How to determine saddle point for a function?

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

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

then

• 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 $$t ^2<4$$.

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