Example 3 Given that $\alpha, \beta$ are two distinct real roots of the equation $4 x^{2}-4 t x-1=\theta(t \in \mathbf{R})$, and the domain of the function $f(x)=\frac{2 x-t}{x^{2}+1}$ is $[\alpha, \beta]$.
(1) Find $g(t)=\max f(x)-\min f(x)$.
(1) Let $\alpha \leqslant x_{1}t\left(x_{1}+x_{2}\right)-2 x_{1} x_{2}+\frac{1}{2}>0$

So $f\left(x_{2}\right)-f\left(x_{\mathrm{r}}\right)>0$, hence $f(x)$ is an increasing function on the interval $[\alpha, \beta]$.
Since
$$\alpha+\beta=t, \alpha \beta=-\frac{t}{4}, \beta-\alpha=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}=\sqrt{t^{2}+1}$$

Therefore
$$\begin{array}{l} 
g(t)= \max f(x) - \min f(x) = f(\beta) - f(\alpha) = -\frac{2 \beta-t}{\beta^{2}+1} - \frac{2 \alpha-t}{\alpha^{2}+1} = \\
\frac{(2 \beta-t)\left(\alpha^{2}+1\right)-(2 \alpha-t)\left(\beta^{2}+1\right)}{\left(\alpha^{2}+1\right)\left(\beta^{2}+1\right)} = \\
\frac{(\beta-\alpha)[t(\alpha+\beta)-2 \alpha \beta+2]}{\alpha^{2} \beta^{2}+\alpha^{2}+\beta^{2}+1} = \\
\frac{\sqrt{t^{2}+1}\left(t^{2}+\frac{5}{2}\right)}{t^{2}+\frac{25}{16}} = \frac{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}{16 t^{2}+25}
\end{array}$$