Applications of Derivatives
Mean Value Theorem
Grade 12
Question:
<p>If \(f(x)\) is continuous and derivable on \([-2, 5]\) and \(-4 \leq f'(x) \leq 3\) \(\forall\, x \in (-2, 5)\), then difference of maximum and minimum value of \(f(5)\) is equal to:</p>
<p>(a) 7</p>
<p>(b) 21</p>
<p>(c) 28</p>
<p>(d) 49</p>
Step-by-Step Solution
Key Concept: Use the Mean Value Theorem constraint: since f'(x) is bounded, the maximum change in f occurs over the interval when f' takes its extreme values throughout [-2,5]. The difference between maximum and minimum value of f(5) is determined by the range of possible values f(5) can take given the derivative bounds and the fixed interval length of 7.
<p><strong>Step 1:</strong> By the Mean Value Theorem, for any x ∈ [-2, 5]:</p><p>f(5) - f(-2) = f'(c) · (5 - (-2)) = 7f'(c) for some c ∈ (-2, 5)</p><p><strong>Step 2:</strong> Since -4 ≤ f'(x) ≤ 3 for all x ∈ (-2, 5), we have:</p><p>7(-4) ≤ f(5) - f(-2) ≤ 7(3)</p><p>-28 ≤ f(5) - f(-2) ≤ 21</p><p><strong>Step 3:</strong> This means:</p><p>f(5)_min = f(-2) - 28</p><p>f(5)_max = f(-2) + 21</p><p><strong>Step 4:</strong> The difference between maximum and minimum value of f(5) is:</p><p>f(5)_max - f(5)_min = [f(-2) + 21] - [f(-2) - 28] = 21 - (-28) = 49</p><p>∴ Answer: D (49)</p>
Correct Answer: D