For the function \(f(x) = x^{4} - 4x^{3}\), \(x \in R\):
(A) x = 0 is a point of local maximum.
(B) x = 3 is a point of local minimum.
(C) The local minimum value of f is −27.
(D) \(f''(0) = 0\), so the second derivative test gives no decision at x = 0.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (B), (C) and (D) only
The critical points are where \(f'(x) = 0\):
\( f'(x) = 4x^{3} - 12x^{2} = 4x^{2}(x - 3) \)
So x = 0 and x = 3. The factor \(4x^{2}\) is never negative, so the sign of f′ is the sign of x − 3:
f′ < 0 on both sides of 0: no extremum at x = 0
f′ changes from − to + at 3: local minimum at x = 3
\( f(3) = 81 - 108 = -27 \)
The second derivative:
\( f''(x) = 12x^{2} - 24x \), so \( f''(0) = 0 \)
So x = 0 is neither a maximum nor a minimum, x = 3 is the local minimum with value −27, and f″(0) = 0.
Checking the statements
(A) x = 0 is a point of local maximum. — Incorrect. f′ is negative on both sides of 0, so f keeps decreasing there.
(B) x = 3 is a point of local minimum. — Correct. f′ changes from − to + at 3.
(C) The local minimum value of f is −27. — Correct. f(3) = 81 − 108 = −27.
(D) \(f''(0) = 0\), so the second derivative test gives no decision at x = 0. — Correct. f″(0) = 0 − 0 = 0, and the test needs f″ ≠ 0.