TITLE: Finding maxima and minima of $f(x,y)=x^4+y^4-2x^2$
QUESTION [7 upvotes]: Finding maxima and minima of $f(x,y)=x^4+y^4-2x^2$
I tried studying this exam problem but I need help in understanding it.
I found $f_{x}=4x^3-4x \\
f_y=4y^3$
From this, I get the stationary points $A(0,0)$, $B(1,0)$ and $C(-1,0)$
$f_{xx}=12x^2-4 \\
f_{yy}=12y^2 \\
f_{xy}=0$
Now after computing  $D(x,y)$ I got $D(A)=D(B)=D(C)=0$ which is inconclusive.
In the graph, it shows that the point $B$ and $C$ are local minimum.
I would really appreciate some good explanation because I have an exam soon and I would like some help.

REPLY [4 votes]: You can analyze the function by completing the square. 
Note that $$x^4-2x^2+y^4=(x^2-1)^2+y^4-1$$ 
The minimum of $-1$ occurs at $$(x^2-1)^2+y^4=0$$
That is $x=\pm 1,y=0$ 
The critical point $(0,0)$ is a saddle point because you get both positive and negative values near that point where the value of your function is $0$ 
There is no local or absolute maximum.