I'm interested in working out whether the quotient of continuous functions
\begin{equation*}
\frac{f(y, z, \eta, p)}{f(x, z, \eta, p)}
\end{equation*}
is increasing or decreasing in $z$.
My problem is that the functions, which have explicit functional forms, are quite complex:
\begin{equation*}
f(y, z, \eta, p)=\frac{p[2+\eta(1+y)]-\eta z(1-y)}{p[3+\eta(2+y)]-z[1+\eta(2-y)]}
\end{equation*}
and
\begin{equation*}
f(x, z, \eta, p)=\frac{p[2+\eta(1+x)]-\eta z(1-x)}{p[3+\eta(2+x)]-z[1+\eta(2-x)]}.
\end{equation*}
The restrictions on the variables are:
\begin{equation*}
x>y>1 \ \text{ and } \ p>z>-p \ \text{ and } \ p, \eta>0.
\end{equation*}
Thinking about this in general terms, the sign can be determined by
\begin{equation}
f(x, z, \eta, p)\cdot\frac{\partial}{\partial z}f(y, z, \eta, p)-f(y, z, \eta, p)\cdot\frac{\partial}{\partial z}f(x, z, \eta, p) \ \ \ \ (1).
\end{equation}
I know that
\begin{equation*}
1>f(x, z, \eta, p)>f(y, z, \eta, p)>0
\end{equation*}
and
\begin{equation*}
\frac{\partial}{\partial z}f(x, z, \eta, p)
=\frac{2(1+\eta)p(1+\eta x)}{\big[p[3+\eta(2+x)]-z[1+\eta(2-x)]\big]^2}>0,
\end{equation*}
\begin{equation*}
\frac{\partial}{\partial z}f(y, z, \eta, p)
=\frac{2(1+\eta)p(1+\eta y)}{\big[p[3+\eta(2+y)]-z[1+\eta(2-y)]\big]^2}>0.
\end{equation*}
However, I'm not sure which term is larger here.
One way I have thought about this is writing $y=x-c$, and plotting this in Desmos shows that (1) is negative.
I'm not sure on the best way to proceed: should I just try and use the functional forms and deal with the very intense algebra, or should I try and use some kind of general results or principles to analyse the terms? Any help is most appreciated.
2026-04-01 12:55:16.1775048116