Prove g(0)=0 from f'(x).(fg)(f(x))=g'(f(x)).f'(x)

[Stevela]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[Stevela]

posted this in a question and answers forum , and someone proved that g(0)=0 does not always hold?here's the link http://math.stackexchange.com/questions/463839/prove-that-g0-0
this is fishy , spivak didn't even include the solutions of this question after i cehcked the solutions book

 

[Theorem.]

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

 

[Theorem.]

I think you need to add the assumption that $f'(x)\neq 0$ I think then it works, as someone posted on your stack exchange (link you posted)

 

[haruspex]

I think you need to add the assumption that $f'(x)\neq 0$ I think then it works, as someone posted on your stack exchange (link you posted)

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.

 

[Theorem.]

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.

Yeah I know, I wasn't too sure about that either...
It seems like the question is missing something: maybe the requirement that $f$ be surjective. Does anyone else have any comments on this?