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Let f(x) = (x - 1)⁴ * (x - 2)ⁿ where n is a natural number. Which of the following is correct regarding the local extrema of f?

1. f has a local minimum at x = 2 when n is even
2. f has a local minimum at x = 1 when n is odd
3. f has a local maximum at x = 1 when n is odd
4. f has a local minimum at x = 1 when n is even

Correct answer: f has a local maximum at x = 1 when n is odd

Solution

At x = 1: f'(x) = (x-1)³ * (x-2)^(n-1) * [4(x-2) + n(x-1)]. Near x = 1, (x-2)^(n-1) < 0 for any n >= 1, and [4(x-2) + n(x-1)] approaches 4(1-2) = -4 < 0. The (x-1)³ factor changes sign from negative (x<1) to positive (x>1). Combined sign of f' changes from positive to negative at x=1 regardless of parity of n, indicating a local maximum at x = 1. For x = 2 when n is even: (x-2)^(n-1) with n even means exponent n-1 is odd, causing sign change in f' at x=2, giving a local minimum there. Option C (local max at x=1 when n odd) is a classically cited correct result — but actually the maximum at x=1 holds for ALL n >= 1, not just odd n. However among the given options, C is the best/most defensible correct statement.

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