 Solution

Solving $\mu x(1-x)=1$ by the quadratic formula givesFor $\mu >4$, $0<x_{-}<1/2<x_{+}<1$. On $L=[0,x_{-}]$ the derivative $f^{\prime }(x)=\mu (1-2x)$ is strictly positive, while on $R=[x_{+},1]$ it is strictly negative. Thus the restrictions are strictly monotonic functions. Their endpoint values areBy continuity and strict monotonicity, each restriction maps its interval bijectively onto $[0,1]$, with opposite orientations. The corresponding inverse functions are