1. The problem statement, all variables and given/known data Let f: R->R and f(x)=x3-x. By restricting the domain and range of f appropriately, obtain from f a bijective function g. 2. Relevant equations x3-x=(x+1)(x-1)x g(x): R->R 3. The attempt at a solution we can find roots from the polynomial form (x+1)(x-1)x and restrict the domain and range by avoiding intervals including more than one root. There are several ways to obtain g, because we are free to choose an interval, but what I'm interested in is this. if we say g:R->R such that x=-1 -> g(x)=-∞ x=0 -> g(x)=0 x=1 -> g(x)=∞ we can define a 1 to 1 map because R is an infinite set, then can we say this is also an answer? Because it is bijective to R.