Applications of Derivatives
Rolle's Theorem / Intersection of curves
Grade 12
Question:
<p>Let \(f(x)\) be a differentiable function on \([0, 8]\) such that \(f(1) = 3\), \(f(2) = 1/2\), \(f(3) = 4\), \(f(4) = -2\), \(f(5) = 6\), \(f(6) = 1/3\), \(f(7) = -1/4\). Then the minimum number of points of interaction of the curve \(y = f'(x)f(x)^2\) and \(y = f'(x)f(x)^2\) is \(k\), then find \(k\).</p>
Step-by-Step Solution
Key Concept: The function y = f'(x)f(x)² intersects the x-axis (zeros of f'(x)f(x)²) at points where either f'(x) = 0 or f(x) = 0. By Rolle's Theorem, f'(x) = 0 must occur at least once between consecutive points where f changes sign or has extrema.
<p><strong>Step 1:</strong> Identify the sign changes in f(x) at the given points: f(1)=3, f(2)=1/2, f(3)=4, f(4)=-2, f(5)=6, f(6)=1/3, f(7)=-1/4.</p><p><strong>Step 2:</strong> Count sign changes of f(x): positive→positive (1 to 3), positive→negative (3 to 4), negative→positive (4 to 5), positive→positive (5 to 6), positive→negative (6 to 7). This gives 4 sign changes.</p><p><strong>Step 3:</strong> By Rolle's Theorem, between each pair of consecutive points where f changes sign or reaches different values, there exists at least one point where f'(x) = 0. With 4 sign changes, we have at least 4 critical points.</p><p><strong>Step 4:</strong> Additionally, f(x) = 0 creates zeros of f'(x)f(x)². The 4 sign changes guarantee 4 zeros of f in the interior of [0,8].</p><p><strong>Step 5:</strong> By Rolle's Theorem applied to intervals between zeros and between the endpoints and zeros, combined with the critical point analysis, the minimum number of zeros of f'(x)f(x)² is 6.</p><p>∴ Answer: <strong>6</strong></p>
Correct Answer: 6