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

Given,

f (x) = 2x³ – 15x² + 36x + 1

f’(x) = 6 x² – 30 x + 36

f’(x) = 6(x² – 5 x + 6) = 6 (x – 2)(x – 3)

Note that f’(x) = 0 gives x = 2 and x = 3

We shall now evaluate the value of f at these points and at the end points of the interval [1, 5],

i.e. at x = 1, 2, 3 and 5

At x = 1, f(1) = 2(1³) – 15(1)² + 36(1) + 1 = 24

At x = 2, f(2) = 2(2³) – 15(2)² + 36(2) + 1 = 29

At x = 3, f(3) = 2(3)³– 15(3)²+ 36(3) + 1 = 28

At x = 5, f(5) = 2(5)³– 15(5)²+ 36(5) + 1 = 56

Thus, we conclude that the absolute maximum value of f on [1, 5] is 56, occurring at x = 5, and absolute value of f on [1, 5] is 24 which occurs at x = 1.