# 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

- 1977: K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*... (previous) ... (next): $\S 11.7$