Proving $f'$=$g'$ has some c such that $g=f+c$ Suppose that $f$ and $g$ are differentiable functions on $(a,b)$ and suppose that $g'(x)=f'(x)$ for all $x \in (a,b)$. Prove that there is some $c \in \mathbb{R}$ such that $g(x) = f(x)+c$.
So far, I started with this:
Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then $h'(c)=0 \implies h(c)=c$
After this i'm not sure where to go, or if this is correct at all, any hints?
This is also my first post in Latex so sorry if there's any mistakes!
 A: Fixing your proof a little bit, I can extend your proof.
Suppose that your domain of function is (a,b). And let's choose any $\beta\in (a,b)$.
So for any $x\in (a,b)$ $\exists \alpha\in (\beta,x)$ (or $\alpha \in(x,\beta)$ if $x<\beta$) such that $h'(\alpha)=\frac{h(x)-h(\beta)}{x-\beta}=0$.
Thus,
$$\frac{h(x)-h(\beta)}{x-\beta}=0 \implies h(x)-h(\beta)=0\\ \implies f(x)-g(x)=f(\beta)-g(\beta)\\ \implies f(x)=g(x)+f(\beta)-g(\beta)$$
Thus, the $c$, we were finding is $f(\beta)-g(\beta)$.
A: You say that 

Let $h'(x)=f'(x)-g'(x)=0$, then MVT implies $\exists$ c $\in
> \mathbb{R}$ such that $h'(c) = \frac{h(b)-h(a)}{b-a} =0$. Then
  $h'(c)=0 \implies h(c)=c.$

This is not correct. If $h'(c)=0$ then you have $h(b)=h(a).$ 
But you are in the correct way. Instead of $a,b$ consider $x,y\in [a,b], x\ne y.$ Then you have 
$$h'(c) = \frac{h(y)-h(x)}{y-x} =0.$$ So $h(x)=h(y)$ from where you get that $h$ must be constant.
A: If $F(x)$ is a differentiable function on $(a, b)$ with
$F'(x) = 0, \; \forall x \in (a, b), \tag 1$
then 
$\exists c \in \Bbb R, \; F(x) = c; \tag 2$
for, picking any $x_1, x_2 \in (a, b)$ with $x_1 < x_2$, we have, by the fundamental theorem of calculus,
$F(x_2) - F(x_1) = \displaystyle \int_{x_1}^{x_2} F'(s) \; ds = \int_{x_1}^{x_2} 0 \; ds = 0; \tag 3$
it follows that
$F(x_1) = F(x_2), \; \forall x_1, x_2 \in (a, b); \tag 4$
thus we may take
$c = F(y) \tag 5$
for any $y \in (a, b)$.  Now setting
$F(x) = f(x) - g(x), \tag 6$
we find
$F'(x) = f'(x) - g'(x) = 0, \; \forall x \in (a, b); \tag 7$
we see from the above that
$f(x) - g(x) = F(x) = c, \tag 8$
whence
$f(x) = g(x) + c, \; \forall x \in (a, b). \tag 9$
If one wishes, for whatever reason, to avoid integration and the fundamental theorem, then of course a demonstration based upon the mean value theorem is also possible:  again, letting $F(x)$ be differentiable on $(a, b)$, suppose that there were $x_1, x_2 \in (a, b)$, $x_1 < x_2$, with
$F(x_1) \ne F(x_2); \tag{10}$
then by the MVT, there exists $\xi \in (x_1, x_2)$ such that
$F(x_2) - F(x_1) = F'(\xi)(x_2 - x_1), \tag{11}$
whence
$F'(\xi) = \dfrac{F(x_2) - F(x_1)}{x_2 - x_1} \ne 0 \tag{12}$
by virtue of (10), contradicting the hypothesis that $F'(x) = 0$ for $x \in (a, b)$.  Thus
$F(x_2) = F(x_1) = c, \; \text{a constant}, \; \forall x_1, x_2 \in (a, b); \tag{13}$ 
the desired result now follows by taking $F(x)$ as in (6).
As for our OP Brad Scott's effort to invoke the MVT, though it seems to be headed in the right general direction, it does erroneously conclude that $h(c) = c$ for some $c \in \Bbb R$; it is almost as if the MVT has been confused with the intermediate value theorem here.  But as pointed out by my colleagues mfl and Lev Ban in their answers, the correct approach is to use the MVT in the present context is to affirm it implies that $h(x_1) = h(x_2)$ for $x_1, x_2 \in (a, b)$, or as I have phrased it, $F(x) = f(x) - g(x)$ is constant.
A: Assume f-g is non-constant. Then f'-g' is not identically zero. But f'=g' by hypothesis. Contradiction. Hence f-g is constant.  
