Section: AP Calc AB

19) Find the absolute minimum of f(x) = 3x⁴ – 4x³ on [-2,3]

1. f_min = 135
2. f_min = 80
3. f_min = 0
4. f_min = 1

f’(x) = 12x³ – 12x² ⇒ Critical numbers: 0 and 1. On the interval [-2,3], we have: f(-2) = 80, f(0) = 0, f(1) = -1 and f(3) = 135. Therefore, f_min = -1.