant, and in this case we have a saddle because the Hessian determinant at this point is $H=-1$.

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On the line $y=x$, the function becomes $f(x)=x^2$ and therefore $(0,0)$ is a minimum.

On the other hand, when you take $y=-x$, the function becomes $f(x)=-x^2$ and hence $(0,0)$ is a maximum.

These imply that $(0,0)$ is neither a maximum nor a minimum.

However, you have already found that it is a critical point since $f_x(0,0)=f_y(0,0)=0$.

Thus it must be a saddle point.

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