= Positive invariant interval for a cubic biological population map
{title2=$F((0,3/2))\subset(0,3/2)$}
For $F(N)=(N-1/2)(N-1)(2-N)+1/2$, every positive initial value with $F(N_0)>0$ has positive subsequent <discrete dynamical system>[iterates]. Indeed $F(N)<3/2$ for every $N>0$, since $F(N)-3/2=-N(2N^2-7N+7)/2<0$. On $[0,3/2]$, the negative part of the <cubic polynomial>[cubic product] has magnitude at most $3/8$, so $F(N)\ge1/8$. Thus the first positive image enters the <forward-invariant set>[forward-invariant interval] $(0,3/2)$.
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