Example 7 Given that $\alpha, \beta$ are two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, and the domain of the function $f(x)=\frac{2 x-t}{x^{2}+1}$ is $[\alpha, \beta]$, find $g(t)=\max f(x)-\min f(x)$.
First, consider the monotonicity of $f(x)$ on $[\alpha, \beta]$.
$$
f^{\prime}(x)=\frac{2\left(x^{2}+1\right)-2 x(2 x-t)}{\left(x^{2}+1\right)^{2}}=\frac{2\left(-x^{2}+t x+1\right)}{\left(x^{2}+1\right)^{2}}
$$

Given that $\alpha, \beta$ are the two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, when $x \in[\alpha, \beta]$, it must be that $4 x^{2}-4 t x-1 \leqslant 0, x^{2}-t x-\frac{1}{4} \leqslant 0$. Therefore, $-x^{2}+t x+1=-\left(x^{2}-t x-\frac{1}{4}\right)+\frac{3}{4} \geqslant \frac{3}{4}>0$, so $f(x)$ is monotonically increasing on $[\alpha, \beta]$. Hence,
$$
g(t)=f(\beta)-f(\alpha)=\frac{2 \beta-t}{\beta^{2}+1}-\frac{2 \alpha-t}{\alpha^{2}+1}=\frac{(\beta-\alpha)[t(\alpha+\beta)-2 \alpha \beta+2]}{\alpha^{2} \beta^{2}+\alpha^{2}+\beta^{2}+1}=\frac{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}{16 t^{2}+25} .
$$