How to find all fixed points of $x^3-x^2-4x+5$?

Question

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?

You have a cubic equation and you know one of the roots. Factor that out and your are left with a quadratic. — dxiv, Oct 01 '21 at 20:34
Well if you identify the correct cubic, you have a root and can reduce it to a quadratic ... — Mark Bennet, Oct 01 '21 at 20:34
By definition, a fixed point $x^$ of a function $f(x)$ satisfies $f(x^) = x^*$. — heropup, Oct 01 '21 at 20:35
In fact it isn't hard to just see that the other two fixed points are $\pm\sqrt{5}$. — Servaes, Oct 01 '21 at 20:37

score 2 · Accepted Answer · answered Oct 01 '21 at 20:35

2

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.

answered Oct 01 '21 at 20:35

Servaes

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How to find all fixed points of $x^3-x^2-4x+5$?

1 Answers1