For the function \(f(x) = x^{2} + \displaystyle\frac{54}{x}\), \(x \ne 0\):
(A) f is strictly decreasing in (0, 3).
(B) The local minimum value of f is 27.
(C) f is strictly decreasing in \((-\infty, 0)\).
(D) f has a local maximum at some point x < 0.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
Differentiating:
\( \displaystyle f'(x) = 2x - \frac{54}{x^{2}} \)
\( \displaystyle = \frac{2(x^{3} - 27)}{x^{2}} \)
Since \(x^{2} \gt 0\), the sign of f′(x) is the sign of \(x^{3} - 27\), which is negative for every x < 3 and positive for x > 3:
\( f'(x) \lt 0 \) in \((-\infty, 0)\) and in \((0, 3)\)
\( f'(x) \gt 0 \) in \((3, \infty)\)
The only critical point is x = 3, where f′ changes from − to +:
\( f(3) = 9 + 18 = 27 \)
So f decreases in (−∞, 0) and in (0, 3), increases in (3, ∞), and its only local extremum is a local minimum 27 at x = 3.
Checking the statements
(A) f is strictly decreasing in (0, 3). — Correct. For 0 < x < 3, \(x^{3} \lt 27\), so f′(x) < 0.
(B) The local minimum value of f is 27. — Correct. f(3) = 9 + 54/3 = 27.
(C) f is strictly decreasing in \((-\infty, 0)\). — Correct. For x < 0, \(x^{3} - 27 \lt 0\), so f′(x) < 0.
(D) f has a local maximum at some point x < 0. — Incorrect. f′(x) < 0 for every x < 0, so there is no critical point there.