How to find all fixed points of $x^3-x^2-4x+5$ if has exactly 3 and $x=1$ is one of them?
I found this problem, and normally one has a reduced domain and range say $[a,b]$ to $[a,b]$ then if it's Lipschitz ($\lambda <1$) for example there is only one fixed point.... However if I already have one, how to find the other two? Is there a theorem or procedure?
The fixed points are the solutions to the cubic $$x^3-x^2-5x+5=0.$$ You are given the solution $x=1$, so you can factor out the factor $x-1$, leaving a quadratic, which you surely know how to solve.