real analysis - Using unbounded derivative to show function is not uniformly convergent

I'm confused how to use the following theorem:

19.6 Theorem.
Let $f$ be a continuous function on an interval $I$ [$I$ may be bounded or unbounded]. Let $I^◦$ be the interval obtained by removing from $I$ any endpoints that happen to be in $I$. If $f$ is differentiable on $I^◦$ and if $f′$ is bounded on $I^◦$, then $f$ is uniformly continuous on $I.$

So far, I have encountered examples.
$f(x)= \sqrt{x}, g(x)= \frac{1}{x}, h(x)= x^2$
They are each on the interval $(0,\infty)$

I know $f$ is uniformly continuous, but $g$ and $h$ are not.
However, the derivatives for each of these functions is unbounded on $(0,\infty)$

To show that a continuous function is not uniformly continuous on $(0, \infty)$, do I need to show the derivative is unbounded for every interval $[a, \infty )$ , where $a>0$?

If so, how would I prove the function is unbounded from $[a, \infty )$?

I would appreciate a worked out example with one of the functions above or one of your choosing.

Answer

If $f$ is differentiable on $I^\circ$ and $f'$ is bounded on
$I^\circ$, then $f$ is uniformly continuous on $I$

is correct.

If $f$ is uniformly continuous on $I$, then $f$ is differentiable on $I^\circ$ and $f'$ is bounded on $I^\circ$

is not correct. The counterexample is exactly $f(x)=\sqrt x$.

To show that $f$ is not uniformly continous on $I$, it is enough to show that it is not uniformly continuous on $(0,1]$. And uniformly continous functions on a bounded interval are always bounded.