Mark M. answered 10/15/23
Retired math prof. Very extensive Precalculus tutoring experience.
f(x) = (x-6)(x-7)(x-9)
f is differentiable on (6,9) and is continuous on [6,9] and f(6) = f(9) = 0. So, Rolle's Theorem is applicable.
f'(x) = (x-7)(x-9) + (x-6)(x-9) + (x-6)(x-7) = x2 - 16x + 63 + x2 - 15x + 54 + x2 - 13x + 42
f'(c) = 3c2 - 44c + 159 = 0
c = [44 ± √28] / 6 = (22 + √7) / 3 or (22 - √7) / 3 ≈ 8.22, 6.45