Advanced Functions Question?

Apparently there can be different answers. A sample answer is f(x) = (x + 3)(x - 4) + 1 and g(x) = 1

How would i approach this question to get that equation?

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  • 2 years ago
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    f(x)/g(x) > 1

    f(x)/g(x) - 1> 0

    (f(x) − g(x)) / g(x) > 0

    Since f(x) is a quadratic function and g(x) is a linear function, then f(x)−g(x) is a quadratic function.

    If f(x)−g(x) has no roots, then f(x)−g(x) > 0 for all x or f(x)−g(x) < 0 for all x

    But then (f(x) − g(x)) / g(x) > 0 only where g(x) > 0 or g(x) < 0

    This would create just 1 interval, not 2

    So f(x)−g(x) must have roots.

    Let a, b be roots of f(x)−g(x) (where a ≤ b), and let c be the root of g(x)

    Then (f(x)−g(x)) / g(x) = k(x−a)(x−b)/(x−c) > 0 on the intervals (−∞,−3) and (4,∞)

    The function k(x−a)(x−b)/(x−c) is > 0 on the same intervals as k(x−a)(x−b)(x−c)

    But k(x−a)(x−b)(x−c) is a cubic function that has opposite behaviour at opposite ends.

    So it cannot possible by > 0 on the interval (−∞,−3) and (4,∞)

    Therefore, g(x) must be linear, and a = −3, b = 4, giving:

    g(x) = c

    f(x)−g(x) = k (x − a) (x − b)

    f(x) − c = k (x + 3) (x − 4)

    f(x) = k (x + 3) (x − 4) + c

    (f(x) − g(x)) / g(x) > 0

    k (x + 3) (x − 4) / c > 0

    This is true when k/c > 0

    So general solution is:

    g(x) = c

    f(x) = k (x + 3) (x − 4) + c, where k and c are both positive or both negative

    Example: k = 2, c = 12

    g(x) = 12

    f(x) = 2 (x + 3) (x − 4) + 12 = 2x² − 2x − 12

    f(x)/g(x) > 1

    (2x² − 2x − 12) / 12 > 1

    (x² − x − 6) / 6 > 1

    x² − x − 6 > 6

    x² − x − 12 > 0

    (x + 3) (x − 4) > 0

    x < −3 or x > 4

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