prove $f$ is a constant Lets's say we have a differentiable function
$f:[a,b]\to \mathbb{R}$
with
$f^\prime\equiv0$
How do I show that 
$f\equiv C$
by using the mean value theorem?

For $x\in (a,b]$, we have
$$\tag{1} f(x)-f(a)=f'(\zeta)\cdot(x-a)\mbox{ for some }\zeta\in (a,x)$$
by the mean value theorem. Now, since $f'\equiv 0$, it follows from $(1)$ that
$$f(x)-f(a)=0,$$
that is
$f(x)=f(a)$. 
