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Proof showing f(x) = ce^x has a constant

  1. Sep 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that if f is infinitely differential and f'(x)=f(x), then theres a constant "c" with f(x)=ce^x.

    2. Relevant equations



    3. The attempt at a solution

    I think i'm supposed to begin with an arbitrary function and show that it looks like f(x) = ce^x... but that's as far as I've gotten.
     
  2. jcsd
  3. Sep 20, 2009 #2

    Office_Shredder

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    start with c=f(0). Then f(x) = cex is certainly A solution. Suppose g(x) is another solution. What can you conclude about g(x)-cex?
     
  4. Sep 20, 2009 #3
    Well.. since f(x)=ce^x is a solution and g(x) is also a soltion, then g(x)-ce^x is the same as g(x)-f(x). Is that right so far?
     
  5. Sep 20, 2009 #4

    Office_Shredder

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    What can you determine about g(x)-f(x) though?
     
  6. Sep 20, 2009 #5

    HallsofIvy

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    For example, what is (g(x)- f(x))' ?
     
  7. Sep 20, 2009 #6
    (g(x)- f(x))' is the change in slope of graphs g(x) and f(x). Am i getting closer?
     
  8. Sep 20, 2009 #7

    statdad

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    Use a rule for simplifying [tex] (g(x) - f(x))' [/tex] and then apply what you know about the two functions [tex] f [/tex] and [tex] g [/tex].
     
    Last edited: Sep 20, 2009
  9. Sep 20, 2009 #8
    Using the difference rule, if g and f are differentiable functions, then the derivative of the difference of g(x) and f(x) is equal to the derivative of g(x) minus the derivative of f(x).
    As for applying what I know about the two functions... they are the same, so the derivative would be zero. Better so far?
     
  10. Sep 20, 2009 #9

    Office_Shredder

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    Why are they the same? I thought that's what we were trying to prove.

    You DO know that
    f'(x)=f(x)
    g'(x)=g(x)

    So (g(x) - f(x))' = g'(x) - f'(x) = ?

    Also, what is g(0)-f(0)?
     
  11. Sep 20, 2009 #10
    How do i know that a function is equal to its derivative? (referring to f'(x) = f(x), and same with g(x))
    Sorry..
     
  12. Sep 20, 2009 #11
    Oh, like e^(x) is the derivative of e^(x)?
     
  13. Sep 20, 2009 #12
    [tex] f'(x) = f(x) [/tex]

    [tex] \frac{f'(x)}{f(x)} = 1 [/tex]

    [tex] \int \frac{f'(x)}{f(x)}dx = \int dx [/tex]

    [tex] ln|f(x)| = x + C [/tex]

    [tex] f(x) = e^{x+C}[/tex]

    [tex] f(x) = Ae^x [/tex]

    where A is just a constant. [tex] A = \pm e^C [/tex]
     
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