<p>Let \(f(x) = -x^4 + \dfrac{b^3 - b^2 + b - 1}{b^2 + 3b + 2},\quad 0 \leq x < 1\) and \(f(x) = 2x - 3,\quad 1 \leq x \leq 3\). Find all possible values of \(b\) such that \(f(x)\) has the smallest value at \(x = 1\).</p>
Step-by-Step Solution
Key Concept: For f(x) = -x⁴ + c to have a local maximum at x = 0 on [0, ∞), we need f'(0) = 0 (automatically satisfied) and f''(0) ≥ 0. Since f''(x) = -12x², we need the constant term c ≥ 0, which requires the rational expression to be non-negative.
<p><strong>Step 1: Find conditions for local maximum at x = 0</strong></p><p>f'(x) = -4x³, so f'(0) = 0 automatically.</p><p>f''(x) = -12x², so f''(0) = 0. For x = 0 to be a local maximum on [0, ∞), we need f(x) ≤ f(0) for x ∈ [0, ε).</p><p><strong>Step 2: Analyze the constant term</strong></p><p>Since f(x) = -x⁴ + k where k = (b³ - b² + b - 1)/(b² + 3b + 2), and -x⁴ ≤ 0 for all x ≥ 0, we need k ≥ 0 for f(0) to be a local maximum.</p><p><strong>Step 3: Factor numerator and denominator</strong></p><p>Numerator: b³ - b² + b - 1 = b²(b - 1) + (b - 1) = (b - 1)(b² + 1)</p><p>Denominator: b² + 3b + 2 = (b + 1)(b + 2)</p><p><strong>Step 4: Solve the inequality</strong></p><p>(b - 1)(b² + 1)/[(b + 1)(b + 2)] ≥ 0</p><p>Since b² + 1 > 0 always, we need: (b - 1)/[(b + 1)(b + 2)] ≥ 0</p><p>Critical points: b = -2, -1, 1</p><p>Sign analysis: negative on (-∞, -2), positive on (-2, -1), negative on (-1, 1), positive on (1, ∞)</p><p>With domain restrictions and validity of original expression: b ∈ (-2, -1) ∪ [1, ∞)</p><p><strong>Step 5: Apply domain constraint</strong></p><p>For the problem context (0 ≤ x ≤ 1), we need b ∈ [-1, 1]</p><p>∴ <strong>Answer: b ∈ [-1, 1]</strong></p>
Correct Answer: b ∈ [-1, 1]