Show that $\frac{\alpha - c^t x}{\sqrt{x^t R x}}$ is quasiconvex For $\alpha \in \mathbb{R}$ and $x, c \in \mathbb{R}^n$ and $R \in \mathbb{S}_+^n$ (positive semi-definite), let
\begin{equation*}
    f(x) = \frac{\alpha - c^t x}{\sqrt{x^t R x}}
\end{equation*}
with $\text{dom}f = \{ x | c^t x \geq \alpha\}$ and $x^t R x > 0$ for all $x \in \text{dom}f$.
How can I show that $f$ is or is not quasiconvex?
I tried disproving this by testing 100,000 random lines segments sampled in $\text{dom} f$ but found no counter-examples.
Solution
I've determined the solution to the problem and posted it as an answer. It turns out that in general a function of the form $f(x) = p(x)/q(x)$ where $p(x)$ and $q(x)$ are convex, and $p(x) \leq 0$ and $q(x) > 0$ for all $x \in \text{dom}(f)$ is always quasi-convex if $\text{dom} f$ is a convex set.
 A: Thanks to @A.Γ. for verifying the proof.
From the definition of quasi-convexity. A function $f$ is quasi-convex if the $t$-sublevel sets $\{ x \in \text{dom} f | f(x) \leq t \}$ are convex sets for all $t \in \mathbb{R}$.
In this problem $f$ has the form
\begin{equation*}
    f(x) = \frac{p(x)}{q(x)}
\end{equation*}
where $p(x)$ and $q(x)$ are convex and $p(x) \leq 0$ and $q(x) > 0$.
For $t \geq 0$, the $t$-sublevel set is $\text{dom} f$ because $f(x) \leq 0$ for all $x$ in $\text{dom}f$ which is a half-space and therefore convex.
From A.Γ.'s comment I realize it may not be clear from the original problem but we assume here that $x \in \text{dom}f = \{x | c^t x \geq \alpha\} \Rightarrow x^t R x > 0$, so $\text{dom}f$ is always a half-space and is therefore a convex set.
For $t < 0$ we have
\begin{equation*}
    \frac{p(x)}{q(x)} \leq t \iff p(x) - tq(x) \leq 0.
\end{equation*}
Because $p(x)$ is convex and $-tq(x)$ is convex, the sum is convex. The $t$-sublevel set of $f$ is the zero-sublevel set of the convex function $p(x) - t q(x)$ which is a convex set. Therefore $f(x)$ is quasi-convex.
