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Critical values

  1. Feb 10, 2005 #1
    I'm supposed to find all critical numbers of the given function.
    Book defines critical number c as the place where f'(c)=0 or where f is not differentiable.

    1. g(x)= x + 1/x
    2. f(x)= x ln (x)

    work:
    1. [tex] g'(x) = \frac{x^2 - 1}{x^2} [/tex]

    c= 1, -1, 0

    Zero is wrong however. I put it in the answer because the function is not differentiable at that point. So I'm guessing I have the definition of differntiable at a point wrong. The limit doesn't exist as x approaches 0 so I thought the function wouldnt be differentiable there.

    2. f'(x) = lnx + x/x

    c= 1/e and all numbers less than or equal to zero.

    The book only lists 1/e so same problem here.
     
  2. jcsd
  3. Feb 10, 2005 #2
    to add to your definition of a critical number, c must also be in the domain of the original function, thus for g(x), 0 is not in the domain of the original function and so it is not a critical number.
     
  4. Feb 10, 2005 #3

    dextercioby

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    Why...?Are those numbers in the domain of the function "f"...?If so,is the derivative zero...?

    Daniel.
     
  5. Feb 10, 2005 #4
    Thx, math student I understand now. I was used to my pre-cal teacher teacher telling me to find critical points to solve rational inequalties which included numbers not in the domain.
     
  6. Feb 10, 2005 #5

    dextercioby

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    I'm glad you figured out this is something totally different and that the domain of the function is essential.

    Daniel.
     
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