Find the absolute maximum and minimum values of a function f is given by
f (x) = 2x³ – 15x² + 36x + 1 on the interval of [1, 5].

f (x) = 2x³ – 15x² + 36x + 1
∴   f ' (x) = 6x² – 30x + 36 = 6 (x² – 5x + 6) = 6 (x – 2) (x – 3)
f ' (x) = 0 ⇒ 6 (x – 2) (x – 3) = 0 ⇒ x = 2, 3 ∊ [1,5]
Now f (1) = 2 – 15 + 36 + 1 = 24
f (2) = 2 × 8 – 15 x 4 + 36 x 2 + 1 = 16 – 60 + 72 + 1 = 29
f (3) = 2 × 27 – 15 x 9 + 36x3 + 1 = 54 – 135 + 108 + 1 = 28
f (5) = 2 × 125–15 × 25 + 36 x 5 + 1 = 250–375+ 180+ 1 = 56
∴  absolute maximum value = 56 at x = 5
and absolute minimum value = 24 at x = 1.