jmc

We start by finding the inverse function of $f$. By definition, we know that $f(f^{-1}(x))=x$, so $$\frac{f^{-1}(x)-3}{f^{-1}(x)-4}=x.$$We can solve this equation for $f^{-1}(x)$. First we multiply both sides by $f^{-1}(x)-4$: $$f^{-1}(x)-3=x\cdot (f^{-1}(x)-4).$$Then we expand: $$f^{-1}(x)-3=x\cdot f^{-1}(x)-4x.$$Then we rearrange so as to group all terms involving $f^{-1}(x)$ on the left side: $$f^{-1}(x)-x\cdot f^{-1}(x)=3-4x.$$We can factor on the left side: $$f^{-1}(x)\cdot (1-x)=3-4x.$$Finally, we divide both sides by $1-x$ to obtain our inverse function, $$f^{-1}(x)=\frac{3-4x}{1-x}.$$This function is defined for all $x$ except $\boxed{1}$.