3. Let the function $y=$ $f(x)$ with domain and range both $\mathbf{R}$ have the inverse function $y=f^{-1}(x)$. If $f(x)+f(1-$ $x)=2$ holds for all $x \in \mathbf{R}$, then the value of $f^{-1}(x-2)+$ $f^{-1}(4-x)$ is $\qquad$
3. Solution: From $f(x)+f(1-x)=2$, we know that the graph of the function $y=f(x)$ is symmetric about the point $\left(\frac{1}{2}, 1\right)$, so the graph of $y=f^{-1}(x)$ is symmetric about the point $\left(1, \frac{1}{2}\right)$, hence $f^{-1}(x)+f^{-1}(2-x)=1$,

Substituting $x-2$ for $x$ in the above equation, we get: $f^{-1}(x-2)+$ $f^{-1}(4-x)=1$.