= Solution
A <horseshoe for an interval map> consists of an <open interval> $J$ containing two disjoint open subintervals each mapped onto $J$. <Glendinning chaos> means that some positive iterate has a horseshoe.
Solving on the appropriate branches of the <tent map> gives
$$
\boxed{x_0=\frac{\mu}{\mu+1},\quad x_{-1}=\frac1{\mu+1},\quad x_{-2}=1-\frac1{\mu(\mu+1)}.}
$$
Here $F(x_{-2})=x_{-1}$ and $F(x_{-1})=x_0$. On $J=[x_{-1},x_0]$ the second iterate is
$$
G(x)=\begin{cases}\mu-\mu^2x,&x\le1/2,\\\mu-\mu^2+\mu^2x,&x\ge1/2.\end{cases}
$$
Both endpoints map to $x_0$, and the minimum is $v=\mu-\mu^2/2$. Both monotone branches cover $J$ when $v\le x_{-1}$, equivalently $(\mu-1)(\mu^2-2)\ge0$. Restricting them to preimages of the interior proves \b[a horseshoe for $F^2$ when $\mu\ge\sqrt2$].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-2-tent-iterates.png]
{title=The tent map and its second iterate with the fixed point and its indicated preimages}
For $1<\mu\le\sqrt2$, $G(J)\subseteq J$. Set $d=(\mu-1)/(\mu+1)$ and $h(x)=(x_0-x)/d$. This maps $J$ onto $[0,1]$; substitution in each linear piece gives $hGh^{-1}=F_{\mu^2}$. The <renormalization of the tent map near its fixed point> therefore makes $F^4$ conjugate on $J$ to $F_{\mu^2}^2$. Applying the preceding threshold proves its horseshoe when $2^{1/4}\le\mu<\sqrt2$.
For $\mu\ge\sqrt2$, the inverse branch of $G$ fixing $x_0$ contracts distance to $x_0$ by $\mu^{-2}$. Compose $m$ copies of it with the other inverse branch. This contraction has a <fixed point> with a periodic itinerary, distinct from $x_0$ and at distance $O(\mu^{-2m})$. Hence nontrivial periodic points accumulate at $x_0$.
For $\mu<\sqrt2$, the conjugate tent parameter $\nu=\mu^2<2$ has every nonzero periodic point in $[\nu-\nu^2/2,\nu/2]$. This interval is forward invariant; below it, a nonzero point increases until entering it. The positive lower bound separates those periodic points from zero, which corresponds to $x_0$. Thus no nontrivial periodic points are sufficiently close to $x_0$ from the left. A point sufficiently close on the right maps to such a left-hand point; periodicity would pass to its image and is impossible. Thus \b[$x_0$ is isolated among periodic points], even though a fourth-iterate horseshoe exists elsewhere.
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