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A simple question about maxima/minima

  1. Apr 22, 2008 #1
    I'm perpetually confused on this topic.

    i) We all know that stationary points in 1D are either minima, maxima or inflection points, but consider y=|x|. x=0 is not a stationary point, and yet it is clearly the point at which y is the smallest. Am I technically correct in calling x=0 a 'minimum'? -or should I use some other terminology?

    ii) consider y=1/x. In the limit x-> infinity, the derivative dy/dx -> 0. Does that mean that there exists a stationary point in this limit? I know it's an asymptote, but can we speak of a horizontal asymptote as also being a stationary point? (Stationary line?)
  2. jcsd
  3. Apr 22, 2008 #2


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    1) critical points also occur when the derivative of the function is undefined, which it is for |x|. (verify by taking the limit from both sides).

    2) It's not a stationary line because in reality the value of the function will change forever, but will get arbitrarily close to 0 for extremely large numbers (or not so large for the derivative since it's x^2).
  4. Apr 22, 2008 #3
    1) Hmm, well then I don't understand what a critical point is. It was my understanding that a critical point is the point at which the derivative is zero.

    2) Of course, I understand that it never actually reaches the stationary point for finite values of x, but what would a mathematician say? Would it be OK to claim that it approaches the stationary point in the limit x-> infinity?
  5. Apr 23, 2008 #4

    Gib Z

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    1) Look up Critical point in Wikipedia: http://en.wikipedia.org/wiki/Critical_point_(mathematics).

    As exk said, a Critical point occurs either when the derivative is zero or where the function ceases to be differentiable. We could neatly sum that up in a definition that I would use if I am ever to write a text book:

    "A critical point of the function f(x) occurs at [itex](x_0, f(x_0))[/itex] if and only if the function [itex]g(x) = 1/f'(x)[/itex] is discontinuous at [itex]x=x_0[/itex]."

    2) Only if the domain of your functions was the extended reals and you defined it as such. In the usual R->R form, this limit approach is not considered correct.
    Last edited: Apr 23, 2008
  6. Apr 23, 2008 #5

    1) Thanks for the reply. I think half my problem is being confused over the strict technical definition of terms. But yes, that's very clear, I was wrong about the definition of 'critical points'.

    2) So, the normal domain of functions is assumed to exclude those nasty infinities? But, doesn't all talk about asymptotes implicitly require the existence of those infinities? If I restrict a function to finite numbers only, then can it really be said to have an asymptote?
  7. Apr 23, 2008 #6

    Gib Z

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    When a function is defined from R to R, it does restrict them to finite numbers, but not a finite number of them =] When we take the limit it just says the value of 1/x goes to 0 as x continues along the sequence of the integers etc.
  8. Apr 23, 2008 #7
    Thanks. That sort-of makes sense. OK, I give in! I'll accept that in common usage, a horizontal asymptote is not regarded as the same sort of beast as a stationary point.

    (This problem came up in a tutorial I was giving to an undergraduate. I'm a physicist, not a mathematician, and I want to make sure that I'm telling him something approaching the truth. It may sound dumb, but I do want to be careful that I'm using the right terminology, so your help is much appreciated.)
  9. Apr 23, 2008 #8

    Gib Z

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    Well at least you are one of those good physicists that make attention to make sure they are doing their math properly :D I love those [tex]f(x) = 1 + x + \frac{x^2}{2} + O(x^3), \rightarrow f(x) = e^x[/tex] jokes =D
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