2. $f(x)$ has an inverse function $f^{-1}(x)$, if $f(x)+f(-x)=2$, then $f^{-1}(x-2)+f^{-1}(4-x)=$
2. 0 Detailed Explanation: Substitution $t=f^{-1}(x-2)$, we get $x=f(t)+2$, thus, $f^{-1}(4-x)=f^{-1}(2-f(t))=f^{-1}(f(-t))$ $=-t$, so $f^{-1}(x-2)+f^{-1}(4-x)=0$.

