What to do when the second derivative test fails?

Question

What do we do when the second derivative test fails?

For example, I'm asked to find all the critical points of the function $$f(x,y)=x^{2013}−y^{2013}$$

And determine the nature of the critical points.

The critical point that I have found is at $(0,0)$, but I'm unable to determine its nature as the second derivative test fails here.

Answer 1

Can be seen easily without second derivatives that (0,0) is a saddle point because $f$ takes values $>f(0,0)$ and $<f(0,0)$ arbitrarily near of (0,0):

For $(x,y)=(\epsilon,0)$, $\epsilon>0$: $f(x,y)=\epsilon^{2013}>0$.

For $(x,y)=(0,\epsilon)$, $\epsilon>0$: $f(x,y)=-\epsilon^{2013}<0$.

Answer 2

Let $y=0$ and see what your function looks like. Then let $x=0$ and see what the function looks like.

Alternatively: Here's a very similar function (in terms of behavior around $(0,0)$): $g(x,y) = x^3-y^3$. It might help to see what this does if it's not clear to you what $f$ does. See here if you still have difficulty.

Here's a more rigorous argument for why it is a saddle point: Let $y=cx$. Then

$$f(x,y) = x^{2013}-c^{2013}x^{2013} = (1-c^{2013})x^{2013}.$$

For any value of $c$, this does not have concave-up nor concave-down behavior. Hence it cannot be a local min or max and hence is a saddle point.

Answer 3

Hint:

• Take into consideration higher-order derivatives.
• Note the parity of the first non-zero derivative.
• What are the similarities among $x^3, x^5, x^7,\ldots$ and similarities among $x^2,x^4,x^6,\ldots$ (e.g. how the graphs would look like, and what is the parity of the first non-zero derivative)?
• You can read more about it here.

I hope this helps $\ddot\smile$

Answer 4

Note that $f(x,y) > 0$ whenever $x>0$ and $y<0$. Moreover, whenever $x<0$ and $y>0$ we have $f(x,y)<0$. Thus, as $f(0,0)=0$, we have points $(x_1,y_1)$ and $(x_2,y_2)$ in every neighborhood of $(0,0)$ such that $f(x_1,y_1) < f(0,0) < f(x_2,y_2)$. As $f$ is defined on $\mathbb{R}^2$ and hence on the just constructed points, we see that $(0,0)$ satisfies the definition of a saddle point of $f$.

Answer 5

(I was having a bit of indecision as to whether this merited being an answer, or just a comment, since it is related to the answers already posted.)

If we note the "cross-sections" in the $ \ yz- \ (x = 0) \ $ and $ \ xz- \ (y=0) \ $ planes, we note that the function $ \ f(x,y) \ $ reduces to $ \ f(0,y) \ = \ -y^{2013} \ $ and $ \ f(x,0) \ = \ x^{2013} \ $ . Both of these "reduced" functions have odd symmetry about the $ \ z-$ axis.

For a function of two variables, a point cannot be a local extremum, but will be a saddle point, if either one or both of these "cross-sectional" functions is odd about the axis through that point. If both of the functions are even, both must have the same direction of concavity about that point in order for it to be a local extremum. If the two functions have opposite concavities, the point will be a saddle point. (This is related to the concept of Gaussian curvature -- in that view, a local extremum has $ \ K \ > \ 0 \ $ .)

So at the origin, we will find that $ \ x^{odd} \ \pm \ y^{odd} \ $ will produce a saddle point there, $ \ \pm \ (x^{even} \ + \ y^{even}) \ $ has a local extremum there, and will be a saddle point for $ \ \pm \ (x^{even} \ - \ y^{even}) \ $ . This can be extended to more complicated symmetrical functions and to functions with more variables. (A local extremum requires more care to construct in larger numbers of dimensions, since the directions of concavity must agree in all of those dimensions.)

[EDIT -- In addition, it might be noted that we will also have a saddle point at the origin if the parities of the "cross-sectional" functions differ, as with, for instance $ \ x^{odd} \ \pm \ y^{even} \ $ . ]

The "second-derivative test", like a lot of mathematical "tools", has a limit to its utility. We must expect to do some extra investigation on critical points when we deal with power functions with exponents greater than two.