Let f: R R be a twice differentiable function such that f''(x) 0 for all x R and f'(5) = 0 . Let g(x) = f(x^2 - 6x + k) , where k is a real number. If g(x) has exactly one point of local extremum, then the interval of possible values of k

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Let f: R R be a twice differentiable function such that f''(x) > 0 for all x R and f'(5) = 0 . Let g(x) = f(x^2 - 6x + k) , where k is a real number. If g(x) has exactly one point of local extremum, then the interval of possible values of k is

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Correct answer: B. [14, )

Given f''(x) > 0 for all x R , the function f'(x) is strictly increasing. We are given f'(5) = 0 . Since f'(x) is strictly increasing, f'(x) 0 for x > 5 . The function is g(x) = f(x^2 - 6x + k) . Differentiating with respect to x using the chain rule: g'(x) = f'(x^2 - 6x + k) (2x - 6) = 2(x - 3) f'(x^2 - 6x + k) . For g(x) to have exactly one point of local extremum, g'(x) must change sign exactly once. The term 2(x - 3) changes sign at x = 3 . Therefore, the term f'(x^2 - 6x + k) must not change sign for any x R .

Let f: R R be a twice differentiable function such that f''(x) > 0 for all x R and f'(5) = 0 . Let g(x) = f(x^2 - 6x + k) , where k is a real number. If g(x) has exactly one point of local extremum, then the interval of possible values of k is

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