The function \(f(x) = x^{3} + \displaystyle\frac{1}{x^{3}}\), \(x \ne 0\), is strictly decreasing in
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \((-1, 0) \cup (0, 1)\)
Differentiating,
\( \displaystyle f'(x) = 3x^{2} - \frac{3}{x^{4}} = \frac{3\left(x^{6} - 1\right)}{x^{4}} \)
Since \(x^{4} \gt 0\) for x ≠ 0, the sign of f′(x) is the sign of \(x^{6} - 1\):
\(x^{6} - 1 \lt 0\) when \(|x| \lt 1\)
\(x^{6} - 1 \gt 0\) when \(|x| \gt 1\)
So f is strictly decreasing in (−1, 0) and in (0, 1), and strictly increasing in \((-\infty, -1)\) and in \((1, \infty)\).
Each of the other options contains \((-\infty, -1)\) or \((1, \infty)\), where f is increasing.