9. Given the function
$$
f(x)=\left\{\begin{array}{ll}
3 x-1 & x \leqslant 1 ; \\
\frac{2 x+3}{x-1} & x>1
\end{array}\right.
$$

If the graph of the function $y=g(x)$ is symmetric to the graph of the function $y=f^{-1}(x+1)$ about the line $y=x$, then the value of $g(11)$ is
9. $\frac{3}{2}$.
The graph of $y=f^{-1}(x)$ is shifted one unit to the left to get the graph of $y=f^{-1}(x+1)$. The graph of $y=f(x)$ is symmetrical to the graph of $y=f^{-1}(x)$ about the line $y=x$, and the graph of $y=f^{-1}(x+1)$ is symmetrical to the graph of $y=g(x)$ about the line $y=x$ as well. Therefore, the graph of $y=f(x)$, when shifted one unit downward, gives the graph of $y=g(x)$, i.e., $g(x)=f(x)-1$.
Thus, $g(11)=f(11)-1=\frac{25}{10}-1=\frac{3}{2}$.