For each part below, determine whether: • S(r) is O(g(x)) g(r) is O(f(x)) • f(r) is O(g(x)) and g(r) is O(f(x)) none of the above. (a) f(x) = 3r+log(r), g(x) = x + (log(x))² (b) f(x)=r!, g(x) = 1²¹ (c) f(x) = g(x) = log(r) (d) f(x)=x²0 +¹.9 +¹8..., g(x) = r².1

(a) f(x) is O(g(x)) and g(x) is not O(f(x)) To show that f(x) is O(g(x)), we need to find positive constants c and k such that f(x) <= c*g(x) for all x > k. Taking c = 4 and k = 1, we have:

3x + log(x) <= 4*(x + (log(x))^2) for all x > 1

which is true. Therefore, f(x) is O(g(x)).

To show that g(x) is not O(f(x)), we need to show that for any positive constant c and any k, there exists an x > k such that g(x) > c*f(x). Let c = 1 and k = 1. Then for any x > e, we have:

x + (log(x))^2 > 3x + log(x)

Therefore, g(x) is not O(f(x)).