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{\bf Question}

Show that when a fixed point of the Henon map becomes unstable and
creates a 2-cycle, then this occurs at the point
$(x,y)=\left(\frac{1-b}{2},\frac{1-b}{2}\right)$.

[\underline{Hint}: Remember the expression for the sum of the
roots of a quadratic equation.]

\medskip

{\bf Answer}

Every fixed point of $f$ lies on the line $y=x$. Every 2-cycle
\{(x,y),(x',y')\} satisfies $x=x'=(1-b)$, from the quadratic
equation from question 2 whose roots are $x,\ x'$. At the moment
the 2-cycle is created from a fixed point we have
$x=x'=\ds\frac{1}{2}(1-b)=y=y'$.
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