Zero Derivative implies Constant Function

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

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

But:
 * $f\,^{\prime} \left({\xi}\right) = 0$

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.