= Solution
Solving $\mu x(1-x)=1$ by the <quadratic formula> gives
$$
x_\pm=\frac12\pm\sqrt{\frac14-\frac1\mu}.
$$
For $\mu>4$, $0<x_-<1/2<x_+<1$. On $L=[0,x_-]$ the <derivative> $f'(x)=\mu(1-2x)$ is strictly positive, while on $R=[x_+,1]$ it is strictly negative. Thus the restrictions are strictly <monotonic functions>. Their endpoint values are
$$
f(0)=0,\quad f(x_-)=1,\qquad f(x_+)=1,\quad f(1)=0.
$$
By <continuity> and strict <monotonicity>, \b[each restriction maps its interval bijectively onto $[0,1]$], with opposite orientations. The corresponding <inverse functions> are
$$
\boxed{g_L(v)=\frac{1-\sqrt{1-4v/\mu}}2,\qquad
g_R(v)=\frac{1+\sqrt{1-4v/\mu}}2,\qquad 0\leq v\leq1.}
$$
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