Let f (x) = 2x³ – 24x + 107, x ∈ [1, 3]

⇒ f’(x) = 6x² – 24

=6(x² - 4)

Now, f’(x) = 0

⇒ 6(x² - 4) = 0

⇒ x² = 4

⇒ x = �2

Therefore, we will only consider the interval [1,3]

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

f(2) = 2(2)³ – 24(2) + 107

= 2(8) – 24(2) + 107

= 75

f(1) = 2(1)³ – 24(1)+ 107

= 2 – 24 +107

= 85

f(3) = 2(3)³ – 24(3) + 107

= 2(27) -24(3) +107

= 89

Therefore, we have the absolute maximum value of f on [1,3] is 89 occurring at x =3.

Now, we will only consider the interval [-3, -1]

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

f(-3) = 2(-3)³ – 24(-3) + 107

= 2(-27) – 24(-3) + 107

= 125

f(-1) = 2(-1)³ – 24(-1)+ 107

= -2 + 24 +107

= 129

f(-2) = 2(-2)³ – 24(-2) + 107

= 2(-8) -24(-2) +107

= 139

Therefore, we have the absolute maximum value of f on [-3, -1] is 89 occurring at x = -2.