I am having some difficulty with how to think about this question. Graph theory questions like this are not my strong point.
Let $n\geq2$ be an even integer. Define $f(n)$ to be the smallest integer $k$ such that every graph $G$ on $n$ vertices with minimum degree at least $k$ contains a perfect matching. Determine $f(n)$ and justify your answer.
To begin with, I thought that complete bipartite graph on n vertices, where $n=2k$ (so as to ensure $k$ vertices in each independent set), would definitely contain a perfect matching. Therefore, $f(n)=\frac{n}{2}$. Can this answer be improved upon?
This is not the right way to think about the problem: just because a complete bipartite graph $K_{n/2,n/2}$ has a perfect matching, and satisfies $k = \frac n2$, doesn't mean that every graph with minimum degree $\frac n2$ would have a perfect matching.
You should think of it this way: you tell me a $k$, and I try as hard as I can to come up with a graph with minimum degree $k$, but no perfect matching, because I'm mean and don't like perfect matchings. You like perfect matchings, so you want to find a $k$ such that no matter what I do, I can't come up with such a graph.
If you want to take a moment to try to think about the problem again from this perspective, here is the point to pause reading.
Specific examples are good for the other direction. An unbalanced complete bipartite graph $K_{n/2-1, n/2+1}$ has minimum degree $\frac n2 - 1$, so if your choice of $k$ is $\frac n2 - 1$ or less, I can give you $K_{n/2-1,n/2+1}$, and you won't be able to find a perfect matching. So $f(n)$ must be at least $\frac n2$.
In this case, we do actually have $f(n) = \frac n2$, but we need a different argument: one that applies to all graphs with minimum degree $\frac n2$. A theorem you may be familiar with is Dirac's theorem: that any graph with minimum degree $\frac n2$ has a Hamiltonian cycle. If you can find a Hamiltonian cycle, then taking every other edge of that cycle will produce a perfect matching. This completes the proof.