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Math Help - Neighborhoods/differentiability

  1. #1
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    Neighborhoods/differentiability

    Let f(x)=x+2x^2*sin(1/x) for x=/=0 and f(0)=0

    Prove that f(x) is differentiable everywhere and that f'(0)>0, but there is no neighborhood of 0 on which f is increasing.


    I differentiate using the standard rules to get:

    f'(x)=-2cos(1/x)+4xsin(1/x)+1 for x=/=0

    And when x=0, I look at

    lim(x->0) x+2x^2*sin(1/x)/x, which upon factoring, gives that the limit is equal to 1>0, and so f'(0)>0.

    The part I can't get is the last part, show there is no neighborhood of 0 on which f is increasing. I know the sin(1/x) term gives wild oscillation, and 1/x is unbounded, but I don't know how to formally show the conclusion that there is no neighborhood of 0 on which f is increasing.

    Thanks.
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  2. #2
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    Quote Originally Posted by zhupolongjoe View Post
    Let f(x)=x+2x^2*sin(1/x) for x=/=0 and f(0)=0

    Prove that f(x) is differentiable everywhere and that f'(0)>0, but there is no neighborhood of 0 on which f is increasing.


    I differentiate using the standard rules to get:

    f'(x)=-2cos(1/x)+4xsin(1/x)+1 for x=/=0

    And when x=0, I look at

    lim(x->0) x+2x^2*sin(1/x)/x, which upon factoring, gives that the limit is equal to 1>0, and so f'(0)>0.

    The part I can't get is the last part, show there is no neighborhood of 0 on which f is increasing. I know the sin(1/x) term gives wild oscillation, and 1/x is unbounded, but I don't know how to formally show the conclusion that there is no neighborhood of 0 on which f is increasing.

    Thanks.

    For any neighborhood I_\epsilon:=(-\epsilon,\epsilon) of zero there exists n\in\mathbb{N}\,\,\,s.t.\,\,\,x_n:=\frac{1}{2\pi n}\,,\,\,y_n:=\frac{1}{(4n+1)\pi\slash 2}\in I_\epsilon , but:

    f'(x_n) =-2\cos(2\pi n)+8\pi n\sin(2\pi n)+1 = -1<0

    f'(y_n)=-2\cos\left(\frac{(4n+1)\pi}{2}\right)+2(4n+1)\pi\s  in\left(\frac{(4n+1)\pi}{2}\right)+1=2(4n+1)\pi+1>  0

    Thus, the derivative of the function doesn't keep the same sign in any neighborhood of zero \Longrightarrow f isn't increasing (or decreasing) in any such neighborhood.

    Tonio
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  3. #3
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    Thanks. One question though. Since I found in the problem that f'(0)=1, can I just use the xn you have and that or do I need to find another yn as you did?
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  4. #4
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    Quote Originally Posted by zhupolongjoe View Post
    Thanks. One question though. Since I found in the problem that f'(0)=1, can I just use the xn you have and that or do I need to find another yn as you did?

    I don't understand your question: the xn's and yn's we defined served to show that in any neighborhood of zero f ' has positive and negative values, so it does not keep one single sign in it ==> f isn't increasing or decreasing in no neighborhood of zero. What does this have to do with the actual value of f'(0)?

    Tonio
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  5. #5
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    Okay, nevermind...that answered my question, thanks.
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