Is (0,0) a minimum or saddle point of $f(x,y)=2x^4+y^2−3yx^2$?

Question

I saw that $f(x,y)=2x^4+y^2−3yx^2$ has a critical point at $(0,0)$ calculating the gradient of the function and equalizing to (0,0). if I study the behavior of the function at an environment of $(0,0)$ reducing the problem to a single-variable function using straights with the form $y=ax$ I have $f(x,ax)=2x^2+x^2a^2-3ax^3$ then $f'(0,0)=0$ and $f''(0,0)=4+2a^2>0$ so along any straight line that passes through the origin, f has a minimum at $(0,0)$ But if I study too the behavior of the function at an environment of $(0,0)$ reducing the problem to a single-variable function using parabolas with the form $y=ax^2$ namely: $f(x,ax^2)=2x^4+a^2x^4-3ax^4=x^~(2-3a+a^2)$ then, $f'(x,ax^2)=4x^3(2-3a+a^2)$ and $f'(0,0)=0$. Then, $f''(x,ax^2)=12x^2(2-3a+a^2)$ so $f''(0,0)=0$ given that $x^2>0$ for all $x$ pertaining to real numbers, $x≠0$ then: $i)$ if $a=-1, f''(x,-x^2)>0$ $ii)$ if $a=3/2, f''(x,3x^2/2)<0$ so by $i)$ and $ii)$ $(0,0)$ is a saddle point

So I don't know if $(0,0)$ is a minimum or saddle point.

Answer 1

[Answering the edited post]

Your analysis with the parabolas show correctly that $(0,0)$ is a saddle point.

Your situation shows precisely what we must be wary of when considering limits and local behaviour in more dimensions. Even if the point is a minimum along any straight line going through it, it can still be a saddle point, since we must consider all curves going trough the point. This is a very important thing to remember.

EDIT/NOTE: Usually, we would try to use the second partial derivative test (Hessian matrix) to determine the type of critical point. In this case, the test is inconclusive, so we have to get creative.

Answer 2

We have that

$$f(x,y)=2x^4+y^2−3yx^2=-\frac14x^4+\left(y-\frac32x^2\right)^2$$

therefore it is a saddle point indeed for $x\neq 0$ we have

• for $y=0\implies f(x,y)>0$

• for $y=\frac32x^2\implies f(x,y)<0$

There are not general methods when hessian matrix test fails. Excluding numerical methods, by hands the best we can do is to try some good manipulation in order to determine the sign of the function near the point of interest.