Proving product of two functions is uniformly continous

Question

If $f$ is bounded and continuous on $(0,1)$ and $g$ is defined on $(0,1)$ as $$g(x)=x(1-x)f(x)$$ how can we show $g$ is uniformly continuous on $(0,1)$?

Answer 1

Use boundedness of $f$ to observe that $g(x) \to 0$ as $x \to 0$ or $x \to 1$. Hence $g$ can be extended to a continuous function on the compact interval $[0,1]$. This makes $g$ uniformly continuous on $[0,1]$, hence also on $(0,1)$.

Answer 2

Another possible way is that, when function $\phi$ is uniformly continuous on not closed set $A \subset \mathbb{R}^n$, then it can be extended on $\overline{A} \setminus A$ by the unique way and obtained function will be continuous on $\overline{A}$. Now we can use this directly to $g(x)=x(1-x)f(x)$ on $(0,1)$ and obtain $g$ continuous on $[0,1]$, hence, uniformly continuous.