Applications of Derivatives
Local Extrema via Integration
Grade 12
Question:
<p>If m and n are positive integers and <i>f</i>(<i>x</i>) = ∫<sub>x</sub><sup>1</sup> (<i>t</i> − <i>a</i>)<sup>2n</sup>(<i>t</i> − <i>b</i>)<sup>2m+1</sup> d<i>t</i>, <i>a</i> + <i>b</i>, then</p>
<p>(a) <i>x</i> = <i>b</i> is a point of local minimum</p>
<p>(b) <i>x</i> = <i>b</i> is a point of local maximum</p>
Step-by-Step Solution
Key Concept: To determine whether x = b is a local extremum, we need to find f'(x) using the Fundamental Theorem of Calculus and analyze the sign change of f'(x) around x = b. The exponent on (t - b) being odd (2m + 1) is crucial for determining whether f'(x) changes sign.
<p><strong>Step 1: Find f'(x) using Fundamental Theorem of Calculus</strong></p><p>Given: f(x) = ∫₁ˣ (t − a)²ⁿ(t − b)²ᵐ⁺¹ dt</p><p>By FTC: f'(x) = (x − a)²ⁿ(x − b)²ᵐ⁺¹</p><p><strong>Step 2: Analyze the sign of f'(x) near x = b</strong></p><p>Since m and n are positive integers:</p><p>• (x − a)²ⁿ ≥ 0 for all x (even exponent), and equals 0 only when x = a</p><p>• (x − b)²ᵐ⁺¹ is odd exponent, so it changes sign at x = b</p><p><strong>Step 3: Apply First Derivative Test</strong></p><p>For x < b: (x − b)²ᵐ⁺¹ < 0, so f'(x) < 0 (function decreasing)</p><p>For x > b: (x − b)²ᵐ⁺¹ > 0, so f'(x) > 0 (function increasing)</p><p><strong>Step 4: Conclusion</strong></p><p>Since f'(x) changes from negative to positive as x passes through b, the function has a local minimum at x = b.</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A