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**redsoxfan325** A bounded derivative implies uniform continuity. However, there are certain instances of functions (most notably $\displaystyle \sqrt{x}$) that are uniformly continuous and have unbounded derivatives. I attempt to strengthen the theorem below:

__Theorem__: An everywhere continuous function $\displaystyle f$ is uniformly continuous on $\displaystyle \mathbb{R}$ if its derivative is bounded on $\displaystyle (-\infty,-m]$ and $\displaystyle [n,\infty)$ ($\displaystyle m,n\in\mathbb{R}$).

__Proof__: On $\displaystyle [-m,n]$ we have a continuous function defined on a compact set, so it is uniformly continuous and $\displaystyle \exists~\delta_1$... On $\displaystyle (-\infty,-m]$ and $\displaystyle [n,\infty)$, $\displaystyle f$ has a bounded derivative so there exist $\displaystyle \delta_2$ and $\displaystyle \delta_3$ such that... Therefore taking $\displaystyle \delta=\min\{\delta_1,\delta_2,\delta_3\}$ proves that the entire function is uniformly continuous on $\displaystyle \mathbb{R}$.