jmc

Writing out the equation $f(f(x))=x$, we have $$f\left(\frac{ax+b}{cx+d}\right)=x⟹\frac{a\cdot \frac{ax+b}{cx+d}+b}{c\cdot \frac{ax+b}{cx+d}+d}=x⟹\frac{a(ax+b)+b(cx+d)}{c(ax+b)+d(cx+d)}=x$$or $$(a^2+bc)x+(ab+bd)=(ac+cd)x^2+(bc+d^2)x.$$Since this equation holds for infinitely many distinct values of $x$, the corresponding coefficients must be equal. Thus, $$ab+bd=0,a^2+bc=bc+d^2,ac+cd=0.$$Since $b$ and $c$ are nonzero, the first and last equations simplify to $a+d=0$, so $d=-a$, and then the second equation is automatically satisfied. Therefore, all we have from $f(f(x))=x$ is $d=-a$. That is, $$f(x)=\frac{ax+b}{cx-a}.$$Now, using $f(19)=19$ and $f(97)=97$, we get $$19=\frac{19a+b}{19c-a}\text{and}97=\frac{97a+b}{97c-a}.$$These equations become $$b=19^{2}c-2\cdot 19a=97^{2}c-2\cdot 97a.$$At this point, we look at what we want to find: the unique number not in the range of $f$. To find this number, we try to find an expression for $f^{-1}(x)$. If $f(x)=\frac{ax+b}{cx+d}$, then $cxf(x)+df(x)=ax+b$, so $x(a-cf(x))=df(x)-b$, and so $x=\frac{df(x)-b}{a-cf(x)}$. Thus, $$f^{-1}(x)=\frac{dx-b}{a-cx}.$$Since $x=a/c$ is not in the domain of $f^{-1}(x)$, we see that $a/c$ is not in the range of $f(x)$.

Now we can find $a/c$: we have $$19^{2}c-2\cdot 19a=97^{2}c-2\cdot 97a,$$so $$2\cdot (97-19)a=(97^2-19^2)c.$$Thus $$\frac{a}{c}=\frac{97^2-19^2}{2\cdot (97-19)}=\frac{97+19}{2}=\boxed{58}$$by the difference of squares factorization.