Boundary, stationary, and singular points

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This is a topic in multi-variable calculus, extrema of functions.

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
 

Answers and Replies

  • #2
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This is a topic in multi-variable calculus, extrema of functions.
Since the topic is multi-variable calculus, why are many of your examples single-variable functions?
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.
Boundary point - 0 is not in the interior of the domain, which is [0, inf). Also, you're talking about the plain derivative, not a partial derivative.
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?
Why do you say that the partials don't exist? You're not being very clear here - is the function f(x, y) = sqrt(x^2 - y^2 -1)?
#2 if it is a polynomial function, it should always be the stationary point?
Yes, because polynomials are continuous and differentiable everywhere, so there are no boundary points to consider and no singularities.
#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?
As already stated, 0 is a boundary point of this function.
Thank you
 
  • #3
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Thank you Mark. :)
 

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