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Separate fixed points from roots of $F^2(x)=x$. Fixed points are zero, and $\pm \sqrt{\mu -1}$ for $\mu \ge 1$. The remaining factors in the given factorization yield first $x^2=\mu +1$. For $\mu >-1$ these form the symmetric two-cycle $\{\sqrt{\mu +1},-\sqrt{\mu +1}\}$ because $F(x)=-x$. Its multiplier is $(-2\mu -3{)}^2>1$, so this cycle is always unstable.

The quartic factor gives $s^2-\mu s+1=0$ for $s=x^2$, hence $s_{\pm }=(\mu \pm \sqrt{{\mu }^2-4})/2$. Positive roots require $\mu \ge 2$. At $\mu =2$ they give the fixed points $\pm 1$, not genuine cycles. For $\mu >2$, $s_{+}{s}_{-}=1$ and $F(\sqrt{s_{+}})=\sqrt{s_{-}}$, giving exactly two additional cycles:There are therefore at most three two-cycles. For either of the latter two, its multiplier isThus they are stable for $2<\mu <\sqrt{5}$ and unstable for $\mu >\sqrt{5}$, with a flip at $\sqrt{5}$. The nonzero fixed points have multiplier $3-2\mu$, hence are stable for $1<\mu <2$. Zero is stable for $-1<\mu <1$. At $\mu =1$ there is a supercritical pitchfork; at $\mu =2$ both nonzero fixed points lose stability to these two-cycles. The symmetric unstable cycle terminates at zero at the subcritical flip $\mu =-1$. These are the cubic map period-two branches.

Beyond $\sqrt{5}$ one expects period-four branches and further period doublings, potentially a cascade toward chaotic behaviour. This local argument does not claim that every larger parameter is chaotic; periodic windows and escape may also occur.