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Let 'a' be a real number such that the function f(x) = ax² + 6x − 15, x ∈ R is increasing in (−∞, 3/4) and decreasing in (3/4, ∞). Then the function g(x) = ax² − 6x + 15, x ∈ R has a

1. local maximum at x = −3/4
2. local minimum at x = −3/4
3. local maximum at x = 3/4
4. local minimum at x = 3/4

Correct answer: local maximum at x = −3/4

Solution

f increasing then decreasing means a < 0 with vertex at -6/(2a) = 3/4, giving a = -4. Then g(x) = -4x^2 - 6x + 15, g'(x) = -8x - 6 = 0 -> x = -3/4. Since a = -4 < 0, this is a local maximum at x = -3/4.

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