<p>Suppose a function \(f:[0,10]\to R\) is continuous and differentiable everywhere in its domain. If \(f(10) = 19\) and \(|f'(x) - 5| \leq 4\) \(\forall\, x\) in domain. Find maximum value of \(f(0)\).</p>
Step-by-Step Solution
Key Concept: The constraint |f'(x) - 5| ≤ 4 means 1 ≤ f'(x) ≤ 9. To maximize f(0), minimize f(10) - f(0) by choosing f'(x) as small as possible (f'(x) = 1) throughout [0,10].
<p><strong>Step 1:</strong> Interpret the constraint |f'(x) - 5| ≤ 4.</p><p>This gives us: -4 ≤ f'(x) - 5 ≤ 4, which means 1 ≤ f'(x) ≤ 9 for all x ∈ [0,10].</p><p><strong>Step 2:</strong> Apply Mean Value Theorem interpretation.</p><p>By MVT, f(10) - f(0) = f'(c)(10 - 0) for some c ∈ (0,10).</p><p>Therefore: 19 - f(0) = 10·f'(c) where 1 ≤ f'(c) ≤ 9.</p><p><strong>Step 3:</strong> Find maximum f(0).</p><p>To maximize f(0), we minimize [f(10) - f(0)] = 19 - f(0).</p><p>This means we want the smallest possible value of 10·f'(c).</p><p>The minimum occurs when f'(x) = 1 throughout [0,10] (the lower bound).</p><p>Then: 19 - f(0) = 10(1) = 10</p><p>∴ f(0) = 19 - 10 = <strong>9</strong></p>
Correct Answer: 1