Let f(x) = 2x3 − 24x + 107.
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(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 at the 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.
