Application of Derivatives Part 1

S = {x ∊ R, x2 + 30 – 11x ≤ 0}
= {x ∊ R, 5 ≤ x ≤ 6}
Now ƒ(x) = 3x3–18x2 + 27x – 40
⇒ ƒ'(x) = 9(x – 1)(x – 3),
which is positive in [5, 6]
⇒ ƒ(x) increasing in [5, 6]
Hence maximum value = ƒ(6) = 122