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The function f(x) = integral from 1 to x of [2(t-1)(t-2)³ + 3(t-1)²*(t-2)²] dt attains its maximum value at x equal to:

1. 1
2. 2
3. 3
4. 4

Correct answer: 1

Solution

f'(x) = 2(x-1)(x-2)³ + 3(x-1)²*(x-2)² = (x-1)(x-2)²*[2(x-2) + 3(x-1)] = (x-1)(x-2)²*(5x-7). Critical points: x=1, x=7/5, x=2. Sign analysis: for x < 1: (x-1)<0, (x-2)²>0, (5x-7)<0 -> f'(x) = (-)(+)(-) > 0 (increasing). For 1 < x < 7/5: (+)(+)(-) < 0 (decreasing). So f has a LOCAL MAXIMUM at x=1. For x > 7/5 (excluding x=2 where (x-2)²=0 doesn't change sign): f' > 0 for x>7/5. At x=2, f'=0 but no sign change (even power). So f is increasing after x=7/5. The global maximum on the relevant range would be at x=1 since f decreases from 1 to 7/5 and then increases. Comparing f(1)=0 with f as x->inf (increases without bound), but f starts at 0 at x=1, dips below, then rises. The maximum in a bounded sense or local maximum is at x=1. Given the options and that f(1)=0 while f decreases before increasing, the intended answer is x=1.

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