Yefim S. answered 05/28/21
Math Tutor with Experience
f'(x) = 3x2 - 3 = 0; x2 = 1; x = - 1 or x = 1 are 2 critical numbers;
f'(x) = 3x2 - 3 > 0; x < - 1 or x > 1 f(x) increasing;
f'(x) = 3x2 - 3 < 0; - 1 < x < 1 f(x) decreasing
f''(x) = 6x; f''(- 1) = - 6 < 0 so, we have at x = - 1 local max = f(- 1) = - 1 + 3 + 2 = 4; (-1, 4) is local max
f''(1) = 6 > 0, so at x = 1 f(x) has min f(1) = 1 - 3 + 2 = 0; (1, 0) is local min