For \(x \in \mathrm{R} - \{0\}\), the function \(f(x) =\displaystyle\frac{3}{x}+ 7\) is decreasing when
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(x \in \mathrm{R}-\{0\}\)
The sign of the derivative decides where f is decreasing:
\(\displaystyle f'(x) = -\frac{3}{x^2}\)
Since \(x^2 \gt 0\) for every \(x \ne 0\), \(f'(x) \lt 0\) at every point of the domain \(\mathrm{R} - \{0\}\). So f is decreasing for all \(x \in \mathrm{R} - \{0\}\), that is, on each of the intervals \((-\infty, 0)\) and \((0, \infty)\).
The domain has a gap at 0, so f decreases on each interval separately: a value on the left of 0 need not be larger than one on the right, for example f(−1) = 4 and f(1) = 10.
Why the other options are incorrect
Option (1) \(x \in \mathrm{R}\) — Incorrect. It includes x = 0, where f is not defined.
Option (3) \(x = \{0\}\) — Incorrect. f is not defined at 0 at all.
Option (4) \(x \in \emptyset\) ; \(\emptyset\) is empty set — Incorrect. \(f'(x) \lt 0\) at every point of the domain, so the set where f decreases is not empty.