Question

Find the maximum value of 2x ^3 − 24x + 107 in the interval [1, 3]. Find the maximum value of the same function in [−3, −1].

Answer 1

before getting maximum value of function , we want to see which interval function is increasing and which interval it is decreasing.
so, let's differentiate with respect to x,
f'(x) = 6x² - 24
now, f'(x) = 0 = 6x² - 24
x = ± 2

when x > 2 or x < -2 , f'(x) > 0
so, f(x) is increasing .
now, case 1 :- we have to find maximum value of f(x) in the interval [1, 3]
but we see in (2, 3] function is increasing.
hence, maximum value of f(x) = f(3)
= 2(3)³ - 24(3) + 107
= 54 - 72 + 107
= -18 + 107 = 89

and when -2 ≤ x ≤ 2 , f'(x) < 0
so, f(x) is decreasing .
case 2 :- we have to find maximum value of f(x) in the interval [-3, -1].
but we see in [-2, -1] , f(x) is decreasing.
so, maximum value of f(x) = f(-2)
= 2(-2)³ - 24(-2) + 107
= -16 + 48 + 107
= 32 + 107
= 139