Determine whether $x^3$ is $O(g(x))$ for certain functions $g(x)$. a) $g(x) = x^2$ 
b) $g(x) = x^3$
c) $g(x) = x^2 + x^3$
d) $g(x) = x^2 + x^4$
e) $g(x) = 3^x$ 
f) $g(x) = (x^3)/2$
Do you guys have any ideas?  Thanks!
 A: Since you asked in a comment about (e), let's discuss that.
Saying "$x^3$ is $O\bigl(3^x\bigr)$" means the following:

There is some constant $c$, such that for all sufficiently large $x$, $$x^3 < c\cdot3^x.$$

Consider the functions $x^3$ and $3^x$.  Clearly, $3^x$ increases much faster than $x^3$.  For $x=10$, $3^x$ is already 59,049 and $x^3$ is only 1,000.  And $3^x$ triples every time $x$ becomes  $x+1$, whereas $x^3$ does not triple so easily; to triple $x^3$ you have to multiply $x$ by something.
So there is no trouble finding the constant $c$ that we want; $c=1$ will do.  And indeed, for all $x>10$, it is the case that $x^3 < 1\cdot 3^x$.  So $x^3$ is $ O\bigl(3^x\bigr)$.
Now on the other hand, $x^3$ is not $O\bigl(x^2\bigr)$.  Why not?  Well, if it were, then

There is some constant $c$, such that for all sufficiently large $x$, $$x^3 < c\cdot x^2$$

But clearly that's not true, since no matter what $c$ is, $x^3 > c\cdot x^2$ whenever $x>c$.  So $x^3$ is not $O\bigl(x^2\bigr)$. 
Does that help?
A: As $x\to\infty $ :
Well , for polynomials you should just look at the term with highest degree.
Positive coefficients are negligible.
And  $a^x$   always dominate polynomials in the long run for $a>1$..
So , for your question :all possible except a...
A: You can use the following result

$$f=O(g) \iff \limsup_{x \to \infty}\frac{|f(x)|}{|g(x)|} =c,$$

where $c$ is finite.
