Is x^3 O(x^4) but x^4 not O(x^3)?

[nicnicman]

Homework Statement

To establish a big - O relationship, find witnesses C and k such that |f(x)| <= C|g(x)| whenever x > k.

Show that x^3 is O(x^4) but that x^4 is not O(x^3).

Homework Equations

If f(x) is O(g(x)) then there is a C such that f(x) <= Cg(x).

The Attempt at a Solution

If f(x) is O(g(x)) then there is a C such that f(x) <= Cg(x).
Thus, if x^3 is O(x^4), then x^3 <= C(x^4).
This is true for all C > 0 when C is a real number.

Similarly, if x^4 is O(x^3) then
x^4 <= Cx^3.
x <= C (divide both sides by x^3)
However, x <= C can not be true for any C. Thus x^4 is not O(x^3).

I'm not sure about this proof because I didn't provide witnesses C and k as the instructions called for. Big-O problems confuse me and I'm not sure if this proof is correct. Any suggestions are appreciated.

 

[nicnicman]

Any suggestions?

 

[clamtrox]

If f(x) is O(g(x)) then there is a C such that f(x) <= Cg(x).
Thus, if x^3 is O(x^4), then x^3 <= C(x^4).
This is true for all C > 0 when C is a real number.

What is true? x³≤Cx⁴ if and only if x ≥ 1/C. Perhaps this can help you to find some values of k and C.