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Critical points

  1. Dec 7, 2013 #1
    What are the critical numbers of f(x) = (x2+3)/(x-1)

    I took the derivative and found f'(x) = (x+1)(x-3)/(x-1)

    so I thought the critical numbers were 3, ±1

    ...sadly, this is incorrect. I always thought the definition of a critical number was where f'(x) = 0 or DNE. But for this to be applicable, must the value for x firstly exist in the domain of f(x) or not? Since for all values of x not in the domain of f(x), it would also not exist in f'(x) and that includes vertical asymptotes. So are values for vertical asymptotes not considered critical numbers?
     
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  3. Dec 7, 2013 #2

    Ray Vickson

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    Things vary a bit from source-to-source, but I don't think any of them would regard the vertical asymptote in this case a a critical point.

    The source http://en.wikipedia.org/wiki/Critical_point_(mathematics) regards a critical point c as having f'(c) = 0---and does not allow for non-existence of f'(c). (However, one paragraph therein does mention that other authors define it slightly differently). The source http://tutorial.math.lamar.edu/Classes/CalcI/CriticalPoints.aspx allows for the non-existence of f'(c), but only in cases where f(c) itself does exist---and it emphasises this last point. In your case f(c) does not exist at c = 1, so c = 1 would not be called critical by either of these two sources. The source http://www.sosmath.com/calculus/diff/der13/der13.html emphasizes the relationship between critical points and local optima---even when f'(c) does not exist. Here, too, the point c=1 in your example would not be regarded as 'critical'.
     
  4. Dec 7, 2013 #3

    vela

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    The Wikipedia article cites Stewart's calculus text as the source for the definition, yet I just checked and that text defines a critical point c as a point in the domain of f where f'(c) = 0 or f'(c) does not exist.


    Edit: I see the Wikipedia article refers to only differentiable functions so the non-existence of f'(c) is automatically ruled out.
     
  5. Dec 7, 2013 #4
    Haha yep. I'm using Stewart's Calculus right now and it essentially states if f'(x) = 0 or DNE, then it is considered a critical point. I was actually doing a previous exam, and the answer was that the vertical asymptote is not included since it is not defined for f(x)--it's not in the domain of the original function. Any further thoughts or comments to add?
     
  6. Dec 8, 2013 #5

    Ray Vickson

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    So, that is exactly what the second source I quoted says and the third one implies.
     
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