<p>Given <em>f</em>(<em>x</em>) = <em>x</em>√(<em>kx</em> − <em>x</em>²). If <em>f</em>(<em>x</em>) is an increasing function on [0, 3] and <em>M</em> is the maximum value of <em>f</em>(<em>x</em>), find the minimum value of <em>k</em> (denoted <em>m</em>) such that <em>f</em> is increasing on [0,3], and find <em>M</em>. What is the value of <em>M</em> (approximately)?</p>
Step-by-Step Solution
Key Concept: For f(x) to be increasing on [0,3], we need f'(x) ≥ 0 throughout the interval. The critical constraint occurs at the endpoint where the derivative condition becomes tightest, which determines the minimum value of k.
<p><strong>Step 1: Find the domain constraint.</strong> For f(x) = x√(kx - x²) to be defined on [0,3], we need kx - x² ≥ 0, so x(k - x) ≥ 0. For x ∈ [0,3], this requires k ≥ 3.</p><p><strong>Step 2: Compute f'(x).</strong> Using the product rule: f'(x) = √(kx - x²) + x · (k - 2x)/(2√(kx - x²)) = (kx - x² + (kx - 2x²)/2)/√(kx - x²) = (2kx - 2x² + kx - 2x²)/(2√(kx - x²)) = (3kx - 4x²)/(2√(kx - x²)) = x(3k - 4x)/(2√(kx - x²))</p><p><strong>Step 3: Apply increasing condition.</strong> For f to be increasing on [0,3], we need f'(x) ≥ 0. Since x ≥ 0 and √(kx - x²) > 0 for x ∈ (0,3), we need 3k - 4x ≥ 0, i.e., x ≤ 3k/4 for all x ∈ [0,3]. This requires 3 ≤ 3k/4, so k ≥ 4. Thus m = 4.</p><p><strong>Step 4: Find maximum value M.</strong> When k = 4, f(x) = x√(4x - x²) = x√(x(4-x)). At x = 3: M = f(3) = 3√(12 - 9) = 3√3 ≈ 5.196</p><p>∴ Answer: 5.196</p>
Correct Answer: 5.196