= Solution
At a non-axis fixed point the determinant and trace are
$$
\det J=x(1-x)(2x-β-1),\qquad \operatorname{tr}J=βx(1-x)-1.
$$
These formulas determine the solid stable and dashed unstable branches in the <bifurcation diagram>. The pair $x_±$ is born in a <saddle-node bifurcation> at $μ=-(β+1)^2/4$. For $β\ne1$, it exchanges branches with $x=1$ at $μ=-β$ and with $x=0$ at $μ=0$, both <transcritical bifurcations>. If $0<β<1$, the saddle-node lies to the left of $x=1$; if $β>1$, it lies to the right. For $β=1$, the saddle-node and the collision at $x=1$ combine into a <pitchfork bifurcation normal form>[pitchfork bifurcation] at $μ=-1$.
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