Why Can't the Gradient Vector Be Undefined at a Minimum?

[quasar987]

I have few questions about extrema of fonctions of two variables. It is written in my textbook: "At a local maximum, the gradient vector must be nul or undefined. A similar reasoning shows that the gradient must be nul at a local minimum." Actually there was no preceeding reasoning to this statement so I don't understand.

- Why couldn't the gradient vector be undefined at the min?

- If one of the partial derivative is undefined at a certain point, does it automatically means the point is a max? If no, how do you tell analytically?

- If (a,b) is a critical point because the gradient at (a,b) is 0 and if the test of the second order partial derivative fails (i.e. =0). How can I conclude analytically to the nature of the critial point?

 

[arildno]

The author has been sloppy!
- Why couldn't the gradient vector be undefined at the min?
It certainly can be!
.

 

[quasar987]

Ok, then I will rewrite question #2:

- If one of the partial derivatives is undefined at a certain point, how do I conclude analytically to the nature of the critical point?

- Could it be that a derivative in a direction other than the x or y-axis is undefined while it is defined in the direction of the x and y axis? In this case wouldn't the method of analysis of the partial derivatives fail to detect the critical point?

 

[Hurkyl]

Remember that the definition of optima don't involve derivatives at all -- when differential techniques fail, you often have to resort to the definitions to get your answers.

 

[arildno]

Remember that the definition of optima don't involve derivatives at all -- when differential techniques fail, you often have to resort to the definitions to get your answers.

And, I would think, some functions might be perverse enough to refuse yielding up where its extrema are, despite our best efforts..

 

[Hurkyl]

Even worse, there are functions that don't even have local extrema!

example: (here, p and q are relatively prime)


f(x) = 0 if x is irrational
f(p/q) = 1 - 1/q if q is even
f(p/q) = -1 + 1/q if q is odd