Suppose that we have a function $f(x)$ that is continuous and bounded on $(0,1)$. Let us then define a function $g(x)=x(1-x)f(x)$. Prove that g(x) is continuous. I solved it through $\delta,\varepsilon$ definition of uniform continuity. However, I didn't need to use the fact that $f(x)$ was continuous; I basically implied through my proof that for $g(x)$ to be uniformly continuous, $f(x)$ only needed to be bounded. Which I think is wrong. Here is what I tried $$\vert g(x)-g(x_0)\vert$$$$\Rightarrow\vert x(1-x)f(x)-x_0(1-x_0)f(x_0)\vert$$$$\Rightarrow\vert x(1-x)f(x)-x(1-x)f(x_0)+x(1-x)f(x_0)-x_0(1-x_0)f(x_0)\vert$$$$\leq\vert x(1-x)\vert\vert f(x)-f(x_0)\vert+\vert f(x_0)\vert\vert x(1-x)-x_0(1-x_0)\vert$$
Since $f(x)$ is bounded we have that $\vert f(x)\vert\leq M$ for all x with M being some number. Thus we have that $$\vert x(1-x)\vert\vert f(x)-f(x_0)\vert+\vert f(x_0)\vert\vert x(1-x)-x_0(1-x_0)\vert$$$$\leq2M\vert x(1-x)\vert+M\vert x(1-x)-x_0(1-x_0)\vert$$ Afterwards, I used calculus to find the maximum of $x(1-x)$. I took derivative and set it equal to 0. This yielded me that $x(1-x)$ maximum is at $x=1/2$. Hence, $$2M\vert x(1-x)\vert+M\vert x(1-x)-x_0(1-x_0)\vert$$$$<2M\vert1/4\vert+M\vert1/4\vert=3M/4$$ Hence $\delta=4\varepsilon/3M$. Since $\delta$ does not depend on $x_0$, $g(x)$ is uniformly continuous. What did I do wrong? What are other methods to prove this?
Hint: Let $\varepsilon > 0$ be given. Assume that $|f(x)| \le M$. Since $f$ is uniformly continuous on $\left[\dfrac{\varepsilon}{4M}, 1-\dfrac{\varepsilon}{4M}\right]$ you can choose $\tilde\delta$ such that $|f(x)-f(y)| < \varepsilon$ when $x,y \in \left[\dfrac{\varepsilon}{4M}, 1-\dfrac{\varepsilon}{4M}\right]$ and $|x-y| < \tilde\delta$. Now put $\delta := \min\left\{\tilde\delta, \dfrac{\varepsilon}{4M}\right\}$.