Question -
Answer -
Let f(x) = 2x3 −24x + 107.
We first consider the interval [1, 3].
Then, we evaluate the value of f atthe critical point x = 2 ∈ [1, 3] andat the end points of the interval [1, 3].
f(2) = 2(8)− 24(2) + 107 = 16 − 48 + 107 = 75
f(1) = 2(1)− 24(1) + 107 = 2 − 24 + 107 = 85
f(3) = 2(27)− 24(3) + 107 = 54 − 72 + 107 = 89
Hence, the absolute maximum value of f(x)in the interval [1, 3] is 89 occurring at x = 3.
Next, we consider the interval [−3, −1].
Evaluate the value of f atthe critical point x = −2 ∈ [−3, −1]and at the end points of the interval [1, 3].
f(−3) = 2(−27) − 24(−3) + 107 = −54 + 72 + 107 = 125
f(−1) =2(−1) − 24 (−1) + 107 = −2 + 24 + 107 = 129
f(−2) =2(−8) − 24 (−2) + 107 = −16 + 48 + 107 = 139
Hence, the absolute maximum value of f(x)in the interval [−3, −1] is 139 occurring at x = −2.