Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>The point of extremum of \(f(x) = \int_0^x (t-2)^2(t-1)\,dt\) is a</p>
<p>(a) maximum at \(x=1\)</p>
<p>(b) maximum at \(x=2\)</p>
<p>(c) minimum at \(x=1\)</p>
<p>(d) minimum at \(x=2\)</p>
<p>(e) none of these</p>
Step-by-Step Solution
Key Concept: Find extrema by computing f'(x) = (x-2)²(x-1) and analyzing sign changes. The function has a local minimum where f'(x) changes from negative to positive, occurring at x = 1 where the derivative is zero but doesn't change sign at x = 2 (even root).
<p><strong>Step 1:</strong> Apply Leibniz rule to find the derivative:</p><p>f'(x) = (x-2)²(x-1)</p><p><strong>Step 2:</strong> Find critical points by setting f'(x) = 0:</p><p>(x-2)²(x-1) = 0 ⟹ x = 1 (simple root) or x = 2 (double root)</p><p><strong>Step 3:</strong> Analyze sign of f'(x):</p><p>• For x < 1: (x-2)² > 0, (x-1) < 0 ⟹ f'(x) < 0</p><p>• For 1 < x < 2: (x-2)² > 0, (x-1) > 0 ⟹ f'(x) > 0</p><p>• For x > 2: (x-2)² > 0, (x-1) > 0 ⟹ f'(x) > 0</p><p><strong>Step 4:</strong> Determine extremum type:</p><p>At x = 1: f'(x) changes from (−) to (+) ⟹ <strong>local minimum</strong></p><p>At x = 2: f'(x) doesn't change sign ⟹ inflection point (not an extremum)</p><p>∴ Answer: C (local minimum at x = 1)</p>
Correct Answer: C