9. If the domain of functions $f(x)$ and $g(x)$ is the set of non-negative real numbers, for any $x \geqslant 0$, it is defined that $f(x) \cdot g(x)=$ $\min \{f(x), g(x)\}$. If $f(x)=3-x, g(x)=\sqrt{2 x+5}$, then the maximum value of $f(x) \cdot g(x)$ is $\qquad$ .
9. Given $x \geqslant 0$, let $3-x>\sqrt{2 x+5}$, solving this yields $0 \leqslant x<4-2 \sqrt{3}$,

so $f(x) * g(x)=\left\{\begin{array}{l}\sqrt{2 x+5}, 0 \leqslant x<4-2 \sqrt{3}, \\ 3-x, x \geqslant 4-2 \sqrt{3} .\end{array}\right.$
Since $3-x$ is monotonically decreasing on $\mathbf{R}$, when $x \geqslant 4-2 \sqrt{3}$, $f(x) * g(x) \in f(4-2 \sqrt{3}) * g(4-2 \sqrt{3})=3-$ $(4-2 \sqrt{3})=2 \sqrt{3}-1$

When $0 \leqslant x \leqslant 4-2 \sqrt{3}$, $\sqrt{2 x+5}$ is monotonically increasing, so when $x \in[0,4-2 \sqrt{3}]$, $f(x) * g(x)<$ $\sqrt{2 \cdot(4-2 \sqrt{3})+5}=2 \sqrt{3}-1$.
In summary, the maximum value of $f(x) * g(x)$ is $2 \sqrt{3}-1$, fill in $2 \sqrt{3}-1$.