Let \(f(x) = |x|^{3}\). Consider the following statements:
(A) f is differentiable at \(x = 0\).
(B) \(f'(x) = 3x|x|\) for all real x.
(C) \(f''(x) = 6x\) for all \(x \ne 0\).
(D) \(f''(0) = 0\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (B) and (D) only
Removing the modulus:
\(x \ge 0: \ f(x) = x^{3}\)
\(x \lt 0: \ f(x) = -x^{3}\)
At x = 0, by the definition of the derivative:
\( \displaystyle \frac{f(h) - f(0)}{h} = \frac{|h|^{3}}{h} = h|h| \to 0 \)
So f′(0) = 0. For x > 0, f′(x) = \(3x^{2}\); for x < 0, f′(x) = \(-3x^{2}\). Together:
\(f'(x) = 3x|x|\) for all x
Differentiating again: for x > 0, f″(x) = 6x; for x < 0, f″(x) = −6x. So \(f''(x) = 6|x|\) for x ≠ 0. At 0:
\( \displaystyle \frac{f'(h) - f'(0)}{h} = \frac{3h|h|}{h} = 3|h| \to 0 \)
So f′(0) = 0, f′(x) = 3x|x|, f″(x) = 6|x| for x ≠ 0, and f″(0) = 0.
Checking the statements
(A) f is differentiable at \(x = 0\). — Correct. The difference quotient h|h| tends to 0.
(B) \(f'(x) = 3x|x|\) for all real x. — Correct. It gives \(3x^{2}\) for x > 0, \(-3x^{2}\) for x < 0 and 0 at x = 0.
(C) \(f''(x) = 6x\) for all \(x \ne 0\). — Incorrect. For x < 0, f″(x) = −6x, so f″(x) = 6|x|; for example f″(−1) = 6, not −6.
(D) \(f''(0) = 0\) — Correct. The quotient 3|h| tends to 0.