Answer: Yes, f is differentiable but f'' may not exist at 0 in the naive cubic sense, yet checking shows f''(0)=0 exists.
- A No, f is not even continuous at the origin because the cube of |x| is mistakenly assumed to blow up there
- B No, f is not differentiable at 0 because of the sharp absolute-value kink visible in the graph near zero
- C Yes, f is differentiable but f'' may not exist at 0 in the naive cubic sense, yet checking shows f''(0)=0 exists
- D f is not defined at x=0 since the cube of an absolute value is wrongly treated as an undefined operation
Correct answer: C. Yes, f is differentiable but f'' may not exist at 0 in the naive cubic sense, yet checking shows f''(0)=0 exists
Explanation: f(x) = x<sup>3</sup> for x>=0 and -x<sup>3</sup> for x<0, so f'(x) = 3x<sup>2</sup> for x>=0 and -3x<sup>2</sup> for x<0; f'(0)=0 from both sides, and f''(x) = 6|x|, giving f''(0) = 0, so f is twice differentiable at 0.
f(x)=|x| is perfectly continuous at x=0 (no break, no jump), but the curve has a sharp corner there: approaching from the left gives slope -1 and from the right gives slope +1 - since these one-sided derivatives disagree, the function is not differentiable at that point.
Concept context
When a function has no breaks (continuity) and when it has a well-defined slope (differentiability), plus rules for differentiating implicit, inverse trig, exponential, log, and parametric functions.