For our earlier example with \(f(x) = x^2/3\) on R = [-1,2], the 50th percentile (i.e. the median) is found by starting with p = 0.5 and then solving
\begin{equation*}
F(c) = 0.5
\end{equation*}
or
\begin{equation*}
c^3/9 + 1/9 = 1/2
\end{equation*}
or
\begin{equation*}
c^3 + 1 = 9/2.
\end{equation*}
After solving for c, you find
\begin{equation*}
\text{median} = \sqrt[3]{7/2} \approx 1.518.
\end{equation*}