Find the absolute maximum and absolute minimum values of f on the given interval. f(x) = 4 + 81x − 3x^3, [0, 4]

Answer 1

Let find derivative: f'(x) = 81 - 9x² = 0, x² = 9, x = ± 3;

Only critical number 3 belong to given closed interval.

To get absolute maximum and minimum we evaluate our function at end points 0, 4 and at critical number 3.

f(0) = 4, f(4) = 136, f(3) =166.

Answer: absolute maximum is f (3) = 166 and absolute minimum is f(0) = 4 .

Answer 2

No absolute maximum or minimum exists. A local minimum is at x = -3 and a local maximum is at x = 3.