I have a cubic function:
\begin{equation\*}
h(x)\triangleq \eta+x-\frac{V(\eta-x)^3}{c\eta}
\end{equation\*}
we know that $x\in[0,\eta)$ and all letters are positive and $V>c/\eta$. Hence we know that $h(0)<0$ and $h(\eta)>0$ and $h(x)$ is concave increasing for $x\in[0,\eta)$. So we can infer that there will be exactly one root of $h(x^\*)=0$ for $x^\*\in(0,\eta)$. Question is: how to characterize the monotonicity of $x^\*$ as a function of $\eta$?



Characterize the Monotonicity of a root of a cubic equation
Differentiating $h(x, \eta) = 0$ implicitly, we get


$$ \dfrac{dx}{d\eta} = \dfrac{-c\eta^2 + V(2\eta+x)(\eta-x)^2}{\eta (3 V (\mu - x)^2 + c \eta)}$$


The denominator is always positive.
On any interval where the numerator doesn't change sign, 
$x$ is monotone. 


For the numerators of $dx/d\eta$ and $h(x,\eta)$ to both be $0$, we need
$$ 108 v^2 - 36 v - 1 = 0$$
where $v = V \eta/c$. But the roots of that are approximately $-.0258$ and $0.3591$, so this won't happen if $V \eta/c > 1$.


