# Find and classify fixed points

#1

Find and classify the fixed points of f(x) = x^2-3x+3.

#2

Since x^2-3x+3=x gives only x=1,3 as solutions. \frac{d}{dx} f^1(x) = 2x-3, so 1 has multiplier -1, so is neutral. 3 has multiplier 3, so is repulsive.

Orbit 2 is similar, with only x=1,3 as solutions. Since \frac{d}{dx} f^2(x) = 4 x^3 - 18 x^2 + 24 x - 9, then 1 has multiplier 1, so is neutral, and 3 has multiplier 9, so is repelling.

The same is true for f^3(x), it only has fixed points 1 neutral and 3 repelling.

I suspect this has to do with the fact that f(x) = x^2-3x+3 can be rewritten as (x - 3) x + 3. It seems logical that a function of the form (x - a) x + a has only the fixed points of 1 and a.