= Root selection by bisection with multiple roots in the initial bracket
{title2=$f(x)=x(x-1)(x-2)(x-5),\quad[a,b]=[-1,b]$}
The <bisection method> preserves some root in a sign-changing bracket but does not select a fixed root when several are present. For the displayed <polynomial>, endpoint and midpoint roots are returned immediately. If $2<b<3$, the first midpoint lies in $(1/2,1)$ with negative function value, so the retained bracket contains only zero. If $b=3$, the first midpoint is exactly one. If $3<b\leq4$, that midpoint lies in $(1,3/2]$ with positive function value, so the retained bracket contains only two. At $b=2$ the right endpoint itself is the root two.
Thus arbitrarily small changes of the initial endpoint across $b=3$ change the limiting root from zero to two, with one selected exactly at the transition. The guarantee concerns distance to a root inside the retained interval, rather than continuous dependence of that root's identity on the starting data.
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