Critical points of multi-variable function.

determinanthessian-matrixmultivariable-calculusstationary point

I have the expression:

$(x^3-y^2 )(x-1)$

and have to find the critical points and their nature. The points found are:

$(1,1) = $ saddle point

$(1,-1) = $ saddle point

$(\frac 3 4 ,0) = $ global minimum

$(0,0) = $ inconclusive, however, when I take $0.1$ and $-0.1$ as $2$ points slightly higher or lower to find the nature of the point, I get a value positive or negative value for both the $fxx$ and the Hessian Determinant i.e. $D < 0$ at $0.1$ and $D > 0$ at $-0.1$.

I understand, for a single variable function, that two different signs for values higher or lower than the points, would mean the point $(0,0)$ is an inflection point. But since it is a multi-variable function, how can I determine the nature of $(0,0)$ since the Hessian Determinant must be $< 0 $ for the points to be saddle points?

Best Answer

Note that the section of the surface over the line $y = ax$ is $$z = (x^3 - a^2x^2)(x-1) = x^2(x - a^2)(x - 1)$$ Near $(0,0)$ this is approximately $z = a^2x^2$. If $a$ is any non-zero value, this is a parabola, opening upward, with $x = 0$ as the apex. I.e., if you leave $(x,y) = (0,0)$ in any direction other than along the lines $x = 0$ or $y = 0, z$ is going to increase. Setting $x = 0$ also gives you the parabola $z = y^2$, so the same is true for travel along the $y$-axis.

But when $y = 0$, the section becomes $z = x^3(x - 1)$, which near $x = 0$ looks like $z=-x^3$, increasing in the negative $x$-direction, but decreasing in the positive $x$-direction.

Hence $(0,0)$ is a saddle point, since it decreases in one direction and increases in others, though it does not have the classic saddle shape.