Consider a map:
$g\left ( x \right )=x+\lambda x\left ( 2x-1 \right )\left ( x-1 \right ) $for $[0,1)$ and $\lambda \in (0,8]$.
It is given that $x_{-}$ and $x_{+} $ forms a period -2 orbit.
It is also given that $ g'\left ( x_{-} \right )g'\left ( x_{-} \right )=\left ( \lambda-5 \right )^{2}$.
Observe that $\frac{\mathrm{d} }{\mathrm{d} t} g^{2}\left ( x \right )=\frac{d}{dt}\left ( g\left ( g\left ( x \right ) \right ) \right ) =g'\left (g\left ( x_{+} \right ) \right ) g'\left ( x_{+} \right )=g'\left ( x_{-} \right )g'\left ( x_{+} \right )= \left ( \lambda -5 \right )^{2}$
Let $\bar{\lambda}$ denote the eigenvalue $g'\left (g\left ( x_{+} \right ) \right ) g'\left ( x_{+} \right )=g'\left ( x_{-} \right )g'\left ( x_{+} \right ) \left ( \lambda -5 \right )^{2}$
The stability of a period-2 orbit can be evaluated by taking the absolute value of the time derivative of the eigenvalue.
Hence, $\left | \bar{\lambda} \right |=\left | \left ( \lambda-5 \right )^{2} \right |< or >1 $
To check for stable period 2 orbit: $\left | \left ( \lambda-5 \right )^{2} \right |<1$
This is
$\left ( \lambda-5 \right )^{2} <1$ which gives root values of 6 and 4 (which should be picked?)
We should also check for $\left ( \lambda-5 \right )^{2}>-1$
But the roots are imaginary; in particular, they are $10\pm i$ (again, which should be picked?)
I would like to know how I can reconcile the two real root values of which one must be picked and the imaginary roots so that I can determined the stability of the period-2 orbit and sketch a bifurcation diagram.
Edit:
$x_{\pm }=\frac{1}{2}\left ( 1\pm \sqrt{1-\frac{4}{\lambda}} \right )$