Is $(0,0)$ a local minima,local maxima or the saddle point?

Question

$f(x,y)=5x^2+xy^3-3x^2y$

My attempt:I calculated

$$f_{xx}(0,0)=10, \hspace{0.3cm}f_{xy}(0,0)=0, \hspace{0.3cm}f_{yy}(0,0)=0$$

So $$f_{xx}(0,0)f_{yy}(0,0)-(f_{xy}(0,0))^2=0$$

I concluded nothing from this test.I do not know other method to solve this problem.Please help me.

Thanks.

Similar question here: http://math.stackexchange.com/questions/416432/proving-the-origin-is-a-saddle-point — , Oct 01 '14 at 06:11

Christian Blatter · Answer 1 · 2014-10-01T18:02:37.040

3

It is true that $\nabla f(0,0)=(0,0)$ and that the Hessian $d^2f(0,0).(X,Y)=10X^2$ is positive semidefinite. The latter implies that the second derivative test is inconclusive in this example.

Fact is that our $f$ has an intricate zero at $(0,0)$. Writing $f$ in the form $$f(x,y)=(5-3y)\>x\left(x+{y^3\over5-3y}\right)$$ shows that near the origin we have $f(x,y)=0$ along two curves having the $y$-axis as common tangent, namely $x=0$ and $$x=-{y^3\over 5-3y}\ ,$$ the latter resembling the cubic parabola $x=-{y^3\over5}$ near the origin. Crossing either of these two curves changes the sign of $f$. This shows that $f$ assumes as well negative as positive values near $(0,0)$. It follows that we don't have a local extremum of $f$ at $(0,0)$.

edited Oct 01 '14 at 18:02

answered Oct 01 '14 at 11:28

Christian Blatter

226,825

Nice solution! If you don't mind, would you please explain how you got $x=-\frac{1}{5}y^3$ from your factorization of $f(x,y)$. (Did you just set $y=0$, and then set the second factor equal to 0?) – user84413 Oct 01 '14 at 16:35
@user84413: See my edit. Now nothing is "out of the blue". – Christian Blatter Oct 01 '14 at 18:03
Thanks for the explanation. – user84413 Oct 01 '14 at 18:05

Is $(0,0)$ a local minima,local maxima or the saddle point?

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