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Math Help - differentiable function

  1. #1
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    differentiable function

    Hi!

    I must prove that the function f(x) is continuously differentiable function (just first derivate) on interval (-1,1).
    f(x)=exp(\frac{1}{x^2-1}) \ |x|<1 , \ 0 for all other values.

    I do that:

    f(x)=e^{\frac{1}{x^2-1}}=e^u \ [u(x)=\frac{1}{x^2-1}]
    \frac{d}{dx}f(x)=\frac{df}{du}\frac{du}{dx}=-e^u \frac{2x}{(x^2-1)^2}=-\frac{2x}{(x^2-1)^2}e^{\frac{1}{x^2-1}

    Is it ok?

    Thanks in advance
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by tom27 View Post
    Is it ok?
    If the question it is only related to (-1,1) then, it is right (evidently f'(x) is continuous in (-1,1) ).


    Fernando Revilla
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  3. #3
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    The basic question is if the f(x) is a part of C_0^1(\mathbb{R}) space?
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  4. #4
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    That's actually a different question from what you first asked. In your first post you said "continuously differentiable function (just first derivate) on interval (-1,1)". Now you are saying on all of \mathbb{R}. You had already answered the first question- f(x) is continuously differentiable on (-1, 1). Since f is defined to be a constant (0) for x\le -1 or x\ge 1, it is certainly continuously differentiable there. But now the question is about x= -1 and x= 1. What are the derivatives at those points? What are the derivatives on either side? Is the derivative continuous at x= -1 and x= 1?
    Last edited by HallsofIvy; January 8th 2011 at 03:52 PM.
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by tom27 View Post
    The basic question is if the f(x) is a part of C_0^1(\mathbb{R}) space?
    In that case, you should prove f'_{-}(1)=f'_{+}(-1)=0 and f' continuous at x=\pm 1 (for the rest is almost trivial)


    Fernando Revilla

    P.S. I suppose C_0^1(\mathbb{R}) means C^1(\mathbb{R}) (the standard notation for the set of continuosly differentiable functions on \mathbb{R} ) .


    Edited: Sorry, I didn't see HallsofIvy's post
    Last edited by FernandoRevilla; January 8th 2011 at 09:23 AM.
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  6. #6
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    How can I prove f'_{-}(1)=f'_{+}(-1)=0?

    I tried with limits:

    \displaystyle \lim_{h \to 1}\frac{e^{\frac{1}{(x+h)^2-1}}-e^{\frac{1}{x^2-1}}}{h} and \displaystyle \lim_{h \to -1}\frac{e^{\frac{1}{(-x-h)^2-1}}-e^{\frac{1}{x^2-1}}}{h}

    but this certainly is not the correct way.
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  7. #7
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    While derivatives are not necessarily continuous, they do obey the "intermediate value property"- f'(x) takes on all values between f'(a) and f'(b) for x between a and b. In particular, that means that f'(c) exists if and only if \displaystyle \lim_{x\to c^-} f'(x)= \lim_{x\to c^+}f'(x). That is, determine the derivatives for x< -1, and x>-1 and determine if the limits at x= -1 are the same. Do the same for x= 1.

    By the way, you don't want to prove that " f'_-(1)= f'_+(1)= 0". You want to prove that f'_-(1)= f'_+(1) and f'_-(-1)= f'_+(-1). You don't need to use the limit definition of the derivative- but, of course, h goes to 0, not 1 and -1.
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  8. #8
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    So f'_-(-1)=f'_+(1)=0 and i must calculate:

    f'_-(1)=\displaystyle \lim_{x \to 1} -\frac{2x}{(x^2-1)^2}e^{\frac{1}{x^2-1}}
    f'_+(-1)=\displaystyle \lim_{x \to -1} -\frac{2x}{(x^2-1)^2}e^{\frac{1}{x^2-1}}

    but how can I do that? The exp function is not defined in those two points!
    Last edited by tom27; January 9th 2011 at 04:52 AM.
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  9. #9
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    Fuction with exp is not define in those two points so limit does not exist. But wikipedia says: the function is smooth so is continuosly differentiable on \mathbb{R}. But I dont know how to prove that if one limit does not exist so f'_-(1)\ne f'_+(1)?
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