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Math Help - Show that g is differentiable and bounded

  1. #1
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    Show that g is differentiable and bounded

    Given the piecewise function

    g(x) =

    { x^(3)e^(-x^(2)/4)sin(4/x^(2)) if x=/=0
    { 0 if x = 0

    Prove that g is differentiable and g' is bounded on (-infinity,infinity).

    I know that I have to use the definition of a derivative for the first part (not too sure about how to do that). How to show it's bounded, I am not sure.

    This is the first time I have ever seen anything like this.

    Thanks for any help.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by katielaw View Post
    Given the piecewise function

    g(x) =

    { x^(3)e^(-x^(2)/4)sin(4/x^(2)) if x=/=0
    { 0 if x = 0

    Prove that g is differentiable and g' is bounded on (-infinity,infinity).

    I know that I have to use the definition of a derivative for the first part (not too sure about how to do that). How to show it's bounded, I am not sure.

    This is the first time I have ever seen anything like this.

    Thanks for any help.
    g(x)=\left\{\begin{array}{lr}x^3e^{-x^2/4}\sin(4x^{-2}):&x\neq0\\0:&x=0\end{array}\right\}

    f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})

    Do you know where to go from here?

    See Tonio's post below for proving g' is bounded.
    Last edited by redsoxfan325; November 20th 2009 at 02:37 PM.
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  3. #3
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    Quote Originally Posted by katielaw View Post
    Given the piecewise function

    g(x) =

    { x^(3)e^(-x^(2)/4)sin(4/x^(2)) if x=/=0
    { 0 if x = 0

    Prove that g is differentiable and g' is bounded on (-infinity,infinity).

    I know that I have to use the definition of a derivative for the first part (not too sure about how to do that). How to show it's bounded, I am not sure.

    This is the first time I have ever seen anything like this.

    Thanks for any help.

    g(x)=\left\{\begin{array}{cc}x^3e^{-x^2\slash 4}\sin\frac{4}{x^2}&\,\mbox{ if }\,x\neq 0\\0&\,\mbox{ if }\,x=0\end{array}\right.

    g'(x)=\left\{\begin{array}{cc}e^{-x^2\slash 4}\left(3x^2\sin\frac{4}{x^2}-\frac{x^4}{2}\sin\frac{4}{x^2}-8\cos \frac{4}{x^2}\right)&\,\mbox{ if }\,x\neq 0\\4&\,\mbox{ if }\,x=0\end{array}\right.

    Pay attention to the fact that the derivative is an even function and thus it's enough to show boundness in, say (0,\infty),:

    |g'(x)|=\left|e^{-x^2\slash 4}\left(3x^2\sin\frac{4}{x^2}-\frac{x^4}{2}\sin\frac{4}{x^2}-8\cos \frac{4}{x^2}\right)\right|= \left|e^{-x^2\slash 4}\right|\left|12\,\frac{\sin\frac{4}{x^2}}{\frac{  4}{x^2}}-2x^2\frac{\sin\frac{4}{x^2}}{\frac{4}{x^2}}-8\cos\frac{4}{x^2}\right|

    But 12\,\frac{\sin\frac{4}{x^2}}{\frac{4}{x^2}}\xright  arrow [x\to \infty] {}12 , -8\cos\frac{4}{x^2}\xrightarrow [x\to \infty] {}-8, so the only problem is with the middle summand, but in fact:

    \frac{2x^2}{e^{x^2\slash 4}}\,\frac{\sin\frac{4}{x^2}}{\frac{4}{x^2}}\xrigh  tarrow [x\to \infty] {} 0 (Applying L'Hospital to the first factor we get zero, whereas the second one converges to 1) , so the whole thing's bounded.

    Tonio
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  4. #4
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    Quote Originally Posted by redsoxfan325 View Post
    g(x)=\left\{\begin{array}{lr}x^3e^{-x^2/4}\sin(4x^{-2}):&x\neq0\\0:&x=0\end{array}\right\}

    \frac{f(x)-f(0)}{x-0}=\frac{f(x)}{x}=x^2e^{-x^2/4}\sin(4x^{-2})=\frac{4e^{-x^2/4}\sin(4x^{-2})}{4x^{-2}}

    So f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{4e^{-x^2/4}\sin(4x^{-2})}{4x^{-2}}

    Do you know where to go from here?

    See Tonio's post below for proving g' is bounded.
    I just take the limit as x -> 0 and from that can determine what f'(0) = .. ? Right? Not confident in that, lol.
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  5. #5
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by katielaw View Post
    I just take the limit as x -> 0 and from that can determine what f'(0) = .. ? Right? Not confident in that, lol.
    Sorry, I actually mislead you the way I wrote that. I wrote \frac{\sin(4x^{-2})}{4x^{-2}} thinking that it would fall under the \lim_{\theta\to0}\frac{\sin\theta}{\theta}=1 category, but it doesn't because \lim_{x\to0}4x^{-2}\neq0.

    So the way you should actually do the limit is to use the squeeze theorem.

    \lim_{x\to0}-x^2e^{-x^2/4}\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq\lim_{x\to0}x^2e^{-x^2/4}

    Thus, 0\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq0, so the limit is zero.

    Sorry for any confusion I may have caused.
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  6. #6
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    Quote Originally Posted by redsoxfan325 View Post
    Sorry, I actually mislead you the way I wrote that. I wrote \frac{\sin(4x^{-2})}{4x^{-2}} thinking that it would fall under the \lim_{\theta\to0}\frac{\sin\theta}{\theta}=1 category, but it doesn't because \lim_{x\to0}4x^{-2}\neq0.

    So the way you should actually do the limit is to use the squeeze theorem.

    \lim_{x\to0}-x^2e^{-x^2/4}\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq\lim_{x\to0}x^2e^{-x^2/4}

    Thus, 0\leq\lim_{x\to0}x^2e^{-x^2/4}\sin(4x^{-2})\leq0, so the limit is zero.

    Sorry for any confusion I may have caused.
    That's more than OK. Thank you so much for your help.
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