Theorem (0.4 in notes)
If \(f'(x) = 0\) for all \(x \in (a,b)\), then \(f\) is a constant function.

Proof

Let \(x_1, x_2 \in (a,b)\) with \(x_1 < x_2\). By the Mean Value Theorem, there exists a point \(c \in (x_1, x_2)\) such that

$$ \begin{align*} f(x_2) - f(x_1) &= f'(c)(x_2 - x_1). \end{align*} $$

Since \(f'(c) = 0\), then

$$ \begin{align*} f(x_2) - f(x_1) &= 0. \end{align*} $$

Hence

$$ \begin{align*} f(x_2) = f(x_1). \end{align*} $$

Thus, \(f\) is constant on \((a,b)\). \(\ \blacksquare\)

References