<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>
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 length by using f(5) - f(-2) = f'(c)·(5-(-2)) for some c ∈ (-2,5). The extreme values of f(5) are found by considering the bounds on f'(x) applied over the full interval of length 7.
<p><strong>Step 1:</strong> By the Mean Value Theorem, for any x ∈ (-2, 5), there exists c ∈ (-2, x) such that:</p><p>f(x) - f(-2) = f'(c)(x - (-2)) = f'(c)(x + 2)</p><p><strong>Step 2:</strong> For x = 5, we have:</p><p>f(5) - f(-2) = f'(c)·7 for some c ∈ (-2, 5)</p><p><strong>Step 3:</strong> Since -4 ≤ f'(x) ≤ 3 for all x ∈ (-2, 5):</p><p>f(5) - f(-2) has minimum value: (-4)·7 = -28</p><p>f(5) - f(-2) has maximum value: 3·7 = 21</p><p><strong>Step 4:</strong> If f(5) - f(-2) ranges from -28 to 21, then:</p><p>Maximum value of f(5) = f(-2) + 21</p><p>Minimum value of f(5) = f(-2) - 28</p><p><strong>Step 5:</strong> The difference between maximum and minimum values of f(5):</p><p>(f(-2) + 21) - (f(-2) - 28) = 21 - (-28) = 49</p><p>∴ Answer: <strong>49</strong></p>
Correct Answer: D