["# Evaluate at ( x = 2 ): ( f'(2) = 9(2)^2 - 10(2) + 2 = 18 )", "When analyzing functions in calculus, finding the derivative at a specific point helps determine the slope of the tangent line and the instantaneous rate of change. In this article, we evaluate ( f'(2) ) using the expression ( f'(x) = 9x^2 - 10x + 2 ).", "## Step-by-Step Evaluation of ( f'(2) )", "Start with the given derivative:", "[
\nf'(x) = 9x^2 - 10x + 2
\n]", "Substitute ( x = 2 ):", "[
\nf'(2) = 9(2)^2 - 10(2) + 2
\n]", "Calculate each term:", "- ( 9(2)^2 = 9 \ imes 4 = 36 )
\n- ( -10(2) = -20 )
\n- Constant term: ( +2 )", "Add these together:", "[
\nf'(2) = 36 - 20 + 2 = 18
\n]", "## Why This Matters", "Evaluating the derivative at ( x = 2 ) provides key insight into the behavior of the function at that point. A derivative value of ( 18 ) indicates a strong upward slope, reflecting a steep positive rate of change. This calculation is essential in optimization, motion analysis, and change rate modeling across engineering, economics, and physics.", "## Conclusion", "Evaluating ( f'(2) ) for ( f'(x) = 9x^2 - 10x + 2 ) steps simply involves plugging in ( x = 2 ) and computing the expression:", "[
\nf'(2) = 9(2)^2 - 10(2) + 2 = 36 - 20 + 2 = 18
\n]", "This result confirms that the function’s slope at ( x = 2 ) is 18, a critical value for understanding its rate of change and tangent line behavior.", "---", "Keywords: evaluate f’(2), derivative at x = 2, f’(x) = 9x² - 10x + 2, find f’(2), calculus derivative evaluation, instantaneous rate of change, slope of tangent line, function analysis, math problem solution."]
