Big-O and Big Omega of a function - Homework - analysis

Question

So I have been given a function, and I'll change the function since it is homework, and I want to learn HOW to do this instead of being told what the answer is.
Using the definitions of big-Oh and Ω, find the upper and lower bounds for
the following expressions. Be sure to state appropriate values for c and k.
c13n + c2n4, where the constants are positive integers.
Now, I understand how to determine if a function f(n) ∈ O(g(n)) or a f(n) ∈ Ω(g(n) from class.
What I don't understand is how to determine the g(n) if all you have is f(n). I hope that makes sense!
Edit: I'm sure you could brute force it an plug in a bunch of functions for g(n), but that isn't really what I want if there is a better solution.
Edit2: We can't use the limit methods for this, they want us to use the basic definitions somehow.
Edit 3: Here are the definitions we have been given:
Here is what I have for Big O:
For T(n) a non-negatively valued function, T(n) is in set O(f(n))
if there exist two positive constants c and k such that T(n)<=c*f(n)
for all n > k.
And for Ω:
For T(n) a non-negatively valued function, T(n) is in set Ω(g(n)) if there exist two positive constants c and k such that T(n) >= c*g(n) for all n > k

Answer 1

The intuition is f ∈ O(g) implies g is somehow equal or bigger than f; and f ∈ Ω(g) implies g is somehow equal or smaller than f. In my answer, I won't be too precise/picky on how to choose a constant.
First to warm up, you should convince yourself that
f ∈ O(f) and f ∈ Ω(f). (let c=1, k=1 in the definitions).
If f ∈ O(g), then g ∈ Ω(f) and vice versa. (if you find constants (c,k) for one, then (1/c, k) are the constants you need for the other)
If f ∈ O(g) then f ∈ O(P*g) and Q*f ∈ O(g) for any positive constant P,Q. This means multiplying functions by positive constants don't matter. Similarly for Ω.
If f ∈ O(g) and f ∈ O(h), then f ∈ O(MIN(g,h)).
If f ∈ Ω(g) and f ∈ Ω(h), then f ∈ Ω(MAX(g,h)).
When you are faced with trying to find O or Ω of f+g, you normally would guess O(f) or O(g) or Ω(f) or Ω(g).
In your case of 3^n + n^4, we know 3^n ∈ O(3^n), n^4 ∈ O(n^4), and 3^n + n^4 ∈ O(3^n + n^4). But we want to do better. We want to prove 3^n + n^4 ∈ O(3^n + 3^n) = O(3^n). We can do this if we can show n^4 ∈ O(3^n).
We should do exactly as the definition says we should do: show there are (c,k) such that for all n>k
n^4 ≤ c3^n
4log(n) ≤ log(c) + nlog(3)
4log(n) - nlog(3) ≤ log(c)
One way of showing that this c always exists is with calculus: show that 4log(n) - nlog(3) is a decreasing function eventually. The derivative is 4/n - log(3) and we can see for sufficiently large n, it is negative. Therefore for sufficiently large n, 4log(n) -nlog(3) is decreasing. Therefore there is a positive constant c for which the inequality is true. Therefore n^4 ∈ O(3^n). And 3^n + n^4 ∈ O(3^n + 3^n) = O(3^n).
Because 3^n + n^4 ≥ 1*3^n, 3^n + n^4 ∈ Ω(3^n). To illustrate that constants don't matter, let's use the c_1 and c_2 you had: c_1*3^n + c_2*n^4. Let d := min(c_1, c_2). Then
c_1*3^n + c_2*n^4 ≥ d(3^n + n^4) ≥ d*3^n
So c_1*3^n + c_2*n^4 ∈ Ω(3^n). Similarly, with O(3^n): let d := max(c_1, c2). Then for sufficiently large n,
c_1*3^n + c_2*n^4 ≤ d(3^n + n^4) ≤ d(c*3^n) = (dc)*3^n
We know this c exists because 3^n + n^4 ∈ O(3^n). Therefore c_1*3^n + c_2*n^4 ∈ O(3^n).
Not sure if I answered sufficiently but hope it helps.