11. Find the minimum value of the function
$$
y=2 x+\sqrt{4 x^{2}-8 x+3}
$$
II. 11. From
$4 x^{2}-8 x+3 \geqslant 0 \Rightarrow x \leqslant \frac{1}{2}$ or $x \geqslant \frac{3}{2}$.
Therefore, the domain of the function is
$$
\left(-\infty, \frac{1}{2}\right] \cup\left[\frac{3}{2},+\infty\right) \text {. }
$$

Let $y=f(x)=2 x+\sqrt{4 x^{2}-8 x+3}$. Then
$$
\begin{array}{l}
f^{\prime}(x)=2+\frac{8 x-8}{2 \sqrt{4 x^{2}-8 x+3}} \\
=2+\frac{4 x-4}{\sqrt{4 x^{2}-8 x+3}} .
\end{array}
$$

When $x>\frac{3}{2}$, it is easy to see that $f^{\prime}(x)>0$.
Thus, $f(x)=2 x+\sqrt{4 x^{2}-8 x+3}$ is an increasing function on the interval $\left[\frac{3}{2},+\infty\right)$.
Therefore, when $x \geqslant \frac{3}{2}$, the minimum value of $f(x)$ is
$$
f\left(\frac{3}{2}\right)=3 \text {. }
$$

When $x<\frac{1}{2}$,
$$
\begin{array}{l} 
f^{\prime}(x)=2+\frac{4 x-4}{\sqrt{4 x^{2}-8 x+3}} \\
= 2-\frac{4(1-x)}{\sqrt{4(x-1)^{2}-1}} \\
<2-\frac{4(1-x)}{2(1-x)}=0 .
\end{array}
$$

Then $f(x)$ is a decreasing function on the interval $\left(-\infty, \frac{1}{2}\right]$.
Hence, when $x \leqslant \frac{1}{2}$, the minimum value of $f(x)$ is $f\left(\frac{1}{2}\right)=1$.
In summary, the minimum value of the function $y=2 x+\sqrt{4 x^{2}-8 x+3}$ is 1.