For the function \(f(x) = 2x^3 - 9x^2 + 12x + 1\), \(x \in R\):
(A) f is strictly decreasing in the interval (1, 2)
(B) \(x = 1\) is a point of local minimum
(C) the local maximum value of f is 6
(D) f is strictly increasing in the interval \((2, \infty)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A), (C) and (D) only
Differentiating:
\(f'(x) = 6x^2 - 18x + 12\)
\(f'(x) = 6(x^2 - 3x + 2)\)
\(f'(x) = 6(x - 1)(x - 2)\)
The sign of f’(x):
\(x \lt 1\): \(f'(x) \gt 0\)
\(1 \lt x \lt 2\): \(f'(x) \lt 0\)
\(x \gt 2\): \(f'(x) \gt 0\)
So f’ changes from + to − at x = 1 (local maximum) and from − to + at x = 2 (local minimum).
\(f(1) = 2 - 9 + 12 + 1 = 6\)
\(f(2) = 16 - 36 + 24 + 1 = 5\)
So f decreases on (1, 2), increases on \((-\infty, 1)\) and \((2, \infty)\), has local maximum value f(1) = 6 and local minimum value f(2) = 5.
Checking the statements
(A) f is strictly decreasing in the interval (1, 2) — Correct. \(f'(x) = 6(x - 1)(x - 2) \lt 0\) there.
(B) \(x = 1\) is a point of local minimum — Incorrect. f’ changes from + to − at x = 1, so it is a local maximum.
(C) the local maximum value of f is 6 — Correct. f(1) = 2 − 9 + 12 + 1 = 6.
(D) f is strictly increasing in the interval \((2, \infty)\) — Correct. Both factors x − 1 and x − 2 are positive there.