Show that f(x) = (x−1)2(x−3)3satisfies Rolle’s Theorem on the intervalD= [1,3] and find all values of c which satisfy the conclusion of Rolle’s Theorem.

Answer 1

Rolle's Theorem is simply the MVT for a function on an interval [a,b] such that f(b) = f(a) (ie the 2 y-values are the same). When such is the case, the MVT guarantees at least one c with a < c < b such that f^'(c) = 0 (in other words, the function must have a turning point, local min or max, between 2 = y-values).

f(x) = (x-1)²(x-3)³ so f(1) = f(3) = 0. f(x) a polynomial, so differentiable on reals and MVT applies.

f'(x) = 2(x-1)(x-3)³ +3(x-1)²(x-3)² = (x--1)(x-3)²(2x - 6 + 3x - 3) = (x--1)(x-3)²(5x-9)

f^'(9/5) = 0 and c = 9/5 is the only x-value between 1 and 3 satisfying the conclusion.