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Below is my solution. Is this sufficient enough to prove?

$$|x^3+5x| \le |x^3+5x^3|$$

$$x^3 + 5^x \le 6x^3, x > 1, C=6$$

Let's take $x = 2$.

$$f(x) = (2)^3 + 5(2) = 18$$

$$g(x) = 6\cdot 8 = 48$$

For when $x > 1 |f(x)| \le C|g(x)|$ is false hence proving $x^3 + 5x$ is not $O(x^2)$.

Raphael
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user86867
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  • We discourage "please check whether my answer is correct" questions, as only "yes/no" answers are possible, which won't help you or future visitors. See here and here. Can you edit your post to ask about a specific conceptual issue you're uncertain about? As a rule of thumb, a good conceptual question should be useful even to someone who isn't looking at the problem you happen to be working on. If you just need someone to check your work, you might seek out a friend, classmate, or teacher. – Raphael Apr 07 '18 at 15:14
  • Big-O = "there is an N0 and a c > 0 such that". Not-Big-O = "there is no N0 and c > 0 such that"... You have a failed proof that f = O(g). That's not a proof that f ≠ O (g). – gnasher729 Apr 07 '18 at 15:25

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