Zero Derivative implies Constant Function

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Theorem

Let $f$ be a real function which is continuous on the closed interval $\left[{a \,.\,.\, b}\right]$ and differentiable on the open interval $\left({a \,.\,.\, b}\right)$.

Suppose that:

$\forall x \in \left({a \,.\,.\, b}\right): f' \left({x}\right) = 0$


Then $f$ is constant on $\left[{a \,.\,.\, b}\right]$.


Proof

Let $y \in \left[{a \,.\,.\, b}\right]$.

Then $f$ satisfies the conditions of the Mean Value Theorem on $\left[{a \,.\,.\, y}\right]$.

Hence:

$\exists \xi \in \left({a \,.\,.\, y}\right): f' \left({\xi}\right) = \dfrac {f \left({y}\right) - f \left({a}\right)} {y - a}$

But:

$f' \left({\xi}\right) = 0$

which means:

$f \left({y}\right) - f \left({a}\right) = 0$

and hence:

$f \left({y}\right) = f \left({a}\right)$

As $y$ is any $y \in \left[{a \,.\,.\, b}\right]$, 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 $\forall x: f' \left({x}\right) = 0$.


Sources