If $f^{\prime}(x)=0$ at each point $x$ of an open interval $(a, b)$, then $f(x)=C$ for all $x \in(a, b)$, where $C$ is a constant. #### Proof We want to show that $f$ has a constant value on the interval $(a, b)$. We do so by showing that if $x_{1}$ and $x_{2}$ are any two points in $(a, b)$ with $x_{1}<x_{2}$, then $f\left(x_{1}\right)=f\left(x_{2}\right)$ Now $f$ satisfies the hypotheses of the Mean Value Theorem on $\left[x_{1}, x_{2}\right]$ : It is differentiable at every point of $\left[x_{1}, x_{2}\right]$ and hence continuous at every point as well. Therefore, $\frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}=f^{\prime}(c)$ at some point $c$ between $x_{1}$ and $x_{2}$. Since $f^{\prime}=0$ throughout $(a, b)$, this equation implies successively that $ \frac{f\left(x_{2}\right)-f\left(x_{1}\right)}{x_{2}-x_{1}}=0, \quad f\left(x_{2}\right)-f\left(x_{1}\right)=0, \quad \text { and } \quad f\left(x_{1}\right)=f\left(x_{2}\right) $