Question

Find the absolute maximum and absolute minimum of the function \( f(x) = 2x^3 - 15x^2 + 36x + 1 \) on \( [1, 5] \).

Hint:

To find the absolute maximum and minimum of a function on a closed interval, first find the critical points by setting the first derivative equal to zero. Then evaluate the function at the critical points and the endpoints of the interval.

Answer 1

The given function is: \[ f(x) = 2x^3 - 15x^2 + 36x + 1. \] First, find the first derivative of \( f(x) \): \[ f'(x) = 6x^2 - 30x + 36. \] Set \( f'(x) = 0 \) to find the critical points: \[ 6x^2 - 30x + 36 = 0. \] \[ x^2 - 5x + 6 = 0. \] \[ (x - 2)(x - 3) = 0. \] Thus, \( x = 2 \) and \( x = 3 \) are critical points. Next, evaluate \( f(x) \) at the critical points and the endpoints of the interval \( [1, 5] \): \[ f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24, \] \[ f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29, \] \[ f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 10 + 1 = 2, \] \[ f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 250 - 375 + 10 + 1 = 56. \] The absolute maximum value is \( f(5) = 56 \) and the absolute minimum value is \( f(1) = 24 \).