For the function \(f(x) = (x - 1)^{2}(x + 2)\), \(x \in R\):
(A) f has a local maximum at x = −1.
(B) The local minimum value of f is 0.
(C) f has an absolute maximum value on R.
(D) The absolute maximum value of f on the interval [−2, 2] is 4.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (D) only
Differentiating by the product rule:
\( f'(x) = 2(x - 1)(x + 2) + (x - 1)^{2} \)
\( = (x - 1)(3x + 3) = 3(x - 1)(x + 1) \)
By the second derivative test, with \(f''(x) = 6x\):
\( f''(-1) = -6 \lt 0 \): local maximum, \( f(-1) = 4 \times 1 = 4 \)
\( f''(1) = 6 \gt 0 \): local minimum, \( f(1) = 0 \)
On [−2, 2], the end-point values are
\( f(-2) = 9 \times 0 = 0,\ f(2) = 1 \times 4 = 4 \)
So the local maximum is 4 at x = −1, the local minimum is 0 at x = 1, f grows without bound on R, and on [−2, 2] the largest value is 4.
Checking the statements
(A) f has a local maximum at x = −1. — Correct. f′(−1) = 0 and f″(−1) = −6 < 0.
(B) The local minimum value of f is 0. — Correct. f(1) = 0 × 3 = 0.
(C) f has an absolute maximum value on R. — Incorrect. f(10) = 81 × 12 = 972, and f keeps growing as x grows.
(D) The absolute maximum value of f on the interval [−2, 2] is 4. — Correct. The values at −2, −1, 1, 2 are 0, 4, 0, 4.