Prove, using Rolle’s theorem, that for any value of there is at most one
such that
Proof. The argument is by contradiction. Suppose there are two or more points in for which
. Let
be two such points with
. Since
are both in
we have
Since is continuous on
and differentiable on
(all polynomials are continuous and differentiable everywhere), we may apply Rolle’s theorem on the interval
to conclude that there is some
such that
But this contradicts that (since
and
). Hence, there can be at most one point
such that
I think reasoning in a contrapositive manner (rather than via contradiction) provides an easier solution. Let f(x) = x^3 – 3x + b, with x in [-1, 1]. We want to solve f(x) = 0. Note that f'(x) < 0 for all x in (-1, 1). So f'(c) = 0 for no c in (-1, 1). This is the denial of the conclusion of Theorem 4.4 (Rolle). But f is continuous on [-1, 1] and f has a derivative everywhere in (-1, 1). So the first condition of Theorem 4.4 is true here. Therefore, f(c) equals f(d) for no pair (c, d) in [-1, 1] with unequal c and d. That is, the second condition of Theorem 4.4 is false. Therefore, if f(c) = f(d) = 0, we have c = d, and x^3 – 3x + b has at most one solution.
Yes, but with a direct proof, I am not sure how Rolle’s Theorem applies, since we do not have end point equality.