Example 12 (2002 Hunan Provincial Competition Question) Let the two roots of the quadratic equation $2 x^{2}-t x-2=0$ with respect to $x$ be $\alpha, \beta(\alpha<\beta$.
(1) If $x_{1} 、 x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove: $4 x^{2} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(2) Let $f(x)=\frac{4 x-t}{x^{2}+1},$ and the maximum and minimum values of $f(x)$ in the interval $[\alpha, \beta]$ be $f_{\max }$ and $f_{\min }$, respectively. Let $g(t)=f_{\max }-f_{\min }$. Find the minimum value of $g(t)$.
$$
\begin{aligned}
& \text { (1) From the conditions, we have } \alpha+\beta=\frac{t}{2}, \alpha \beta=-1. \\
& \text { Without loss of generality, assume } \alpha \leqslant x_{1}4 x_{1} x_{2}-2(\alpha+\beta)\left(x_{1}+x_{2}\right)+4 \alpha \beta=4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4 .
\end{aligned}
$$

Thus, $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-40, f_{\min }=f(\alpha)<0$.
$$
\begin{aligned}
g(t) & =f(\beta)-f(\alpha) \\
& =|f(\beta)|+|f(\alpha)| \\
& \geqslant 2 \sqrt{|f(\beta)| \cdot|f(\alpha)|}=4 .
\end{aligned}
$$

Equality holds if and only if $f(\beta)=-f(\alpha)=2$,
i.e., $\frac{8}{\sqrt{t^{2}+16}+t}=2$, which implies $t=0$.
Therefore, the minimum value of $g(t)$ is 4.
$$
$$