For the function \(f(x) = x^{3} - 3x^{2}\) on the interval [−1, 4], match List-I with List-II.
| List-I | List-II | ||
| (A) | Point of local maximum | (I) | 2 |
| (B) | Point of local minimum | (II) | −4 |
| (C) | Absolute maximum value on [−1, 4] | (III) | 0 |
| (D) | Absolute minimum value on [−1, 4] | (IV) | 16 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
The critical points are where \(f'(x) = 0\):
\( f'(x) = 3x^{2} - 6x = 3x(x - 2) \)
So x = 0 and x = 2, with \(f''(x) = 6x - 6\). The absolute values come from comparing f at the critical points and the end points:
\( f(-1) = -1 - 3 = -4,\ f(0) = 0 \)
\( f(2) = 8 - 12 = -4,\ f(4) = 64 - 48 = 16 \)
(A) Point of local maximum
This matches (III) 0.
\( f''(0) = -6 \lt 0 \), so x = 0 is the point of local maximum
(B) Point of local minimum
This matches (I) 2.
\( f''(2) = 6 \gt 0 \), so x = 2 is the point of local minimum
(C) Absolute maximum value on [−1, 4]
This matches (IV) 16.
The largest of −4, 0, −4, 16 is f(4) = 16
(D) Absolute minimum value on [−1, 4]
This matches (II) −4.
The smallest is −4, reached at both x = −1 and x = 2