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Do we include asymptotes as critical points?

  1. Nov 30, 2007 #1
    I was wondering if we include asymptotes as criticals:

    For example in y = x sqrt(4-x^2)

    when u find the critical points it gives u root 2 and negative root 2

    But when I draw the graph im missing the 2 and -2 which were the asymptotes which help define the graph?

    So are they needed?

    Thanks
     
  2. jcsd
  3. Dec 1, 2007 #2
    In general, a point that is not differentiable should be considered a critical value. However, the domain should be [tex]{ x,-2 \leq x \leq 2 }[/tex]. In other words, [tex]\sqrt{0}[/tex] is defined.
     
  4. Dec 1, 2007 #3

    arildno

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    You may use the vertical lines x=2 and x=-2 to help you draw the curve in that the graph of this function becomes vertical at those points.
     
  5. Dec 1, 2007 #4
    yay thank you!
     
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