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I want to prove the following statement:

Let $f$ and $g$ be two real functions continuous in some interval $[a, b]$ and differentiable in $(a, b)$. If $f' = g'$, then $f(x) = g(x) + c$, where $c$ is a constant.

I thought I could argue like this: Consider $h(x) = f(x) - g(x)$. Knowing that $h' = 0$, we want to prove that $h$ is constant. Pick any $k \in (a,b)$, then $h'(k) = 0$. Therefore (and this is the part I'm not sure about), by Rolle's theorem we can find $d, e \in (a, b)$ such that $d < k < e$ and $h(d) = h(e)$. Since this must be true for any $d, e, k$, $h$ is a constant.

Is this a correct application of Rolle's theorem?

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No, that's not what Rolle's theorem says. What you want here is the mean value theorem. – Chris Eagle Apr 30 '12 at 16:36
Your hypothesis are the wrong way: you want continuity in the closed interval, differentiability in the open interval. – Arturo Magidin Apr 30 '12 at 16:40
Hint. Let $r\in (a,b]$. By the Mean Value Theorem, there exists $c$ in $(a,r)$ such that $f'(c) = (f(r)-f(a))/(r-a)$. Therefore... – Arturo Magidin Apr 30 '12 at 16:41
@ArturoMagidin: I fixed the intervals. – Javier Apr 30 '12 at 17:19
up vote 6 down vote accepted

No, this is not correct. That is, the converse of Rolle's theorem does not hold (if this is what you're asking). For instance, let $f(x)=x^3$ on $[-1,1]$. Then $f'(0)=0$ but there are not two points $c$ and $d$ in $[-1,1]$ with $c\ne d$ and $f(c)=f(d)$.

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As Chris Eagle said. If $x, y \in (a,b)$ by mean value theorem, we have $$ 0 = h'(c) = \dfrac{h(x) - f(y)}{x-y}.$$ Hence, $ h(x) = h(y)$ and $h$ is a constant.

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Thank you for your answer. While it helped me with the problem, David's answered the point of the question, so I accepted his. Thank you anyway. – Javier Apr 30 '12 at 17:20

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