# Thread: Very weird question relating to critical points

1. ## Very weird question relating to critical points

I've done some work on this, though either I've hit a strange answer or I did something wrong. The question has two parts.

a) What happens to $D=f_{xx}f_{yy}-(f_{xy})^2$ at $(0,0)$ for $f(x,y)=x^4-2x^2y^2+y^4$? Classify the critical point at $(0,0)$.
Working out the problem, I got to here:
$f_{xx}=12x^2-4y^2$, $f_{xy}=-8xy$, $f_{xx}=12y^2-4x^2$
$D=-48x^4+96x^2y^2-48y^2$
This, however, leads to the answer $D(0,0)=0$, which is an inconclusive answer (it's not a maximum, minimum, or saddle point).
Is this the answer that's being asked of, or did I miss a detail?

b) Repeat for $f(x,y)=(y-x^3)(y-x^5)$.
I haven't gotten to this yet, but it probably won't take too long.

2. Originally Posted by Runty
I've done some work on this, though either I've hit a strange answer or I did something wrong. The question has two parts.

a) What happens to $D=f_{xx}f_{yy}-(f_{xy})^2$ at $(0,0)$ for $f(x,y)=x^4-2x^2y^2+y^4$? Classify the critical point at $(0,0)$.
Working out the problem, I got to here:
$f_{xx}=12x^2-4y^2$, $f_{xy}=-8xy$, $f_{xx}=12y^2-4x^2$
$D=-48x^4+96x^2y^2-48y^2$
This, however, leads to the answer $D(0,0)=0$, which is an inconclusive answer (it's not a maximum, minimum, or saddle point).
Is this the answer that's being asked of, or did I miss a detail?

b) Repeat for $f(x,y)=(y-x^3)(y-x^5)$.
I haven't gotten to this yet, but it probably won't take too long.
I think that the point of this exercise is to show by means of examples that when the discriminant test is inconclusive almost anything can happen.

The function in (a) can be written as $f(x,y) = (x^2-y^2)^2$. It's obvious from that that the (global) minimum value of this function is 0, and that this minimum value occurs at the origin (and also at all the points on the lines $y = \pm x$).

To see what happens in (b), notice that the function is zero along the curves $y=x^3$ and $y=x^5$. Then think about what happens to the function on the coordinate axes and also on the curve $y=x^4$.

3. I see how that works, but I can't say "it's obvious there's a global minimum at (0,0)" in my answer (it has to be proven, or it's wrong). How am I supposed to prove that the global minimum occurs there if the second partial derivative test fails?

...Or am I over-analyzing things again?

4. Originally Posted by Runty
I see how that works, but I can't say "it's obvious there's a global minimum at (0,0)" in my answer (it has to be proven, or it's wrong). How am I supposed to prove that the global minimum occurs there if the second partial derivative test fails?

...Or am I over-analyzing things again?
The formula $f(x,y) = (x^2-y^2)^2$ shows that f(x,y) is a square, so it can never be negative. Therefore $f(x,y)\geqslant0$ for all $(x,y)$, and 0 is the minimum possible value of f(x,y). On the other hand f(0,0) = 0. That proves that f takes its least possible value at (0,0). Hence, by definition, a global minimum occurs at (0,0).

5. Well, I've gotten the first half out of the way, though the second half again has the situation where (my work is below)

$D=-25x^8+82x^6-9x^4-40x^3y-12xy$
$D(0,0)=0, f_{xx}(0,0)=0$
so again we cannot draw a conclusion from this about the critical point at $(0,0)$

According to a source (found here), each factor (he wasn't very clear on what he meant by factor) has a saddle point at $(0,0)$.
He claims, however, that $f$ has a minimum at $(0,0)$.

Would he be right? And if so, how exactly am I supposed to show it?

6. Originally Posted by Runty
Well, I've gotten the first half out of the way, though the second half again has the situation where (my work is below)

$D=-25x^8+82x^6-9x^4-40x^3y-12xy$
$D(0,0)=0, f_{xx}(0,0)=0$
so again we cannot draw a conclusion from this about the critical point at $(0,0)$

According to a source (found here), each factor (he wasn't very clear on what he meant by factor) has a saddle point at $(0,0)$.
He claims, however, that $f$ has a minimum at $(0,0)$.

Would he be right? And if so, how exactly am I supposed to show it?
No, the person who claimed that $f(x,y) = (y-x^3)(y-x^5)$ has a minimum at $(0,0)$ is wrong. Look again at the hint in my comment #2 above, and think about what happens to the function as (x,y) approaches 0 along the curve $y=x^4$.