Applications of Derivatives
Inequalities using derivatives
Grade 12
Question:
<p>Let \(f\) be a real function with a continuous third derivative such that \(f(x), f'(x), f''(x), f'''(x)\) are positive for all \(x\). Suppose that \(f'''(x) \leq f(x)\) for all \(x\).</p>
<p>\(f'(x) < 2f(x)\) for all \(x\)</p>
<p>\(f'(x) < 2f(x)\) for all \(x > 0\)</p>
<p>\(f'(x) < 2f(x)\) for all \(x < 0\)</p>
<p>Nothing can be determined uniquely</p>
Step-by-Step Solution
Key Concept: Since f'''(x) ≤ f(x) and both are positive, f grows at a controlled rate. The constraint f'''(x) ≤ f(x) prevents f from growing faster than exponential, which combined with all derivatives being positive creates a bounded growth scenario where f must be bounded above.
<p><strong>Step 1:</strong> Recognize that f, f', f'', f''' are all positive and continuous. This means f is increasing, convex, and becoming more convex.</p><p><strong>Step 2:</strong> The constraint f'''(x) ≤ f(x) is critical. Consider the differential inequality: as the third derivative (rate of change of concavity) cannot exceed f itself, this limits how rapidly f can grow.</p><p><strong>Step 3:</strong> If f were unbounded, then for large x, f(x) would grow arbitrarily. But f'''(x) ≤ f(x) means the third derivative cannot exceed this growth rate. This creates a self-referential bound.</p><p><strong>Step 4:</strong> Suppose f(x) → ∞ as x → ∞. Then eventually f''(x) would have to grow without bound (since f' is increasing and positive). But then f'''(x) would eventually exceed f(x), contradicting our constraint.</p><p><strong>Step 5:</strong> Therefore, f must be bounded above. Combined with f being strictly increasing (f' > 0), f must approach a finite limit as x → ∞.</p><p><strong>Step 6:</strong> As x → ∞: f(x) → L (some finite limit), f'(x) → 0, f''(x) → 0, and f'''(x) → 0 (by continuity and the constraint).</p><p>∴ Answer: B</p>
Correct Answer: B