4. Find the minimum value of the function $f(x)=\sqrt{4 x^{2}-12 x+8}+\sqrt{4+3 x-x^{2}}$.
4. $f(x)=\sqrt{4\left(x-\frac{3}{2}\right)^{2}-1}+\sqrt{\frac{25}{4}-\left(x-\frac{3}{2}\right)^{2}}$, due to $4 x^{2}-12 x+8 \geqslant 0$ and $4+3 x-x^{2} \geqslant 0$, we have $\frac{1}{4} \leqslant\left(x-\frac{3}{2}\right)^{2} \leqslant \frac{25}{4}$, thus, let $\left(x-\frac{3}{2}\right)^{2}=6 t+\frac{1}{4}$, then $t \in[0,1]$, and $f(x)=g(t)=\sqrt{24 t}+\sqrt{6-6 t}=$ $\sqrt{24} \cdot \sqrt{t}+\sqrt{6} \cdot \sqrt{1-t}$, by equation (5-3) we know that the maximum value of $f(x)$ is $\sqrt{30}$, and the minimum value is $\sqrt{6}$.