Zero Derivative means 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.

$\blacksquare$

Notes

This is the converse of Differentiation of a Constant.

Thus we see that $f$ is a constant function iff $\forall x: f^{\prime} \left({x}\right) = 0$.

Sources

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