Zero Derivative implies Constant Function

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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:

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

$\blacksquare$


Also see

This is the converse of Derivative of Constant.

Thus we see that $f$ is a constant function if and only if $\forall x: \map {f'} x = 0$.


Sources