= Solution
A <fixed point> $x_*$ satisfies $f(x_*,\mu)=x_*$. A <perturbation> obeys $u_{n+1}=m u_n+O(u_n^2)$, where the <periodic-orbit multiplier> is $m=f_x(x_*,\mu)$. It is linearly <attracting> for $|m|<1$ and <repelling> for $|m|>1$; at $m=\pm1$ nonlinear terms decide <Lyapunov stability>. A crossing through $+1$ can produce a saddle-node, transcritical or <pitchfork bifurcation>, depending on the <mathematical model parameter> dependence and <symmetry>. A crossing through $-1$ generically gives a <period-doubling bifurcation>. A real one-dimensional <function> cannot have the <complex conjugate> pair needed for a <Neimark–Sacker bifurcation>.
For a genuine <period-two orbit>, study the <return map> $f\circ f$ at either point. By the <chain rule> its <periodic-orbit multiplier> is
$$
\boxed{m_2=f_x(x_1,\mu)f_x(x_2,\mu).}
$$
Its modulus determines <linear stability> by the same criterion, now per two <iterates>.
For the <real quadratic iteration map>, the fixed-point equation gives
$$
x_\pm=\frac{-1\pm\sqrt{1+4\mu}}2,\qquad \mu\ge-\frac14.
$$
The <derivative> is $-2x$. The lower branch $x_-$ is <repelling> whenever $\mu>-1/4$. The upper branch satisfies $|-2x_+|<1$ exactly for
$$
\boxed{-\frac14<\mu<\frac34.}
$$
At $\mu=-1/4$ the two <fixed points> merge in a <saddle-node bifurcation>. With $u=x+1/2$ at that <mathematical model parameter>, $u'=u-u^2$, so the point attracts from the right and is <repelling> from the left. At $\mu=3/4$, $x_+=1/2$ has <periodic-orbit multiplier> $-1$ and undergoes a <period-doubling bifurcation>. At the threshold itself $u'= -u-u^2$ and the two-step return is $u''=u-2u^3-u^4$, proving weak <asymptotic stability> there despite the inconclusive linear test.
Factor the second-iterate equation:
$$
f(f(x),\mu)-x=-(x^2+x-\mu)(x^2-x+1-\mu).
$$
Removing the fixed-point factor yields the <periodic orbit> points
$$
x_1=\frac{1-\sqrt{4\mu-3}}2,\qquad x_2=\frac{1+\sqrt{4\mu-3}}2.
$$
They are distinct and exchanged by $f$ exactly when $\mu>3/4$; at equality they coincide with the <fixed point>. Their product is $1-\mu$, hence their <periodic-orbit multiplier> is $4(1-\mu)$. Thus
$$
\boxed{\text{the two-cycle exists for }\mu>\frac34,\quad\text{and is linearly attracting for }\frac34<\mu<\frac54.}
$$
It is <repelling> for $\mu>5/4$ and reaches its next <period-doubling bifurcation> threshold at $\mu=5/4$. At that exact endpoint it is weakly <attracting>: if $f^2$ is expanded about either <periodic orbit> point as $v'=-v+Av^2+Bv^3+O(v^4)$, then $A=2\mu-6x_i^2$, $B=-4x_i$, and $A^2+B=20\pm10\sqrt2>0$. Its second return is $v''=v-2(A^2+B)v^3+O(v^4)$.
At $\mu=2$ put $h(y)=4\sin^2(\pi y)-2=-2\cos(2\pi y)$. The double-angle identity gives
$$
f(h(y),2)=2-4\cos^2(2\pi y)=-2\cos(4\pi y)=h(2y\bmod1).
$$
This verifies the printed relation to the <Bernoulli shift>. Precisely, it is a <semiconjugacy>: $h(y)=h(1-y)$, so it is not invertible on the circle. To obtain a genuine interval <topological conjugacy>, use the <homeomorphism> $H(u)=-2\cos(\pi u)$ from $[0,1]$ onto $[-2,2]$, and the full <tent map>
$$
T(u)=\begin{cases}2u,&0\le u\le1/2,\\2-2u,&1/2\le u\le1.\end{cases}
$$
The same identity proves $f(H(u),2)=H(T(u))$.
The full <tent map> has <dense periodic points>, is <topologically mixing> and has <sensitive dependence on initial conditions>. To see these properties directly, its $n$th iterate has $2^n$ monotone branches on dyadic intervals of length $2^{-n}$, each mapping onto $[0,1]$. Every nonempty open interval contains such a branch domain, and a later iterate covers all of $[0,1]$, proving <topological mixing>. On any branch, the <function graph> of the iterate has a <fixed point> by the <intermediate value theorem>, so <periodic points> occur in every open interval. Given any initial point and any neighborhood, some iterate of that neighborhood covers the interval; one can choose a nearby point whose image differs from the initial point's image by at least a fixed positive distance. This proves <sensitive dependence on initial conditions>. <Uniform continuity> of the inverse <homeomorphism> transfers these properties to $f$ on $[-2,2]$.
Consequently the <real quadratic iteration map> is <chaotic> on its <invariant interval> in the Devaney sense: \b[<topological transitivity>, <dense periodic points> and <sensitive dependence on initial conditions>], with all these orbits bounded. There are also aperiodic orbits: <periodic points> form a countable union of finite sets of <roots of an equation> of the equations $f^n(x)=x$, whereas the interval is uncountable. The <semiconjugacy> alone would not justify treating $h$ as invertible; the tent-map argument supplies the missing rigorous transfer.
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