Solution: We evaluate $ I(x) = \frac{1}{x^2 + 4x + 5} $ at $ x = 1, 2, 3, 4 $:
["Evaluating the Function ( I(x) = \frac{1}{x^2 + 4x + 5} ) at Integer Points: A Detailed Analysis", "Understanding how functions behave at specific values is fundamental in mathematics, science, and engineering. This article evaluates the rational function ( I(x) = \frac{1}{x^2 + 4x + 5} ) at key integer points: ( x = 1, 2, 3, 4 ), and explores its properties, applications, and educational value.", "---", "### Overview of ( I(x) )", "The function ( I(x) = \frac{1}{x^2 + 4x + 5} ) is a rational function where the numerator is constant and the denominator is a quadratic expression. Completing the square on the denominator reveals important features like its minimum value and domain.", "Step 1: Complete the Square", "[\nx^2 + 4x + 5 = (x^2 + 4x + 4) + 1 = (x + 2)^2 + 1\n]", "This shows the denominator is always positive (since ( (x + 2)^2 \geq 0 ), so the smallest value is 1), guaranteeing ( I(x) > 0 ) for all real ( x ).", "---", "### Evaluating ( I(x) ) at ( x = 1, 2, 3, 4 )", "We now compute the function values step by step.", "#### At ( x = 1 )", "[\nI(1) = \frac{1}{(1 + 2)^2 + 1} = \frac{1}{3^2 + 1} = \frac{1}{9 + 1} = \frac{1}{10} = 0.1\n]", "#### At ( x = 2 )", "[\nI(2) = \frac{1}{(2 + 2)^2 + 1} = \frac{1}{4^2 + 1} = \frac{1}{16 + 1} = \frac{1}{17} \approx 0.0588\n]", "#### At ( x = 3 )", "[\nI(3) = \frac{1}{(3 + 2)^2 + 1} = \frac{1}{5^2 + 1} = \frac{1}{25 + 1} = \frac{1}{26} \approx 0.0385\n]", "#### At ( x = 4 )", "[\nI(4) = \frac{1}{(4 + 2)^2 + 1} = \frac{1}{6^2 + 1} = \frac{1}{36 + 1} = \frac{1}{37} \approx 0.0270\n]", "---", "### Results Summary", "| ( x ) | ( I(x) = \frac{1}{x^2 + 4x + 5} ) | Approximate Value |\n|--------|-------------------------------------|-------------------|\n| 1 | ( \frac{1}{10} ) | 0.1 |\n| 2 | ( \frac{1}{17} ) | 0.0588 |\n| 3 | ( \frac{1}{26} ) | 0.0385 |\n| 4 | ( \frac{1}{37} ) | 0.0270 |", "The values show a decreasing trend as ( x ) increases, consistent with the function's properties: since the denominator increases quadratically, ( I(x) ) decreases as ( x ) grows.", "---", "### Why This Evaluation Matters", "1. Understanding Function Behavior:\n Evaluating rational functions at discrete points helps visualize how rapidly they decay and understand their asymptotic behavior (as ( x \ o \infty ), ( I(x) \ o 0 )).", "2. Applications in Physics and Engineering:\n Functions like ( I(x) ) often model phenomena such as signal decay, electric field strengths, or concentration gradients diminishing with distance.", "3. Educational Value:\n This exercise builds skills in algebraic manipulation (completing the square), function evaluation, and data interpretation — essential for advanced math and STEM disciplines.", "---", "### Conclusion", "Evaluating ( I(x) = \frac{1}{x^2 + 4x + 5} ) at ( x = 1, 2, 3, 4 ) reveals clear, predictable behavior suitable for classroom study and applied modeling. The results highlight how quadratic denominators control function growth and decay, offering both mathematical insight and practical utility.", "For students and professionals alike, mastering such evaluations strengthens analytical thinking and supports deeper engagement with continuous functions.", "---", "Keywords:\n( I(x) = \frac{1}{x^2 + 4x + 5} ), evaluate at ( x = 1, 2, 3, 4 ), rational function, quadratic denominator, function analysis, math education, decay behavior.", "---", "Further Reading:\n- Analyzing rational functions and their graphs\n- Applications of completing the square in calculus\n- Discrete evaluation techniques in numerical analysis"]
