Let f(x) = x^2 + x + 1 x - 2 be defined for x > 2 . If the minimum value of k for which f(x) is one-one on the interval [k, ) is k = 4 , then the value of is

Step-by-step solution

To find the intervals where f(x) is one-one, we examine its derivative. Using the quotient rule: f'(x) = (2x + )(x - 2) - (x^2 + x + 1)(1) (x - 2)^2 f'(x) = 2x^2 - 4x + x - 2 - x^2 - x - 1 (x - 2)^2 f'(x) = x^2 - 4x - 2 - 1 (x - 2)^2 For f(x) to be one-one on [k, ) , f'(x) must not change sign on this interval. Since _ x f'(x) = 1 > 0 , we require f'(x) 0 for all x k . The minimum value of k for which this holds is the largest root of the numerator x^2 - 4x - 2 - 1 = 0 . Given that this minimum value is k = 4 , x =