What would be the smallest/lowest Big-O set possible of these $2$ functions Here i have $2$ functions and i was just wanting to know what the smallest big-oh set possible is for these functions, my guess would be the lowest possible of 1 would be the big-ph set $O(n^3)$ but was unsure if it may be $O(1)$, and for the 2nd function i was unsure if it was $O(n \log n)$ or $O(n)$

*

*$\frac {n^3} 4 + 6n^2 – 10000$

*$5n\log 2n + 6n$
Any help and explanation is appreciated.
 A: You have:
$\begin{align*}
    n^3 / 3 + 6 n^2 - 10000
      &= O(n^3) \\
    5 n \log 2 n + 6 n
      &= O(n \log n)
\end{align*}$
You prove the above and take it from here.
A: It seems that you have struggles with understanding what Big-O means in the first place, so I've decided to add some further details, although the comments and the already given answer are sufficient.
The Big-O notation was first introduced by Paul Bachmann in "Analytische Zahlentheorie (1894)" for a way of expressing approximation. On page $401$ is the first occurrence of Big-O in history. He states (in german)

[...] (es) findet sich $\tau(n)=n\ log\ n\ +\ O(n)$ wenn wir durch das Zeichen $O(n)$ eine Grösse ausdrücken, deren Ordnung in Bezug auf $n$ die Ordnung von $n$ nicht überschreitet; ob sie wirklich Glieder von der Ordnung $n$ in sich enthält, bleibt [...] dahingestellt.

This translates to

[...] (it) yields $\tau(n)=n\ log\ n\ +\ O(n)$ if we express by the sign $O(n)$ a quantity whose order with respect to $n$ does not exceed the order of $n$; whether it really contains members of order $n$ in itself remains undecided.

By this definition, we could write $1 + n + n^2=O(n^4)$, since the order of the "quantity" on the left hand side does not exceed the order of $n^4$. But furthermore, the following is also valid $1+n+n^2=O(n^3)$ as the "quantity" on the left hand side does not exceed the order of $n^3$. Obviously, the second statement is much stronger than the first statement. Being a little bit informal, we could say that we have introduced a notation which allows us to replace "$\approx$" with "$=$".
Being more precise now in todays terms, I am going the state the definition of Big-O as given by Donald E. Knuth in his famous book "The Art of Computer Programming".
Def.: The symbol $O(f(n))$ stands for the set of all functions $g$ of integers such that there exist constants $M$ and $n_0$ with $|g(n)| \leq M|f(n)|$ for all integers $n \geq n_0$. We usually write $g(n)=O(f(n))$ or $g(n) \in O(f(n))$.
Taking your first function as an example. You need to find constants $M$ and $n_0$ such that for $n \geq n_0$ it holds that $\frac{1}{4}n^3 + 6n^2-10000 \leq Mn^3$, and if we can show that, $\frac{1}{4}n^3 + 6n^2-10000 \in O(n^3)$, and this is by definition the best we can do.
(once we are more familiar with the notion we could prove a much stronger result, which states that given any polynomial $P(n)=a_0+a_1n+\cdots+a_mn^m$, we have $P(n)=O(n^m)$. Then the solution would follow directly.)
For the second function, the following rules might help:

*

*$c \cdot O(f(n)) = O(f(n))$, $c$ is constant

*$O(f(n))+O(f(n))=O(f(n))$

*$O(f(n)) O(g(n))=O(f(n)g(n))$
A: I am not sure that the smallest $O$ does exist.
For example,
$$\frac{n^3} 3 + 6 n^2 - 10000= O(n^3) $$
but it is also true that
$$\frac{n^3} 3 + 6 n^2 - 10000= O((n-1)^3) $$
and
$$\frac{n^3} 3 + 6 n^2 - 10000= O(T_n)$$
while $T_n<(n-1)^3<n^3$. [$T_n$ denotes the $n^{th}$ tetrahedral number.]

For the second,
$$5n\log 2n + 6n=O(\log_{2021}(n!))$$ is the same as
$$O(n\log(n)).$$
