Exercise 1.4. A DS F:SS is invertible if there is a function F1:SS such that F1(F(X))=X and F(F1(X))=X for all XS. In other words, the functions F,F1 satisfy FF1=F1F= Id where Id : SS is the identity function. You can think of F1 as a function which takes states backward in time by one time step. In this case we can define the negative iterates Fn=F1F1 where n>0. It's easy to see that the associative laws (3) continue to hold. Show that the DSF(x)=1+x1 is invertible by finding a formula for F1(x). Then find the iterates F2(x),F3(x). Finally, check the associative laws by showing that F1=F2F3=F3F2=FF2. Hint: You can find F1(y) by setting y=F(x) and solving for x. For the moment, ignore the discontinuity at x=1