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Our professor wrote:

Boundary points: points on the edges of the domain if only such points

stationary: points in the interior of the domain such that f is differentiable at x,y and gradient x,y is a zero vector.

singular points - points in the interior of the domain such that f is not differentiable at x,y

I understood that to classify the type, we have to look at the domain.

For example, sqrt of x gives you a domain x>= 0. It is clear that the partial derivative does not exists at (0,0), because there is a square root of x in the bottom. So I say that it is a singular point.

Our professor said it was also a boundary point. I agreed with him in class, but I didn't remember my reason... lmao.

In the case of x^2, it is defined across the x-axis, so when we take f ' (x), we get 2x = 0, and the critical point is x = 0.

This is a stationary point.

So what is the best and most convinent way to distinguish the three types? Sometime the function can be quite ugly?

My approach would be #1 check the domain, and if it only appears to be the points on the boundary, such as sqrt(x^2-y^2-1), then i suppose the critical points are boundary, because fx and fy does not exists for this function?

#2 if it is a polynomial function, it should always be the stationary point?

#3 the trouble point is the singular point.

But it seems like in the case of sqrt(x), 0 is both singular and boundary, just how one looks at it. Then how do we justify it?

Thank you