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

But $$f^{\prime} \left({\xi}\right) = 0$$ and hence $$f \left({y}\right) = f \left({a}\right)$$.

As $$y$$ is any $$y \in \left[{a \,. \, . \, b}\right]$$, the result follows.