Classification of critical point

Question

The critical point of this function are $(0,0),(-1/3,-2/9),(-1/3,2/9)$. And for $(0,0)$, I get the difference that $△f=f(0+a,0+b)-f(0,0)=a^4 + (3a+1)b^2$ will always greater than zero where a,b are small values around $(0,0)$, so the point $(0,0)$ has local min. But for the other two points, I have no idea how to classify them, that is hard to expand the difference.

Answer 1

We are given:

$$f(x, y) = x^4 + 3 x y^2 + y^2$$

We have three critical points for consideration:

$$(x, y) = (0,0), \left(-\dfrac{1}{3},\dfrac{2}{9}\right), \left(-\dfrac{1}{3},-\dfrac{2}{9}\right)$$

A 3D-Plot shows:

A Contour-Plot shows:

The partial derivatives are given by:

$$f_{xx} = 12 x^2,~f_{xy} = f_{yx} = 6y, ~ f_{yy} = 2 + 6x$$

The Hessian determinant is given by:

$$\det(H) = \begin{vmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy}\end{vmatrix} \ $$

If you are using the Hessian, there are four conditions you need to test for each critical point:

• $(01) ~f_{xx} \gt 0$ and $\det H \gt 0 \rightarrow$ local minimum
• $(02) ~f_{xx} \lt 0$ and $\det H \gt 0 \rightarrow$ local maximum
• $(03) ~\det(H) \lt 0 \rightarrow$ saddle point
• $(04) ~\det(H) = 0 \rightarrow$ no statement can be made using this approach

$(x, y)=( 0, 0):$

We have $\det(H) = 0 \rightarrow$ no statement can be made using this approach. The function value is $f(0,0) = 0$.

$(x, y)= \left(-\dfrac{1}{3},\pm ~\dfrac{2}{9}\right):$

We have $\det(H) = - \dfrac{16}{9} ~ \lt 0 \rightarrow$ saddle point. The function value is $f\left(-\dfrac{1}{3},\pm ~\dfrac{2}{9}\right) = \dfrac{1}{81}$.