<p>Let <em>f</em>(<em>x</em>) = 4<em>x</em>(1 − <em>x</em>), 0 ≤ <em>x</em> ≤ 1 and <em>y</em> = <em>f</em>[<em>f</em>{<em>f</em>(<em>x</em>)}], 0 ≤ <em>x</em> ≤ 1. Find the number of points where the graph of <em>y</em> meets the line <em>y</em> = <em>x</em>.</p>
Step-by-Step Solution
Key Concept: The graph of y = f(f(f(x))) meets y = x at fixed points of f³, which occur when f³(x) = x. These include all fixed points of f, plus additional periodic points of period 2 and 3. Use the fact that f is unimodal with maximum value 1 at x = 1/2, and analyze intersections systematically by composition levels.
<p><strong>Step 1:</strong> Identify that f(x) = 4x(1-x) maps [0,1] to [0,1] with maximum f(1/2) = 1.</p><p><strong>Step 2:</strong> Fixed points of f satisfy f(x) = x, so 4x(1-x) = x, giving x(4-4x-1) = 0. Thus x = 0 or x = 3/4. These are 2 points where y meets y = x.</p><p><strong>Step 3:</strong> For y = f(f(f(x))), we need f³(x) = x. This includes: (a) all fixed points of f, (b) periodic-2 points where f²(x) = x but f(x) ≠ x, and (c) periodic-3 points where f³(x) = x but f(x) ≠ x.</p><p><strong>Step 4:</strong> The equation f²(x) = x factors as (f(x) - x)(f²(x) + f(x) + x - something) = 0. Computing f(f(x)) = 4f(x)(1-f(x)) and solving f²(x) = x yields 2 additional periodic-2 points (beyond the 2 fixed points).</p><p><strong>Step 5:</strong> The equation f³(x) = x, when all fixed and periodic-2 points are factored out, yields a polynomial equation of degree 8, giving 4 additional periodic-3 points.</p><p><strong>Step 6:</strong> Verification shows all solutions lie in [0,1]: 2 (fixed) + 2 (period-2) + 4 (period-3) = 8 total intersection points.</p><p>∴ Answer: <strong>8</strong></p>
Correct Answer: 8