Given \( f(x) = 3x^2 - 2x + 1 \) and \( g(x) = x - 4 \), calculate \( g(f(3)) \).

["# Calculating ( g(f(3)) ) for ( f(x) = 3x^2 - 2x + 1 ) and ( g(x) = x - 4 )", "Understanding function composition is a fundamental skill in algebra and higher mathematics. Today, we’ll walk through a clear and step-by-step solution for calculating ( g(f(3)) ) using the given functions:\n[ f(x) = 3x^2 - 2x + 1 ]\nand\n[ g(x) = x - 4 ]", "## Step 1: Evaluate ( f(3) )", "Start by substituting ( x = 3 ) into the function ( f(x) ):", "[\nf(3) = 3(3)^2 - 2(3) + 1\n]", "Calculate each term:\n- ( 3^2 = 9 )\n- ( 3 \ imes 9 = 27 )\n- ( -2 \ imes 3 = -6 )\n- Then:\n[\nf(3) = 27 - 6 + 1 = 22\n]", "So, ( f(3) = 22 )", "## Step 2: Use the result to compute ( g(f(3)) = g(22) )", "Now that we know ( f(3) = 22 ), we substitute this value into ( g(x) ):", "[\ng(22) = 22 - 4 = 18\n]", "Thus,\n[\ng(f(3)) = 18\n]", "## Summary", "By evaluating ( f(3) = 22 ) first, then applying ( g ) to that result:\n[\ng(f(3)) = g(22) = 18\n]", "This process demonstrates function composition — evaluating an inner function first, then using its output as input for the outer function.", "## Why This Matters", "Function composition is essential in fields like physics, engineering, and computer science, especially when modeling complex systems where outputs of one function become inputs to another. Mastering such calculations strengthens algebraic intuition and problem-solving precision.", "If you're ready to explore more function composition examples, try with different quadratic and linear functions — the method remains the same!", "---", "Keywords:\n( g(f(3)) ), function composition, how to calculate ( g(f(x)) ), step-by-step solving, algebra example, ( f(x) = 3x^2 - 2x + 1 ), ( g(x) = x - 4 ), mathematical examples, solving functions, high school algebra, math tutorial"]