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Math Help - Squeeze theorem with differentiation

  1. #1
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    Squeeze theorem with differentiation

    I've attached the problem typed up to be more clear. I'm basically having trouble on how to prove that g'(x) exists using only that I know that f(x)<=g(x)<=h(x) for some neighborhood around x_0. I'm pretty sure I have to use some definition of the differentiation, but can't quite prove to myself that it exists without having more information.

    Any help would be appreciated.

    Squeeze theorem with differentiation-problem-15.bmp
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  2. #2
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    Quote Originally Posted by Dagger2006 View Post
    I've attached the problem typed up to be more clear. I'm basically having trouble on how to prove that g'(x) exists using only that I know that f(x)<=g(x)<=h(x) for some neighborhood around x_0. I'm pretty sure I have to use some definition of the differentiation, but can't quite prove to myself that it exists without having more information.

    Any help would be appreciated.

    Click image for larger version. 

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    First note that by the squeeze theorem that

    \displaystyle \lim_{x \to x_0} f(x)\le \lim_{x \to x_0} g(x) \le \lim_{x \to x_0} h(x) but since f(x_0)=h(x_0) they must also equal g(x_0)

    Now in any neighborhood of x_0 you have

    \displaystyle f(x_0+k) \le g(x_0+k) \le h(x_0+k)

    Now use the fact above that f(x_0)=h(x_0)=g(x_0) are equal to get

    \displaystyle f(x_0+k)-f(x_0) \le g(x_0+k)-g(x_0) \le h(x_0+k)-h(x_0)

    Can you finish from here?
    Last edited by TheEmptySet; February 23rd 2011 at 07:40 AM. Reason: Used the letter h to repesent two different things
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  3. #3
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    If \displaystyle f(x) \leq g(x) \leq h(x) in some neighbourhood of \displaystyle x_0, then

    \displaystyle f(x_0 + \Delta x) \leq g(x_0 + \Delta x) \leq h(x_0 + \Delta x), where \displaystyle \Delta x is some small change in \displaystyle x in the neighbourhood of \displaystyle x_0...

    \displaystyle \frac{f(x_0 + \Delta x)}{\Delta x} \leq \frac{g(x_0 + \Delta x)}{\Delta x} \leq \frac{h(x_0 + \Delta x)}{\Delta x}

    \displaystyle \lim_{\Delta x \to 0}\frac{f(x_0 + \Delta x)}{\Delta x}\leq \lim_{\Delta x \to 0}\frac{g(x_0 + \Delta x)}{\Delta x} \leq \lim_{\Delta x \to 0}\frac{h(x_0 + \Delta x)}{\Delta x}

    \displaystyle f'(x_0) \leq g'(x_0) \leq h'(x_0).


    I believe this proves \displaystyle g'(x_0) exists...
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    Quote Originally Posted by TheEmptySet View Post
    First note that by the squeeze theorem that

    \displaystyle \lim_{x \to x_0} f(x)\le \lim_{x \to x_0} g(x) \le \lim_{x \to x_0} h(x) but since f(x_0)=h(x_0) they must also equal g(x_0)

    Now in any neighborhood of x_0 you have

    \displaystyle f(x_0+h) \le g(x_0+h) \le h(x_0+h)

    Now use the fact above that f(x_0)=h(x_0)=g(x_0) are equal to get

    \displaystyle f(x_0+h)-f(x_0) \le g(x_0+h)-g(x_0) \le h(x_0+h)-h(x_0)

    Can you finish from here?
    Maybe you should use a different symbol to represent the change in \displaystyle x, since \displaystyle h is the name of a function...
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  5. #5
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    You are absolutely correct. Thanks
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  6. #6
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    Thanks for all the help. You guys are stating pretty much as far as I got. My worry is that in the final line where you take the limit of everything. Does the limit of the g function being between the f and h functions enough to prove that it exists? I might be over thinking this, but just proving that the limit, if it exists, would be between the f'(x) and h'(x) might not be enough to actually prove that it exists? Or would it?

    More specifically in terms of slope. You can have a undefined slope between a very steep positive slope and a very steep negative slope?

    Thanks again for all the help and fast feedback!
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