Limits, Continuity & Differentiability
Rolle's Theorem
Grade 12
Question:
<p>Let \(f\) be continuous on \([a, b]\), differentiable on \((a, b)\), and \(f(a)/a = f(b)/b\). Then by Rolle's theorem applied to \(g(x) = f(x)/x\), there exists \(x_0 \in (a, b)\) such that:</p>
<p>(a) \(x_0 f'(x_0) = f(x_0)\)</p>
<p>(b) \(f'(x_0) = 0\)</p>
<p>(c) \(x_0 f'(x_0) = 2f(x_0)\)</p>
<p>(d) \(f'(x_0) = f(x_0)\)</p>
Step-by-Step Solution
Key Concept: To apply Rolle's theorem to g(x) = f(x)/x, you must verify that g is continuous on [a,b] and differentiable on (a,b), then use the given condition g(a) = g(b) to guarantee g'(x₀) = 0 for some x₀ ∈ (a,b).
<p><strong>Step 1:</strong> Define g(x) = f(x)/x on [a,b] where a,b > 0.</p><p><strong>Step 2:</strong> Verify conditions for Rolle's theorem:</p><ul><li>g(x) is continuous on [a,b] (quotient of continuous functions, x ≠ 0)</li><li>g(x) is differentiable on (a,b) (quotient of differentiable functions, x ≠ 0)</li><li>Given: g(a) = f(a)/a = f(b)/b = g(b)</li></ul><p><strong>Step 3:</strong> By Rolle's theorem, ∃ x₀ ∈ (a,b) such that g'(x₀) = 0.</p><p><strong>Step 4:</strong> Compute: g'(x) = [xf'(x) - f(x)]/x² = 0</p><p>This gives: xf'(x) - f(x) = 0, or <strong>x₀f'(x₀) = f(x₀)</strong></p><p>∴ Answer: <strong>x₀f'(x₀) = f(x₀)</strong> (or equivalently, f'(x₀) = f(x₀)/x₀)</p>
Correct Answer: A