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Critical point

  1. Nov 13, 2004 #1
    let f(x)=sin(1/x)*x^2 for x not 0, and f(0)=0. show that x=0 is a critical point for f which is neither a local minimum, a local maximum, nor an inflection point.

    well I tried differentiating this, and got f'=-cos(1/x) +2xsin(1/x). to find a critical point i make f'=0. Not sure how to do this. Do I take the limx->0?

    Any hints or tips would be great
     
  2. jcsd
  3. Nov 13, 2004 #2

    arildno

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    What you've done here, is to find the derivative of f at all points EXCEPT at x=0!
    But you are to find f'(0)...
    Use the definition of the derivative.
     
  4. Nov 13, 2004 #3
    k thanks i'll try that
     
  5. Nov 14, 2004 #4

    arildno

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    If you're a bit unsure what I mean, the definition of f'(0) is:
    [tex]f'(0)=\lim_{h\to{0}}\frac{f(0+h)-f(0)}{h}[/tex]
     
  6. Nov 14, 2004 #5
    yeh i got that to work, now how do I show that it's not a local min,max or inflection. Would I look at the second derivative? If that's not defined it's not anything?
     
  7. Nov 15, 2004 #6

    arildno

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    The standard second-derivative fails, since the first derivative is discontinuous at x=0 (the 2.derivative is not defined).

    It remains to be shown that f(0) is not a local maximum/minimum.
    This should be fairly easy to show..

    Use, for example, the following definition of local maximum:
    We say that a function f has a local maximum at [tex]x_{0}[/tex], iff there exists a [tex]\delta>0[/tex] so that for all [tex]x\in{D}(x_{0},\delta),f(x)\leq{f}(x_{0})[/tex]
    I've assumed that the x's in the open [tex]\delta[/tex]-disk are in the domain of f, as is the case in your problem.

    Note that this definition makes no assumption of differentiability or continuity of f.
     
    Last edited: Nov 15, 2004
  8. Nov 16, 2004 #7
    ok i finished that part of the question.( this is a 4 part question)

    I can't figure out these 2 parts. Any tips would be fantastic
    f(x)=x^2*sin(1/x)
    1.let g(x)=2x^2 +f(x) (f from the first question i asked)

    Show g has a global minimum at x=0 but g'(x) changes sign infinitely often on both (0,e) and (-e,0) for any e>0.

    For this question I can easily show 0 is a critical point. But when I show it's a minimum is what's difficult, when I differentiate twice I cannot see that f''(0)>0


    2. Let h(x)=x+2f(x). Show h'(0)>0, but h is not monotone increasing on any interval that includes 0.

    I know how to show h'(0)>0 but have no idea how to show it's monotone increasing.



    Again any help would be fantastic
    thanks in advance
     
  9. Nov 17, 2004 #8

    arildno

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    Show that for 1., g(x)>=x^2 for ALL x.
    How can that help you in showing that x=0 must be a global minimum?
     
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