Other people have mentioned the Adjoint Functor Theorems. Here's a different perspective.

There's a famous Cambridge exam question set by Peter Johnstone:

Write an essay on
(a) the usefulness, or
(b) the uselessness,
of the Adjoint Functor Theorems.

I agree with the undertone of the question: the Adjoint Functor Theorems (AFTs) aren't as useful as you might think when you first meet them. They're not *useless*: but my own experience is that the range of situations in which I've had no easy way of constructing the adjoint, yet have been able to verify the hypotheses of an AFT, has been very limited.

Perhaps more useful than knowing the AFTs is knowing some large classes of situation where an adjoint is guaranteed to exist. Here are two such classes.

${}$1. **Forgetful functors between categories of algebras.** Any time you have a category $\mathcal{A}$ of algebras, such as **Group**, **Ring**, **Vect**, ..., the forgetful functor $\mathcal{A} \to \mathbf{Set}$ has a left adjoint. What's not quite so well-known is that you don't have to forget *all* the structure; that is, the codomain doesn't have to be **Set**.

For example, the functor $\mathbf{AbGp} \to \mathbf{Group}$ forgetting that a group is abelian automatically has a left adjoint. The functor $\mathbf{Ring} \to \mathbf{Monoid}$ forgetting the additive structure of a ring automatically has a left adjoint. The forgetful functor $\mathbf{Assoc} \to \mathbf{Lie}$, sending an associative algebra to its underlying Lie algebra (with bracket $[a, b] = a\cdot b - b \cdot a$) automatically has a left adjoint. (That might not look so much like a forgetful functor, but that's only because the bracket on an associative algebra isn't given as a primitive operation in the usual definition of associative algebra: it has to be derived from the other operations.)

The same can be said if you talk about *topological* groups, rings, etc, basically because **Top** has all small limits and colimits.

All that is a consequence of the General AFT (= 'the' AFT in some people's usage). To my mind it's the principal reason why it's worth learning or teaching the General AFT.

${}$2. **Kan extensions.** Let $F: \mathbf{A} \to \mathbf{B}$ be any functor between small categories. Then there's an induced functor
$$
F^{*}: {[\mathbf{B}, \mathbf{Set}]}
\to
{[\mathbf{A}, \mathbf{Set}]}
$$
defined by composition with $F$. (Here ${[\mathbf{B}, \mathbf{Set}]}$ means the category of functors from $\mathbf{B}$ to $\mathbf{Set}$, sometimes denoted ${\mathbf{Set}}^{\mathbf{B}}$.)

The fact is that $F^{*}$ always has both a left and a right adjoint. These are called left and right Kan extension along $F$. The same is true if you replace $\mathbf{Set}$ by any category with small limits and colimits.

This is really useful, though that might not be obvious. For example, suppose we're interested in representations of groups. A group can be regarded as a one-object category, and the category of representations of a group $G$ is just the functor category $[G, \mathbf{Vect}]$. Now take a group homomorphism $f: G \to H$. The induced functor
$$
f^{*}: [H, \mathbf{Vect}] \to [G, \mathbf{Vect}]
$$
sends a representation of $H$ to a representation of $G$ in the obvious way. And it's guaranteed to have both left and right adjoints. These adjoints turn a representation of $G$ into a representation of $H$, in a canonical way. I believe representation theorists call these the 'induced' and 'coinduced' representations, at least in the case that $G$ is a subgroup of $H$ and $f$ is the inclusion.

*Exercise*: let $G$ be a group. There are unique homomorphisms $G \to 1$ and $1 \to G$, where $1$ is the trivial group. Each of these two homomorphisms induces a functor "$f^{*}$" between the category $[G, \mathbf{Set}]$ of $G$-sets and the category $[1, \mathbf{Set}] = \mathbf{Set}$ of sets. These two functors each have adjoints on both sides. So we end up with six functors and four adjunctions. What are they?

The existence of Kan extensions is best derived from the theory of ends. In fact, ends allow you to describe them explicitly.