If f(x) = cases x3 ( 1x ), & x 0 \ 0, & x = 0 cases, then
- f''(0) = 0
- f''(0) = 1
- f'' ( 2 ) = 24 - 22
- f'' ( 2 ) = 12 - 22
Answer
(3) f'' ( 2 ) = 24 - 22
Solution
x 0: f'(x) = 3x2 1x - x 1x; f'(0) = h 0 h2 1h = 0
f''(0) = h 0 f'(h) - 0h = h 0 (3h 1h - 1h ) does not exist
1h oscillates (A), (B) are false
x 0: f''(x) = (6x - 1x ) 1x - 4 1x
At x = 2 : 2 = 1, 2 = 0
f'' ( 2 ) = 12 - 2 = 24 - 22 (C) true, (D) false
Answer: (3)