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Given $g(x,y)=y^2 - x^3$

find the critical points and classify them

$$\nabla g(x,y) = \begin{pmatrix} -3x^2 \\ 2y \\ \end{pmatrix}$$


$\implies -3x^2=0,2y=0$

$\implies x=0,y=0$

$$Hf(x,y) = \begin{pmatrix} -6x & 0 \\ 0 & 2 \\ \end{pmatrix}$$

$$Hf(0,0) =\begin{pmatrix} 0 & 0 \\ 0 & 2 \\ \end{pmatrix}$$

$det( Hf(0,0) ) = 0$

Is this correct?

How would I finish this and interpret the results? For example, whether or not it's positive-semi definite.


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+1 for showing what you have done. Others please note, the response is much more on point because of this. – Ross Millikan Nov 11 '12 at 4:02
up vote 2 down vote accepted

A critical point $(x,y)$ for $g$ is such that the vector $(\nabla g)(x,y)$ equals to $\vec{0}$. That is, both coordinates must be equal to zero. In your case, you must have both $-3x^2 = 0$ and $2y = 0$. This means that the only critical point is $(0,0)$. Plugging it into the Hessian, and taking the determinant you obtain $0$. So the determinant doesn't provide you enough information on the type of the critical point.

Notice that the value of $g$ at the critical point is $0$. What you can do to determine the type of the critical point is to plot the contour $g(x,y) = 0$ on which the critical point lies. It looks like: enter image description here

On the contour, the function $g$ is $0$. What is the value of $g(x,y)$ on a point $(x,y)$ that lies to the right of the contour? You can take any point, and just plug it into $g$ to check the sign of $g$. Say, $g(1,0) = -1$, so on the right of the contour the function $g$ is negative. Similary, checking $g(-1,0) = 1$, we see that on the right of the contour the function is positive. This means that the critical point is a saddle point. This is confirmed by the graph of $g$: enter image description here

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****Thanks!**** – user42538 Nov 11 '12 at 3:09

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