Proving a continuous function is a constant Let $f$ be a function that is continuous on an interval $[a,b]$ and differentiable  on $(a,b)$. Assume $f'(x)=0$ for all $x\in (a,b)$. Show f is a constant.

I feel like the obvious set up is by contradiction. If I assume $f$ is not a constant, then it must have a variable, but I'm unsure how to construct an $f$ (without loss of generality) which I can then differentiate to contradict $f'(x)=0$.

any suggestions to start this proof?
Let, $x,y\in [a,b]$ then by Mean value theorem we have, 
$$f(x)-f(y)=(x-y)f'(\xi)$$ for some $x&lt;\xi&lt;y$ and as given $f'(t)=0,\forall t\in(a,b)$ so, $f'(\xi)=0\implies f(x)=f(y),\forall x,y\in[a,b]$

Hence, $f$- is constant.
Well, $f'(x)$ gives the rate of change of $f$ at a point $x$. If $f'(x)=0 \forall x \in (a,b)$, then $f(x)$ isn't changing, i.e. it is constant, on $(a,b)$.

That's the idea, you just need to show it mathematically.