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Prove g'(x) = g(x)

  1. Oct 2, 2012 #1
    1. The problem statement, all variables and given/known data
    Let f and g be two functions on ℝ such that:
    1. [itex]g(x) = xf(x) + 1[/itex] for all x,
    2. [itex]g(x + y) = g(x)g(y)[/itex] for all x,y,
    3. [itex]\mathop {\lim }\limits_{x \to 0} f(x) = 1[/itex].
    Prove that: [itex]g'(x) = g(x)[/itex] for all x.


    2. Relevant equations



    3. The attempt at a solution
    Sorry for asking this question without showing my workings but I just don't know how to get started. Any hints would be much appreciated. TIA!
     
  2. jcsd
  3. Oct 2, 2012 #2

    SammyS

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    Just use the "limit" definition of the derivative.
     
  4. Oct 2, 2012 #3
    So I have to prove [itex]\mathop {\lim }\limits_{h \to 0} \frac{{g(x + h) - g(x)}}{h} = g(x)[/itex]?
     
  5. Oct 3, 2012 #4

    SammyS

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    Yes.

    It works out quite nicely.
     
  6. Oct 3, 2012 #5
    Yup, remember the 3 properties while you use the precise definition of derivative.
     
  7. Oct 3, 2012 #6
    Thank you all, please check my working below:

    [itex]g'(x) = \mathop {\lim }\limits_{h \to 0} \frac{{g(x + h) - g(x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{g(x).g(h) - g(x)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{g(x).((g(h) - 1)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{g(x)hf(h)}}{h} = \mathop {\lim }\limits_{h \to 0} g(x)f(h) = \mathop {\lim }\limits_{h \to 0} g(x).\mathop {\lim }\limits_{h \to 0} f(h) = \mathop {\lim }\limits_{h \to 0} g(x) = g(x)[/itex]
     
  8. Oct 3, 2012 #7
    That looks correct to me.
     
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