14. Let the quadratic equation in $x$ be $2 x^{2}-t x-2=0$ with two roots $\alpha$ and
(Problem 13 interval) $\beta(\alpha<\beta)$.
(1) If $x_{1} 、 x_{2}$ are two different points in the interval $[\alpha, \beta]$, prove: $4 x_{1} x_{2}-t\left(x_{1}+x_{2}\right)-4<0$;
(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 $g(t)$.
14. (I) From the conditions, we have: $\alpha+\beta=\frac{t}{2}, \alpha \beta=-1$.

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

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