When does an optimization problem have a unique solution, no solution, or infinitely many? What conditions are required for an optimization problem (e.g., given  $f(x) = a \cdot x^2 + b \cdot x + c $, find $\min f(x)$) to have a unique solution, no solution, or infinitely many?
I tried to analyze the function by taking its derivative(s) and to look for minima (or maxima). But is that the only way to approach such a problem? Or is there a faster – and a more efficient – way?
 A: Generalities
Let us introduce $X \subseteq \mathbb{R}$, the set of admissible solutions of your optimization problem.

*

*If $X$ is closed and convex ($X = \mathbb{R}$ or $X = [a, b]$), if $f$ is strictly convex on $X$ ($\forall x \in X, \; f''(x) > 0$), then you have existence and uniqueness of a global minimum.


*When $X = \mathbb{R}$, if $f$ is coercive ($f(x) \to \infty$ when $\vert x \vert \to \infty$), then there exists a minimum (but you have not uniqueness).


*When $X$ is compact (bounded and closed, e.g. a closed interval $X = [a, b]$), if $f$ is continuous, then there exists a minimum (but you have not uniqueness).


*If $X$ is open ($X = \mathbb{R}$ for example), any minimizer $x_0 \in X$ verifies $f'(x_0) = 0$ (it is a necessary condition: you need convexity at $x_0$ to ensure it to be a local minimum, i.e $f''(x_0) > 0$). In general, the minimum is not ensured to be global.
All these properties can be generalized when optimizing in $\mathbb{R}^n$ or a subset of $\mathbb{R}^n$.
In general, apart for strict convexity, it may be hard to prove uniqueness of a minimum, or to check that a minimum is global.
Answer to your question
On your example $X = \mathbb{R}$, you can divide cases depending on $a$ and $b$:

*

*when $a > 0$, you have a unique global minimizer because of strict convexity

*when $a = 0$, $b \neq 0$, you don't have any minimizer, since $f'(x) = 0$ has no solution and $X$ is open (see point 4)

*when $a = 0$, $b = 0$, you have an infinite number of minima as $f$ is identically equal to $c$

*when $a < 0$, $f$ is strictly concave and goes to $-\infty$ when $\vert x \vert \to \infty$: you can check that no points verify $f'(x) = 0$ in this case apart at some $x_0$ which is a global maximizer

A: Let
$$f(x) = ax^2 + bx +c.$$
Suppose $a \ne 0.$
Then
$$f(-\frac{b}{2a})=-\frac{b^2}{4a}+c.$$
Hence we have
$$f(x) = ax^2 +bx + f(-\frac{b}{2a})+\frac{b^2}{4a} =a(x+\frac{b}{2a})^2 + f(\frac{-b}{2a}).$$
If $a > 0$ then we have
$f(x) \ge f(\frac{-b}{2a})$  for all  $x.$
$f(\frac{-b}{2a})$ is the global minimum.
Same way for $a < 0.$
