 Solution

If the starting length is $L=b-a$, then after $k$ interval halvings the current midpoint is at most $L/2^{k+1}$ from some root remaining in the bracket. Thus, with the midpoint returned after those halvings, it suffices to takeIf iterations instead count midpoint evaluations starting with the initial midpoint as evaluation one, the usual bound is $n\ge \lceil {\log }_2(L/ϵ)\rceil$, with at least one evaluation unless an endpoint root has already been found. Stating the counting convention avoids an apparent one-step discrepancy.