Let f(x) = x^2 + 6x + 6 x^2 + 2x + 2 . Let a be the smallest integer such that f(x) b for all x R . The value of a - b is:

Step-by-step solution

For f(x) x^2 + 6x + 6 x^2 + 2x + 2 Since x^2 + 2x + 2 = (x+1)^2 + 1 > 0 for all x R , we can cross-multiply without changing the inequality sign: x^2 + 6x + 6 (1-a)x^2 + 2(3-a)x + 2(3-a) For this quadratic to be strictly negative for all real x , we require: 1) Leading coefficient 1 2) Discriminant D 4(3-a)[(3-a) - 2(1-a)] 4(3-a)(a+1) (a-3)(a+1) > 0 This gives a (- , -1) (3, ) . Taking the intersection with a > 1 , we get a > 3 . The smallest integer a satisfying this is a = 4 . For f(x) > b for all x R : x^2 + 6x