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{\bf Question}

Find the fixed points and a period 2 orbit for $f(x)=2x^2-5x$.
Decide in each case if they are attracting or repelling.
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{\bf Answer}

$f(x)=2x^2-5x$. Fixed points where $2x^2-5x=x: {\underline
x=0,3}$.
$f^2(x)=2(2x^2-5x)^2-5(2x^2-5x)=(2x^2-5x)(2(2x^2-5x)-5)=x(2x-5)(4x^2-10x-5)$.
Fixed points of $f^2$ where $x(2x-5)(4x^2-10x-5)=x$, i.e.
$x^3-5x^2+5x+3=0$.  We know $x=3$ is a solution, so $(x-3)$ is a
factor of LHS:

$$x^3-5x^2+5x+3=(x-3)(x^2-2x-1)=0$$

giving $x=1 \pm \sqrt 2$.

(Check $f(1 \pm \sqrt 2)=2(1 \pm \sqrt 2)^2-5(1 \pm \sqrt 2)=2(3
\pm 2\sqrt 2)-5(1 \pm \sqrt 2)=(1 \pm \sqrt 2)$.)

ALSO: $f'(x)=4x-5$ so fixed points repelling; $f'(1 + \sqrt 2)f'(1
- \sqrt 2)=-31$ so 2-cycle repelling too.
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