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Math Help - continuity and differentiability

  1. #1
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    continuity and differentiability

    Hello I would just like you to tell me what would be the best method to demonstrate that the function g is of class C^1 on the interval J(I was spoken to apply Taylor Lagrange or Taylor-Young but I did not understand much) I thank:
    Here is the statement:

    Let f be a function of class C^2 over an interval J, such that f (0) = 0.
    It is assumed that 0 is an interior point to J. We define the function g on J by:
    g (x) = \frac{f (x)}{(x)} if x  \neq 0
    and g (0) = f '(0).

    What I have done yet below:

    In a first time I apply Taylor Young to see where that leads me

    -The formula for Taylor Young in 0 to order 2 of f gives us

      f (x) = f (0) + xf '(0) + x ^ 2 \frac{f"(0)}{2} + o (x ^ 2)

    or f (0) = 0 therefore
     f (x) = xf '(0) + x ^ 2 \frac{f"(0)}{2}
    and
      \ frac{f(x)}{x} = f '(0) + x \frac{f"(0)}{2} = g (x)

    I wonder now if this is fair and whether it will indeed serve me something.
    (I just want to say that I'm French and I think you'll find my language can be a little too basic, I hope that this will not prevent you from answering me)
    Last edited by boolean; February 1st 2009 at 02:29 PM. Reason: lack of grammar
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  2. #2
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    The function g is clearly differentiable at all points of J other than 0. In fact, g'(x) = \frac{xf'(x)-f(x)}{x^2} by the usual quotient rule for differentiation.

    To see what happens at x=0, you have to find the derivative from first principles: g'(0) = \lim_{x\to0}\frac{g(x)-g(0)}x. Using the definitions of g(x) when x≠0 and when x=0, you see that this is equal to \lim_{x\to0}\frac{f(x)-xf'(0)}{x^2}. Thus g is differentiable at 0 (with derivative \tfrac12f''(0)).

    Finally, to show that g is of class C^1, you must check that the derivative g' is continuous at 0. To do this, evaluate \lim_{x\to0}\frac{xf'(x)-f(x)}{x^2} (using l'H˘pital's rule) and verify that it is equal to \tfrac12f''(0).
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