let f(x) = 2x³ – 24 x + 107

∴ f'(x) = 6 x² – 24 = 6(x² – 4)

Now,

f'(x) = 0

⇒ 6(x² – 4) = 0

⇒ x² = 4

⇒ x = ±2

We first consider the interval [1, 3].

Then, we evaluate the value of f at the critical point x = 2 [1, 3] and at the end points of the interval [1, 3].

f(2) = 2 (2³)– 24 (2) + 107 = 75

f(1) = 2(1)³ – 24(1) + 107 = 85

f(3) = 2(3)³– 24 (3) + 107 = 89

Hence, the absolute maximum value of f(x) in the interval [1, 3] is 89 occurring at x = 3,

Next, we consider te interval [ – 3, – 1].

Evaluate the value of f at the critical point x = – 2 [1, 3]