for problem 2. (6 points) - this is the correct answer right? - R is reflexive. f(x) = O(f(x)). Therefore, R is reflexive. R is not antisymmetric. It is not necessary for f(x) = O(g(x)) and g(x) = O(f(x)). However, it can be (as an example): f(x) = x and g(x) = 2x. Then f(x) = O(g(x)) and g(x) = O(f(x)). (f, g) R and (g,f) R, but f and g are not equal. Therefore, R is not antisymmetric. R is not symmetric. f(x) = x and g(x) = x2. f(x) = O(g(x)), but g(x)= O(f(x)). Therefore, R is not symmetric. R is transitive f(x) = O(g(x)) and g(x) = O(h(x)) |f(x)| C1|g(x)| and |g(x)| C2|h(x)| |f(x)| C1|g(x)| C1C2|h(x)| Thus, f(x) = O(h(x)). Therefore, R is transitive