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!

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.
