Rolle's Theorem is normally proved using the extreme value theorem, but we can in fact implement Rolle's Theorem as a bisection algorithm, similar to the intermediate value theorem.
Given a function f and its derivative f', defined
over some interval [a, b], we can find an increasing sequence of
arguments l(n) and a decreasing sequence of arguments
r(n), so that at least one of them is infinite, elements of
l are always less than elements of r, and the
distance between them gets arbitrarily small as both sequences progress, along
with any of the following properties:
f'(l(n)) is eventually always positive, and
f'(r(n)) is eventually always negative, or vice versa, both
converging to zero,
f'(l(n)) is eventually always positive, and
f'(r(n)) is eventually always negative, or vice versa, but at
least one is always at least distance h from zero,
f'(l(n)) and f'(r(n)) are always positive, and
f(l(n)) never decreases as n increases, and
is eventually greater than any value f(r(n)), or
f'(l(n)) and f'(r(n)) are always negative, and
f(r(n)) never increases as n increases, and
is eventually less than any value f(l(n))/
In case 1. we have an approximation of a turning point. In case 2. we have a
discontinuity in f'. In case 3. f(l(n)) and
f(r(n)) must converge to a discontinuity in f, or
else r(n) will converge to a place where f'(x) is
zero or negative, giving either a turning point or a discontinuity in
f'. Similar will happen symmetrically in case 4. If
f and f' are continuous, then only outcomes that
converge to a turning point will be possible.
In order to achieve one of these outcomes, we compare the current left and
right extremes l and r to their midpoint
m=(l+r)/2, and if f'(m) is zero, we terminate with an
exact value, otherwise we observe the following cases:
f'(l) > 0 and f'(m) < 0 or vice versa, then take
m to be the new r,
f'(r) > 0 and f'(m) < 0 or vice versa, then take
m to be the new l,
f'(l), f'(r), and f'(m) are all
positive, and f(m) >= f(l), then take m to be the
new l,
f'(l), f'(r), and f'(m) are all
positive, and f(m) < f(l), then take m to be the
new r,
f'(l), f'(r), and f'(m) are all
negative, and f(m) <= f(r), then take m to be the
new r,
f'(l), f'(r), and f'(m) are all
negative, and f(m) > f(r), then take m to be the
new l.
In the cases where all of our derivatives are the same, we aim to keep one
sequence increasing or decreasing, since it must contain some kind of
s
bend in order to turn the other way and then recover the same
sign in its derivative. Then we will either find a different sign, or yield one
of outcomes 3. or 4.