Finding maxima and minima of $f(x,y)=x^4+y^4-2x^2$ 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.
 A: Your function can be written as $$f(x,y)=(x^2-1)^2+y^4-1$$ and we have $$(x^2-1)^2+y^4-1\geq -1$$. And we can conclude that there is no maximum.($\infty$)
A: 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. 
A: When can we say a point $v$ provides a local minimum on a smooth function $f$?
we can say so, if for every vector $w\neq 0$, we can say there exists $\epsilon>0$ such that for any $t\in\mathbb{R}$ such that $|t|<\epsilon$, $f(v+w)>f(v)$.
What you calculated is based on the fact that this can be said when the second directional derivative on $f$ is strictly positive everywhere.
However, the points $B,C$ have one bad direction, which is $y$-axis direction that gives us second derivative $0$. (it is strictly positive for every other direction)
Thus, we only need to concentrate on that direction.
Since $f(B+t(0,1))>f(B)$ for every $t\neq 0$, we can say $B$ is a local minimum point, and similar for $C$.
Now, $A$ is not a local minimum, since $f(\epsilon,0)<f(0,0)$ for every small $\epsilon>0$.
