The interval in which \(f(x) = \displaystyle\frac{x}{2} + \frac{2}{x}\) is a decreasing function of x is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((-2, 0) \cup (0, 2)\)
The function is defined for all x except x = 0. It is decreasing where \(f'(x) \lt 0\).
\( \displaystyle f'(x) = \frac{1}{2} - \frac{2}{x^2} \)
\( \displaystyle = \frac{x^2 - 4}{2x^2} \)
Since \(2x^2 \gt 0\) for x ≠ 0, the sign of \(f'(x)\) is the sign of \(x^2 - 4\):
\(f'(x) \lt 0 \iff x^2 \lt 4,\ x \ne 0\)
\(\iff -2 \lt x \lt 0 \text{ or } 0 \lt x \lt 2\)
So f is decreasing on each of the intervals (−2, 0) and (0, 2). The interval (−2, 2) cannot be the answer because f is not defined at x = 0.