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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?
  2. jcsd
  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


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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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