$\underline{\text{Motivation}:-}$
I started thinking about this question a few days ago while checking on my old questions. I thought about a similar question to this one, but instead of roots this time I want to know about singularities.
$\underline{\text{Details}:-}$
Let, $\mathcal{S}(f)$ be the set of all singularities of $f$ where $f$ is a real valued function.
For example,
$\mathcal{S}(\frac{1}{x})=\{0\}$
$\mathcal{S}(\Gamma(x))=\mathbb{Z}_{<1}$
And so on.
I was wondering whether there exists a real function for which $\mathcal{S}(f)$ is dense.
Maybe a function which have singularities at the rationals.
An easy answer to my question would be $\frac{1}{\sin(\frac{1}{x})}$, since $\sin(\frac{1}{x})$ has dense, disconnected roots.
$\underline{\text{My questions}:-}$
We can write a function $f$ with dense singularities as $\frac{1}{g}$, where $g$ is a function with dense roots, so does there exist a function $g$ which has the following properties-
●The roots of $g$ are dense and uncountable in an interval $[a,b]$.
●$g$ has no singularities in $[a,b]$.
●$g$ is continuous in $[a,b]$.
If there does exist such a function, I would like to have an example and the graph of $g$ and $f$ both.
(I just can't imagine how cool the graph of such a function would look like!)