Prove that if f'(x) = 0 for all x in (a,b), then f(x) =

Prove that if $f'(x) = 0$ for all $x \in (a,b)$, then $f(x) = \text{const}$ on $(a,b)$.

Step 1: Fix two arbitrary points

Let $x_{1}, x_{2} \in (a,b)$ with $x_{1} \lt x_{2}$. It suffices to show $f(x_{1}) = f(x_{2})$, since $x_{1}$ and $x_{2}$ are arbitrary.

Step 2: Apply the mean value theorem

Being differentiable on $(a,b)$, $f$ is in particular continuous on $[x_{1},x_{2}]$ and differentiable on $(x_{1},x_{2})$ — exactly the hypotheses of Lagrange's mean value theorem. Hence there is a point $c \in (x_{1},x_{2})$ with

\[ f(x_{2}) - f(x_{1}) = f'(c)\,(x_{2}-x_{1}). \]

Step 3: Use the hypothesis

Since $c$ lies in $(x_{1},x_{2}) \subset (a,b)$, the hypothesis gives $f'(c) = 0$, so

\[ f(x_{2}) - f(x_{1}) = 0\cdot(x_{2}-x_{1}) = 0 \;\Longrightarrow\; f(x_{1}) = f(x_{2}). \]

Step 4: Conclude

The two points were arbitrary, so $f$ takes one and the same value throughout $(a,b)$: $f(x) = C \qquad\text{for all } x \in (a,b)$.

Step 5: Why the interval matters

The hypothesis must hold on an interval, that is, on a connected set. The function

\[f(x) = \operatorname{sgn}(x) = \begin{cases} 1, & x \gt 0, \\ -1, & x \lt 0 \end{cases}\]

\[\text{satisfies } f'(x) = 0 \text{ on } \mathbb{R}\setminus\{0\}, \text{ yet is not constant there.}\]

The mean value theorem cannot be applied across the gap, because $f$ is not defined (let alone continuous) at $x = 0$.