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Do derivatives belong to unique functions?

  1. Sep 13, 2008 #1
    I need to show that f(x)=cexp(-cx) is the only solution to c-[integral from 0 to x](f(x)) = f(x). Is this trivial??? If not how would you suggest I go about showing that functions have unique derivatives (after taking into account constants). Thanks in advance for suggestions!
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  3. Sep 13, 2008 #2


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    Is that equation really

    [tex]f(x) = c - \int_0^x f(t) \, dt?[/tex]

    Because f(x)=cexp(-cx) isn't a solution to this for all c.
    Last edited: Sep 13, 2008
  4. Sep 14, 2008 #3
    the integral should be multiplied by c as well. This is actually a stats problem...the entire question is to show when the hazard/failure rate is constant -> c=f(x) / (1-F(x)) where f=pdf, F=cdf. Its obvious that f(x)=cexp(-cx) is a solution to this. I just don't know how to show its the only solution.
  5. Sep 14, 2008 #4


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    The fact that "derivatives belong to unique functions" or, more correctly, that if f(x) and g(x) have the same derivative then f(x) and g(x) differ at most by a constant, is given in most calculus books and follows from the mean value theorem:

    Lemma: Suppose f(x) is continuous on [a, b], is differentiable on (a, b), and f'(x)= 0 for all x in (a, b). Then f(x)= C (f(x) is a constant) for all x in (a,b).

    Let x be any point in (a, b). By the mean value theorem, (f(x)- f(a))/(x- a)= f '(c) for some c between a and x. Since f'= 0 between a and b, f'(0)= 0 from which it follows that f(x)- f(a)= 0 or f(x)= f(a). That is, f(x) is equal to the number f(a) for all x between a and b and so f(x) is a constant there.

    Theorem: If f(x) and g(x) are both continuous on [a, b], differentiable on (a, b), and f'(x)= g'(x) for all x in (a, b), then f(x)= g(x)+ C, a constant.

    Let H(x)= f(x)- g(x). Then H(x) is continuous on [a,b] and differentiable on (a, b). Further, for all x in (a, b), H'(x)= f'(x)- g'(x)= 0 because f'(x)= g'(x). By the lemma, H(x)= f(x)+ g(x)= C for some number C. Then f(x)= g(x)+ C.
  6. Sep 14, 2008 #5
    thanks...that makes sense. Do you think this logic still holds if we only know that F(x) is right continuous? This is one of the properties of a cdf.
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