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.