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Prove that g(0)=0?

  1. Aug 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Hello,
    Suppose that f and g are differentiable functions satisfying
    ##\displaystyle \int_{0}^{f(x)} (fg)(t) \, \mathrm{d}t=g(f(x))##
    Prove that g(0)=0
    now if f(x)=0 in some point then it's straigh forward that g(f(x))=g(0)=0 anyways:
    differentiating the first formula we get the following equation :

    f'(x).(fg)(f(x))=g'(f(x)).f'(x)

    let's suppose that f'(x)=0 , thus f is constant i.e f(x)=c, if c=0 we are done , g(0)=0 , if c=/=0 then :

    ##\displaystyle \int_{0}^{c} fg(t) \, \mathrm{d}t##=g(c)

    ##\displaystyle \int_{0}^{c} g(t) \, \mathrm{d}t##=g(c)/c, **i'm stuck here** , how can we prove that g(0)=0 (or get a contradiction) from this equation?



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 9, 2013 #2
  4. Aug 9, 2013 #3
    Are you sure you mean f(t)g(t) on the left hand side and not f(g(t))? what does the actualy question state
     
  5. Aug 9, 2013 #4
    I think you need to add the assumption that [itex]f'(x)\neq 0[/itex] I think then it works, as someone posted on your stack exchange (link you posted)
     
  6. Aug 9, 2013 #5

    haruspex

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    I'm not convinced yet. If f(x) = 0 for some x then the result is trivial, so suppose f(x) >= c > 0 for all x. The equation f′(x)f(f(x))g(f(x))=g′(f(x))f′(x), after cancelling f'(x), leads to f(f(x))g(f(x))=g′(f(x)). The deduction that f(y)g(y) = g'(y) is only valid for the range of f. In particular, it's not valid for 0 < y < c.
     
  7. Aug 9, 2013 #6
    Yeah I know, I wasn't too sure about that either...
    It seems like the question is missing something: maybe the requirement that [itex]f[/itex] be surjective. Does anyone else have any comments on this?
     
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