Applications of Derivatives
Mean Value Theorem
Grade 12
Question:
<p>Using mean value theorem, if \(f(x) = \log x\) on \([1, 3]\), find the value of \(c\) such that \(f'(c) = \frac{f(3) - f(1)}{3 - 1}\). What is \(c\)?</p>
<p>\(c = \log_3 e\)</p>
<p>\(c = 2\log_3 e\)</p>
<p>\(c = \log_e 3\)</p>
<p>\(c = 2\log_e 3\)</p>
Step-by-Step Solution
Key Concept: Mean Value Theorem states that there exists c ∈ (1,3) where f'(c) equals the average rate of change. For f(x) = log x, we need f'(c) = (log 3 - log 1)/(3-1), then solve for c using f'(x) = 1/x.
<p><strong>Step 1:</strong> Apply the Mean Value Theorem condition.</p><p>We need: f'(c) = [f(3) - f(1)]/(3 - 1)</p><p><strong>Step 2:</strong> Calculate the right side.</p><p>f(3) = log 3, f(1) = log 1 = 0</p><p>Average rate of change = (log 3 - 0)/(3 - 1) = log 3/2</p><p><strong>Step 3:</strong> Find f'(c).</p><p>f(x) = log x ⟹ f'(x) = 1/x</p><p>So f'(c) = 1/c</p><p><strong>Step 4:</strong> Solve for c.</p><p>1/c = log 3/2</p><p>c = 2/log 3</p><p><strong>Step 5:</strong> Verify c ∈ (1, 3).</p><p>Since log 3 ≈ 1.099, we get c ≈ 1.82 ✓</p><p>∴ Answer: c = <strong>2/log 3</strong> or equivalently <strong>2/ln 3</strong> (depending on logarithm base used)</p>
Correct Answer: B