I'm given a function which is defined based on a condition, for example
$$ f(n) = \begin{cases} 4n+1, \ \text{n is even}\\ 3n^2+2, \ \text{n is odd} \end{cases}. $$
I have to prove or disprove that $f(n)$ is $\mathcal{O} (n^2)$. Do I have to show $f(n) < c n^2$ for all $n > n_0$ for both expressions or do I just select the $3n^2+2$ expression?
Your intuition tells you correctly that it should suffice to check the bound for the term $3n^2+2$, since $3n^2+2$ dominates the function $4n+1$. nir shahar has pointed that out in their answer. However, for a formal proof that $f(n) = \mathcal{O}(n^2)$, you will need to formalize your intuition that $3n^2+2$ dominates $4n+1$ and you will also have to argue why this reduces your problem to showing $3n^2+2 < cn^2$.
I advise you not to prove $3n^2+2 \ge 4n+1$ and then use this statement. I think it will be easier to instead prove both $3n^2+2 \le cn^2$ and $4n+1 \le c'n^2$ for $n > n_0$, with suitable constants $c, c' > 0$, and $n_0$. Then, with $C = \max(c, c')$, you will have $f(n) \le Cn^2$ for all $n>n_0$.
Maybe this way will cost you two more lines of a proof, but you will be rewarded with a more straight-forward proof.
you have to do for both expressions. But in this case, one expression is strictly smaller than the other one, so you can do it just for the $3n^2+2$ and say $4n+1<3n^2+2$ and therefore also $4n+1<cn^2$.
