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For a period-$n$ orbit of $g(x)=\mu -x^2$ that avoids zero, the chain rule gives its multiplier of a periodic orbit of an iteration asThe sign of $g^{\prime }$ is positive on the negative half-axis and negative on the positive half-axis, exactly as the parabola sketch shows. ConsequentlyThis product is independent of the chosen starting point because changing $\hat{x}$ only cyclically reorders the factors. If the orbit contains zero, the derivative product is zero and its orientation is undefined. One must then use the actual zero periodic-orbit multiplier, rather than infer a positive or negative orientation from the special assignment $\mathrm{sgn}(0)=1$. The nonzero-periodic-orbit multiplier cases needed in part (d) avoid this exception automatically.

For the iteration of a map $f(x)=\lambda x(1-x^2)$, a fixed point satisfies $x[1-\lambda +\lambda x^2]=0$. If $0<\lambda <1$, the bracket is strictly positive, so the sole fixed point is $0$. Its multiplier of a periodic orbit of an iteration is $f^{\prime }(0)=\lambda$, whose modulus is below one. Thus $\boxed{x=0\text{ is locally asymptotically stable}}$ by fixed point stability for an iteration.

The exact second-iterate equation factors asThe positive nonfixed period-two points come from the final factor. If $y=x^2$, their squared values areThey are real and distinct for $\lambda >2$.

The multiplier of a periodic orbit of an iteration isFor $\lambda =2+\mu$ with $0<\mu \ll 1$,whose modulus is less than one. The new two-cycle is therefore locally asymptotically stable.

The family $x_{n+1}=\mu -x_n^2$ has fixed points $(-1\pm \sqrt{1+4\mu })/2$. Its upper branch attracts linearly for $-1/4<\mu <3/4$ and loses stability in a period-doubling bifurcation. The resulting two-cycle has points $(1\pm \sqrt{4\mu -3})/2$ and multiplier $4(1-\mu )$, attracting for $3/4<\mu <5/4$. At $\mu =2$, the map is conjugate on $[-2,2]$ to the full tent map.