Zero Derivative implies Constant Function

Theorem
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.

Suppose that:
 * $\forall x \in \openint a b: \map {f'} x = 0$

Then $f$ is constant on $\closedint a b$.

Proof
When $x = a$ then $\map f x = \map f a$ by definition of mapping.

Otherwise, let $x \in \hointl a b$.

We have that:
 * $f$ is continuous on the closed interval $\closedint a b$


 * $f$ is differentiable on the open interval $\openint a b$

Hence it satisfies the conditions of the Mean Value Theorem on $\closedint a b$.

Hence:
 * $\exists \xi \in \openint a x: \map {f'} \xi = \dfrac {\map f x - \map f a} {x - a}$

But by our supposition:
 * $\forall x \in \openint a b: \map {f'} x = 0$

which means:
 * $\forall x \in \openint a b: \map f x - \map f a = 0$

and hence:
 * $\forall x \in \openint a b: \map f x = \map f a$

Also see

 * Derivative of Constant: This is the converse. Thus we see that $f$ is a constant function $\forall x: \map {f'} x = 0$.
 * Zero Derivative implies Constant Complex Function