Let \(f(x) = x^{1/3}\) (the real cube root of x) for all real x. Consider the following statements:
(A) f is continuous at \(x = 0\).
(B) f is differentiable at \(x = 0\).
(C) \(f'(-1) = \displaystyle\frac{1}{3}\)
(D) \(f'(x) = \displaystyle\frac{1}{3x^{2/3}}\) for all \(x \ne 0\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (C) and (D) only
Since \(|x^{1/3}|\) becomes as small as we like when x is close to 0, \(\lim_{x \to 0} x^{1/3} = 0 = f(0)\).
At x = 0, by the definition of the derivative:
\( \displaystyle \frac{f(h) - f(0)}{h} = \frac{h^{1/3}}{h} = \frac{1}{h^{2/3}} \)
which grows without bound as h → 0, so f′(0) does not exist (the graph has a vertical tangent at the origin).
For x ≠ 0, by the power rule:
\( \displaystyle f'(x) = \frac{1}{3}x^{-2/3} = \frac{1}{3x^{2/3}} \)
\( \displaystyle f'(-1) = \frac{1}{3\left((-1)^{1/3}\right)^{2}} = \frac{1}{3(-1)^{2}} = \frac{1}{3} \)
So f is continuous at 0 but not differentiable there, f′(x) = 1/(3x2/3) for x ≠ 0, and f′(−1) = 1/3.
Checking the statements
(A) f is continuous at \(x = 0\). — Correct. \(x^{1/3} \to 0 = f(0)\).
(B) f is differentiable at \(x = 0\). — Incorrect. The quotient \(\frac{1}{h^{2/3}}\) has no finite limit.
(C) \(f'(-1) = \frac{1}{3}\) — Correct. \((-1)^{2/3} = 1\), so \(f'(-1) = \frac{1}{3}\).
(D) \(f'(x) = \frac{1}{3x^{2/3}}\) for all \(x \ne 0\). — Correct. This is the power rule with exponent \(\frac{1}{3}\).