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Step 1: Find the derivative of the function.
Given \(f(x) = x^3 - 3x^2 + 12x - 18\), we find its derivative \(f'(x)\).
\(f'(x) = 3x^2 - 6x + 12\)
Step 2: Analyze the derivative.
We want to determine if \(f'(x)\) is always positive, always negative, or changes sign.
\(f'(x) = 3(x^2 - 2x + 4)\)
Complete the square: \(f'(x) = 3((x - 1)^2 + 3)\)
Step 3: Determine the sign of the derivative.
Since \((x - 1)^2\) is always non-negative, \((x - 1)^2 + 3\) is always greater than or equal to 3. Therefore, \(3((x - 1)^2 + 3)\) is always positive.
Thus, \(f'(x) > 0\) for all \(x \in \mathbb{R}\).
Step 4: Conclude the nature of the function.
Since \(f'(x) > 0\) for all \(x\), the function \(f(x)\) is strictly increasing on \(\mathbb{R}\).
Correct Answer: strictly increasing on R