Example 17 Given that $a, b, c, d$ are non-zero real numbers, $f(x)=\frac{a x+b}{c x+d}, x \in \mathbf{R}$, and $f(19)=$ 19, $f(97)=97$. If for any real number $x$, when $x \neq-\frac{d}{c}$, $f[f(x)]=x$, try to find the only number outside the range of $f(x)$.
Given the problem, for any real number $x \neq-\frac{d}{c}$, we have $f[f(x)]=x$, so
$$
\frac{a \cdot \frac{a x+b}{c x+d}+b}{c \cdot \frac{a x+b}{c x+d}+d}=x .
$$

Simplifying, we get $\quad(a+d) c x^{2}+\left(d^{2}-a^{2}\right) x-b(a+d)=0$.
Since the above equation holds for $x \neq-\frac{d}{c}$, we have $a+d=0$, and $d^{2}-a^{2}=0$, so $d=-a$. Given $f(19)=19, f(97)=97$, i.e., 19 and 97 are roots of the equation $\frac{a x+b}{c x+d}=x$, which means
19 and 97 are roots of the equation $c x^{2}+(d-a) x-b=0$. Therefore, by Vieta's formulas, we get
$$
\frac{a-d}{c}=116, -\frac{b}{c}=1843 .
$$

Combining $d=-a$, we get $a=58 c, b=-1843 c, d=-58 c$, thus
$$
f(x)=\frac{58 x-1843}{x-58}=58+\frac{1521}{x-58} .
$$

Therefore, $f(x)$ cannot take the value 58, i.e., 58 is the only number outside the range of $f(x)$.