Bifurcation values for logistic map To find the bifurcation values for $$x_{i+1}=f(x_i) = rx_i(1-x_i)$$first I set $rx(1-x) = 0$ and found the x values and then used the x values to find $r = 0$ and $r = 1$. 
Do you think what I did here is correct? If not, can you help me find the mistakes here?

To study bifurcations of maps, begin by looking for fixed points. The logistic equation is
$$x_{i+1} = f(x_i) = rx_i(1-x_i)$$
so a fixed point satisfies
$$x = rx(1-x) \Rightarrow x(rx - r+1) = 0$$
which implies that there are fixed points at $x=0$ and $x=1-1/r$. To analyze bifurcations as $r$ varies, consider the linearization of the map around these fixed points $x_i=x^*+y_i$
$$y_{i+1} = f'(x^*)y_{i}$$
Now $f'(x)=r(1-2x)$ so linearizing about $x=0$ gives
$$y_{i+1} = ry_i$$
so there is a clear loss of stability when $r=1$ and a period-doubling bifurcation when $r=-1$.
Linearizing about $x=1-1/r$ gives
$$y_{i+1} = (2-r)y_i$$
so there is a bifurcation from instability to stability as $r$ passes through 1 from below, and then a period doubling bifurcation as $r$ passes through $3$.
That gives you three bifurcations, at $r=-1$, $r=1$ and $r=3$. To find further bifurcations you must analyze the second iteration of the map,
$$x_{i+2} = f(f(x_i))$$
and search for fixed points, which correspond to period-2 orbits. This isn't as hard as it sounds - you end up having to solve a quartic equation, but you already know two of the solutions, as any fixed point trivially satisfies this equation. That gives you further period-doubling bifurcations, from 2-cycles to 4-cycles. Finding the location of the next set of period-doubling bifurcations (from 4-cycles to 8-cycles) is hard.
