I have a function $M: [0,1] \rightarrow R_{\geq 0}$ defined as
$M(p)=\frac{F_1^{-1}(p) +F_2^{-1}(p)}{2}$,
with $F_1(t)=\int_0^t f_1(\zeta) d\zeta$ for some continuous $f_1: R_{\geq 0} \rightarrow [0,1]$ and similarly for $f_2$. Also I know that $\lim_{t\rightarrow \infty} F_i(t)=1$. That is, $f_1$ and $f_2$ are probability distributions on $R_{\geq 0}$.
Furthermore, I can assume that $F_1(t)$ is strictly monotonous on some interval $I_1=(T_{1,0},T_{1,1})$ (i.e. $f_1(t)>0$ on this interval) and likewise for $F_2(t)$ for the interval $I_2=(T_{2,0},T_{2,1})$, such that $F_i$ is bijective on this interval and $F_i^{-1}(p)$ can be interpreted as the inverse restricted to this interval. Also let $F_i^{-1}(0) = T_{i,0}$ and likewise for $F_i^{-1}(1)$.
I am looking for an expression of $M^{-1}: [\frac{T_{1,0}+T_{2,0}}{2},\frac{T_{1,1}+T_{2,1}}{2}]\rightarrow [0,1]$.
Since $f_i(t)>0$ and continuous on $I_i$ we know that the $F_i$ are strictly monotonous and hence the $F_i^{-1}$ are as well. Thus the inverse as defined above exists. I can also find this inverse numerically by defining $\delta_i(p)=F_i^{-1}(p)-M(p)$ and then solving for $\delta_1(p)+\delta_2(p)=0$.
However, I am not sure if I could find an expression for this inverse using only known terms, i.e. $f_i$ and their integrals, inverses of integrals etc. Is there a way to get this inverse or is there some additional information missing?