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Homework Help: Boundary, stationary, and singular points

  1. Jun 28, 2010 #1
    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
  2. jcsd
  3. Jun 28, 2010 #2


    Staff: Mentor

    Since the topic is multi-variable calculus, why are many of your examples single-variable functions?
    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.
    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)?
    Yes, because polynomials are continuous and differentiable everywhere, so there are no boundary points to consider and no singularities.
    As already stated, 0 is a boundary point of this function.
  4. Jul 5, 2010 #3
    Thank you Mark. :)
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