Earlier this semester in my Multivariable Calculus course we were discussing the second derivative test. Recall the pesky condition that if is a critical point and , then the test fails.

A student emailed me after class and asked the following question. Suppose a function has a critical point at and . Moreover, suppose that as we approach along we have when and when . Is that enough to say that the critical point is a not a maximum or a minimum? His thought process was that if we look at slices we get curves that are concave up when and curves that are concave down when —surely that could not happen at a maximum or minimum.

I understood his intuition, but I was skeptical. Indeed, after a little playing around I came up with the following counterexample. The function is

The first partial derivatives are

Clearly is a critical point. The second partial derivatives are

Thus . So the second derivative test fails. But observe that when we have

So when and when . Yet it is easy to see that is a minimum: and for all . A graph of the function is shown below. You can see the concave down cross sections for and .

I think the functions and also work. They are inspired by the pitchfork bifurcation (compare with the shape of the contour ).

By:

Jan Van lenton April 19, 2012at 1:15 pm

I just realised that the examples I gave are closely related to the Rosenbrock function [1].

[1] http://en.wikipedia.org/wiki/Rosenbrock_function

By:

Jan Van lenton April 19, 2012at 3:52 pm

Wow. Cool. Thanks for sharing that!

By:

Dave Richesonon April 19, 2012at 4:09 pm

I actually meant and , but the other two also work. From the contour plots for all these functions it is clear that they use the same idea.

By:

Jan Van lenton April 19, 2012at 1:35 pm

Great! Thanks. I haven’t checked the derivatives, but it looks good on Grapher.

By:

Dave Richesonon April 19, 2012at 2:29 pm