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

I found the stationary points of this function using the equations -

$$\frac{\partial f}{\partial x} = 0 \qquad \frac{\partial f}{\partial y} = 0$$

I got $(0,0)$.

Now we calculate $R$, $S$, and $T$ at $(0,0)$ where

$$R = \frac{\partial^{2} f}{\partial x^{2} } \qquad S = \frac{\partial^{2} f}{\partial x\partial y} \qquad T = \frac{\partial^{2} f}{\partial y^{2} }$$

then, $$RT - S^{2} = 0$$

This does't lead to any result. How to find nature of this stationary point?

I know what local minima, local maxima, and saddle points are, but how to find the nature of $(0,0)$?

  • $\begingroup$ Look at this post math.stackexchange.com/questions/2156583/… $\endgroup$ – Paul Dec 11 '18 at 8:28
  • $\begingroup$ You should show what you got for all of the derivatives, just in case there was a stray sign error or something. $\endgroup$ – Blue Dec 11 '18 at 8:45

$$f(x,y)=2x^4-3x^2y+y^2=\left( y-2 {{x}^{2}}\right) \, \left( y-{{x}^{2}}\right)$$ $$f\left( \varepsilon ,\frac{3 {{\epsilon }^{2}}}{2}\right) =-\frac{{{\varepsilon }^{4}}}{4}<0$$ $$f\left( \varepsilon ,\frac{{{\varepsilon }^{2}}}{2}\right) =\frac{3 {{\varepsilon }^{4}}}{4}>0$$

Then $(0,0)$ is saddle point.


Along the line $y=0$ is $$f(x,0)=2x^4>0$$ in a neighborhood of $(0,0),$

while along $y=\frac 32 x^2\;$ it is $$f(x,\frac 32 x^2)=-\frac 14 x^4<0$$ in a neighborhood of $(0,0).$
Therefore is $(0,0)$ a saddle point.

  • 1
    $\begingroup$ $f(x,x) = 2x^4 - 3x^3 + x^2$. Also $f(x,x)>0$ is $x$ is small $\endgroup$ – Dylan Dec 11 '18 at 10:18

From $f(x,y)=2(x^2-\frac34y)^2-\frac18y^2$ it's easy to see that $(0,0)$ is a saddle.


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