Answer :

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.


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