Period-doubling of the logistic map at r = 3

Step 1 (find the fixed points). Solve $rx(1-x)=x$, i.e. $x[r(1-x)-1]=0$. This gives $x=0$ or the nontrivial fixed point $x^* = 1-\tfrac{1}{r}$, which lies in $(0,1)$ precisely when $r>1$.

Step 2 (linearize). Differentiating, $f'(x)=r-2rx$. Evaluating at the nontrivial fixed point, $f'(x^*) = r - 2r\left(1-\tfrac1r\right) = r - 2r + 2 = 2-r$, so $f'(x^*) = 2-r$.

Step 3 (stability criterion). A fixed point is locally stable exactly when $|f'(x^*)|<1$, i.e. $|2-r|<1$, which rearranges to $1<r<3$. So $x^*$ attracts nearby orbits throughout this whole range.

Step 4 (the bifurcation at r = 3). At $r=3$ exactly, $f'(x^*)=-1$: the linearization is marginal. For $r$ slightly above 3, $|f'(x^*)|>1$ and $x^*$ becomes repelling. Meanwhile the second-iterate map $f\circ f$ has, by the chain rule, derivative $f'(x^*)^2=1$ at $x^*$ exactly at $r=3$; expanding $f\circ f$ to higher order shows this degenerate point splits into two genuine new fixed points of $f\circ f$ (a stable 2-cycle of $f$) as $r$ increases past 3 — the period-doubling bifurcation.