13. Let $a \in R$, find the maximum value of the function $f(a)=2 a \sqrt{x}-\frac{1}{x}$ on the interval $(0,1]$.
13. First prove: when $a>-1$, the function $f(x)$ is monotonically increasing. Let $0<\sqrt[4]{x_{1} x_{2}}<\sqrt{x_{1}}+\sqrt{x_{2}}<1+x_{1}+x_{2}<2$ (since $x_{1}, x_{2} \in(0,1)$), then $\sqrt[4]{x_{1} x_{2}}>x_{1} x_{2}>-a x_{1} x_{2}$ (because $x_{1}, x_{2} \in(0,1)$) $\Rightarrow \frac{1}{x_{1} x_{2}}>$ $\frac{-2 a}{\sqrt{x_{1}}+\sqrt{x_{2}}} \Rightarrow \frac{1}{x_{1} x_{2}}+\frac{2 a}{\sqrt{x_{1}}+\sqrt{x_{2}}}>0 \Rightarrow f\left(x_{1}\right)-f\left(x_{2}\right)<0$. Therefore, when $a>-1$, $f(x)$ is monotonically increasing. When $x=1$, $f(x)$ reaches its maximum value of $2 a-1$.
When $a-1$, $f(x)=-\frac{1}{x}+2 a \sqrt{x}=-\frac{1}{x}+a \sqrt{x}+a \sqrt{x}$
$$
\begin{array}{l}
=-\left[\frac{1}{x}+(-a \sqrt{x})+(-a \sqrt{x})\right] \\
\leqslant-3 \sqrt[3]{(-a \sqrt{x})(-a \sqrt{x}) \cdot \frac{1}{x}}=-3 \sqrt[3]{a^{2}}
\end{array}
$$

Equality condition: $\quad-a \sqrt{x}=\frac{1}{x} \Rightarrow x=\frac{1}{\sqrt[3]{a^{2}}} \in(0,1]$
Conclusion: When $a>-1$, $f(x)_{\max }=2 a-1$.
When $a-1$, $f(x)_{\operatorname{trax}}=-3 \sqrt[3]{a^{2}}$.