I have a cubic function: \begin{equation*} h(x)\triangleq \eta+x\frac{V(\etax)^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 $\eta$?
Differentiating $h(x, \eta) = 0$ implicitly, we get
$$ \dfrac{dx}{d\eta} = \dfrac{c\eta^2 + V(2\eta+x)(\etax)^2}{\eta (3 V (\mu  x)^2 + c \eta)}$$
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 $V \eta/c > 1$.

$\begingroup$ Could you please elaborate a little bit on how you get $108v^2−36v−1=0$? And so in this case $x$ is monotone increasing in $\eta$? $\endgroup$ – FTXX Sep 28 '16 at 0:24

$\begingroup$ The resultant of $c \eta (\eta + x)  V (\eta  x)^3$ and $c\eta^2 + V (2\eta+x)(\etax)^2$ with respect to $x$ is $V^2 \eta^5 c^2 (108 V^2 \eta^2  36 V c \eta  c^2) = V^2 \eta^5 c^4 (108 v^2  36 v  1)$ where $v = V\eta/c$. $\endgroup$ – Robert Israel Sep 28 '16 at 20:45
In reply to the followup questions from FTXX, here is an argument. Using Israel's notations, suppose $h(x,\eta)=0$ and $\frac{dx}{d\eta}=0$. That means $$V(\etax)^3=c\eta(\eta+x) \tag1$$ and $$c\eta^2+V(2\eta+x)(\etax)^2=0. \tag2$$ Multiply (2) by $\etax$ to get $$c\eta^2(\etax)+V(2\eta+x)(\etax)^3=0.$$ Now, replace (1) into this last equation: $$c\eta^2(\etax)+c\eta(2\eta+x)(\eta+x)=0.$$ Divide through by $c\eta$ and expand to get the quadratic equation $$\eta^2+4\eta x+x^2=0$$ which is impossible since $x\geq0$ and $\eta>0$.