Prove $g : (0, 1) \rightarrow \mathbb{R}$ , $g_{(x)}=\frac{f_{(x)}}{x}$ is an increasing function over $(0,1)$ $f:[0,1] \rightarrow \mathbb{R}$ is continuous over $[0,1]$ and $f_{(0)} = 0$, if $f^{\prime}$ exists over $(0,1)$ and is an increasing function then prove $g : (0, 1) \rightarrow \mathbb{R}$ , $g_{(x)}=\frac{f_{(x)}}{x}$ is an increasing function over $(0,1)$.
My attempt:
Suppose we approximate $f^{\prime}$ using polynomials, it's an increasing function so it has a positive leading coefficient and degree of at least 0, so: $f_{(x)}^{\prime} = a_0x^{n} + ...$ where $a_0 > 0$ and $n \geqslant 0$ then after integration: $f_{(x)} = b_0x^{n+1} + ...$ where $b_0 > 0$ and $n+1 \geqslant 1$ so dividing by $x$: $g_{(x)}=\frac{f_{(x)}}{x} = b_0x^{n} + ...$ where $b_0 > 0$ and $n \geqslant 0$ so $g_{(x)}$ is an increasing function.
Is this proof correct? If YES, what could go wrong? If NO, why? and what's the correct way to prove this?
and finally, is there a better way you can think of? and can you simply explain the reason for open and closed intervals and $f_{(0)} = 0$ conditions?

Use the mean value theorem: for $x \in (0,1)$
$$f(x) - f(0) = xf'(\xi)$$
where $0 < \xi < x$.  Then for $x \in(0,1)$
\begin{align}
g'(x) &= \frac{f'(x)}{x} - \frac{f(x)}{x^2} \\
&= \frac{1}{x^2}\Big(xf'(x) - f(x)\Big) \\
&= \frac{1}{x^2}\Big(xf'(x)-xf'(\xi)\Big)
\end{align}
and because $f'$ is increasing, the term on the right is non-negative, so that $g'(x) \geqslant 0$ and $g$ is increasing.
