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?

$\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left( #1 \right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$$
\mbox{With the Logistic Map}\qquad x_{n + 1} = \mu x_{n}\pars{1 - x_{n}}\,,\qquad
0 < \mu < 4\tag{1}
$$ you must study $\mu$-behaviour ( 'fixed points' ) of
$\pars{1}$ and $\mu$-behaviour of
\begin{align}
x_{n + 1} &= \mu^{2}x_{n - 1}\pars{1 - x_{n - 1}}
\bracks{1 - \mu x_{n - 1}\pars{1 - x_{n - 1}}}
\\
x_{n + 1} &=
\mu^{3}x_{n - 2}\pars{1 - x_{n - 2}}\bracks{1 - \mu x_{n - 2}
\pars{1 - x_{n - 2}}}\braces{1 - \mu^{2} x_{n - 2}\pars{1 - x_{n - 2}}
\bracks{1 - \mu x_{n - 2}\pars{1 - x_{n - 2}}}}
\end{align}
and so on.
