9. (16 points) Find the range of the function
$$
f(x)=2 x^{2}+x \sqrt{4 x^{2}-3}
$$
9. The domain of the function is easily obtained as
$$
\left(-\infty,-\frac{\sqrt{3}}{2}\right] \cup\left[\frac{\sqrt{3}}{2},+\infty\right) \text {. }
$$
(1) It is easy to see that the function
$$
f(x)=2 x^{2}+x \sqrt{4 x^{2}-3}
$$

is increasing on the interval $\left[\frac{\sqrt{3}}{2},+\infty\right)$.
Therefore, when $x \geqslant \frac{\sqrt{3}}{2}$, we have $f(x) \geqslant \frac{3}{2}$.
(2) When $x \leqslant-\frac{\sqrt{3}}{2}$,
$$
\begin{array}{l}
f(x)=x\left(2 x+\sqrt{4 x^{2}-3}\right) \\
=\frac{x\left(2 x+\sqrt{4 x^{2}-3}\right)\left(2 x-\sqrt{4 x^{2}-3}\right)}{2 x-\sqrt{4 x^{2}-3}} \\
=\frac{3 x}{2 x-\sqrt{4 x^{2}-3}}=\frac{3}{2+\sqrt{4-\frac{3}{x^{2}}}} .
\end{array}
$$

Since $x \leqslant-\frac{\sqrt{3}}{2}$, we have
$$
0 \leqslant 4-\frac{3}{x^{2}}<1$, which means the range of the function $f(x)$ is $(1,+\infty)$.