I need to find maxima, minima/saddle point for the function

$f(x,y) = x^4 + y^4 -2x^2 -2y^2 + 4xy$. I have figured out that critical point of the functions are $(\sqrt{2}, -\sqrt{2}), (-\sqrt{2} , \sqrt{2}), (0,0)$

My main question is about the point $(0,0)$ here the discrminant $rt-s^2 =0$ so derivative test fails

Now I need to know nature of $f(x)$ at origin.

Here is what I did:

$f(x,y) = x^4 + y^4 -2x^2 -2y^2 + 4xy$ , take the line $y = 0$ then

$f = x^4 - 2x^2 = x^2\left(x^2 - 2\right)$. Then for $x \lt \sqrt{2}$

$f \lt 0$ and for $x \gt \sqrt{2}$ $f \gt 0$

Since we have both positive and negative values in neighborhood of $(0,0)$ hence $(0,0)$ is a saddle point .

I have two questions at this point :

(a) Is the above method to find saddle point at origin correct ?

(b) I have solved a couple of problems like these where you have to solve for maxima/minima but the derivative test fails . Does there exist any technique/algorithm to solve such kind of questions easily ?

Can anyone answer these doubts ?

Thank you.


There are not general methods, in this case we have that

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


  • for $x=y=t$

$$f(t,t)=2t^4 \ge 0$$

  • for $x=-y=t$

$$f(t,-t)=2t^4-8t^2 =2t^2(t^2-4)$$

which is negative for $t ^2<4$.

  • $\begingroup$ Thanks for your answer,I have just one doubt : Is My method to determine the nature of this function at $(0,0)$ correct ? $\endgroup$
    – zeroflank
    Oct 13 '19 at 14:13
  • 1
    $\begingroup$ No your method fails when you assume $x>\sqrt 2$. We need to look for the sign near $(0,0)$. $\endgroup$
    – user
    Oct 13 '19 at 14:47

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