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

Let $y \in \closedint a b$.

Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint a y$.

Hence:

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

But:

$\map {f'} \xi = 0$

which means:

$\map f y - \map f a = 0$

and hence:

$\map f y = \map f a$

As $y$ is any $y \in \closedint a b$, the result follows.

$\blacksquare$


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