Increasing and differentiable implies nonnegative derivative that is not identically zero on any interval

Statement

On an open interval

Suppose $f$ is a function on an open interval $I$ that may be infinite in one or both directions (i..e, $I$ is of the form $\!(a,b)$, $(a,\infty )$, $(-\infty ,b)$, or $(-\infty ,\infty )$). Suppose the derivative of $f$ exists everywhere on $I$. Suppose further that $f$ is an increasing function on $I$, i.e.:

$x_{1},x_{2}\in I,x_{1}<x_{2}\implies f(x_{1})<f(x_{2})$

Then, $\!f'(x)\geq 0$ for all $x\in I$. Further, there is no sub-interval of $I$ such that $f'(x)=0$ for all $x$ in the sub-interval.

On a general interval

Suppose $f$ is a function on an interval $I$ that may be infinite in one or both directions and may be open or closed at either end. Suppose $f$ is a continuous function on all of $I$ and that the derivative of $f$ exists everywhere on the interior of $I$. Further, suppose $f$ is an increasing function on $I$, i.e.:

$x_{1},x_{2}\in I,x_{1}<x_{2}\implies f(x_{1})<f(x_{2})$

Then, $\!f'(x)\geq 0$ for all $x$ in the interior of $I$. Further, there is no sub-interval of $I$ such that $f'(x)=0$ for all $x$ in the sub-interval.