<p>Let \(f\) be differentiable for all \(x\). If \(f(1) = -2\) and \(f'(x) \geq 2\) for \(x \in [1, 6]\), then</p>
Step-by-Step Solution
Key Concept: Use the Mean Value Theorem: if f'(x) ≥ 2 on [1,6], then the average rate of change f(6)-f(1))/(6-1) must be at least 2, giving a lower bound for f(6).
<p><strong>Step 1:</strong> Given: f(1) = -2 and f'(x) ≥ 2 for all x ∈ [1, 6]</p><p><strong>Step 2:</strong> By the Mean Value Theorem, there exists c ∈ (1, 6) such that:</p><p>f'(c) = [f(6) - f(1)]/(6 - 1) = [f(6) - (-2)]/5 = [f(6) + 2]/5</p><p><strong>Step 3:</strong> Since f'(c) ≥ 2, we have:</p><p>[f(6) + 2]/5 ≥ 2</p><p><strong>Step 4:</strong> Multiply both sides by 5:</p><p>f(6) + 2 ≥ 10</p><p><strong>Step 5:</strong> Solve for f(6):</p><p>f(6) ≥ 8</p><p>∴ The minimum value of f(6) is 8 (Answer: A)</p>
Correct Answer: A