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Answer: f is increasing (for all \(x \ne 0\)).

\(f'(x) = 1 + \frac{1}{x^2}\).

For every \(x \ne 0\), \(x^2 \gt 0\), so \(\frac{1}{x^2} \gt 0\) and \(f'(x) = 1 + \frac{1}{x^2} \gt 0\). Hence \(f\) is an increasing function (on each of the intervals \(x \lt 0\) and \(x \gt 0\)).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report
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