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Math Help - MVT differentiation proof question

  1. #1
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    MVT differentiation proof question

    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)]

    ??
    Last edited by transgalactic; January 20th 2009 at 12:08 PM.
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  2. #2
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    Quote Originally Posted by transgalactic View Post
    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)]

    ??

    Note, \frac{g(x)-g(x_0)}{x-x_0} = \frac{(x-x_0)f(x)}{x-x_0} = f(x).
    Also, \lim_{x\to x_0}f(x) exists because f is continous (in neighborhood) at x_0.
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  3. #3
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    but they dont say that g(x) is differentiable on x_0
    from where you came to this conclusion
    <br /> <br />
\frac{g(x)-g(x_0)}{x-x_0} = \frac{(x-x_0)f(x)}{x-x_0} = f(x)<br />

    ??


    and regarding the limit you mentioned
    <br /> <br />
\lim_{x\to x_0}f(x) <br />
    that means
    lim f(x)=[f(x+x_0) -f(x_0)]/[x-x_0]
    as x->x_0

    that helps??
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  4. #4
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    i did the definition of one sided derivative
    <br />
 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 }} \\ <br />
    <br />
 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 }} <br />

    how to prove that the limits from positive side and from negative side are equal??
    Last edited by transgalactic; January 30th 2009 at 06:49 AM.
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  5. #5
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    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.
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  6. #6
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    Quote Originally Posted by transgalactic View Post
    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.
    That is exactly what I said in post #2.
    Once you cancel by x-x_0 you are left with f(x).
    Now you know that \lim_{x\to x_0}f(x) = f(x_0) by continuity.
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