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Mvt differentiation proof question

  1. Jan 20, 2009 #1
    suppose f is a continues function on point x_0
    prove that g(x)=(x-x_0)*f(x) differentiable on x_0??
    calculate g'(x_0)

    i tried to think like this:
    if f(x) is continues on x_0 then lim f(x) as x->x_0 equals f(x_0)

    mvt says f'(c)=[f(a)-f(b)]
    cauchys mvt says f'(c)/g'(c)=[f(a)-f(b)]/[g(a)-g(b)]

    ??
     
  2. jcsd
  3. Jan 20, 2009 #2

    quasar987

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    Obviously those mvt formula you wrote don't make sense a priori since f is only assumed continuous and not differentiable.

    Try doing something with the definition of the derivative of g instead.
     
  4. Jan 21, 2009 #3
    the definition of a derivative is
    lim f(x)=[f(x+x_0) -f(x_0)]/[x-x_0]
    as x->x_0

    what should i do with it?
     
  5. Jan 21, 2009 #4

    quasar987

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    If you write that limit out for g, what do you get?
     
  6. Jan 21, 2009 #5
    by definition if x->x_0
    then the limit formula is
    [g(x)-g(x_0)]/[x-x_0]

    i only know this
    but i dont know
    what to do next?
    how to combine things into a logical thing
    that proves what it asks
     
  7. Jan 21, 2009 #6

    quasar987

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    Well, yes, this is the formula for the derivative of g at x_0, but you also know that

    g(x)=(x-x_0)*f(x).

    So what do you get when you substitute this expression into the definition of g'(x_0) and take the limit?
     
  8. Jan 30, 2009 #7
    i did the definition of one sided derivative
    [itex]
    g_ - '(x) = \mathop {\lim }\limits_{x \to x_0 - } \frac{{(x - x_0 )f'(x) - (x_0 - x_0 )f'(x)}}{{x - x_0 }} = \frac{{(x - x_0 )f(x)}}{{x - x_0 }} \\
    [/itex]
    [itex]
    g_ + '(x) = \mathop {\lim }\limits_{x \to x_0 + } \frac{{(x - x_0 )f'(x) - (x_0 - x_0 )f'(x)}}{{x - x_0 }} = \frac{{(x - x_0 )f(x)}}{{x - x_0 }}
    [/itex]

    can i cut the numenaor and the denominator by x-x_0
    and then each side will equal f(x)
    so the limit is f(x_0) wich is known that its continues so it equals from both side
    so the limits equal

    is this a correct proof.
     
    Last edited: Jan 30, 2009
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