591. Verify Roll's Theorem for the function :f(x) = x (x - 3)² in 0 ≤ x ≤ 3

f(x) = x (x - 3)² = x (x² - 6x + 9) = x³ - 6 x² + 9 x.
It is a polynomial in x.
(a)    Since every polynomial in x is a continuous function for every value of x.
∴ f(x) is continuous in [0, 3].
(b)    f'(x) = 3 x² - 12 x + 9, which exists in (0, 3)
∴ f(x) is derivable in (0, 3).
(c)    f(0) = 0, f(3) = 0
∴ f(0) = f(3)
∴ f(x) satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that f'(c) = 0 where 0 < c < 3.
f'(c) = 0 gives us 3 c² - 12 c + 9 = 0 or c² - 4 c + 3 = 0, i.e., (c - 1) (c - 3) = 0 i.e.. c = 1, 3
Now 3 ∉ (0, 3), but 1 ∈ (0, 3)
∴ Rolle's Theorem is verified.