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Answer: \(f\) is increasing for all \(x \in R\).

\(f'(x) = 3x^2 + 12x + 12 = 3(x^2 + 4x + 4) = 3(x + 2)^2\).

Since \((x + 2)^2 \ge 0\) for all real \(x\), \(f'(x) \ge 0\) for all \(x \in R\) (it is zero only at the single point \(x = -2\)). Hence \(f\) is an increasing function for all \(x \in R\).

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