Results 1 to 11 of 11

Math Help - derivative of a piecewise function

  1. #1
    Junior Member
    Joined
    Jun 2007
    Posts
    63

    derivative of a piecewise function

    define f(x) to be 0 at 0 and exp(-1/x^2) everwhere else.

    i'm having trouble showing that f'(0) = 0.

    can anyone help?

    thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,966
    Thanks
    350
    Awards
    1
    Quote Originally Posted by kkoutsothodoros View Post
    define f(x) to be 0 at 0 and exp(-1/x^2) everwhere else.

    i'm having trouble showing that f'(0) = 0.

    can anyone help?

    thanks!
    What is \lim_{x \to 0^-}\frac{df}{dx}?
    What is \lim_{x \to 0^+}\frac{df}{dx}?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2007
    Posts
    63
    it's 0. i just can't show this using the definition of derivative.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    (Note: since the function is even, f_{+}'(0)=0 is enough. But L'Hospital's rule goes here, so no need to distinguish.)


    I'm sure there's a slicker way than this:

    {\rm lim}_{h\rightarrow0}f(h)/h={\rm lim}_{h\rightarrow0}{\rm e}^{-1/h^2}/h={\rm lim}_{h\rightarrow0}(1/h)/{\rm e}^{1/h^2}

    and apply L'Hospital's rule to get this equal to {\rm lim}_{h\rightarrow0}h/{\rm e}^{1/h^2}={\rm lim}_{h\rightarrow0}h{\rm e}^{-1/h^2}=0.


    Just a quick point here. This function is smooth at 0, but not analytic at 0!
    This is a classic. Only seen such misbehaving functions, when creating partitions of unity... Are you studying Differential Geometry by the way?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2007
    Posts
    63
    no, not studying differential geometry yet. i'm actually preparing for the math subject gre. i know i should resist these sort of problems for now and concentrate on gre type problems. but i just can't help myself. this problem was a harder problem in a standard calculus book.

    by the way. technically, isn't it circular reasoning to use L'Hopital's rule since you are differentiating the function we are interested in? or is it ok since you've differentiated somewhere near zero using the standard differentiation formulas?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jun 2007
    Posts
    63
    i guess it works using the following logic:

    f'(0) = lim (x-->0) (1/x)*exp(-1/x^2). since we know the derivative of
    exp(-1/x^2) at a p;oint different from 0 we are okay.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    Technically, isn't it circular reasoning to use L'Hopital's rule since you are differentiating the function we are interested in? or is it ok since you've differentiated somewhere near zero using the standard differentiation formulas?

    f(h)/h is inside a limiting process, so we are only interested in values of f(h) in the vicinity of zero.


    f'(0) = lim (x-->0) (1/x)*exp(-1/x^2)
    Careful, not needed minus in the denominator.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jun 2007
    Posts
    63
    thanks!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Let
    f(x) = \left\{ \begin{array}{c}e^{-1/x^2} \mbox{ for }x\not =0\\ 0 \mbox{ for }x=0\end{array}\right.

    Now,
    f'(0) = \lim_{x\to 0}\frac{f(x)-f(0)}{x-0} = \lim_{x\to 0}\frac{e^{-1/x^2}}{x}

    Using the limit composite rule we can write,
    \lim_{x\to \infty} xe^{-x^2}
    But,
    e^x \geq x^2/2 \mbox{ for }x>0
    Thus,
    xe^{-x^2} \leq \frac{2}{x} \to 0.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member Rebesques's Avatar
    Joined
    Jul 2005
    From
    At my house.
    Posts
    538
    Thanks
    11
    <br />
xe^{-x^2} \leq \frac{2}{x} \to 0<br />

    I thought you squared both sides


    Anyway, the title of this topic isn't very accurate, this function is smooth.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Jun 2007
    Posts
    63
    thanks for that. i'll have to keep the "limit composite rule" in mind.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 14
    Last Post: October 7th 2011, 09:45 PM
  2. derivative of a piecewise function
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 25th 2011, 12:56 AM
  3. Replies: 2
    Last Post: August 28th 2010, 09:31 AM
  4. Derivative of a Piecewise Function
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 2nd 2009, 06:23 PM
  5. Derivative of piecewise functions...halp!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 13th 2008, 07:16 PM

Search Tags


/mathhelpforum @mathhelpforum