Final answer:
To find the absolute maximum and absolute minimum values of f(x) = 4 + 81x − 3x³ on the interval [0, 4], we need to check the critical points and endpoints of the interval. The absolute maximum value is 166 which occurs at x = 3, and the absolute minimum value is 4 which occurs at x = 0.
Step-by-step explanation:
To find the absolute maximum and absolute minimum values of f(x) = 4 + 81x − 3x³ on the interval [0, 4], we need to check the critical points and endpoints of the interval.
Step 1: Find the critical points by taking the derivative of f(x) and setting it equal to zero:
f'(x) = 81 - 9x² = 0
9x² = 81
x² = 9
x = ±3
Step 2: Evaluate f(x) at the critical points and endpoints:
f(0) = 4
f(3) = 4 + 81(3) - 3(3)³ = 4 + 243 - 81 = 166
f(4) = 4 + 81(4) - 3(4)³ = 4 + 324 - 192 = 136
Step 3: Compare the values obtained. The absolute maximum value is 166 which occurs at x = 3, and the absolute minimum value is 4 which occurs at x = 0.