Do derivatives belong to unique functions?

  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!
     
  2. jcsd
  3. morphism

    morphism 2,020
    Science Advisor
    Homework Helper

    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. 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. HallsofIvy

    HallsofIvy 40,795
    Staff Emeritus
    Science Advisor

    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. 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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