The function \(f(x) =\displaystyle\frac{x}{2}+\frac{2}{x}, x \ne 0\) is increasing on
(A) \((-\infty, -2)\)
(B) \((-2, 2)\)
(C) \((2, \infty)\)
(D) \((-1, 1)\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (C) only
f is increasing where \(f'(x) \gt 0\):
\( \displaystyle f'(x) = \frac{1}{2} - \frac{2}{x^2} = \frac{x^2 - 4}{2x^2} \)
Since \(2x^2 \gt 0\) for \(x \ne 0\), the sign of f'(x) is the sign of \(x^2 - 4\):
\(f'(x) \gt 0\) when \(x^2 \gt 4\), that is \(x \lt -2\) or \(x \gt 2\)
\(f'(x) \lt 0\) when \(-2 \lt x \lt 2,\ x \ne 0\)
So f is increasing on \((-\infty, -2)\) and \((2, \infty)\), and decreasing on \((-2, 0)\) and \((0, 2)\).
Checking the statements
(A) \((-\infty, -2)\) — Correct. Here \(x^2 \gt 4\), so \(f'(x) \gt 0\).
(B) \((-2, 2)\) — Incorrect. Here \(f'(x) \lt 0\) (and f is not defined at 0), so f is decreasing on each side of 0.
(C) \((2, \infty)\) — Correct. Here \(x^2 \gt 4\), so \(f'(x) \gt 0\).
(D) \((-1, 1)\) — Incorrect. This lies inside \((-2, 2)\), where \(f'(x) \lt 0\).