Applications of Derivatives
Mean Value Theorem
Grade 12
Question:
<p>We have \(f(1) = -2\), \(f'(x) \geq 2\) for all \(x \in [1, 6]\). By LMVT, there exists \(c \in (1, 6)\) such that \(f'(c) = \frac{f(6) - f(1)}{5}\). Since \(f'(x) \geq 2\) for all \(x \in [1, 6]\), what is the minimum value of \(f(6)\)?</p>
<p>\(f(6) \geq 6\)</p>
<p>\(f(6) \geq 8\)</p>
<p>\(f(6) \geq 10\)</p>
<p>\(f(6) \geq 4\)</p>
Step-by-Step Solution
Key Concept: Apply LMVT to establish that f'(c) = [f(6) - f(1)]/5 for some c ∈ (1,6), then use the constraint f'(x) ≥ 2 everywhere to find that f'(c) ≥ 2, which directly bounds f(6) from below.
<p><strong>Step 1:</strong> By LMVT, there exists c ∈ (1,6) such that:</p><p>f'(c) = [f(6) - f(1)]/(6 - 1) = [f(6) - f(1)]/5</p><p><strong>Step 2:</strong> Since f'(x) ≥ 2 for all x ∈ [1,6], this condition holds at x = c as well:</p><p>f'(c) ≥ 2</p><p><strong>Step 3:</strong> Substitute the LMVT equation into this inequality:</p><p>[f(6) - f(1)]/5 ≥ 2</p><p><strong>Step 4:</strong> Substitute f(1) = -2:</p><p>[f(6) - (-2)]/5 ≥ 2</p><p>[f(6) + 2]/5 ≥ 2</p><p>f(6) + 2 ≥ 10</p><p>f(6) ≥ 8</p><p><strong>Step 5:</strong> The minimum value of f(6) is achieved when f'(x) = 2 throughout [1,6], giving f(6) = f(1) + 2(6-1) = -2 + 10 = 8.</p><p>∴ Answer: B (Minimum value of f(6) is 8)</p>
Correct Answer: B