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Critical Point when f is undefined

  1. Apr 21, 2015 #1
    1. The problem statement, all variables and given/known data
    I am being asked to find the absolute extrema of the function

    [tex] f(x) = 3x^{2/3}- 2x, [-1,1] [/tex]

    2. The attempt at a solution
    I have taken the derivative and simplified it to:

    [tex] \frac {1- \sqrt [3] x}{\sqrt [3]x} = 0 [/tex]

    Here is my dilemma: 0 lies in the interval, and I know that f'(x) is undefined at x=0. Does this make x=0 a critical point? Is it wrong to multiply both sides by the denominator and simply be left with [tex] 1- \sqrt [3]x = 0 [/tex] leaving x = 1 as my only CP? Thanks!
     
  2. jcsd
  3. Apr 21, 2015 #2

    Orodruin

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    What branch of ##x^{2/3}## are you using for negative x? Regardless, if f' is not defined you will simply have to check those points explicitly.
     
  4. Apr 21, 2015 #3

    MarcusAgrippa

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    Answers: 1. Ask: what is the definition of a critical point? Then check whether "undefined" is classified as a "critical point".
    2. No, it is not wrong. And 1 is a critical point.
     
  5. Apr 21, 2015 #4
    What do you mean what branch of ##x^{2/3}## am I using for negative x? Sorry to ask what is most likely a silly question.
     
  6. Apr 21, 2015 #5

    Orodruin

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    Your quoted interval is [-1,1]. What is ##f(-1)##?
     
  7. Apr 21, 2015 #6
    5, unless I'm horribly mistaken somewhere.
     
  8. Apr 21, 2015 #7

    Orodruin

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    Is ##(-1)^{2/3} = 1##?
     
  9. Apr 21, 2015 #8

    PeroK

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    Rational powers can be well defined on negative Real numbers as long as the denominator is odd.

    ##(-1)^{3/2}## would be problematic.
     
  10. Apr 21, 2015 #9
    Yes.

    I'm pretty sure I understand what you're saying. My question here was whether or not I should keep the denominator when I get a derivative in a form such that, say, x is in the denominator, making the derivative undefined when x=0. OR, should I always make it a point to multiply both sides by the denominator in order to only have a numerater = 0.
     
  11. Apr 21, 2015 #10

    Orodruin

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    This is true only for a very particular interpretation, which also breaks the relation ##a = b^c \ \Rightarrow \ b = a^{1/c}##. Generally, the solution to an equation involving ##x## must be taken with that in mind. How do you interpret ##x^{1/3}## in your derivative? It must be taken with the same interpretation as you use for the function!
     
  12. Apr 21, 2015 #11

    PeroK

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    Y
    Drawing a graph of f(x) will answer your question about whether 0 is a critical point.

    I can't see that it matters about having a denominator.

    a/b = 0 iff a = 0, where b not = 0
     
  13. Apr 21, 2015 #12

    PeroK

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    ##x^{1/3}## and ##x^{2/3}## look like well defined real functions to me. Differentiable everywhere except 0. There's no ambiguity in the function values or derivatives that I can see.

    ##x^{2/3}## is not invertible unless you restrict it to non-negative x, but that's not relevant here.
     
  14. Apr 21, 2015 #13

    Mark44

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    @Cosmophile, your thread title ('Critical Point when f is undefined') might be an indication of your misunderstanding. Per wikipedia, a critical point is a point in the domain of f at which f' = 0 or is undefined. For the function in this problem, f(x) is defined for all real numbers. It is true that f' is not defined at x = 0, but the function itself is defined at x = 0.
     
  15. Apr 21, 2015 #14
    Nope, that's just a silly typo on my end that I didn't catch. Thanks! Haha
     
  16. Apr 21, 2015 #15
    I thought they were well-defined, too.

    So, I do need to include 0 as one of my critical numbers because f'(x) is undefined at x = 0. That makes sense to me. I just hadn't remembered the entire definition of a critical point, so I thought to just multiply both sides by the denominator.

    Thank you, everyone!
     
  17. Apr 21, 2015 #16

    Orodruin

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    Did you ever try feeding it into Mathematica or Wolfram alpha? The analytic continuation from the real axis certainly is not a real function on the same branch as it is real on the positive axis. In order to make it real on the negative axis you need to pick a different branch.
     
  18. Apr 22, 2015 #17

    SammyS

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    In the problem at hand, and at the level it's presented It's a real function over a real domain.

    I believe all this discussion about branches, etc. has just confused OP.
     
  19. Apr 22, 2015 #18

    Orodruin

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    You are probably correct. Let us drop that part of the discussion.
     
  20. Apr 22, 2015 #19

    Orodruin

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    Yes, you need to include it. The function can have an extremum where the derivative is not defined.
     
  21. Apr 22, 2015 #20
    Yep yep, solved! And the discussion of the branches didn't necessarily confuse me; in fact, it all sounds extremely interesting, and I'd love to know more about it. I'm going through Serge Lang's text and am using a different textbook's practice problems so that I can check answers (my copy of Lang's text is a PDF, and skipping to answers is tedious), and I was just eager to figure this out so I can move on. I appreciate the insight - even the insights that weren't necessarily tuned for my question.
     
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