Past exam of the probability, statistics and options I course of the University of Oxford / 2011 / U / 1 / c / Solution

At the left endpoint $f(-1)=36>0$. If $b=2$, the endpoint is exactly a root and the function above immediately returns $2$. For $2<b\le 4$, $f(b)<0$, so the first midpoint is $m=(b-1)/2$.

If $2<b<3$, then $1/2<m<1$ and $f(m)<0$. The first retained bracket is $[-1,m]$, containing only the root zero, so the iterates converge to zero. If $b=3$, the first midpoint is exactly the root one and is returned. If $3<b\le 4$, then $1<m\le 3/2$ and $f(m)>0$. The retained bracket is $[m,b]$, containing only the root two, so the iterates converge to two. HenceThis illustrates root selection by bisection with multiple roots in the initial bracket: bisection guarantees a root, but which root it chooses depends discontinuously on the starting endpoint.