9. (16 points) Find the range of the function $f(x)=2 x^{2}+x \sqrt{4 x^{2}-3}$.

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9. (16 points) Find the range of the function $f(x)=2 x^{2}+x \sqrt{4 x^{2}-3}$.
9. It is easy to find that the domain of the function is $\left(-\infty,-\frac{\sqrt{3}}{2}\right] \cup\left[\frac{\sqrt{3}}{2},+\infty\right)$.
(1) It is easy to know that the function $f(x)=2 x^{2}+x \sqrt{4 x^{2}-3}$ is increasing on $\left[\frac{\sqrt{3}}{2},+\infty\right)$, so when $x \geqslant \frac{\sqrt{3}}{2}$

we can get $f(x) \geqslant \frac{3}{2}$.
(2) When $x \leqslant-\frac{\sqrt{3}}{2}$
$$
\begin{aligned}
f(x) & =x\left(2 x+\sqrt{4 x^{2}-3}\right)=\frac{x\left(2 x+\sqrt{4 x^{2}-3}\right)\left(2 x-\sqrt{4 x^{2}-3}\right)}{2 x-\sqrt{4 x^{2}-3}} \\
& =\frac{3 x}{2 x-\sqrt{4 x^{2}-3}}=\frac{3}{2-\frac{\sqrt{4 x^{2}-3}}{x}}=\frac{3}{2+\sqrt{4-\frac{3}{x^{2}}}} .
\end{aligned}
$$

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