Applications of Derivatives
Increasing/Decreasing functions and extrema
Grade 12
Question:
<p>If \(f(x) = \dfrac{e^x}{x^2}\), then which of the following is correct?</p>
<p>A. \(f\) is increasing on \((0, \infty)\)</p>
<p>B. \(f\) is decreasing on \((0, 2)\) and increasing on \((2, \infty)\)</p>
<p>C. \(f\) is increasing on \((0, 2)\) and decreasing on \((2, \infty)\)</p>
<p>D. \(f\) has a minimum at \(x = 2\)</p>
Step-by-Step Solution
Key Concept: Find f'(x) using the quotient rule, then analyze the sign of f'(x) to determine where f is increasing/decreasing. The critical point occurs when the numerator of f'(x) equals zero.
<p><strong>Step 1:</strong> Apply quotient rule to find f'(x).</p><p>$$f'(x) = \frac{d}{dx}\left(\frac{e^x}{x^2}\right) = \frac{e^x \cdot x^2 - e^x \cdot 2x}{x^4}$$</p><p><strong>Step 2:</strong> Factor the numerator.</p><p>$$f'(x) = \frac{e^x(x^2 - 2x)}{x^4} = \frac{e^x \cdot x(x-2)}{x^4} = \frac{e^x(x-2)}{x^3}$$</p><p><strong>Step 3:</strong> Analyze the sign of f'(x).</p><p>Since e^x > 0 always, the sign depends on (x-2)/x³:</p><p>• For x < 0: denominator x³ < 0, (x-2) < 0, so f'(x) > 0 (f is increasing)</p><p>• For 0 < x < 2: denominator x³ > 0, (x-2) < 0, so f'(x) < 0 (f is decreasing)</p><p>• For x > 2: denominator x³ > 0, (x-2) > 0, so f'(x) > 0 (f is increasing)</p><p>∴ f is increasing on (-∞, 0) and (2, ∞); decreasing on (0, 2)</p>
Correct Answer: D